It is believed that atomic clocks are accurate, even when they are moving.

Scientists claim that results of the measurements of moving atomic clocks indicate that Einsein was correct, and that time really does slow down the faster an object is moving.

However, after reading about the construction of caesium atomic clocks, I found that there are three factors that would influence a moving atomic clock, independently of the time that it is measuring:

a) The motion and direction of the magnetic fields which are used to seperate the different caesium atoms.

b) The motion and direction of the caesium atoms themselves.

c) The motion and direction of the microwaves used to excite the caesium atoms.

As the motion of the atomic clock increases, one or more of the above factors would influence the clock to give a false reading. This would mean that atomic clocks are only accurate if there at rest(at least relatively).This would also mean that time is constant, but the speed of the atomic clock changes based on its motion.

Any comments are appreciated.

Tom

""The motion and direction of the microwaves used to excite the caesium atoms.""

--> microwaves are light and isn't it so that its motion is absolute?

I believe that the answer to the question is that the motion of the clock (or of the atoms for that matter) do not at influence the accuracy because the velocity of photons (that make up the facors (a) and (c) in your post) are indepemdent of the velocity of the clock.

edited to add: oh, C'est moi, you beat me again. ;)

Merlijin and c'est moi,

What about the red or blue shift of the microwaves. Doe these shifted microwaves have the same effect on caesium atoms as unshifted microwaves?

What about the magnetic fields that separate the caesium atoms. Since the magnetic fields travel light speed, do they have less of an effect on the caesium atoms the faster the clock is traveling???

Tom

Just to correct my previous post, when I said red and blue shifted microwaves, I meant microwaves that have frequencies shifted above or below their normal frequencies.

Tom

Merlijn,


I believe that the answer to the question is that the motion of the clock (or of the atoms for that matter) do not at influence the accuracy because the velocity of photons (that make up the facors (a) and (c) in your post) are indepemdent of the velocity of the clock.

That is the problem, the velocitiy of the photons are independent. If the velocities of the photons were dependent on the clock, everything would be synchronised regardless of the clock's speed.

But since the velocity of the photons are independent of the velocity of the clock, the faster the clock travels, the less the photons are synchronised with the clock which, in turn, results in faulty readings.

Tom

The atoms in the clock experience no Doppler shifts, since the microwave source is moving with the rest of the clock.

James R,

Good one!!! You got me there. I'll have to think about it a little more.

I'm also considering something c'est moi told me. He claims that increased velocity causes the Bohr radius of atoms to shrink. This reduction in the radius could cause a change the frequency between orbitals. Since atomic clocks count the frequency emitted by atoms, this change in frequency could give a false impression that time is slowing down(or speeding up).

But I still get the feeling that there are too many factors in atomic clocks that can lead to false readings.


Tom

The photon frequencies emitted by orbital electrons is a quantum mechanical affect. What matters is the total spin, angular momentum and energy of the electron. If the source was moving with respect to the detector the frequency would indeed then be shifted, by doppler shifting, by an exact amount that can be determined. I refer you also to <a href="http://rd11.web.cern.ch/RD11/rkb/PH14pp/node16.html">Bremsstrahlung radiation and synchrotron radiation</a> so you can see that these affects are well known and accounted for. But as the emitter and detector are at rest relative to each other this should not be required.

The Bohr Radius of an atom is a classical approximation of what atoms really are and should not be used in this case. You will get misleading results from using it.

Also, as the entire experiment is moving it is in the same frame of reference and so special relativistic corrections do not apply between each element of the apparatus. What does matter and needs accounting for, is a another clock moving relative to your 'stationary' clock and general relativistic corrections which affect both.

Thed,


Also, as the entire experiment is moving it is in the same frame of reference and so special relativistic corrections do not apply between each element of the apparatus.

You have to take into consideration that the magnetic fields, and the microwaves, in the atomic clock travel independently of the frame of reference of the clock. Both the magnetic fields and the microwaves travel in their own frame of reference, which may cause problems as the clock travels faster.


Tom

Where is your evidence for your above statement?

Give me a reference that clearly shows the magnets and lasers are moving independant of the source. That is, all moving on differnet vectors and not together.

Thed,


Where is your evidence for your above statement?

Give me a reference that clearly shows the magnets and lasers are moving independant of the source. That is, all moving on differnet vectors and not together.

The magnets and lasers are not moving independent of the source, but the magnetic fields and microwaves that they produce are.

This is one of the characteristics of light: It travels independently of any frame of reference. Since microwaves are photons as well, this characteristic applies to microwaves as well. I also believe that this characteristic applies to magnetic fields, but I'm not sure.

Tom

You seriously misunderstand what has been said. The speed of light is invariant between inertial frames of reference. That is not the same as saying that light in independant of the frame.

You would do well to understand the difference and what invariance means. Clue, it is not affected by rotations (changes) between inetial frames of reference.

Thed,

Are you saying that light's speed and direction is dependent on the frame of reference of the object producing it, or not???

As far as I know, the moment a beam of light is produced, that beam of light's characteristics become independent of the source producing it. It's speed and direction become constant, regardless of the speed and direction of the source that produced it.

If I'm wrong, please explain.

Tom

Let's look at what you said above


You have to take into consideration that the magnetic fields, and the microwaves, in the atomic clock travel independently of the frame of reference of the clock. Both the magnetic fields and the microwaves travel in their own frame of reference, which may cause problems as the clock travels faster.

And


That is the problem, the velocitiy of the photons are independent. If the velocities of the photons were dependent on the clock, everything would be synchronised regardless of the clock's speed.

But since the velocity of the photons are independent of the velocity of the clock, the faster the clock travels, the less the photons are synchronised with the clock which, in turn, results in faulty readings

The point here is that the entire apparatus is in the same frame of reference. There is no relativistic correction between the emitter (the photons) and the detector because there is no relative motion between them.

The important thing to realise, if I see where you are going wrong, is that even if the clock was moving at 0.99c all elements of the clock are travelling at the same speed. The photons are still emitted at c and traverse the distance to the emitter in the same time as if the thing where stationary.

Only to an observor outside that frame (point of view or whatever) would you need to worry about the relative velocities.

Going back to something I said I while back. From the frame of reference of a distant observor we are travelling at near c. We can't tell that and apply no corrections. If that observor could measure our atomic clock they should be able to correctly use it to measure time, after applying the Lorentz transform.

My wrist-watch is atomic, runs off a thirty-pound nuclear reactor backpack. The crappy piece of so-and-so loses about three minutes each day; have to reset it all when I get up in the morning.

Thed,


The important thing to realise, if I see where you are going wrong, is that even if the clock was moving at 0.99c all elements of the clock are travelling at the same speed. The photons are still emitted at c and traverse the distance to the emitter in the same time as if the thing where stationary.

Imagine that you had two fixed plates, one an emitter and the other a detector, and they were one meter apart. While the two are at rest, if a laser beam shoots out of the emitter towards the detector the the time it would take the laser beam to hit the detector would be:

Time = 1/c

However if the two plates were traveling at .99c away from the laser beam, it would take much longer for the laser beam to hit the detector because the speed of light is always constant regardless of any frame of reference.

Now if it were possible for the two plates to travel at c, the laser beam would NEVER reach the detector. In order for the laser beam to reach the detector in this case, the laser beam would have to travel faster than light. And as you know, this is impossible.

Tom

Tom,

<i>Imagine that you had two fixed plates, one an emitter and the other a detector, and they were one meter apart. While the two are at rest, if a laser beam shoots out of the emitter towards the detector the the time it would take the laser beam to hit the detector would be:

Time = 1/c</i>

Yes. I agree.

<i>However if the two plates were traveling at .99c away from the laser beam, it would take much longer for the laser beam to hit the detector because the speed of light is always constant regardless of any frame of reference.</i>

Wrong. In the reference frame of the plates, the light would still take time 1/c to travel between them. The fact that the whole apparatus is moving is completely irrelevant.

BUT (and this is a big "but") according to a "stationary" observer watching the plates go past, the distance between the plates would be length contracted, and the time frame of the plates would appear to run slower. The net effect of this is that this observer would measure the speed of light (distance between plates as seen by him divided by time taken as seen by him) to be c, still. That is what is meant by the invariance of the speed of light.

<i>Now if it were possible for the two plates to travel at c, the laser beam would NEVER reach the detector.</i>

Yes, but that isn't possible.

James R and Thed,

Here is a link I posted on the "Galileo & Einstein - second thoughts" thread:

http://www.physics.wustl.edu/~visser/physics-216/notes-light-clock.html

It illustrates one of the problems a moving atomic clock can experience. As you can tell from the link, the vertical motion of light can change in an object that is moving. This change may result in an atomic clock giving false readings.

Tom

Read the last line of the third paragraph of that link. The analysis is done from an external frame of reference. That makes a world of difference and is the point you are consistently or deliberately missing.

Nothing affects the rate the clock 'ticks' from the frame of reference of the clock. Only to another observor, in another frame of reference, would the clock appear to run slower. That is what they measured.

Thed,

I posted this from my response to James R on another thread. The same applies here:


If the speed of light is constant, and if you increase the horizontal velocity of the light in the device by moving it, the vertical velocity of the light must slow down in order for the light's speed to remain constant.

If you increase the horizontal velocity of light without decreasing the vertical velocity, the light would be travelling faster than c.


Tom

What is the relevance of it?
It seems totally misconceived.... as thed wrote "The analysis is done from an external frame of reference. That makes a world of difference and is the point you are consistently or deliberately missing. "

Merlijn,

The velocity of light is independent of any frame of reference.

Therefore, whether the observer is moving with the clock or whether the observer is stationary, the results are the same.

Tom

Hi Tom,

"The velocity of light is independent of any frame of reference. Therefore, whether the observer is moving with the clock or whether the observer is stationary, the results are the same."

That is exactly what your intuition tells you, and it would seem that it has to be like that. Unfortunately, it isn't.

You have to very carefully dissect the problem and keep the observations of the two frames of reference seperate. On one hand you have the observer travelling with the clock, and on the other you have a stationary observer. It has been said before in this thread, but I'll repeat it once more:


The observer that moves with the clock has the most easily understandable observation. He sees light, emitted from a laser, flipping between two plates that are seperated by one meter. Hence the time required for the light to flip back and forth is, as you mentioned, t = 2/c seconds. This is exactly what that observer will say, and irregardless of his perception, that is also what is really happening.
The outside observer seems a completely different scenario. Assume the apparatus travels in a horizontal direction while the light bounces upwards and downwards (vertical direction). For that observer, the light follows a curved path (triangular) and hence for him the time required for the light to pass that longer distance is a fraction more than 2/c. This is exactly what he sees, and because we assume that the speed of light is c, regardless of the motion of the source producing it, his measurement t' = t + Dt = 2/c + Dt, where Dt is the extra time required for the light. The faster the light apparatus is moving, the larger Dt will be.


The important point I think you are missing is that there is no right or wrong in this situation. Both observers are right when they say the time required is t or t' respectively ... even though t does not equal t'.

So to conclude: both observers measure different times. It is not because the speed of light varies, or because there is some magical fairy appearing that somehow changes the perception of one observer. From that simple experiment, one can comprehend that time measurement is relative to the observer.

The experiment with the atomic clocks confirms this discrepance of time measurement between two observers that move relative to eachother. An atomic clock is just a sophisticated way of performing the experiment with light bouncing up and downwards, but the conclusion is the same.

I'm sorry it doesn't fit into ones daily perception. I admit that it sounds incredible but this is how nature appears to work.

Bye!

Crisp

Crisp,


The observer that moves with the clock has the most easily understandable observation. He sees light, emitted from a laser, flipping between two plates that are seperated by one meter. Hence the time required for the light to flip back and forth is, as you mentioned, t = 2/c seconds. This is exactly what that observer will say, and irregardless of his perception, that is also what is really happening.

You are wrong. This is what would really be observed:

a) The observer that moves with the clock would see the laser light bouncing up and down between the two mirrors. As the observer and the clock move faster, the observer would notice that it takes longer and longer for the light to go between the two mirrors. The observer wouldn't understand because he/she does not see the horizontal motion of the light. The lights vertical motion would decrease in order for it's speed to remain constant as it's horizontal motion increased.

b) The outside observer would see the(almost) true path of the light(that it is moving horizontally and vertically) but he/she would also notice that it takes longer for the light to bounce of the mirrors. In other words, the time it would take for the light to bounce of the mirrors would be the same for the outside observer and the observer traveling with the clock. The only difference is is that the outside observer would know why the vertical speed of the light is decreasing as the clock travels faster, while the observer traveling with the clock would not.

Let me remind you, the speed of light is always constant regardless of the frame of reference. The direction of light is dependent on a frame of reference, but the speed of light is not.

If you were right, and the observer traveling with the clock did not see it slow down, that would mean that the light would be traveling faster than c, and as you know, that's not possible.

One last thing: Since the speed of light is constant regardless of any frame of reference, it means that the light travels in an absolute frame of reference. This would mean that the speed of light in a moving object could be used to determine the absolute motion of that object.

Tom

You claimed a while ago that you wish to understand the truth. Yet every time some one tries to explain how modern physics describes things you 'correct' them, as if you and only you understand what is really going on.

Why is that?

Bit of a clue here. Velocity has direction, speed does not. The speed of light is invariant. The important word is invariant.

Hi Tom,

"The observer wouldn't understand because he/she does not see the horizontal motion of the light. The lights vertical motion would decrease in order for it's speed to remain constant as it's horizontal motion increased. "

Why should this happen ? I think Thed pointed you in the good direction for the answer to this paradox.

Bye!

Crisp

Tom,

You're contradicting yourself. Compare your two statements:

<b>1</b>. "As the observer and the clock move faster, the observer would notice that it takes longer and longer for the light to go between the two mirrors."

and

<b>2</b>. "Let me remind you, the speed of light is always constant regardless of the frame of reference."

Now, I suppose you agree that for the observer moving with the clock, the distance between the mirrors doesn't change - right?

If that is the case then we can calculate the speed of light by taking the distance between the mirrors divided by the time it takes the light to bounce between them. The distance is constant, but your statement 1 says that the time increases with speed. Therefore the calculated speed of light decreases with increasing speed.

But then you make statement 2, which says the speed of light is independent of the frame of reference (i.e. the speed of the clock).

For consistency, you need to give up one of your two statements. Either you let the speed of light change, or you agree that the time taken to bounce between the two mirrors is constant for the moving observer.

Which is it to be?

Seems pertinent to the discussion.

“Ultra-precise clocks on the International Space Station and other space missions may determine whether Albert Einstein's Special Theory of Relativity is correct and could dramatically change our understanding of the universe.”

http://newsinfo.iu.edu/news/page/normal/406.html

James R, Thed, and Crisp

I don't know why I have to keep repeating myself.

The speed of light is always constant. As the clock travels faster, the light has to keep up with the horizontal motion of the clock. So the light must increase it's horizontal motion in order to keep up with the clock. If the light's horizontal motion increases, its vertical motion must decrease in order for the light's total speed to remain constant. The mirrors do not measure the total speed of the light, they only measure the vertical motion of the light.

To repeat myself again, the total speed of light does not change, but it's horizontal and vertical motion does.

Tom

Yes, Tom. Your last post is perfectly correct. If you follow your argument through to its conclusion, you must therefore conclude that both space and time depend on the observer's frame of reference in order to keep the speed of light constant.

Congratulations. You're now a relativist.

James R,

If I follow my argument to its conclusion, this is what I get:

1. The speed of light is always constant, therefore it travels in the absolute frame of reference.

2. The light clock appears to be indicating that time is slowing down as it travels faster, when in reality time remains constant.

3. The absolute velocity of any object can be determined using light, because the speed of light is always constant regardless of the frame of reference it is in.

James, it appears that I'm an anti-relativist.:)

Tom

ImaHamster2,

Thanks for the interesting link.

The only problem would be the actual construction of the clocks. The more complex the structure of the clocks, the more likely there will be a clash between the relativistic and the non-relativistic components of the clocks.

These clashes may lead to false readings.

Tom

How can the constant speed of light be used to determine absolute velocity? No matter what velocity and direction you're moving, and no matter what velocity and direction the source of light it, you'll still measure it to be <i>c</i>. How do you determine absolute velocity if you measure the speed of light to be <i>c</i> no matter what? (This is assuming a non-accelerating frame of reference, of course.)

Tom,

<i>1. The speed of light is always constant, therefore it travels in the absolute frame of reference.</i>

Clearly you still do not understand the term "frame of reference", despite repeated, careful explanation. Nothing exists only in one frame of reference, because a frame of reference is only a point of view. Many observers can look at the same beam of light, or baseball or whatever from different viewpoints with different states of motion. It is meaningless to make statements like "light travels in an absolute frame of reference".

I suggest you review the earlier posts to this thread.

<i>2. The light clock appears to be indicating that time is slowing down as it travels faster, when in reality time remains constant.</i>

You have given no cogent argument to support that conclusion.

<i>3. The absolute velocity of any object can be determined using light, because the speed of light is always constant regardless of the frame of reference it is in.</i>

I have told you why this is mistaken already. Please review the thread.

<i>James, it appears that I'm an anti-relativist.</i>

Yes, but not a <b>reasoned</b> anti-relativist. What you have is a faith, not a rational position. I can't argue you out of that. Your faith is not based on anything rational, but only on a gut feeling driven by who-knows-what psychological motivation. Obviously, rational explanation is wasted on you.

SpyFox_the_KMeson,


How can the constant speed of light be used to determine absolute velocity? No matter what velocity and direction you're moving, and no matter what velocity and direction the source of light it, you'll still measure it to be c. How do you determine absolute velocity if you measure the speed of light to be c no matter what? (This is assuming a non-accelerating frame of reference, of course.)

I posted a link earlier on this thread to a device called a light clock. A light clock is just two parallel mirrors, one on top of each other, that has light reflecting between them. By counting how many times light bounces between these two mirrors per second, you can measure the absolute velocity of the clock.

In the case that the light clock is in absolute rest, the vertical velocity(up and down) of the light bouncing between the mirrors is equal to c.

But if the clock is in motion, the light's horizontal(forward) velocity has to increase in order for the light to keep up with the motion of the clock. Because of the increased horizontal motion of the light, the vertical motion of the light must decrease in order for the total velocity of the light to remain constant.

Because the mirrors of the light clock measure only the vertical motion of the clock, and because the lights vertical motion decreases the faster the clock travels as a result of the light's increased horizontal motion, the amount of time it takes for the light to bounce of the mirrors would increase the faster the clock is moving.

Therefore since the speed of light is alway constant regardless of the frame of reference it is in, the clock would be able to measure the absolute motion of any object.

Tom

James R,

If the speed of light is constant, in which frame of reference would the speed of light be c????

The answer is the absolute frame of reference.

The fact is that the speed of light proves that there is an absolute frame of reference. As I described earlier in this thread, a light clock can measure the absolute motion of any object.

What you can't face is the fact that light came along and crapped on Einsteins theory of relativity. I guess everything was going fine for Einsein until he found that the speed of light doesn't care what frame of reference it's in.

As I explained before, the light clock measures the vertical velocity of the light between it's mirrors, and NOT the total velocity of the light. If a stupid physicist would assume that the light clock measures the total velocity of the light, he/she would come to the conclusion that time slows down the faster the clock is traveling. This could be one of the many errors associated with atomic clocks.

Tom

Hi Tom,

"If the speed of light is constant, in which frame of reference would the speed of light be c???? The answer is the absolute frame of reference."

Wrong answer, the correct answer is: in ALL possible frames of reference. The direction the light is moving in is irrelevant, we're not talking about a vector component here, but the total size of the velocity vector.

"As I explained before, the light clock measures the vertical velocity of the light between it's mirrors, and NOT the total velocity of the light."

This is not correct. For the observer standing next to the clock, the light has no horizontal component. Since the speed of light is constant in every frame of reference, the speed would have a vertical component equal to c, which in this case happens to be the size of the velocity vector.

For an observer watching the lightclock passing by, the vertical component would indeed be different, and this is exactly the reason why that external observer would perceive the clock of the moving observer tick slower.

For you this seems to be a convincing argument for an absolute frame of reference. I don't have the slightest clue where that could possible ever fit in.

But there's even more. Assume we have two lightclocks that move relative to eachother. Each one will conclude that the other clock is ticking slower because the light will follow a triangular path from one clock's point of view. Both clocks are right: there is no experiment you can perform that shows that one clock should be preferred over the other, because they both undergo the same effects when moving relative to eachother. In fact, for ANY observer with a light clock this is true. This means that all frames of references are equal, or in other words: there is no prefered frame of reference.

Light cannot be used as a frame of reference: any frame of reference is bound to travel at a speed smaller than lightspeed, since in a frame of reference travelling at lightspeed, all laws of physics break down (all distances become zero, everything takes an infinite amount of time to happen - not quite an interesting frame of reference to observe reality in).

"What you can't face is the fact that light came along and crapped on Einsteins theory of relativity. I guess everything was going fine for Einsein until he found that the speed of light doesn't care what frame of reference it's in. "

Have you ever read any of the posts on frames of reference ? It was exactly Einstein who postulated that the speed of light was invariant for all frames of reference, and he started to work on his theory from there.

"If a stupid physicist would assume that the light clock measures the total velocity of the light, he/she would come to the conclusion that time slows down the faster the clock is traveling. This could be one of the many errors associated with atomic clocks. "

No, it is not the observer standing next to the lightclock that will see time slow down, but an external observer. For the observer standing next to the lightclock, nothing will change, regardless of the speed he is moving at (from the point of view from the external observer).

Bye!

Crisp

Crisp,

You're not making sense.


"If the speed of light is constant, in which frame of reference would the speed of light be c???? The answer is the absolute frame of reference."

Wrong answer, the correct answer is: in ALL possible frames of reference. The direction the light is moving in is irrelevant, we're not talking about a vector component here, but the total size of the velocity vector.

If I'm in a frame of reference that is not moving, and I shine a beam of light, the light travels at the speed of c.

However, if I'm travelling at .99 c and I shine a beam of light forward, the light CANNOT be traveling at c in my frame of reference. If it were the total speed of the light would be .99c+c. This would mean that the light would be travelling faster than c in the stationary frame of reference.

Let me say again that speed of light is constant compared to an absolute frame of reference. If you are right, and the speed of light is constant in all frames of reference, then the speed of light would NOT be constant at all.

You can't say that the speed of light is constant and then say that it's relative to a frame of reference. The two statements contradict each other.

Tom

Hi Tom,

Ok, it looks like we're getting to the fundamental difference here...

"However, if I'm travelling at .99 c and I shine a beam of light forward, the light CANNOT be traveling at c in my frame of reference. If it were the total speed of the light would be .99c+c. This would mean that the light would be travelling faster than c in the stationary frame of reference."

Wrong. If you travel at .99c and you shine a beam forward, it will move away from you at c. That is the beauty of the constancy of the speed of light, regardless of your own movement, the light travels at the same speed.

For an outside observer, the "total" speed of light would not be .99c+c. You are using the Galilean addition of velocities, which does not work for light, the proper velocity addition formula (which is based on the Lorentz transformations) takes the constancy of the speed of light into account. The resulting velocity for the outside observer would also be c.

Bye!

Crisp

Wrong. If you travel at .99c and you shine a beam forward, it will move away from you at c. That is the beauty of the constancy of the speed of light, regardless of your own movement, the light travels at the same speed.
Yes, that is what I was trying to say before. No matter <b>how</b> you're moving, as long as you're moving at a constant velocity, you'll measure the speed of light to be c. Period. If you witness a person moving at .5c turn on a light beam that travels away from them at c, from their frame of reference the speed of light is c. From your frame of reference, the speed is c.

BTW, there is no absolute frame of reference. This was proven by the Michaelson-Morley (sp?) experiment. They shine light in different directions and measure the speed via and interference pattern, and guess what? It measures the speed of light to be c in any direction, even though the earth is moving at some thousands of miles an hour around the sun.

Also, Prosoothus, the speed of light does not "speed up" in the light clock if it's moving horizontally, it stays the same. Hence time dialation, it takes longer for the light to travel between the two mirrors.

SpyFox_the_KMeson, Crisp, Thed, and James R,

Unfortunately, nothing that you guys are saying is making any sense to me. Maybe I'm just stupid. Therefore I pose a question to all of you, so that I can understand what's really going on:

There are to observers. Observer1 is (relatively) stationairy. Observer2 is traveling towards observer1 at .90 c(270,000 km/s). As soon as observer2 flys over over observer1, observer2 turns on his flashlight and shines the light in the same direction that he is travelling. Observer2 continues to travel at .90 c away from observer 1. One second after observer2 flew over observer1 where is the light??

a) The light is 300,000 km away from observer1. The light is 30,000 km away from observer2. Observer2 is 270,000 km away from observer 1.

b) The light is 300,000 km away from observer1. The light is 300,000 km away from observer2. Observer2 is 270,00 km away from observer1.

As you can see, if the answer is a), then observer2 is seeing the light travelling at only 30,000 km/s or only .10 c, because he is traveling at .90 c.

However, all of you are claiming that the answer is b). This would mean that there are two beams of light: one beam which is 300,000 km from observer1, and the other which is 570,000 km away from observer1.

Which is the correct answer???

Tom

Hi Tom,

There is no point in using intuition here (as you did when formulating both your answers). I am pretty confident that both possibilities are wrong, but I'll do all the math in an hour or two and post it once I worked it out.

Bye!

Crisp

Hi Tom,

I've done the maths for the problem you mentioned, here are the results:

Problem

Consider two observers O and O', where O' moves horizontally with a speed v = 0.9c with respect to O. When the two observers meet, O turns on a flashlight in the direction of motion of O'. After one second, measured in the frame of reference of O, where is the light for O' ?

Solution

For O, the solution is simple: since the speed of light is c for him, after one second in his frame of reference, the light will be 3 * 10⁶ meters in front of him. Symbolic notation:

t = 1 s
d = 3*10⁶ m (distance covered by light)

Hence, for observer O, after one second, the light is located at spacetime point ( t, d, 0, 0 ), i.e. a distance of d along his x-axis. We assume that the light is stopped there (eg. it is absorbed or something)

This same event can be viewed from the perspective of O', by using the Lorentz transformation between the two observers. This transformation is given by:

t' = gamma * ( t - V*d / c² )
d' = gamma * (d - V*t)

V = 0.9c
gamma = 1 / sqrt(1 - (V/c)²) = 1 / sqrt( 1 - 0.9*0.9 )

Hence, for O', the light will stop at (t', d', 0, 0) with:

t' = 0,2294 s
d' = 6,8824 * 10⁴ m

This difference arises because there are two important relativistic effects to consider when looking from the perspective of O': Lorentz contraction and time dilatation. The Lorentz transformation takes these two effects into account.

As you can see, the speed of light from the frame of reference of O' is given by:

speed = distance / time = d' / t' = c

Explanation

The speed of light is invariant and thus the same for the two observers. The distance the light covers and the time it requires to do so differ.

You'll probably will question the Lorentz transformation at this point, but to be completely honest, that is something I cannot explain without writing a 20 page document and attaching it to a message. In case you are wondering: No, I will not do that :).

Bye!

Crisp

Crisp, nice work!

Prosoothus, if you want to read something (relatively accessible, but still quite hard to read) on the subject, maybe Roger Penrose's "The Emperor's new Mind" is good. Somewhere halgway the volumunous book is a piece on the subject.
(Does anybody have an alternative?)

Merlijn

Prosoothus, it would be answer B I believe. Now you understand why relativity is so weird, and why so many people refused to accept it.

Crisp,

First of all, light travels at 3*10^8 m/s and not 3*10^6 m/s as you indicated.

Secondly, your formulas provide no proof. You assumed that time slows down, and you calculated the distance light travelled so that the speed of light would be constant to the second observer. That's very bad logic.

To prove you wrong, let's look at distance of the light travelled compared to the stationary observer:

If you change c in your formulas to the correct value, then:

d2=6.8824 * 10^6 m

However, the total distance the light would pass in that relative time period would be:

Total distance=distance of the moving observer from the stationary observer+distance light travelled from the moving observer

Which is:

Total distance = .90c + 6.8824 * 10^6
=2.768824 * 10^8

That means that if you add the distance of the moving observer from the stationary observer, with the distance that the light travelled from the moving observer, you get 2.768824 *10^8 m.

However, you said that the light moved 3 * 10^8 meters from the stationary observer.

There is a discrepency of 8% in your formulas.

As you can see, your formulas give two different results in the stationary observer's frame of reference.

Tom

SpyFox_the_KMeson,

It can't be b. Answer b gives two different results, just like Crisp's formulas. Logic dictates that the same photon can't be at two different places at the same time.

Tom

Tom,

Here's your problem:

<i>There are to observers. Observer1 is (relatively) stationairy. Observer2 is traveling towards observer1 at .90 c(270,000 km/s). As soon as observer2 flys over over observer1, observer2 turns on his flashlight and shines the light in the same direction that he is travelling. Observer2 continues to travel at .90 c away from observer 1. One second after observer2 flew over observer1 where is the light??</i>

The first thing to ask is: according to whom? Who is trying to determine the position of the light? Observer 1 or observer 2? It makes a difference, since we need to know who the one second time interval is measured by. Since you have not specified, I will do both for you.

First, <b>observer 1's frame of reference, assuming that observer 1 sees observer 2 switch on the light and then looks at its position 1 second later as measured by observer 1's clocks.</b>

After 1 second, the light is 300,000 km away from observer 1 by this measure. Observer 2 is 270,000 km away from observer 1. The light is therefore 30,000 km from observer 2 <b>as measured by observer 1</b>. If observer 1 calculates the speed of light using the measured distance and time he finds it is 300,000 km/1 second = 300,000 km/s.

Now, in the 1 second that observer 1 measures, observer 2 observes a time interval of 0.44 seconds due to time dilation. In observer 2's frame of reference, in that amount of time observer 1 travels 270,000 km/s &times; 0.44 s = 117690 km. In the same time, the light travels 130767 km, but in the opposite direction. The distance between the light and observer 1 is therefore 130767+117690 = 248457 km, as measured by observer 2. If observer 2 calculates the speed of light by dividing the distance by the time he gets 130767/0.44 = 300,000 km/s.

Second, <b>observer 2's frame of reference, assuming that observer 2 switches on the light and then looks at its position 1 second later as measured on observer 2's clocks.</b>

Observer 2, from his own frame of reference, is stationary, so he sees observer 1 moving at 0.9c relative to him in the opposite direction to the beam of light. He sees the light moving at speed c in front of him. So, after 1 second as measured by him, he is 270,000 km away from observer 1, the light is 300,000 km away from him, and the light is 570,000 km away from observer 1. Observer 2 measures the speed of light to be the distance divided by the time, again giving 300,000 km/s.

Now, in the 1 second that observer 2 measures, observer 1 observes a time interval of 2.29 seconds due to time dilation. In observer 1's frame of reference, in that amount of time observer 2 travels 270,000 km/s &times; 2.29 s = 619422 km. In the same time, the light travels 688247 km, in the same direction. The distance between the light and observer 2 is therefore 688247 - 619422 = 68825 km, as measured by observer 1. If observer 1 calculates the speed of light by dividing the distance by the time he gets 688247/2.29 = 300,000 km/s.

Notice that in both cases, observers 1 and 2 measure different travel distances and times for each other and for the light, but their calculations of the speed of light always give 300,000 km/s. In other words, the speed of light is the same in both reference frames, but their ideas of distances and times change.

Tom,

One more thing...

<i>Secondly, your formulas provide no proof. You assumed that time slows down, and you calculated the distance light travelled so that the speed of light would be constant to the second observer. That's very bad logic.</i>

It's exactly the same logic you are using but with different assumptions. Crisp and I are assuming that relativity is correct. You are assuming that Galilean relativity is correct, even though you probably don't realise that that's the name for your particular world-view in this example.

The relativistic formulae give one set of answers. The "common-sense" Galilean formulae which you insist on using give another set of answers. This thought experiment you've put forward won't tell us which set of formulae are correct and which are wrong. The only thing that will tell us that is real experiments.

Guess what? The real experiments show that the relativistic formulae are correct and Galileo's common sense is wrong.

End of story.

Hi Tom,

"First of all, light travels at 3*10^8 m/s and not 3*10^6 m/s as you indicated."

LOL, indeed... This is the way you get punished for always putting c = 1 in your calculations :).

"It can't be b. Answer b gives two different results, just like Crisp's formulas. Logic dictates that the same photon can't be at two different places at the same time."

You have to be careful saying that: in relativity both time and space are relative to the observer, so a photon can be at more places in the sense that two observers might not agree on the distance traveled. This gives two different results, but ofcourse does not mean that the photon has split into two and is at two places.

Bye!

Crisp

James R,


After 1 second, the light is 300,000 km away from observer 1 by this measure.


The distance between the light and observer 1 is therefore 130767+117690 = 248457 km, as measured by observer 2.

You made the same mistake as Crisp, only more severe. The distance between observer 1 and the light should be equal for both observers.

But 300,000 km does not equal 248,457 km.

You cannot claim that the discrepancy is the result of time dilation either, since you included time dilation in your calculations already.

Tom

Tom,

<i>You made the same mistake as Crisp, only more severe. The distance between observer 1 and the light should be equal for both observers.</i>

No, wrong again, I'm afraid. Length contraction is another well-known result of special relativity. Notice that the different distances are measured by <b>different</b> observers.

James R,

It's kind of funny how the time dilation formula is so similiar to the length contraction formula that they balance each other out in the formula for velocity.

As you pointed out in the "Galileo & Einstein - second thoughts" forum, time dilation and length contraction are superfluous, therefore time dilation and length contraction don't exist.:)

Tom

Tom,

<i>It's kind of funny how the time dilation formula is so similiar to the length contraction formula that they balance each other out in the formula for velocity.</i>

Call it funny all you want. That's the way the world works.

<i>As you pointed out in the "Galileo & Einstein - second thoughts" forum, time dilation and length contraction are superfluous, therefore time dilation and length contraction don't exist.</i>

I don't like being misquoted. Please quote me where I said time dilation and length contraction are superfluous, provide a link to the relevant post, or admit that I said no such thing.

In the Galileo & Einstein - second thoughts" thread I explained how absolute motion does exist, but that absolute motions cancel each other out in a frame of reference. Your response was:


Yes, that's right, Tom. Therefore, the concept of a is superfluous. There is no absolute reference frame.

Therefore, since the relativistic effects of time dilation and length contraction cancel each other out in the velocity formula, I guess you would imply that they don't exist either???


I don't know about other people, but I get the impression that when Einstein tested his "invariant property of light" theory, he found that his results did not match reality. So instead of admitting that he was wrong, he made up length contraction so that his results would match reality.

The fact is that if you assume that there is no time dilation and no length contraction you get the same result as if there is time dilation and length contraction.

In layman's terms, Einstein made something up. Because what he made up didn't match reality, he made something else up that cancels the first thing that he made up.

It's such a shame when scientists shape the universe based on their theories, instead of the other way around.

Tom

Hi Tom,

I am sure that 137 will pop into this thread and explain you nicely how the concepts of time dilatation and length contraction evolved. I'll give it a brief shot without too much details:

It has for long been known that Newton's theory of mechanics does not work well for high speeds. Several experiments around the 1850's suggested that the speed of light was an upper limit for electrons that are accelerated in a high electric field. Around the 1880's, the idea had grown that light propagated through aether, and that perhaps that aether, at that time thought as the propagating medium for light, had something to do with the strange behaviour experiments brought forward (Newton simply didn't work out).

The Michelson-Moreley experiment, performed around that time (1880's) hinted that there probably was no such thing as aether, but at that time, the aether was firmly believed in and the result of the experiment was explained by other means.

One idea came from Lorentz, who, around the year 1900, proposed that the aether actually tampered with measurement devices, shortening them in the direction of movement. This idea could explain why the Michelson-Moreley experiment did not detect the presence of aether. The effect is still known today as length contraction or Lorentz contraction.

However, that meant that aether was undetectable (it made itself undetectable by tampering with measurement equipment), and this idea annoyed physicists: there appeared to be some unknown "force" that played with the experiments.

Einstein in 1905, after the concept of Lorentz contraction was introduced, proved that with the constancy of the speed of light, no mysterious force was required to explain Lorentz-contraction: it was a property of nature, a consequence of the invariance of the speed of light. (This also made the concept of aether superfluous, and there was more than just this reason).

Hence, Einstein did not come up with the idea of Lorentz contraction. He did not tamper with some laws to get things right.

Also, time dilatation and lorentz contraction do not nicely compensate in the formula's for moving observers. The proper relations are (if O' is moving with respect to O):

t = gamma t'
d = d' / gamma

gamma > 1, hence t is larger than t' (moving clocks tick slower for the stationary observer) and d is smaller than d' (moving objects contract in length for the stationary observer). Where do the two effects compensate ? The fact that everything works out nicely for light is not because the formula's were tampered with, it is merely an indication that the theory of relativity is self-consistent: you start with the invariance of the speed of light, this induces the existance of time dilatation and lorentz contraction, and by taking those two effects into account (as James R nicely did), you get back to the invariance of the speed of light.

Bye!

Crisp

Tom,

<i>In the Galileo & Einstein - second thoughts" thread I explained how absolute motion does exist, but that absolute motions cancel each other out in a frame of reference.</i>

In other words, absolute motion (if it exists) has no observable consequences at all. Therefore, by Occam's razor it is superfluous, as I said.

On the other hand, time dilation and length contraction is observable, so those concepts are not superfluous.

<i>Therefore, since the relativistic effects of time dilation and length contraction cancel each other out in the velocity formula, I guess you would imply that they don't exist either???</i>

No. They are measurable in other ways, unlike an absolute standard of rest.

<i>I don't know about other people, but I get the impression that when Einstein tested his "invariant property of light" theory, he found that his results did not match reality. So instead of admitting that he was wrong, he made up length contraction so that his results would match reality.</i>

1. Einstein did not invent length contraction. You might want to look up Fitzgerald.
2. The invariance of the speed of light has been tested over and over again, not just by Einstein. It matches reality.

<i>The fact is that if you assume that there is no time dilation and no length contraction you get the same result as if there is time dilation and length contraction.</i>

Wrong. If you throw out time dilation and length contraction, you are forced to adopt the position that the speed of light is different in different reference frames That is contrary to observation.

<i>In layman's terms, Einstein made something up.</i>

I think it is you who is making something up.

<i>It's such a shame when scientists shape the universe based on their theories, instead of the other way around.</i>

It's such a shame when people comment on things they don't understand.

James R,

First I'd like to thank you for attempting to explain the question I asked a few posts ago. It must have taken a lot of time.

Unfortunately, after reading your post I still don't understand something: Where did you get the 130767 km, number from?

I looked at results I got from the formulas for time dilation and length contraction but wasn't able to get that number in any way.

Please explain the formula you used to get this number.

Thanks

Tom

The speed of light is 300,000 km/s. It is a fundamental postulate of special relativity that the speed is the same in all reference frames. You have previously agreed that that is the case.

In a time of 0.44 seconds, light travels a distance given by

s = vt = 300,000 &times; 0.44 = 130767 km.

Thanks to Crisp, JamesR, Prosoothus, and everyone else who takes part in these Physics discussions. I've actually learnt quite a lot out of it all, many things have been made clear.

Crisp, James R, Thed, and Adam,

Both James R and Crisp have proved that the speed of light is constant in all frames of reference by including that assumption in their formulas. To me this is isn't proof at all.

I attempted to prove that the speed of light was constant in all frames of reference by only using the formulas for time dilation and length contraction, since they are the only formulas that would "influence" the speed of light relative to the moving observer.

Unfortunately, I never came to the result that the speed of light is c for a moving observer. I tried using only time dilation, then only length contraction, then both, but the result was never c for a moving observer.

Here is a simplified question with only one observer:

An observer is traveling at .90c. The observer turns on his/her flashlight, which is pointing in the same direction as his/her motion, while continuing to travel at that speed. After one second of a stationairy clock, how far is the light from the observer, and using only the formulas for time dilation and length contraction, explain why the light is travelling at c relative to the observer.

Do not make the assumption that light is travelling a c relative to the observer, instead, use only the formulas for time dilation and length contraction to prove that the light is traveling at c relative to the observer.

Thanks to anyone who can explain this to me.

Tom

Tom,

You're obviously a little mixed up about all this stuff. Let me see if I can make it a bit simpler for you...

1. It is a <b>postulate</b> of special relativity that the speed of light is constant in all reference frames.
2. The word "postulate" means, essentially, "assumption". A postulate is not based on anything <i>a priori</i>. It is assumed. The consequences are then derived theoretically.
3. To see whether a postulate is ultimately justifiable or not, <b>the only true arbiter is experiment and observation</b>. We must look at the real world and compare the theory based on the postulates with what really happens. If the theory matches the reality, the postulate is correct (or at least useful). If not, the theory is wrong, and hence the postulate is wrong since the theory is derived from it.
4. In the case of special relativity, <b>all experiments and observations</b> support the constancy of the speed of light in all inertial reference frames, with no exceptions. Hence, we conclude that the postulate is true.
5. Time dilation and length contraction are <b>inevitable consequences</b> of the constancy of the speed of light in all inertial reference frames. Since they are directly observed, they provide further evidence for the constancy of the speed of light.
6. If, instead of the postulate of the constancy of the speed of light you choose to assume that time dilation and length contraction exist in the same mathematical form that Einstein et al. proposed, then the constancy of the speed of light follows inevitably from that alternative assumption.
7. It is incorrect that there is any circularity in relativity. The theory is justified by experiment, not by other theory or itself.

Now, to your comments...

<i>I attempted to prove that the speed of light was constant in all frames of reference by only using the formulas for time dilation and length contraction, since they are the only formulas that would "influence" the speed of light relative to the moving observer.</i>

See point 6, above.

<i>Unfortunately, I never came to the result that the speed of light is c for a moving observer.</i>

Then your maths is not very good. See point 6, above.

<i>Here is a simplified question with only one observer:

An observer is traveling at .90c. The observer turns on his/her flashlight, which is pointing in the same direction as his/her motion, while continuing to travel at that speed. After one second of a stationairy clock, how far is the light from the observer, and using only the formulas for time dilation and length contraction, explain why the light is travelling at c relative to the observer.

Do not make the assumption that light is travelling a c relative to the observer, instead, use only the formulas for time dilation and length contraction to prove that the light is traveling at c relative to the observer.</i>

1 second on a stationary clock corresponds to 2.29 seconds for the observer, using time dilation.

In the stationary reference frame, the speed of light is 300,000 km/s. The speed of the observer is 270,000 km/s, so after 1 second (measured in the stationary frame), the light is 30,000 km from the observer <b>as measured in the stationary frame</b>.

In the observer's frame, in the 2.29 seconds the point which the light started from moves backwards from the observer (who considers himself stationary) a distance of 270,000 &times; 2.29 = 619422 km. Due to the length difference the observer measures from the stationary frame, after 2.29 seconds, the light is a distance of 30,000 &times; 2.29 = 68,825 km from the observer at the end of the time interval, but in the opposite direction to the start point.

To calculate the speed of the light, the observer takes the total distance measured by him for the travel of the light divided by the total time measured by him.

The total distance is: 619422 + 68825 = 688247 km
The total time is: 2.29 seconds.

The speed of light, according to the observer is therefore 688247/2.29 = 300,000 km/s

Note that the only assumptions made in getting this solution were:

1. The speed of light in the stationary frame is 300,000 km/s.
2. The time dilation and length contraction formulae of special relativity apply.

From that, the constancy of the speed of light for the moving observer follows inevitably. See point 6, above.

Surely you must be satisfied now?

(I'm betting you'll now change your tune and claim that the speed of light is not constant in all frames of reference after all.)

James R,

Thanks for your explanation, but I don't understand this part:


In the observer's frame, in the 2.29 seconds the point which the light started from moves backwards from the observer (who considers himself stationary) a distance of 270,000 × 2.29 = 619422 km.

The light at all times is moving in the same direction as the observer, not in the opposite direction.

Here is my explanation. Please correct my errors:

After 1 second of the stationairy clock the observer is 30,000 km away from the light. Therefore:

v=30,000km/1s=30,000km/s

For that 1 second of the stationairy clock, 2.29 seconds passed for the observer. Therefore:

v=(30,000 km/s)/2.29=13,100 km/s

where 2.29 is equal to 1/sqrt(1-(v^2/c^2))

Because of space contraction:

v=13,100 km/s *.436=5711.6 km/s

where .436 is equal to sqrt(1-(v^2/c^2))

As you can see, I took the stationairy distance between the light and the observer and divided it by the stationairy time in order to get the stationairy velocity(relative to the observer). Then I divided this velocity with the time dilation of the observer. Finally I multiplied it with the length contraction relative to the observer. As you can see, I don't get 300,000 km/s.

Please point out any errors I have made.

Tom

Hi Tom,

I'll attempt to clarify the part you did not understand. This will probably give you some insights in where your reasoning goes wrong; it is always more fun and more educational to find the error yourself...

"Thanks for your explanation, but I don't understand this part:


In the observer's frame, in the 2.29 seconds the point which the light started from moves backwards from the observer (who considers himself stationary) a distance of 270,000 × 2.29 = 619422 km.
The light at all times is moving in the same direction as the observer, not in the opposite direction."

That is also what James assumes: the light travels in the same direction as the observer, from the stationary observer's point of view, or forward for the moving observer. With the notation O' for the moving observer and O for the stationary observer: O' sees the light advance straight ahead of him (at c) and O sees the light advance in the direction of motion of O'.

For O' to calculate the speed of light, he has to take into account the total distance the light has travelled. At a certain point, he turns on the light, let's mark this event (1). Then, for a certain time (1s for O, 2.29s for O'), he travels forward. In that second, the location where event (1) happened is left behind. In ASCII art, the situation after that 1s/2.29s is:

--- (1) --------------------- O' ----- P -->

where (1) is the location the flashlight was turned on, O' is the location of observer O' and P is the location of the photon. The total distance the light has travelled is hence not just the distance between P and O', but you also have to take the distance between O' and (1) into account. This is the additional factor 270,000 × 2.29 = 619422 km.

Bye!

Crisp

Crisp,


The total distance the light has travelled is hence not just the distance between P and O', but you also have to take the distance between O' and (1) into account. This is the additional factor 270,000 × 2.29 = 619422 km.

Why would I have to take it into consideration???

To the observer the light only traveled from O' to P.

Everyone knows what the stationairy distance is between O' and P in your diagram. Why can't I just take this stationairy distance and convert it to relative distance(for the observer) using the formula for length contraction.

Then take the stationairy time and convert it to relative time (for the observer) using the formula for time dilation.

Finally, I divide the relative distance by the relative time to get the speed of light for the observer.

30,000km * length contraction = 13,080 km

1 second * time tilation = 2.29 seconds

Therefore the speed of light to the observer is:

v=d/t=13,080/2.29= 5711.79 km/s

In other words, to the observer, since the light is 13,080 km away in 2.29 seconds this means that the light is traveling 5711.79 km/s in the observers frame of reference.

As you can see, 5711.79 km/s comes nowhere close to 300,000 km/s.

Tom

Tom,

The way you're doing it you're not measuring the speed of light. To measure the speed your must mark off on your space axis the point where the light started and where it ended up. The distance between those two points is the distance the light has travelled.

What you're doing is marking off the position of the <b>observer</b> and the position the light ends up at and pretending that the distance between those points is the only distance the light has travelled.

Use your common sense.

James R,

I AM using common sense. I don't care what the total distance of the light is, I just care about the distance the light travelled relative to the observer.

Fact: The light is 30,000 km away from the observer after 1 second in the stationairy frame of reference.

Question 1: How far away is the light from the observer after
1 stationairy second in the observers frame of reference?

Question 2: How much time passed by for the observer during that one second of stationairy time?

Question 3. By dividing the answer from question 1 by the answer from question 2, how fast is the light traveling in the observer's frame of reference?

It doesn't get any simpler or more logical than this.

Let me repeat myself again, I'm not interested in the speed of light in the stationairy frame of reference, I'm only interested in the speed of light in the observer's frame of reference. To calculate the speed of light in the observer's frame of reference, you do not need to know the total path of the light, you only need to know how far the light is from the observer, and how much time it took to get there.

Tom

Hi Tom,

"you only need to know how far the light is from the observer, and how much time it took to get there."

I guess that James and I, and quite some other people on the forum, eagerly await the answer to how you would calculate that.

Bye!

Crisp

To calculate the speed of light in the observer's frame of reference, you do not need to know the total path of the light, you only need to know how far the light is from the observer, and how much time it took to get there.

And by some strange coincidence, it *is* the total path of the light... in any reference frame. ;)

Crisp,


"you only need to know how far the light is from the observer, and how much time it took to get there."

I guess that James and I, and quite some other people on the forum, eagerly await the answer to how you would calculate that.



I already did the calculation in two of my previous posts. I'm sure not going to do it a third time.

Tom

Tom,

<i>Fact: The light is 30,000 km away from the observer after 1 second in the stationairy frame of reference.

Question 1: How far away is the light from the observer after
1 stationairy second in the observers frame of reference?</i>

68,700 km. The stationary frame sees a contracted distance of 30,000 km.

<i>Question 2: How much time passed by for the observer during that one second of stationairy time?</i>

0.229 seconds. The stationary frame sees the moving clocks as running slow.

<i>Question 3. By dividing the answer from question 1 by the answer from question 2, how fast is the light traveling in the observer's frame of reference?</i>

68,700 km/ 0.229 seconds = 300,000 km/second.

<i>To calculate the speed of light in the observer's frame of reference, you do not need to know the total path of the light, you only need to know how far the light is from the observer, and how much time it took to get there.</i>

That is correct. Here's how I worked it out, in detail:

Consider 2 events:
Event A: The observer turns on the light.
Event B: The position of the light pulse is measured after 1 stationary second.

In the <b>stationary</b> frame, the spacetime co-ordinates of the two events are:
Event A: (x,t) = (0 km, 0 seconds)
Event B: (x,t) = (300,000 km, 1 second)

The speed of light in this case is the change in the distance divided by the change in the time, which gives 300,000 km/s.

In the <b>observer's</b> frame, we must use the Lorentz transformations to get the observer's spacetime co-ordinates for events A and B. The Lorentz transformations from the (x,t) co-ordinate system to the (x',t') co-ordinate system are:

x' = <font face="symbol">g</font>(x - vt)
t' = <font face="symbol">g</font>(t - vx/c²)

(which you can confirm by looking them up on any relativity web site).

For event A, we find:
(x',t') = (0 km, 0 seconds).

For event B, we find:

x' = 2.29 (300,000 - 270,000 &times; 1) = 68,700 km
t' = 2.29 (1 - (0.9c)(300,000)/c²)
= 2.29 (1 - 270,000/300,000) = 2.29 &times; 0.1 = 0.229 seconds.

Again, the speed of light in this case is the change in the distance divided by the change in the time, which gives:

68,700 km/ 0.229 seconds = 300,000 km/s

Lorentz transformations!! Ahhh I love Physics!

James R, I'm curious, how far in the education ladder did you climb?

You know alot, intelligent :D

Have you watch the Mechanical Universe tapes (MUS) by Caltech? They are great for conceptualizing of science and the basics.

The Chosen,

You never stop climbing the education ladder. There's always more to learn.

I haven't seen the MUS tapes, but thanks for the tip.

James R,

I learned to accept just about anything from you, but I didn't expect you to outright lie.


"Question 1: How far away is the light from the observer after 1 stationairy second in the observers frame of reference?"

68,700 km. The stationary frame sees a contracted distance of 30,000 km.

The stationairy observer does not see a contracted distance, because he/she is at rest. The moving observer sees the contracted distance. Therefore the distance is 13,080 km for the moving observer.


Question 2: How much time passed by for the observer during that one second of stationairy time?

0.229 seconds. The stationary frame sees the moving clocks as running slow.

Wrong again. If the the time for the moving observer is running slower than the stationairy frame, then 1 second in the stationairy frame is 2.29 seconds for the moving observer.

Funny, just a few posts ago, you said it was 2.29 seconds as well. You must have changed your calculations when they didn't give you the correct result, right?? :)

And how exactly did you get 0.229 seconds from the formula:

t = t0/sqrt(1-(v^2/c^2))

Was it those magic fairies again???

Tom

The stationairy observer does not see a contracted distance, because he/she is at rest.

The stationary observer will calculate the contracted distance of the moving observer relative to his own position.

The moving observer sees the contracted distance.

The moving observer will see everything normal and will only experience the effects of contracted distance relative to the stationary observer.

Hi Tom,

"Was it those magic fairies again???"

LOL, you just gotta love that sarcasm :). Too bad I don't have time at the moment to go deeper into the calculations (exam of field theory closing in awfully fast), but if you don't get everything settled and discussed before 2nd of July, be sure to remind me to look at this thread again.

Bye!

Crisp

Q,

You're right.

Thanks for correcting my statements. That's what I get for being impatient.

Tom

Crisp,

Good luck on the exam.

Just don't let them brainwash you too much.:)

Tom

Tom,

I've thought this problem through properly now. My previous post (where I used the Lorentz transformations) is the correct solution to your problem. Please disregard my previous posts on that issue. There's mostly truth there, but it's mixed in with a couple of mistakes. Relativity is tricky. You need to be very careful about which reference frame you are using at all times. I played it a bit fast and loose initially, but now I've tightened up the reasoning. As I say, my post immediately before this one is the definitive and correct answer, so let's just concentrate on it and go from there.

I said the observer measures a distance of 68,700 km. The stationary observer measures a shorter distance of 30,000 km.

You said: <i>The stationairy observer does not see a contracted distance, because he/she is at rest. The moving observer sees the contracted distance. Therefore the distance is 13,080 km for the moving observer.</i>

Remember that relativity is symmetrical. The stationary observer sees the moving observer's clocks running slow and the moving observer's lengths contracted. The moving observer sees the stationary observer's clocks running slow and the stationary lengths contracted. This can cause a lot of confusion when you try to apply the length contraction and time dilation formulae. As I said, you need to be VERY careful that you get things the right way round. I stuffed it up in my first couple of posts on this problem, which is why I decided to use the Lorentz transformations instead. That way, I could be guaranteed I was keeping track of everything properly.

Anyway, what all this boils down to is that your statement above is incorrect, and my statement is correct.

<i>If the the time for the moving observer is running slower than the stationairy frame, then 1 second in the stationairy frame is 2.29 seconds for the moving observer.</i>

No, that statement would be wrong regardless of my calculation. Think about it. The stationary observer sees the moving observer's clocks as running slow. Therefore, in the time that the stationary clocks tick off 1 second, we'd expect that the moving clocks, according to the stationary observer, would tick off <b>less</b> than 1 second, since the moving clocks are ticking slower from this point of view. Therefore, it cannot possibly be the case that the stationary observer sees the moving clock tick off 2.29 seconds in the time that the stationary observer's clock ticks off 1 second.

<i>Funny, just a few posts ago, you said it was 2.29 seconds as well. You must have changed your calculations when they didn't give you the correct result, right??</i>

Yes. Right. I was wrong the first time, but now I am correct. My previous paragraph above makes clear where I made my mistake.

<i>And how exactly did you get 0.229 seconds from the formula:

t = t0/sqrt(1-(v^2/c^2))

Was it those magic fairies again???</i>

I didn't use that formula. I used the Lorentz transformations, which are clearly set out above. The formula you have quoted turns out to be not directly applicable in your example. That fact in itself had me confused at first, which is another reason why I made some mistakes initially. In fact, you need the full Lorentz formulae to handle your problem. I have used the correct formulae to come up with the correct answer. No magic fairies required. Just careful thought.

Thankyou for continuing to push me until I got my thinking correct on this. I am now 100% confident that my solution is correct and we can discuss it from there.

So, I imagine you will now take issue with the Lorentz transformation. Can you see any problem with what I have done there? (There isn't any, I assure you.)

James R,

Now I'm confused.

After thinking about what you said about time slowing down in moving clocks, I've also come to the conclusion that 2.29 seconds can't pass for every 1 second of a stationairy clock. In other words the formula should be:

t = t0*sqrt(1-(v^2/c^2))

But even in this case it would mean that for every one second of stationairy time, 0.436 seconds pass for the moving observer.

But 0.436 still does not equal .229 seconds, which you obtained from the Lorentz transformations.

If you are correct, then the time dilation formula I've indicated above must be wrong.

The questions now are:

Is the time dilation formula wrong or is it correct only under certain circumstances??

What about the formulas for increasing mass and length contraction?? Are they also wrong, or are they correct only under certain circumstances??(Since they are similiar to the time dilation formula)

It appears that the two of us can agree upon one thing: Using only the relativity formulas for length contraction and time dilation, the speed of light is not c to a moving observer.

However, if you are correct, and using Lorentz transformations the correct results are obtained, then relativity just got whole lot messier.

Tom

Tom,

I'll have to think about this some more. I <b>think</b> this problem can be solved by a correct application of the usual time dilation formula, but I'll need to think about the situation a bit. I'll get back to you about this when I have some more time.

<i>If you are correct, then the time dilation formula I've indicated above must be wrong.</i>

The time dilation formula is derived from the Lorentz transformations, which are the basic result of special relativity. Your time dilation formula is an <b>application</b> of the Lorentz transformations to a specific set of situations, and therein lies the problem I suspect. The particular example you've come up with is not one of the situations for which the time dilation formula you gave can be used, I think. My thoughts are still tentative on this right now, though.

<i>Is the time dilation formula wrong or is it correct only under certain circumstances??</i>

Only under certain circumstances, I would say. It's a "special case" of the more general Lorentz transformations.

<i>What about the formulas for increasing mass and length contraction?? Are they also wrong, or are they correct only under certain circumstances??</i>

Length contraction is a special case too - only valid under certain circumstances.

I think we've covered relativistic mass in a previous thread.

<i>It appears that the two of us can agree upon one thing: Using only the relativity formulas for length contraction and time dilation, the speed of light is not c to a moving observer.</i>

I don't agree with that.

I'll get back to you soon.

Hi all,

James R,

The "traditional" formula for time dilatation is valid for the situation where the event for the moving observer occurs at x' = 0. The formula is usually derived for a lightclock that ticks back and forth at x' = 0. Inverting the Lorentz-transformation for time yields:

t = <font face="symbol">g</font>(t' + vx'/c²)

which gives the traditional formula if x' = 0.

The correction term "vx/c²" takes the motion of the observer relative to the studied object into account, which I think is exactly what you did by adding the moving observer's travelled distance in your previous posts.

Tom,

I believe the classical method of introducing Lorentztransformations in textbooks on relativity is the following:

1) Postulate that c is constant for all frames of reference
2) This immediatelly leads to the "traditional" formulas for time dilatation and Lorentz contraction (t = <font face="symbol">g</font>t' and d' = <font face="symbol">g</font>d).
3) Derive the Lorentztransformation (for techies: the boost Lorentztransformation) using the traditional time dilatation/length contraction formulas and taking relative motion into account.

This scheme works 1 => 2 => 3, so it must be perfectly possible to invert the scheme: 3 => 2 in the way denoted above (by taking the correct situation) and 2 => 1 in the way James explained.

I think the reason why James had to "complicate" things (adding the travelled distance) is indeed because the time dilatation formula used first is only valid in special cases. So my conclusion would be that it *is* possible to calculate the problem Tom proposed "the hard way" (only using traditional formulas) and compensating for motion correctly, or do it the easy way and use the Lorentz transformations.

Just some thoughts... :)

Bye!

Crisp

Xev, Crisp, James R, Thed, and Q,

This post will summarize the example I have posted earlier in this thread in the event that anyone forgot what the discussion was about.

I posted this problem:

There are two observers with two clocks. One observer is (relatively) stationairy with his clock, while the second observer is travelling at .90 c towards the first observer. When the moving observer reaches the stationairy observer, he turns on his flashlight and points it forward(in the direction that he is travelling). The moving observer passes the stationairy observer and continues travelling at .90c. After 1 second of the stationairy clock:

1) How far is the light from the moving observer in the stationary observer's frame of reference.

2) Use the formula for length contraction and the result from answer 1 to calculate how far the light is from the moving observer in the moving observer's frame of reference.

3) Use the formula for time dilation to calculate how much time passed for the observer in the 1 second of stationairy time.

4) Divide the answer from 2 with the answer from 3 to calculate the speed of the light for the moving observer.


If you take into consideration that time dilation and length contraction do not exist, then after one second the light is 30,000 km away from the moving observer in both frames of reference.

d=c-.90c/1 second
d=300,000 km/s-270,000 km/s/1 second
d=30,000 kms/1 second
d=30,000 km


However, if you assume that time dilation and length contraction occur then the light is only 30,000 km away from the moving observer, after one second, in the stationairy frame of reference.

To figure out how far the light is from the moving observer in the moving observer's frame of reference then:

Using time dilation:

t=t0*sqrt(1-(v^2/c^2))
t=0.436 seconds

In other words for 1 second of the stationairy observer, 0.436 seconds passed for the moving observer.

Using length contraction:

d=30,000 km* .436
d=13,080 km

Now if you divide the length by the time you get:

v=d/t
v=13,080 km/.436 seconds
v=30,000 km/s

In other words, the light is travelling away from the moving observer at a speed of 30,000 km/s in the moving observer's frame of reference. This is only 1/10 the speed of light. Relativity dictates that the answer should be 300,000 km/s.


James R and Crisp claimed that my formulas don't apply in this case and that the answers are(according to James):

d=30,000 km/.436
d=68,700 km

while

t= .229 seconds(don't ask me how he got this number, I have no idea)

Therefore:

v=68,7000 km/0.229 seconds
v=300,000 km/s

Even though I don't agree with James R's result, let's pretend that it is correct for the sake of argument.

So if the moving observer was to shine the flashlight in the opposite direction of his motion when he passed over the stationairy observer then the light would be(after 1 second):

d=300,00 km+270,000 km
d=570,000 km

The light would be 570,000 km away from the moving observer after 1 second in the stationairy observers frame of reference.

Using James R formulas you get:

d=570,000/.436
d=1307339 km

The distance between the light and the moving observer in the moving observers frame of reference is 1307339 km/s.

Since the observer is still moving at .90 c , time dilation according to James is still:

t=.229

Therefore:

v=d/t
v= 1307339/.229
v= 5708905 km/s

According to James R's formula the speed of light in the moving observer's frame of reference is 5,708,905 km/s. That is almost 20 times faster than c.

I'm sure that James R and Crisp don't even want to think about what the results would be if the light is going up or down, instead of forwards and backwards in the moving observer's frame of reference. :)

Tom

Cute, I'll think about it later today. However, I can already point out that you are using the formulas that James R and I called invalid in this scenario. But there must be some way to work around that... That will take some thinking :)

Bye!

Crisp

Crisp,

Just to remind you, when you find the time dilation and length contraction ratios, using the Lorentz transformations or the traditional formulas, you must use the same results for both cases (where the light is shining forward, and where the light is shining back.)

And in both cases, the result has to be 300,000 km/s.

Note: You will find that it is a mathematical impossibility.

Tom

while

t= .229 seconds(don't ask me how he got this number, I have no idea)

Is it not the answer calculated using the time dilation formula from the moving observer relative to the stationary observer ?

Q,

I'm not sure how James R obtained the .229 figure. He either:

1) Obtained the figure using the Lorentz transormations.( I'm not familiar with these formulas)

2) Obtained the formula by assuming that the speed of light is c, relative to the moving observer. He then calculated it using the formula t=v/d.

According to the traditional formula:

t=t0/sqrt(1-(v^2/c^2))

the result is .436 seconds or 2.29 seconds if you look at it in the other frame of reference.

In no cases, is t equal to .229 seconds.

Tom

Prosoothus

According to the traditional formula:

t=t0/sqrt(1-(v^2/c^2))

the result is .436 seconds or 2.29 seconds if you look at it in the other frame of reference.

In no cases, is t equal to .229 seconds.

I think 2.29 is the Gamma factor.

The moving observer would view the stationary observers clock as running slow. He would view the stationary observers clock as .229 seconds relative to his own clock which ticked one second.

Q,

Actually, for every second of the stationairy observer's clock, 0.436 seconds would pass for the moving observer's clock.

How did you get .229????

Tom

Prosoothus

You're not following me here. The moving observer cannot tell if he is moving or the stationary observer is moving, therefore he views the stationary observers clock as running slow. If one second passes for the moving observer, and he views the stationary observers clock ticking slower, he would view the stationary observers clock as ticking .229 seconds.

Q,

I am following you, I just don't know where you got the .229 figure from.

Tom

Prosoothus

James R already worked it out:

t' = 2.29 (1 - (0.9c)(300,000)/c2)
= 2.29 (1 - 270,000/300,000) = 2.29 × 0.1 = 0.229 seconds

Q,

[QUOTE]James R already worked it out:

t' = 2.29 (1 - (0.9c)(300,000)/c2)
= 2.29 (1 - 270,000/300,000) = 2.29 × 0.1 = 0.229 seconds[QUOTE]

Don't be silly. Time dilation is not influenced by how far something travels, it is only influenced by the speed of an object.

The formula you posted basically says that the farther something goes, the more time slows down.

Take a good look at the formula you posted.

Tom

Prosoothus

Don't be silly. Time dilation is not influenced by how far something travels, it is only influenced by the speed of an object.

The formula you posted basically says that the farther something goes, the more time slows down.

Take a good look at the formula you posted.

That is correct. The formula indicates .9c which is a gamma factor of 2.29. Therefore the time is .229 seconds.

Q,

If you reread my lengthy post, you will find that I did assume that the figure .229 is correct.

However, when I used the .229 figure in my second example(where the light is moving in the opposite direction of the observer), I don't get the result that light is travelling at c in the moving observer's frame of reference.

Tom

Prosoothus

I don't get the result that light is travelling at c in the moving observer's frame of reference.

You're gonna hate me for saying this but, light travels at c in every reference frame. That is the invariance of light.

Q,

The example I have provided is proof that light DOES NOT travel at the speed of c in all frames of reference.

In other words, the principle of invariance of light is wrong.

Tom

Prosoothus

The example I have provided is proof that light DOES NOT travel at the speed of c in all frames of reference.

I went back and checked over your figures. One thing you're doing that is incorrect is making calculations while assuming one frame of reference will produce the same results as the other frame of reference. Again, we go back to defining reference frames.

Q,

You misunderstood my example. I am not saying that one frame of reference will produce the same results as another frame of reference, I am saying that the result of one frame of reference can be converted to the result of the second frame of reference by incorporating the formulas for time dilation and length contraction.

Read my post again, you will see that I am doing conversions between two different frames of reference.

Tom

James R, Crisp, Thed, Q, and anyone else who cares,

The proof that I have provided in my previous post may have been a little too complicated for some people to understand. Therefore, here is the same problem, only more simplified:

Case 1) An observer, moving at .90c, flys over a stationairy clock. When he is directly above the stationairy clock, he turns on his flashlight and points it forward. After 1 second of the stationairy clock, how far away is the light from the moving observer?

Case 2) An observer, moving at .90c, flys over a stationairy clock. When he is directly above the stationairy clock, he turns on his flashlight and points it behind him(in the opposite direction of his motion). After 1 second of the stationairy clock, how far away is the light from the moving observer?

In the first case, since the light is moving with the observer, the distance between the light and the observer, in the stationairy frame of reference is:

v= c - .90c
v=300,000 km/s - 270,000 km/s
v=30,000 km/s
d=v/t
d=30,000 km/s/1 second
d=30,000 km

Therefore, the the light is 30,000 km away from the moving observer, in the stationairy frame of reference, after one second.

In the second case, since the light is moving away from the observer, the distance between the light and the observer, in the stationairy frame of reference is:

v= c + .90c
v= 300,000 km/s + 270,000 km/s
v= 570,000 km/s
d= v/t
d= 570,000 km/s/1 second
d= 570,000 km

Therefore, in the second case, the light is 570,000 km away from the moving observer, in the stationairy frame of reference, after one second.

Let's review, after 1 second, in the stationairy frame of reference, the distance between the light and the observer is:

Case 1) 30,000 km
Case 2) 570,000 km

Unfortunately, these values can't be used in the moving observer's frame of reference because of the length contraction and time dilation that the moving observer experiences. Therefore, we have to convert these figures so that they are valid in the moving observer's frame of reference.

Let's assume that the length contraction ratio for the moving observer is equal to x, and that the time dilation ratio for the moving observer is equal to y.

Therefore, to get the speed of light in the moving observer's frame of reference, you need to multiply the stationairy length by x, to get the length in the moving observer's frame of reference. Then you need to multiply the stationairy time by y to get the time in the moving observer's frame of reference. You would then divide the length by the time to get the speed of light in the moving observer's frame of reference.

Since relativity claims that the speed of light is always c in a moving observer's frame of reference, the mathematical formula for case 1 would be:

30,000 km/1 second * x/y = 300,000 km/s

In case 2 it would be:

570,000 km/1 second * x/y = 300,000 km/s

Because the speed of the observer is equal in both cases, the length contraction ratio(x) is equal in both cases, and the time dilation ratio(y) is also equal in both cases.

The conclusion: If you adjust the time dilation and length contraction ratios in one case so that the speed of light is equal to c, the same time dilation and the length contraction ratios would not give c as the result in the other case. There are no two numbers(x and y) that would give a result of 300,000 km/s in both cases.

In other words, the speed of light is not always equal to c in all frames of reference. The principle of invariance of light is incorrect.

Crisp: I noticed that you still didn't respond to my earlier post. Am I right or am I wrong??

Tom

Hi Tom,

"Crisp: I noticed that you still didn't respond to my earlier post. Am I right or am I wrong??"

Neither, I am busy :) ... Will get to it as soon as I have the time, at the moment I just drop in for brief comments and keeping up with what happens. Perhaps friday 28th of June I can do some of the calculations myself (yes, it requires scheduling in my agenda at the moment :)).

Bye!

Crisp

Tom,

What you have is an incorrect way of measuring light speed. It's not enough for you to sit on top of the "stationary" clock and observe the moving wavefronts. The problem is that once the flashlight emits its photons, it's never going to see them again. So how is the "moving" flashlight supposed to calculate the speed of light when it can't make the measurement?

What it would have to do, is perhaps bounce the light off a mirror. But you will observe that any such "bouncing" would completely cancel out the contributions of the flashlight's "velocity", in the end always coming out to be c even from the flashlight's perspective.

So... As far as you are concerned, the speed of light is c. And... As far as the "moving" flashlight is concerned, the speed of light is also c! See? :D

Overdoze,

You're right. If you want the flashlight to truely measure the speed of the light, then the light would have to bounce off a mirror and return to the flashlight. And you are correct, in that case the discrepencies would balance each other out.

However, we don't need to do that because we already know the formulas for time dilation and length contraction. All we have to do is use these formulas to obtain the resulting speed. And as you can see from my example, when the light does not bounce off any mirrors, it's relative speed to the observer is not c.

I hope that you see, from my example, why I believe that light does not travel at c in all frames of reference. This is what I was talking about on the other thread when I was referring to the unrelative components of atomic clocks.

Tom

Tom,

Let's rephrase. Forget for a moment about length contraction and time dilation, as these are not the problem in your case and you can ignore them. Let's say three bodies, A B and C are arranged along their velocity vector, like this:

<tt><pre>
C
/\
A | (velocity)

B
</pre></tt>

Assume AB = AC.

What you're saying is, looking from a "stationary" FOR, if A emits a spherical light pulse the light will reach B before C. That is absolutely true -- from a particular FOR registering that particular velocity for A, B and C. But you aren't looking from the FOR of A. If you were co-moving with A, you would see B and C light up at the same time (simultaneously). Remember that as far as A is concerned, it's standing still while you are flying away in the opposite direction.

That's one of the more mind-bending consequences of relativity -- relativity of simultaneity. IOW, there is no such thing as absolutely simultaneous events. Depending on your FOR, for two events X and Y you can either perceive X occur before Y, X and Y occur simultaneously, or Y occur before X. You will always observe events, but in what order depends entirely on how you're moving relative to them.

For example, if you were looking from an FOR that actually travels in the same direction as A but even faster (so that your measured velocity of A is negative), you would in fact perceive C light up before B. It's perfectly symmetrical, and depends only on the velocity that you are measuring given your FOR.

So now take your claim:



I hope that you see, from my example, why I believe that light does not travel at c in all frames of reference.


How does this follow? In your "stationary" frame of reference the light is travelling at c. In A's frame of reference the light is still travelling at c (as measured by A). You have yet to provide a frame of reference from which the speed of light would not be observed to be c.

So for example, even if in an imaginary "absolute" reality B truly lights up before C, A would never know it -- it would have no way of ever finding out. Of course, if you can come up with a way for A to find out, you'd become the most celebrated scientist of this century.

I really don't know what to say.

I took the stationairy distance and the stationairy time that the light travelled in two cases:

1) If the observer shines the flashlight forward.
2) if the observer shines the flashlight back.

In case 1, the light is 30,000 km away from the observer after 1 second in the stationairy frame of reference.

In case 2, the light is 570,000 km away from the observer after 1 second in the stationairy frame of reference.

Relativists will argue that the moving observer experiences time dilation and length contraction, so I included these in my calculations.

The fact is the when the results of one frame of reference are adjusted for time dilation and length contraction, they should give the results for the other frame of reference. In other words, the results of two frames of reference should be equal to each other when they are adjusting for time dilation and length contraction.

The fact is that if you give time dilation and length contraction a value of z, then there is no mathematical solution for both of the equations:

30,000 km * z = 300,000 km/s

and

570,000 km * z = 300,000 km/s

As I stated before, if you modify z to give 300,000 km/s for one formula, then using the same z will not give you 300,000 km/s for the second formula. In other words, only one formula, at most, can give the result of 300,000 km/s. The other formula will give a result larger or smaller than c.

If you still don't believe me, give me what you believe is the length contraction and time dilation ratios so that I may incorporate them in my formulas to demonstrate that light can't travel at c under both circumstances.

By the way, a few posts back I included the time dilation and length contraction figures provided by James R in my formulas, and proved that the speed of light is not c in both cases. Crisp and James then told me that the time dilation and length contraction figures that James R provided were incorrect.

I am still waiting for James R or Crisp to provide me with the correct time dilation and length contraction ratios, so that I can prove that Einstein was wrong once and for all.

If you do not wish to wait for James R or Crisp, please feel free to give me your length contraction and time dilation figures.

Tom

I am still waiting for James R or Crisp to provide me with the correct time dilation and length contraction ratios, so that I can prove that Einstein was wrong once and for all.

You will of course be publishing your results so that others will peer review your theory. There is a Nobel Prize waiting for you. ;)

Tom,

A couple more corrections are in order:



The fact is the when the results of one frame of reference are adjusted for time dilation and length contraction, they should give the results for the other frame of reference. In other words, the results of two frames of reference should be equal to each other when they are adjusting for time dilation and length contraction.


That is not true. In addition to time dilation/length contraction, you still have to take into account the relative motion of the FORs. The latter goes back to Newton, and Relativity is not exempt (remember, it is merely a generalization of Newtonian mechanics.)



In other words, only one formula, at most, can give the result of 300,000 km/s. The other formula will give a result larger or smaller than c. (emphasis: overdoze)


That is incorrect. Please ponder this question: from which FOR is the measurement being made? What is the actual speed of light in that FOR? Take my diagram above, and imagine yourself sitting on A. What do you observe? (And remember, to observe anything information must be incoming as opposed to outgoing.)



If you still don't believe me, give me what you believe is the length contraction and time dilation ratios so that I may incorporate them in my formulas to demonstrate that light can't travel at c under both circumstances.


I believe you! We all believe you. There is nothing wrong with your math. But you are drawing the wrong conclusion because you are misinterpreting the math. That's what I'm trying to point out.

Overdoze,


"The fact is the when the results of one frame of reference are adjusted for time dilation and length contraction, they should give the results for the other frame of reference. In other words, the results of two frames of reference should be equal to each other when they are adjusting for time dilation and length contraction."

That is not true. In addition to time dilation/length contraction, you still have to take into account the relative motion of the FORs. The latter goes back to Newton, and Relativity is not exempt (remember, it is merely a generalization of Newtonian mechanics.)


You're wrong. The only thing we are measuring is distance and time. Relative motion does not effect distance or time(excluding the effects of length contraction and time dilation).

If you are positive that relative motion effects this experiment in addition to length contraction and time dilation, please state your formulas so that I can include them in by calculations. :)


"If you still don't believe me, give me what you believe is the length contraction and time dilation ratios so that I may incorporate them in my formulas to demonstrate that light can't travel at c under both circumstances."

I believe you! We all believe you. There is nothing wrong with your math. But you are drawing the wrong conclusion because you are misinterpreting the math. That's what I'm trying to point out.

If you throw a ball and after 1 second it goes 5 meters, how fast is the ball moving from you. Let me guess 5 m/s??? Opps, I'm wrong, I'm misinterpreting math again, right? :)

The simplist way to calculate the speed of an object is to measure the distance it travelled in a certain amount of time. This is just not logical, it is common sense as well. After all, if you have a distance and the time that distance was travelled, how would you calculate speed??

I'm so sorry if using the basic formula for speed contradicts Einstein's theories. But don't blame me, it's not my fault, blame reality. :)

Tom

Tom,

Here we go again:



In case 1, the light is 30,000 km away from the observer after 1 second in the stationairy frame of reference.

In case 2, the light is 570,000 km away from the observer after 1 second in the stationairy frame of reference.


Think, which observer are you talking about, and how exactly is this "observer" observing the lightfront?

The distances you calculated are in the frame of reference of someone who is sitting "still". That "stationary" observer could have placed little markers along the way that light up as they are reached by the lightfront, and that is how this observer would actually observe these unequal distances with respect to the emitter following an elapsed time of 1 sec (plus whatever additional time needed for reflected light from the markers to reach this observer).

Now, how does the emitter observe the distance the lightfront has travelled? If the emitter looks at those markers, it would see them light up at exactly the same time, progressively, front and back, and so would conclude they were equidistant from it at the time they reflected the propagating spherical lightfront it emitted. Remember, the effects of the emitter's relative velocity with respect to the markers are cancelled out as the light bounces back toward the emitter. Remember also that emitter assumes speed of light to be equal in all directions, and has no way of determining that it isn't.

It seems you are finding it impossible to picture yourself as the emitter rather than the "stationary" observer, and so you are perpetually stuck in the "stationary" FOR. Well great, in the stationary FOR, the distance light travels in either direction is not measured with respect to the travelling emitter, but with respect to the precise point in space where the light was emitted. Where the emitter has gone since then doesn't matter. So in the stationary FOR the light travels exactly the same distance forward and backward from the point of emission in 1 second, namely the roughly 300,000 km.

However, as far as the emitter is concerned the point of emission is located on its premises at all times, and does not move. Rather, it's the coordinate "markers" flying past the emitter that move. So the emitter is going to emit the light forward and backward, and record the lidar return from the markers at what it believes is 300,000 km front and back after exactly 2 seconds. The emitter can easily calculate that from the "stationary" FOR the marker in the back was a lot closer to it than the marker in the front (since the emitter knows that the "stationary" FOR is receding from it at the same exact velocity as the coordinate markers.) But the emitter has no reason to suspect that the "stationary" FOR really is stationary and somehow superior to its own.

[edit: corrected some mistakes (I was somewhat tired when I first wrote this) -- such as changing "farther" to "closer" and such in a couple of places.]

Overdoze,

Let's get down to basics:

If the light in case 1 is 30,000 km away after 1 second in the stationairy frame of reference:

1) What is the 30,000 km of stationairy distance equal to in the moving observer's frame of reference??

2) For 1 second of stationairy time, how much time passed in the moving observers frame of reference??

3) Divide the answer from question 1 with the answer from question 2 to get the speed of light for the moving observer.

If the light in case 2 is 570,000 km away after 1 second in the stationairy frame of reference:

1) What is 570,000 km of stationairy distance equal to in the moving observer's frame of reference??

2) For 1 second of stationairy time, how much time passed in the moving observers frame of reference??

3) Divide the answer from question 1 with the answer from question 2 to get the speed of light for the moving observer.

Do the calculations and you will find that answer 3 from the first case can't equal answer 3 from the second.

You keep on implying that the only way to measure the speed of the light is to bounce it off something. I agree with you that if you take the reflected light into effect, the speed of light averages out to give c. However, Einstein did not claim that the speed of light averages out to c after reflection, he claimed that it always travels at c in every frame of reference.

Through your suggestion that the speed of light averages out to c as a result of reflection, you are actually admitting that the speed of light is not equal to c in the moving observer's frame of reference. If it were, it wouldn't have to "average out". :)

Tom

Sure, let's get down to basics.



If the light in case 1 is 30,000 km away after 1 second in the stationairy frame of reference:

1) What is the 30,000 km of stationairy distance equal to in the moving observer's frame of reference??


In your case, with relative velocity of 0.9c, we can use the length contraction formula in reverse to figure it out:

L' = L/sqrt(1-v^2/c^2) = 30000/sqrt(1-0.81)

which is approximately 68824.720 (km).



2) For 1 second of stationairy time, how much time passed in the moving observers frame of reference??


Using the time dilation formula:

t' = t*sqrt(1-v^2/c^2) = 1*sqrt(1-0.81)

which is approximately 0.43588989 (s).



3) Divide the answer from question 1 with the answer from question 2 to get the speed of light for the moving observer.


approximately 157894.74 (km/s)



If the light in case 2 is 570,000 km away after 1 second in the stationairy frame of reference:

1) What is 570,000 km of stationairy distance equal to in the moving observer's frame of reference??


approximately 1307669.7 (km)



2) For 1 second of stationairy time, how much time passed in the moving observers frame of reference??


approximately 0.43588989 (s)



3) Divide the answer from question 1 with the answer from question 2 to get the speed of light for the moving observer.


approximately 3000000.1 (km/s)



Do the calculations and you will find that answer 3 from the first case can't equal answer 3 from the second.


You're absolutely correct. However, what we've been calculating is not the speed of light as the moving observer would measure it.



You keep on implying that the only way to measure the speed of the light is to bounce it off something.


There are plenty of other ways, but they are all equivalent to bouncing light off something (to which the distance is known). If you have a non-equivalent way, I'm waiting to hear of it.



However, Einstein did not claim that the speed of light averages out to c after reflection, he claimed that it always travels at c in every frame of reference.


If we used a proper method to measure lightspeed for the moving observer (i.e. using lidar), we would indeed come away with c as the answer. Therefore, we would claim that light travels at speed c in the moving frame of reference.



Through your suggestion that the speed of light averages out to c as a result of reflection, you are actually admitting that the speed of light is not equal to c in the moving observer's frame of reference.


I've admitted from the get-go that relativity does not exclude absolute frames of reference. What you have to admit, is that it doesn't matter whether such an absolute frame of reference exists or not. What matters is your own FOR, and what you measure in it.



If it were, it wouldn't have to "average out". :)


The point is, it does average out. Always. 100%. No matter what experimental setup you contrive. That's the real message of the theory -- not that there is a separate universe for each distinctly moving observer, but that there is precisely 1 universe that we all share, and no matter what (inertial) perch you look from you still see that same universe with the same constants and the same laws of physics.

Overdoze, Crisp, Thed, Q, James R, Xev, C'est Moi, and 137,

Overdoze:

Thank you for your posts. I finally understand what the principle of invariance of light really means. I just wish that someone explained this to me sooner.

Everyone Else:

My formulas were correct in both cases. It is assumed in relativity that the observer actually has to recieve the light he/she transmitted, from a reflection, in order for the observer to measure the speed of light. If the observer measures the speed of light by reflection(the speed of light to a mirror or object, and back), the observer will always measure the speed of light to be c. This is because, as in case 1, the transmitted light is slower than c for the observer, but the reflected light is faster than c, averaging out the total speed of light to c. This "averaging out" to c happens in all cases, regardless of the speed of the observer. However, this does not mean that the one-way speeds of light are equal to c. Logic dictates that the one-way speed of light is always more or less than c in a moving frame of reference.

In other words, the observer perceiving the round trip of the light to be c, does not mean that all one-way speeds of the light are equal to c in a moving frame of reference.

So finally, what are the one-way speeds of light in a moving frame of reference?? We can't measure them directly, so what could they possibly be??The answer is something that none of you (except Overdoze, C'est Moi, and 137) want to hear:

The one-way speeds of the light in a moving frame of reference are related to the speed that the relative frame of reference is traveling in relation to the absolute frame of reference. Only in an absolute frame of reference is the one-way speed of light always equal to c.

As I stated before, there isn't a direct way to measure the one-way speed of light in a moving frame of reference. However, there is an indirect way. It is called a light clock:

http://www.physics.wustl.edu/~visser/physics-216/notes-light-clock.html

I already posted this link on another thread.

This light clock slows down the faster the clock is traveling in relation to the absolute frame of reference. It can be used to measure it's own absolute motion. Once the absolute motion of the clock is known, it can be used to derive the one-way speed of light that is traveling with the clock.

One more important thing to consider is that sometimes the round trip speed of light is not as important as the one-way speed of light. This is especially true in atomic clocks(where the electromagnetic radiation stimulates and seperates the caesium atoms, this same radiation DOES NOT need to get reflected back in order for it to influence the caesium atoms). This effect may cause the traveling atomic clock to give the appearance that time is slowing down, when in reality, it isn't.

Tom

Hi Tom,

I think you jumped to your conclusion too fast, overdoze also said:



If we used a proper method to measure lightspeed for the moving observer (i.e. using lidar), we would indeed come away with c as the answer. Therefore, we would claim that light travels at speed c in the moving frame of reference.


Also, you said:

"In other words, the observer perceiving the round trip of the light to be c, does not mean that all one-way speeds of the light are equal to c in a moving frame of reference."

I disagree and (for now) proclaim without further evidence that the "one way" speeds are also c. I'll get back to this as soon as I have more time (quantum field theory eats all time at the moment :)). I will also attempt to prove that the formula's you used to formulate your contradiction ("that no two numbers x and y can be found to ... ") are not valid. That should be fun to do :), just a little patience...

Bye!

Crisp

[edited to fix bold/italic statements]

Crisp and Overdoze,


If we used a proper method to measure lightspeed for the moving observer (i.e. using lidar), we would indeed come away with c as the answer. Therefore, we would claim that light travels at speed c in the moving frame of reference.

A lidar would not be able to measure the one-way speed of light. It would only be able to measure the average speed of the transmitted and the reflected light.

Crisp: I will be looking forward to your evidence. :)


Tom

Can someone explain lidar briefly to me please?

Adam,

A lidar is a laser radar. Instead of bouncing low frequency electromagnetic radiation off an object to determine it's location like a regular radar, a lidar would bounce a pulse of light off the object.

Tom

Why use light part of spectrum? Why not just use radar like normal?

Adam,

I'm not sure.

Maybe it's more accurate because the wavelength of light is shorter.

Tom

Tom,

I have been very busy for the last week or so, and still am, so I'll get back to you in detail later.

I used the Lorentz transformations to calculate the distances and times. My calculations are clearly set out in my post where I gave the correct result. You can check them yourself.

Crisp has already told you why your calculations are incorrect. The formulae for time dilation and length contraction you are using are for specific situations. For example, the time dilation formula assumes that the object whose time the "stationary" observer is measuring stays at space co-ordinate x'=0 at all times. It moves relative to the "stationary" co-ordinate x.

The particular problem you posed involves a light pulse moving relative to both the x and x' co-ordinate systems. In that case, the specific time dilation formula you keep using is not applicable. Instead, you have to use the general formulae from which that result is derived - the Lorentz tranformations. That is what I used to get 0.229 seconds, and my result is correct.

Your recent calculations repeat the same mistake multiple times.

I only prefer it to radar because Einstein's original arguments were concerned specifically with visible light (or at least that's what he used in his popular explanations.) It's more natural since we intuitively know what visible light looks like and it's easier for us to visualize it. But really, if you are equally comfortable with radio waves, infrared, or gamma rays it's all the same.

Frequency/energy doesn't matter, what matters is that it has to bounce back. That's because Einstein uses light as his measuring stick -- to measure lengths. To do that, you have to bounce the light; there is no "unidirectional" length measurement using light.

Tom,

Perhaps you should re-read that link you posted. The light-clock is merely a thought experiment they use to derive the time dilation equation, very similar to what I argued in <a href="http://www.sciforums.com/showthread.php?s=&threadid=8184&perpage=20&pagenumber=3#post122007">this post</a> on the "Unrelative Relativity 2" thread (and I remember you nodding in agreement.)

It is not, and cannot, be used to measure a "one-way" speed of light. All it demonstrates, is that when it moves relative to an observer, the observer is going to see it "tick" slower (i.e. once again, time dilation.)

Keep on thinkin', you almost have it. :)

With all due respect (and having read a few of your posts I do have a lot of respect for you), you are on the wrong track this time. You cannot address Tom's problem with a single silver bullet like the Lorentz transformations, because the problem itself is not stated correctly.

Tom was attempting to "measure" distance without actually doing it all the way. That's the real root of the problem. If he allowed the light to bounce back to the emitter, then the estimated distance between the emitter and the point of reflection would be amenable to Lorentz transformations. Taking the Lorentz-contracted round-trip time and dividing it into the Lorentz-expanded roundtrip distance as seen from the moving frame would give c for the moving frame, as it should.

Hi overdoze,

I can think of some ways of measuring the speed of a lightbeam without having to bounce it back and forth... You could use a semi-transparant mirror to split the beam in two, resulting in two pulses over a given distance (the distance between the first detector and the second detector).

But anyway, where exactly did the principle of relativity state that "light is invariant for all inertial frames, but only if you bounce it of a mirror and let it travel the same distance back" ? You will find it doesn't (unless you have had some really extraordinary course :)). Hence, the speed of light is c, even in forward and backward directions, that's really what relativity is about.

A theoretical result to prove this is the velocity addition formula in special relativity: insert c as the velocity of a moving observer and c comes out as the "added velocity". Note that this formula can be deduced from the Lorentz transformations (and I think even of the "simplified" length contraction and time dilatation formula's, but I'd have to look that one up).

Some experiments that support this: decay of [name particle that decays into photons]. Photondetectors (PMD) can be placed to detect the exact time of the photon, and using a trajectory detector for the original particle, the location of decay can be established. One finds that the distance travelled between the decay location and the photondetector is exactly the distance light travels in that time. I don't see any "bouncing of mirrors" here.

However, this was not really the original problem.

Bye!

Crisp

Hi Crisp

I have to admit I don't remember seeing what I'm saying explicitly stated anywhere, but I sort of arrived at it after trying to resolve the various "paradoxes" like the one Tom came up with, and also after reading Einstein's popular book on the subject (where he starts off explaining how he wants to measure all distances via the lidar method.)

So, maybe I'm totally wrong and what I thought I understood I really am misunderstanding. I, for one, want to find out. :) So let me just continue to argue along the same line, and let's see if you (or anyone here) can shoot down my argument.



I can think of some ways of measuring the speed of a lightbeam without having to bounce it back and forth... You could use a semi-transparant mirror to split the beam in two, resulting in two pulses over a given distance (the distance between the first detector and the second detector). (emphasis: overdoze)


Ok, first question is: how do you measure that distance (between detectors) to begin with? According to me, you'd have to bounce light (or something very equivalent.)

Second question is: why do you want to split the beam? What are you gaining by doing that? Forgive me, but I'm not following you there.



But anyway, where exactly did the principle of relativity state that "light is invariant for all inertial frames, but only if you bounce it of a mirror and let it travel the same distance back" ?


Well, it depends on how you interpret the principle of relativity. You can take it "absolutely", or you can view it as regards actual measurement. I take the latter interpretation.



Hence, the speed of light is c, even in forward and backward directions, that's really what relativity is about.


Well if so, but how would you experimentally determine that?



A theoretical result to prove this is the velocity addition formula in special relativity: insert c as the velocity of a moving observer and c comes out as the "added velocity". Note that this formula can be deduced from the Lorentz transformations (and I think even of the "simplified" length contraction and time dilatation formula's, but I'd have to look that one up).


No contest. But that doesn't really say much about how events are measured in SR (or even GR.) For example, if you had some k-calculus intro to SR you would have drawn some 2-D spacetime diagrams, where moving bodies are bouncing light off each other.



Some experiments that support this: decay of [name particle that decays into photons]. Photondetectors (PMD) can be placed to detect the exact time of the photon, and using a trajectory detector for the original particle, the location of decay can be established. One finds that the distance travelled between the decay location and the photondetector is exactly the distance light travels in that time. I don't see any "bouncing of mirrors" here.


Again the "distance" you mention is one that has already been properly adjusted for your frame of reference (IOW, you can't establish that distance without bouncing light first.) As well, the internal "clock" of the particle is already adjusted to your frame (and to that particular unidirectional motion on top of that, might I add.) Finally, there's the problem of "exact time". How do you synchronize two clocks in a Relativistic universe, with its relativity of simultaneity? You may think they're synchronized, but are they really? (And who's to say what "really" means; with Tom we assumed a single absolute reference frame -- if you observe the clocks to be synchronized and you are not in that reference frame, then from that reference frame the clocks won't be synchronized...)



However, this was not really the original problem.


I agree, but I think it's related (I may be wrong, but that's what I think.)

TIA for any and all input.

James R,


The particular problem you posed involves a light pulse moving relative to both the x and x' co-ordinate systems. In that case, the specific time dilation formula you keep using is not applicable. Instead, you have to use the general formulae from which that result is derived - the Lorentz tranformations. That is what I used to get 0.229 seconds, and my result is correct.

If you recall, a few posts ago I accepted your .229 seconds figure for the sake of argument. If you also recall, using this figure, the speed of the forward moving light comes out to c.

However, when I use your figure on the backwards moving light the result does not equal c.

Tom

Overdoze,


Perhaps you should re-read that link you posted. The light-clock is merely a thought experiment they use to derive the time dilation equation, very similar to what I argued in this post on the "Unrelative Relativity 2" thread (and I remember you nodding in agreement.)

It is not, and cannot, be used to measure a "one-way" speed of light. All it demonstrates, is that when it moves relative to an observer, the observer is going to see it "tick" slower (i.e. once again, time dilation.)

If you reread the details about the light clock, you will find that the light slows down because the vertical motion of the light in the clock decreases.

The vertical motion of the light must decrease as it's horizontal motion increases in order for it's total speed to remain c.

The slowing down of the clock has nothing to do with time dilation, it just gives the false impression that it does.

Tom

Originally posted by Prosoothus



The vertical motion of the light must decrease as it's horizontal motion increases in order for it's total speed to remain c.


That's what happens when you are the "stationary" observer looking at the clock as it whizzes by. However, if you were inside the clock, you wouldn't be aware of any horizontal motion.



The slowing down of the clock has nothing to do with time dilation, it just gives the false impression that it does.


It has everything to do with time dilation. In fact it's synonymous with time dilation. I'm beginning to think you didn't understad anything I've said in that post I linked to...

Overdoze,

You don't understand what I am saying. I am saying that time dilation DOES NOT exist.

The stationairy observer and the moving observer will both see the clock slow down. The stationairy observer will know why(because he/she sees the horizontal motion of the clock), while the moving observer will not(because he/she is moving with the clock).

The reason the clock slows down as it travels faster is because the speed of light is always c in only the absolute frame of reference. As the clock begins to move, the vertical speed of the light decreases in order to compensate for the increasing horizontal motion of the light so that the speed of light in the absolute frame of reference remains c.

I believe something similiar happens to moving atomic clocks as well.

Therefore, let me state it again:

I believe that the one-way speed of light is only equal to c in the absolute frame of reference. If this is really the case, then all measurements anyone has ever taken over the years using moving atomic clocks are corrupt because atomic clocks have electromagnetic components that become out of sync with the rest of the clock as the clocks travel faster. These false readings imply that time is slowing down as the clock travels faster, when in reality time remains constant.

Let me repeat myself: I don't believe in relative mass, length contraction, or time dilation. I just sometimes use these in my examples to prove that Einstein was wrong.

Tom

Originally posted by Prosoothus



You don't understand what I am saying. I am saying that time dilation DOES NOT exist.


Heh, that's what we both seem to be saying.



The stationairy observer and the moving observer will both see the clock slow down.


No. Only the "stationary" observer will see the clock slow down. The moving observer will be just as slowed-down as the clock, so the clock will tick at normal rate for the moving observer.



The stationairy observer will know why(because he/she sees the horizontal motion of the clock), while the moving observer will not(because he/she is moving with the clock).


That's actually correct.



The reason the clock slows down as it travels faster is because the speed of light is always c in only the absolute frame of reference. As the clock begins to move, the vertical speed of the light decreases in order to compensate for the increasing horizontal motion of the light so that the speed of light in the absolute frame of reference remains c.


That's good enough as well.



I believe something similiar happens to moving atomic clocks as well.


But not just atomic clocks: everything. Speed of light is the speed of propagation of information. All fields -- electromagnetic, weak, strong, gravitational -- propagate disturbances at lightspeed (one of the big hints toward unification.) Which is why the following is not correct (the effect is not merely electromagnetic):



I believe that the one-way speed of light is only equal to c in the absolute frame of reference. If this is really the case, then all measurements anyone has ever taken over the years using moving atomic clocks are corrupt because atomic clocks have electromagnetic components that become out of sync with the rest of the clock as the clocks travel faster.




These false readings imply that time is slowing down as the clock travels faster, when in reality time remains constant.


Moving systems are slower with regard to the observer. Universally. As for the "absolute" time, you're right, it remains constant. But nobody has a way to measure this absolute time.



Let me repeat myself: I don't believe in relative mass, length contraction, or time dilation. I just sometimes use these in my examples to prove that Einstein was wrong.


Perhaps what you should believe in, is relative measurement of mass, length, or time. And proving Einstein wrong is a tall order indeed.

Overdoze,


Perhaps what you should believe in, is relative measurement of mass, length, or time. And proving Einstein wrong is a tall order indeed.

Well, first I'll start by proving that the one-way trip of light in a moving frame of reference is not always equal to c (we both agree about this).

Then I will take that fact and apply it to atomic clocks to prove that time does not slow down for the moving observer, only the clock does.

And, finally everything else will fall into place. :)

Note: C'est Moi and I already have a theory that explains the decreasing acceleration of particles in particle accelerators as the particles travel faster, without having to take into consideration any increase in mass. It's a theory that you might find very interesting. If you wish, I can post it on the appropriate thread.

Tom

Tom,

Going back to your example...

<b>My analysis</b>

<i>Case 1) An observer, moving at .90c, flys over a stationairy clock. When he is directly above the stationairy clock, he turns on his flashlight and points it forward. After 1 second of the stationairy clock, how far away is the light from the moving observer?</i>

We've already done this one.

x' = <font face="symbol">g</font>(x - vt)

so the answer is:

2.29(300,000 - 270,000 &times; 1) = 68,700 km.

Here, and below, all speeds are in km/s and distances are in km.

The time calculated by the moving observer is

t' = <font face="symbol">g</font>(t - vx/c²)

or t' = 2.29(1 - 270,000/300,000) = 0.229 seconds.

The speed of the light according to the observer, for this "forwards" case is therefore 68,700/0.229 = 300,000 km/s.

<i>Case 2) An observer, moving at .90c, flys over a stationairy clock. When he is directly above the stationairy clock, he turns on his flashlight and points it behind him(in the opposite direction of his motion). After 1 second of the stationairy clock, how far away is the light from the moving observer?</i>

First, the distance co-ordinate of the light in the stationary frame after 1 stationary second will be -300,000 km. The speed of the observer relative to the stationary observer is +270,000 km/s.

After 1 stationary second the light will be at an x' co-ordinate of:

2.29(-300,000 - 270,000 &times; 1) = -1,305,300 km.

(i.e. 1,305,300 km behind the moving observer.)

According to the moving observer, the time elapsed from emission of the pulse until 1 stationary second later is:

t' = <font face="symbol">g</font>(t - vx/c²)

So t' = 2.29(1 - 270,000 &times; (-300,000)/300,000²) = 4.35 seconds.

The calculated speed of light for the "backwards" case is therefore 1,305,300/4.35 = 300,000 km/s

The observer measures the speed of light to be 300,000 km/s in both the forwards and backwards cases.

<b>Your analysis</b>

<i>In the first case, since the light is moving with the observer, the distance between the light and the observer, in the stationairy frame of reference is [...]

d=30,000 km</i>

Correct.

<i>Therefore, the light is 30,000 km away from the moving observer, in the stationairy frame of reference, after one second.</i>

Yes. That's one stationary second.

<i>In the second case, since the light is moving away from the observer, the distance between the light and the observer, in the stationairy frame of reference is ...

d= 570,000 km</i>

Also correct.

<i>Therefore, in the second case, the light is 570,000 km away from the moving observer, in the stationairy frame of reference, after one second.</i>

Yes, according to the stationary observer in one stationary second.

<i>Unfortunately, these values can't be used in the moving observer's frame of reference because of the length contraction and time dilation that the moving observer experiences. Therefore, we have to convert these figures so that they are valid in the moving observer's frame of reference. ...</i>

Up to this point you're ok, but then you go off the rails for the rest of the post. You're failing to use the correct formulae. Instead, you're substituting what you wrongly think are correct formulae. For the correct working, see above.

James R,

Funny, if the observer points his flashlight forward, the time is 0.229 seconds, but if he points his flashlight back, the time is 4.35 seconds.

I didn't realize that time dilation is dependent on the direction that the observer decides to point his flashlight. :)

Tom

Tom,

The time in the stationary frame is the same regardless of the direction. You specified that.

The times in the moving frame depend on the direction of motion of the light. The effect arises because the light is moving relative to both observers in <b>both time and space</b>, not just one or the other. That, as I've already said, is also why your formulae don't apply here.

However, there are certain invariant quantities which observers always agree on. One is the spacetime interval between events, defined as:

Interval = sqrt[(c<font face="symbol">D</font>t)² - (<font face="symbol">D</font>x)²]

The interval is always the same in any reference frame, provided that the space and time intervals are both measured in that frame. In this case we have, for the space and time intervals between the light pulse being emitted and looked at 1 stationary second later:

<b>"Stationary" frame</b>

<font face="symbol">D</font>t = 1 second.
<font face="symbol">D</font>x = 300,000 km.
Interval = 0 km

<b>Observer's frame - forwards</b>

<font face="symbol">D</font>t = 0.229 seconds.
<font face="symbol">D</font>x = 68,700 km.
Interval = 0 km

<b>Observer's frame - backwards</b>

<font face="symbol">D</font>t = 4.35 seconds.
<font face="symbol">D</font>x = 1,305,300 km.
Interval = 0 km

It is no coincidence that the interval is zero in all three cases. In fact, any situation involving the measurement of a light pulse will give an interval of zero. That value of the spacetime interval is called a <b>light-like</b> interval, for that reason.

You know straight away you've made a mistake if you calculate the distance and time between any two events on the path of a light ray and the spacetime interval does not work out to be zero.

BTW, your gut feeling that there is something wrong with my calculations is neither here nor there. The universe doesn't care about satisfying your common sense, Tom.

James R,

There is no way that you are going to convince me that the same event, occuring in the same frame of reference, takes two different amounts of time.

I'm really surprised that you can be arguing this point. It is not only illogical, it is unsensable. Time dilation is not effected by the direction an observer points his flashlight.

You are going to extremes just to prove that light travels at c in all frames of reference. For the sake of science, don't put a man's theory above logic and reality.

It is far more likely, as Overdoze suggested, that the roundtrip of light (to an object and back) was measured by physicists and divided by 2 to get the speed of the light to equal c. It has probably always been assumed that since the speed of the roundtrip of the light averages out to c, that the one-way trips of light are c as well.

I don't think that the scientific community ever attempted to measure the one-way speed of light, even if they knew how to in the first place.

Tom

Tom,

It has been measured and it has been found to be c. This is an undergraduate experiment at most universities.

Bye!

Crisp

Crisp,


It has been measured and it has been found to be c. This is an undergraduate experiment at most universities.

Was it the one-way speed of light that was measured, or the roundtrip speed of light??

Tom

Hi Tom,

Well, you fire a laser that is controlled with an electronic switch through 300 meters of optical fiber and place a detector at the end. Then you switch on the laser, look how much time it takes to travel to your detector (photomultiplier tube = photo-electric effect, no mirrors or bouncing involved) and calculate c. This is the "one-way" measurement of the speed of light.

"but hey, we're not moving"

Yes we are, at 30 km/s through space ;)

Bye!

Crisp

JR, Crisp, Tom,

The spacetime interval is precisely the description of lidar (think what it means to "see the propagating wavefront".) Point is, the picture is consistent no matter which frame of reference you take up. The xv-dependent corrections in Lorentz transforms that JR used are essentially what is "simulating" the bouncing of light.

Finally, Crisp, once again when you've measured a length of anything, including fiber-optic cable, you've already bounced light, in effect. As a matter of fact, any time you move any piece of equipment in your laboratory, you are propagating light (sound waves propagating through the thing once you push it from one end to get the other end moving are ultimately dependent on interplay of fields which is dependent on lightspeed, etc.) Your rulers are already Lorentz-dilated to match your reference frame, and the only reason you perceive them to be of exactly the same length regardless of orientation, is because you keep bouncing light off their ends! (You have no other way; no matter how complicated your setup it will always reduce to traveling information.) If you bend the fiber-optic cable to bring the light back to you, you've effectively bounced light inside the cable. If you synchronize two clocks and move them to opposite ends of the cable, you've already in effect sent light one way along the cable (the clocks have already lost their synchronization as a result of that movement in precisely such a way as to give you c for your "one-way" measurement.)

With regard to mass: I've been under impression that there actually are two distinct types of mass:
<ol>
<li>inertial</li>
<li>gravitational</li>
</ol>

They turn out to be mathematically identical, or at least proportional (within a relatively stationary frame of reference) but nobody knows why. What I wonder, is whether in particle accelerators it's only the inertial mass that grows (as seen from the "stationary" reference frame of the accelerator), without increasing the gravitational mass. Admittedly, I am thin on theory here, so more knowledgeable people are invited to educate.

Tom,

Thankyou once again for pushing me on this. It has forced me to think a little deeper about the problem. I now have a better explanation for you, since I have realised something else you've missed in your problem. To explain that, I'm going to re-phrase the problem a little and work around to the point I want to make. Before I say anything, though, let me say that my previous post is 100% correct. The Lorentz tranformations, as used above, give exactly the right results. What you want is a way of interpreting those results which makes some sense to you, and I will try to explain here.

First, let's add a couple of things to the situation. We have the stationary observer sitting at co-ordinate x=0. Let's say that the stationary observer sets up two targets for the light pulses to hit. One, which we will call the <b>forward target</b>, he places at x=300,000 km. The other, the <b>backwards target</b>, he places at x=-300,000 km. In other words, they are placed so that light, according to him, will hit the targets in both directions after 1 second, as measured by him. Finally, he also positions a gate at x=270,000 km.

Things are set up so that when the moving observer passes through the gate an alarm rings. Similarly, when a light pulse hits either of the targets a similar alarm goes off.

Finally, we introduce your observer. The observer, carrying a two-directional torch, moves past x=0 and at 0.9c = 270,000 km/s, travelling in the positive x direction. At the instant he passes x=0, which we will designate t=0, he switches on his torch and two pulses of light move away in the positive and negative x directions, towards the two targets. Your observer keeps moving in the positive x direction at constant speed. The observer also sets his watch to time t'=0 as he turns on the torch to release the light pulses.

Now, let us consider 3 events:

Event A: The forwards alarm goes off (it is hit by a light pulse).
Event B: The backwards alarm goes off (it is hit by a light pulse).
Event C: The gate alarm goes off (the observer passes through the gate).

Let's look at these events both from the point of view of the stationary observer (x,t frame) and the moving observer (x',t' frame).

<b>Stationary frame</b>

Event A co-ordinates: x = 300,000 km, t = 1 second.
Event B co-ordinates: x = -300,000, t = 1 second.
Event C co-ordinates: x = 270,000, t = 1 second.

Note that to the stationary observer, all three alarms go off at the same time, t = 1 second.

<b>Moving frame</b>

We've already calculated the co-ordinates of events A and B (see my previous post). To calculate the co-ordinates of event C, we again use the Lorentz transformations:

x' = <font face="symbol">g</font>(x - vt) = 2.29(270,000 - 270,000 &times; 1) = 0 km.
t' = <font face="symbol">g</font>(t - vx/c²) = 2.29(1 - (270,000)(270,000)/(300,000)²) = 0.435 seconds.

We expect, of course, that x'=0 for this event, since the moving observer is stationary in his own reference frame. He remains at x'=0 at all times.

So, to summarise:

Event A co-ordinates: x' = 68,700 km, t' = 0.229 seconds.
Event B co-ordinates: x' = -1,305,300 km, t' = 4.35 seconds.
Event C co-ordinates: x' = 0 km, t' = 0.435 seconds.

Now for the solution to your difficulties.

Notice that <b>for the moving observer</b>, the events A, B and C are <b>not</b> simultaneous any more.

In your reasoning, you have implicitly assumed simultaneity, but simultaneity is relative. What one observer considers to be simultaneous is not considered that way by an observer moving relative to the first observer.

The stationary observer here sees all three alarms go off at the same time. So, to him, the moving observer passes through the gate at the same time as the forwards and backwards light pulses hit their targets (namely, at t=1 second). But the moving observer sees something quite different. He first sees the forwards light pulse hit its target, setting off alarm A (at 0.229 seconds as measured on his watch). Then, <b>a little later</b>, he passes through the gate (at t'= 0.435 seconds). Finally, <b>later again</b>, he sees the backwards pulse hit its target and set off its alarm (at t' = 4.35 seconds).

Since the backwards pulse had a longer time to travel from his point of view, it is not surprising that it went further in that time than the forwards pulse did in the time it had to travel to its target.

Once again, we learn from all this that you have to be <b>very</b> careful when you start trying to interpret relativistic results. Distances change, time intervals change, and things that used to be simultaneous cease to be so any more.

Tom,

A reply to your comments...

<i>There is no way that you are going to convince me that the same event, occuring in the same frame of reference, takes two different amounts of time.</i>

1. The forwards and backwards pulses hitting their respective targets are two different events, not the same event.
2. The same events occur in all frames of references. Observers only disagree on when and where they occur.
3. If there is no way to convince you, I may as well give up now. If this is some kind of religious belief for you, then I won't be able to convince you with science. Please let me know if I'm wasting words here. I'm quite happy to let you believe in whatever fantasy you want. But if you want to know what real science says I'm also happy to explain that, too.

<i>I'm really surprised that you can be arguing this point. It is not only illogical, it is unsensable. Time dilation is not effected by the direction an observer points his flashlight.</i>

1. It's logical.
2. It's settled in my previous post.
3. Time dilation isn't, but the relativity of simultaneity is.

<i>You are going to extremes just to prove that light travels at c in all frames of reference. For the sake of science, don't put a man's theory above logic and reality.</i>

1. I thought you agreed that the speed of light was constant in all inertial frames.
2. For the sake of science, don't put your own mistaken beliefs on a pedestal when you don't know enough about what you're talking about.

<i>It is far more likely, as Overdoze suggested, that the roundtrip of light (to an object and back) was measured by physicists and divided by 2 to get the speed of the light to equal c. It has probably always been assumed that since the speed of the roundtrip of the light averages out to c, that the one-way trips of light are c as well.</i>

This is wrong. See below.

<i>I don't think that the scientific community ever attempted to measure the one-way speed of light, even if they knew how to in the first place.</i>

Of course it's been done. Don't be silly. You can do it in high school with a laser beam and a ruler. And lots of other ways too.



overdoze:

<i>The xv-dependent corrections in Lorentz transforms that JR used are essentially what is "simulating" the bouncing of light.</i>

Wrong, I'm afraid. There is no simulated bouncing in the Lorentz tranformations.

<i>Finally, Crisp, once again when you've measured a length of anything, including fiber-optic cable, you've already bounced light, in effect.</i>

No. If you want to measure the length of something, get your standard ruler, put it next to the object and work out how many rulers fit into the length of the object. No light involved.

<i>Your rulers are already Lorentz-dilated to match your reference frame</i>

Rulers in your reference frame have no length contraction or dilation. They are at their rest length.

<i>...and the only reason you perceive them to be of exactly the same length regardless of orientation, is because you keep bouncing light off their ends!</i>

That's a bizarre thing to say, and it is also wrong.

<i>If you synchronize two clocks and move them to opposite ends of the cable, you've already in effect sent light one way along the cable (the clocks have already lost their synchronization as a result of that movement in precisely such a way as to give you c for your "one-way" measurement.)</i>

Please explain your understanding of the word "synchronisation". How can you tell if two clock are synchronised?

<i>With regard to mass: I've been under impression that there actually are two distinct types of mass:

inertial
gravitational

They turn out to be mathematically identical, or at least proportional (within a relatively stationary frame of reference) but nobody knows why.</i>

They are exactly equal, and we know why. Because gravity isn't a force. It's due to curving of spacetime.

<i>What I wonder, is whether in particle accelerators it's only the inertial mass that grows (as seen from the "stationary" reference frame of the accelerator), without increasing the gravitational mass.</i>

The two are indistinguishable.

Crisp,


Well, you fire a laser that is controlled with an electronic switch through 300 meters of optical fiber and place a detector at the end. Then you switch on the laser, look how much time it takes to travel to your detector (photomultiplier tube = photo-electric effect, no mirrors or bouncing involved) and calculate c. This is the "one-way" measurement of the speed of light.

You forgot one important thing: the timer that measure the time the laser is pulsed and the time the laser hits the target.

Because the electric timer has to measure both of these events, both of the signals (from the laser and from the target) have to be sent to the target using electricity. The speed of the electrical signals is a result of the speed of the electric field inside the wires. The electric field in the wires slows down or speeds up just like the light of the laser.

Therefore, the delay of the laser reaching it's target is averaged out by the electric signals in the wires sending the information back to the timer.

It's the same as bouncing a laser off a target and back, the only difference is that light goes to the target and an electric field (electric signal) comes back.

Tom

Hi Tom,

"The electric field in the wires slows down or speeds up just like the light of the laser."

It won't if you place them perpendicular to the direction of motion. But I agree that there are signal delays in the wire, that can be compensated by making the wires exactly the same length.

But anyway, I was just mentioning the experiment as one possible way to measure the speed of light without directly using mirrors.

Bye!

Crisp

Hi James,

Great post, the oh so easily made assumption of simultanity once again slipped into the original analysis of the problem :)... Probably wouldn't have thought of that one.

Bye!

Crisp

James R,


"There is no way that you are going to convince me that the same event, occuring in the same frame of reference, takes two different amounts of time."

1. The forwards and backwards pulses hitting their respective targets are two different events, not the same event.

You're wrong. If the observer had a regular light bulb instead of two flashlights, it would be one event.


"I don't think that the scientific community ever attempted to measure the one-way speed of light, even if they knew how to in the first place."

Of course it's been done. Don't be silly. You can do it in high school with a laser beam and a ruler. And lots of other ways too.


Read my post just before this one. It shows why the "laser and ruler" experiment doesn't give a valid result either.

Finally, let me say again that if I turn on a lightbulb while I'm travelling at .90 c, each beam of light does not have it's own time dilation. Relativity is supposed to be science, not religion. This isn't religiousforums.com.

I can only be convinced of something if you provide logical and mathematical proof. You can't throw logic out the window, and expect to convince me.

Tom

Crisp,


It won't if you place them perpendicular to the direction of motion. But I agree that there are signal delays in the wire, that can be compensated by making the wires exactly the same length.

I don't think you understand the significance of my post. Let me explain in detail:

Example 1: The electric timer is by the laser.

The electric timer triggers the laser pulse. The pulse triggers the laser immediately because it is next to the laser. If the whole assembly is moving forward, the light travels slower than c because the target is moving away from the laser pulse. The target get's hit by the laser and sends an electric pulse back to the timer. The electric pulse travels faster than c back to the timer and compensates for the decreased speed of the laser pulse. The timer measures the speed of light to be c.

Example 2: The electric timer is by the target.

The electric timer sends an electric pulse to the laser to trigger it. The electric pulse travels faster than c because the laser is moving towards the electric pulse. The electric pulse triggers the laser. The laser pulse travels towards the target at slower than c, which compensates for the faster than c electric pulse. The laser pulse hits the target. The timer at the target measures the speed of light to be c.

Even if you put the timer in between the laser and the target, you will get the same result. There is nowhere that you can put the timer where "averaging" of the speed of light does not occure.

Tom

Overdoze,

I posted a new thread called "Relativistic Increase of Mass at High Speeds". It gives an alternate explanation for a particles decreasing acceleration as it approaches light speed in a particle accelerator, without having to take into account "relativistic mass".

Tom

JR

*sighs*

Sometimes I get the impression people just aren't paying attention to what I'm writing.

Take, for example, the following you said in reply to Tom:



Time dilation isn't, but the relativity of simultaneity is.


And then you turn around and say to me (in the same post!):



Wrong, I'm afraid. There is no simulated bouncing in the Lorentz tranformations.


Well shucks man, how did you imagine we arrive at relativity of simultaneity to begin with? Information gets emitted by an event, and travels to the observers at speed of light, and because observers move differently they receive information out of step with each other. That's how. In our particular case, the "event" itself is triggered by one of the observers, which in other words means a transmission of information from the observer to the locus of the event. IOW, to make the math work out information (in the form of light) makes a roundtrip. That's what you have really been calculating around.

You said:



Things are set up so that when the moving observer passes through the gate an alarm rings. Similarly, when a light pulse hits either of the targets a similar alarm goes off.


Well, this "alarm" you're talking about is the same thing as light being bounced back from our event of choice to the observer. You and I have been talking along identical lines (at least since you've posted a correct version of your calculation), yet you seem intent on denying it for some reason. ???

Ok, now to the misconception of "unidirectional" lightspeed measurements you and Crisp are promulgating.



If you want to measure the length of something, get your standard ruler, put it next to the object and work out how many rulers fit into the length of the object. No light involved.


I expected better than that from such a physics/math buff. When you move a ruler holding it at one end, how does the other end catch up? There's propagation of information through space involved, in this case along the constituent atoms of the ruler, which is ultimately constrained and governed by lightspeed. In fact not just the ruler, but your arm and even your nerve impulses are bound by the same constraints. Everything is. Slinging rulers around is the same thing as shooting light beams. Is that really so hard to comprehend, especially for someone like you or Crisp?



Your rulers are already Lorentz-dilated to match your reference frame

Rulers in your reference frame have no length contraction or dilation. They are at their rest length.


I was taking perspective from an imaginary absolute reference frame I postulated for Tom's conceptual convenience. You must not have followed our conversation very closely.

The point was that no matter what inertial reference frame you look from, all effects of relative motion on measurements of space and time cancel out precisely so as to result in lightspeed of c within the moving frame as judged by the "stationary" frame. IOW, even while the stationary frame has its own "rest" measurement of c, looking at the workings of a moving frame a stationary observer understands exactly how and why the moving frame would also measure the speed of light as c even in spite of its movement. There's no magic or abstract mathematics involved; it's very straightforward and even intuitive once you grasp it.



...and the only reason you perceive them to be of exactly the same length regardless of orientation, is because you keep bouncing light off their ends!

That's a bizarre thing to say, and it is also wrong.


Riddle me this: how do you detect where the end of your ruler is at? Moreover, as you displace the ruler, your actions are equivalent to beaming light (as explained above.) Finally, as the ruler changes its orientation within your "rest" frame, as observed by a relatively moving frame the ruler's length actually changes (and that would be the case from the imaginary "absolute" reference frame -- which is what I was talking about.)



Please explain your understanding of the word "synchronisation". How can you tell if two clock are synchronised?


Exactly. Relativity of simultaneity. Notice I said it first in this thread, before you did (but I get the distinct impression you weren't paying attention.)

The only way you can be sure two clocks are synchronized, is when they are exactly superimposed on each other in space. As soon as there is any distance between them (and I'm not talking spacetime intervals here, just regular space interval), any observed synchronization is relative and not absolute. If you synchronize the clocks by bringing them together, when you take them apart you violate that synchronization by accelerating one or both of the clocks. If you "synchronize" clocks by senting signals between them, you're just shooting light again (or doing an equivalent thereof) and from an imaginary absolute reference frame it will be observed that your velocity with respect to the lightfront will skew the "synchronization" -- though not in any way you could detect from your rest frame.

I must give credit to Tom here. He started out way behind, basically not understanding relativity at all -- yet he already knows exactly what I'm talking about even while you and Crisp are still struggling to interpret the reasons behind why the math works out.

Ok, enough polemic. Now I wish to learn something new, if you would so kindly oblige. :)



They [inertial and gravitational mass] are exactly equal, and we know why. Because gravity isn't a force. It's due to curving of spacetime.


Sorry, I don't follow. Gravitational mass isn't a force either. In Newtonian formulation, it's just a factor in the force.

Unless you want to deny the existence of the force that provides weight (resistance of the ground to our tendency of plunging toward the center of the planet), you have to acknowledge that mathematically there needs to be a balance of forces in order for us to remain at rest on the surface. IOW, there is a balancing force pushing us into the ground. Arguably, the magnitude of that force is still dependent on gravitational mass, regardless of why you think this force exists. According to GR, it's because we want to travel along the shortest spacetime interval...but why is that of relevance to the issue at hand? QM, from what I've gathered, tends to think the force is due to exchange of virtual photons, or gravitons or whatever -- just like electrostatic interactions. Under such a rubric, it's pretty hard to see why travelling faster would cause one to emit/absorb more gravitons with respect to other "stationary" reference frames (especially as one's internal rate of information propagation slows down as observed from those frames.) Not to say that either the GR or the QM take on gravity is entirely correct -- since we obviously still can't marry the two theories.

But anyway, conceptually, the two types of mass are hardly identical. Inertial mass resists acceleration. Gravitational mass warps spacetime or whatever the next theory will claim. These effects are not the same thing, and I fail to see an obvious connection. Could you please elaborate a bit, as an alternative to write-offs that assume I already know everything you do? Thanks in advance.

Tom:

Seeing as you haven't replied regarding my post on the relativity of simultaneity, I take it you agree with my explanation. Good.

<i>You're wrong. If the observer had a regular light bulb instead of two flashlights, it would be one event.</i>

Wrong. Different photons go off in different directions. Hitting two different targets is two different events.

<i>Finally, let me say again that if I turn on a lightbulb while I'm travelling at .90 c, each beam of light does not have it's own time dilation.</i>

Right. Because, as I've told you so many times, time dilation has nothing to do with the properties of a light beam. It is to do with relative motion of an observer. Good to see you're getting the hang of it.

<i>Relativity is supposed to be science, not religion. This isn't religiousforums.com.</i>

Yes. And your point is...?

<i>I can only be convinced of something if you provide logical and mathematical proof.</i>

I've done that. Didn't you read my posts? Please re-read the thread. It's all there in black and white.

<i>You can't throw logic out the window, and expect to convince me.</i>

I've seen no contrary argument from you which stands up to any scrutiny.

overdoze,

<i>Sometimes I get the impression people just aren't paying attention to what I'm writing.</i>

You and me both.

<i>Well shucks man, how did you imagine we arrive at relativity of simultaneity to begin with?</i>

By some very simple arguments, coupled with the assumption that the speed of light is constant. Any introduction to relativity text will show you.

<i>Information gets emitted by an event, and travels to the observers at speed of light, and because observers move differently they receive information out of step with each other.</i>

Yes. As you will be aware from your careful reading of the previous posts, relativity has nothing to do with the limited speed of information transfer. I've already explained that many times to Tom, and I'm sure you've read and digested my explanations on this point. Time dilation is an effect which is still apparent even <b>after</b> you've taken light travel times into account.

<i>Well, this "alarm" you're talking about is the same thing as light being bounced back from our event of choice to the observer.</i>

No it isn't. Sure, to hear the alarm from a distance the sound (or light or radio wave or whatever) has to travel from the alarm to you. You take that into account, subtract off the travel time of the light (or sound or whatever) and thereby deduce the time at which the alarm was triggered. <b>Then</b> you see only the residual relativistic effects.

<i>You and I have been talking along identical lines (at least since you've posted a correct version of your calculation), yet you seem intent on denying it for some reason. ??? </i>

It should be clear from this post and others that we have different ideas on this. If you think we're saying the same thing, you're missing something.

<i>When you move a ruler holding it at one end, how does the other end catch up? There's propagation of information through space involved, in this case along the constituent atoms of the ruler, which is ultimately constrained and governed by lightspeed.</i>

Yes. That is entirely irrelevant to the length measurement you make with the ruler in the end.

<i>...even while the stationary frame has its own "rest" measurement of c, looking at the workings of a moving frame a stationary observer understands exactly how and why the moving frame would also measure the speed of light as c even in spite of its movement. There's no magic or abstract mathematics involved; it's very straightforward and even intuitive once you grasp it.</i>

There is, in fact, absolutely no <i>a priori</i> reason why you'd expect the speed of light to be the same for all inertial observers. In fact, it took until 1905 for anybody to even consider the idea. It is <b>very</b> counter-intuitive, and goes against everything we see in our daily lives.

<i>Relativity of simultaneity. Notice I said it first in this thread, before you did (but I get the distinct impression you weren't paying attention.)</i>

While an argument over priority in this matter might be entertaining, I think I'll skip that for now. It's a bit petty.

<i>The only way you can be sure two clocks are synchronized, is when they are exactly superimposed on each other in space. As soon as there is any distance between them (and I'm not talking spacetime intervals here, just regular space interval), any observed synchronization is relative and not absolute. If you synchronize the clocks by bringing them together, when you take them apart you violate that synchronization by accelerating one or both of the clocks. If you "synchronize" clocks by senting signals between them, you're just shooting light again (or doing an equivalent thereof) and from an imaginary absolute reference frame it will be observed that your velocity with respect to the lightfront will skew the "synchronization" -- though not in any way you could detect from your rest frame.</i>

Let me explain a synchronisation process usually used in relativistic experiments (real and thought experiments). You take one clock and put it where you want to measure an event. You take the second clock and put it somewhere else, where it is stationary relative to the first clock. You pull out your ruler and measure the distance between the clocks. Then, you send a light signal from the first clock to the second clock, carrying the time on the first clock. Knowing how far away the second clock is, and knowing that the speed of light is constant, the second clock then takes the received reading, adds the light travel time and voila! The two clocks are synchronised in the same reference frame.

<i>I must give credit to Tom here. He started out way behind, basically not understanding relativity at all -- yet he already knows exactly what I'm talking about even while you and Crisp are still struggling to interpret the reasons behind why the math works out.</i>

I'm not struggling at all. I've completely explained all aspects of this problem in detail now. No mysteries remain. If you disagree with my answers, please post your own detailed analysis, as I have done, and we'll compare the two.

<i>Sorry, I don't follow. Gravitational mass isn't a force either.</i>

I didn't say it was.

<i>Unless you want to deny the existence of the force that provides weight (resistance of the ground to our tendency of plunging toward the center of the planet), you have to acknowledge that mathematically there needs to be a balance of forces in order for us to remain at rest on the surface.</i>

General relativity says that the surface of the Earth is accelerating upwards continuously. The force of the ground pushing on your feet is what you call weight. There is no force of gravity. The Earth's surface is a non-inertial frame of reference in the context of general relativity.

<i>IOW, there is a balancing force pushing us into the ground.</i>

According to GR, that is false.

<i>QM, from what I've gathered, tends to think the force is due to exchange of virtual photons, or gravitons or whatever -- just like electrostatic interactions.</i>

No quantum gravity theory exists at present.

<i>But anyway, conceptually, the two types of mass are hardly identical.</i>

It is one of the triumps of GR to show that they are indistinguishable and to explain why. What you perceive as a gravitational force is nothing more than acceleration in spacetime. Therefore, the resistance to gravitational force which you call "gravitational mass" is really just the well-known resistance to acceleration, commonly known as "inertial mass". Thus, gravitational mass is no different from inertial mass in the GR picture.

Crisp,

I am looking forward to your proof on why the one-way speed of light is equal to c in a moving frame of reference. I hope that you can find some time in your schedule to share what you think.

Out of all the crazy ideas I've heard in my life, none of them are as crazy as James R's assumption that, to a moving observer, every beam of light has it's own time dilation. His idea hits the "10" on the illogic scale.

I hope that your proof will be different than James R's.

Tom

There's a lot of discussion here about the speed of light and reference frames. I admit I'm not so well educated in physics yet, but I fail to see what the big problem is.

1)

There is no known universal absolute point or direction of reference. Until or unless such a thing is determined, all motion must be relative.

2)

Light (EM emissions) travel at a certain speed regardless of the motion of any objects or viewers or anything. That speed is C modified by the matter the emission is passing through (yes, this includes space, since there is heaps of stuff even in deep space).

3)

If you are moving in a direction with a signal, away from the source, and faster than the signal, its frequency lessens. If you are moving toward the source, or away from it slower than the signal, the frequency increases. Something like that.

Result of 1, 2, and 3

To me, these three points seem to indicate that the action of a signal travelling to and meeting any observer is relative to that observer, but the signal does not change depending on the motion of the observer. Like the propagation of sound waves through air remains pretty constant for any given temperature/pressure and sound, but how I hear it depends on the motion of me and the source (change in frequency). Like that nifty redshift.

I'm thinking that any calculations which you might apply to any signal will have a changing result based on the duration/distance of that signal, which is dependent on the relative motion of source and reciever. Which means that if time dilation is occuring (I'm still not entirely convinced on that one, but then I've never studied it) then there will indeed be a differenc occurring in different signals because the distances, times, and relative motions are different.

I think. At least that's how it is seeming to me.

PS: I'm not sure, I might be doing some more physics next year.

James R,


Let me explain a synchronisation process usually used in relativistic experiments (real and thought experiments). You take one clock and put it where you want to measure an event. You take the second clock and put it somewhere else, where it is stationary relative to the first clock. You pull out your ruler and measure the distance between the clocks. Then, you send a light signal from the first clock to the second clock, carrying the time on the first clock. Knowing how far away the second clock is, and knowing that the speed of light is constant, the second clock then takes the received reading, adds the light travel time and voila! The two clocks are synchronised in the same reference frame.

After reading your post, I've concluded that you are correct, and that it would be possible to measure the one-way speed of light using two clocks and a laser.

However, I don't think that anyone has ever tried this approach. I believe that the scientific community always believed that since the roundtrip speed of light is equal to c, that the one-way speed of light is equal to c as well.

Does anyone here have the resources or the "pull" to get this experiment done???

Tom

Hi Tom,

I did some thinking and researching (read: dispite my extreme lazyness I actually *did* lean over to grab a book off my shelf). The time dilatation formula t = <font face="symbol">g</font>t' is indeed only valid when the event for the moving observer occurs at x' = 0. For an introductory derivation on why only at x' = 0, I can only suggest reading "Spacetime physics" (Taylor and Wheeler, W.H Freeman New York, 1998, page 99, formula L-3). The derivation is too long to type.

That leaves no choice but to use the more general Lorentztransformations (assuming movement along the x-axis):

t' = <font face="symbol">g</font>(t - vx/c²)
x' = <font face="symbol">g</font>(x - vt)
y = y' / z = z'

This basically reduces the problem to the already mentioned solutions. If you'd like some more information, be sure to let me know.

Bye!

Crisp

Hi JR

Let's see if I can demonstrate that we are both, after all, talking about the same thing...



... assumption that the speed of light is constant. Any introduction to relativity text will show you.


I do not dispute that, and I do not seek to invalidate the assumption (I wouldn't know how.) However, it is an assumption. We'll see if it's mathematically equivalent to lightspeed constancy in only a single, "absolute" reference frame, where for all other relatively moving reference frames the above assumption corresponds to mere illusion (but a very persistent illusion indistinguishable from reality.)



As you will be aware from your careful reading of the previous posts, relativity has nothing to do with the limited speed of information transfer. ... Time dilation is an effect which is still apparent even <b>after</b> you've taken light travel times into account.


Well, you have a certain approach where limited speed of information transfer wouldn't matter. I agree. And from that perspective time dilation is indeed an entirely separate phenomenon. I do not dispute the fact that the math works out.

What I argue, on the other hand, is that time dilation is an effect that arises precisely from the fact of limited speed of information transfer. Not due to travel times, but due to movement of observers with respect to the "absolute" lightfront.

I've already derived time dilation and length contraction for Tom using the absolute reference frame and constancy of lightspeed within that frame. Granted, I did not finally put my results in the classical form, but I can easily do that for you. For convenience, I will drop the quotes around "absolute" in the subsequent discussion. Note that does not mean I have a way of discovering the absolute reference frame, or that it is mathematically preferred over any other frame in the end. I'm not familiar with the form of Lorentz transforms that you used; I will work with the following time dilation and length contraction formulae:

t' = t*sqrt(1-v^2/c^2) [time dilation]
L' = L/sqrt(1-v^2/c^2) [length contraction]
where:
<ul>
<li>t' is the time as measured by the moving frame</li>
<li>t is the time as measured by the absolute frame</li>
<li>L' is the length as measured by the moving frame</li>
<li>L is the length as measured by the absolute frame</li>
<li>v is the (constant) speed at which the moving frame is travelling with respect to the absolute frame; of course |v|&lt;c.</li>
</ul>

I'm going to re-derive the above equations from an absolute reference frame, which I've previously done (though not all the way) in the "Unrelative Relativity 2" thread. This time I'm going to be quite a bit more explicit and rigorous. Here goes:

There are three bodies A, B, and C, all travelling with constant speed v&lt;c along parallel vectors (the distance between them doesn't change.) The bodies are arranged as follows: A and B are arranged along a line orthogonal to their direction of motion, while A and C are arranged precisely along the direction of motion. |AB| == |AC| == L (L is the distance as measured by the absolute observer), so that ABC is an equilateral right triangle in the absolute frame. Here's an illustration for visual reference:

<tt><pre>
C /\
| (direction of motion)
A B |
</pre></tt>

<p>
<p>

<h4>Case (1)</h4>

A wants to measure the distance to B. To perform the measurement, A bounces a photon off B and uses the speed of light to calculate the distance (remember, A assumes the speed of light to be constant in all directions.) But as A bounces a photon off B, in the absolute reference frame the photon is actually describing an equilateral triangle with height L since A and B are co-moving. The trip from A to B will take exactly the same amount of time as the trip from B to A, and the total roundtrip time will be twice that. To solve for the one-way-trip time, we take advantage of the right triangle formed by A at the moment of light emission, A at the moment B receives the light and B at the moment B receives the light:

v^2t^2 + L^2 = c^2t^2.

Solving for t, we have:

t^2(v^2 - c^2) = -L^2

t = L/sqrt(c^2 - v^2)

The roundtrip time t_r(AB) is just twice that:

(1) t_r(AB) = 2L/sqrt(c^2-v^2).

As a check, if v was 0, then we would have 2t = 2L/c which is what we would expect. Also,

(2) |v| &gt; 0 =&gt; t_r(AB) &gt; 2L/c

The reason I constructed ABC the way I did, is that AB will correspond to a unit on an X axis and AC to a unit on the Y axis of the moving reference frame. The Z axis sticks out of the page, but with respect to the movement behaves identically to the X axis in this case (hopefully, you can see that.) Any interaction in the moving frame will have to occur along a vector in the XYZ coordinate system I just defined. The vector will either be parallel to the Y axis or have a component orthogonal to the Y axis.

<h4>Case (2)</h4>

A wants to determine the distance to C. To perform the measurement, A is going to bounce a photon off C and use the speed of light to calculate the distance (remember, A assumes the speed of light to be constant in all directions.) In the absolute frame, going from A to C the photon is going to take L/(c - v) seconds to reach C, and then from C to A it's going to take L/(c+v) seconds. Add these fractions and simplify to get the total round-trip time t_r(AC):

t_r(AC) = L/(c-v) + L/(c+v) = (Lc + Lv + Lc - Lv)/(c^2 - v^2)

(3) t_r(AC) = 2Lc/(c^2 - v^2)

Rewriting t_r(AC) as 2L/(c - v^2/c), we note that

(4) |v| &gt; 0 =&gt; t_r(AC) &gt; 2L/c

Now let's compare t_r(AB) and t_r(AC) by subtracting the former from the latter:

2Lc/(c^2 - v^2) - 2L/sqrt(c^2 - v^2) = (2Lc - 2Lsqrt(c^2 - v^2))/(c^2 - v^2)

The denominator is always positive. The numerator can be re-written as 2L(c - sqrt(c^2 - v^2)). From which we see that:

(5) |v| &gt; 0 =&gt; t_r(AC) &gt; t_r(AB)

<h4>discussion</h4>

For conceptual simplicity, assume that any mechanism in the moving frame is going to operate via exchange of information between its components through the electromagnetic field (the argument can be identically repeated for all other fields.) Note that "exchange" means information flows in both directions. I cannot overemphasize enough the significance of the word "exchange"; it is inherent in the notion of interaction. As can be concluded from (2) and (4) above, such information exchanges will take longer in the moving frame than they would if the mechanism was stationary. This is uniformly true of all mechanisms, including clocks. This means that all moving clocks are slowed down as a function of their speed relative to the absolute frame.

Therefore, let us define t' to be the time as measured by the moving frame. We now know that t' lags behind t, so it makes sense for us to define it separately. We're going to define t' using the results of (1) and (3) above. But because of (5), we see that the moving frame's time is not distorted equally along all vectors of interaction. We have a choice here. We can define t' via two separate equations and apply each of those equations to corresponding components of interaction vectors. Or we can choose to define t' via one equation for all interaction vectors, and define a separate correction factor to allow for (5) above based on interaction vector components. Einstein, it turns out, takes this latter approach.

Define t = t_r(AB)/2 = L/sqrt(c^2-v^2).

To express t' in terms of t, we know L=ct' as measured by the moving frame, so:

t = t'c/sqrt(c^2-v^2) = t'/sqrt(c^2/c^2-v^2/c^2) = t'/sqrt(1-v^2/c^2)

(6) t' = t*sqrt(1-v^2/c^2)

Which is exactly the time dilation equation given waay above. Keep in mind that L=ct' used here is the measurement of distance by the moving frame in a direction orthogonal to the moving frame's velocity (since we are working with t_r(AB), or in the XZ plane of the moving frame using the coordinate axes I defined. We know that t_r(AB) is less than t_r(AC) so we'll need to correct t' as just defined for interactions along the moving frame's Y axis.

We know t_r(AC) should be equal to L/c from the perspective of the moving frame. But t_r(AC) is larger than it should be, if L = ct'. Since we allow the moving frame to assume that speed of propagating lightfronts with respect to it is constant no matter the direction, and we've already fixed t', we are forced to conclude that the moving frame will measure a distance L' such that L' &gt; L (where L is the distance the moving frame would measure in directions orthogonal to its velocity.)

Note that this effective lengthening of the Y axis in the moving frame is not real in absolute terms; it is merely an artifact of the way we choose to define t'. Rather than saying that for the moving frame time is stretched out globally and distance stretched out in the direction of motion, we could have said that for the moving frame time is stretched out, but more so along the direction of motion. However, since the moving frame cannot perceive such anisotropy, and to make it more mathematically homologous to the absolute reference frame, we define a global notion of time as per (6) above but then are forced to introduce an anisotropy of distance to compensate. Additionally, actual solid bodies will indeed experience contraction along line of motion as observed from the absolute reference frame, due to (5), as they must maintain geometry within the moving reference frame (as per principle of relativity.) And rather than going through a circumlocution of anisotropic directional interaction rates, it is more convenient for the absolute observer to simply describe the effect as a virtual compression of the moving frame along its velocity.

An interesting question arises: why did we choose t_r(AB) to define t' as opposed to t_r(AC) or some combination of the two? The answer is that if from the moving frame's perspective it took a certain amount of time for light to bounce back from B, it is going to take longer for light to bounce back from C (as per (5) above), forcing the moving observer to conclude that |AC|, as set up from the absolute frame, is greater than |AB| due to the fact that the moving observer assumes constant velocity of light, regardless of direction, and has no reason for suspecting otherwise. By this argument t_r(AB) is the "norm", while the additional elapsed time in t_r(AC) is the "aberration".

Now all that's left, is to express the measurement of |AC| from the moving frame (L') in terms of L. To do this, we first determine what t_r(AC) is in terms of t' (using (6)), multiply the result by c (to get roundtrip distance) and divide by 2:

L' = (t_r(AC)*sqrt(1-v^2/c^2))*c/2 = Lc^2/(c^2-v^2)*sqrt(1-v^2/c^2) = L*sqrt(1-v^2/c^2)/(1-v^2/c^2)

(7) L' = L/sqrt(1-v^2/c^2)

Which, of course, is the length contraction formula.

<hr>

Well, ok. As you can hopefully see, the time dilation/length contraction effects arise naturally due to the fact that the speed of light is constant omnidirectionally in the absolute reference frame. Note that from the absolute perspective, the speed differential between a moving reference frame and a traveling light front makes the speed of light variable with respect to the moving observer -- it varies in the interval (0,2c). However, this is in principle as well as in practice not observable from the moving frame, and that's the key to the consistency of the illusion for the moving frame that it's not moving.

Everything else in SR can be derived in terms of length contraction and time dilation, so hopefully I have demonstrated that the entire SR is derivable from an absolute perspective.

So... on to the rest of your reply.



Sure, to hear the alarm from a distance the sound (or light or radio wave or whatever) has to travel from the alarm to you. You take that into account, subtract off the travel time of the light (or sound or whatever) and thereby deduce the time at which the alarm was triggered. <b>Then</b> you see only the residual relativistic effects.


Which is precisely what your generalized Lorentz transformations do. What did you think the "x" term was doing in there?



<i>When you move a ruler holding it at one end, how does the other end catch up? There's propagation of information through space involved, in this case along the constituent atoms of the ruler, which is ultimately constrained and governed by lightspeed.</i>

Yes. That is entirely irrelevant to the length measurement you make with the ruler in the end.


No, it is not irrelevant. You should be able to picture it as observed from the absolute reference frame, to see what I mean.



There is, in fact, absolutely no <i>a priori</i> reason why you'd expect the speed of light to be the same for all inertial observers. In fact, it took until 1905 for anybody to even consider the idea. It is <b>very</b> counter-intuitive, and goes against everything we see in our daily lives.


But you see, in an absolute reality the speed of light is not the same relative to all moving objects in all directions. Problem is, observers have no way of determining whether they are "absolute" or "moving"; they have no way of "seeing" the light actually propagating without this light (or something it triggered) hitting them back. That, perhaps, is the only counter-intuitive thing -- but even that is quite obvious mathematically just due to sheer symmetry.



Let me explain a synchronisation process usually used in relativistic experiments (real and thought experiments). You take one clock and put it where you want to measure an event. You take the second clock and put it somewhere else, where it is stationary relative to the first clock.


Ok, no problem so far.



You pull out your ruler and measure the distance between the clocks.


Problem number one. Your ruler is appropriately contracted, depending on its orientation, as viewed from the absolute reference frame.



Then, you send a light signal from the first clock to the second clock, carrying the time on the first clock.


Problem number two. From the absolute reference frame, the light will reach the second clock either sooner or later than the moving observer estimates -- due to the difference between the moving observer's velocity and the velocity of the propagating light front along the vector connecting the clocks.



Knowing how far away the second clock is


Which is probably closer than the moving observer thinks.



, and knowing that the speed of light is constant


Assuming, not knowing.



, the second clock then takes the received reading, adds the light travel time and voila!


Except the "light travel time" is not what the second clock thinks it is -- but this error cancels out perfectly with the error in distance.



The two clocks are synchronised in the same reference frame.


Or that's the illusion the observer is under. Of course, from the absolute perspective the clocks won't be synchronized, unless the reference frame in question just happens to coincide with the absolute reference frame.

With all of that said, let me reiterate that even if the absolute reference frame might exist in some physically meaningful way, I won't necessarily know how based on modern physics. As far as I am concerned, we will never know who's "absolute" vs. who is "relative". I did mention my own provisional method of defining the "absolute" reference frame as such a state of motion that all cosmic background radiation is measured to have precisely equal average wavelength in all directions (after compensation for local gravitational anisotropies). The idea was that CBR is everywhere, like an electromagnetic "fog". As you move through that fog, CBR should be blue-shifted in forward direction of your motion and red-shifted in the opposite direction (due to simple Doppler effect.) If you do not observe such a shift, then you are at rest with respect to the "fog", and, putatively, with respect to space itself.



General relativity says that the surface of the Earth is accelerating upwards continuously. The force of the ground pushing on your feet is what you call weight. There is no force of gravity. The Earth's surface is a non-inertial frame of reference in the context of general relativity.


I am aware of the derivation of spacetime curvature in terms of acceleration (as per Principle of Equivalence.) However, don't you think it a little ridiculous to claim that "the surface of the Earth is accelerating upwards continuously" when the distance between that surface and the point in space it's "accelerating" away from is constant?

Your statement might make sense in the context of spacetime intervals as opposed to space intervals, but I'm not sure the notion of "acceleration" is so easily translatable to time. Besides, from where I sit (the imaginary absolute perspective), there is no spacetime. There is only space, constancy of information propagation speed, and symmetry. See the <a href="http://www.sciforums.com/t8381/s/thread.html#post123891">"Relativist Increase of Mass at High Speeds"</a> thread for my admittedly very vague as of now alternative to Einstein's spacetime curvature.



<i>QM, from what I've gathered, tends to think the force is due to exchange of virtual photons, or gravitons or whatever -- just like electrostatic interactions.</i>

No quantum gravity theory exists at present.


I know I've heard about gravitons from QM buffs. I don't believe I've been hallucinating on those multiple occasions.



... the resistance to gravitational force which you call "gravitational mass" is really just the well-known resistance to acceleration, commonly known as "inertial mass". Thus, gravitational mass is no different from inertial mass in the GR picture.


Umm, no. What I call "gravitational mass" is that which, in GR, curves spacetime. What I call "inertial mass" is that which resists motion. It's all nice and well to recast gravity as acceleration, but then you still have to answer the question of why your "inertial mass" creates a spacetime gradient whose curvature is exactly proportional to that inertial mass. You haven't really answered that question.

Hi overdoze,

"We'll see if it's mathematically equivalent to lightspeed constancy in only a single, "absolute" reference frame, where for all other relatively moving reference frames the above assumption corresponds to mere illusion (but a very persistent illusion indistinguishable from reality.)"

When an illusion is indistinguishable from reality, it *is* reality from a physical point of view.

"In the absolute frame, going from A to C the photon is going to take L/(c - v) seconds to reach C, and then from C to A it's going to take L/(c+v) seconds."

I only partially agree... If you take a "relative" frame of reference that from which the ABC object moves at a speed of v, this formula is (in a corrected form) valid. This does not prove anything about an absolute frame of reference, except that you assumed its existance to define your velocity v. However, you need to make the following correction to this formula: it is not going to take L/(c+v) seconds: the distance to cover increases by a factor vt for a stationary observer (since the entire object moves forward). The time required for the light to cross this distance is (L + vt)/c. The trip back takes (L-vt)/c seconds because the object closes in for the stationary observer. Hence the total time required equals L/c.

The rest of your post seems to assume the existance of an absolute frame of reference. The formula's you derived do not require its existence.

Bye!

Crisp

Hi Crisp,

I've been wondering when you'll chime in. :)

To answer the question posed in the title of your post, we share a single universe, so there simply has to be an absolute perspective of some sort. Think of a traveling lightwave. If you imagine yourself as "being" the lightwave (just ignore the physical impossibility for the sake of the image), then you know that your velocity and direction of movement are absolute in a sense. Perhaps absolute with respect to whatever medium whose traveling disturbance you, in fact, are (e.g. the "electromagnetic field".) It doesn't matter if one inertial observer says you are at position X1 while another claims you are at X1. You are you, as a lightwave, in an absolute sense, and you have only a single set of real parameters. The disparate judgements of various observers are, in comparison, mere illusions, or measurement errors -- pick your favorite terminology. I feel that to truly form a "theory of everything", we'll have to step back into this absolute perspective and postulate a fine structure of this medium that forms the base of existence for and enables transmission of light, matter and everything in between.



Originally posted by Crisp

When an illusion is indistinguishable from reality, it *is* reality from a physical point of view.


Mathematically, no contest. Pragmatically, you bet. Conceptually, though...



"In the absolute frame, going from A to C the photon is going to take L/(c - v) seconds to reach C, and then from C to A it's going to take L/(c+v) seconds."

I only partially agree... If you take a "relative" frame of reference that from which the ABC object moves at a speed of v, this formula is (in a corrected form) valid.


Actually, the formulation is correct as-is. You misunderstood the problem; see below.



This does not prove anything about an absolute frame of reference, except that you assumed its existance to define your velocity v.


A major objective was to show that it is possible to define the concept of an absolute reference frame -- e.g. that of a universal field -- within which the observed relativity arises as a result of the mechanics and symmetries of that field.

Another objective was to demonstrate that relativity does not exclude the possibility of an absolute reference frame, which is what I started with on this thread.

Finally, I know that for a lot of people it is easier to grasp relativity if they start with an absolute perspective -- even if it is imaginary and impractical in reality. So I embraced the opportunity to explain relativity in absolute terms, for the benefit of Tom and anyone else who wishes to learn.



However, you need to make the following correction to this formula: it is not going to take L/(c+v) seconds: the distance to cover increases by a factor vt for a stationary observer (since the entire object moves forward). The time required for the light to cross this distance is (L + vt)/c. The trip back takes (L-vt)/c seconds because the object closes in for the stationary observer. Hence the total time required equals L/c.


The objects are both moving, so with respect to each other they are stationary. We are interested in determining the distance between these objects, from the "absolute" stationary perspective, as well as from perspective of the objects. So L/(c+v) and L/(c-v) is indeed correct in this case.



The rest of your post seems to assume the existance of an absolute frame of reference. The formula's you derived do not require its existence.


No, they do not. However, they are in this case derived using it. The very feasibility of such a derivation is significant, especially when people start taking their interpretations of relativity to nonsensical extremes.

Which is, of course, the biggest problem with math. For us humans, it is much less important to arrive at an answer than it is to interpret the answer. So depending on which planet you're from, you may or may not care... :p

Crisp,

Actually your formulation is as valid as mine. I misinterpreted your position, thinking you were talking of measuring distance to a stationary point from a moving object. I should obviously pay more attention. *slaps self across face* To see that both forms are correct:

t = (L + vt)/c

ct = L + vt

ct - vt = L

t(c - v) = L

t = L/(c - v)

Same goes for (L-vt)/c vs. L/(c+v).

Sorry about the mixup! :o

Hi overdoze,

About the maths: my mistake, the formula you used made me believe that you used an incorrect velocity addition (c+v) and (c-v) for the speed of light. In this case, it seems to work out fine though :).

"To answer the question posed in the title of your post, we share a single universe, so there simply has to be an absolute perspective of some sort."

I disagree. There could perhaps be some absolute perspective if somehow you could place yourself outside the universe and observe what is happening there. But since per definition, the universe is "everything" (even what is outside of it :)) you are always bound to participate in what you observe.

And that is exactly what I believe to be the limit: from the moment you take part in what you are observing, there is, I think, no way of getting an absolute view.

"Think of a traveling lightwave. If you imagine yourself as "being" the lightwave (just ignore the physical impossibility for the sake of the image), then you know that your velocity and direction of movement are absolute in a sense."

That your speed is absolute with respect to another observer, ok. Direction of motion is by definition relative to the outside observer. An inertial observer itself cannot tell what way it is going. For example: you sit in the train, you cannot tell whether it is the train moving forward, or the trees you see outside the window going the opposite way.

One lightwave, or all lightwaves together, can never form a frame of reference. Photons get created and destroyed all the time, there's too much going on to make it a "stationary" frame of reference. For example: if all photons were all of the sudden taken away (eg. they condense to matter), what would you take as a frame of reference then ?

"Perhaps absolute with respect to whatever medium whose traveling disturbance you, in fact, are (e.g. the "electromagnetic field".)"

We are basically talking about aether here, since once again, electromagnetic fields are subject to drastic changes, making it nearly impossible to take it as a frame of reference. Now, there are two possible scenario's we can follow:

1) We assume that aether exists as an absolute frame of reference. It has not been detected, IMHO it probably never will, and the theories we have prove that an absolute frame of reference is not necessary to describe the physical reality.

2) We assume that aether does not exist, and have nothing to worry about something we can't detect and basically don't need anyway. The theories also succeed in explaining effects that could be attributed to aether, so from that perspective there is no interaction between the real world and the aether.


We're walking on the borderline of physics and metaphysics here. A physicist will clearly choose scenario 2: the models work, and it work well, without aether, so why complicate things and take it into account ? (believe me, the math is already complicated enough :)). A metaphysicist will ask the question: ok, but what if there is something we can't detect, that could possible have an influence that we can explain with other theories, what is it really ? I am afraid I can't give a motivated scientific answer to that. All I can say is that I think aether does not exist. Note the use of "I" (as in personal opinion) and "think" (as in "i could be proven wrong at any moment").

Bye!

Crisp

Well, I was deliberately trying to avoid saying "ether", just because it gives a false impression. I'm not trying to talk about an ideal gas that permeats the universe (which is what ether is, classically.) I'm more concerned with a universal medium within which everything happens.

Kind of like a block of steel through which sound propagates. The universe would be a block of steel, while light would be the sound. The block of steel would be the "absolute frame". (It's only an analogy, so it's not perfect -- but you know what I mean.)

So I'm not somehow trying to jump outside the universe; I'm just trying to express (poorly) that there is a single underlying reality to everything (including observers and FORs). That's what I want to associate with an absolute reference frame (even if it's not possible with current knowledge to actually detect this, and potentially not ever.)



A physicist will clearly choose scenario 2: the models work, and it work well, without aether, so why complicate things and take it into account ?


Well, I am just not satisfied with models that define light as a wave but refuse to postulate a medium in which the wave arises and travels in the first place. As well, note the "I". :) As they say, whatever gets it up...:D

Hi overdoze,

"Well, I was deliberately trying to avoid saying "ether", just because it gives a false impression. I'm not trying to talk about an ideal gas that permeats the universe (which is what ether is, classically.)"

Ideal gas ? What has matter got to do with aether ? The aether is the "medium" through which electromagnetic waves propagate (well, that's what the common believe was until somewhere around 1905-1910). It is not matter-based at all.

"Well, I am just not satisfied with models that define light as a wave but refuse to postulate a medium in which the wave arises and travels in the first place."

Looks like you are ready for some quantum mechanics then, were ordinary matter is given a wave character too :) ... After that you'll see that the current theoretical believe is that all matter and light have a wave-particle duality that "live" in spacetime. (Sidenote: I say "current", I should rather say: somewhere around the 1970's, in the meanwhile string theory has popped up and is gaining popularity, even though I think the majority of the scientific community is rather sceptic against it).

Bye!

Crisp

Overdoze,

First, I'd like to say that your post was enlightening. I'll read it a few more times so that I can fully understand its value. If I'm correct, you may have brought relativity to it's knees.

Next, I'd like to say that I believe that a device can be built that would measure the one-way speed of light in a moving frame of reference. This is how it would work:

You have two stationairy clocks on a platform. The clocks are 1 meter apart. One clock has a laser and the other clock has a detector. The clocks DO NOT have to be synchronized.

While at rest (relative to the Earth), the first clock fires a laser pulse towards the second clock. The first clock records the time when the laser is fired (t1), in its own time. When the laser pulse hits the target, the second clock records the time in it's own time (t2).

After t1 and t2 are obtained, the platform, with both clocks, is accelerated to a certain speed (relative to the Earth). When the speed is reached, the first clock fires a laser pulse towards the second clock. The first clock records the time when the laser pulse is fired, in it's own time (t3). When the pulse hits the target, the second clock records the time in it's own time (t4).

Conclusion:

In this example, since both clocks are in the same frame of reference, time dilation and length contraction(if they exist) do not have to be taken into consideration.

Therefore, if the speed of the laser pulse is the same in both cases then:

t1-t2=t3-t4

However, if t1-t2 does not equal t3-t4 then the difference in time between the two can be used to calculate the absolute speeds of the two clocks, and of the Earth (of course to calculate the total velocity, six clocks would have to be used).

Also, if it was found that the speed of light in a moving frame of reference is dependent on the speed of the frame of reference relative to the absolute frame of reference, relativity is in serious trouble.

Tom

Crisp,

I see we're getting a little off-track here, so I probably won't argue along this direction much longer. But just to clarify my position:



Ideal gas ? What has matter got to do with aether ? The aether is the "medium" through which electromagnetic waves propagate (well, that's what the common believe was until somewhere around 1905-1910). It is not matter-based at all.


Classically, aether was modelled as an ideal gas that filled the universe. That's the reason why Maxwell's original electromagnetic equations are full of concepts such as flow and flux. People thought this "gas" was stationary in an absolute sense, so moving through it would produce detectable effects -- which, of course, failed to be detected (e.g. Michelson-Morley.) Following which, it was concluded that there is no aether at all -- which IMHO is throwing the baby out with the bathwater. Rather, I think the situation is that "aether" is real but it's not a gas of some sort with respect to which you can move. Rather, it is a medium from which light and matter and everything else arises (kind of like the ocean, with everything we know being just waves on the ocean's surface.) If you want to think of moving through aether, the more proper analogy would be a soundwave "moving" through a medium. Yes, you may indeed be moving with respect to the fundamental texture of the medium, but you are an emergent phenomenon on top of that medium, and you can't directly interact with it to even so much as perceive it (no more than a sound wave can interact with air to determine its velocity with respect to the average velocity of molecues in the air.) You can't see the medium; you just presume its existence based on your own existence and features thereof.

Tom,



First, I'd like to say that your post was enlightening. I'll read it a few more times so that I can fully understand its value. If I'm correct, you may have brought relativity to it's knees.


Thank you, I'm flattered :o

However, I would prefer to think that I didn't bring relativity to its knees. On the contrary, I think I've shown where it comes from and why it is valid. And Einstein is still my hero :p



Next, I'd like to say that I believe that a device can be built that would measure the one-way speed of light in a moving frame of reference. This is how it would work:


Great minds think alike, eh? :D

I remember my own struggles with relativity, and I've gone through exactly the same thing you're going through right now. A physics "puberty" of sorts; anyone with any genuine interest in relativity pays this price. But once you do get to the other side, there's no regrets because the clarity is awesome. So a fair warning and an encouragement: don't despair no matter what :)



You have two stationairy clocks on a platform. The clocks are 1 meter apart. One clock has a laser and the other clock has a detector. The clocks DO NOT have to be synchronized.

While at rest (relative to the Earth), the first clock fires a laser pulse towards the second clock. The first clock records the time when the laser is fired (t1), in its own time. When the laser pulse hits the target, the second clock records the time in it's own time (t2).

After t1 and t2 are obtained, the platform, with both clocks, is accelerated to a certain speed (relative to the Earth). When the speed is reached, the first clock fires a laser pulse towards the second clock. The first clock records the time when the laser pulse is fired, in it's own time (t3). When the pulse hits the target, the second clock records the time in it's own time (t4).


Astonishingly enough, this is almost exactly my own line of thought a few years back. Just one of those many little thought experiments I’ve struggled with.



Conclusion:

In this example, since both clocks are in the same frame of reference, time dilation and length contraction(if they exist) do not have to be taken into consideration.


This is where your (and mine, previously) subtle mistake lies. Time dilation and length contraction do play a role. Since you are accelerating both clocks identically, they can be considered to be in a special reference frame of their own. Remember that in all moving FORs the length is shrunk along direction of motion. That means that while you think you’ve accelerated both clocks equally, the actual distance between then has shrunk. Additionally, this means one of the clocks has traveled a slightly longer distance in process of acceleration, meaning at some point its speed was greater than the other clock’s and its time a little more slowed down. It turns out these seemingly minor effects grow in exact correspondence with the added differential between the speed of the clocks and the speed of the propagating light front – thereby precisely canceling out.



Also, if it was found that the speed of light in a moving frame of reference is dependent on the speed of the frame of reference relative to the absolute frame of reference, relativity is in serious trouble.


It won’t affect relativity really all that much. What it will do, is confirm my analysis and generate a more encompassing perspective where relativity would be just a special case – in a manner similar to how Newtonian physics is a special case of relativistic physics.

But personally, so far I have not been able to come up with a method to do what you're trying to do, that would stand up to scrutiny.

overdoze,

Your derivation of the time dilation and length contraction formulae appears to be correct. However, it does not require or assume an absolute reference frame. It simply follows a standard derivation of those equations which consider two frames - a "stationary" one and a "moving" one. There is nothing in the argument which says that the frame labelled "stationary" is absolutely stationary. That extension is not required, and you have not shown that it is in any way required, as far as I can see.

You say:

<i>Note that this effective lengthening of the Y axis in the moving frame is not real in absolute terms; it is merely an artifact of the way we choose to define t'.</i>

The assumption that it is not "real" depends on your assumption that there is an absolute observer who sees "reality" and knows that his reality is the "real" one, whereas the perceptions of other observers are wrong. That assumption is an unsupported one.

<i>Well, ok. As you can hopefully see, the time dilation/length contraction effects arise naturally due to the fact that the speed of light is constant omnidirectionally in the absolute reference frame.</i>

No. The effects arise becaus the speed of light is constant omnidirectionally in <b>any inertial frame of reference</b>. No absolute frame is required.

<i>But you see, in an absolute reality the speed of light is not the same relative to all moving objects in all directions.</i>

You haven't shown that. (Remember, BTW, that I am only saying the speed is constant in <b>inertial</b> frames.)

<i>Problem is, observers have no way of determining whether they are "absolute" or "moving"</i>

Exactly. Hence the concept of an absolute frame is superfluous.

I said, in the context of clock synchronisation: <i>You pull out your ruler and measure the distance between the clocks.</i>
You replied: <i>Problem number one. Your ruler is appropriately contracted, depending on its orientation, as viewed from the absolute reference frame.</i>

There is no absolute reference frame. Also, my example uses only one reference frame. In that frame, both the ruler(s) and clocks are stationary, so there are no time dilation or length contraction effects.

<i>I did mention my own provisional method of defining the "absolute" reference frame as such a state of motion that all cosmic background radiation is measured to have precisely equal average wavelength in all directions (after compensation for local gravitational anisotropies). The idea was that CBR is everywhere, like an electromagnetic "fog". As you move through that fog, CBR should be blue-shifted in forward direction of your motion and red-shifted in the opposite direction (due to simple Doppler effect.) If you do not observe such a shift, then you are at rest with respect to the "fog", and, putatively, with respect to space itself.</i>

The CBR is just one more thing which is in the universe. It provides a reference frame. You can certainly measure the speeds of objects relative to the CBR if you like, but that in no way means that the CBR is somehow the special "absolute" frame you're so fond of.

<i>I am aware of the derivation of spacetime curvature in terms of acceleration (as per Principle of Equivalence.) However, don't you think it a little ridiculous to claim that "the surface of the Earth is accelerating upwards continuously" when the distance between that surface and the point in space it's "accelerating" away from is constant?</i>

It is counter-intuitive, but certainly not ridiculous.

An object in gravitational free-fall is in an inertial reference frame, according to GR. When you are standing on the Earth's surface you are not in free-fall. The ground prevents you from free-falling to the Earth's centre by continuously providing an upward force.

<i>I know I've heard about gravitons from QM buffs. I don't believe I've been hallucinating on those multiple occasions.</i>

Gravitons are hypothetical particles at present. They have not been detected and no coherent theory of them currently exists (with the possible exception of string theories, which also have not been verified).

<i>What I call "gravitational mass" is that which, in GR, curves spacetime. What I call "inertial mass" is that which resists motion. It's all nice and well to recast gravity as acceleration, but then you still have to answer the question of why your "inertial mass" creates a spacetime gradient whose curvature is exactly proportional to that inertial mass. You haven't really answered that question.</i>

Nobody has answered that question. Science does not answer these questions of "final causes", in the Aristotlean sense. Nobody knows why matter bends spacetime, but GR describes the results very well. Nobody knows exactly why inertia exists, but we can describe the effects very well. What we do know, however, is that there is no requirement for two different types of mass, as I said. That idea is as superfluous as the idea of an absolute reference frame.

<i>First, I'd like to say that your post was enlightening. I'll read it a few more times so that I can fully understand its value. If I'm correct, you may have brought relativity to it's knees.</i>

'fraid not, Tom. Essentially, the derivation given is a relativistic one.

I find it interesting that you display such a fervent hope that relativity will be shown to be wrong. It is clouding your perspective and preventing you from learning. Good luck in overcoming this problem.

James R,

I find interesting your religious belief that relativity must be correct. Your closed-mindedness goes against the core of what science stands for.

Your closed-mindedness is preventing you from learning. I hope one day you get over YOUR problem.

I bet all of you are wondering why I'm so much against relativity. The reason is is that I believe that the whole universe is built from quantums of space, energy, matter, and time. Einstein's theories claim that you can have fractions of "quantas" at relativistic speeds. These "fractions" go against the concept of quantum physics.

Tom

Overdoze,

Your entire post, including your formulas for length contraction and time dilation, is based on the assumption that the omnidirectional speed of light is c in all frames of reference.

However, if you assume that there is an absolute frame of reference, and that the omnidirectional speed of light is only c in the absolute frame of reference, then time dilation and length contraction DO NOT exist.

The time dilation and length contraction percieved by the moving observer is the result of an illusion caused by the increase or decrease in the speed of electromagnetic radiation, for the observer, that the observer uses to measure time and distance.

In other words, this means that relativity is wrong.

Please correct me if my conclusions are incorrect.

Tom

Hi Tom,

"I bet all of you are wondering why I'm so much against relativity. The reason is is that I believe that the whole universe is built from quantums of space, energy, matter, and time. Einstein's theories claim that you can have fractions of "quantas" at relativistic speeds. These "fractions" go against the concept of quantum physics."

Well, I am sorry to tell you this but this makes no sense. Quantum physics tells nothing about the structure of spacetime, the "quanta" refer to the minimal packets of energy exchange, charge, momentum, angular momentum, ... Quantum physics prohibits the use of "fractions" of quanta.

Special relativity does not describe smaller items than eg. energy quanta (photons) going at relativistic speeds. As a matter of fact, quantum physics and (special) relativity work well in what is called quantum field theory, i.e. they are compatible and a theory that takes both into account already exists.

Considering the unification of gravity (general relativity) and quantum physics: I think a Nobel prize is waiting for the person who can accomplish that. If you can work out your theory of quantum spacetime, make it consistent, explain previously unexplained effects, or at least produce some results with it, then please feel free to share those thoughts with us. Until then, I am afraid it is nothing more but speculation. Interesting speculation though ;).

"However, if you assume that there is an absolute frame of reference, and that the omnidirectional speed of light is only c in the absolute frame of reference, then time dilation and length contraction DO NOT exist."

Hrmmm... Have you proven this somewhere ?

"The time dilation and length contraction percieved by the moving observer is the result of an illusion caused by the increase or decrease in the speed of electromagnetic radiation, for the observer, that the observer uses to measure time and distance. In other words, this means that relativity is wrong."

There is more to it. As James R said, "even when compensating for the finite transmission speeds of information, i.e. the speed of light, the effects of time dilatation and length contraction still remain".

Now you keep on claiming that relativity is "wrong". You can't claim that, since that would require:
a) an inconsistency in the theory of relativity. So far none has been discovered (yes, we had troubles replying to some of your problems, but in the end they worked out)
b) a better alternative that explains reality in a better way.

You claim that relativity is wrong based on "logic" and the claim that time dilatation is merely an illusion. I have no problem with that, but if you want to convince some people, then I suggest you explain... hrmm... for example why elementary particles can live longer than they should when going at relativistic speeds. You can use your absolute frame of reference, but you are not allowed to use time dilatation and length contraction formulas since those are concepts from the theory of relativity.

Bye!

Crisp

Crisp,


"I bet all of you are wondering why I'm so much against relativity. The reason is is that I believe that the whole universe is built from quantums of space, energy, matter, and time. Einstein's theories claim that you can have fractions of "quantas" at relativistic speeds. These "fractions" go against the concept of quantum physics."

Well, I am sorry to tell you this but this makes no sense. Quantum physics tells nothing about the structure of spacetime, the "quanta" refer to the minimal packets of energy exchange, charge, momentum, angular momentum, ... Quantum physics prohibits the use of "fractions" of quanta.


You didn't understand what I was saying. The more we know about the universe, the more we find that things are composed of quanta. Physicists have so far found that energy and charge are "quantumized", but it is very likely that scientists will find that everything in the universe is "quantumized", including time and distance.

If time and distance are "quantumized", then Einstein's formulas for time dilation and length contraction would give fractions of time and distance. This would be contrary to quantum physics.


"However, if you assume that there is an absolute frame of reference, and that the omnidirectional speed of light is only c in the absolute frame of reference, then time dilation and length contraction DO NOT exist."

Hrmmm... Have you proven this somewhere ?


Yes I did. If you recall the example with the observer and his flashlight.

If you recall, you and James came to the illogical conclusion that there is a seperate time dilation for each beam of light in order to preserve Einstein's assumption of the invariance of light.

I have to say, you really disappointed me. I was expecting James to choose Einstein over logic, but I wasn't expecting you to do the same.

Did you ever think about what you and James were claiming, instead of blindly following the formulas in your textbook??

C'mon try to. Close your eyes and imagine it. You will also find it illogical.


I have no problem with that, but if you want to convince some people, then I suggest you explain... hrmm... for example why elementary particles can live longer than they should when going at relativistic speeds.

I'll have to think about that. The thing that bothers me the most is that it is so hard to find "raw data" from experiments. It's easy to find theories, but the data that was used to derive the theories is absent. Example: If physicists found that some unstable particles DO NOT live longer at relativistic speeds, would we ever here of it?? Or would they assume that it's a flaw in their readings?

Note: Since the speed of electromagnetic radiation is affected by the speed of a frame of reference, perhaps this may also prevent an unstable particle travelling at relativistic speeds from decaying when it should.

Tom

Tom,

Closed-minded? Moi? Surely not.

<i>I bet all of you are wondering why I'm so much against relativity. The reason is is that I believe that the whole universe is built from quantums of space, energy, matter, and time.</i>

So you have a belief. Based on what, exactly? I, too, have a belief. I believe that relativity correctly describes space and time to a certain level of accuracy (I admit that it is an incomplete theory of spacetime). My belief is backed up by experiments, observations and mathematical, physical theory.

Your belief seems to be based on nothing better than your own gut feeling, combined with a wilful blindness to facts which are inconvenient for you. Nevertheless, as I scientist I am happy to hear you out on your ideas, because there's just a tiny chance that you may be onto something useful. So, let's see your maths. Let's compare your theory to Einstein. Who knows? Maybe you're right and 100 years of physics is wrong. At this stage, I have to admit to a certain level of scepticism, I'm afraid. Forgive me.

<i>Einstein's theories claim that you can have fractions of "quantas" at relativistic speeds.</i>

Relativity is not a quantum theory.

<i>However, if you assume that there is an absolute frame of reference, and that the omnidirectional speed of light is only c in the absolute frame of reference, then time dilation and length contraction DO NOT exist.</i>

What replaces them? Does it accord with what we observe?

<i>The time dilation and length contraction percieved by the moving observer is the result of an illusion caused by the increase or decrease in the speed of electromagnetic radiation, for the observer, that the observer uses to measure time and distance.</i>

So you've changed your mind about the constancy of the speed of light, just as I predicted you would way back earlier in this thread? That's a predictable fall-back position for you.

<i>Please correct me if my conclusions are incorrect. </i>

Already have, multiple times. You respond like a child with his hands over his ears shouting "I'm not listening!" Grow up, Tom.

<i>If time and distance are "quantumized", then Einstein's formulas for time dilation and length contraction would give fractions of time and distance. This would be contrary to quantum physics.</i>

Yes. If spacetime is quantised, relativity will turn out to be a classical limit of a more complex quantum theory. That doesn't make it wrong.

<i>Yes I did. If you recall the example with the observer and his flashlight.

If you recall, you and James came to the illogical conclusion that there is a seperate time dilation for each beam of light in order to preserve Einstein's assumption of the invariance of light.</i>

Please re-read all the relevant posts again. And again. And again, until their meaning gets through to you. I explained it all in detail so you could understand, yet you completely misrepresent my position here. That's either serious misunderstanding or intellectual dishonesty on your part. I hope it's the former, since that is the lesser evil.

<i>Did you ever think about what you and James were claiming, instead of blindly following the formulas in your textbook??</i>

Look at my series of posts again, Tom. I considered your problem. I had a stab at a solution. That solution was partly right and partly wrong. I corrected my mistake and got the correct solution. I then realised I didn't have a good interpretation of my solution, so I had a long hard think about it until I understood it properly. In the process, I learned something which will no doubt be useful to me in the future.

At no time in this intellectual exercise did I consult a textbook. When I found mistakes, I admitted them, put them behind me and moved on.

Compare your behaviour. You're still back at square one, refusing to budge like a stubborn donkey. Who is likely to be the more successful scientist here - you or me?

<i>If physicists found that some unstable particles DO NOT live longer at relativistic speeds, would we ever here of it?? Or would they assume that it's a flaw in their readings?</i>

They would check carefully for flaws. If they could find none, they would report their findings, probably gaining considerable fame in the process for finding a new phenomenon. Scientists love proving theories wrong. The more established the theory is, the more the kudos for debunking it. Relativity has so far withstood all attacks - and there's no shortage of people attempting to take shots at it, I assure you.

Hi Tom,

"If time and distance are "quantumized", then Einstein's formulas for time dilation and length contraction would give fractions of time and distance. This would be contrary to quantum physics."

As James R pointed out, special relativity is not a quantum theory and hence incompatible. However, they can be made compatible (as I said in my previous post, using quantum field theory). So if further quantisations can be found, there probably will also be a new theory that is relativistically invariant.

"If you recall, you and James came to the illogical conclusion that there is a seperate time dilation for each beam of light in order to preserve Einstein's assumption of the invariance of light."

Correction: that is what you made of it :). And also, there are no attempts to "presever the invariance of the speed of light": this is automatically taken care of.

"I have to say, you really disappointed me. I was expecting James to choose Einstein over logic, but I wasn't expecting you to do the same."

There is a *huge* difference between logic and daily intuition, which I think you are confusing with logic. The theory of relativity is a good one, I like it a lot since it is the first major advancement over Newtonian mechanics that still has more or less the same structure. One of the cute things of SR is that it is logically consistent, but counterintuitive things are derived from its postulates. That does not make the theory illogical, just harder to get a grip of.

I've already advised it, but you should really pick up a textbook on relativity if you want to go deeper into it. Logical deductions in words simply won't do the job in physics, logical math is also required.

"Did you ever think about what you and James were claiming, instead of blindly following the formulas in your textbook??"

Hrm... I am actually wondering whether I should question the mind's capability to grasp counterintuitive things and just rely on the math. But then again, I still have to meet the first person that purely on reason (and without math) can answer the question "if I am going at nearly lightspeed and I turn out my car's headlights...". Believe me, I sometimes wish that physics could also be done without the math :) ... things don't just always go the way you want.

IMHO, what should be thought about are the postulates, the rest is math. You don't want to question math, do you ? As far as the interpretation of the results is concerned, I can only say: illogical does not necessarily mean not true. However, since we assume math is consistent, and that we accept the postulates, then all results are correct within that framework. From the moment you step out of that framework (eg. you throw away the constancy of the speed of light for all frames of reference like you do), you are no longer allowed to work in that framework. Hence, as a convinced pro-relativity person, I accept the results, even when they seem illogical. There is not really much to think about (though I should note that eg. the "shut up and calculate" quantum mechanical approach is also not what I am referring to).

Bye!

Crisp

Crisp and James,

Quote from James:


"If you recall, you and James came to the illogical conclusion that there is a seperate time dilation for each beam of light in order to preserve Einstein's assumption of the invariance of light."-me

Please re-read all the relevant posts again. And again. And again, until their meaning gets through to you. I explained it all in detail so you could understand, yet you completely misrepresent my position here. That's either serious misunderstanding or intellectual dishonesty on your part. I hope it's the former, since that is the lesser evil.


Quote from Crisp:


"If you recall, you and James came to the illogical conclusion that there is a seperate time dilation for each beam of light in order to preserve Einstein's assumption of the invariance of light."-me

Correction: that is what you made of it . And also, there are no attempts to "presever the invariance of the speed of light": this is automatically taken care of.



Ok. Let's see what both of you said.

First James:


The time calculated by the moving observer is

t' = g(t - vx/c2)

or t' = 2.29(1 - 270,000/300,000) = 0.229 seconds.


According to the moving observer, the time elapsed from emission of the pulse until 1 stationary second later is:

t' = g(t - vx/c2)

So t' = 2.29(1 - 270,000 × (-300,000)/300,0002) = 4.35 seconds.

Finally, Crisp's response:



That leaves no choice but to use the more general Lorentztransformations (assuming movement along the x-axis):

t' = g(t - vx/c2)
x' = g(x - vt)
y = y' / z = z'

This basically reduces the problem to the already mentioned solutions. If you'd like some more information, be sure to let me know.



In other words, the two of you believe that the moving observer experiences two time dilations, one for each beam of light. As James put it t=0.229 seconds AND t=4.35 seconds at the same time.

However, if you extend your reasoning then almost every photon of light coming from the observers flashlight has its own time dilation. I'm really surprised that the two of you can even consider that the observer observers more than one time dilation at the same time. What puzzles me even more is that the two of you believe that time speeds up for the observer, depending on which way he is pointing his flashlight.

If I recall correctly, Einstein claimed that time slows down for the observer, it doesn't speed up.

If I misunderstood the two of you somehow, please enlighten me.

James R,

I never claimed that light travels at c in all frames of reference. You mentioned this for the second time in your last post, and I really don't know what you are referring to. I have always claimed that light doesn't travel at c in a moving frame of reference, and I tried to illustrate this in my example (the observer and his flashlight). Maybe you have me confused with someone else.

Crisp and James,

I see that the two of you don't respect logic(or you respect it until it gets in the way of relativity).

I've concluded that no matter what logical proof I provide that Einstein is wrong, I will never be able to change your minds. The fact is that we have different priorities: The two of you place theories above logic, while I place logic over any theory.

As long as the two of you practice this "philosophy", our discussions on sciforums will never be resolved.

Tom

Hi Tom,

"In other words, the two of you believe that the moving observer experiences two time dilations, one for each beam of light. As James put it t=0.229 seconds AND t=4.35 seconds at the same time."

Ah, an unfortunate misunderstanding from my side of what you meant by "time dilatation". Usually the "time dilatation" factor is referred to as the value of <font face="symbol">g</font>, and this is ofcourse the same in both situations. As a matter of fact, from the moment you know an observers speed relative to what is considered to be a stationary observer, <font face="symbol">g</font> is known and remains the same.

But more to the point: the reason why the two times are different. I'm having some trouble putting it to words at the moment, I'll think about it and get back later today/tomorrow. I'm gonna work out a slightly different situation first for comparison.

Bye!

Crisp

Hi Tom,

Quick question, I have some ideas on how to explain why the two times are different, but will you accept an answer where the speed of light is assumed to be invariant for all observers ?

Bye!

Crisp

Crisp,


Quick question, I have some ideas on how to explain why the two times are different, but will you accept an answer where the speed of light is assumed to be invariant for all observers ?

I can't. This whole discussion on this thread is about questioning the validity of the principle of invariance of light. I used the example of the moving observer with the flashlight as proof that it can't be correct. If I were to assume that it is correct, then there would be nothing to argue about.

Therefore, I can only accept an explanation that proves that the principle of invariance of light is correct, using only the length contraction and time dilation formulas.

And, by the way, prove it in a way where the moving observer experiences only one time dilation.

Tom

Tom,

<i>In other words, the two of you believe that the moving observer experiences two time dilations, one for each beam of light. As James put it t=0.229 seconds AND t=4.35 seconds at the same time.</i>

In this context, "time dilation" is a bad term to use to refer to the two light pulses. As Crisp has said, the gamma factor is the same in both situations. Notice that the two times I calculated were for two different pulses moving in opposite directions. Even from a common sense perspective, we would not expect the times to be the same, because the observer is moving in the same direction as one of the pulses and in the opposite direction to the other (as seen from the "stationary" frame).

<i>What puzzles me even more is that the two of you believe that time speeds up for the observer, depending on which way he is pointing his flashlight.</i>

That is not true. The observer always measures his own time as normal. It is a misunderstanding of relativity to think that any observer's own time slows down or speeds us from his own point of view. That isn't what happens. It is when the observer looks at <b>other</b> objects moving relative to him that he sees <b>their</b> time slowed down. Also, bear in mind that you've picked a somewhat special case, since you used light in your example.

<i>If I recall correctly, Einstein claimed that time slows down for the observer, it doesn't speed up.</i>

You have a very superficial understanding of what Einstein claimed.

<i>I never claimed that light travels at c in all frames of reference. You mentioned this for the second time in your last post, and I really don't know what you are referring to.</i>

I might sort through your previous posts later to find examples showing how you've changed your tune on this, just as I predicted you would, but I don't have the time right now. You probably don't remember what you said before, but I do.

<i>I see that the two of you don't respect logic(or you respect it until it gets in the way of relativity).</i>

I'm not a big fan of your brand of logic.

<i>I've concluded that no matter what logical proof I provide that Einstein is wrong, I will never be able to change your minds.</i>

You haven't provided a shred of evidence that Einstein is wrong so far.

<i>As long as the two of you practice this "philosophy", our discussions on sciforums will never be resolved.</i>

Give me one good reason, other than your gut feeling, as to why I should throw away 100 years of physics in favour of your half-baked ideas.

Hi Tom,

"Therefore, I can only accept an explanation that proves that the principle of invariance of light is correct, using only the length contraction and time dilation formulas."

I am afraid that this is not possible (at least not in the way you want to hear it). The classical time and length dilatation formula's are only correct under certain circumstances (eg for time dilatation this would mean that the two events that mark the time interval have to occur at x' = 0, i.e. in the origin of the moving observer). Hence if we would want to use that formula, we would need to bounce the lightbeam of a mirror to get it back at the moving observer's origin. Unfortunately, you will then argue that some "averaging" out occurs, and hence the discussion will not be settled.

But once again, you can object all you want, from relativity - if properly applied - you will always get that the speed of light is c for all observers. It is a postulate in the theory, everything is built on top of that. And because of mathematical consistency, this result will pop back out every time you calculate it. At this point I actually wonder why we are still debating the invariance of the speed of light, the only thing that can say something about it is the experiment, and those are in favor of the invariant lightspeed position (i.e. the relativist position). It has already been said, you cannot disprove a consistent theory from within its own framework.

"And, by the way, prove it in a way where the moving observer experiences only one time dilation."

There is only one time dilatation factor, however, different times/events transform in different ways.

Bye!

Crisp

Crisp and James,

Why don't the two of you face the fact that there is no solution for the example that I have given, where the light travels at c in both cases.

I'm still surprised by your belief that if the observer points his flashlight forward, for every stationairy second the observer experiences .229 seconds, but if the observer points his light behind him, then for every stationairy second, the observer experiences 4.35 seconds.

Obviously, the observer can't experience two time dilations at the same time, can he?? Why don't the two of you think about what you are saying???

The fact is that both of you can't accept that the speed of light in a moving frame of reference is only c if you reflect the light off a mirror, and then average the speed. The two of you assumed that Einstein is correct regarding the invariance of light, even though it hasn't been experimentally proven. Now, it turns out that it can't be mathematically proven either.

Well, I guess there's always faith. :)

Tom

Hi Tom,

The difference in the two times can be nicely explained if you accept the invariance of the speed of light. This is not suprising since you indirectly use that same invariance in the Lorentz transformations.

Also, we are talking about two different events here (flashlight shining forward/backward). James R already noted that the notion of simultanity if not necessarily preserved when transforming from one observer to another. Hence, for the stationary observer, the two flashlights might simultaniously hit -300.000 km and +300.000 km after one second, this does not necessarily mean that for the moving observer they do.

The times we (or rather: James, but I completely agree with his calculations) calculated denote the time required for the MOVING observer to reach the two points established above (respectivily -300.000 km and +300.000km for the stationary observer). As the concept of simultanity is no longer necessarily valid, the times for the moving observer can differ. This seems to be the case here (note that we're talking about an extremely extraordinary problem here, observers at 0.9c are not really everyday stuff ;)).

That kinda solves the time difference debate I think...

Bye!

Crisp

Crisp,


The times we (or rather: James, but I completely agree with his calculations) calculated denote the time required for the MOVING observer to reach the two points established above (respectivily -300.000 km and +300.000km for the stationary observer).

I'm sorry I can never accept your explanation. The observer has only one clock. For every stationairy second, x seconds passes for the moving observer.

x can't be equal to 0.229 and 4.35 seconds at the same time. The observer's clock can't indicate that 0.229 seconds and 4.35 seconds passed at the same time. It is an impossibility. The moving observer's clock can only indicate ONE time.

How can you or James possibly prove, using atomic clocks, that 1 stationairy second can equal both 0.229 and 4.35 seconds for the observer?? Where would you place the clocks to prove this??You can't move them with the light, because then they would not be in the observer's frame of reference. If you put both clocks next to the observer, you will find that they will both have the same time.

Tom

Hi Tom,

This is what James tried to explain by placing an alarmbell in the situation (when the observer passes the 270.000 km barrier in the stationary frame of reference). What we are comparing here are the following two events:

Event 1: For the stationary observer, the light passes at +300.000 km. This is ofcourse after one second for the stationary observer.

Event 2: For the stationary observer, the light passes at -300.000 km. This is ofcourse after one second for the stationary observer.

Conclusion: for the stationary observer the two events are simultaneous.

For the moving observer, our claim is that he sees event 1 happen after 0.229s (which intuitively can be understood because that event "happens closer to him"), while on his clock, event 2 only happens after 4.35s.

Conclusion: for the moving observer the events are no longer simultaneous.

This does not mean that for every stationary second, a different amount of time passes for the moving observer, depending on whether the light moves forward or backward. As a matter of fact, regardless of events we are observing, the "classical" time dilatation formula t = <font face="symbol">g</font>t will do just fine in that case (that formula does not apply for all events though). If I remember correctly, that was exactly the 0.435s we used a while back. Conclusion: for one stationary second, only 0.435 seconds pass for the moving observer, or, if we inverse the reasoning: for one second on the moving observer's watch, 2.29s pass on the stationary clock: the moving clock ticks slower.

Hope this explains why there are two different times.

Bye!

Crisp

Crisp,


For the moving observer, our claim is that he sees event 1 happen after 0.229s (which intuitively can be understood because that event "happens closer to him"), while on his clock, event 2 only happens after 4.35s.

Your conversions from one frame of reference to the other are wrong. Let me explain:

After 1 stationairy second, one beam of light is 30,000 stationairy km away from the observer, while the other beam of light is 570,000 stationairy km away from the observer.

When you apply time dilation:

After 0.435 observer seconds, one beam of light is is 30,000 stationairy km away from the observer, while the other beam of light is 570,000 stationairy km away from the observer.

When you apply length contraction:

After 0.435 observer second, one beam of light is 68,965 observer km away from the observer, while the other beam of light is 1,310,344 observer km away from the observer.

Therefore:

v1=68,965 km/0.435 s = 158,540 km/s

v2=1,310,344/0.435 s = 3,012,285 km/s


I don't know why you insist on using two times when we know that one stationairy second equals .435 observer seconds in both cases. It is a fact that both events take .435 seconds for the moving observer because both events took 1 second in the stationairy frame of reference.

Tom

Hi Tom,

My explanation was perhaps not as clear as I had hoped. Allow me to largely repeat myself, but to indicate where your and my reasoning differ.

First of all, I hope you agree that the lightbeam arriving at -300.000 km and at +300.000 km (for the stationary observer) are two different events, clearly seperated in space for the stationary observer. They are not seperated in time for this observer, they both occur at time t = 1s (if we take t = 0s the time the lightbeams are emitted, when the two observers coincide).

Hence: Stationary observer: Event 1 and 2 occur at two different places, at the same time.

Now we take these two events fixed and calculate how the moving observer thinks of those events. We need to use the Lorentz transformations to do that, and these transformations transform time and space in a coupled way. One direct result is that the concept of simultanity for one observer is no longer preserved. For the moving observer the two events are no longer simultaneous, which means that one event happens before the other, i.e. that the times they occur at are different (in numbers: t = 0.229s for event 1 and t = 4.35s for event 2). The Lorentztransformations break simultanity, that is the key to the apparant paradox.

The time dilatation formule t = <font face="symbol">g</font>t' is only valid for events that occur at the origin of the moving observer. For example, when he looks at the clock he has on his wrist. For this clock, time will run slower for a stationary observer, who is comparing it to a clock at his origin. The formula is no longer valid for events that do no occur at the origin of the moving observer (i.e. the lightbeam positions).

"I don't know why you insist on using two times when we know that one stationairy second equals .435 observer seconds in both cases. It is a fact that both events take .435 seconds for the moving observer because both events took 1 second in the stationairy frame of reference."

Nope, this assumes that simultanity is preserved by the Lorentztransformations. Purely intuitively, you can deduce that the times HAVE to be different for a moving observer. One event happens closer to him (the seperation is 30.000km in the stationary observer's frame of reference). The other event happens quite a bit further away (570.000 km in the stationary observer's FOR). A not entirely correct reasoning would be that it takes longer for the moving observer to "see the light of the furthest event" arrive at the position -300.000 km (for the stationary observer). The moving observer's clock will keep on ticking until he can "know" the light has arrived.

The above reasoning is not entirely correct (there is more to it than the finite transmission speed of signals) but it might help you understand why there is a difference in times for the lightbeam shining forward or backward.

Bye!

Crisp

Crisp,

Let me put it another way:

Assume that there are two targets 300,000 km away from the stationairy observer. We both agree that the light hits both targets at the same time in the stationairy frame of reference.

Question: After 0.435 seconds in the moving observers frame of reference(which is 1 second in the stationary observers frame of reference), did the two beams of light hit their targets??

Notice, I'm not asking if the observer percieved the beam of lights to hit their targets at the same time, I'm asking did the two beams of light hit there targets after 0.435 seconds in the moving frame of reference.

You keep on implying that the observer percieves that they hit their targets at different times. You are correct. Even if they did hit the targets at the same time, as I'm suggesting, it would take longer for the light to return to the observer the farthur the target is from the observer. So the observer would not see both beams hit their targets at the same time, even if they actually did.

It's like two stars exploding at the same time. One star is 100 light years away and the other is 1000 light years away. You will not see them blow up at the same time, but that doesn't mean that they didn't blow up at the same time.

Therefore, let's take the observers perception out of the example so that it is simpler.

So I ask again: After 0.435 seconds in the moving observers frame of reference(which is 1 second in the stationary observers frame of reference), did the two beams of light hit their targets??

Tom

Hi Tom,

"Question: After 0.435 seconds in the moving observers frame of reference(which is 1 second in the stationary observers frame of reference), did the two beams of light hit their targets??" Notice, I'm not asking if the observer percieved the beam of lights to hit their targets at the same time, I'm asking did the two beams of light hit there targets after 0.435 seconds in the moving frame of reference.

Hrm, we're walking on the edge of physics and metaphysics again ;). What considers the theory of special theory of relativity: no, the beams did not both hit their target, only the beam going along with the observer did (after 0.229s < 0.435s).

"You keep on implying that the observer percieves that they hit their targets at different times. You are correct. [...] So the observer would not see both beams hit their targets at the same time, even if they actually did."

I should add that even taking the finite time required for the observer to be informed of this event into account still leaves some "residual time dilatation". I've been thinking about that, and I thought I figured it out but there seemed to be a flaw in my reasoning. I hope James will clarify this a bit.


I can already feel where this discussion is going to end up (does special relativity describe reality or not). There is no way to experimentally verify this ofcourse, but there is a philosophical counter argument: if the beams DID actually hit their targets, but the observer has not yet perceived it, then the principle of causality breaks down. He might not have perceived the second beam hitting the target, but because he witnessed the first beam hitting the target, he knows deep inside that also the second beam did hit the target, but that he just did not perceive it yet. However, that way, information was transferred faster than the speed of light, and we all know where that leads to (people getting killed by bullets before they were even fired etc). You can question the principle, but I still have to meet someone who observed the effect before the cause had happened.

Bye!

Crisp

Oh no done it agin - shifted frames of refernce.

Lets see if we can undo this. Two clocks staionary 1 meter apart.
cool and groovy speed of C ok.

Now speed both of them up. Now to an observer traveling with either clock. (allowed since both are in the same FOR) For this guy the distance apart is still 1 meter and the time taken is the same - cool and groovy.

Now shift FOR back to the earth, in the first situation we are in the same FOR as the clocks so we see the same thing as the traveler with the clocks.

Situation two from earth what do we see ? First the distance the clocks are apart "appears" to be shorter ! But then so does the time taken for the light to travel from one clock to the other by the same amount. So let this factor be y ( y will be less than 1) then Distance Travelled as seen from earth = 1 Meter * y ; Time taken as seen from earth = Original time * y ; now devide the two we have (1 * y)/(t *y) or 1/t * y/y or 1/t; Now the original speed in the first event = 1/t so you orginal equation was wrong.

your equation
t1-t2 = t3-t4
is not correct it should be
t1-t2/d1 = t3-t4/d2

The first is wrong because it assume d1 = d2. Which is true for an observer with the clocks but not true for the observer on the earth. but distance per time i.e speed of light IS the same.


Some people cant understand relativity
Some can do the maths
Very few understand relativity. - ( Can't remeber who - maybe Feynman ?)

Tom,

<i>Why don't the two of you face the fact that there is no solution for the example that I have given, where the light travels at c in both cases.</i>

I've given you the solution. It's right there in front of you.

<i>I'm still surprised by your belief that if the observer points his flashlight forward, for every stationairy second the observer experiences .229 seconds, but if the observer points his light behind him, then for every stationairy second, the observer experiences 4.35 seconds.</i>

That's not true. For every stationary second, the stationary observer sees the moving observer's clock tick off 0.44 seconds. That is a constant.

Time dilation is not something an object <b>has</b>, as you seem to think, but something an observer <b>sees</b>.

<i>Obviously, the observer can't experience two time dilations at the same time, can he?? Why don't the two of you think about what you are saying???</i>

Why don't you think about what we're saying? I'm sick of repeating myself.

<i>The two of you assumed that Einstein is correct regarding the invariance of light, even though it hasn't been experimentally proven.</i>

It has been experimentally verified, over and over again in many different ways. Wake up and smell the coffee, Tom.

<i>Now, it turns out that it can't be mathematically proven either.</i>

Of course not. That's what the word <b>assumption</b> means.

<i>I'm sorry I can never accept your explanation.</i>

That's because of you unwillingness to learn something new.

<i>The observer's clock can't indicate that 0.229 seconds and 4.35 seconds passed at the same time.</i>

It doesn't. In fact, it indicates 4.35 seconds 4.121 seconds after it indicates 0.229 seconds. Duh!

<i>How can you or James possibly prove, using atomic clocks, that 1 stationairy second can equal both 0.229 and 4.35 seconds for the observer??</i>

That's not what I proved. Please review my previous posts.

<i>Your conversions from one frame of reference to the other are wrong.</i>

'fraid not.

<i>After 1 stationairy second, one beam of light is 30,000 stationairy km away from the observer, while the other beam of light is 570,000 stationairy km away from the observer.</i>

Yes.

<i>After 0.435 observer seconds, one beam of light is is 30,000 stationairy km away from the observer, while the other beam of light is 570,000 stationairy km away from the observer.</i>

Wrong.

<i>After 0.435 observer second, one beam of light is 68,965 observer km away from the observer, while the other beam of light is 1,310,344 observer km away from the observer.</i>

Wrong.

<i>Therefore:....</i>

Wrong.

<i>Assume that there are two targets 300,000 km away from the stationairy observer. We both agree that the light hits both targets at the same time in the stationairy frame of reference.</i>

Ok.

<i>Question: After 0.435 seconds in the moving observers frame of reference(which is 1 second in the stationary observers frame of reference), did the two beams of light hit their targets??</i>

One of them did, prior to 0.435 seconds. It hit at 0.229 seconds. The other is still going, from this point of view.

<i>Notice, I'm not asking if the observer percieved the beam of lights to hit their targets at the same time, I'm asking did the two beams of light hit there targets after 0.435 seconds in the moving frame of reference.</i>

The questions are indistinguishable.

You can't ask, in relativity, when an event "really" happened. Each observer has his own reality. The time he measures is the "real" time, as far as he is concerned. Nothing he can do will show things to be otherwise. Time and space are relative. That's the whole point.

<i>It's like two stars exploding at the same time. One star is 100 light years away and the other is 1000 light years away. You will not see them blow up at the same time, but that doesn't mean that they didn't blow up at the same time.</i>

That's why you have to take the light travel time into account. Once you've done that, you're left with the Lorentz transformations or the time dilation and length contraction formulae. They deal with events separate from issues of light travel time, as I've already explained.

James R,


"The two of you assumed that Einstein is correct regarding the invariance of light, even though it hasn't been experimentally proven."

It has been experimentally verified, over and over again in many different ways. Wake up and smell the coffee, Tom.


Haven't you been listening to anything that Overdoze of I have been saying on this thread. There has NEVER been an experiment that measured ONLY the one-way speed of light in a moving frame of reference. ALL experiments that have measured the speed of light in a moving frame of reference, have been measuring the AVERAGE of the round-trip speed of light to a mirror, or object, and back. If you believe that there was ever a device made that measures only the one-way speed of light, please share.


"Now, it turns out that it can't be mathematically proven either."

Of course not. That's what the word assumption means.

Well then, since it's an assumption let's assume that it isn't correct.

I noticed that you haven't responded to Overdoze's post. The one where he proves that if you assume that the omnidirectional speed of light is only c in a abslolute frame of reference, you would get the same results as you would from Einstein's relativistic formulas.

Therefore, thanks to Overdoze, there are now two models:

1) Relativity model:

a) There is no absolute frame of reference.
b) Time dilation occures as an object travels faster.
c) Length contraction occures as an object travels faster.

2) Absolute model:

a) There is an absolute frame of reference.
b) The omnidirectional speed of light is c only in the absloute frame of reference.

Pick one. Which one is more logical to you? Overdoze proved that they both give the same results, therefore they are both valid.


Tom

Tom.

<i>Haven't you been listening to anything that Overdoze of I have been saying on this thread. There has NEVER been an experiment that measured ONLY the one-way speed of light in a moving frame of reference. ALL experiments that have measured the speed of light in a moving frame of reference, have been measuring the AVERAGE of the round-trip speed of light to a mirror, or object, and back. If you believe that there was ever a device made that measures only the one-way speed of light, please share.</i>

Read this, please, and get back to me:

<a href="http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html" target="_blank">Experimental basis of special relativity</a>

<i>Well then, since it's an assumption let's assume that it isn't correct.</i>

That assumption does not accord with experimental evidence. Relativity does.

<i>I noticed that you haven't responded to Overdoze's post. The one where he proves that if you assume that the omnidirectional speed of light is only c in a abslolute frame of reference, you would get the same results as you would from Einstein's relativistic formulas.</i>

You didn't look hard enough, then.

<i>Pick one. Which one is more logical to you? Overdoze proved that they both give the same results, therefore they are both valid.</i>

'fraid not.

I'm on this board more or less continuously for a week, and discussion crawls. I leave for two days, and return to find a ton of new posts! :eek: Arrrgh....

Ok, I'll just have to go back in space and time to where my last post was addressed by James R and Tom, and pretend none of what followed exists at the moment... First James R:



Your derivation of the time dilation and length contraction formulae appears to be correct. However, it does not require or assume an absolute reference frame.


First, thank you for stamp of approval. :) However, if you re-read my derivation then you will discover it indeed assumes, even while not requiring, an absolute reference frame. The reasoning used in conceptualizing time dilation relied very heavily indeed on an absolute frame. In fact, you'll notice that under my formulation the absolute frame is the one with the fastest flow of time among all frames of reference. You will also note that the derivation was based lock stock and barrel on bidirectional flows of information. This latter is a concept to which you and Crisp seem to be opposed. Well, I hope I demonstrated the concept is entirely consistent with relativity and all of its experimental confirmations to date.



... your assumption that there is an absolute observer who sees "reality" and knows that his reality is the "real" one, whereas the perceptions of other observers are wrong. That assumption is an unsupported one.


Unsupported, true. Unreasonable? If so, then what is the reasonable alternative? Each moving frame exists in its own parallel universe, would be more reasonable? Under what common context do the various FORs co-exist and interact? That unifying context is precisely the "absolute" FOR in my derivation.



Problem is, observers have no way of determining whether they are "absolute" or "moving"

Exactly. Hence the concept of an absolute frame is superfluous.


I disagree. The concept is only superfluous if there was a less superfluous alternative. There is none to my knowledge, however.



I said, in the context of clock synchronisation: You pull out your ruler and measure the distance between the clocks.
You replied: Problem number one. Your ruler is appropriately contracted, depending on its orientation, as viewed from the absolute reference frame.

There is no absolute reference frame. Also, my example uses only one reference frame. In that frame, both the ruler(s) and clocks are stationary, so there are no time dilation or length contraction effects.


Come on JR, I'm sure you've outgrown the typical toddler difficulty of switching perspective. We are, both of us, talking about the same exact effects -- just from two different perches. In your example, the rulers and clocks are stationary with respect to the frame, but it's improbable that in real life they would be stationary in an absolute sense. What your inertial observer would perceive as "no time dilation or length contraction effects" would be precisely the outcome of time dilation and length contraction effects applied to the observer itself. The fact that two distinct FORs observe the same experimental outcome does not imply that there is only one distinct FOR.



An object in gravitational free-fall is in an inertial reference frame, according to GR. When you are standing on the Earth's surface you are not in free-fall. The ground prevents you from free-falling to the Earth's centre by continuously providing an upward force.


I wasn't going to go into this to such a depth, but since you insist...

With respect to the first sentence, the GR equivalence of gravitational fields to simple acceleration is in fact flawed at all scales above the differential limit. In real life, gravitational fields are spatially anisotropic, and any gravitational field will induce tidal stresses, however weak, in an otherwise "inertial" observer -- which is physically distinct from straightforward acceleration. So in GR, you cannot contemplate any inertial "objects" directly (because there's no such thing); instead you must subdivide it into an infinity of point-like reference frames, and integrate over the result to derive real-world behavior.

With respect to the second sentence, even while you are standing on the earth surface, you are still both in freefall as part of a combined you-earth body. If you would recall, this whole thing started with mention of a QM-centric perspective of gravity in terms of gravitons. Well, from this particular perspective the "forces" binding you and the earth together look no different than the forces binding your constituent atoms together.

With respect to the last sentence, if the ground were providing an upward force with no force to counteract it, you would fly off into space. The fact that you remain on the ground means there is zero net force, or a balance of forces. Either that, or stop using the "force" terminology altogether.



Gravitons are hypothetical particles at present. They have not been detected and no coherent theory of them currently exists (with the possible exception of string theories, which also have not been verified).


Well, space-time is also a hypothetical entity at present. I don't believe anyone has detected "space-time" yet. Even detection of hypothetical gravitational waves might amount to nothing more than detection of graviton radiation. Precession of perihelion of Mercury is derivable using light-speed-limited gravitons just as easily as using space-time. Ditto for "frame-dragging", whenever it does actually get measured.



Science does not answer these questions of "final causes", in the Aristotlean sense.

LOL What causes the plague?


Nobody knows why matter bends spacetime, but GR describes the results very well. Nobody knows exactly why inertia exists, but we can describe the effects very well.

Nobody knew what caused the plague, but they could describe the symptoms very well. I suppose, with 20-20 hindsight, they were right and it was after all pointless to delve deeper into the problem.


What we do know, however, is that there is no requirement for two different types of mass, as I said. That idea is as superfluous as the idea of an absolute reference frame.


No requirement, other than the observed facts that remain to be explained (as you said, "nobody knows why...") Are explanations really that superfluous?

For example, using the conceptual framework I used to re-derive the Lorentz transforms, the concept of time travel suddenly disintegrates. There is only a single, unidirectional, global flow of absolute time, but relatively slower matter-energy reaction rates within travelling reference frames. Time ceases to be a dimension, and returns to its historical role as a mere parameter. So, are conceptual frameworks really so irrelevant when trying to interpret mathematical results? Especially vis a vis what is possible and what is not?

Ok, now for Tom's reply:



Your entire post, including your formulas for length contraction and time dilation, is based on the assumption that the omnidirectional speed of light is c in all frames of reference.


'fraid not. That is JR's misinterpretation or misrepresentation; choose your favorite terminology. It does imply, via its re-derivation of the Lorentz transforms, that for moving frames it will appear as if speed of light is c omnidirectionally.



However, if you assume that there is an absolute frame of reference, and that the omnidirectional speed of light is only c in the absolute frame of reference, then time dilation and length contraction DO NOT exist.


Go over my derivation again; you'll see that you're wrong. That is, time doesn't really slow down in a sense of it being a somehow independent property of the FOR; rather all processes slow down in traveling FORs, in effect leading to their clocks ticking slower. Similar for length contraction.



The time dilation and length contraction percieved by the moving observer is the result of an illusion caused by the increase or decrease in the speed of electromagnetic radiation, for the observer, that the observer uses to measure time and distance.


That is correct enough. That illusion is precisely the reason behind the consistency of the mathematical formulation of Special Relativity, both internally and with respect to confirmatory experiments.



In other words, this means that relativity is wrong.


No, relativity is quite right. What is wrong, is the popular interpretation (or rather, misinterpretation ;)) of the mathematical and experimental results.

BTW JR, Crisp, et al:

From the website JR posted, at this subsection:

<a href="http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#one-way tests">One-Way Tests of Light-Speed Isotropy</a>

find the following text:



Note that while these experiments clearly use a one-way light path and find isotropy, they are inherently unable to rule out a large class of theories in which the one-way speed of light is anisotropic. These theories share the property that the round-trip speed of light is isotropic in any inertial frame, but the one-way speed is isotropic only in an ether frame. In all of these theories the effects of slow clock transport exactly offset the effects of the anisotropic one-way speed of light (in any inertial frame), and all are experimentally indistinguishable from SR. All of these theories predict null results for these experiments.


(emphasis mine.)

I see I wasn't the first one to think along these lines... But then it's truly rare to be the first, especially in an issue that is a century old.

overdoze,

<i>If you re-read my derivation then you will discover it indeed assumes, even while not requiring, an absolute reference frame.</i>

Your derivation would be no different if I labelled you "absolute" frame as an arbitrary "stationary" frame, as in the usual relativistic derivations.

You postulate an absolute frame. It is unnecessary. In this situation, Occam's razor says that we should accept the explanation with the fewest number of required entities. Your theory has one more than relativity, so relativity should be provisionally accepted unless your theory makes predictions which differentiate it from relativity (supported by evidence).

<i>With respect to the first sentence, the GR equivalence of gravitational fields to simple acceleration is in fact flawed at all scales above the differential limit. In real life, gravitational fields are spatially anisotropic, and any gravitational field will induce tidal stresses, however weak, in an otherwise "inertial" observer -- which is physically distinct from straightforward acceleration.</i>

Yes, I know. How is that relevant here?

<i>With respect to the second sentence, even while you are standing on the earth surface, you are still both in freefall as part of a combined you-earth body.</i>

But I'm talking about your interaction with the Earth, not the interaction between the Earth-you system and something else.

<i>With respect to the last sentence, if the ground were providing an upward force with no force to counteract it, you would fly off into space.</i>

That would be true in flat spacetime, but this spacetime isn't flat.

<i>Well, space-time is also a hypothetical entity at present.</i>

I can walk from here to the other side of the room. There seems to be observable "space" here. I'm sure you have it where you are too. But I don't see any gravitons here.

<i>Precession of perihelion of Mercury is derivable using light-speed-limited gravitons just as easily as using space-time.</i>

Link or reference, please.

<i>Nobody knew what caused the plague, but they could describe the symptoms very well. I suppose, with 20-20 hindsight, they were right and it was after all pointless to delve deeper into the problem.</i>

The cause of the plague which we now know of is not a "final" cause in the Aristotlean sense.

Overdoze : If understand what you are saying it is the interpretation of SR and GR not the derivation you are arguing.

This is based on an idea of there being a fastest time rate, minimum mass, this defining the basis FOR. However it is important to note this is a local FOR and not a global one. Two objects both in this state but at two different places can be moving relative to each other and have a different time rate. This is caused by gravity - the curve of space.

Nor is it possible to choose the slowest since it is possible to find three places call then A , B and C. where A is faster than B, B faster than C and C faster than A. This may sound anti-common sense but then nature never agreed to common sense.

JR,

It seems I was wrong. Somehow, I've been under impression that I've seen a lightspeed-limited graviton derivation of perihelion precession. I couldn't find it anywhere after an hour's search, so I don't know what happened exactly. Maybe I saw it in a dream? :D

Well, ok, you got me there. Hands down, no contest. However...



You postulate an absolute frame. It is unnecessary. In this situation, Occam's razor says that we should accept the explanation with the fewest number of required entities. Your theory has one more than relativity, so relativity should be provisionally accepted unless your theory makes predictions which differentiate it from relativity (supported by evidence).


Ok, hold your horses. I don't have a theory (at least not yet), only an interpretation. The interpretation is an answer to the question of "where does relativity come from". In absense of an answer, all you have is an unanswered question, so I'm not sure what Occam's razor has to do with that. As far as I know, Occam also preferred answers to shrugs. (and the answer I'm giving is not just an arbitrarily tacked-on one; it's an integral part of the interpretation.)

Now, this "entity" I'm "introducing" -- the absolute reference frame -- can we really do without it? Suppose you have two distinct reference frames A and B. One measures time as one thing, another as something else -- and similar for length. Are we to postulate that there really are two distinct time flows in the universe -- one for each frame -- merely by the virtue of their relative movement? If you have a googol FORs, do you have a googol different time axes? And how exactly does the universe support such a wild abundance of distinct coordinate systems at the same time, while simultaneously allowing all of them to interact in a consistent fashion? Or perhaps there is a single time, and a single space, and all the FORs are just measuring them differently? IMHO, it's the latter proposal that has the fewest number of entities.

And actually, let's count some entities:

SR:
<ol>
<li>All inertial observers are equivalent (Principle of special relativity)</li>
<li>The velocity of light is the same in all inertial systems</li>
</ol>

SR according to me:

<ol>
<li>There exists a cosmic medium within which all observable events take place</li>
<li>Within this cosmic medium, the velocity of light is the same in all directions</li>
</ol>

Lessee... Einstein:2, me:2. Seems like a tie, don't it? But IMHO, my premises are more intuitive, simpler, and somehow have a more fundamental flavor.



That would be true in flat spacetime, but this spacetime isn't flat.


Well, there's the rub. Curved spacetime is a mathematical abstraction. Or would you suppose that quantum particles are in reality traveling probability distributions, rather than the latter being merely a mathematical model of the former? Question is, what is behind the model? When you computationally model a fluid, you make do with vector fields. However, they don't exactly correspond to the real-life nature of fluids, do they. Granted, my thought on GR is not nearly as advanced as my SR comprehension, and neither is the requisite math on my side of things. I'm working on it though in my spare time, however slowly... But meanwhile, we should keep in mind that a mathematical model does not necessarily a valid description make. A model, yes. An explanation... it would have to make sense first, won't it?

Newton had a mathematical model of gravity too, and it held up pretty well in his time. However, he personally despised his own model from the outset (he couldn't grok invisible forces acting over empty distances.) Einstein came up with curved spacetime, but he fudged it to agree with Newtonian gravity; the precise relationship between mass and curvature was never derived from first principles. I don't have a very good understanding of this, but I've seen on multiple occasions people complaining that GR has otherwise inexplicable factors built in that magically compensate for lightspeed-limited propagation of gravitational interaction just so as to conserve energy. As far as I understand, graviton-based models also fail to conserve energy in absense of additional fudge factors (e.g. orbiting bodies react to "retarded" parameters of each other, and spiral out of orbit as a result.) To top it off, Einstein made time into a dimension, which IMHO is a non-starter. Time has no business being a dimension. And so the math works out, but why? Is it because the original premises are correct as stated, or is it because the original premises are in turn a consequence of something that actually makes sense for a change? At least in the case of SR, I've shown the latter to be a distinct (if not an outright obvious) possibility. Apparently, I wasn't the first one either.

You can make all the claims you want about how spacetime is curved; that doesn't necessarily give you a handle on what is really going on. Computationally, maybe. Exegetically, no. Generally speaking, you've got to be willing to entertain alternatives, as your sig says. Otherwise you end up sounding like a modern-day flat-earther (which in your case would me more like warped-earther :D)



I can walk from here to the other side of the room. There seems to be observable "space" here. I'm sure you have it where you are too. But I don't see any gravitons here.


Ok, I don't see any gravitons either. :p

Though I can't help but notice your usage of "space" rather than "spacetime". Fine by me, as I have yet to be startled by the spectacle of anyone strolling along time.

Overdoze,


"In other words, this means that relativity is wrong."

No, relativity is quite right. What is wrong, is the popular interpretation (or rather, misinterpretation ) of the mathematical and experimental results.

I disagree. Relativity states that time slows down. However, both of us agree that time does not slow down, only reactions slow down. There is a huge difference.

It's like me saying that catalysts speed up time, instead of speeding up chemical reactions.

Tom

James R,

From the link you provided:


Note that while these experiments clearly use a one-way light path and find isotropy, they are inherently unable to rule out a large class of theories in which the one-way speed of light is anisotropic. These theories share the property that the round-trip speed of light is isotropic in any inertial frame, but the one-way speed is isotropic only in an ether frame. In all of these theories the effects of slow clock transport exactly offset the effects of the anisotropic one-way speed of light (in any inertial frame), and all are experimentally indistinguishable from SR. All of these theories predict null results for these experiments.

Since, we can't rule out the fact that the one-way speed of light in a moving frame of reference is not c, we can assume that the principle of invariance of light could be incorrect.

After all, it's just an assumption anyway.

As I stated before, and nobody listened to me except overdoze, a light clock proves that the principle of invariance of light is wrong. However, knowing you, you would probably argue that the light clock slows down because time slows down, instead of taking the more rational approach.

Tom

Tom,

This might be splitting hairs, but relativity as a mathematical framework in fact doesn't say what you claim it says. It's the people interpreting the math that keep "saying" things.

So while I'm entirely in agreement with you when it comes to interpreting the time dilaiton results (in terms of decreased reaction rate as opposed to stretching some imaginary time metric), the fact remains that the formulas predict all moving mechanisms (including clocks), regardless of specifics, will slow down. That's the unambiguous and indisputable message of the math.

Well anyway, this is about as far as I'll take this particular debate. So if you want to have the last word, go right ahead. :)

Overdoze,


This might be splitting hairs, but relativity as a mathematical framework in fact doesn't say what you claim it says. It's the people interpreting the math that keep "saying" things.

I didn't know that. The only people I've ever discussed relativity with are on this board. I've been on this board for months, and everyone has been arguing with me that relativity states that time is slows down.

If they said in the beginning that clocks slow down, but that doesn't necessarilly mean that time slows down, I would've saved everyone alot of time. :)

But then, I wouldn't know what I know now. :)

Tom

overdoze,

<i>I don't have a theory (at least not yet), only an interpretation. The interpretation is an answer to the question of "where does relativity come from".</i>

Interpretations abound. You are welcome to yours, of course. If your interpretation has no observably different consequences to mine, it doesn't matter much, does it? Compare, for example, the Copenhagen and Many Worlds interpretations of quantum mechanics (assuming you're familiar with QM). Neither interpretation changes the usefulness of the theory. Both are compatible with it.

<i>Are we to postulate that there really are two distinct time flows in the universe -- one for each frame -- merely by the virtue of their relative movement? If you have a googol FORs, do you have a googol different time axes?</i>

Yes.

<i>And how exactly does the universe support such a wild abundance of distinct coordinate systems at the same time, while simultaneously allowing all of them to interact in a consistent fashion?</i>

A coordinate system is not a physical thing. It is a mathematical construct. The universe has no trouble support however many you want.

<i>There exists a cosmic medium within which all observable events take place</i>

I ask you, then: can we detect this medium? How is having this medium different from not having it?

<i>Within this cosmic medium, the velocity of light is the same in all directions</i>

In that case, if you move relative to the medium the velocity of light should change, right? If the medium is "absolutely stationary", it can't also move so as to maintain a constant speed of light.

<i>Curved spacetime is a mathematical abstraction. Or would you suppose that quantum particles are in reality traveling probability distributions, rather than the latter being merely a mathematical model of the former? Question is, what is behind the model?</i>

We don't have access to the kind of underlying reality you seem to want. All we have is models for processes. We don't know if an atom really contains quarks, as opposed to the quark model merely being a good description for nuclear processes. We <b>cannot</b> know that. I might postulate that quarks are "really" tiny little jelly beans, or little dragons which fly around in predictable ways. Either theory <b>could</b> be correct. But such a theory cannot be tested, so it isn't much use.

<i>Einstein came up with curved spacetime, but he fudged it to agree with Newtonian gravity; the precise relationship between mass and curvature was never derived from first principles.</i>

That's right. Maybe there's a theory out there which does do that. GR isn't necessarily the last word. String theories may have things to say about this, but I can't tell you since string theory isn't my thing.

<i>I don't have a very good understanding of this, but I've seen on multiple occasions people complaining that GR has otherwise inexplicable factors built in that magically compensate for lightspeed-limited propagation of gravitational interaction just so as to conserve energy.</i>

Not as I understand it. It's a very neat theory. There's very little in there which is tacked on in an <i>ad hoc</i> fashion. It's much better than QM in that regard, if you ask me - and I think QM is a pretty good theory too.

<i>To top it off, Einstein made time into a dimension, which IMHO is a non-starter.</i>

It obviously is a starter. It gives the right answers. Actually, more than that, it generates some amazingly beautiful mathematical symmetries, and leads to new physical insights. And GR just wouldn't work without it. You really have to understand the maths to get the full picture. When you do, it's one of the most elegant in physics.

<i>You can make all the claims you want about how spacetime is curved; that doesn't necessarily give you a handle on what is really going on.</i>

Right, but nothing will. We just don't have access to the ultimate reality. Maybe someday we will, but we don't right now, and if we did we couldn't know we did.

<i>Though I can't help but notice your usage of "space" rather than "spacetime". Fine by me, as I have yet to be startled by the spectacle of anyone strolling along time.</i>

You're strolling along time whilst you sit there reading this post. :)

Tom,

<i>As I stated before, and nobody listened to me except overdoze, a light clock proves that the principle of invariance of light is wrong.</i>

If it proved that, you would have been able to show it proved that. I listened to you. I just don't find your arguments compelling.

<i>I've been on this board for months, and everyone has been arguing with me that relativity states that time is slows down.

If they said in the beginning that clocks slow down, but that doesn't necessarilly mean that time slows down, I would've saved everyone alot of time. :)</i>

Relativity is a theory about space and time, not clocks. Nothing about the internal workings of a clock changes when you move past it at high speed. If that happened, it would indeed be a strange universe. What changes is your perception of the clock.

James R,

The main difference between you and me is that you believe illusions are reality, while I do not.

Tom

Hi James

I see we're actually getting somewhere, for a change :eek: :D



Interpretations abound. You are welcome to yours, of course. If your interpretation has no observably different consequences to mine, it doesn't matter much, does it? Compare, for example, the Copenhagen and Many Worlds interpretations of quantum mechanics (assuming you're familiar with QM). Neither interpretation changes the usefulness of the theory. Both are compatible with it.


Precisely. However, I've observed that my interpretation is far more digestible. As such, a lot more people would understand relativity -- and be at peace with it -- if they were taught it from my point of view (witness Tom, for example.)



A coordinate system is not a physical thing. It is a mathematical construct. The universe has no trouble support however many you want.


But the universe itself isn't just a mathematical construct. I think my interpretation is a little less abstract and therefore somehow closer to reality.



<i>There exists a cosmic medium within which all observable events take place</i>

I ask you, then: can we detect this medium? How is having this medium different from not having it?


As far as detection, it's all around (and within) you. :) You are, in fact, just a part of it. As for not having this medium, I suppose that would equal non-existence of everything (or should I say anything?)



<i>Within this cosmic medium, the velocity of light is the same in all directions</i>

In that case, if you move relative to the medium the velocity of light should change, right? If the medium is "absolutely stationary", it can't also move so as to maintain a constant speed of light.


Ah-ha! I see you're catching on. ;) Now go back to my derivation and read over it. Indeed, the unidirectional velocity of light changes with respect to moving observers. However, the observers have no way of measuring such unidirectional velocity without, in essense, their experimental setup balancing it out to the good old value of c -- which has been my thesis all along. This shouldn't be too hard for you to accept, since the math actually turns out identically (i.e. Lorentz transforms.)

Point is, Einstein's two core premises (vis a vis inertial reference frames) actually "drop" out of my two core premises very naturally. But not vice versa. Which is why I insist I have a leg up on simplicity here (not to mention comprehensibility.)



We don't have access to the kind of underlying reality you seem to want. All we have is models for processes. We don't know if an atom really contains quarks ... But such a theory cannot be tested, so it isn't much use.


Do you think Einstein had testability in mind as he was crafting SR (or GR, for that matter?) I very much doubt it. He was merely trying to build up a conceptual framework that would make sense. Testability could only be determined once the math was complete and actual predictions could be derived.

Similarly, any new grand theory of everything would have to start out not as a bunch of testable propositions, but as a coherent and elegant conceptual framework. It is to be hoped that the testable propositions would become apparent once the framework is completed.



GR isn't necessarily the last word. String theories may have things to say about this, but I can't tell you since string theory isn't my thing.


Yes, I've been following string theory at a layman's level (reading popular science books and Scientific American articles on the topic) -- but not much beyond that. So far from the little I've seen (and obviously, most of it was oversimiplified, overgeneralized, or both), I'm both impressed and unimpressed with the theory. Impressed because it at least has a promise of unifying all forms of matter/energy, and all forces, under a single umbrella. Unimpressed because it typically builds on enigmatic vibrating structures of impressive complexity (not to mention all of those dimensions...) One yearns for something simpler...but then again, maybe the universe isn't that simple after all.



It gives the right answers. Actually, more than that, it generates some amazingly beautiful mathematical symmetries, and leads to new physical insights. And GR just wouldn't work without it.


Well, as I've said before I don't have the toolkit to attack this particular problem just yet. Though from analogy with SR, if it can be re-cast with time being just a parameter (as opposed to a dimension), then why shouldn't the same be possible with GR?



You really have to understand the maths to get the full picture. When you do, it's one of the most elegant in physics.


I've heard plenty about the beauty and elegance. I've even seen a little bit of them myself. But beauty and elegance must be put in context. What if, reinterpreted the way I wish it was, the mathematics would appear even more beautiful and elegant? And even if not, what if it would nevertheless become much more comprehensible? One would never know until one tried...

For example, at a risk of sounding conceited, I find my own derivation and interpretation of SR even more beautiful than the original. The perfect symmetry of illusion is indeed breathtaking.



We just don't have access to the ultimate reality. Maybe someday we will, but we don't right now, and if we did we couldn't know we did.


Well shoot, with attitude like that we never will. I don't think we are justified in resting on our laurels, no matter what heights of elegance and beauty we have achieved to date.



You're strolling along time whilst you sit there reading this post.

More like being transported than strolling. I can only travel in one direction, and at a constant speed (as long as I'm inertial) to boot. Hardly what I would call "strolling".

overdoze,

<i>I see we're actually getting somewhere, for a change</i>

I'm not so sure...

<i>Precisely. However, I've observed that my interpretation is far more digestible.</i>

I'm not finding it very digestible.

<i>As far as detection, it's all around (and within) you. You are, in fact, just a part of it.</i>

You've dodged the question. I asked how we could detect it. If we can't, Occam's razor says we don't need it. Just like Einstein said we don't need the aether.

<i>Now go back to my derivation and read over it. Indeed, the unidirectional velocity of light changes with respect to moving observers. However, the observers have no way of measuring such unidirectional velocity without, in essense, their experimental setup balancing it out to the good old value of c -- which has been my thesis all along. This shouldn't be too hard for you to accept, since the math actually turns out identically (i.e. Lorentz transforms.)</i>

All I have to do to your original derivation to get rid of the concept of an absolute reference frame is to introduce a third observer moving relative to the first two.

<i>Do you think Einstein had testability in mind as he was crafting SR (or GR, for that matter?) I very much doubt it.</i>

I can't recall, actually. I'd need to re-read his original papers.

<i>...One yearns for something simpler [than string theory]...but then again, maybe the universe isn't that simple after all.</i>

I agree.

<i>Though from analogy with SR, if it can be re-cast with time being just a parameter (as opposed to a dimension), then why shouldn't the same be possible with GR?</i>

You'd need to understand the maths to see why that is not possible for GR. The main mathematical objects in GR are 4-vectors and 4 dimensional tensors. A 4-vector consists of one "time" component and three "space" components. 4-vectors and tensors obey very strict symmetry rules and conservation laws. Try to turn the time component into a parameter and you break the symmetry and destroy Lorentz invariance.

<i>For example, at a risk of sounding conceited, I find my own derivation and interpretation of SR even more beautiful than the original. The perfect symmetry of illusion is indeed breathtaking.</i>

I find that your interpretation introduces an unnecessary complication which does not mesh with what we can observe.

<i>I don't think we are justified in resting on our laurels, no matter what heights of elegance and beauty we have achieved to date.</i>

I don't either.

<i>I can only travel in one direction [in time], and at a constant speed (as long as I'm inertial) to boot. Hardly what I would call "strolling".</i>

Your perception of time is a little different from the relativistic description of it, remember. GR lays out time as just another dimension. It's all there at once, so to speak, whereas you perceive a past, present and future at any particular time. The question of who's right is a philosophical one.

Tom,

You can say things like that until the cows come home, but they're empty statements unless you can back them up.

Overdoze and James,

I can prove that aether exists!!! Listen very closely:

It takes light 1 second to travel 300,000 km in a vacuum, but it takes light 10 seconds to travel 3,000,000 km in a vacuum!!!

If aether didn't exist, then in both cases the light would take the same amount of time, because in both cases the light would be passing through nothing.

Is that proof enough?? :)

Tom

eeh...that is proof of something allright.

James,


You can say things like that until the cows come home, but they're empty statements unless you can back them up.

Let me explain what I meant. After reading your posts on several different threads, it appears that your core problem is not that you can't accept an alternative model that explains relativistic phenomena, your problem is much deeper and more troubleing.

Your core problem is that you think that the things any observer percieves in any frame of reference is correct. This sounds like the philosophy of a madman. :)

On another thread, 137 asked you if a person one mile away is 1 inch tall because he looks like he's only one inch tall. You said that he is.

I asked you previously, on another thread, whether the sun revolves around the Earth or whether the Earth revolves around the sun. You said that both points of view are correct. They do appear correct, until you take gravity into consideration.:)

What you don't understand is that the things an observer sees in his frame of reference are almost all illusions. These illusions are the result of distance(an object appears smaller the farthur away it is), point of view(objects look differently when percieved from different angles), velocity(an object appears to be travelling slower if you are travelling with it), and delay( two beams of light can hit there targets at the same time, but they are not seen by the observer to hit the targets at the same time unless the observer is an equal distance from them both).

The facts are that time, distance, mass, and energy are all constant in all frames of reference, but an observer will percieve them as being changed or distorted depending on the observers frame of reference. Just because the observer "sees" these properties change, doesn't mean that they actually do change.

Therefore, science is absolute, while relativity explains the illusions a moving observer "percieves".

Tom

Prosoothus

It takes light 1 second to travel 300,000 km in a vacuum, but it takes light 10 seconds to travel 3,000,000 km in a vacuum!!!

Therefore, light travels at 300,000 km per second in a vacuum.

If aether didn't exist, then in both cases the light would take the same amount of time, because in both cases the light would be passing through nothing.

How can light take the same amount of time if you've just shown that light travels at 300,000 km per second in a vacuum ? You've just contradicted yourself.

Q,

You misunderstood my post. :confused:

What is the difference between 300,000 km of vacuum, and 3,000,000 km of vacuum?? If you were to assume that aether does not exist, then the answer is that there is no difference.

If that were the case, the light would travel both distances in the same amount of time, since in both cases, nothing would be in the way of the light.

Since light takes ten times longer to travel 3,000,000 than it takes to travel 300,000 km, I'm suggesting that 3,000,000 km has ten times more of something than 300,000 km does. In other words, if you were to take away 90% of the thing that makes up the 3,000,000 km, the 3,000,000 km of space would collapse into 300,000 km of space.

Tom

Prosoothus

What is the difference between 300,000 km of vacuum, and 3,000,000 km of vacuum??

2,700,000 km of vacuum space which is the distance light would travel in 9 seconds.

If you were to assume that aether does not exist, then the answer is that there is no difference.

Regardless of whether there is an aether or not, or whether that space is comprised of matter or is a vacuum, it is still a difference in distance.

If that were the case, the light would travel both distances in the same amount of time, since in both cases, nothing would be in the way of the light.

If light travels at 300,000 km per second, and there was nothing in the way, light would take 10 seconds to travel 3,000,000 km and 1 second to travel 300,000 km.

Since light takes ten times longer to travel 3,000,000 than it takes to travel 300,000 km, I'm suggesting that 3,000,000 km has ten times more of something than 300,000 km does.

Correct, the answer is "distance."

In other words, if you were to take away 90% of the thing that makes up the 3,000,000 km, the 3,000,000 km of space would collapse into 300,000 km of space.

Again correct, if you were to take away 90% of the distance between 3,000,000 km. and 300,000 km., you would be taking away 2,700,000 km. of distance, the distance light would travel in 9 seconds.

Sorry man, I really must be missing something tangible in your posts if I'm still not understanding.

Q,

What is distance???

What makes 1000 m, 1000 m, and not 1 m.

What is the difference between 1 m and 1000 m??

Is distance just a "number", or does it represent something physical??

If you were God, what would you do to convert 1000 m to 1 m???

Tom

Tom,

<i>Your core problem is that you think that the things any observer percieves in any frame of reference is correct.</i>

They are correct <b>for that observer</b>.

<i>This sounds like the philosophy of a madman.</i>

It's the philosophy you live by every day of your life. Don't you trust your own senses?

<i>On another thread, 137 asked you if a person one mile away is 1 inch tall because he looks like he's only one inch tall. You said that he is.</i>

Yes, as far as the observer is concerned. Did you read the rest of my explanation? You know, the bit where I explained that to transform a measurement <b>there</b> to a measurement <b>here</b> you need to do some maths? The fact that you can do the required calculations automatically to estimate the size of an object at a distance doesn't mean that those calculations aren't happening. And your brain doesn't always get them right, either. Ever seen an optical illusion, Tom? That's where your brain messes up the calculations.

<i>I asked you previously, on another thread, whether the sun revolves around the Earth or whether the Earth revolves around the sun. You said that both points of view are correct. They do appear correct, until you take gravity into consideration.</i>

Either way you need to take gravity into consideration. Either view is correct when you do that properly.

<i>What you don't understand is that the things an observer sees in his frame of reference are almost all illusions.</i>

I prefer the term "perceptions". For the observer, his perceptions are reality.

<i>The facts are that time, distance, mass, and energy are all constant in all frames of reference, but an observer will percieve them as being changed or distorted depending on the observers frame of reference.</i>

You contradict yourself. When you say "frames of reference" and "observer", you're talking about the same thing. A frame of reference, as I've explained <i>ad nauseam</i> is no more than the point of view of an observer. If two observers see two different things, then those things are different in their two frames of reference. The statements are equivalent and interchangeable.

<i>Just because the observer "sees" these properties change, doesn't mean that they actually do change.</i>

Depends what you mean by "actually".

<i>Therefore, science is absolute, while relativity explains the illusions a moving observer "percieves".</i>

No. Observations are relative. Relativity tells us how to convert one set of observations to another point of view.

<i>What is distance???

What makes 1000 m, 1000 m, and not 1 m.

What is the difference between 1 m and 1000 m??</i>

Suppose you have a standard 1 metre ruler. You need just one of those to measure out a distance of 1 metre. You need 1000 of them end-to-end to measure out 1000 metres.

Do that, and place markers at the ends of the lines of rulers. Now take the rulers away. Between the markers you now have an aetherless vacuum. But the distance between the markers hasn't changed. If you want to lay out your rulers between the markers again, there's no way you can cover the 1000 m with a single ruler.

You're really off into fairy land if you believe that 1000 m = 1 m.

Hi Everybody,

I am a physicist student from England. I found a site with the following title:

Particle and atomic physics
- instead of theories -
according to experimental
results and laws of the
classical physics by
Gabor Fekete

Please view this site at www.physics.uw.hu address.

Thou
Shalt
Not
Commiteth
Thread
Necromancy!!!
Thus Sayeth The Prince!!!!!!!!!!!!!!!!

The atoms in the clock experience no Doppler shifts, since the microwave source is moving with the rest of the clock.

So, you are telling us that the atoms experience no Doppler shift. OK (?).

Since the atoms are their own observers, in Einstein Relativity, I think you can make a case for that.

Just in case you are faltingly trying to say that the microwaves take on the velocity of their source, I am prepared to argue with you, in the unlikely event that you might make a clear statement of your position on the matter.

It is believed that atomic clocks are accurate, even when they are moving.

Scientists claim that results of the measurements of moving atomic clocks indicate that Einsein was correct, and that time really does slow down the faster an object is moving.

However, after reading about the construction of caesium atomic clocks, I found that there are three factors that would influence a moving atomic clock, independently of the time that it is measuring:

a) The motion and direction of the magnetic fields which are used to seperate the different caesium atoms.

b) The motion and direction of the caesium atoms themselves.

c) The motion and direction of the microwaves used to excite the caesium atoms.

As the motion of the atomic clock increases, one or more of the above factors would influence the clock to give a false reading. This would mean that atomic clocks are only accurate if there at rest(at least relatively).This would also mean that time is constant, but the speed of the atomic clock changes based on its motion.

Any comments are appreciated.

Tom

Atomic clocks operate on the basis of the two-way travel of the photons emitted by the excited atom.

The emitted photon's total two way travel time will always be affected by the direction and velocity of the photons relative to direction and velocity of the clock as a whole. A clever experimentalist would be able to cook the books by deliberately orientating the clock one way or the other in respect to the trajectory of the jet plane or satelite carrying it.

CANGAS,


The emitted photon's total two way travel time will always be affected by the direction and velocity of the photons relative to direction and velocity of the clock as a whole.

So, you also believe that the speed of light changes in a moving clock. Your statement seems to imply that you believe that the speed of light is equal to c only relative to a specific medium. What do you believe that this medium is?

How accurate are atomic clocks??

You can set your watch by them.:)

They add a leap second about every year and a half, on average, because the Earth's spin rate is slowing down that much.

Whether one speaks in terms of motion in absolute space, ... This reference to absolute space almost made me stop reading, but the next paragraph confirmed that I should, so I read no more of this nonsense. Your "next paragraph", plausible sounding to the ignorant, but entirely, false was:

Therefore an observer who is stationary with respect to the source of a photon may observe several different things; if the source and the photon have velocity in the same direction then the photon may slowly pull away from its emitter, in which case the observer would see a RELATIVE velocity between the source and the photon of somewhat less than c....
I have no idea what you stated after that as I do have this policy about not eating all of a rotten egg.

At the current rate of the slowdown of the Earth's spin, about 30 millions years ago, a day would have been about four hours long, another reason that the Earth and universe are much younger than commonly advertised.

Why do you suspect that the average rate of the last 30 million years is equal to the current rate?

Seems reasonable.

This reference to absolute space almost made me stop reading, but the next paragraph confirmed that I should, so I read no more of this nonsense. Your "next paragraph", plausible sounding to the ignorant, but entirely, false was:

I have no idea what you stated after that as I do have this policy about not eating all of a rotten egg.


And I have no idea what any of your post was, for the same reason.

CANGAS:

I have no intention of replying to your comments about a post I wrote 6 years ago. Frankly, you're not worth the effort.

CANGAS:

I have no intention of replying to your comments about a post I wrote 6 years ago. Frankly, you're not worth the effort.

If you ignore me, that works for me. Will you start soon?

At the current rate of the slowdown of the Earth's spin, about 30 millions years ago, a day would have been about four hours long, another reason that the Earth and universe are much younger than commonly advertised.

So, the Earth has been slowing down for 30 million years, in which the day was 4 hours long and is now 24 hours long? Yeah, that makes sense.:rolleyes:

Actually, the slowing of Earth's spin is well documented by a number of sources that agree relatively closely.

The following web-sites detail this, and gives varying estimates ranging from .00001 seconds/year to .005 seconds/year.

The fossil records correlate well with the projected past day-length, which might have been as short as 12 hours 4.5 billion years ago.

www.creation-answers.com/slowing.htm

http://www.talkorigins.org/indexcc/CE/CE011.html

Perhaps Ice Age meant they add a leap millisecond every 1 1/2 years, which is the way I usually hear it reported.

Actually, the slowing of Earth's spin is well documented by a number of sources that agree relatively closely.

I agree, but Ice's numbers are silly in the extreme.

Perhaps Ice Age meant they add a leap millisecond every 1 1/2 years, which is the way I usually hear it reported.

Even then, this only accounts for a difference of about 5.5 hrs over 30,000,000 years, for a day of a length of 19.5 hrs, not 4.

Also, it should be noted that leap-seconds are not a measure of the rate at which the Earth is slowing, but are due to the fact that the new defintion of a second differs from the old defintion of 1/86400 of a solar day. The only time this was true was around 1820. Rather than constantly changing the length of the second, a fixed length was chosen that confirms closely to 1/86400 of a 1820 solar day. Since 1820, the solar day has increased by .002 seconds, which means that every solar day is now 86400.002 seconds long. And it is this difference of .002 seconds a day that results in the addition of the leap second every so often, not the slowing of the Earth's rotation from day to day.(Remember, it took almost 2 centuries for this 2 millisecond difference to form due to the Earth's slowing, but once it formed, it is the accumulation of this difference over the days that causes Solar time and standard time to drift apart. After one day they will be .002 sec apart, after two 0.004sec, after three .006 sec etc)

Thanks for the clarification, Janus.

So, the Earth's daily rotation slowed down by .002 seconds in 200 years.

Multiply that by 1,000,000, and projecting backwards, means that 200 million years ago, the Earth's daily rotation rate would have been about 2,000 seconds shorter than nowadays. That is, instead of 24 hours, it would have been 24 hours minus 2,000 seconds, or 24 hours minus 33 minutes; i.e. about 23 1/2 hours. Of course, that is assuming the decrease in the rate of spin would have remained absolutely constant, though that would not be the case due to the varying ice-packs at the poles, which would have impacted the tidal influence of the slow-down.

So, Ice Age was correct that we constantly insert a leap-second every year or two, but his reasoning as to why such is done is faulty. It's not because the Earth is rapidly slowing down its spin, but because we're using the 1820 A.D. year-length as the original standard (which is now incorporated as an atomic time), which is only off by .002 seconds in the daily spin rate after almost 200 years