The purpose of this thread is to discuss whether length contraction is a required conclusion if the speed of light is the same in all non-accelerating reference frames.

Here is a simple device that emits two flashes, one to the North and one to the East.
The flashes bounce off mirrors that are each one metre away from the device, then return to the device.
If the flashes return at the same time, the device pops up a flag to indicate the fact:

http://home.teegee.com.au/byrnes/Pete/Relativity/Images/RestFrame.gif

Here is the same situation in a reference frame in which the device is moving rapidly Northward.
EDIT - ie the device hasn't changed its motion, but our viewpoint (the "camera") is different. This time, we're moving rapidly southward. Thanks Cangas.

I've shown it twice... in the second image, I've hidden the device for clarity, showing only the flashes, along with their paths and some key points. These diagrams are not drawn to scale.
<table width=100%><tr><td width=50% align=center>http://home.teegee.com.au/byrnes/Pete/Relativity/Images/MovingFrameA1.gif</td><td align=center width=50%>http://home.teegee.com.au/byrnes/Pete/Relativity/Images/MovingFrameA2.gif</td></tr></table>
Take Note - in all animations, the device is securely attached to the Earth. In the diagrams in which the device appears to be moving, our viewpoint (the camera) is moving South relative to the Earth and the device.


Consider the frame in which the device is moving. If the light flashes are moving at a steady speed, what does that tell us about the distances W and N from the device to the mirrors?


(Its also interesting to think about what this tells us about simultaneity, but one thing at a time.)

Pete,

For clarity and courtesy I don't think I will debate the issue in your thread but I will start my own thread showing how and why length contraction conclusions are invalid.

I will refer readers to your thread to see the SRT arguement.

You just want to muddy the waters, MacM.

Just my two cents...

The fact that Mac refers to this as a debate gives away his agenda. Given the observed constancy of 'c' in all frames, this is a matter of demonstrable logic, not debate.

For clarity and courtesy I don't think I will debate the issue in your thread but I will start my own thread showing how and why length contraction conclusions are invalid.
Well, if you've got an agenda to push, go ahead.

This thread is for open investigation into the logical conclusions of a postulate. I invite you to participate, but you're welcome not to if you so wish.

You just want to muddy the waters, MacM.

If that were true I would ost my arguement here.

Just my two cents...

The fact that Mac refers to this as a debate gives away his agenda. Given the observed constancy of 'c' in all frames, this is a matter of demonstrable logic, not debate.

Thanks for showing you don't have a clue. The fact is my arguement does not challenge an invariant 'c' and is also based on demonstable logic. So there. :p

Well, if you've got an agenda to push, go ahead.

This thread is for open investigation into the logical conclusions of a postulate. I invite you to participate, but you're welcome not to if you so wish.

Funny how trying to get people to understand the reality is an agenda and to repeatedly post threads arguing the SRT view is not an agenda. Hmmm.

I am clueless and glueless. Whatever you say, bounces off of me and sticks to you! Ha! :p

Funny how trying to get people to understand the reality is an agenda and to repeatedly post threads arguing the SRT view is not an agenda. Hmmm.
Has anyone argued the SRT view in this thread?

Again... this thread is about open logical investigation. No agendas.

Cool animations Pete!

One thing that I have been thinking about is that if you start with the perpendicular clock you wind up with the Lorentz transform, and if you start with the parallel clock you wind up with the Voigt transform (http://en.wikipedia.org/wiki/Woldemar_Voigt). In the Lorentz transform the time dilation is gamma and in the Voigt transform it is gamma&sup2;. So I wonder what would happen if you started with a clock at a 45º angle. Could you get a time dilation between gamma and gamma&sup2;?

It seems that the actual time dilation follows the Lorentz transform, but I wonder if there is some logical reason to choose it over the Voigt.

-Dale

Has anyone argued the SRT view in this thread?

Again... this thread is about open logical investigation. No agendas.

Now you sound like an ID'er. Mincing with words does not alter the basis of your arguement. Or will you state here that your presentation is somehow in conflict with the assumptions and/or conclusions of SRT. :D

The only assumption (clearly identified) is that the speed of light is the same in all frames.

No conclusions have been presented, so it's not in conflict with anything, yet.

Come one, Mac, put up or shut up. You've said that invariant c doesn't imply length contraction... now's your chance to prove it.

The only assumption (clearly identified) is that the speed of light is the same in all frames.

No conclusions have been presented, so it's not in conflict with anything, yet.

Come one, Mac, put up or shut up. You've said that invariant c doesn't imply length contraction... now's your chance to prove it.

Like I said I have posted my view of this situation in another thread to deal with realities.

*Sigh*

It looks like Mac doesn't want to play. Would anyone else like to have a go?


http://home.teegee.com.au/byrnes/Pete/Relativity/Images/MovingFrameA.gif
We need to come up with some way to relate the distances the light flashes travel to the distances between the device and the mirrors.

We know that each flash has the same round trip time, and we're assuming that each flash moves at the same speed for the whole trip.

In addition to the constant c, I assume the variables are named as follows...

v (velocity)
t_sub (time increments at rest), also t'_sub (time increments in motion)
N (vertical dimension of light clock at rest), also N' (for motion)
W (horizontal dimension of light clock), also W' (for motion)

...Plus these lengths which only apply to the clock in motion
AC
CD
AB
BD

Hi Neddy, welcome along!
N and W specify the dimensions in motion. At rest, the dimensions are 1m by 1m.

Also, we really don't need to consider the device's rest frame at all.

The only question to be answered is "are N and W the same?", and we can do that directly in the frame in which the device is moving.

One thing that I have been thinking about is that if you start with the perpendicular clock you wind up with the Lorentz transform, and if you start with the parallel clock you wind up with the Voigt transform (http://en.wikipedia.org/wiki/Woldemar_Voigt). In the Lorentz transform the time dilation is gamma and in the Voigt transform it is gamma&sup2;. So I wonder what would happen if you started with a clock at a 45º angle. Could you get a time dilation between gamma and gamma&sup2;?

Hi Dale,
How does the parallel clock work?

</em>Hi Neddy, welcome along!
N and W specify the dimensions in motion. At rest, the dimensions are 1m by 1m.

Also, we really don't need to consider the device's rest frame at all.

The only question to be answered is "are N and W the same?", and we can do that directly in the frame in which the device is moving.

Thanks, Pete. So, in addition to the constant c, we have...

v
t
N
W
AC
CD
AB
BD

I'm going to go out on a limb here and postulate that, at v=0...

v = 0
N = 1m
W = 1m
AC = N
CD = N
AB = W
BD = W

Good stuff.
What if v is not zero?

Good stuff.
What if v is not zero?

I prefer for v to be zero; its easier that way. :p

Believe me, you don't want me to fill this thread with an amatuer derivation by me.

Trying not to assume anything which is not already assumed in this thread, on a scrap paper nearby, I have already constructed a mess that looks like the W dimension is getting shorter (should that happen?...)

Later, by edit:
I bet there is a sticky somewhere that would teach me the codes for displaying math symbols, for example, some better way than this for displaying the square root

= ((c^2 - v^2)^.5)/c

Any hints would be appreciated. Thankyou

Aer posted a primer somewhere for symbols.

Let's start with the AB segment. What do we know about it? Can we use Pythagorus?

AB segment? Use Pythagorus? Well, I did.

What's your next question?

Can I use a lifeline? How about asking the audience? I really hope I do good on this one. Will you give hints?

I bet there is a sticky somewhere that would teach me the codes for displaying math symbols, for example, some better way than this for displaying the square root

= ((c^2 - v^2)^.5)/c

Any hints would be appreciated. Thankyou

One simple approach used by some here is simply sqrt(X).

But by code (leave out the " sysmbols:

(X)"<"s"u"p">"0.5"<"/"s"u"p">"

for

(X)^0.5

or use "sub" for

(X)₅

AB segment? Use Pythagorus? Well, I did.

What's your next question?

Can I use a lifeline? How about asking the audience? I really hope I do good on this one. Will you give hints?
Mocker ;)

Aer posted a table of math symbols in the FAQ.

&amp;radic; gives a square root sign: &radic;(....)
&amp;sup2; gives a raised two: (....)&sup2;
Or use superscripts or text shorthand like Mac showed.

Anyway. AB segment. Pythagorus.

Variables:
t = the time it takes the flash to get from A to B

The length of AB is ct.
The east-west component of AB is W.
The north-south component of AB is vt.

...

Hi Dale,
How does the parallel clock work?Using your notation the time, t_AC, for the light pulse to go from A to C is the solution of:
c t_AC = v t_AC + N --> t_AC = N/(c-v)

The time, t_CD, for the light pulse to go from C to D is the solution of:
N - c t_CD = v t_CD --> t_CD = N/(c+v)

So the total time, t_AD, is:
t_AD = t_AC + t_CD = 2 c N/(c&sup2;-v&sup2;) = 2 γ&sup2; N/c
where γ = 1/sqrt(1-v&sup2;/c&sup2;)

Correspondingly, the total distance traveled by the light pulse, d_AD, is:
d_AD = c t_AD = 2 γ&sup2; N

-Dale

Pete, the REASON lengths are contracted and clocks beat at variable rates is because those values are defined by the assumed invariant speed of light. The speed of light in a vacuum is DEFINED as 299,792,458 meters per second. Special Theory extends that further to postulate that the SOL is invariant for all observers, regardless of the speed of the source emitting the light. By necessity, if the SOL is invariant, the meter must change when measured by that ruler, and the second must vary when timed by that light clock. Many pulsars beat at a very steady rate due to the fact that their pulse is directly connected to their rotation. Use the beat of the pulsar to reference your clock and 'time' does not vary with velocity and lengths do not contract. The clock will beat at the same rate in all reference frames. The question is, does 'time' itself slow when velocity is increased according to the speed of light clock, or does the speed of light clock run slow when compared with 'universal' time?

Pete, I assume you know what the 'Hubble sphere' refers to. It is a reference in the universe beyond which everything is receeding from our location at greater than the speed of light. If the Lorentz transforms were applied to objects at the Hubble sphere distance, then those objects would appear to be right next to us because of distance contraction. Of course, they do not appear to be nearby. A new modification is made to the original Special Theory in this case. It is said that the wavelength of light is 'stretched' by the expansion of the universe. What does that mean? Does it mean the objects 'really are' nearby, but they 'appear' to be far away because the light emitted from them was 'stretched'? Isn't it more likely our ruler based on the invariance of light speed FOR ALL OBSERVERS is wrong and length contraction is an illusion? The modern interpretation of the speed of light is that it is invariant in CO-MOVING reference frames only. In other words, the speed of light is invariant ONLY with regard to frames that are 'at rest' relative to your own frame.

Do I think the 'travelling twin' in the twin paradox will age less than his stay-at-home counterpart? No, they will age approximately equally. If the they both used a clock based on the speed of light for its beat, the travelling twins clock will accumilate less time that the stay-at-home twins. If they both used a clock timed by the beat of the pulsar, the clocks will agree and the distance travelled will be calculated the same for both frames. The pulsar will beat the same total number of times for both twins during the trip. Time is universal, clocks can vary according to what reference they use for 'timekeeping'.

Pete, the REASON lengths are contracted and clocks beat at variable rates is because those values are defined by the assumed invariant speed of light. The speed of light in a vacuum is DEFINED as 299,792,458 meters per second. Special Theory extends that further to postulate that the SOL is invariant for all observers, regardless of the speed of the source emitting the light. By necessity, if the SOL is invariant, the meter must change when measured by that ruler, and the second must vary when timed by that light clock. Many pulsars beat at a very steady rate due to the fact that their pulse is directly connected to their rotation. Use the beat of the pulsar to reference your clock and 'time' does not vary with velocity and lengths do not contract. The clock will beat at the same rate in all reference frames. The question is, does 'time' itself slow when velocity is increased according to the speed of light clock, or does the speed of light clock run slow when compared with 'universal' time?

Pete, I assume you know what the 'Hubble sphere' refers to. It is a reference in the universe beyond which everything is receeding from our location at greater than the speed of light. If the Lorentz transforms were applied to objects at the Hubble sphere distance, then those objects would appear to be right next to us because of distance contraction. Of course, they do not appear to be nearby. A new modification is made to the original Special Theory in this case. It is said that the wavelength of light is 'stretched' by the expansion of the universe. What does that mean? Does it mean the objects 'really are' nearby, but they 'appear' to be far away because the light emitted from them was 'stretched'? Isn't it more likely our ruler based on the invariance of light speed FOR ALL OBSERVERS is wrong and length contraction is an illusion? The modern interpretation of the speed of light is that it is invariant in CO-MOVING reference frames only. In other words, the speed of light is invariant ONLY with regard to frames that are 'at rest' relative to your own frame.

Do I think the 'travelling twin' in the twin paradox will age less than his stay-at-home counterpart? No, they will age approximately equally. If the they both used a clock based on the speed of light for its beat, the travelling twins clock will accumilate less time that the stay-at-home twins. If they both used a clock timed by the beat of the pulsar, the clocks will agree and the distance travelled will be calculated the same for both frames. The pulsar will beat the same total number of times for both twins during the trip. Time is universal, clocks can vary according to what reference they use for 'timekeeping'.

Of course as 2inq has said that if you hold to the view that lights behaviour is as postulated time and length contraction is the only solution that comes close to explaining observation.
However one must agree that light travels as postulated.

By granting light the nature we have determined, SRT is the only outcome.
Which includes relative time and distance.
So if I was wantirg to argue against SRT the only arguement must be with our understanding of the nature of light 'c'.

Hi 2inquisitive and QQ,
I appreciate your thoughts and do take note that it is an assumption of the exercise that c is invariant.

If you wanted to open another thread entitled "Is the speed of light the same in all frames?" then I'd be glad to participate.

Hi 2inquisitive and QQ,
I appreciate your thoughts and do take note that it is an assumption of the exercise that c is invariant.

If you wanted to open another thread entitled "Is the speed of light the same in all frames?" then I'd be glad to participate.
Pete, thanks, however I have no real desire to debate the validity of light postualtes, I was merely saying that if one subscribes to those postulates that relative time and distance must be the outcome.

Do you realize that this thread gedanken is a strong arguement against SR and in favor of absolute space?

No mock. Dead serious. Are you in favor of proving SR or are you really trying to disprove it?

[QUOTE=Pete]Hi 2inquisitive and QQ,
I appreciate your thoughts and do take note that it is an assumption of the exercise that c is invariant. (.....)

Sure, no problem Pete. We will measure the rotation of the pulsar at 60 per second in the Earth frame. Light travels at 299,792,458 meters per second. Assume an astronaut is travelling at .866c relative to the Earth. In the Earth frame, the pulsar 'blinked' 60 times per second, one time for each revolution. How far, in astronaut meters, will the Earth travel in 60 'blinks' of the pulsar, while the Earth is travelling at .866c relative to the astronaut? How much time will pass on the astronaut's onboard 'clock' during the SAME 60 'blinks'? I am assuming 'c' is invariant. And, yes, of course I have a followup.

Let the height of the top mirror be 'N' in the frame where the system appears to be moving. If 't<sub>1</Sub>' is the time that the flash takes to reach the mirror, then, since it has to travel the distance 'N', as well as the additional distance vt<Sub>1</Sub>:

ct<Sub>1</Sub> = N+vt<Sub>1</Sub>

.'. t<Sub>1</Sub> = N/(c-v)

On the return journey, it has to travel a smaller distance, since the device is moving towards it. Therefore, if t<Sub>2</Sub> is the time taken for the return:

ct<Sub>2</Sub> = N-vt<Sub>2</Sub>

.'. t<Sub>2</Sub> = N/(c+v)

So the total time is:

t= t<Sub>1</Sub> + t<Sub>2</Sub> = N/(c-v) + N/(c+v)

t = N(c+v+c-v)/(c<Sup>2</Sup>-v<Sup>2</Sup>)
t = 2Nc/(c<Sup>2</Sup>-v<Sup>2</Sup>)

N = (c<Sup>2</Sup> - v<Sup>2</Sup>)t/2c

Now what about the second flash of light? Within this time, the total distance it will have travelled will be 'ct'. And it takes the same time to travel back and forth, since the system is not moving horizontally. So that it in the first part of its journey, the distance covered by it, namely AB, is 'ct/2'. Now use Pyhtagoras theorem (a wonderful theorem really, everything from the Special Theory to the General Theory relies heavily upon it):

(AD/2)<Sup>2</Sup> + W<Sup>2</Sup> = AB<Sup>2</Sup>

(vt/2)<Sup>2</Sup> + W<Sup>2</Sup> = (ct/2)<Sup>2</Sup>

W<Sup>2</Sup> = (c<Sup>2</Sup> - v<Sup>2</Sup>)t<Sup>2</Sup>/4

W = (c<Sup>2</Sup> - v<Sup>2</Sup>)<Sup>1/2</Sup>t/2

Divide our previous equation for N by this equation, and we have:

N/W = (c<Sup>2</Sup> - v<Sup>2</Sup>)<Sup>1/2</Sup>/c

N/W = [1 - (v/c)<Sup>2</Sup>]<Sup>1/2</Sup>

N = W[1 - (v/c)<Sup>2</Sup>]<Sup>1/2</Sup>


Which is our good old length contraction equation. by other arguments, we can establish that W remains unaffected by the movement so that it will still be 1m. So we have a relation between the length of N in the devices frame (1m), and the the contracted length in this frame.

We will measure the rotation of the pulsar at 60 per second in the Earth frame. Light travels at 299,792,458 meters per second. Assume an astronaut is travelling at .866c relative to the Earth. In the Earth frame, the pulsar 'blinked' 60 times per second, one time for each revolution. How far, in astronaut meters, will the Earth travel in 60 'blinks' of the pulsar, while the Earth is travelling at .866c relative to the astronaut? How much time will pass on the astronaut's onboard 'clock' during the SAME 60 'blinks'? I am assuming 'c' is invariant. And, yes, of course I have a followup.
Is the pulsar at rest with respect to Earth?

A non-mocking serious question:

Does the velocity vector of a light source have any effect on the velocity vector of the emitted light?

Who is willing to make a guess?

Yes, in any frame of reference, they are relative; classic redshift/blueshift, unless I'm misinterpreting the general question.

Hi Dale,
How does the parallel clock work?

Graphically:

-----> (velocity)
_
_ (parallel mirrors)

| | (perpendicular mirrors)

The parallel mirrors are usually used to derive SRT time dilation since the distance between the mirrors are perpendicular to the velocity vector - you don't have to worry about length contraction.

As Dalespam pointed out, if you have the mirrors perpendicular to the velocity vector, so that light bounces back and forth parallel to the velocity vector, you will (unless you take into account length contraction) get a different transform.

Yes, in any frame of reference, they are relative; classic redshift/blueshift, unless I'm misinterpreting the general question.You are correct, but I think it very useful for full understanding to add the velocity of the source has effect ONLY DURING THE INTERVAL while the photon emission is in progress.

Normally the motion of the source has very little acceleration, so the blue or red shift remains constant, but if the source is accelerting, say towards us, then the blue shift is continuously increasesing.

It is important to undestand that (if source's gravity field is neglected) once the photon leaves the source behind, there is NO EFFECT upon the photon by the subsequent motion of the source. For example, some of the blue shifted light now coming to Earth may be from ONE of a relatively near-by star pair that is CURRENTLY going away from Earth.

Yes, in any frame of reference, they are relative; classic redshift/blueshift, unless I'm misinterpreting the general question.

Depends ... the assumption at the start of the thread was invariance of c - I assume that meant with respect to the observer?

Yes, Zephyr, that is the SPECIAL theory of relativity; it isn't perfect, but like Maxwell- it works for now.

...

Excellent point, Billy, about the interval (or instance) of photon emission.

Greetings

A non-mocking serious question:

Does the velocity vector of a light source have any effect on the velocity vector of the emitted light?

Who is willing to make a guess?

Definitely! The direction of the light flash depends upon the velocity of the source, as we can see from Pete's animations.

*Sigh*

It looks like Mac doesn't want to play. Would anyone else like to have a go?


http://home.teegee.com.au/byrnes/Pete/Relativity/Images/MovingFrameA.gif
We need to come up with some way to relate the distances the light flashes travel to the distances between the device and the mirrors.

We know that each flash has the same round trip time, and we're assuming that each flash moves at the same speed for the whole trip. Nice animation but seriously flawed.

Incase U didnt notice, look at how fast the light is reflected back compared to how long it takes to reach the mirror in the direction the experiment is traveling.

As Dalespam pointed out, if you have the mirrors perpendicular to the velocity vector, so that light bounces back and forth parallel to the velocity vector, you will (unless you take into account length contraction) get a different transform.I think we are calling different things parallel, you are refering to the orientation of the mirror and I was refering to the orientation of the axis of the light clock.

But we agree on the results. You get a different transform depending on which you start with. Empirical data follows the Lorentz transform, but the Voigt transform also gives a frame-invariant c.

-Dale

Nice animation but seriously flawed.

Incase U didnt notice, look at how fast the light is reflected back compared to how long it takes to reach the mirror in the direction the experiment is traveling.The trip out is longer than the trip back, as derived above. The animation is qualitatively correct.

-Dale

I think we are calling different things parallel

Sorry, yes, I mixed yours up :p

That Voigt transform looks interesting - especially since you can derive it without assuming length contraction in the direction of motion, but the wiki article says it predicts length expansion in the perpedicular directions.

So it wouldn't be able to explain muon lifetime? The Lorentz transform gives a nicely symmetric explanation in terms of time dilation for observers on earth and length contraction for the muon. Whereas the Voigt transform predicts even greater time dilation but no length contraction in the muon's frame...

The animation appears to show the two photons emitted simultaneously and received simultaneously. That doesn't happen without Lorentz contraction. The premise is proved.

Sorry, yes, I mixed yours up :pHehe, no problem. I don't think there is any standard terminology one way or the other anyhow.

That Voigt transform looks interesting - especially since you can derive it without assuming length contraction in the direction of motion, but the wiki article says it predicts length expansion in the perpedicular directions.

So it wouldn't be able to explain muon lifetime? The Lorentz transform gives a nicely symmetric explanation in terms of time dilation for observers on earth and length contraction for the muon. Whereas the Voigt transform predicts even greater time dilation but no length contraction in the muon's frame... I thought of the muon problem also. In the muon's frame perpendicular distances are expanded under the Voigt transform, but, as you said, that doesn't help explain how it reaches the ground in it's own frame. Symmetry seems to be such an important concept in general that it seems like there should be some theoretical reason to choose the Lorentz over the Voigt. But if there is I certainly can't think of it myself. It feels kind of inelegant to rely purely on experiment in choosing something as important as your transform between inertial frames.

-Dale

To rephrase my previous serious question in hope of it not being misunderstood:

If we have a source of photons, attached to a device capable of movement, which is precisely aimed and focused in a direction perpendicular to the intended direction of movement of the device, and we initiate movement, does the motion of the device provide a velocity vector to the photons being emitted?

Do the photons then have only their expected velocity vector perpendicular to the device movement?

Or do they also have an additional velocity component in the direction of device movement?

Please consider this question in regard to current mainstream physics theory, not in respect to the rejected Emitter Theory of over a century ago.

In all the excitement, I lost count, myself, of how many questions I had asked. Do I have one more? Yes, I do.

In the nice diagrams, the diagram representing the stationary device shows two light paths with an included angle of exactly 90 degrees.

The diagram of the moving device shows two light paths with an included angle of significantly less than 90 degrees.

Does this mean that the device automatically changes the mirror angle when in motion? Or does it mean that the angle of the light source changes its angle automatically when the device is in motion? Perhaps the mirror automatically changes its location when the device is in motion?

What did you intend to show us, Pete?

The trip out is longer than the trip back, as derived above. The animation is qualitatively correct.

-Dale And U think thats correct, can U tell me why ?

U r Dale dont giveup this question so easily. U have a slight higher awarness that rest of these morons.

And U think thats correct, can U tell me why ?

U r Dale dont giveup this question so easily. U have a slight higher awarness that rest of these morons.

Asshole.

In all the excitement, I lost count, myself, of how many questions I had asked. Do I have one more? Yes, I do.

In the nice diagrams, the diagram representing the stationary device shows two light paths with an included angle of exactly 90 degrees.

The diagram of the moving device shows two light paths with an included angle of significantly less than 90 degrees.

Does this mean that the device automatically changes the mirror angle when in motion? Or does it mean that the angle of the light source changes its angle automatically when the device is in motion? Perhaps the mirror automatically changes its location when the device is in motion?
Hi Cangas,
Why do you think the mirror angles or locations have to change? Do you understand where the mirrors are?

Nice animation but seriously flawed.

Incase U didnt notice, look at how fast the light is reflected back compared to how long it takes to reach the mirror in the direction the experiment is traveling.
Nice observation, Anomalous. What do you conclude, and why?

Your diagrams show radically different light paths in the stationary situation compared to the moving situation and I'm trying to determine how you think that can be possible in the real world.

Do you understand how light really moves?

To rephrase my previous serious question in hope of it not being misunderstood:

If we have a source of photons, attached to a device capable of movement, which is precisely aimed and focused in a direction perpendicular to the intended direction of movement of the device...
Hi Cangas,
The two diagrams show exactly the same situation.
In the first, the "camera" stays directly above the device.
In the second, the "camera" is moving southward relative to the device.

I guess I should have made that clearer at the start - in fact, I'll edit the OP now.

Nice observation, Anomalous. What do you conclude, and why? Given the illusion that humans have that Light is at constant speed for all observers, thats not the case in your animations.

And U think thats correct, can U tell me why ?Sure. Following my notation above:
t_AC = N/(c-v)
t_CD = N/(c+v)

so
t_AC/t_CD = (N/(c-v))/(N/(c+v))
t_AC/t_CD = (c+v)/(c-v)

and since
(c+v)/(c-v) > 1

therefore
t_AC/t_CD > 1
t_AC > t_CD

-Dale

Sure. Following my notation above:
t_AC = N/(c-v)
t_CD = N/(c+v)

so
t_AC/t_CD = (N/(c-v))/(N/(c+v))
t_AC/t_CD = (c+v)/(c-v)

and since
(c+v)/(c-v) > 1

therefore
t_AC/t_CD > 1
t_AC > t_CD

-Dale
Thanks for letting me know your thoughts regardless of if I understood them.

Given the illusion that humans have that Light is at constant speed for all observers, thats not the case in your animations.
Well I did say they weren't to scale. But interestingly, the speed of the light flash is the same on the way up as it is on the way down (14 pixels per frame, I think... can't remember exactly). I thought you post-humans were beyond being fooled by optical illusions?

What I thought you were commenting on was that the flash takes longer on the way up than it does on the way back...

I am so old that I remember Arlo Guthrie's Motorcycle Song, in which his motorcycle ( including its rider ) is going over a cliff travelling 135 miles per hour sideways and 80 miles per hour straight down.

In your stationary scenario the light is going 186,000 miles per second sideways and no miles per hour forward.

In your moving scenario the light is going 186,000 miles per second sideways and X miles per second forward. The resultant velocity is FTL.

...
In your moving scenario the light is going 186,000 miles per second sideways and X miles per second forward. The resultant velocity is FTL. In the side ways scenario the light has to go diagonally and has to go extra miles total lenght yet the speed is supposed to remain same for the both, thats a paradox.

Why do you say that the light HAS to go diagonally?

Do you say that the velocity of its source influences its velocity? Do you say that the forward velocity of its source gives it a forward velocity component?

In your moving scenario the light is going 186,000 miles per second sideways and X miles per second forward. The resultant velocity is FTL.
Are your sure the light is going 186,000 miles a second sideways in the moving point of view?

You seem to be assuming two things...
1) That the distance from the source to the side mirror is the same, and
2) that the time it takes the flash to reach the side mirror is the same.

What would you conclude if you put those assumptions aside for moment? Just for curiosity's sake?

The thread start specified that a light flash is going North ( forward in your diagram ) and a light flash is going West ( sideways in your diagram ). I am assuming these thread postulates.

I am assuming that light speed is 186,000 miles per second.

"1) That the distance from the source to the side mirror is the same, and": Can I call this a sentence fragment? Same as what?

"2) that the time it takes the flash to reach the side mirror is the same." Can I also call this a sentence fragment? Same as what?

Your postulate that a light flash goes West makes the sideways flash initially be moving perpendicular to the velocity vector of the moving device. If it goes diagonally, what is acting upon it to alter its perpendicular sideways motion and change it to a diagonal vector?

Why do you say that the light HAS to go diagonally?

Do you say that the velocity of its source influences its velocity? Do you say that the forward velocity of its source gives it a forward velocity component? If The forward velocity doesnot add to the velocity of the light , then its very difficult to imagine how only the direction is affacted and not the speed.
http://home.teegee.com.au/byrnes/Pete/Relativity/Images/MovingFrameA.gif

Secondly U cant ignore the fact that distance between AB in the restframe is more than W meters. So imagine what happens if we have premarked fixed distanced stripes in AB and DB. The light from Flags frame will be seen to come at a faster rate parallel to the stripes traced on ABD. ABD is a real physical trace.

In your stationary scenario the light is going 186,000 miles per second sideways and no miles per hour forward.

In your moving scenario the light is going 186,000 miles per second sideways and X miles per second forward. The resultant velocity is FTL.You are exactly correct that, by using the Galilean transform, you get a variant c. That is the whole point. You cannot use the Galilean transform if c is invariant.

-Dale

Correct, DaleSpam, but if you use the Lorentz transform to contract the distance the light travels in order to maintain an invariant 'c', there is no difference in the time it takes the photon to complete the trip wrt the rest frame. No time dilation. The whole point of the 'V' gedankin is to illustrate how 'time' is slowed wrt the rest frame.

Correct, DaleSpam, but if you use the Lorentz transform to contract the distance the light travels in order to maintain an invariant 'c', there is no difference in the time it takes the photon to complete the trip wrt the rest frame. No time dilation. The whole point of the 'V' gedankin is to illustrate how 'time' is slowed wrt the rest frame.

Correct. This is the flip side of the coin to my arguement. My position is if you clain time dilation you cannot claim length contraction.

That is both cannot exist physically. To argue that which is a matter of frame means neither is to be considered physical reality. The fact that time dilation has supporting emperical data strongly suggests it is physical and the failure of the claim for length contraction.

Correct, DaleSpam, but if you use the Lorentz transform to contract the distance the light travels in order to maintain an invariant 'c', there is no difference in the time it takes the photon to complete the trip wrt the rest frame. No time dilation. The whole point of the 'V' gedankin is to illustrate how 'time' is slowed wrt the rest frame.

No. Lorentz contraction keeps the time equal for the parallel and perpendicular paths. However, they are both increased by the time dilation factor. Both time dilation and Lorentz contraction are required by isotropic invariance of c.

No. Lorentz contraction keeps the time equal for the parallel and perpendicular paths. However, they are both increased by the time dilation factor. Both time dilation and Lorentz contraction are required by isotropic invariance of c.

Uh, kevinalm, we are only discussing the perpendicular path at this time. You quoted my post, then disputed something I did not state. I never stated that IF the speed of light was invariant, the Lorentz contraction was not required.

Yes, but the two must be considered together in order to resolve the question. There are two aspects of interest. Is there a difference in the trip time of the two paths and does the trip time increase for either/both? And this must be considered for both the apparatus stationary and moving cases. Lorentz contraction _and_ time dilation (both) are the only answer that is consistant.

If the path length is increased AND the speed of light is invariant with the rest frame of the distant observer, the photon will take a bigger time to complete the path. If the path length is contracted, say by 1/2, AND the clock is slowed by 1/2, both relative to the distant observer, then the speed of light is constant in both frames, true. The photon will travel 1/2 meter in 1/599,584,916 seconds of the distant observer's clock, consistent with an invariant 'c'. How long will it take the photon to travel one meter according to the distant observer's clock?

Edited time. Again. Sorry.

See, if the meter is contracted in the direction of travel ONLY, the 'V' shaped path is still longer for the photon. The photon must travel faster than 'c' to transit the bigger distance in the same length of time. If 'c' is invariant, one of two things must happen. The path along the 'V' must contract, perpendicular the the direction of travel of the frame, OR time must slow giving the photon a bigger time to complete the 'V' path. Not both at the same time.

Correct, DaleSpam, but if you use the Lorentz transform to contract the distance the light travels in order to maintain an invariant 'c', there is no difference in the time it takes the photon to complete the trip wrt the rest frame. No time dilation. The whole point of the 'V' gedankin is to illustrate how 'time' is slowed wrt the rest frame.CAGNAS' point to which I was responding was about the perpendicular clock (the W one). For the W clock there is no length contraction, only time dilation, according to the Lorentz transform.

-Dale

So the total time, t_AD, is:
t_AD = 2 γ&sup2; N/c
where γ = 1/sqrt(1-v&sup2;/c&sup2;)

Correspondingly, the total distance traveled by the light pulse, d_AD, is:
d_AD = c t_AD = 2 γ&sup2; NFor completeness I thought I would finish the derivation. Above are the results for the parallel (N) clock. Here is the derivation for the perpendicular (W) clock:

If t is the time required for the light pulse to go from A to B to D, then by the Pythagorean theorem:
(c t/2)&sup2; = (v t/2)&sup2; + W&sup2;
(c t)&sup2; = (v t)&sup2; 4 W&sup2;
t&sup2; (c&sup2;-v&sup2;) = 4 W&sup2;
t&sup2; = 4 W&sup2;/(c&sup2;-v&sup2;)
t&sup2; = 4 W&sup2;/(c&sup2; (1-v&sup2;/c&sup2;))
t&sup2; = 4 γ&sup2; W&sup2;/(c&sup2;)
t = 2 γ W/c

Correspondingly, the total distance traveled by the light pulse, d, is:
d = t c = 2 γ W

The important thing to note is that the results are different, specifically
parallel: t = 2 γ&sup2; N/c
perpen: t = 2 γ W/c

-Dale

The thread start specified that a light flash is going North ( forward in your diagram ) and a light flash is going West ( sideways in your diagram ). I am assuming these thread postulates.

I am assuming that light speed is 186,000 miles per second.

"1) That the distance from the source to the side mirror is the same, and": Can I call this a sentence fragment? Same as what?
The same in the two points of view. Ie that the distance from the source to the side mirror in the device's rest frame is the same as the distance fro the source to the side mirror in the frame in which our point of view is moving South.


"2) that the time it takes the flash to reach the side mirror is the same." Can I also call this a sentence fragment? Same as what?
Same as before :)


Your postulate that a light flash goes West makes the sideways flash initially be moving perpendicular to the velocity vector of the moving device. If it goes diagonally, what is acting upon it to alter its perpendicular sideways motion and change it to a diagonal vector?
Nothing is acting on it at all. The two animations are two camera views of exactly the same thing. In the second view, the camera is moving southward.

If The forward velocity doesnot add to the velocity of the light , then its very difficult to imagine how only the direction is affacted and not the speed.
Hi Anomalous,
In both animations, the device is fixed to the Earth. In the second animation, the "camera" is moving south.
Does that explain why the apparent direction is affected?

http://home.teegee.com.au/byrnes/Pete/Relativity/Images/MovingFrameA.gif


Secondly U cant ignore the fact that distance between AB in the rest frame is more than W meters.
Correct (unless there is some transverse length contraction)... so if the distance is longer but the velocity is the same, what does that suggest about the time taken?

Correct, DaleSpam, but if you use the Lorentz transform to contract the distance the light travels in order to maintain an invariant 'c', there is no difference in the time it takes the photon to complete the trip wrt the rest frame. No time dilation. The whole point of the 'V' gedankin is to illustrate how 'time' is slowed wrt the rest frame.
Hi 2inquisitive,
In the perpendicular path, only infinite length contraction in the Northward direction would suffice to reduce the distance enough to keep the speed of the flash invariant without time dilation.

Transverse length contraction would do it, of course... but is transverse length contraction possible? Think about a wide train entering a tunnel only just wider than the train.

Yes, I agree with the points in your post, Pete. And, no, I do not believe transverse length contraction is possible.

So... you agree that if light-speed is invariant, then both length contraction and time dilation are implied?

So... you agree that if light-speed is invariant, then both length contraction and time dilation are implied?

I will agree both are implied by Light clocks and Special Theory. But remember, in the transverse direction of photon travel we just illustrated, only time dilation is implied, no length contraction.

Pete, do you believe light clocks that are based on an invariant 'c' give true 'time'? Time is simply a measurement which will vary with the type of clock used. Do you want an example?

Hi 2inquisitive,

Pete, do you believe light clocks that are based on an invariant 'c' give true 'time'?
Yes, if light speed is frame invariant. Otherwise no, of course.

Not that it's really relevant in this case, since we're not using the device to measure time except to affirm that both flashes leave the emitter at the same time and return to the emitter at the same time.


Time is simply a measurement which will vary with the type of clock used. Do you want an example?
Sure, go ahead.

Yes, I agree with you DaleSpam. The 'W' gedankin with the perpendicular clock supposes time dilation only. That prediction of time dilation is a product of a light clock in the 'W' frame in which 'c' is invariant wrt the distant observer frame.

Here is my example, Pete.

Remember my earlier post about using the 'beat' of a pulsar as a reference signal for both frames? Pete was the only one that responded to me, and that was to ask if the pulsar was 'at rest' wrt the Earth frame. It really makes no difference, the frequency of the 'beat' is due to rotation of the pulsar itself, not Doppler shift of the light frequency.

There are over 1400 known pulsars today, most within the Milky Way galaxy. If I am to choose a particular pulsar, let's choose the one in the Crab Nebula. For those that are unaware of its background, I will give a short overview. It is believed the Crab Nebula was formed from a Super Nova explosion on July 4, 1054 A.D. At that time, it was called the 'guest star', a star that suddenly appeared that could be seen even in the daytime, like the planet Venus, for 23 days. The pulsar is the neutron star that was formed from the Super Nova. It rotates at the rate of 30 times per second. The 'pulses' we see are because the poles of the very intense magnetic field do not line up with the poles of the neutron star's rotational axis. That results in a beam that rotates with the star, emitted directly outward from the magnetic pole. It is often described as similar to the beam of a lighthouse as the beam circles around the lighthouse. We see the beam only when it points in Earth's direction, 30 times a second. This pulse is at least as steady as the reference signals used in the best atomic clocks, cesium or rubidium. Many pulsars are called cosmic clocks because of this steady beat.

My question boils down to this. Assume a twin scenario in which Bob, the travelling twin, shakes hands with Rob, the stay-at-home twin, before leaving on his trip at, say an average of .866c. Bob shakes hands with his twin Rob again upon completion of the trip. According to Rob, Bob was gone a total of 10,000 seconds. That corresponds to 300,000 rotations of the common clock, the pulsar. How many times will the pulsar rotate according to Bob, who has one clock synchronized with the pulsar's beat?

It really makes no difference, the frequency of the 'beat' is due to rotation of the pulsar itself, not Doppler shift of the light frequency.
The beat frequency also depends on the pulsar's motion, for two reasons.

1) If the pulsar is moving away, then each beat has further to travel than the previous one, so the time between beats being received is longer than the time between beats being generated.

2) (According to SR) In a reference frame in which the pulsar has relativistic speed the pulsar's rotation rate will be slower due to time dilation.


My question boils down to this. Assume a twin scenario in which Bob, the travelling twin, shakes hands with Rob, the stay-at-home twin, before leaving on his trip at, say an average of .866c. Bob shakes hands with his twin Rob again upon completion of the trip. According to Rob, Bob was gone a total of 10,000 seconds. That corresponds to 300,000 rotations of the common clock, the pulsar. How many times will the pulsar rotate according to Bob, who has one clock synchronized with the pulsar's beat?
300,000, of course.
This is precisely equivalent to both Rob and Bob agreeing that 10,000 seconds elapsed on Rob's clock during Bob's absence.

The 300,000 rotations of the pulsar was counted by both Rob and Bob. It doesn't matter what gymnastics Bob's clock perform, the time interval was the same for both twins. The pulsar is a clock independant of reference frames. Bob and Rob will age the same.

There are some pulsars detected in other galaxies. Are you stating they appear to rotate slower if they are receeding from us? Pete, you seem to be stating the pulses will become slower and slower as the pulsar gets farther and farther away. I can agree with your logic, but Bob would see the pulsar's rate increase as he traveled in the pulsar's direction, then slow by an equilavent rate as he traveled away from the pulsar. There would be no total difference in the count of rotations.

Observations of steady beat pulsars indicate they may slow by a slight amount over a great many years. That has been attributed to a natural loss of energy. Are the pulsars 'at rest' with respect to the Earth, or they very gradually slowing because of recession?

The pulsar is a clock independant of reference frames.
Why do you think that?


Pete, you seem to be stating the pulses will become slower and slower as the pulsar gets farther and farther away.
No, that's not the case. The rate depends on proper rotation rate and relative velocity, not distance.


Observations of steady beat pulsars indicate they may slow by a slight amount over a great many years.
Not relevant here.

by Pete:

1) If the pulsar is moving away, then each beat has further to travel than the previous one, so the time between beats being received is longer than the time between beats being generated.....


No, that's not the case. The rate depends on proper rotation rate and relative velocity, not distance.
================================================== ===========

I understood you to mean that each successive pulse has farther to travel due to an increasing distance from the position of the previous pulse. Now tell me how relative velocity can have an affect on the rotation rate of the pulsar. Remember, the pulsar must rotate the same total number of times in both reference frames between beginning handshake and ending handshake.

I understood you to mean that each successive pulse has farther to travel due to an increasing distance from the position of the previous pulse.
You understood correctly.

This has nothing to do with the total distance to the pulsar, only the difference in distance from one beat to the next, right?


Now tell me how relative velocity can have an affect on the rotation rate of the pulsar. Remember, the pulsar must rotate the same total number of times in both reference frames between beginning handshake and ending handshake.
Time dilation. If the pulsar is at rest in Earth's frame, how is it any different to Rob's clock?

Do you really want another twin paradox discussion?

Why does Rob's clock have to be inaccurate? It is Bob's light clock that does not keep time because it is based on flawed gedankins. 'Time' itself is universal. That is proven by way of the total number of rotations of the pulsar in both frames. Clocks that move are subject to all kinds of errors, both conceptional and enviromental.

If you want to prove time dilation in one frame, then you must prove the pulsar rotates fewer times between handshakes.

I came up with my idea while studing how atomic clocks work. Cesium and Rubidium clocks both generate a reference signal, for instance the transition of cesium 133 from ground state to excited state, and then back to ground state. When the cesium atom drops from the excited state to the ground state, it emitts a photon of a certain frequency. An atomic clock cannot be 'adjusted' in space to this transition occur at a different frequency. The clocks are free-running and the frequency of the reference signal is unchanging once the clocks have been given sufficient time to adjust to the space enviroment. A cesium clock does not 'age' in its entire 7-10 year lifetime, it keeps generating the same frequency, with small variations due to enviromental factors. How are GPS clocks synchronized with Earth clocks? The next step in how the clocks operate is that a quartz crystal oscillator is locked on to the frequency generated by the cesium atoms. The locking frequency also cannot change or the clocks (cesium and quartz oscillator) will be out of phase. So, synchronizing GPS clocks with Earth clocks is not possible by 'adjusting' the beat of the cesium clock. What IS adjusted is the frequency broadcast by the GPS satellite. That is done by a frequency generator onboard the satellite. The GPS satellites broadcast a frequency that is slightly lower than the desired frequency wanted on Earth. The satellite does not have to keep broadcasting a lower and lower frequency because the satellite clock reference frequency changes over time. It doesn't. The cesium transition may occur at a slightly higher frequency in orbit, but it maintains the same relative difference in frequency wrt a cesium clock on Earth as when it stabilized in orbit. Again, you cannot change the reference frequency of a cesium, or rubidium, clock because that is set by quantum mechanics, the change of state from excited to ground state when the photon is emitted.

So, you see how a pulsar generates a similar reference frequency as an atomic clock. That is what gave me the idea in my post. Simply use a quartz crystal oscillator to lock onto the pulsar's frequency and you have a very accurate clock unaffected by local enviromental factors. I thought it was kind of neat, but I know ya'll won't agree. No problem.

So, you see how a pulsar generates a similar reference frequency as an atomic clock. That is what gave me the idea in my post. Simply use a quartz crystal oscillator to lock onto the pulsar's frequency and you have a very accurate clock unaffected by local enviromental factors. I thought it was kind of neat, but I know ya'll won't agree. No problem.


Well some of us certainly do agree and thanks. Excellent work. :D

Why does Rob's clock have to be inaccurate?
Who said it was?

It is Bob's light clock that does not keep time because it is based on flawed gedankins.
Can you back up this assertion?


'Time' itself is universal. That is proven by way of the total number of rotations of the pulsar in both frames.
It doesn't follow. You're assuming that pulsars spin at the same rate in both frames... ie you're assuming that time is universal.
Many great scientists have tried to prove the universality of time and space (including Newton).. without success.

The 300,000 rotations of the pulsar was counted by both Rob and Bob. It doesn't matter what gymnastics Bob's clock perform, the time interval was the same for both twins. The pulsar is a clock independant of reference frames. Bob and Rob will age the same.

There are some pulsars detected in other galaxies. Are you stating they appear to rotate slower if they are receeding from us? Pete, you seem to be stating the pulses will become slower and slower as the pulsar gets farther and farther away. I can agree with your logic, but Bob would see the pulsar's rate increase as he traveled in the pulsar's direction, then slow by an equilavent rate as he traveled away from the pulsar. There would be no total difference in the count of rotations.



Using the relativistic doppler shift formula:

fobserved = fsource sqrt((1+v/c)/(1-v/c))

We will find that heading towards the pulsar at 0.866c, Bob will see the pulsar beat at a rate of 111.9 beats per sec and heading away at 8.1 beats per sec.

Now, if we call the time Bob spends( according to Bob) heading towards the pulsar T1, we can say that the number of pulses he sees during this time is

N1 = 111.9 x T1

If T2 is the time spent heading away then the number of pulses seen during this time is 8.1 x T2

Then the total number of Pulses seen by Bob is N(total) =N1 + N2 or
N(total) = 111.9(T1) + 8.1(T2)

Since Bob spends an equal amount of time heading out ans returning, T1= T2, so we can say:

N(total) = 111.9(T) + 8.1(T)
N(total) = T(111.9 + 8.1)
N(total) = T(120)

Now we also know that Bob has to see an equal number of pulses as Rob (300,000), so :

300,000 =120T
T = 300,000/120 = 2500 sec.

Since this is the time for one leg, the total trip takes 2*2500 = 5000 secs for Bob. Bob ages 5,000 sec while Rob ages 10,000 sec.

The fact that Bob counts the same total number of Pulsar beats as Rob does not mean that he ages the same as Rob, since for him, the average(over the whole trip) beat rate of the pulsar is twice that which Bob sees. He counts the same 300,000 beats, but they took half the time to occur.

"He counts the same 300,000 beats, but they took half the time to occur."
================================================== ===========

According to his light clock. Remember, his 'other' clock is synchronized with the pulsar. One 'tick' equals one pulse. This clock must have a total of 300,000 ticks, the same as Rob's clock on Earth. That was my point about different types of clocks.

Ou, Janus the 58 years old crap, I remember U crystal clear.

Remember, his 'other' clock is synchronized with the pulsar.
...which is a clock in a moving frame. Bob will find that his clock does not stay synchronized with the pulsar... he'll have to adjust it for the return trip.


This clock must have a total of 300,000 ticks, the same as Rob's clock on Earth. That was my point about different types of clocks.
Your point applies to moving clocks, rather than different types of clocks. The pulsar as a clock is no different to any other type of clock. Bob can synchronize his other clock with any clock in Rob's frame, with exactly the same result.

Pulsars as clocks are no more universal than any other clock.

...which is a clock in a moving frame. Bob will find that his clock does not stay synchronized with the pulsar... he'll have to adjust it for the return trip.


Your point applies to moving clocks, rather than different types of clocks. The pulsar as a clock is no different to any other type of clock. Bob can synchronize his other clock with any clock in Rob's frame, with exactly the same result.

Pulsars as clocks are no more universal than any other clock.

Really? I thought Bob sees clocks in Rob's frame running at 1/2 the rate of his own. If Bob slows his clocks by 1/2, how much time will pass on his clock during the trip?

As a beam of light, emitted from a source, enters your measuring equipment you are able to measure only two items of data before calculating the relative velocity: apparent distance between peaks and the time interval between peaks.

This leads to a very fundamental oversight in that our speed of communication (including how we record each wave as it enters the equipment) limits our ability to record the CORRECT and TRUE wavelength of the beam of light.

Imagine that you are sitting inside a box and you have a stream of light photons entering the box. Your job is to record the relative velocity to the best of your ability. So how do you do it?

Well, I guess you don’t have many choices. You simply record the frequency of the wave peaks and jot it down on a piece of paper. Great, got that bit. Next, we need to establish the wavelength. To do this you decide to set up a simple arbitrary method of determining distance between peaks. A good way would be to place a conveyer belt behind you and set this conveyor moving at a constant rate. The conveyor contraption is working in YOUR frame of reference.

Ok, lets start recording. As a wave peak enters the box you say “hey, there’s one” and at this moment you reach out to the conveyor and place a chalk mark on the belt. Then another peak comes in and you say “hey, there’s another” and you reach around to place the next mark, and so on.

When you’ve got a few bits of data you can calculate the velocity using the arbitrary length you got from the conveyor and the frequency that your clock recorded.

Now, someone asks you to repeat the experiment but unbeknown to you, your box has sped up and now travels very fast towards the source.

You start your recording only to find that the frequency has doubled. Then you start placing chalk marks on the conveyor. They’re coming in much faster now so you have your job cut out trying to record them. You reach round as fast as you can and place the chalk mark on the conveyor.

You notice that the chalk marks are closer together than before – in fact they are half the distance.

You now proceed to calculate the velocity and this determines the same velocity as before. Weird!

So, as an observer in the box, and with a limited speed of communication, will always observe the same relative velocity.

This is no different to measuring the speed of light for real using a piece of equipment which records time between wave peaks and the distance between them.

So how can we have faith in EVER measuring the CORRECT relative velocity of light. It aint gonna happen! Light will always APPEAR invariant. Length contraction is an illusion!!!

Very interesting.

The observer-in-a-box might reasonably assume that the light frequency, in the frame of reference of its source, has has not changed. Then, the observer would conclude that the relative velocity of light versus box has doubled. However, the OIAB would still have no basis for determing absolute box velocity.

An observer knowledgeable only of Newtonian relativity could conclude equally logically that the box velocity had remained constant and that the speed of light had doubled. :eek:

If you divide the distance between chalk marks by time between chalk marks, all you're measuring is the speed of the conveyer belt :confused:

If the conveyor belt speed has been kept constant according to the OIAB's stopwatch, the collection of twice as many waves per unit time, according to the OIAB's stopwatch, means they are coming in twice as fast. This is prama facie evidence that something is moving twice as fast. Either the box is twice as fast or the light is twice as fast.

If the box is assumed to have speeded up, light is then assumed to have stayed at the old speed.

Or vice versa.

Oops! dav57 did not say anything about contraband such as a stopwatch! Maybe the OIAB had it hidden in a boot.

Guys, your equipment can only record two things: an event and a time interval between events.

You know nothing about the original source, absolutely nothing.

The observer is NOT, repeat NOT measuring the speed which each wave peak enters the box because this is impossible. You cannot measure the velocity of each photon using this method. Remember, as soon as you look at a photon, its information disappears. He is forced to RECORD them onto the conveyor as he sees each one.

It's a bit like counting cars on a race track as they speed past. You are allowed to measure the number of cars passing, NOT by using a speed gun, but by using some arbitrary length measuring method (in your frame of reference) and also the time interval between cars.

Remember that in every frame of reference you require an ARBITRARY way of determining length (the conveyor) and frequency (stop watch).

The calculated velocity based on wavelength and frequency always remains, sorry APPEARS, unchanged.

If you divide the distance between chalk marks by time between chalk marks, all you're measuring is the speed of the conveyer belt :confused:

No you're not. You're simply using the conveyor as an arbitrary, local measuring system which operates relative to you. It quite simply provides our observer in a box with a constant and reliable means of measuring lengths in HIS frame of reference.

How else can the observer try to establish the TRUE wavelength between peaks? I mean, he can only take each event one at a time. And by the time he's dealt with recording one event, the box has moved by the time he records the next.

“ It is Bob's light clock that does not keep time because it is based on flawed gedankins. ”


Can you back up this assertion?(by Pete)
================================================== ==============

Yes, I can give a logical explaination. I don't make 'assertions' I can't back up. James R taught me that.

Pete, the following is in regard to your 'W' animation, the one in which the 'photon' is emitted perpendicular to direction of frame. The 'flaw' in the gedankin is assuming a SINGLE photon will reflect back to the mirror in the moving frame. The single photon is assumed to follow a 'V' trajectory in the frame moving relative to the distant observer. Do you have any evidence the photon will change its direction of travel and follow the longer 'V' path? The explaination is simple. Light travels in a wave of MANY photons. The wavefront is moving at 'c' away from the emitter, more or less a straight line perpendicular to the direction of the frame. The wavefront, the 'line', extends enough to reflect off the mirror regardless if the mirror is at the 'rest' position 90 degrees opposite the emitter, or if the mirror has moved upward as in your animation. The wavefront moves outward at the same 90 degree angle regardless if viewed from the frame of the emitter or viewed from the moving frame of the distant observer. The wave front will hit, and reflect, of the mirror in the SAME amount of travel time and travel distance in either frame. No time dilation because the wavefront does not travel at the angle illustrated. It is similar to what MacM is talking about when he says a DIFFERENT photon is reflected. That is what I mean by flawed gedankins.

As a beam of light, emitted from a source, enters your measuring equipment you are able to measure only two items of data before calculating the relative velocity: apparent distance between peaks and the time interval between peaks.
Hi Dav,
We've had this discussion before, remember?
Speed can be measured in other ways which do not lead to the stated limitations.

If you really want to discuss it again, perhaps you'd like to start a thread for that purpose? In this thread, the invariant speed of light is a specifically stated assumption... if light speed is invariant, is length contraction required?

Hi 2inquisitive,

I have been thinking about your pulsar clock. I think one thing to do in order to make things clearer is to have the traveler go in a direction more or less perpendicular to the pulsar. If the pulsar is far enough away then the distance to the pulsar will be more or less constant and you can more or less ignore any discussion of Doppler effects.

I think the other thing to do is to think about accelerating the pulsar rather than having Bob and Rob journey off into twin-paradox-land. If the pulsar beat frequency slows (relative to inertial clocks) as it accelerates then it is a frame-dependent clock like any other.

-Dale

Pete, the following is in regard to your 'W' animation, the one in which the 'photon' is emitted perpendicular to direction of frame. The 'flaw' in the gedankin is assuming a SINGLE photon will reflect back to the mirror in the moving frame.
It's not a single photon, it's a flash of light. The emitter isn't sensitive enough to emit only a single photon, nor is the detector sensitive enough to detect one.


The single photon is assumed to follow a 'V' trajectory in the frame moving relative to the distant observer. Do you have any evidence the photon will change its direction of travel and follow the longer 'V' path?
The flash's movement noes not change, of course, because relative to Earth, the device never moves. The apparent change in direction is because the "camera" is moving southward.

Do you see? These two animations show exactly the same flashes. In the second case, we're in a frame that is southbound relative to the device... ie the "camera" is moving southward.

<table width=100%><tr><td width=50% align=center>http://home.teegee.com.au/byrnes/Pete/Relativity/Images/RestFrame.gif</td><td width=50% align=center>http://home.teegee.com.au/byrnes/Pete/Relativity/Images/MovingFrameA.gif</td></tr></table>

Pete, your mirror, the 'B' in the animation is not moving along with the emitter of the photon. Instead of a dot moving outward from 'A' to the mirror at 'B' and then back to the emitter, now located at the 'D' position, have an extended line moving outwards and back towards a moving 'A' and a 'B' moving in the same direction. 'A' and 'B' both travel north, the 'wavefront' travels east-west from 'A' to 'B' and back to 'A'. I don't know how to do animations, I am almost computer illiterate!

By the way, it will be easier to illustrate if you use something like .5c for the north-moving emitter and mirror. That way, the distance traveled north by the emitter/mirror will be half the distance of the east-west wavefront travel and be to scale.

Hi 2inquisitive,

I have been thinking about your pulsar clock. I think one thing to do in order to make things clearer is to have the traveler go in a direction more or less perpendicular to the pulsar. If the pulsar is far enough away then the distance to the pulsar will be more or less constant and you can more or less ignore any discussion of Doppler effects.

I think the other thing to do is to think about accelerating the pulsar rather than having Bob and Rob journey off into twin-paradox-land. If the pulsar beat frequency slows (relative to inertial clocks) as it accelerates then it is a frame-dependent clock like any other.

-Dale

One of the reasons I thought about the pulsar-clock was because of local enviromental influences on standard atomic clocks, such as temperature, solar wind, and gravitational potential affects of the enviroment the clocks are located within. There are many other influences too, of course. NASA's deep space network uses three bases with the recieving/broadcasting antennas located at different points around the globe, so a distant satellite is always in view of one of the locations. If the time were synchronized at all stations, as it is done in the GPS network, the pulsar could be used as a reference signal anywhere on Earth. And there are millisecond pulsars too. An example is the pulsar PSR 1937 + 21, which rotates once every 1.56 milliseconds. Such would make a good reference signal for a clock. The pulses can be recieved by radio telescopes.

Pete, your mirror, the 'B' in the animation is not moving along with the emitter of the photon.
It's a flash, not a photon. The B is attached to the frame which is moving south relative to the device. It's purpose (along with A, C, and D) are to identify key points in that southbound frame.


Instead of a dot moving outward from 'A' to the mirror at 'B' and then back to the emitter, now located at the 'D' position, have an extended line moving outwards and back towards a moving 'A' and a 'B' moving in the same direction. 'A' and 'B' both travel north, the 'wavefront' travels east-west from 'A' to 'B' and back to 'A'. I don't know how to do animations, I am almost computer illiterate!
Why? The wavefront idea is a red herring. The moving dot (flash) is the same dot in both viewpoints... it's just being recorded by different cameras. It doesn't even need to be a flash of light... say it's a marble moving at 0.9999c instead, if you like.

Pete, in your 'W' animation, the unseen observer is located to the south. Does this observer not 'see' the emitter and mirror in motion relative to him? The wavefront is 'seen' moving to this observer's east from the emitter toward the mirror.

2IQ...,

Correct me if I am wrong but I got the impression that you were talking about the spherical wavefront and the fact that light would appear to have taken the angular path with motion but in reality it was just different photons which were projected at a different angle which got reflected back.

Yes, that is correct, MacM. When the emitter and mirror are in motion, a different 'part' of the wavefront is reflected back to the emitter.

On review, I've found a couple of unanswered posts.


Really? I thought Bob sees clocks in Rob's frame running at 1/2 the rate of his own.
What Bob actually sees is highly affected by Doppler shifts, which is affected not only by what frame the clocks are in but also their direction from Bob.
Bob will see a clock next to Rob run very slowly for the trip out, and very fast for the trip back.

If Bob makes allowances for Doppler effects, he will still find that clocks in Rob's frame didn't run at 1/2 the rate of his own for the whole trip... they will run at 1/2 the rate while Bob has a constant speed of 0.866c relative to Rob, but will run extremely fast during the turnaround.

That's enough twin paradox discussion for this thread, I think. Open another twin paradox thread, if you dare! :)



I came up with my idea while studing how atomic clocks work. Cesium and Rubidium clocks both generate a reference signal, for instance the transition of cesium 133 from ground state to excited state, and then back to ground state. When the cesium atom drops from the excited state to the ground state, it emitts a photon of a certain frequency. An atomic clock cannot be 'adjusted' in space to this transition occur at a different frequency. The clocks are free-running and the frequency of the reference signal is unchanging once the clocks have been given sufficient time to adjust to the space enviroment. A cesium clock does not 'age' in its entire 7-10 year lifetime, it keeps generating the same frequency, with small variations due to environmental factors. How are GPS clocks synchronized with Earth clocks? The next step in how the clocks operate is that a quartz crystal oscillator is locked on to the frequency generated by the cesium atoms. The locking frequency also cannot change or the clocks (cesium and quartz oscillator) will be out of phase. So, synchronizing GPS clocks with Earth clocks is not possible by 'adjusting' the beat of the cesium clock. What IS adjusted is the frequency broadcast by the GPS satellite. That is done by a frequency generator onboard the satellite. The GPS satellites broadcast a frequency that is slightly lower than the desired frequency wanted on Earth. The satellite does not have to keep broadcasting a lower and lower frequency because the satellite clock reference frequency changes over time. It doesn't. The cesium transition may occur at a slightly higher frequency in orbit, but it maintains the same relative difference in frequency wrt a cesium clock on Earth as when it stabilized in orbit. Again, you cannot change the reference frequency of a cesium, or rubidium, clock because that is set by quantum mechanics, the change of state from excited to ground state when the photon is emitted.
I'm not sure what your point is here. Do you think that relativity predicts that the satellite reference frequency should steadily change over time in the ECEF or ECI frame?

Pete, in your 'W' animation, the unseen observer is located to the south.
What?? In the "W" animation, the "camera" (observer, if you like) is above the device, moving south. The animation is what is seen through the observer's eyes.


Does this observer not 'see' the emitter and mirror in motion relative to him? The wavefront is 'seen' moving to this observer's east from the emitter toward the mirror.
Forget the wavefront. There is no wavefront. It's a precisely directed laser flash, or even a bullet.

Do you think that the laser/bullet will miss the mirror just because we record this event on a moving camera? How could a moving camera record a miss while the stationary camera record a hit?

by Pete:

"Bob will see a clock next to Rob run very slowly for the trip out, and very fast for the trip back."
================================================== =============

I thought the Lorentz transforms always had the clock in the moving frame beating slower, regardless if approaching or receeding. Einstein stated all moving clocks run slow when compared to the observer's rest frame.

by Pete:

"Bob will see a clock next to Rob run very slowly for the trip out, and very fast for the trip back."
================================================== =============

I thought the Lorentz transforms always had the clock in the moving frame beating slower, regardless if approaching or receeding. Einstein stated all moving clocks run slow when compared to the observer's rest frame.
Did you read the whole post? Bob will see the clock run fast for the trip back because of Doppler effects.
Fuller explanation here (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html).

If Bob allows for Doppler (ie he calculates how much closer the clock each from one tick to the next, and allows for the shortened light-travel time), he will find that Rob's clock is running slower as long as he (Bob) has a constant speed of 0.866c relative to Rob.

Pete, are you familiar with the Huygens wavefront? Electromagnetic waves do not only propagate in the direction of motion, they propagate transverse to that motion also. Here is an excerpt and a link to the explaination and diagrams on page 38 of this pdf. paper. I have read better descriptions, but this is what I found on the spur of the moment.

"For an infinite plane wave, the sideways propagation from each point is balanced by the propagation
from its neighbors, so the wave continues on as a plane. However, when a wave encounters
an object, the effect at the edges of the object is that the path of the radiation is slightly bent.
•
•
•
Direction of
Propagation
Plane waves
•
•
Every point on plane wave
propagates energy according
to the inverse-square law
Huygens’ Plane Waves
http://www2.jpl.nasa.gov/radioastronomy/Chapter4.pdf

By the way, if the camera is directly above the 'W' animation and moving with the frame, as you say, then the waves, or photons, will appear to travel directly outward and back with NO 'V' shape, unless you want to introduce a 'luminferious ether' that sweeps the photons from their projected path. In that case, the 'V' path would be reverse of the way you have it illustrated. Again, if the 'B', the mirror, is not in motion, the frame is not moving wrt your 'camera'.

2inquisitive, you are confusing yourself with wavefronts. They are irrelevant to this scenario. We can do exactly the same exercise with bullets if you like.


=By the way, if the camera is directly above the 'W' animation and moving with the frame, as you say, then the waves, or photons, will appear to travel directly outward and back with NO 'V' shape, unless you want to introduce a 'luminferious ether' that sweeps the photons from their projected path.
I have no idea why you think that would be the case.

A laser is fixed to the ground. It fires a brief pulse, which strikes a mirror and returns to the laser.
We record this with two cameras - one is also fixed to the ground, the other is moving rapidly perpendicular to the laser's line of fire.

Do you think that the moving camera will show that the laser misses the mirror?


Again, if the 'B', the mirror, is not in motion, the frame is not moving wrt your 'camera'.
'B' is not the mirror. 'B' is a fixed point in the southbound camera's reference frame.

Which frame and camera do you mean? There are two of each. Each camera defines a frame.

One of the reasons I thought about the pulsar-clock was because of local enviromental influences on standard atomic clocks, such as temperature, solar wind, and gravitational potential affects of the enviroment the clocks are located within. There are many other influences too, of course. NASA's deep space network uses three bases with the recieving/broadcasting antennas located at different points around the globe, so a distant satellite is always in view of one of the locations. If the time were synchronized at all stations, as it is done in the GPS network, the pulsar could be used as a reference signal anywhere on Earth. And there are millisecond pulsars too. An example is the pulsar PSR 1937 + 21, which rotates once every 1.56 milliseconds. Such would make a good reference signal for a clock. The pulses can be recieved by radio telescopes.I think it is an interesting idea, but I don't think that pulsar-clocks are immune to relativistic effects. I think they are just a good steady clock in one reference frame. I suspect that an accelerating pulsar would slow down.

I don't know exactly how to do the analysis. I think I would assume that the pulsar was emitting a pure monochromatic light at e.g. 1 GHz (proper frequency) with a perfect sinusoidal beat of e.g. 1 kHz. Then, even if some frames disagree on the exact frequency they will still agree that every beat corresponds to exactly 1 million peaks of the RF signal. You could probably then use standard Doppler experiments to predict the results.

-Dale

We can do exactly the same exercise with bullets if you like.Only with bullets that have a frame-invariant speed.

-Dale

Or close enough to it... how about bullets at 0.9999c in the device frame?

2inquisitive,
Here's the animations from the two cameras again, with backgrounds to make sure you know what's going on. The background is supposed to represent the concrete floor to which the device is bolted. The cameras are hanging from the ceiling looking down. One camera is fixed directly above the device, the other is moving southward on a track.

Remember, these two animations are two "films" of exactly the same event. It is completely impossible for the flash to strike the mirror in one frame and miss in another. The flash hitting the mirror is not an ambiguous event!

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OK, Pete, I was thinking of an animation in which the frame of the emitter/mirror is moving and the observer/camera is at rest. You state the emitter/mirror is at rest and the observer/camera is in motion above the scene.

In that case, you cannot 'see' the path of a bullet or a photon. Both travel to the reflector and back in a straight line if the emitting device is not moving. If a laser beam is seen travelling to the reflector and back, it still appears as a straight line from above, just the straight line is itself moving relative to the observer's motion. No 'V's and still no time dilation. Light reflects back at the same angle wrt the reflector as it struck the reflector. It cannot change to a 'V' because the observer/camera is in motion. If light strikes a reflector at a 90 degree angle relative to the reflector surface, it will be reflected back at a 90 degree angle regardless of how one assumes it 'should' appear. A bullet will do the same if it is elastic. I have watched tracer rounds fired at my helicopter in Vietnam and they 'appear' to CURVE in their flight, an ellipse 'bowed' according to my direction of travel. That does not mean the tracers really curve in their flight and it does not mean they appear to leave the weapon at an angle wrt the direction the barrel is pointed. Again this is can be referred to as a Sagnac effect, not a confirmation for 'time dilation'.

In that case, you cannot 'see' the path of a bullet or a photon.
You can't see a bullet? Why not!
As for the photon, we know where it was emitted and where it was reflected... so put a straight line in between those two points.


If a laser beam is seen travelling to the reflector and back, it still appears as a straight line from above, just the straight line is itself moving relative to the observer's motion.
We're talking about a very short pulse from the laser beam... say a 30 picosend flash.


I have watched tracer rounds fired at my helicopter in Vietnam and they 'appear' to CURVE in their flight, an ellipse 'bowed' according to my direction of travel. That does not mean the tracers really curve in their flight and it does not mean they appear to leave the weapon at an angle wrt the direction the barrel is pointed.
Now we're at the heart of it.
In the reference frame of your helicopter, the path of the tracers is curved. This is because your reference frame is moving.

Whether or they are "really" curved is beside the point... they are curved in the helicopter frame. In this thread, we're simply examing paths in different reference frames, without worrying about whether one reference frame is more real than another. We can do that later if you like.

So, in the scenario it doesn't matter at this time whether the light path is "really" a V or not. We're only concerned with the shape of the path in the reference frame of the southbound camera.

Do you agree that in this moving frame, the light flash follows a V shape?

The "reality of reference frames" issue might be worth discussing in a separate thread... but we'll try it here and see how we go.

You're riding shotgun in a jeep. The jeep is travelling at 30m/s parallel to a long wall, which is 10m away on your side.

You are carrying a rifle loaded with subsonic rounds, muzzle velocity 300m/s.

You aim and fire the rifle directly at the passing wall, at 90° to your direction of travel. To your horror, at very instant that the bullet leaves the rifle a man standing against the wall flashes past your sights... Ie at the moment of firing, the rifle was aimed directly at the man.

Does the bullet strike the man?

Well, give me some factors here how high is the jeep of the ground? what part of the body was the gun aimed at?

Sure, a ball of light might seem to follow a curved path if viewed from a moving frame.

Well, give me some factors here how high is the jeep of the ground? what part of the body was the gun aimed at?
The rifle is chest-high to the man.
When the bullet left the muzzle, the rifle was aimed at the center of the man's chest.

Sure, a ball of light might seem to follow a curved path if viewed from a moving frame.
OK... that's sufficient for this thread, which makes no assumptions on whether any reference frame is better than any other.

If the speed in the moving frame (ie on the moving camera's film) is the same for each flash at all times, we can conclude that:

In the moving frame, the East-west mirror is further from the device than the North-South mirror

Right?
We can easily conclude some things about the time taken for the flashes to return to the device, and the simultaneity of the flashes striking the mirrors, but those are peripheral to the original purpose of this thread.

We can now proceed to the other idea that appears to be bothering you...
Are reference frames equivalent?
If the device and first camera were moving Northward on tracks, and the second camera fixed in one place, would the films look the same?

Do you want to continue that discussion in this thread, or a new thread?

Bullets and other massive objects are acknowledged to participate in the momentum and the velocity vector of the moving object from which they are ejected.

According to the old Emitter Theory of light, rejected by Einstein and others over a century ago, light participated in the momentum and velocity vector of the moving object from which it was ejected.

In modern science theory, including SR, does light participate in the momentum and velocity vector of the moving object from which it is ejected?

Does light which is ejected in a forward direction from a moving object have its own velocity of c plus the velocity of the object?

Does light which is ejected rearward from a moving object have its own velocity c minus the velocity of the object?

Does light which is ejected at a perpendicular or diagonal angle wrt the velocity vector of a moving object have its own velocity c plus ( or minus ) a component of velocity provided by the object?

Suppose instead of a flashbulb and a nonfocused spherical wavefront, which is really muddy water, we have one laser and a double action beam splitter which directs a tightly focused thread of light both in the North direction and in the West direction.

Alternatively suppose we have a small diameter hollow tube surrounding the light path from Source to West Mirror.

Back to the animations, Cangas.
It's clear that in the moving frame, the path of the flash is diagonal, right?

Why is that?

Is it because the flash is carried along by the device?
Or is it just that the flash is (obviously) not carried along by the moving camera?

"Why is that" is what I am trying to get YOU to tell ME.

Back to my most recent post prior to this one.

How does light get "carried along" by a moving device?

Einstein in his elevator gedankin explained that light would NOT get "carried along" by the moving elevator.

Suppose instead of a flashbulb and a nonfocused spherical wavefront, which is really muddy water, we have one laser and a double action beam splitter which directs a tightly focused thread of light both in the North direction and in the West direction.
Precisely.
What do you think would be the result?
Do you think that the moving camera would record that the flash missed the mirror?

"Why is that" is what I am trying to get YOU to tell ME.

OK... It's because the camera is moving south.



How does light get "carried along" by a moving device?
Does it get "carried along"? Or is its initial direction determined by the motion of the emitter?

In the case of our rifle firing at the wall, is the bullet "carried along" by the rifle? Or does it simply maintain its original forward motion?

Pete, not CANGAS, introduced the concept of light getting "carried along" by a moving device. I think Pete knows more about how light gets "carried along" than I do.

"Or is its initial direction determined by the motion of the emitter?"

CANGAS claims that modern science, including SR, says NO.

What does Pete say?

Pete, not CANGAS, introduced the concept of light getting "carried along" by a moving device.
I think not.
Are you still misinterpreting those animations?

These animations show two films of the same event. One camera is fixed to the ceiling, one camera is moving rapidly south:
<table width=100%><tr><td width=50% align=center>http://home.teegee.com.au/byrnes/Pete/Relativity/Images/RestFrameBackground.gif</td><td width=50% align=center>http://home.teegee.com.au/byrnes/Pete/Relativity/Images/MovingFrameA1Background.gif</td></tr></table>


Do you think the flash is moving diagonally in the camera's frame because it is carried along by the device?
Or do you think it is because the camera frame is moving south?

Look at your post of 11:41 to refresh your memory of your introduction of the infamous "carried along" phrase.

I think that the light is not going diagonally unless the laser is aimed diagonally.

I think that I have asked a number of important questions and made a number of comments that you have chosen to ignore rather than to confront.

I respectfully request you to addres my backlogged questions and comments before their quantity becomes hopelessly immense.

I need a post number... times are local time. Hold your mouse over the reply link, and look at the number at the end of the link shown in the status bar at the bottom of you explorer window


I think that the light is not going diagonally unless the laser is aimed diagonally.
If the camera filming the event is moving south, does the light flash appear to be going diagonally when the film is replayed?

I think that all your questions are labouring undering a misinterpretation of the scenario. In the scenario, the device never moves. It is fixed to the floor. It is our viewpoint that moves.

Cangas,
You may have missed this response to an earlier post of yours... I don't think you acknowledged or replied to it.

"Or is its initial direction determined by the motion of the emitter?"

CANGAS claims that modern science, including SR, says NO.

What does Pete say?
Pete says CANGAS is wrong, that galilean relativity and special relativity both say that the direction of travel of anything (including a light flash) is frame dependent.

I think there is a lot of confusion here about what a frame of reference is. A frame of reference is just a coordinate system. You can talk about clocks and cameras and which is moving, but an inertial frame does not have to coincide with any physical object.

If you have anything move back and forth in a straight line in one coordinate system it will zigzag in another coordinate system which is moving with a constant velocity relative to the first. It has nothing to do with things getting "carried along" or with things being "aimed diagonally". It is just a mathematical substitution of variables.

-Dale

The "reality of reference frames" issue might be worth discussing in a separate thread... but we'll try it here and see how we go.

You're riding shotgun in a jeep. The jeep is travelling at 30m/s parallel to a long wall, which is 10m away on your side.

You are carrying a rifle loaded with subsonic rounds, muzzle velocity 300m/s.

You aim and fire the rifle directly at the passing wall, at 90° to your direction of travel. To your horror, at very instant that the bullet leaves the rifle a man standing against the wall flashes past your sights... Ie at the moment of firing, the rifle was aimed directly at the man.

Does the bullet strike the man?

No but the bullet has mass and forward momentum. I have seen some data that claims light doesn't have such lateral momentum. (Granted most says it does however).

I think some may be confused because most (all?) animations I have seen portray the observer (camera) at rest and the other frame moving. I had forgotten Pete's early statement, and different animation, that the observer is the one moving in this animation. I guess I just overlooked his statements to the contrary in regard to his animations.

No one else is commenting, so I will state the obvious: although the photon 'appears' to take a longer path in the moving frame....it doesn't. Time of travel is the same in both frames.

Bullets and other massive objects are acknowledged to participate in the momentum and the velocity vector of the moving object from which they are ejected.

According to the old Emitter Theory of light, rejected by Einstein and others over a century ago, light participated in the momentum and velocity vector of the moving object from which it was ejected.

In modern science theory, including SR, does light participate in the momentum and velocity vector of the moving object from which it is ejected?...
http://www.u.arizona.edu/~lilley98/

Last things first.

Thank you Anomalous, for attempting to be helpful.

The link was a nice description of radiation pressure and directly related matters. It had no relation to the point I am making, however.

My point is that the motion of a photon source has no effect upon the velocity of the photon.

.... My point is that the motion of a photon source has no effect upon the velocity of the photon.

If that was so then why will the photon seem to be at speed of light to the source ?

I wont be much help to U from your point of view but U hold too much potential for me.

Now to get started with the thread topic:

A source of possible confusion is the switch from the original thread description of a pair of devices, one stationary and one moving, now changed to a pair of devices, one bolted to the floor and the other one bolted to the floor, too.

Another source of possible confusion is the original thread statement of light being emitted in a Northward path and also in a Westward path ( a clear description of two light paths separated bt exactly 90 degrees ) having been changed to spherical wave fronts ( which, radiating 360 degrees, have no formal separation at all ).

Having made comments based on the ORIGINAL thread statement which, in my opinion, were not responded to satisfactorily, I have persued the original statement. The intervening switches by the thread starter have made it difficult for me and others to keep track of which comments were pertinent to which of the several switches. Of course, It could not be possible that the thread starter has deliberately attempted to muddy the water for any self serving reason.

At this time a question needs to be asked: If the thread has been switched to analysis of two stationary devices, and if we are correctly remembering Lorentz/SR length contraction be a result of an object moving and therefore contracting, what is the present point of this thread?

Anomalous:

According to the old Emitter Theory ( or it may have been Emission Theory ), light had a velocity of c RELATIVE TO ITS SOURCE. Light had a velocity of c plus source velocity or minus source velocity, depending on the circumstances, when observed by an observer who did not share the velocity of the light source. In other words, a stationary observer looking at light emitted by a moving source would measure a light velocity higher or lower than c. Einstein and other scientists considered Emitter Theory and found several problems considered insurmountable. Einstein went on to invent Special Relativity, in which the speed of light is completely independent of the speed of its source.

If the speed of emitted light is completely independent of the speed of its source, not only light emitted straight ahead or straight back must be subject to the rule, but light emitted in any direction.

If light takes on the velocity of its source, suppose an object is traveling at .99999999 c relative to a stationary observer and light is emitted at a 90 degree angle to the path of the object; the light will have its usual velocity of 1.00 c in the perpendicular direction and ALSO an additional contributed velocity of .99999999c in the direction of the motion of the object, all as measured by a stationary observer. The two velocities at right angles are vector components which result in a resultant velocity, conveniently determined by, at least approximately, in this simple example, use of Pythagoras Theorem. The resultant velocity vector of the light is approximately at a diagonal angle to the path of the object and is significantly greater than c.

The terrifying prospect of light traveling faster than light causes many scientists to affirm that light does not take on any of the velocity of its source.

If light emitted by a moving object does not take on any of the velocity of its source, light emitted exactly sideways just keeps on going sideways and does not flip into a diagonal ( or any angle ) path with a forward component.

If light does take on the velocity of its source and so light can travel faster than light, we need go no further to declare Special Relativity dead and can begin today planning a nice funeral. Or a wake; I would like Heinekin beer, please.

Anomalous:......Einstein and other scientists considered Emitter Theory and found several problems considered insurmountable. .....
I hope someone someday will tell me about what thoes problems were.


If the speed of emitted light is completely independent of the speed of its source, not only light emitted straight ahead or straight back must be subject to the rule, but light emitted in any direction.

What made them believe its so is another big mystery.


.... as measured by a stationary observer. The two velocities at right angles are vector components which result in a resultant velocity, conveniently determined by, at least approximately, in this simple example, use of Pythagoras Theorem. The resultant velocity vector of the light is approximately at a diagonal angle to the path of the object and is significantly greater than c.

The terrifying prospect of light traveling faster than light causes many scientists to affirm that light does not take on any of the velocity of its source........ I always wondered who or what terrified them about it. For me its good news because that makes space travel more interesting.

Now to get started with the thread topic:

A source of possible confusion is the switch from the original thread description of a pair of devices, one stationary and one moving, now changed to a pair of devices, one bolted to the floor and the other one bolted to the floor, too.
There was never any mention by me of one device moving and one stationary. I'm sorry you misinterpreted.


Another source of possible confusion is the original thread statement of light being emitted in a Northward path and also in a Westward path ( a clear description of two light paths separated bt exactly 90 degrees ) having been changed to spherical wave fronts ( which, radiating 360 degrees, have no formal separation at all ).
The notion of spherical wavefronts was incorrectly introduced by 2inquisitive. Your interpretation is


At this time a question needs to be asked: If the thread has been switched to analysis of two stationary devices, and if we are correctly remembering Lorentz/SR length contraction be a result of an object moving and therefore contracting, what is the present point of this thread?
One device, not two.
We are examining the same set of events from two different reference frames.
We are viewing exactly the same thing from two different "cameras".

The point of the thread is to take the postulate that the speed of the flashes is the same in each reference frame (camera film), and see if length contraction is required.

If light emitted by a moving object does not take on any of the velocity of its source, light emitted exactly sideways just keeps on going sideways and does not flip into a diagonal ( or any angle ) path with a forward component.
Sideways is frame dependent, CANGAS.
Sideways on the first camera's film (ie in the device's rest frame) is clearly not the same as sideways on the second film (ie in the frame which is moving south relative to the device; the frame in the which the device is (apparently) moving north).


Do you think the flash is moving diagonally on the moving camera's film because it is carried along by the device?
Or do you think it is because the camera is moving south?

No one else is commenting, so I will state the obvious: although the photon 'appears' to take a longer path in the moving frame....it doesn't. Time of travel is the same in both frames.
Flash, 2inq, not photon.

The path recorded on the moving camera's film is the path of the flash in the moving frame. Clearly, it is longer than the path of the flash in the rest frame.

I think what you mean to say is that the longer path is not real, that the rest frame path is the real path. Right?

Interestingly, this thread seems to have become a discussion about wha a reference frame is. It is because of miscommunications that I've begun using the language of cameras and film rather than reference frames... and I think perhaps we're beginning to make progress?

I think I'll use the cameras on tracks approach to reference frames more.

No one else is commenting, so I will state the obvious: although the photon 'appears' to take a longer path in the moving frame....it doesn't. Time of travel is the same in both frames.This is incorrect. You can determine that the path is longer by the Pythagorean theorem.

Again, a frame is just a coordinate system. The length of a path is well defined in any coordinate system (by the Pythagorean theorem for a Cartesian coordinate system). There is no getting around the mathematical fact that it is a different length in the two coordinate systems.

-Dale

Anomalous:

Problems with Emitter/Emission Theory: http://en.wikipedia.org/wiki/Emitter_theory

Pete:

I know that you know that I never thought that light is carried along by any moving device.

YOU were apparently making a heroic effort yourself to claim that your light flash was carried along by your moving device, prior to your changing your device to being bolted down and moving a camera over it.

No one else is commenting, so I will state the obvious: although the photon 'appears' to take a longer path in the moving frame....it doesn't. Time of travel is the same in both frames.

Excellent. That seems to justify my conclusion that the correct view is the moving observer would consider his velocity as being greater.

MacM:

Be careful in this thread: Pete has changed the key factors several times. For instance there is now no movement of the device but rather the observer ( a camera now ) is doing the moving. You need to read the whole thing or you will be saying cat where Pete is now saying dog.

YOU were apparently making a heroic effort yourself to claim that your light flash was carried along by your moving device, prior to your changing your device to being bolted down and moving a camera over it.

The two scenarios are two ways of explaining the same thing. Light isn't 'carried along' by the emitter, but the velocity of the emitter in your frame does change the initial angle of the light as seen from your frame.

Why? Ok, imagine this. Two horizontal boards, one above the other. A light shines vertically down from the centre of the top board to the centre of the lower one, forming a spot of light you can easily see.

Now imagine this whole board-and-flashlight system moving past you at high speed. If you were sitting on the boards (i.e. if you were in their reference frame), the boards would be still, so obviously there would be no change - the light would hit the centre of the lower board.

Since you're watching the boards fly past you, though, you're in a different reference frame. But the light still has to hit the centre of the lower board (do you agree?). So a photon leaves the top board. The lower board moves in the time it takes the photon to descend. The photon still hits the centre of the lower board. Obviously, the angle at which the photon travels appears different in your frame.

Do you agree?

Anomalous:

Problems with Emitter/Emission Theory: http://en.wikipedia.org/wiki/Emitter_theory Sorry dude that link had nothing that says why speed of Light remains constant, may be I misread it.

This is incorrect. You can determine that the path is longer by the Pythagorean theorem.

Again, a frame is just a coordinate system. The length of a path is well defined in any coordinate system (by the Pythagorean theorem for a Cartesian coordinate system). There is no getting around the mathematical fact that it is a different length in the two coordinate systems.

-Dale

DaleSpam, remember my post when I recalled that tracer rounds (bullets) appeared to travel a curved path from the ground to the helicopter? A curved path is longer between the ground and the helicopter than a straight path. Do you claim that is due to 'time dilation'? Why do you think I eased that post in?

Are you postulating that the speed of bullets is the same in all reference frames? I think that is a pretty strange postulate, but if so then the curve must be due to some change in time and/or distance.

In any case, the path length of the bullet is unarguably longer in a frame where it curves than in a frame where it goes straight.

-Dale

MacM:

Be careful in this thread: Pete has changed the key factors several times. For instance there is now no movement of the device but rather the observer ( a camera now ) is doing the moving. You need to read the whole thing or you will be saying cat where Pete is now saying dog.

Thanks for the heads up. I am however accustomed to the shifting sands of relativists. :D

by Pete:

"The notion of spherical wavefronts was incorrectly introduced by 2inquisitive. Your interpretation is...."
================================================== ===========

I need to clear up a point. I introduced a wavefront as a 'line' moving from the emitter to the reflector. MacM stated 'spherical' in his post and I agreed with his overall statement. But there is one stipulation. The Huygens action of a moving EM wave. The light 'ray' is propagated directly away from the source as a wave, although there are many different rays emitted from a blackbody in a spherical pattern all around the body. Speaking of one ray only now, the wave not only propagates in the direction of motion of the wave, it also propagates transverse to the direction of travel as it travels. The result is the wavefront of an individual 'ray' is two crossed lines propagating through space, similar to the '+' notation. You can polarize those 'lines', horizontally, vertically, and both in right-hand circles and left-hand circles so they 'spin' while travelling. If you measure the distance and time travelled by one ray, the measurement applies to that one ray only. In the rest frame of the emitting device, you are measuring a particular set of rays for time and distance travelled. That gives the true time of travel, the emitter's rest frame. A moving observer's time does not change everytime he 'calculates' a different travel time for events happening in 'other' frames. Remember, Pete's example is from the viewpoint of a moving observer/camera, not a moving event. The travel of the flash and reflector is an illusion of your moving frame, the observer/camera frame. The flash is travelling in a frame in which the emitter and reflector are at rest with respect to each other.

I know that you know that I never thought that light is carried along by any moving device.
That's good.

YOU were apparently making a heroic effort yourself to claim that your light flash was carried along by your moving device, prior to your changing your device to being bolted down and moving a camera over it.
No change was made, CANGAS, only repeated clarification. If you misunderstood the original post, that's not my fault.

The speed of tracers is not invariant.


DaleSpam, remember my post when I recalled that tracer rounds (bullets) appeared to travel a curved path from the ground to the helicopter? A curved path is longer between the ground and the helicopter than a straight path. Do you claim that is due to 'time dilation'?
No, its due to the helicopter's moving reference frame. The time of travel is still the same... which implies that the speed of the tracers is higher in the helicopter's reference frame, right?

Are you getting a gut feel for what "reference frame" means?

In the helicopter's reference frame, the tracers go faster along a curved path. There's no question of that.
The questions I think you want to address are - Is this "real"? Do the tracers really go at that speed and follow a curved path?

We can address those questions next, if you like.

by Pete:

"The notion of spherical wavefronts was incorrectly introduced by 2inquisitive. Your interpretation is...."
================================================== ===========

I need to clear up a point. I introduced a wavefront as a 'line' moving from the emitter to the reflector. MacM stated 'spherical' in his post and I agreed with his overall statement. But there is one stipulation. The Huygens action of a moving EM wave. The light 'ray' is propagated directly away from the source as a wave, although there are many different rays emitted from a blackbody in a spherical pattern all around the body. Speaking of one ray only now, the wave not only propagates in the direction of motion of the wave, it also propagates transverse to the direction of travel as it travels. The result is the wavefront of an individual 'ray' is two crossed lines propagating through space, similar to the '+' notation. You can polarize those 'lines', horizontally, vertically, and both in right-hand circles and left-hand circles so they 'spin' while travelling. If you measure the distance and time travelled by one ray, the measurement applies to that one ray only. In the rest frame of the emitting device, you are measuring a particular set of rays for time and distance travelled. That gives the true time of travel, the emitter's rest frame. A moving observer's time does not change everytime he 'calculates' a different travel time for events happening in 'other' frames. Remember, Pete's example is from the viewpoint of a moving observer/camera, not a moving event. The travel of the flash and reflector is an illusion of your moving frame, the observer/camera frame. The flash is travelling in a frame in which the emitter and reflector are at rest with respect to each other.

This sounds a lot like something I raised on this forum a couple of years ago but got no good response here and could not find any internet support for what I had been taught.

In Nuke school we were told as you approch v = c the dimension in the direction of motion contradcts toward zero and there was a traverse wave which was propagating toward infinite and at v = c and object had no dimension in the vector of motion and existed as a two demensional wave infinitely wide and infitely high such that a baseball thrown at v = c in a general direction toward an infinite wall would always go through a single pin hole located somewhere in the wall.

To rephrase my previous serious question in hope of it not being misunderstood:

If we have a source of photons, attached to a device capable of movement, which is precisely aimed and focused in a direction perpendicular to the intended direction of movement of the device, and we initiate movement, does the motion of the device provide a velocity vector to the photons being emitted?

Do the photons then have only their expected velocity vector perpendicular to the device movement?

Or do they also have an additional velocity component in the direction of device movement?

Please consider this question in regard to current mainstream physics theory, not in respect to the rejected Emitter Theory of over a century ago.

Yes they have an additional component. The argument supporting this does not have to invoke any physical theory about the mechanism by which this occurs. Rather it consists of deduction, much as Relativity does. For more reference, I suggest this thread:

<A href= "http://sciforums.com/showthread.php?t=47355">Light Beam Path from a Moving Light Source</A>

The two scenarios are two ways of explaining the same thing. Light isn't 'carried along' by the emitter, but the velocity of the emitter in your frame does change the initial angle of the light as seen from your frame.
Why? Ok, imagine this. Two horizontal boards, one above the other. A light shines vertically down from the centre of the top board to the centre of the lower one, forming a spot of light you can easily see.
Now imagine this whole board-and-flashlight system moving past you at high speed. If you were sitting on the boards (i.e. if you were in their reference frame), the boards would be still, so obviously there would be no change - the light would hit the centre of the lower board.
Since you're watching the boards fly past you, though, you're in a different reference frame. But the light still has to hit the centre of the lower board (do you agree?). So a photon leaves the top board. The lower board moves in the time it takes the photon to descend. The photon still hits the centre of the lower board. Obviously, the angle at which the photon travels appears different in your frame. Do you agree?Yes and No. I think you are basically correct, but need to be more careful to distinguish between the beam of light going from the top board to the lower one and the path of photons. To make this distinction clear, let me have a glass tube, with internal diameter, IDg, slightly bigger than the external diameter of the beam of light ODb. The axis of this glass tube passes thru the center of both boards and the beam does not touch it because IDg > ODb. Furthermore the entire ID of glass tube is coated with a photo sensitive chemical that explodes, if even a single photon is absorbed in it.

From the POV of the observer sitting on the moving lower board, the beam steadily and safely passes between the center of the upper board, and the center of the lower board. He is not injured by pieces of exploding glass. This must be true for all observers in any frame. That is the beam is always passing down the axis of he tube, perpendicularly to both boards, for all observers. However, the path of the photons in the beam is perpendicular to the boards, ONLY for observers at fixed positions on the boards (in the frame where boards are not moving.)

I have not read all the posts, but think much of the confusion is due to not thinking clearly about the difference between the "beam" and the "path."

No, BillyT, it isn't. The confusion is about how the direction of a light can depend on the motion of the source, when its speed does not.

I can show you one of the possible ways in which the direction of a wave (I didn't say light, specifically) can depend on the motion of the source, while its speed depends upon the speed of medium.

Imagine a wave which propogates through a certain medium at a certain velocity. The wave is transmitted from particle to particle in that medium. Now, suppose that the wave is being emitted by a moving source. At the time of emission, it will have a momentum which depends on the direction in which the source is moving. However, the propogation depends on the medium, and hence, it cannot go at a higher or lower speed than that permitted by the medium.

But from the frame of the source, the medium itself is moving, so the velocity of the wave will be different.

The case of light is quite different, but I was only trying to resolve the apparent conflict between the dependence of direction and independence of speed by showing that there are certain ways in which it may be possible.

But even these arguments are unnecessary, because in the experiment considered, there is a specific <I>event</I> happening. This <I>has</I> to occur in both frames. In order for this, the light pulse has to take a diagonal path in one of the frames. This is not essentially a physical argument. It is more of a philosophical one.

No, BillyT, it isn't. The confusion is about how the direction of a light can depend on the motion of the source, when its speed does not. I have not read enough here to make strong claims as to the confusion, but even this sentence IMHO is part of it as I am not sure what you mean. I.e. Is "the direction of a light" the "beam's direction" or the "photon's path's direction"? That was my point, hoping to at least get people speaking clearly to each other, but obviously I failed. Also I was only trying to be helpful about direction concepts, not speed, which will clearly needs care about time dilations etc., effects.


I can show you one of the possible ways in which the direction of a wave (I didn't say light, specifically) can depend on the motion of the source, while its speed depends upon the speed of medium.No need to imply this is not true for light also. The time required for light to pass one meter upsteam in moving water is greater than for it to pass one meter down steam in that same river. (This is because light in any transparent medium is able to cause the medium to radiate coherently with the incident light, atom by atom, but the phase of the locally radiated wave is not the same as the incident wave. For example, the net wave is the sum of two sign waves of same frequency, the reradiation one much weaker of course, which can be shown to be a still different and displaced sin wave, as if the incident wave were no longer existing but replaced by one going slower. (Hint-Work with the E field strength of wave, not the intensity.) How much slower will depend upon the speed of the moving source that produces the reradiation component, LEAVING EACH MOVING ATOM AT THE SPEED OF LIGHT IN ITS REST FRAME. This is easy to demonstrate with a "two path interferomenter" and light passing thru equal lengths of glass tubes with flat closed ends, but side conedtions that premit the same water to flow (or not) thru the two tubes in different directions. (This experiment has a name, but I forget it, just now.)

...The case of light is quite different, ...Did not read you carefully because I think this is wrong.Light, in a moving transparient medium, travels thru the lab space at a different speed when the medium is moving in the lab space. - See the above for why.

PS, by edit - will not reply as will be traveling thru USA to Europe, probably leaving this nite if can not delay one day.

PS, by edit - will not reply as will be traveling thru USA to Europe, probably leaving this nite if can not delay one day.


Have a nice long trip. Don't hurry back. :D Just kidding of course.

I don't understand all this confusion about the direction. A frame of reference is just a coordinate system. It is somewhat like the lines on a football field. If we draw the lines one way the QB throws a touchdown, if we draw it another way the QB throws out of bounds. If the ball drips paint as it is thrown then the resulting lines will go in different directions relative to the yard lines. Nothing really changed except the lines and which regions are named endzone and which are named sideline. Why is it so confusing that the direction of a light pulse is different in different frames? It is just the mathematical result of the change in coordinates.

-Dale

Because Special Theory claims the time required for a light pulse to travel from point A to point B is longer when viewed from a moving frame. Your football analogy would be more correct if you had one observer sitting behind the endzone watching the ball travel to the endzone, and another observer sitting at the 10 yard line claiming the flight of the ball was longer from his diagonal viewpoint, thus taking longer for the time of flight of the ball.

Well, with the football analogy I was purposfully trying to avoid any confusion due to the invariance of c. I get the impression that the current confusion is not restricted to transforms of light. Particularly regarding all of the discussion of how the direction can change in different inertial frames.

If we were talking about bouncing ping-pong balls back and forth instead of bouncing pulses of light off a mirror would you still doubt that in some frames the direction is different than in others?

-Dale

We are not speaking of apparent directions. We are speaking of why it takes light longer to travel from point A to point B when viewed from a moving frame. TIME, DaleSpam, time.

Let's try ping-pong balls bouncing between two paddles instead of light bouncing between two mirrors for a minute first. We can keep velocities low so that we don't have to worry about relativity. To further simplify things let's have this ping-pong game in free space far away from anything else. Lets consider two frames that are initially together with the paddles located on the y-y' axis. The paddles move at a velocity v in the positive x' direction but are stationary in the unprimed frame. Now, lets look at the path of the ping-pong ball.

Do you agree that the direction (angle with the x-x' axis) is different in the two frames?

Do you agree that the distance is different in the two frames?

-Dale

You don't seem to understand, DaleSpam. I figured out long ago, when I first started reading STR gedankins, how they used the observer's rest frame to confuse the issue of time dilation. Do your gedankin as Pete did his last one, place the observer in the moving frame and the ping pong ball/paddles in the stationary frame. All STR's claim there is no difference in which frame is illustrated as the moving frame. When illustrating STR's supposed 'time dilation' based on an invarient 'c', there is a difference in the way the illustration is portrayed. Place the observer in the moving frame, with the path of light between two points in the rest frame, then you have to state light is slowing in the rest frame. I know you know this, so are you trying to confuse the issue to support Special Theory?

I am not trying to confuse the issue, I am trying to find out what the issue is. Is it that you don't understand the idea of transforms between frames of reference? Is it that you don't believe that all frames of reference are equivalent? Or is it that you don't believe that c is invariant? It has to be one of the three and, based on all of the comments about directions, I thought the first was most likely. That is why I asked the two questions (about direction and distance in the two different frames). If you agree with those statements then the issue is one of the two other points. I am not trying to be a pain, I just don't want to argue about something that isn't the key point of disagreement.

-Dale

I don't believe all frames of reference are equivalent when the speed of light is calculated based on 'apparent' paths to give 'real' time dilation. I also believe the speed of light is invariant in local co-moving inertial frames. It is not 'really' invariant when comparing one frame with another moving frame. 'c' is defined as invariant by making time and length variable based on an invariant 'c'. Do you believe that 'c' is invariant when a frame is moving in varing gravitational potential? I realize that isn't a 'STR' inertial frame, but gravitational freefall IS a GR inertial frame. An observer on board a craft couldn't tell the difference as no gravitational effects are felt. But it is easy to reference showing that a clock beats faster in lower potential (further from the gravitating object). The Shapiro effect has been confirmed, light moves more slowly through a gravitational field than expected (slows down). Do you believe a meter expands in all directions as the craft moves away from a gravity source? Do you believe there are frames in which a craft is NOT in gravitational freefall, even in deep space?

No need to imply this is not true for light also. The time required for light to pass one meter upsteam in moving water is greater than for it to pass one meter down steam in that same river.

...

Did not read you carefully because I think this is wrong.Light, in a moving transparient medium, travels thru the lab space at a different speed when the medium is moving in the lab space. - See the above for why.


No, there is no medium here. We're talking about light travelling through vacuum, which is why I mentioned particularly that I wasn't talking about light. I don't intend to bring in any 'ether'-concepts.

The Shapiro effect has been confirmed, light moves more slowly through a gravitational field than expected (slows down).

2inquisitive,

Is this really true? Do you know of a good link to this subject or can you expand on it a little more?

I though light was invariint

Thanks

2inquisitive,

Is this really true? Do you know of a good link to this subject or can you expand on it a little more?

I though light was invariint

Thanks

Even Special Theory only states 'c' is invariant in an inertial frame in vacuum. It is no secret that 'c' is not invariant in General Relativity due to gravitational fields. Clocks slow in increasing gravitational potential. There is no 'Lorentz contraction in direction of motion only' to offset the time dilation in General Relativity. Slow time without contracting the meter and 'c' MUST decrease.
http://en.wikipedia.org/wiki/Shapiro_effect

(added link)

Here are some quick references from Wikipedia. Read some of the associated papers with the RACE, PARCS and SUMO links I already gave to find out more. Read some papers on the Linear Ion Trap Atomic Clock (LITS) and the linear ion trap frequency standard for more. I don't remember always remember at exactly where and which papers, sometimes many, many pages long, to look for specific cut & paste excerpts. From a short source, Wikipedia:

"The time delay effect was first noticed in 1964, by Irwin I. Shapiro. Shapiro proposed an observational test of his prediction: bounce radar beams off the surface of Venus and Mercury, and measure the round trip travel time. When the Earth, Sun, and Venus are most favorably aligned, Shapiro showed that the expected time delay, due to the presence of the Sun, of a radar signal traveling from the Earth to Venus and back, would be about 200 milliseconds, well within the limitations of 1960s era technology.

The first test, using the MIT Haystack radar antenna, was successful, matching the predicted amount of time delay. The experiments have been repeated many times since, with increasing accuracy.
....
Quote by Einstein
"In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Now we might think that as a consequence of this, the special theory of relativity and with it the whole theory of relativity would be laid in the dust. But in reality this is not the case. We can only conclude that the special theory of relativity cannot claim an unlinlited domain of validity ; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena (e.g. of light)." - Albert Einstein (The General Theory of Relativity: Chapter 22 - A Few Inferences from the General Principle of Relativity)

No, there is no medium here. We're talking about light travelling through vacuum, which is why I mentioned particularly that I wasn't talking about light. I don't intend to bring in any 'ether'-concepts.Stange. I could swear I read "medium" some where in your post at 59 past hour of yesterday, which excepted light. The part I commented on was:

I can show you one of the possible ways in which the direction of a wave (I didn't say light, specifically) can depend on the motion of the source, while its speed depends upon the speed of medium.
Imagine a wave which propogates through a certain medium at a certain velocity. The wave is transmitted from particle to particle in that medium.

I don't believe all frames of reference are equivalent when the speed of light is calculated based on 'apparent' paths to give 'real' time dilation. I also believe the speed of light is invariant in local co-moving inertial frames. It is not 'really' invariant when comparing one frame with another moving frame. Obviously if you do not believe that the distance measured in one frame is as valid as the distance measured in another then there is no reason to suggest time dilation. If you don't accept the postulates of SR then you have no reason to accept time dilation, length contraction, and the other conclusions of SR.

One of your statements confused me. What is a "local co-moving inertial frame"? A co-moving frame sounds identical (except possibly for a spatial rotation) to the original frame. So I don't understand your "speed of light invariance" statement. Did you intend to mean that the speed of light is the same in every direction in the local prefered frame or am I misunderstanding?

-Dale

Sorry DaleSpam, I wasn't trying to ignore your question, but it is difficult to answer clearly. I started a post earlier, saw that it was confusing, so I didn't submit it. Then, I had other things to do in my real world.

Let me start by stating there are two ways to look at reference frames. First, the coordinate system way such as a Minkowski diagram and Lorentz transforms. Second, a physical way that conforms to 'the real world'. A 'co-moving inertial frame' is the coordinate system method. In that method, you can 'instantly' transform into the co-moving frame, no signal propogation delays between frames and no Sagnac effects or any other discrepancies as when 'physically' comparing one frame with the other. For instance, in the co-moving coordinate method, the Sagnac delay of EM propagation does not come into play, but it does in the 'physical' method. In STR inertial frames using co-moving frames, gravitational time discrepancies between the two frames do not show up because you instantly 'jump' into the other frame on a sheet of paper. I added the 'local' to my 'co-moving frames' because when considering a distant frame near the edge of the universe, the wavelength of the EM radiation has also been 'stretched' about a thousand times its length when it left its source by universal expansion. That would have to be taken into consideration also. I know this is still not clear, but it is the best I can do with my limited physics knowledge. Hope it does help, though.

Sorry DaleSpam, I wasn't trying to ignore your question, but it is difficult to answer clearly. That's OK, after all, I didn't answer any of your GR questions either :). It's taken me about 7 or 8 years to get to my current understanding of SR, so I anticipate another 14+ years before I can talk reasonably about GR (and just forget ever asking me about QM).

Let me start by stating there are two ways to look at reference frames. First, the coordinate system way such as a Minkowski diagram and Lorentz transforms. Second, a physical way that conforms to 'the real world'. A 'co-moving inertial frame' is the coordinate system method. In that method, you can 'instantly' transform into the co-moving frame, no signal propogation delays between frames and no Sagnac effects or any other discrepancies as when 'physically' comparing one frame with the other. For instance, in the co-moving coordinate method, the Sagnac delay of EM propagation does not come into play, but it does in the 'physical' method. In STR inertial frames using co-moving frames, gravitational time discrepancies between the two frames do not show up because you instantly 'jump' into the other frame on a sheet of paper. I added the 'local' to my 'co-moving frames' because when considering a distant frame near the edge of the universe, the wavelength of the EM radiation has also been 'stretched' about a thousand times its length when it left its source by universal expansion. That would have to be taken into consideration also. I know this is still not clear, but it is the best I can do with my limited physics knowledge. Hope it does help, though.I guess I look at different frames just as different viewpoints, different ways of telling the story. Like in our previous discussion about rotating reference frames, sometimes it's easier to describe things from a different viewpoint. But even if it is more difficult a different way you should be able to get the same answers any way you approach it.

In any case, I have no real justification about why different frames are equally valid. Basically it is just easier to work physics problems if you don't have to worry about drawing the lines exactly right to match some prefered local frame, but there is no real reason that the universe should make my physics problems easier. On the other hand, I think that the fact that the testable conclusions seem correct at least makes it reasonable to accept the postulate.

-Dale

Stange. I could swear I read "medium" some where in your post at 59 past hour of yesterday, which excepted light.

I guess you haven't been following the entire discussion. The example I gave was applicable to waves which travel through media. But this <I>need not be true</I> in the case of light. Therefore, I can't assert that this exactly could be the reason for invariance of light-speed. When I said that the discussion did not involve a medium, I meant the original discussion, in which light was going about in vacuum.