# The relative velocity of photon and moving frame:SR heresy.

Discussion in 'Physics & Math' started by geistkiesel, Jul 13, 2004.

1. ### geistkieselValued Senior Member

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2,471
No but I lived there for three years, 6-8 grade, Augsburg und Bamberg.. Im Augsburg Zweiunfunfzieg Uhland Stasse, vertilich.

Hmmm, maybe so. We'll see. But by then I should have you cured of any relativity theory addictions, which would force your astroids video to something more real, and I add, more useful. Do you know NASA has an anti-astroid-collision program going? Get on as a student/intern . . . [there is no trying to get on as an intern, with pay, there is only GETTING ON].. . .. Promise your mom you'll be a good boy and will drink at least a quart of milk a day, get the internship and dad will buy you something more than appropriate for your educational advancement. Video games and physics, all decisions removed, but you have to maintain a continuosly patient persona with those arrogant NASA scientists when asking the frequent question, "will you please prove it to me?"

quote geistkiesel- " Someone earlier called me arrogant for daring to do what I do here. So, now "arrogant" is a bad thing is it? "

and someone reminded me of this statement. But thinking you can outsmart a century of dead guys is the easy part, they can't argue with you.

Arrogant Message received and understood. As a self-claimed self-starter I can assure you that "we" need external stoppers from time to time. Even though I am a admirer of the country's greatest actor of all time, John "Duke" Wayne, who said, "never apologize, its a sign of weakness.", there are always 'adjustments' that can be made. You might be surprised at the transformation seen in this threader's consideration of possible reference frame modifications of his modus operandi. After all, my favorite baseball player of all time, the maestro pitcher, Leroy "Satchel" Paige with infinite wisdom stated: "When facing an enemy stronger than you, walk him."

You can also review your acceleration history were you thoughtful enough to maintain such.

But this would be more fun. Here is what I would do next.
Code:
M|    |<-----------|o
|--------------->|o
|_______L______|

I would have a device that emits a photon at o in the direction opposite to the velocity of my train to a reflecting mirror at M. M and the photon therefore are on a collision course. Knowing L I can determine the distance equation for the round trip of the photon, or L - vt + L = tc and from this solving for v, and measuring t, I get the velocity of my train, v = (2L - ct)/t. Let's say I calculate 4.6meteres/second. Next I take a ten meter stick and apply some superglue to each end. I hold the stick parallel to the other train's direction of motion and flip the stick across the gap. My trusted and very able assistant down stream from my position measures the time it takes the stick to pass her eye position, say 1 second. So our relative velocity is 10meters/second. Adding the two velocities, or subtracting if the additon comes out contradictory measurements, you determine the complete motion profile of your train system:10 - 4.6 = 5.4meters/second.QED.

How would you do it?

I can give one answer right now regarding approaching astroids: Duck!.

Last edited: Jul 14, 2004

3. ### geistkieselValued Senior Member

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pete I read your post and my post a number of times and I cannot see where your objection comes from regarding accelerating and stationary frames. Could you please be more specific?

Is your dispute at all related to my suggested use of using the planet earth as a common inertial zero velocity reference frame wrt the photon and frame in the designing the experiment?
Thank you for the post.

5. ### geistkieselValued Senior Member

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2,471
Brandon9000 does the following repair any of your objections?

Assuming that there are sufficient experiemntal results justifying the use of the earth as a zero velocity reference frame are there any physical prohibitions against using the earth frame as common to two inertial framesmoving at different velocities wrt each other?

7. ### geistkieselValued Senior Member

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If the stationary moving frame experiments where the photons are emitted at A and B just as the moving frame observer arrives at M the midpoint of the A and B photon source, is it prohibited to use aderived expression such as t'3 =t'1(C + V)/(C - v) in calculating the arrival times of the A and B photons where t1 is the arrival time of the B photon and t3 the arrival time of the A photon by the observer in the moving frame? Didn't you make a similar calculation?

I don't see the violation of any law of physics. I am well aware of objections offered by those supprting SR, but I do not see the physical reasoning as sufficient to come to the conclusions that are arrived at. I am aware of the sequence of experiments and papers resulting in SR but I analogously do not observe sufficient physical rational for the conclusions arrived at. Help me out here.

8. ### Brandon9000Registered Senior Member

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172
I'm not sure what you mean. If an Earth bound observer sees a bullet chasing a photon at speed S, or moving in the opposite direction at speed S, he will see their relative speed as being c - S or c + S respectively. This is not a physical result, just the definition of relative velocity, hence the relativistic speed addition formulas are not relevant. Both the Earth bound observer and another observer riding the bullet will measure the speed of the photon as being c.

9. ### geistkieselValued Senior Member

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2,471
Likewise, observers on multiple numbers of bullets are able to determine all relative velocities, all measured consistently with respect to a common earth inertial frame.That is what I see. Even rabid believers in SR should not be compelled to deny herself the advantage of seeing perspectives in a narrowing focusing sense.

Present scientists immersed in SR Theory interetsted in Sanskrit, for instance, with predictably minimal application to current project and standard model scrutiny, ought not have research activity decided by reference to current theoretical structure and form. One ought best to define their own parameters of activity.

In general the SR guided have not explored the problem directly and remain limited to the maximum extention of applcability of SR, most are drastically short changing themselves and others by omitting to purposefully scrutinize dissident views using the self-consistency and interest of the view as its own standard of value. Otherwise scientific progress is measured against ever more trivially narrowing peels of the information onion, and the tears truly flow in salty abject sadness. Progress becomes inured fictional expansions of limited theoreticl structure. Value left unperturbed by imposed and published SR corrrections, often in arbitrarily censored form, are then allowed to grow with minimum distortion error imposed by theoretical designerror.

10. ### Brandon9000Registered Senior Member

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172
Okay, but experiments conducted all over the world have concurred with the predictions of SR for a century, and I have no reason to go looking for anything else. I prefer to think about areas of science where theory and observation don't agree. All I agreed to above was the definition of relative velocity. I believe that any inertial reference frame is equally as correct as any other, and, hence, that velocity has no meaning until you specify in which frame you are measuring it.

11. ### James RJust this guy, you know?Staff Member

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geistkiesel:

No frame can be determined absolutely to be physically moving. To do so would require a known stationary frame, and no such thing exists. There is no experiment which can be done to detect motion with constant velocity. Only relative velocities can be determined. This is a very basic point.

"Sufficiently" for what purpose? What is sufficient for one kind of measurement may not be at all sufficient for other kinds of measurement. It is possible to detect the motion of Earth relative to the Sun, or the centre of the Milky Way. It is also quite easy to detect the rotation of the Earth. So, what do you mean by "sufficiently at rest"? Relative to what?

And a velocity relative to the galactic centre of 200 km/s. And a velocity relative to other galaxies in the local group considerably higher again. Etc. Earth is not an absolute standard of rest.

Yes, and any other frame would do as well. There's nothing special about the Earth frame.

That is incorrect, since the velocity of a photon in any frame is measured to be c, and never less than c.

Wrong again, for the same reason.

No. You've thrown away the postulate of the constancy of the speed of light for all inertial observers.

You have things backwards. SR is defect-free, compared to whatever idea you have.

Please give an experimental measurement she can make to determine her absolute velocity. And specify your proposed absolute standard of rest.

You have not specified a reference frame here. In certain frames, what you say is true; in others it is false.

Train stations are in constant acceleration as the Earth rotates. Ignoring that, train stations are approximately inertial. Therefore, stations do not "feel" any acceleration, and are not perceived to accelerate from any other inertial frame. From the point of view of a non-inertial observer (such as one in an accelerating train), stations obviously do accelerate. They do so because of the apparent inertial forces acting on them, which only exist in non-inertial frames of course.

SR is only concerned with constant velocities, however, so forget the previous paragraph if you prefer to concentrate on SR. In the case of a train moving at constant velocity relative to a station, the station obviously moves at constant velocity relative to any observer on the train. Moreover (and this is the important part), neither an observer on the train nor one on the station can do an experiment to determine whether it is the station or the train which is "really" moving. The only motion they can talk about meaningfully is relative motion.

Clearly, these processes are physically possible, since they occur all the time.

They are merely applying common experience.

I've seen plenty of stations accelerate, from my point of view sitting on a train as the station pulls away. I have observed that the people on the platform generally tend to accelerate at the same rate as the station, and therefore their footing on the platform is not affected by the acceleration.

I'm so glad you're just kidding.

12. ### geistkieselValued Senior Member

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2,471
I am pleased you brought this up. In looking at special relativity from a number of angles I took another look at Einstein's analysis supporting SR as witnessed in his book "Relatvity", I have the 50th edition..

AE has two fundamental postulates, or laws.
1. The first he calls relativity in a restricted sense: the equivalence of the laws of physics.
2. The second, is in regard to the postlates describing the motion of light, being measured as C in vacuo.
From one the Galilean body of reference , or coordinate system is described and AE spends some time indicating that K0, or a "preferred frame", which you were referring to does not exist. Then he says that all indications are that the motion of the planet earth has not produced any anisotropic properties, and to me this sounds like an effective v = 0 inertial frame. The planet is moving at 30km/sec while making a full 2pi direction change.
The direction arrow then moves at a rate of 2pi/year ( 0 -> 2pi), which calculates to 6.4x10^-8rad/sec effective non-zero motion. The 30km/sec reflects a 30/3x10^5 =10^-4 proportional error in velocity. These data, together with the unambiguous experimental results showing no anisotropic properties qualifies V(e) = effective zero as a "preferred" frame. Therefore, when calculating the relative velocity of K, K' etc wrt K0, we may easily do so in steps.

V(e(c)) = C - V(e(c)) = C, and V(e(m)) = V(m) - V(e(m)), where V(e(m)) is the preferred frame used in determining the relative velocity of matter, nonelectromagnetic. Subtracting V(e(C)) - V(e(m)) = C - V. It is to be noted that the definition of relative velocity means what it says and in no way does the expression V(rel) = C + V imply that a photon emitted in the oppostie direction of the frame adds a velocity to C. The expression is a relative velocity. Photons heading east and mentors heading west (into the sunset) merely 'subtract' their velocities.

Relative to the planet earth. I hadn't gotten this far when I prepared the post you responded to, but I use the words "effective zero" in the same sense that SR theorists use the words that a galilean approximation is sufficient for v << c.Actually Galilean frames are good from v = 0 -> C. But plotting it out indicates a rather unique delta function that is flat (actually 1/2mv^2 for the energy of a an accelerated electron, for example) for a few orders of magnitude. Do not be distracted by a fleeting gamma expression. Try a F = d(mv)/dt = m(dv/dt) + v(dm/dt), and make some substitutions from E = 1/2mv^2 and see if you turn into some interesting territory.

See above. Dayotn Miller has us moving in a sothern pole, almost due south to the Sword Fish.

No so fast, Jamses R. Earth is the planet where we had performed all the experiments that indicated the lack of any anisotropic properties of light in virtually 3pi degrees of freedom. This is special, to me. Second, the planet earth is shrouded in a cloak of that force holding everything to the surface. We are imbedded in a gravity aether, that is if SR is in agreement that the force of gravity is indeed a "real" force, creating its own "aether", which I assume will be taken by some in a knee jerk reaction who will state with convincing emotion that "there is no experimental results indicating the gravity field induces any anisotropic properies in the motion of light."

I agree 100%, photons move at the speed of light, but how does SR gets there? Is it not fundamental that when mesuring he relative velocity of frame and Light the SR calulation always comes up V = C? Why is this? Because the SR calculation assumes the body of reference, the frame, is v = 0, correct? Just be egalitarian on your frame manipulation James R. Take the light and the frame wrt the V = 0 planet earth.

I suggest all SR theorists take aother look at Grounded's post, for a simple schooling on the manner and necessity of measuring the realtive velocty of photon and frame. And of counting unperturbed wave length segments passing through the plane of the eyes of the observer moving at velocity V wrt V=0, the preferred frame, and of course in he proper relative velocity wrtn C..

V ((e(C)) = C and V(e(m)) = V (m)
V(l) - V(m) = C - (-V) = C + V, a relative velocity, not an absolute velocity, except wrt to the inertial frame earth.

I throw nothing away except discarded theories. In all measurements of relative velocity of K' and K'' frames, the relative velocity, if properly calculated, has actually done the calculation as I suggested, i.e. wrt V = 0 planet earth.

This sounds like a challenge. I say the lack of anisotropic properties of light when measured from the planet earth qualifies the planet as a preferred frame. Then in taking the relative velocities of light and matter, wrt the earth, any errors, already small, get subtracted out in a normal use of mathematical discipline. This isn't a contrived "dual nature of light" used to explain electron diffraction through two (or is it one?) holes.

SR is only defective when it makes the assumption that when measuring the relative velocity of light and planet in vacuo, that all reference frames must also measure the relative velocity identically. This makes no sense. This is having the horse in the drivers seat, a step removed downward from the horse being merely 'behind the cart'.

The planet earth is the preferred frame. Use any galilean body of reference model and go for it.

I may have a typo or two, but in this case [if] any "false" statements are shown in the post they are trivially recovered.

Approximately inertial, James R, (this sounds like a bad dream trying to poke through to a real wake up call). Look at the last paragraph of yours and you can see whee I might have some basis for my post.

They are effectively inertial with an effective velocty in its orbital track at 30km/sec, or so some say. Michelson-Morely and Dayton Miller detected an aether drift of approximately 8km/sec, interpreted by the bulk of the pervailing view as a "null" result. The 8km/sec finding increases the closeness of the planet to actual zero, instead of the ersatz "null."

I described such a system, buthee is a new one. A and B both with v > 0, where V(A) > V(B) wrt the planet earth. A sends out oulses of light at his frame of one second intervals. Using gigherz frequencues he codes each pulses uniquely |[0000001]| 0 1 10 100 1000 1000 100000 1000000 |etc. where {#] is a tag for a specific second inteval. The ne=xt pulse is |[00000010]|0 1 10 100 10001 000 ....| and so on. This would do it relativistically wouldn't it? Another method would be to exchange acceleration and frame weight data histories. Both frames send their pulses. If there is any time dilation, which you champion, we should get an answer, post haste. This isn't thought out to any great depth, it is just an intutitive answer.

Well, if it wasn't for the lack of any finding of anisotropic properties of light rel to the earth frame, maybe you would have a valid point. I agree about the talk of relative motion, but again, please see the point I was endeavoring to make. Train stations do not accelerate in the same manner as trains,. You know this. I know this. Everybody knows this. But not to worry, just measure the relative velocities of train and station and what do you get? If you use SR you set the moving frame to zero, when you already have a preferred frame in the staion sitting at zero. The station is attached to the planet earth the last time I looked at at atrain station. Aren't frames supposed to provide he same laws of physics?

Let me rephrase. The mathematically described processes of railway station acceleration is describing an abstraction that has no physical analog. Horses accelerate, boats, space ships and the demise pf SR, all accelerate.

No, James R, there is no common experience, they are applying a learned mathematical technique designed to set the velocity of an inertial platform to zero so all light will be measured with respect to an inertial frame = 0, or the earth, or C. Guaranteed. Why even pretend to make the calculation?.

Were you having a draught, or two, or having a few, when you saw the station accelerate awayyyyy?
You do of course recognize that the people on the train actually feel a jerk when they accelerate. Now if the realtivity postulate is to remain true the laws of physics on both frames should be identical. Look at the accelerometer data attached to all trains and stations. One of the frames will read zero, the other a positive, nonzero, ergo SR mathematical contrivances do not apply, and besides that it is all unnecessary labor.

No, it is Mr. Serious at the present and I am only kidding with those with strong Australian accents, but I am having a rather nice laugh otherwise.

It is just a good day, do you know what I mean?

Good post James R. I had no surprises, and when I saw your name on this post I knew I was going to get a perfect SR response, with no extraneous verbiage.

13. ### geistkieselValued Senior Member

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See my post to James R brandon9000. As experimental results fail to provide any anisotropic properties of light with respect to he planet earth The planet qualifies as a preferred frame effectively at v = 0. The motion of the earth at 30km/sec is approximately 10^-4 of that of light. The velocity vector points from 0 - 2pi/year which comes out to about 6.4 x10^-8radiand/sec. This is an effective v= 0. See the post below.

14. ### James RJust this guy, you know?Staff Member

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geistkiesel:

You talk about "anisotropic properties" a lot in your post. Which properties, in particular, are you thinking of? I can't make much sense of your statement that earth "has not produced any anisotropic properties" unless you can explain what anisotropic properties we might expect under various conditions. Your statement is too vague.

As I pointed out earlier, your 30 km/sec motion of the Earth is relative to the Sun. Relative to the centre of the galaxy, for example, the Earth's velocity is much greater. So, I wonder why you prefer the particular reference frame of the Earth over, say, one centred on the centre of the Milky Way?

Your problem is that you are trying to set up an absolute frame of reference, which both Einstein and Galileo said is impossible. More importantly, such a thing is contrary to all experimental evidence.

Your double standard puzzles me, too. Can you explain to me why it is ok to take the Earth as an "effective zero", but not a reference frame moving at constant velocity relative to the Earth?

That has been disproved by experiment. The Galilean transformations must be replaced by the Lorentz transformations for velocities significant compared to c.

This statement seems to make no sense. I suppose you're using the term "delta function" in a different sense to how a mathematician or physicist uses that term. Perhaps you can explain.

Not specific enough, I'm afraid. Can you please make whatever point you are trying to make with this, explicitly?

How is that relevant to your argument?

This is where you will need to explain what you mean by "anisotropic properties" to me. While you're at it, please explain how anything can have "3pi" degrees of freedom, and what those degrees of freedom are. Your statement seems to be meaningless, as far as I can tell right now.

I'd rather not head off into a discussion of "aether" here, unless you feel that is particularly important. SR does not regard an aether as a necessary concept.

It's a postulate of SR - an a priori assumption. BUT, it's an assumption which accords with all available experimental data.

No. By saying there is an assumed velocity of the frame, you are trying to introduce a universal preferred reference frame to which all velocities must be referred. Such a frame is not required in SR.

You can't be more egalitarian than treating all inertial frames on an equal footing, which is what SR does.

You'll need to explain what lack of anisotropic properties you're thinking of.

Something think about: when we observe the cosmic microwave background radiation from Earth, we see the radiation blue-shifted in the direction of travel of the Earth around the Sun, and red-shifted in the opposite direction. To me, this is an isotropy. What do you think?

I'm not sure how you think that happens.

I'm sorry, but I don't see the relevance of wave-particle duality to your argument (if that's what you're bringing in here).

You're right. Your statement here is incorrect. In fact, SR does not make the assumption that all reference frames measure the same relative velocities between two objects. There is, in fact, a velocity addition formula which handles this very problem.

That doesn't work, though.

Sorry, I can't see it. Perhaps you can explain for me, as I have done for you.

What is this "actual zero" you're thinking of? As I've said, SR dispenses with the concept of an absolute standard of rest.

Once again, all I can see here is unnecessary complication and obfuscation. This would do ... what exactly?

What is a "frame weight data history"?

Did you not read my previous post?

I do not prefer the station frame over the moving frame. It is you who insists on that.

Yes. So?

The equations of physics are form invariant in all inertial frames, yes. What is your point?

If you're sitting on a train, you see the station accelerate. What's abstract about that?

Argument by ridicule is a very weak form of argument, geistkiesel. You might do better to try to form substantive arguments in future.

Yes. This was explained in my previous post. The train is a non-inertial reference frame. The station is inertial. That is true regardless of which one we consider to be moving.

You have made a fundamental error here. The theory of relativity states that the laws of physics are the same in all inertial reference frames, not in all frames. The train is a non-inertial frame, so certain laws of physics need to be modified in that frame (e.g. for Newton's laws to apply, inertial forces must be introduced). The station, on the other hand, is approximately inertial. As I've said, it doesn't matter whether we consider the train or the station to be stationary.

I hope this post has helped clear up a few misconceptions for you.

15. ### geistkieselValued Senior Member

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Read Relativity, AE discuses the matter in detail.. Nonisotropic motion means the motion of the light is the same in 3pi directions when measured with respect to the preferred frame earth., or wst anything else.

No, the earth's relative velocity is actually less than 30 km/sec, its orbital speed. Read Dayton Millers Reviews of Modern Physics 1933 Vol 5 202-243. Where the solar system and wrt Hecules is moving in a southernly direction.

You get you question back the earth's velocy wrt what?

You statement that the earth is moving much faster wrt the center of the galaxy, so what, can you prove it? What is the direction and speed of the center of the galaxy? Read Miller.

I do not determine the laws of physics based on the uterances of dead men. So what that AE and Galilleo said "it couldn't be done". I have been doing just that.

I am not trying to do anything. I use the laws of physics that light is invariant under any motion of the source of the light together with the error < 5xE-12 combined error of the planet earth wrt to V = 0.

James R you can take any reference frame system you desire. If you take the "railway stration > 0" frame and the moving train V = 0, you are reporting physics that is impossible to exhibit. You're assumption of moving frames can be used, but it is silly to muddy up reality with mathmatical slurry.

I use the earth for two reasons and I really would l;ike you to uunderstand this.
1. The motion of light does not exhibit anisotropic properties when originating from earth, and
2. the error in the earth motion wrt V = 0 is such rhat measured errors become trivial.

BS. The fact that accelerated particles exhibit properties that indicate a gowth of ineffiency in the energy exchanges necessary during acceleration does not say that we "must" repalce galilean transformations with lorentzian transformations. When measuring the velocity differences of protons moving at .99c does not mean that light will be measured at gamma(proton). This is because of SR theory, not physics. If light is moving in the +X direction at C and a proton is moving in the -X direction then the relative velocity of proton and light is 1.99C as measured from the proton frame..

No I used it to show how I use the term. I specifically do not parrot nor use the prevailing language of what you term mathematicians or physicists. The effective measure of deviations from a clean 1/2mv^2 energy occurs at a value far removed frame v << c. It is when the difference becomes observable that the mathematical properties of "delta function" arises.

This is a simple differential of the newtonian force term: F = d(mv)/dt. Or F = mdv/dt + vdm/dt. This I know physicists and mathmaticians do not bother to mess with. In any case it sounds like this is the first time you've ever scrutinized the term. Thi is physics 101, you're the professor, you expand the terms.

Read Miller and you will see that he has measured the absolute velocity of the planet earth at approximately 8km/s where we are heading in a southerly direction, constantly. Many other stellar velocities were considered in Miller's deductions. So what is the next step? Got it:trash Miller, right?

Nonisotropic means t
he motion of light in all directions is the same, from whatever source the light originated. Does not 3pi mean x,y,z directions? If not then please amend your copy appropriately.

Me either, except that D. Miller measured an 8km/s relative velocity wrt the "aether", whatever it true nature may ever prove to be.

If it's measured or a priori what's the difference. It is a constant in your SR right? So what?

You don't have to reference all velocities, you can stay on the SR side of he river if that is what you choose to do.

You mean such a frame is "excluded" don't you? I am not concerned with postulates of SR that say sucha frame is not required. It can be used and by using the preferred frame you get back to reality by measuring real relative veocities between light and moving frames. Measure the speed of light wrt V(e) to get C. Then measure the speed of the body of reference wrt V(e) = 0 to determine the relative velocity between frame and photon do the simple math, of subtracting the two relative velocities wrt V(e), a common refeence frame to arrive at,
Vrel = C - Vm,
where both the speed of light and the speed of the material frame Vm were measure wrt the preferred frame V(e) = 0..

Then stay with SR James R, if that is your preference. Take each little frame of reference as it own universe. Discard the concepts of absolute time, insert the sillines of the loss of simultaneity, dilate time, add frame contraction, it is your choice.

Any nonisoptropioc properties are the ones I am thinking of. This is the third or fourth time this question has come up.

I think it merely what you described. Dopplar shifts are what you described. Moving into radiation gives a blue shift, moving away from radiation produces red shifts. Actually this proves more of what I have been saying. The light you are talking about is not light emitted from a moving or accelerating or a velocity = 0 frame of reference. You are aware are you not of the COBE experimental evidence that show an almost perfect distribution of smoothness of the mW background radiation, which to me show that the origins of the MW radiation are effectively hidden in the information free MW field. This smooth field had to be juggled as there was insufficient gradients to explain the formation of matter and hence the would have Big Bang petered out. But the clever scientists moved this and that and just happened to find a tiny variation slightly within the allowed tolerances of measurement to save the BB theory.

Calculating error wrt photons and subtracting out the errors wrt material velocity (the frame of reference) the errors tend to subtract out.

I am using the tems as a comparison that the dual nature of light is a contrivance used to explain how one electron can get through two holes simultaneously.

I stand by my statement. It is fundamental to SR that because C is measured as the SOL in vacuo wrt the earth frame, that moving frames must also measure the SOL wrt the moving frames as C. And isn't this just what occurs?, with time dilation and frame contraction and all the other?

Sure it does. Try it. Use the terra firma as a common reference frame between photon and frame. Your life will become much easier.

Wrt to photon motion I will use the above.Yiour question here is refering to body of references such as trains. Looking at the matter just practically, we do not measure the accelerations that give the train station the velocity
SR uses in calculation involving Einstein gedunkens. using the earth frame as V = 0, then all V > 0 measure wrt to the earth frame has a common V = 0 basis. A measurement taking a millisecond for instance will not include any measurable error due to a direction gradient rotating at < 60, or < 5x10^-12mrad/sec^2 (meter radians / sec squared), nor will a velocty error in the range of 10^-4 be measurable. Using the earth frame as V = 0 is a pactical tool, a reality. Even at the tiniest velocities, the momentum of the measured material wrt V = 0, or any other value, will work every time. The velocity error is only with respect to the measurement of photons and even here no error as measured, the light is nonisotropic in all directions emitted.

Actiual 0 is actual 0. I say it can be achieved, which is not to say that there would be any value in proving the matter. But getting down to tenths or thousandths of m/sec velocities is getting near perfect even for mathematicians.

I am not issuing an SR bulletin in my posts. What SR theory is or isn't I am not not substu=ituting anything fior SR. Take it or leave it.So that SR says anything is not of any guiding use necessarily.

Measue the time a detector measured a radiated photon from another source. It isn't realy critical to understanding of my posts. I introduiced the topic to introduce the topic.Unless your curiosity has reached a burning fever level, I will desist from expanding on the matter. here.

Used in the context of accleration, when observers on frame 1 exchanges accleration and frame weight data one can calculate the others surrent absolute velocity wrt V(e). Knowing the weight of a frame and the acceleration history will provide velocity information will it not?

If time dilation happens then passing ships can determine what the absolute velocity of the other is.

I repeat: train stations do not accelerate for the convenience of mathematicians that have discarded absolute velocity or motion. And James R when is the east time you used all the accelerations of the planet earth to solve some problem concerning the velocity of the earth, and any moving frame? So minute is the error in absolute velocity = 0 for the preferred frame it is immeasurable.

But i always keep the station as V = 0.

The train station is attached to the planet earth.
It doesn't move such that the motion is detecable in measurements of relative velocity.

By using the V = 0 for the planet earth no abstractaions or error are inserted into the physics. By demanding that moving frames must always measure the velocity of light with respect to the moving frame induces an error in the laws of physics. First Einstein makes the point that when measured to the stationary platform that one measures C for the speed of light, when he is really measuring the relative velocity of the speed of light and the V = 0 frame (which he insists is moving and therefore not an inertial frame, except that the emitted light does not exhibit nonisotropic properties when emitted from the "noninertial frame" = a contradiciton, a gross contradiction) .

First you do not see the station accelerate. It is the train that jerks and moves the people. To see the station move, or accelerate is only seen in the minds of an SR theorist. SR is a perception discipline, not a physical discipline. look at the language, the operating pohrase in SR is "The moving observer sees her frame velocity = 0.", but then you close your eyes, paint the windows, take away all the cell phones databases and use observers and crew totally ignorant in physical law.

Your point of view is irrelebvant to the real wiorld. Soyou see what you see. You have never seen a stationaccelrate away from you and you know it. Who has been jerked by a station aaccelerating? Put acclerometers on stations and trains stopped to load and unload and see which ones always read 0? No answer i betcha.
You don't know the difference in an attempt at humor. If you consider that little bit of poking as ridicule, me thinks you might be over reacting.

Here is the "ridicule folks"

quote James R
Me (james R): I've seen plenty of stations accelerate, from my point of view sitting on a train as the station pulls away. I have observed that the people on the platform generally tend to accelerate at the same rate as the station, and therefore their footing on the platform is not affected by the acceleration.

You (geistkiesel): Were you having a draught, or two, or having a few, when you saw the station accelerate awayyyyy

The train is non inertial when accelerating, he station neve accelerates.

I never implied that accelerating frames were inertial. I have been stating the oppsoite. If you read the statement as if I made such a mistake then it is my poor writing that is under fire not my physics.

So the station is approxoimately inertial and so that is how you use it. So do I.

That's funny. I have been blowing the fog from your perceptions. Very ironic, n'cest pas?

16. ### 1100fBannedRegistered Senior Member

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If I have a reference frame attached to a train. The train is not moving wrt it. However if I look at the station and write the position of a point wrt this reference frame I find that its position is x = 1/2at^2. Why do you say that this point is not accelerating wrt the train?

17. ### geistkieselValued Senior Member

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If your reference frame is attached to the train then the train is identical to your reference frame. I don't say that the point is not accelerating wrt the train. As I read your post it is contradictory. Train and reference frame attached, I read that frame is equivalent to body of reference. If there is a point in the stationary frame that is accelerating wrt the frame then the frame and tain are stationary, if you allow accelerating train stations.

A physical impossibility also exists. Train stations do not accelerate wrt trains and frames attached to trains. If SR wants to make such silly assumptions that is certainly their right.

Why don't you simply see that the earth frame, train stations are preferred frames at rest, absolute rest. Then all material frames motion wrt the train station move wrt to an inertial frame that is effectively V = 0. Similarly, light emitted from the earth frame does not exhibit nonisotropic properties, therefore one can consider the earth frame as zero wrt the light motion.

The most embarrassing assumption of relativity is assuming that light measured wrt to the earth frame assumed as the relative vekocity Vr = 0 must also move wrt a moving frame with the same Vr= 0. Therefore, Vrl - Vrm = C - V0, where Vrl is he relative velocity of light wrt the preferred frame and Vrm is the relative velocity of a material frame wrt the V - 0 fame.

If a duck's relative velocity wrt the earth is measured Vd - Vrd = 0, then should a duck's relative velocity Vd - Vf = Vrd?
Should Vd - Vf = Vrd? No, Vd - Vf is not equal to Vrd. So why should light be treated any differently?

18. ### geistkieselValued Senior Member

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No. The earth bound observer measuring the velocity of light Vl wrt the earth, will measure C, which effectively assumes the earth system is V = 0, a preferred frame. Using the same logic on the bullet, and taking the velocity of the bullet with respect to the same preferred frame as the photon, = Vb, then the relative velocity between photons and bullet frame is Vl - Vb = Vrlb, where Vrlb is the relative velocity of the photon and bullet.. Your statement that both stationary and moving observers will measure the same relative velocity follows the fundamental error in SR that assumes that a measure of a Vr = 0, for the earth frame should be applied to Vf moving frames also and that the Vl - Vf should also = 0. This is rudimentary. A duck flying wrt the earth at 5 duck units per sec is not and should not have the same 5 duck units of velocity wrt a frame moving 4 duck units in the same direction as the duck. Vd - Vr = 5 - 4 = 1. Ducks and photons, what is the difference?

You as well as all other SR believers, must prove the assumption that for moving frames with Vf > 0 wrt Vr = 0 that Vl - Vr = C proves unambiguously that Vl - Vf = C. I really hoped to find someSRist knowledgeable enough to prove why one must always measure the speed of light wrt to a moving frame as C.

19. ### geistkieselValued Senior Member

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Let me try answering this again. When i use the word "stratuionary" I mean not moving, stationray. When i determine he erth as an inertial frame with Ve = 0, I mean the earth is a preferred frame wioth a Ve =0. Now, I say, we all know that the earth is moving and cnstantly changing direction.

There are three elements to consider:
1. Light emitted from the earth is measured at C wrt the earth. In oether wiords earth can be considered Ve = 0, not from an assumption bitas a result of observation.
2. similarly the velocity of he earth frame is 10^-4 of the velocity of light, or 4 orders of magnitude less.
3. Light emitted from earth, moving remember, never exhibits any nonisotropic properties.
4. the velocity of light is independent of the motion of the source.
5. Assuming Ve, the velocity of the earth frame for the purpose of measuring the velocity of a frame Vf wrt Ve, if we set Ve = 0 there will be no induced error into the calculations of VL - Vf = Vrlf, even with the known velociy of the earth >0.

The combined motion error of the earth is (dX)(dtheta) < 5 x 10^ -12 mrad/sec^2. whee mrad/sec^2 is meer radiand per seconds squared.

If one set the velocity iof the earth = 0 from an assumoption that the frames are equivalent to the extent that they may be swapped, is to impose physical errors. For example to consider the earth moving, and the Vf velocity = 0, then the real, though measurably insignificant velocity will be erroneous and that error will be infecting SR calculations.

I agree, again.

20. ### James RJust this guy, you know?Staff Member

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geistkiesel:

You are ducking and weaving, and have failed to address many of my direct questions. Nevertheless, I will comment on what you have said.

So, instead of answering, you're referring me to an entire monograph by Albert Einstein. Can't you just give me a brief answer? But wait! Maybe you think you've done that in your last sentence.

You say the "motion" of light is the same "in 3pi directions". I'll get to the "3pi" part below, but what does "motion" mean? Do you mean "speed"? In that case, we are in total agreement, and you are also in agreement with special relativity. Hence, it would seem we have no argument.

I would like you to clarify your position. Could you please answer these questions, for a start?

1. Does special relativity gives the right results mathematically? If so, is it just that you think special relativity is unnecessarily complex?
2. If not, do you believe that Galilean (Newtonian) relativity is correct?
3. If you think neither Galilean nor Einsteinean relativity is correct, what alternative system are you proposing?

From your posts up to the present time, my impression is that you think that that the results of special relativity (such as the one regarding the constancy of the speed of light to all observers) are correct, but you also believe that all the consequences of this results can be reproduced by a simple application of Newtonian physics, referenced to a preferred reference frame (the Earth). What you don't seem to realise is that things like the velocity addition formulae of special relativity and Newtonian physics are different, and incompatible. Your use of the Newtonian formulae in many of your posts is technically incorrect, though a reasonable approximation at low speeds. The bottom line is that special relativity correctly predicts experimental observations, while Newtonian physics does not, except in certain special cases. This is very well verified. Contrary to what you think, special relativity does not introduce unnecessary concepts and fictions; it introduces concepts which physicists have found to be necessary to explain experimental and observational results. If Galilean relativity had been sufficient, of course physicists would have continued to use it, rather than adopting the more complicated system of SR. But Galilean relativity was found to be flawed, first with regard to Maxwell's electromagnetic theory, and then in other respects.

If you want to dispense with the edifice of SR, you will need to also replace Maxwell's equations, and much other physics besides. And you will have to come up with some way to reproduce the predictions of SR, which have been verified by experiment.

Are you saying the Earth's velocity with respect to Hercules is less than 30 km/sec? Fair enough, I am happy to take your word for that. But what now makes you think that Hercules is some kind of absolute standard of rest?

Yes. The rotation rate of our galaxy has been measured, and the motion of the solar system around the galactic centre estimated. The Earth moves at around 200 km/s relative to the galactic centre. I doubt you will find any source which disputes that.

Relative to what? And why is that important?

But your method doesn't work. Earth is not an absolute standard of rest. Even if you're smarter than Galileo and Einstein, you're still wrong.

You're confusing me again. One minute you're claiming that the Earth is an absolute standard of rest, and the next you're saying that the "error" between its actual motion and some arbitrary standard of rest is negligible, and therefore the Earth can be taken to be an absolute standard of rest. So...

4. Which is it? Is the Earth absolutely stationary, or only approximately stationary with respect to some other absolutely stationary thing?
5. If the latter, what is this absolutely stationary thing you're implicitly referring to?

Again, I'd like a few direct answers.

6. If I take a "moving train" to be stationary (v=0), can I make valid physics calculations in that frame and get the right answers? If so, what is wrong with using that frame? And if not, how do you explain the fact that it seems that such calculations work and give useful answers? If the calculations work, doesn't that mean the physics which underlies them is correct?

I dispute (1). If I put a constant light source on the Earth, then sent two observers to Mars and Venus, respectively, each observer would, at most times of the year, measure different properties for the observed light from the Earth. For example, they would see different frequencies. This is an anisotropic property, isn't it? Maybe you can get around this by tightening up your definition. I don't know.

Whether (2) is correct or not depends entirely on what kind of measurement you're making. What is "trivial" for one person may be very significant indeed for another.

Do you have a better explanation?

What evidence do you have for that, other than a gut feeling?

The purpose of language is to communicate, geistkiesel. If you want to communicate effectively, I'm afraid you'll have to agree to conform to some of the conventions of language which are shared by the majority of people. It is fine to define your own "delta function", but realise that practically every mathematician and physicist already has a particular mental concept labelled "delta function". If yours is different, you need to explain to these people, in a way they can understand, exactly how your concept differs from theirs. Without a common definition, there can be no progress in understanding.

Actually, you're wrong. I'm quite familiar with the definition of force in terms of rate of change of momentum in Newtonian physics. I was merely asking you what relevance you attach to this piece of mathematics. I will do as you ask, in the hope that some explanation will be forthcoming.

F = m dv/dt + v dm/dt

or

F = ma + v dm/dt

This is a Newtonian equation says that the force required to accelerate at a constant rate a system whose mass is changing with time, increases as the speed of the system increases, provided the rate of change of the mass (dm/dt) is constant.

Now, this is relevant... why, geistkeisel?

Miller's results were apparently consistent with the hypothesis of no ether drift, though some argue that an ether drift of 8 km/s is not ruled out either.

What's your next step? Are you claiming there is an absolutely stationary ether, and Earth is more or less stationary relative to this ether?

I assume your "3 pi" is an attempt at a fancy description of some kind of angular directional measure. I'm not sure where you got this from, but it seems you have a little learning to do. I'm happy to teach you.

A circle can be divided into various equal-angular segments. The number of segments gives us units of measurement of angles. One such measurement is the degree, defined to be the angle of a segment which is 1/360th of a circle. Another such measure is the radian, which is (1/ 2 pi)th of a circle. Therefore, if you are considering all angles in 2 dimensions, you might refer to 360 degrees or 2 pi radians.

Once you go to 3 dimensions, a new definition of "solid angle" is needed to specify angular regions of the 3 dimensional space. The most common such measure is the steradian. There are (4 pi) steradians in a complete sphere.

Unfortunately, your addition of velocities doesn't work. Take 3 reference frames, A, B and C. If an object in C moves at speed v relative to B, and B moves at speed u relative to A, and all motion is colinear then the speed of C relative to A (denoted as v') would be, in Newtonian physics:

v' = v + u

This is the calculation you keep doing in post after post, attemption to show that if one of the speeds, say u, is the speed of light c, then the speed v' = v + c. (It makes no real difference that you do this in reverse, writing something like v' = v - u and proceding from there).

Unfortunately, this is wrong, although it is intuitive and works well enough at low speeds. In fact, the correct calculation, verified by experiment, is:

v' = (v + u) / (1 + uv/c<sup>2</sup>),

where c is the speed of light.

This point invalidates your entire argument that, if we can find a preferred reference frame, everything will be just fine. It actually doesn't matter whether a preferred standard of rest exists or not - your velocity addition formula would still be wrong, and we would still need special relativity to be able to add velocities in a way which reproduces experimental results correctly.

Thankyou. Will do.

That's not right. Formation of matter wasn't a problem. The observed "clumping" of matter was the problem which led us to expect inhomogeneities in the microwave background. And, very importantly, there was no "juggling" of the data. The data shows what the data shows. The fact that it happens to match what the big bang theory predicted is a vindication of that theory. The data wasn't adjusted to fit the theory. There is no conspiracy, if that is what you're trying to imply.

I'm actually surprised that somebody like you, who seems to have at least a little common sense, would be sucked in by this kind of nonsense.

How do you explain it?

It doesn't work, because of the velocity addition problem I talked about above.

If you're saying that using SR is ok, then you cannot at the same time claim that Galilean relativity is sufficient. The two theories give incompatible results. Both cannot be correct.

You didn't answer the question. Would you like to try again?

Your second sentence is confusing, too. A reference frame is a point of view. How can a point of view have weight?

I have no argument with you that, for many purposes, the acceleration of the Earth can be safely ignored. But that doesn't mean it isn't there. And for some purposes it is very important indeed.

Nor does taking any other frame as v=0.

We need to go back to basics here. Acceleration is defined to be the second derivative of displacement. The station's displacement, relative to the train, changes over time, so the acceleration of the station is very real.

The way that non-physicists use language can be sloppy and misleading. That's why physicists use specific definitions and mathematics to describe things unambiguously. The common understanding of a term like "acceleration" differs quite a bit from what a physicist understands by that term, although the technical meaning is closely related to the common-sense perception in many ways.

I repeat: Many times, I have seen the position of a station with respect to a train change over time, with the station speeding up or slowing down with respect to the train. By the accepted definition of acceleration, the station has accelerated from the point of view of the train. There can be no argument about that. The point of view of somebody on the station is, of course, quite different, but it is not mandatory to prefer one point of view over the other, as you seem to be insisting.

As I have explained previously, the station is approximately inertial, while the train is not. As Einstein realised, it is the difference between inertial and non-inertial motion which is important, not the difference between acceleration and no acceleration.

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22. ### James RJust this guy, you know?Staff Member

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Did you have a point you wanted to make, MacM?

23. ### CrispGone 4everRegistered Senior Member

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He doesn't have a point, that is what makes Mac so charming