# A light speed gedanken

Discussion in 'Physics & Math' started by CANGAS, Apr 28, 2006.

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3. ### CANGASRegistered Senior Member

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JamesR has no books and no references and no memory of anything pertinent, but he knows by divine influence that Relativity, which has failed many, many important tests, just has to be right, because JamesR would lose his job if he did not uphold SR like a fanatic.

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

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Which "important tests" has relativity failed, CANGAS? Name one.

7. ### PeteIt's not rocket surgeryRegistered Senior Member

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Hi Montec,
First, a pedantic note about terminology. Your use of "frame" is a little different to the norm. A frame of reference, as used in SR, is of infinite extent in space and time. All events and all things exist in all frames, not just one or some. The difference between frames is in their origin, their orientation, and their rest velocity.

So strictly, you should say that the observer's rest frame has some velocity with respect to the light source rest frame, or equivalently that the observer has some velocity with respect to the light source.

But that doesn't really matter because your meaning is clear.

Right, so if I understand you correctly, there is some circling detector that moves slowly around the light source, remaining a constant distance from the light source in the light source's rest frame, and there is an inertial observer that is moving at high speed with respect to the light source.

The question is what intensity the circling detector detects throughout the journey, and how the inertial observer accounts for it.

Now, you said:
Allow me to clarify:
The motion of the ship in a given inertial reference frame at the instant of emission determines the direction of emission in the given frame.
Since a detector can only directly detect the intensity and direction of the light in the detector's rest frame, the motion of any other observer is irrelevant. The only thing that matters is the relationship of the detector to the source - the relative motion between them.

However, that might be an unsatisfactory answer... so just for the exercise, let's examine the scenario in the inertial observer's rest frame. Essentially, there are two effects in that frame - an effect on the emitted light due to the motion of the source, and an effect on the detected light due to the motion of the receiver.
Breaking it down, in the observer's rest frame:
• the light source is moving at high speed,
• the light is bluer and more intense in front of the source,
• the light is redder and less intense behind the source,
• in all directions except directly in front and behind the source, the light is not moving directly away from the source, but away from a point some distance behind the source,
• the detector is also moving at high speed, keeping up with the light source and in fact circling it slowly,
• when the detector is in front of the light source, it is intercepting intense, blue light
• however, the detector is moving at high speed away from the light, so it actually measures not so intense, and not so blue light
• when the detector is behind the light source, it is intercepting weak, red light
• however, the detector is moving at high speed into the light, so it actually measures not so weak, and not so red light
• when the detector is beside the light source, it intercepts light which is coming from some distance to the rear
• however, the detector is moving a high speed, so the measured direction is shifted forward
The end result is that the detected intensity will be constant, and always coming directly from the light source.

8. ### MontecRegistered Senior Member

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248
Hi Pete
I think I understand. Let me see if I can explain it.

If you take a hollow sphere with photon/light detectors coating the interior surface. Put a light source at the center. Let the sphere have a velocity. The point of emission is now no longer at the center of the sphere due to the speed of light delays. A uniform intensity cannot be measured unless more photons are directed forward. This would satisfy SR's second postulate. However, this brings up a question of an absolute reference frame since the velocity of the above sphere is relative to what?

9. ### Janus58Valued Senior Member

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No, he is saying that the light intensity hitting the detectors will be the same at all points of the sphere according to both the detectors and the observer the sphere is moving with respect to. However, the light detected by that observer will vary in intensity depending on his position realtive to the motion of the sphere. This is true whether you consider the sphere moving, the observer moving or any combination of both observer and sphere moving.

10. ### MontecRegistered Senior Member

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248
Hi Janus58

I understand that the intensity inside the sphere must be uniform no matter who measures it and no matter what their relative velocity is. I am trying to understand the mechanics of how light does this.

To maintain a uniform intensity the light source must respond by directing more photons to the front. This is because the photons have further to travel in the forward direction when the sphere has a velocity. The loss in intensity through the inverse square law must be made up by increasing the photon density. Just the opposite must be done to the photons going to the reverse direction.

To maintain the direction to the center of the sphere and the light source, the detectors via aberration of light, cancels out both the displacement between the point of emission and actual position of the light source and the intensity maintenance vector.

These two maintenance functions satisfy SR's second postulate. As for an external observer that has a velocity with respect to the sphere then normal SR rules apply with consideration applied for the vector change due to the sphere's velocity (if any), aberration of light at the observer (if any), and the relative velocity between the sphere and observer.

11. ### geistkieselValued Senior Member

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Explain please how the light moves along with the moving frame. Is there a momentum component parallel with the moving frame? If so what experimental results support your contention? I know of no such data. The Michelson-Morely experiment are described as you assert. It is said that the light moving 90 degrees to the motion of the frame moves in a triangular path, and is "carried along" with the frame, even when observed from the ground.
Geistkiesel​
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12. ### geistkieselValued Senior Member

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I was using Einsteins description where he used the embankment as the vacua. There is no0 problem of removing the train wiondows ro assure idenmtical envirnonments of train and embankment. Using the resolution to the Twin PAradox that states that the accelerated twin is the one who was accelerated therefore he will age slower than his twin that remained on earth. This requires that the concept of equivalemnt frames be scrapped.

It is the train in motion, not the embankment, however observed by passengers on the train. The passengers feel the acceleration and see no equivalent forces applied to the embankment.

I did not give the embankment a preferred status. I merely pointed out that Einsteins description of the light measurements were all taken wrt the embankment. AE arbitrarily jumped frames and attempted to use the expression as ikf measured wrt the train as the inertial frqame of reference. AE complained that the expression, Vle = Vte = Vle shows light speed on the train as less than the speed of light on the embankment. The expression merely is a calulation of how much faster is the speed of light than the train, or the relative velocity of frame and photon measured wrt the embankment.

Geistkiesel ​

13. ### przyksquishyValued Senior Member

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The new component velocities can be found by a direct application of the Lorentz transformation. If the photon is moving in the y direction in the emitter frame, then it's motion can be represented in the parametric form (the "e" subscript meaning x, y, and t in the emitter frame):

x<sub>e</sub> = 0
y<sub>e</sub> = ct<sub>e</sub>

(assuming the emitter is located at (0, 0) and the photon is emitted at t<sub>e</sub>=0)

If the emitter itself is moving at velocity v in the x direction in what we'll call the "rest" frame, you can find the photon's motion in this frame by applying the Lorentz transforms to the parametrizations above to get x<sub>r</sub> and y<sub>r</sub> in terms of t<sub>r</sub>:

x<sub>e</sub> = &gamma;(x<sub>r</sub> - vt<sub>r</sub>)
y<sub>e</sub> = y<sub>r</sub>
t<sub>e</sub> = &gamma;(t<sub>r</sub> - vx<sub>r</sub>/c<sup>2</sup>)

[with &gamma; = 1/sqrt(1 - v<sup>2</sup>/c<sup>2</sup>)]

Substituting and a little algebra gives:

x<sub>r</sub> = vt<sub>r</sub>
y<sub>r</sub> = ct<sub>r</sub>/&gamma;

So the photon's velocity in the rest frame is v in the x direction and c/&gamma; in the y direction (in general the x velocity would be the result of the Lorentz velocity addition, and the y velocity would slow by the factor &gamma

I don't know why this result should surprise you - intuitively you'd expect the emitter's motion to affect the direction in which the photon travels.

Last edited: Jul 8, 2006
14. ### Neddy BateValued Senior Member

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1,677

The "train & embankment" gedanken neglects accelerations. The passengers would not feel any frame-forces because the train does not accelerate at any time during the course of the experiment.

The apparent "clock paradox" is that the train clocks must tick slow in the embankment frame, while the embankment clocks must tick slow in the train frame. The resolution to this paradox does not require accelerations. All that is necessary is that the concept of aboslute time be scrapped in favor of "loss of simultanaety". In other words, clocks which are synchronized in one frame are not synchronized in other frames.

Did you ever get to see the ClockGraphic.pdf that I posted a while ago? I think you were away at that time and you might have missed it. Even though I was the one who created the graphics, it was Pete who spent quite a bit of time helping me understand the concept, so I like to give him credit whenever I post the link to it. It is a direct result of the Lorenz transforms, and it resolves the train/embankment "clock paradox".​

Using the embankment as a reference body, it is true that the speed of light relative to the moving train is

w = c - v

If that is all you were saying, then I agree with you. I thought you were trying to say that using the train as a reference body (in other words, from the inertial frame in which the passengers are at rest) that the speed of light would be c-v. Of course the whole point that Einstein was trying to make was that the speed of light should be c, and not c-v, from the inertial frame of the train. He says as much in the following excerpt from Relativity: The Special and General Theory. 1920.

Chapter VII. The Apparent Incompatibility of the Law of Propagation of Light with the Principle of Relativity​

15. ### geistkieselValued Senior Member

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The twin experiment carries the same conditions. The question is, which frame is movi8ng from the condition when both were at the same v elocity.
You are saying that after the train has moved for a bit and strikes a mechanical switch that immediately transfers the current embankment time to the train (say total elapsed number of ticks) and the train transfers its accumulated clock ticks to the embankment that the train observer will see a number associated with the embankment ticks that is less than the number of accumulated train ticks.

Likewise, the embankment observer will see a number of train ticks less than thea accumulated embankment ticks.

Each observer compares the ticks to their respective frames when received. The sxplanation you have submitted is a physical impossibility

OK I agree , but in the book Einsteiun did nid not use the expression w = c - v correctly. Your comment about what you thought I wasd saying was what Einstein said, bur he used the expression w = c - v as if the measurement was taken from the train as the reference, and then he criticized this point stating that from the law of light motion the same laws should apply in all inertial frames and then said without discussion of comment that the error in the w = c - v expression must be corrected with the Lorentz transforms and whatever.

At no time when Einstein was developing the expression is there any reading that the way the gedanken was set up that w = c - v was an expression pertinent to the train as the reference frame. There is no error, as you observed, when the expressiuon is measured from the ewmbankment. Einstein did not justify his criticism that w < c is in error as the speed of light should be measured the same, when the equivalence of frames means only that the laws governing light motion should be the same, as it is with all other entities in the universe, not that one will, or must, always measure c as the relative velocity of frame and photon.
Geistkiesel ​

16. ### DaleSpamTANSTAAFLRegistered Senior Member

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What if they were never at the same velocity? (e.g. in particle annihlation/creation)

-Dale

17. ### CANGASRegistered Senior Member

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Dale: If you have a case to make, then MAKE IT!

Don't just ask a d*mn question. If we are as dumb as you think we are, so we don't get the important point that you understand, how are we going to get it just by hearing you ask us why don't we get it?

Are you Pete's understudy for chief queryist?

Einstein pulled the wool over a lot of people's eyes with the fancy prestidigitation that gk is trying to explain to you in a way that you can understand it. Please, for a moment, stop reacting as if your life or job depended upon your claiming to believe that Relativity can do no wrong, and just examine the presentation to see if it makes sense to you.

Last edited: Jul 10, 2006
18. ### DaleSpamTANSTAAFLRegistered Senior Member

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ROFL! Yes sir, CANGAS sir! As chief of obfuscation, uncited quotes, and vague references to unsupported claims you certainly are the expert on making a case clearly

.

The point is that two objects may have relative velocity without ever having accelerated (e.g. in particle annihlation/creation). The assumption that there was ever some moment "when both were at the same v elocity" is demonstrably wrong in some cases. Any theory that requires such a history in order to make predictions will inevitably fail in such cases.

A weaker, but just as important point, is that even if there is a historical "mutual rest" frame we may not always be able to determine it. Any theory that requires knowledge of such a history will fail in such cases also.

Since SR makes accurate predictions without requiring the history it is preferable to any theory that makes similar predictions and requires the history. Is that clear enough for you, CANGAS?

-Dale

19. ### geistkieselValued Senior Member

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The train leaves the station by acelerating from a dead stop. Before the train leaves, clocks on the embankment are synchronized to the clocks on the train. The train gathers speed and ceases accelerating. The train and ambankment clocks have been maintianing a record of the number of clock ticks each makes. After a few hours the train strikes a mechanical switch located on the embankment. The switch triggers the transfer of the accumulated ticks on the train and embankment to the other.

Question:

A. Will the train number be less than the number from the embankment?
B. Will the train number be greater than the embankment number?
c. Will the numbers be equal?

Geistkiesel ​

20. ### geistkieselValued Senior Member

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2,471
After the train moves for a while after initially synchronizing their clocks the train streikes a mechanical switch located on the embankment which triggers a mutual exchange of the number of accumulated ticks on each clock to the other frame.

A. Will the number generated by the train be greater than the number received from the embankment?
B Will the number generated by the train be less than the number received from the embankment?
C. Will the number generated by the train be the same as the number received from the embankment?

Geistkiesel​

21. ### DaleSpamTANSTAAFLRegistered Senior Member

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If the train accelerates from a (relative) dead stop then the train and embankment were once at the same velocity. What if they were never at the same velocity?

-Dale

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

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Yes.

No.

No.

This is the twin paradox again. Haven't you worked this one out yet?

23. ### MontecRegistered Senior Member

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Hello all

A more acurate "twin paradox" analogy would be two trains, A and B, on parallel tracks moving at some velocity. Train A decelerates to V1 while train B accelerates to V2. Which train has faster clocks?