Phase Change in Light: A & B Signal Reflections

[AlMetis]

I would appreciate knowing if there is any error in the mechanics illustrated in this animation with respect to phase change in the signal reflections shown at A and B, or in the description accompanying it.
I have embedded the description with the animation and included it in this post for reference and quoting purposes.
Thank you.

The rigid rod AB is at rest in the frame K. The constancy of the speed of light and the anisotropy of space supports the following conjecture on the mechanics of light observed by and between A and B:
A emits a light signal toward B at the A time At0, it is reflected at B at the B time Bt0 and returns to A at the A time At1. B emits a signal toward A at the B time Bt0, it is reflected at A at the A time At1 and returns to B at the B time Bt1. A and B each mark the total time (Tt) of their own signals ABA and BAB respectively. Each set the frequency of the signals they emit to Tt/2. A and B each send the first of these signals on the anniversary of the previous emissions. (i.e. Atn and Btn for A and B respectively)
The animations show the Tt/2 signals. When AB is at rest in K the signal emission and reflection times are in phase. When AB is in motion relative to K the signals emission and reflection times are out of phase.

 

[Dale]

This is nonsense. Your emission of the blue flash on the top has nothing at all to do with the emission on the bottom. You need to use the Lorentz transform if you want them to be related to each other. Just randomly animating things doesn’t mean anything.

Phase is a relativistic invariant. All frames agree on the phase at any event

 

[PeterDonis]

I would appreciate knowing if there is any error in the mechanics illustrated in this animation

Yes, there is. If the light signals are in sync when they reach A and B in one frame, they must be in sync in every frame. The light signals reaching A and B are invariant events.

 

[Vanadium 50]

First, I think this shows how easy it is to come up with a clever graphic that doesn't match reality.

Next, @AlMetis , what is your goal? If it is to try and convince us relativity is wrong, PF isn't the right place for it. If it's something else, maybe you should be explicit about it so people don't misunderstand.

 

[Sagittarius A-Star]

View attachment 323055
A emits a light signal toward B at the A time At0, it is reflected at B at the B time Bt0 and returns to A at the A time At1. B emits a signal toward A at the B time Bt0
...
A and B each send the first of these signals on the anniversary of the previous emissions. (i.e. Atn and Btn for A and B respectively)

Your animation shows an mix of Einstein's relativity and Galilean relativity. If you call the rest-frame of the rod $K'$', then you must distinguish the time $t$ in frame $K$ from time $t'$ in frame $K'$.

The first emission of B happens in the top animation and the bottom animation at the same time $t$, but in reality it happens at time $t'$, which is only in the top scenario equal to $t$.

Also, your animation doesn't show length contraction.

Maybe, a calculation of the moving, longitudinal light-clock is helpful.

 

[Ibix]

Your animation shows an mix of Einstein's relativity and Galilean relativity.

...the same as the last thread, then.

 

[AlMetis]

The animations are not a comparison of observations from two different reference frames.

Both animations are relative to K.

The time of emission from B, the blue signal, is the same time in both “animations” whether you choose a clock at rest with K or AB.
To establish this time of emission we divided the round trip times in half when AB is at rest with K, ABA/2 and BAB/2
When the time AB = BA. which is true when AB is viewed from and at rest with K, the emission of blue from B will coincide with the reflection of red at B.
When AB is in motion relative to K, AB≠BA. The first leg of the red signal AB does not equal the first leg of the blue signal BA. This puts the times of reflection of red and blue out of phase.

 

[Dale]

The animations are not a comparison of observations from two different reference frames.

Both animations are relative to K.

Ah, sorry. I misunderstood.

It is certainly possible to build such a pair of devices: one that is in phase and another that is not in phase.

This has nothing whatsoever to do with relativity, and as far as I can tell demonstrates no important physics principle. But it is physically possible to build two such devices.

These are two physically distinct devices, not identical devices operated in relative motion. In any frame the out of phase device produces out of phase signals and the in-phase device produces in-phase signals. Again, phase is a relativistic invariant.

 

[PeterDonis]

When AB is in motion relative to K, AB≠BA.

Yes, but that's irrelevant. See below.

The first leg of the red signal AB does not equal the first leg of the blue signal BA.

So what? The important thing is that the second leg of the red signal does equal the first leg of the blue signal--because both are going in the same direction. So if the blue signal is emitted at B at the same instant as the red signal arrives at B, then they will both remain in sync forever. And if you want the "moving" scenario to be physically equivalent to the "stationary" scenario, then you have to have the blue signal emitted at B at the same instant as the red signal arrives at B, because that is how the phase relationship between the signals is defined.

This puts the times of reflection of red and blue out of phase.

No, what put them out of phase is that you changed the scenario. You defined the "moving" scenario as physically different from the "stationary" scenario by having the red signal emitted from B before the blue signal arrives at B, instead of at the same instant. That's a change you made in specifying the second scenario. It tells you nothing whatever about the laws of physics. All it tells you is that you changed the scenario.

 

[Ibix]

No, what put them out of phase is that you changed the scenario. You defined the "moving" scenario as physically different from the "stationary" scenario by having the red signal emitted from B before the blue signal arrives at B, instead of at the same instant. That's a change you made in specifying the second scenario. It tells you nothing whatever about the laws of physics. All it tells you is that you changed the scenario.

This seemed to me to be the core of the problem in @AlMetis's last thread - defining several different scenarios and trying to see some fundamental physics reason why they were different. There is none - they were just defined differently.

 

[AlMetis]

Yes, but that's irrelevant

Thank you PeterDonis
It appears you understand the animations as I intended, and you agree with them.
But you and everyone else posting here seem to think I have a problem with them.I don’t want the moving scenario to be physically equivalent to the stationary scenario.
At no point have I said this “out of phase” is a problem, or something I want to change, or I expect to be physically equivalent in different frames.
The time of the blue signal emitted from B is the time agreed to by K and AB as equal to 1/2 the round trip timed by both, when at rest relative to each other.
The top animations shows the signal reflection times are in phase using this agreed time when K and AB remain at rest.
When one, or the other is set in motion, AB≠BA therefore when using the time agreed to at rest the signal reflection times go out of phase. Remember that the first emission on this “agreed” time will be from the anniversary of, or a multiple of At1 and Bt1 for A and B respectively.

Using the agreed time is why you see blue emitted from B before A reaches B in the second animation. AB is greater than BA when the motion of AB is in the direction of B, in this case increasing x, which means the “agreed time” is reached before the red signal reaches B, therefore the blue signal is emitted before the red signal reaches B.Maybe that is a better description?
I hope this makes it clearer.

I am not trying to convince anyone relativity is wrong, and as Vanadium 50 said, “PF isn’t the right place for it.” What I am trying to do is make sure I can clearly describe this phase shift to people familiar with relativity.

 

[robphy]

Maybe that is a better description?
I hope this makes it clearer.

I am not trying to convince anyone relativity is wrong, and as Vanadium 50 said, “PF isn’t the right place for it.” What I am trying to do is make sure I can clearly describe this phase shift to people familiar with relativity.

In my opinion, while animations of boxcars in space are nice,
an even better description is a spacetime diagram of the situation.

It's more in the spirit of relativity as a more-invariant geometrical description of spacetime.
Often many solutions to puzzles in relativity are clarified by a good spacetime diagram,
which can be [with practice]
analyzed in ways similar (analogous but not identical) to high geometry methods.

My $0.02.

 

[Dale]

I don’t want the moving scenario to be physically equivalent to the stationary scenario.
At no point have I said this “out of phase” is a problem, or something I want to change, or I expect to be physically equivalent in different frames.

So then what is the point? Do you have a question?

What I am trying to do is make sure I can clearly describe this phase shift to people familiar with relativity.

The phase shift has nothing to do with relativity. So you can describe it by saying “in the second case, on the right hand side we choose to generate a phase shifted signal”.

 

[PeterDonis]

I don’t want the moving scenario to be physically equivalent to the stationary scenario.

Ok, then what's the point?

 

[PeterDonis]

What I am trying to do is make sure I can clearly describe this phase shift to people familiar with relativity.

What phase shift? The only "phase shift" is due to the fact that you chose to specify the two scenarios differently. There's no physics in it at all, just different specifications by you.

 

[AlMetis]

In my opinion, while animations of boxcars in space are nice,
an even better description is a spacetime diagram of the situation.

I will give light cones my best shot, and leave spacetime diagrams to those more capable.

 

[jbriggs444]

It is impossible to accelerate a rigid rod. Or, to put it in less jarring terms, it is impossible to accelerate a body while retaining its rigidity according to both its original rest frame and its new momentary rest frame.

If you initiate the acceleration simultaneously throughout the length of the rod according to the initial rest frame then you will not have initiated the acceleration simultaneously throughout the length of the rod according to the final rest frame -- accordingly, the rod will not have been "rigid" throughout the event.

It is also impossible to transmit compression waves in a rigid rod.

 

[Dale]

I will give light cones my best shot, and leave spacetime diagrams to those more capable.

Those are nice diagrams but there is still no physics content to the scenario, regardless of how you display it. You have simply deliberately introduced a phase shift in one case and not in the other. Why you choose to introduce the phase shift for the “moving” one is a question of your own personal decisions, not a question of physics.

 

[robphy]

I will give light cones my best shot, and leave spacetime diagrams to those more capable.

To address the matter more clearly,
a position-vs-time diagram with worldlines is more probably more useful.
Often this is simpler to construct... and would be more enlightening.

The light-cones would probably be more useful for selected events (not all of them).

The phase-shift would presumably be analogous to the time ticked off a wristwatch, possibly with a different rate and possible discontinuities [like a phase-change on reflection].
It's not clear how light-cones would help at this stage...
since the problem statement itself isn't clear [to me].

UPDATE: Text-renderings in the image, but not the text itself in the post is probably discouraged
(and probably leads to fewer viewers paying attention to your questions).

 

[AlMetis]

You have simply deliberately introduced a phase shift in one case and not in the other.

I wasn’t aware I “chose” to set a phase shift for the moving AB. The time it takes light to travel from A to B is longer (for both observers) than the time to travel from B to A when AB is moving toward the direction of B. Are you saying that is not the case?

 

[Dale]

I wasn’t aware I “chose” to set a phase shift for the moving AB.

Which is why I have pointed it out multiple times.

Notice that in the moving apparatus the emission from the right side occurs before light reaches it. So the emission from the right is not caused by the emission from the left. It simply happens according to some pre-arranged decision. Its timing is not derivable from any of the physics of the scenario.

The time it takes light to travel from A to B is longer (for both observers) than the time to travel from B to A when AB is moving toward the direction of B.

That is not the case.

 

[AlMetis]

It simply happens according to some pre-arranged decision. Its timing is not derivable from any of the physics of the scenario.

The emission time of the Blue light from B is not pre-arranged, it is a calculated time based on c and the Einstein convention of two-way light speed as follows:

The emission of Blue light from B is initiated by the Red light arriving at B, the first time, and as many times as B chooses to test the total time of the light path BAB. “A” performs the calculation to emit Red light with the arrival of Blue light from B coinciding with the reflected Red light as the total time ABA.
Once each has independently confirmed a consistent (two-way) time for their own measures, they send their respective signals at a frequency that is 1/2 the total/two-way time. These signals also arrive synchronized at A and B according to Einstein synchronization when AB is at rest relative to K.

 

[AlMetis]

That is not the case.

When the observer in the middle of a moving train disagrees with the observer on the platform, on the simultaneity of two lightening strikes hitting the front and back of the train, it is because, as Einstein said, it takes longer for the light from the back of the train, than the light at the front of the train to reach the observer on the moving train.

If you think that is not the case for light traversing A to B and B to A, which amounts to the same as back to front and front to back of a moving train, please explain why.

 

[Dale]

The emission of Blue light from B is initiated by the Red light arriving at B

In your animation the Blue light is very clearly emitted from B before the Red light arrives at B. That is precisely the problem. That requires a prearranged emission at B. Otherwise some faster-than-light signal from A would be needed.

 

[Ibix]

In your animation the Blue light is very clearly emitted from B before the Red light arrives at B. That is precisely the problem. That requires a prearranged emission at B. Otherwise some faster-than-light signal from A would be needed.

Note that both blue flashes are emitted simultaneously in the frame in which the animation is drawn. I think the lower rod is, for some reason, using clocks de-synchronised and rate-adjusted so that they match those in the frame where it is moving in order to time the emission of the blue light. Disrupting the functioning of your clocks is a silly thing to be doing if you're trying to test something to do with speed, but it's possible to do.

 

[Dale]

I think the lower rod is, for some reason, using clocks de-synchronised and rate-adjusted so that they match those in the frame where it is moving in order to time the emission of the blue light. … it's possible to do.

Certainly it is possible. That is exactly the sort of mechanism I had in mind by “pre arranged”. They could pre-arrange the clocks to match the “rest frame” clocks and they could further pre-arrange the blue signal to be emitted at a specific time according to those clocks.

@AlMetis is under the misapprehension that this is something that will happen without such prearrangement. They have not explicitly stated the prearranged mechanism, but there must be one because otherwise a FTL signal is required

 

[PeterDonis]

The emission time of the Blue light from B is not pre-arranged, it is a calculated time based on c and the Einstein convention of two-way light speed as follows:

You seem to be under the misapprehension that, in a frame in which A and B are moving, this procedure will lead to signals that look like the second diagram you drew. It won't. It will lead to signals that look like the first. By your description, A and B use their own clocks in the setup process, and then again use their own clocks to determine their emission times. That, combined with the simultaneity convention that results from their using Einstein synchronization (which is basically what your process amounts to), means that everything will look the same in any frame as it does in A and B's rest frame. The signals will always be in sync as they are in your first diagram.

In short, the only way to get the signals to look the way they do in your second diagram is to not use Einstein synchronization, but to use some other pre-arranged procedure to determine when B, by B's clock, emits his first signal.

 

[PeterDonis]

I think the lower rod is, for some reason, using clocks de-synchronised and rate-adjusted so that they match those in the frame where it is moving in order to time the emission of the blue light.

That could be what @AlMetis intended, but it is not what will result from the process described in post #22.

 

[Ibix]

That could be what @AlMetis intended, but it is not what will result from the process described in post #22.

I don't think what's described in #22 is self consistent - the last sentence implies all synchronisation is relative to K, while the rest reads like a textbook Einstein synchronisation. The diagram in the OP certainly isn't consistent with either interpretation. Look at the lengths of the rods - eyeballing the speed at ~0.25c they ought to be about 5% different in length if they're meant to be the same rod, but they aren't, so they represent different systems.

The whole thing is a mess, with the OP unable to see that he keeps getting different results because he keeps defining different experiments and trying to interpret them as if they were the same. Unsurprisingly he ends up with contradictions and confusion.

 

[Sagittarius A-Star]

The time it takes light to travel from A to B is longer (for both observers) than the time to travel from B to A when AB is moving toward the direction of B.

I will give light cones my best shot, ...
...

In your last picture you wrote:
"On the anniversary of At₁, AB is set in uniform motion relative to K on x in the positive x direction".

Did the observers at A and B re-synchronize their watches with reference to their common rest frame after the end of the short proper acceleration of the Born-rigid rod "AB"?

 

[Nugatory]

When the observer in the middle of a moving train disagrees with the observer on the platform, on the simultaneity of two lightening strikes hitting the front and back of the train, it is because, as Einstein said, it takes longer for the light from the back of the train, than the light at the front of the train to reach the observer on the moving train if we choose to analyze the situation using coordinates in which the platform is at rest

The bolded addition is necessary, but even with that addition you are misunderstanding Einstein's position. Yes, it "it takes longer for the light from the back of the train, than the light at the front of the train to reach the observer on the moving train" when using the platform frame, and Einstein did correctly rely on that fact, but that's not Einstein's insight here.

Einstein's insight is that the most reasonable ("only sensible" would not be excessive hyperbole here) definition of "time a flash of light was emitted" is to subtract the light travel time from the arrival time to get the emission time, and that when using that most reasonable definition observers in relative motion to one another will not agree on the simultaneity of events.

The problem with describing the thought experiment as you have is that you have inadvertently privileged the platform observer - the "takes longer" explanation loses the equally reasonable analysis that says both light signals covered the same distance but were emitted at different times.

 

[jedishrfu]

With @Nugatory excellent comment, I think this is a good time to close this thread.

Thank you all for participating here.

Jedi