What Are the Paradoxes of the Light Clock Problem in Special Relativity?

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The discussion centers on the paradoxes arising from light clock diagrams in special relativity (SR) and how light is observed. Participants argue that while the formulas and time dilation concepts of SR are sound, the diagrams can lead to logical inconsistencies, particularly regarding how light is emitted and detected. Key points include the assertion that light detection is a local event, complicating how distant observers can interpret light paths. The conversation also questions the validity of diagramming light at angles, suggesting that such representations may not accurately reflect physical reality. Ultimately, the debate highlights the need for clarity in visualizing light behavior within the framework of SR.
  • #31
My difficulty with that drawing is that the moving emitter is giving sideways momentum to the photon. Is that possible? Shouldn't the photon leave the emitter and miss the mirror on the other side or miss the detector on the way back?
 
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  • #32
OK... so I think what you are saying is that the light clock diagram must be viewed as an inferred and possibly flawed interpretation of what is remotely happening, not an existential representation.

If so, and the paradox stems from the current diagram, can the diagram be saved by altering it or is there no proper diagram possible for representing the remote behavior of the light path?
 
  • #33
I think this diagram and the point it is trying to make can be saved by changing the direction of motion 90o
 
  • #34
altergnostic said:
It is real in all frames, but an "observer" can't assign coordinats of events he is not aware of. Those light paths are not seen by a distant observer, so he can't diagram them. If you want him to know the times of the reflections, you have to send him a signal at each reflection event. As it is, the light clock diagram assumes he can actually see those light paths, which easily leads to paradoxes. Also notice that if he really sees those light paths at an angle he will actually measure light going faster than c, and only after the transforms will he have a dilated time and the speed of light normalized to c. But SR and experiment shows that no observer can measure light going faster (or slower) than c. That's why he should use light that reaches him directly, which he will directly measure going at c, to do the transforms in the first place.

This is not correct. The observed diagonal paths in HarryLin's proposed set up would always produce c as a direct calculation of the distance and detection times (D/dt5) in the LAB frame.
No transformation whatsoever required in that frame.

Exactly as would be the case if the mirror was removed and the photon continued on to a detector at rest in the lab frame.
 
  • #35
Please explain how photons are never given momentum from the source but a change in frequency due to the momentum of the source yet can be given the lateral momentum of the sideways motion of the source to cause them to continue across the moving lab and still hit the mirror and come back to the detector.
 
  • #36
bahamagreen said:
OK... so I think what you are saying is that the light clock diagram must be viewed as an inferred and possibly flawed interpretation of what is remotely happening, not an existential representation.

If so, and the paradox stems from the current diagram, can the diagram be saved by altering it or is there no proper diagram possible for representing the remote behavior of the light path?

It seems clear that the diagram is "an inferred and possibly flawed interpretation" - it's a picture for crissakes! - How could it be anything else?

However, it can be saved easily. Just erase the lines that purport to show the path of the light, so that only three events appear in the picture.
A) Flash of light is emitted from bottom mirror in conformance with Maxwell's equations applied locally at the moment of emission.
B) Flash of light hits upper mirror and is reflected, again in conformance with Maxwell's equations applied locally at the moment of emission.
C) Flash of light is is detected at the lower mirror.

We eliminate the concerns (especially Altergnostic's concern in #24 that we cannot assign coordinates of events we are not aware of) about how these events are observed by using the procedure described in Taylor and Wheeler's "Spacetime Physics" (this procedure is, BTW, one of the better ways of avoiding the "what is a frame?" quagmire). We scatter observers, at rest relative to each other and carrying synchronized clocks, throughout space. Whenever anything happens right under an observer's nose, so only local considerations apply, the observer writes down what happened and the time it happened on a slip of paper. At some later time, we collect all the slips of paper and piece together a complete global picture.

We do all of this, and where's the paradox? The local angle each observer reports is exactly what you get by Lorentz-transforming from what observers at rest relative to the mirrors see.

Do those lines in the diagram that purport to show the path of light between these events represent anything "real"? I think so, for the same reason that if I see a mouse enter one end of a length of pipe, and later see a mouse emerge from the other end, I'm inclined to think that I've observed a mouse scamper from one end of the pipe to the other. But I have no quarrel with anyone who tells me that without observations of the mouse's passage through the pipe I'm just offering an interpretation.
 
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  • #37
marty1 said:
Please explain how photons are never given momentum from the source but a change in frequency due to the momentum of the source yet can be given the lateral momentum of the sideways motion of the source to cause them to continue across the moving lab and still hit the mirror and come back to the detector.

I certainly can't explain it , any more than it can be explained in Newtonian mechanics.
But it appears as if photons act just like Newtonian ballistic particles only with a certain difference due to the limit of c.
In low velocity conservation (of massive particles) the conservation is complete and directly additive.
This cannot be the case with light (or relativistic massive particles), as is clear if you look at a photon emitted along the vector of motion. The forward motion of the source cannot add to the forward velocity of the photon which is already maxed out.
I guess it might be looked at as the forward momentum being expressed as Doppler blueshift (or redshift for anti-parallel emission) but I don't know if that is an accepted view.
On the other hand, photons emitted orthogonally to motion conserve the forward momentum,( as motion ),completely, while the change of momentum through frequency shift of the intrinsic emission frequency is at a minimum (zero at some point close to 90deg.) .
SO the percentage of forward momentum of the source conserved as forward motion is a function of angle.
I hope this might be helpful.
 
  • #38
But it not just the maximum they cannot exceed in the direction of travel. They also cannot be slowed down when emitted from a source moving away. They are instead red shifted. Momentum in every other direction seems to be completely Newtonian. Do they "know" which direction they are heading and follow different laws? Do they have a head and a tail?
 
  • #39
marty1 said:
Please explain how photons are never given momentum from the source but a change in frequency due to the momentum of the source yet can be given the lateral momentum of the sideways motion of the source to cause them to continue across the moving lab and still hit the mirror and come back to the detector.

No such thing happens - the lateral component of the momentum doesn't change during the reflection. The vertical component changes sign, switching directions.
 
  • #40
marty1 said:
But it not just the maximum they cannot exceed in the direction of travel. They also cannot be slowed down when emitted from a source moving away. They are instead red shifted. Momentum in every other direction seems to be completely Newtonian. Do they "know" which direction they are heading and follow different laws? Do they have a head and a tail?

They are not slowed down in any direction. The forward momentum is conserved wrt photons emitted to the rear with the same restrictions and angle dependence as applies to forward emissions. That forward momentum does not change the speed in either frame but it changes the angle of propagation. Put another way it changes the x component of the velocity vector , reducing the motion in the -x direction but increasing it in the y and z directions relative to the lab frame but with a constant speed.
The result is the vector is tilted forward in the direction of motion in the lab frame.

So the amount of motion conservation is zero at 0 and 180 deg. emissions in the source frame, smoothly increasing to maxima (1) at 90 and 270 deg.
While the Doppler shift is roughly the inverse. Maximal at 0 and 180deg. to minimal around 90 and 270deg.
 
  • #41
altergnostic said:
It is real in all frames, but an "observer" can't assign coordinats of events he is not aware of. Those light paths are not seen by a distant observer, so he can't diagram them. If you want him to know the times of the reflections, you have to send him a signal at each reflection event.
Sure, but I have a hard time taking this as a serious objection. That should be an obvious and trivially understood part of the operation of any clock. Any clock has an oscillator of some sort (Caesium atom, quartz crystal, pendulum, etc.) and a mechanism for signalling the results (dial with hands, digital display, broadcast signal, paper printout, etc.). When you analyze the physics of a specific clock, you analyze the oscillator part, as is traditonally done with a light clock. It is understood that the oscillation can be reported in any number of ways.

Once you know that the detection occurred then you may use any reference frame to analyze the clock. There is no requirement to use a frame where the clock or the observer is at rest. Everything else follows.

altergnostic said:
As it is, the light clock diagram assumes he can actually see those light paths, which easily leads to paradoxes.
No true paradoxes have been found in over a century of careful scrutiny by the most brilliant minds in the world. All that have been found are unintuitive things which confuse students, many of which are hyperbolically advertised as paradoxes.
 
  • #42
Austin0 said:
The result is the vector is tilted forward in the direction of motion in the lab frame.

Ok I understand how that all works out nice mathematically, but a photon changing direction of emission depending on velocity of source?
 
  • #43
Austin0 said:
This is not correct. The observed diagonal paths in HarryLin's proposed set up would always produce c as a direct calculation of the distance and detection times (D/dt5) in the LAB frame.
No transformation whatsoever required in that frame.

Exactly as would be the case if the mirror was removed and the photon continued on to a detector at rest in the lab frame.

I agree, see post #25.
 
  • #44
Nugatory said:
We scatter observers, at rest relative to each other and carrying synchronized clocks, throughout space. Whenever anything happens right under an observer's nose, so only local considerations apply, the observer writes down what happened and the time it happened on a slip of paper. At some later time, we collect all the slips of paper and piece together a complete global picture.

We do all of this, and where's the paradox? The local angle each observer reports is exactly what you get by Lorentz-transforming from what observers at rest relative to the mirrors see.

I don't know if I have followed your reasoning, are you saying that the final diagram will agree with the stationary frame diagram?
 
  • #45
This really all made sense to me before, but something is cause me a problem with the photon's momentum in the direction of travel of the lab. If I think classically I can see that setup behaving like that if the observers are moving in the opposite direction. I just can't seem to fit in my brain the moving lab except a small one not moving fast. A photon shot sideways could get to the other side after the mirror is gone without the sideways momentum of the lab given to the photon.
 
  • #46
marty1 said:
Ok I understand how that all works out nice mathematically, but a photon changing direction of emission depending on velocity of source?

It does not change direction relative to the source frame.

It is only a change of direction relative to outside inertial frames. Exactly like a bullet fired laterally in a moving frame has an oblique angled trajectory relative to an observing frame.

perhaps a better way of saying it is it does change the direction relative to the emission angle but this is not perceived in the source frame because that frame is moving right along with the forward displacement due to conserved momentum.

Again comparable to the bullet situation.
 
  • #47
Again you are imparting the sideways momentum of the lab onto the photon (sideways meaning the labs toward direction which is sideways to the sideways fired photon). That is where I get lost.
 
  • #48
marty1 said:
Again you are imparting the sideways momentum of the lab onto the photon (sideways meaning the labs toward direction which is sideways to the sideways fired photon). That is where I get lost.

I am sorry if there has been confusion. I interpreted your question as related to a moving clock (source) as observed from a stationary lab.
All my remarks were addressed to a moving source.

If the clock is considered stationary then of course the forward momentum of a moving observer can have no effect on the propagation of the photon.
In this case it is simply the purely kinematic effect of relative motion.

Once again completely analogous to moving by a ball bouncing straight up and down . If the ball is covered with red ink and you could move a strip of paper along it as you pass the resulting motion chart would be a zig zag right? Well actually a sine wave but with lights instantaneous acceleration in it's case there are sharp turns.
 
  • #49
altergnostic said:
I don't know if I have followed your reasoning, are you saying that the final diagram will agree with the stationary frame diagram?

The final diagram will agree with the stationary frame diagram if the speed of all these observers happens to be the speed of the light clock. Otherwise it will agree with one of the moving frame diagrams.

But nothing prevents me from doing this "scatter observers, at rest relative to each other and carrying synchronized clocks" thing multiple times using a different speed for each flock of observers... And I can still gather up all those slips of paper and correlate them. So I'll get to see the exact same light-hits-mirror event as reported from both the stationary frame and as many different moving frames as I please.

I cannot say this strongly enough: get hold of and understand the relevant chapter of that Taylor/Wheeler book. It will extract you from the "what's a frame?" quagmire that you've fallen into like nothing else can.
 
  • #50
So the emitted photon does carry the momentum of the source in its movement 90 degrees from its direction of travel.
 
  • #51
DaleSpam said:
Sure, but I have a hard time taking this as a serious objection. That should be an obvious and trivially understood part of the operation of any clock. Any clock has an oscillator of some sort (Caesium atom, quartz crystal, pendulum, etc.) and a mechanism for signalling the results (dial with hands, digital display, broadcast signal, paper printout, etc.). When you analyze the physics of a specific clock, you analyze the oscillator part, as is traditonally done with a light clock. It is understood that the oscillation can be reported in any number of ways.

Well, it should be obvious, but (as you point out at the end of your post) many "less brilliant" minds don't find it so obvious, so why not diagram the setup more realistically? Einstein usually tried to do so, and usually using many different thought problems and examples to explain the same thing. He is the one who said that if you can't explain something to a six year old, you have not understood it yourself. The point is that the propagation of the signals, or, the times of observations of each tick of the clock from the moving frame point of view, must enter the equations. As it is, they are left out. There's no consideration of the distance between the light clock and the observer in the moving frame, and neither any consideration of the time it takes for each event to be observed by the moving frame. As it is, the observer is assumed to instantaneously see each event, which is very not-SR.

Once you know that the detection occurred then you may use any reference frame to analyze the clock. There is no requirement to use a frame where the clock or the observer is at rest. Everything else follows.

Exactly! So exactly at what x,y,z,t coordinates does the observer know the detection (reflection in the light clock) has occurred? In other words, at what time and relative position does the observer actually observes the events? Is this irrelevant to this problem? Why is it irrelevant here and relevant in problems like the train and embankment?

No true paradoxes have been found in over a century of careful scrutiny by the most brilliant minds in the world. All that have been found are unintuitive things which confuse students, many of which are hyperbolically advertised as paradoxes.

I actually almost agree with this. There is no true paradox if you remember to send signals to the observer, since the final result is correct regardless. But as it is the assumptions are unreal and impossible. You can't diagram unseen light, it is simple as that. Of course you are bound to find a lot of unintuitive things and confused students since the diagram is unintuitive and confused to start with. It is at least incomplete, and it is never diagrammed realistically and complete. We never see any analysis of how the observer knows about those paths, how he sees them, from what distance, and at what times, etc. Even though we have a lot of confused students, we keep the diagram incomplete. Why? Why not remember them that they need to signal the events to the observer somehow? Why do we allow students to wonder how the motion of the emitter can affect the direction of the beam without changing the beam's velocity (very counter-intuitively), in seemingly contradiction with the postulate of SR that the motion of emitter doesn't affect the speed of light? How are they not to get confused? We have many animations on the web showing how the motion of an emitter compresses the frontal waves and stretches the trailing waves, creating Doppler, and those diagrams show no change in the direction of the emitted light due to the motion of the source, so how are they to believe both diagrams without confusion?

You admitted that we need to send signals at each tick of the light clock if we want the observer to diagram anything. So if just remembering that reflection events are not seen directly helps to make some sense of the diagram, why not do so? If you actually need to signal the reflection events to the moving observer, those signals have to travel some distance until they reach the observer, after some time from the moment of emission. The students would immediately relate the problem to the famous train and embankment, for example, and it would be easier to comprehend. And also, the moving observer wouldn't diagram those diagonal paths in such a setup at all. You would apply the transforms on the incoming signals and the light clock's beam would be diagrammed straight up and down. No angles. No confusion.

If you still think the diagram is perfect and it is the students that fail, you are not trying to make a comprehensive diagram, or to make SR easier to understand, you are just trying to flunk them.
 
  • #52
marty1 said:
So the emitted photon does carry the momentum of the source in its movement 90 degrees from its direction of travel.

Yes. If it is emitted at 90 deg. relative to motion in the source frame the forward momentum will result in the path pointing at some angle forward in an observing frame.
 
  • #53
altergnostic said:
Well, it should be obvious, but (as you point out at the end of your post) many "less brilliant" minds don't find it so obvious, so why not diagram the setup more realistically? Einstein usually tried to do so, and usually using many different thought problems and examples to explain the same thing. He is the one who said that if you can't explain something to a six year old, you have not understood it yourself. The point is that the propagation of the signals, or, the times of observations of each tick of the clock from the moving frame point of view, must enter the equations. As it is, they are left out. There's no consideration of the distance between the light clock and the observer in the moving frame, and neither any consideration of the time it takes for each event to be observed by the moving frame. As it is, the observer is assumed to instantaneously see each event, which is very not-SR.



Exactly! So exactly at what x,y,z,t coordinates does the observer know the detection (reflection in the light clock) has occurred? In other words, at what time and relative position does the observer actually observes the events? Is this irrelevant to this problem? Why is it irrelevant here and relevant in problems like the train and embankment?



I actually almost agree with this. There is no true paradox if you remember to send signals to the observer, since the final result is correct regardless. But as it is the assumptions are unreal and impossible. You can't diagram unseen light, it is simple as that. Of course you are bound to find a lot of unintuitive things and confused students since the diagram is unintuitive and confused to start with. It is at least incomplete, and it is never diagrammed realistically and complete. We never see any analysis of how the observer knows about those paths, how he sees them, from what distance, and at what times, etc. Even though we have a lot of confused students, we keep the diagram incomplete. Why? Why not remember them that they need to signal the events to the observer somehow? Why do we allow students to wonder how the motion of the emitter can affect the direction of the beam without changing the beam's velocity (very counter-intuitively), in seemingly contradiction with the postulate of SR that the motion of emitter doesn't affect the speed of light? How are they not to get confused? We have many animations on the web showing how the motion of an emitter compresses the frontal waves and stretches the trailing waves, creating Doppler, and those diagrams show no change in the direction of the emitted light due to the motion of the source, so how are they to believe both diagrams without confusion?

You admitted that we need to send signals at each tick of the light clock if we want the observer to diagram anything. So if just remembering that reflection events are not seen directly helps to make some sense of the diagram, why not do so? If you actually need to signal the reflection events to the moving observer, those signals have to travel some distance until they reach the observer, after some time from the moment of emission. The students would immediately relate the problem to the famous train and embankment, for example, and it would be easier to comprehend. And also, the moving observer wouldn't diagram those diagonal paths in such a setup at all. You would apply the transforms on the incoming signals and the light clock's beam would be diagrammed straight up and down. No angles. No confusion.

If you still think the diagram is perfect and it is the students that fail, you are not trying to make a comprehensive diagram, or to make SR easier to understand, you are just trying to flunk them.
I think you are right in that the fact that the speed of light is independent of the motion of the source should be expanded to include the fact that the direction of propagation is not also independent. With the elaboration of its effects, the relativistic aberration, that is certainly relevant to the light clock and it's understanding.
As for the rest; you may have a certain point but it should be clear that certain unrealistic aspects of most thought experiments are not really relevant to their aid as constructs for understanding. IMO
 
  • #54
Austin0 said:
Yes. If it is emitted at 90 deg. relative to motion in the source frame the forward momentum will result in the path pointing at some angle forward in an observing frame.

So I post my question again. What is special about the "sides" of a photon that permit the source to give it momentum but any attempt to do that in the direction of propagation results only in red or blue shift?

It seems that from the side a photon is no different from any other particle.
 
  • #55
Nugatory said:
The final diagram will agree with the stationary frame diagram if the speed of all these observers happens to be the speed of the light clock. Otherwise it will agree with one of the moving frame diagrams.

But nothing prevents me from doing this "scatter observers, at rest relative to each other and carrying synchronized clocks" thing multiple times using a different speed for each flock of observers... And I can still gather up all those slips of paper and correlate them. So I'll get to see the exact same light-hits-mirror event as reported from both the stationary frame and as many different moving frames as I please.

I cannot say this strongly enough: get hold of and understand the relevant chapter of that Taylor/Wheeler book. It will extract you from the "what's a frame?" quagmire that you've fallen into like nothing else can.

This setup will yield the same results as harrylin's setup.

Only the observers in direct line with the path of the beam will detect anything, the others can only copy what they wrote down. This is true either if they are moving or at rest relative to the beam.

If they are at rest relative to the light clock, the diagram looks like the stationary one. Also, any observer in line with the path of the beam will measure the same period of time between two detections, and they will know the light is bouncing up and down since they will have as many successive detections as they want, as long as they keep their positions wrt the setup. Notice that each observer in line only sees the beam as it passes through him. They detect the beam by contact. They don't know the times of reflections directly, they only know their local detections. This means that if you stick with one observer, you will diagram the beam up and down like in the stationary diagram, and you don't even need a flock of observers, since you can use the time between detections to calculate the distance between mirrors and your relative position between them.

On the other hand, if the observers are moving, each observer may get only one detection, so no single observer knows the light is bouncing up and down directly. Obviously, no single observer directly knows the time between ticks also, and each observer makes his detection at a later time than the previous observer (who will be some x ahead and some y below). There's a true x distance between the first observer (who detected the beam at y=bottom) and the last (who detected the beam at y=top). This x distance implies a time separation. If you want to diagram what a single observer would see, you have to consider this time separation.

If you want to compress all the observers' detections into one diagram, you will get a zig-zag path only because you are compressing observations from many different observers into one diagram, which is very different from diagramming what a single observer sees and very little SR-like. If you choose one observer (A) from the flock from where to do measurements, you will only detect the beam directly once. Any detection after that is done by another observer (B) and occurs some time later. If you want to know about the detection of B, B has to send a signal to you, and you must consider the time it takes for this signal to reach A, so you can calculate the time of detection on B's clock. Only after you do this you will be able to diagram the beam's path properly. The coordinates of the beam are accessible to you only through information detected by the other observers, and it takes time to receive any information from other observers, so you have to apply the transforms. Since the only direct information you have from the beam is a single detection at a single (x,y,z,t) point, you have to receive data from other observers to calculate the beam's path, and you must apply the transforms to find the coordinates measured by the other observers. In the end, you will graph the local x,y,z,t coordinates from every observer and you will find the path to be just like the stationary diagram.

----------

Either if you want your observers to communicate between them with light signals or with written notes, you can't transmit information faster than c, and the time it takes for that information to reach you must be accounted for.

And you must have some sort of transmission of information between the event and the observer. If the event is light being reflected from one mirror to another, this light doesn't reach the distant observer and he can only know about it by secondary means, either light signals, a flock of observers or scattered light. All of which are more realistic than the original diagram, which I take that many of you have already shown to be at the very least incomplete, since it has been admitted and demonstrated over and over that we need some sort of transmission of information (usually light) between the event and the observer to make sense out of it...
 
  • #56
...
The only thing some of you still seem to disagree (?) with me is that once you use information transmitted directly to you, you must apply the transforms using this directly received light, and not on the bouncing undetected light. And once you apply the transforms, you will find the light clock system's coordinates and use them to diagram the path of a beam between the mirrors, which can only yield on stationary-like diagram.
What some seem to be claiming is that you can actually see those diagonal paths, or that they have some intrinsic reality.
Most seem to be missing my argument that if light is reflected at a right angle wrt the mirrors in one system, it must do so in all frames. And I still can't save this diagram without light signals being emitted at each moment of reflection, and if we do that, we have to apply transforms on those light sources, and after that, the coordinates will be the stationary coordinates, and the beam will bounce like in the stationary frame.
 
  • #57
Austin0 said:
I think you are right in that the fact that the speed of light is independent of the motion of the source should be expanded to include the fact that the direction of propagation is not also independent. With the elaboration of its effects, the relativistic aberration, that is certainly relevant to the light clock and it's understanding.
As for the rest; you may have a certain point but it should be clear that certain unrealistic aspects of most thought experiments are not really relevant to their aid as constructs for understanding. IMO

OK, I can live with that, since we have agreed on the physics, we can have differing opinions on how thought experiments should be done. And yes, most unrealistic aspects of most thought problems are completely acceptable and actually very important as they simplify the problem. I don't think this is the case with the light clock diagrams, though. I really think that leaving a mechanism of transmission between the reflection events and the observer out of the picture is bad, because it allows us to draw the diagonal paths as if they are real data observed by a distant observer in relative motion, and this is not a simplification, it is a wrong assumption. From my experience, wrong assumptions are virulent, and can spread like diseases. We should never allow people to even consider that distant light can be seen. I don't care if the resulting equations are correct. I rather achieve correct equations with correct diagrams.
 
  • #58
marty1 said:
So I post my question again. What is special about the "sides" of a photon that permit the source to give it momentum but any attempt to do that in the direction of propagation results only in red or blue shift?

It seems that from the side a photon is no different from any other particle.

I thought that was clear. Directly aligned with the motion vector the forward momentum of the source cannot affect the motion of the photon in that direction because that speed is limited by the intrinsic properties of spacetime.
In this regard a photon is no different than a massive particle. For instance an electron accelerated laterally would conserve the forward momentum completely but if accelerated along the direction of motion it would not. For example say it was accelerated to 0.99c in a frame that was going 0.8 c relative to the lab. it could not conserve that 0.8c momentum in the forward direction completely and so have a velocity of 1.79 c in the lab frame as would be the case with a low velocity bullet. It instead would have a velocity of .9988 in the lab.
Exceedingly little contribution from the motion of the source. 0.0088c
 
  • #59
You know... part of the problem (the part about the motion of the emitter and wondering why the remote observer wouldn't expect the light go 90 degrees from the point of emission and miss the moving mirror) is that the translation of Einstein's writing confounds two terms.

The translations say that "the velocity of light is not influenced by motion of the emitter", but in German their word for velocity only means speed, not speed and direction like we use velocity as a vector in English.

If you read "velocity", and assume it is a vector sourced at the emission, the remote observer would expect the light to ignore the emitter's motion and travel 90 degrees from the emission point with respect to the rest frame of the remote observer, and miss the mirror...
 
  • #60
Angle is the end result, but the truth is that the sideways component of the momentum of a photon is complete under the influence of the source. I guess this is obvious but something I have never seen. Photons are only special along their axis of propagation.
 
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