Michelson experiment

[Lammaite2]

 

The diagram is pretty clear that the transit time is calculated with c+v and c-v, meaning the signal ahead did acquire the additional speed v when traveling in that direction. In the opposite direction, it is retarded by v.

 

The geometry shows the speed of the transverse beam having both transverse and longitudinal components, with a resultant of c.

 

 

It was calculated using t = distance/speed The distance is L, and the speed is c+v or c-v, depending on the direction of travel.

I can almost hear the big pistol being unlocked ....

Here is a quote from Feynman referring to the situation of travelling apparatus and the "ahead" signal:

"Now, while the ligh is from the B to the mirror, the apparatus moves a distance v*t1, so the light must traverse a distance L+v*t1, at the speed c." Hence, c*t1=L+vt1, hence t1=L/(c-v)

 

Similarly Feynman calculates the journey back L-v*t2, hence c*t2=L-vt2, hence t2=L/(c+v). As you can see Feynman says "at the speed c". And from the equations it appeares that the light did not acquire any additional speed when emitted ahead as well as it did not slow down when emitted back. It just travells different distances at speed c. From Feynman equations for times t1 and t2 the denominators are actually c-v for t1 and c+v for t2 that is opposite to what you suggested.

 

This is why I stubbornly insist the assumptions for Michelson experiment were the light moves at constant c in relation to ether and the speed is independant to the speed of the source. Actually I understand these two are actually unseparable. Or may be all this comes from linguistic confusion as to the use of "independant".

 

Anyway, the constant c versus ether tells me the transverse signal (similarly to ahead signal) should not acquire any additional speed in direction of the apparatus. This should be consistently the assumption for calculating the times and delay of the signals before the experiment.

 

[altergnostic]

 

Anyway, the constant c versus ether tells me the transverse signal (similarly to ahead signal) should not acquire any additional speed in direction of the apparatus. This should be consistently the assumption for calculating the times and delay of the signals before the experiment.

 

You are correct. In his original paper, Michelson clearly states that the beam should miss the mirror at a considerable speed, but that this effect would be too subtle to matter at the speeds actually involved. At the speeds shown in the animation, the beam should miss the mirror or the detector. Also, the return speed for the photon in the x axis should be greater than the outgoing speed, relative to the interferometer.

Edited December 12, 2012 by altergnostic

[swansont]

This is why I stubbornly insist the assumptions for Michelson experiment were the light moves at constant c in relation to ether and the speed is independant to the speed of the source. Actually I understand these two are actually unseparable. Or may be all this comes from linguistic confusion as to the use of "independant".

 

Yes — in the ether's frame. This is not the case in other frames, however. In the lab frame, the speed of light will be c+v or c-v if you are moving.

[altergnostic]

 

Yes â in the ether's frame. This is not the case in other frames, however. In the lab frame, the speed of light will be c+v or c-v if you are moving.

 

The speed of light was c relative to the aether in all frames. In the lab frame, you are moving relative to the aether at v and the light is moving relative to the aether at c, so the light is moving relative to you at c+-v, not relative to the background. If the interferometer is moving to the right in a stationary aether, the aether moves to the left in a stationary interferometer's frame. You can't move both the interferometer and the light to the right. That is adding the speed of the source to the speed of light in the aether stationary frame!

 

Again, Michelson said it with all the words, that the beam would indeed miss the detector at considerable speeds. The op is correct and the animation at wiki is not.

[swansont]

The speed of light was c relative to the aether in all frames. In the lab frame, you are moving relative to the aether at v and the light is moving relative to the aether at c, so the light is moving relative to you at c+-v, not relative to the background. If the interferometer is moving to the right in a stationary aether, the aether moves to the left in a stationary interferometer's frame.

 

That's what I said.

 

You can't move both the interferometer and the light to the right. That is adding the speed of the source to the speed of light in the aether stationary frame!

 

Sure you can. You aren't adding the speed when you do that. When the light moves to the left, it adds.

 

Again, Michelson said it with all the words, that the beam would indeed miss the detector at considerable speeds. The op is correct and the animation at wiki is not.

 

We are not at "considerable speed"

[altergnostic]

 

Sure you can. You aren't adding the speed when you do that. When the light moves to the left, it adds.

 

We are not at "considerable speed"

 

If the light moves to the right, it has gained the speed of the interferometer relative to the aether, if the interferometer moves and the light stays behind it has maintained c relative to the aether, just like any other wave in a medium. Very straightforward.

 

And the experiment was not at considerable speeds, but the animation is. The interfereometer obviously travels at a good fraction of the speed of light in the animation. If it depicted the actual speeds of the experiment, the beams would complete the two-way trip before any noticeable displacement of the interferometer at all.

 

Either way, the photon should never have a horizontal component like that, it would move straight up while the interferometer moves to the right and, considering the speeds shown in the animation, it would completely miss the interferometer upon return. That's what Michelson himself said, did you read his original paper?

[phyti]

If my understanding is correct, with light moving at c relative to the ether, the zigzag path was an incorrect assumption.

I want to be sure before moving on.

[altergnostic]

If my understanding is correct, with light moving at c relative to the ether, the zigzag path was an incorrect assumption.

I want to be sure before moving on.

 

Well, from Michelson's book you can read that the beam should miss the receptor (at great enough speeds). That can only mean that they expected that the beam would be "left behind" as the interferometer moves to the right in an stationary aether. In other words, if the interferometer moves to the right, the beam would zigzag to the left: there would be no velocity component gained from the velocity of the interferometer. "The speed of light is c relative to the aether regardless of the speed of the source".

 

There should be a zigzag as observed from the interferometer's frame, not from the aethers frame (which is what the animation shows). If the background is stationary (and the interferometer moves) then beam would describe a straight line up and down while the interferometer moves to the right. From the point of view of the interferometer, the aether moves to the left, so the aether "wind" should force the beam in the same direction.

[Lammaite2]

Well, from Michelson's book you can read that the beam should miss the receptor (at great enough speeds). That can only mean that they expected that the beam would be "left behind" as the interferometer moves to the right in an stationary aether. In other words, if the interferometer moves to the right, the beam would zigzag to the left: there would be no velocity component gained from the velocity of the interferometer. "The speed of light is c relative to the aether regardless of the speed of the source". There should be a zigzag as observed from the interferometer's frame, not from the aethers frame (which is what the animation shows). If the background is stationary (and the interferometer moves) then beam would describe a straight line up and down while the interferometer moves to the right. From the point of view of the interferometer, the aether moves to the left, so the aether "wind" should force the beam in the same direction.

I suppose, from all the posts, I am getting close to solve my dilemma. Please correct the reasoning if wrong:

1. Existence of stationary ether is assumed

2. The zigzag path of the transverse as seen from the apparatus reference should be to the left (not to the right as on the Wikipedia gif).

3. Taking the low speed of the apparatus and the width of the light beam, it still hits the interferometer and the other returning beam.

4. Zigzag path is therefore used for predicting the travel time of the transverse signal

Is that OK?

Thanks for the link to Michelson

 

 

[altergnostic]

 

I suppose, from all the posts, I am getting close to solve my dilemma. Please correct the reasoning if wrong:

1. Existence of stationary ether is assumed

2. The zigzag path of the transverse as seen from the apparatus reference should be to the left (not to the right as on the Wikipedia gif).

3. Taking the low speed of the apparatus and the width of the light beam, it still hits the interferometer and the other returning beam.

4. Zigzag path is therefore used for predicting the travel time of the transverse signal

Is that OK?

Thanks for the link to Michelson

 

 

You are welcome.

 

The zigzag path is supposed to help us visualize the reason for the expected fringe displacement, and it is based on the expected variation of the speed of light relative to the interferometer - although the animation is completely off on that matter, specially concerning the x-axis beam, but it depicts the beams as having varying speeds relative to the aether, which is nowhere near the original assumptions.

 

But your points are correct regarding MM's own assumptions.

 

What concerns me the most now, after I have read Michelson, is his use of a distance measured in a stationary interferometer's frame (moving aether) divided by a speed measured from the stationary aether frame (moving interferometer).

Edited December 13, 2012 by altergnostic

[Janus]

You are welcome. The zigzag path is supposed to help us visualize the reason for the expected fringe displacement, and it is based on the expected variation of the speed of light relative to the interferometer - although the animation is completely off on that matter, specially concerning the x-axis beam, but it depicts the beams as having varying speeds relative to the aether, which is nowhere near the original assumptions. But your points are correct regarding MM's own assumptions. What concerns me the most now, after I have read Michelson, is his use of a distance measured in a stationary interferometer's frame (moving aether) divided by a speed measured from the stationary aether frame (moving interferometer).

 

 

I fail to see where the beams have varying speeds wrt the aether. If the aether is a rest with respect the the animation frame, it is clear the the dots in the left hand animation travel at the same speed in both directions with respect to that frame. If you only consider the vertical speed of the red dots in both animations, you see that they stay abreast at all times. If you watch the blue dots in both animations, you will notice that during those periods when they are moving in the same direction, the distance between them is constant. In that case, if the dots have a fixed speed with respect to the animation frame (aether) in the left animation, they do so in the right animation,

 

Also, having checked out page 5 of the paper you linked to, I don't see where he has done as you claimed. For example, he gives the time for one leg of the light's trip as being D/(V-v) with V being the speed of light with respect to the Aether and v the speed of the interferometer with respect to the same. D is the length of the interferometer arm and (V-v) would be the speed of light wrt respect to the arm. No frame crossing here that I see.

 

We can also work this out from the aether frame. Here the distance traveled by the light is D+vt and the speed of light with respect to the Aether is V,

 

Thus the time for this leg would be

 

t= (D+vt)/V

 

We want to get t alone, so

 

t= D/V+tv/V

 

t-tv/V=D/V

 

t(1-v/V) = D/V

 

dividing both sides by V:

 

t(V-v) = D

 

t= D/(V-v)

 

The same as above.

[Lammaite2]

You are welcome. The zigzag path is supposed to help us visualize the reason for the expected fringe displacement, and it is based on the expected variation of the speed of light relative to the interferometer - although the animation is completely off on that matter, specially concerning the x-axis beam, but it depicts the beams as having varying speeds relative to the aether, which is nowhere near the original assumptions. But your points are correct regarding MM's own assumptions. What concerns me the most now, after I have read Michelson, is his use of a distance measured in a stationary interferometer's frame (moving aether) divided by a speed measured from the stationary aether frame (moving interferometer).

The zigzag path of the transverse signal, it comes to me more and more certain, is assumed by Michelson (also by Feynman’s explanation) to evaluate the time, only because of the width of the beam. Within the beam the light reflects at a range of angles, producing not only perfectly perpendicular signal, and the signal in consideration is the one that reflects at particular angle. The angle is such that the signal’s path aims exactly at the interferometer at the moment when the twin signal reaches the interferometer. This is why in Feynman’s calculation of the transverse time the signal travels at speed c (!) along the arms of the triangle. Which means it was reflected into direction of the arm of the triangle and again reflected into direction along the other arm to aim the interferometer; it travels this directions versus aether frame with speed c. This seems to be the reasoning behind Michelson’s and Feynman’s calculation of journey time.

 

If what I said was true than the ahead signal and the transverse signal are not the same ray split on the first mirror. They are two different rays from the starting from the source of light. When the speed of the apparatus is low then this may not be a problem and the derived time delay equation is OK. What if the speed of the apparatus is much greater? The zigzag will be wide open and the rays would need to be drawn from the source on different paths – the transverse signal should start at some significant angle to the parallel ray even before reaching the splitting mirror. This is because if we want the transverse signal to reflect little bit forward it needs to hit the splitting mirror at bigger than 45 deg. to normal. What will then happen to the equation of the time delay? All SR mathematics is later based on confronting this equation with the fact of no delay.

 

I can not think otherwise than: if the signal was an Euclidean point and travelled perfectly perpendicular to the apparatus direction then the signal would not be able to meet again the interferometer (also a perfect point). The journey of that point, as seen from apparatus frame, would be a zigzag, yet the opposite direction than shown in Feynman’s and Michelson’s pictures – i.e. back from the apparatus. Also the predicted journey time along that zigzag would need to be calculated using speed value from parallelogram of speed c perpendicular to apparatus and the speed v of the apparatus i.e. SQRT(c2+v2). Also the predicted journey time along that zigzag would equal 2L/c.

 

 

[altergnostic]

 

 

I fail to see where the beams have varying speeds wrt the aether. If the aether is a rest with respect the the animation frame, it is clear the the dots in the left hand animation travel at the same speed in both directions with respect to that frame. If you only consider the vertical speed of the red dots in both animations, you see that they stay abreast at all times. If you watch the blue dots in both animations, you will notice that during those periods when they are moving in the same direction, the distance between them is constant. In that case, if the dots have a fixed speed with respect to the animation frame (aether) in the left animation, they do so in the right animation,

 

Ok, let's see if I can convince you. In the left animation, the distance described by the straight line is smaller than the distance described by the diagonal of the animation to the right, and the time it takes the beam to cross both these distances are the same, so the speeds, relative to the background, are not the same, I hope this is clear.

 

Also, having checked out page 5 of the paper you linked to, I don't see where he has done as you claimed. For example, he gives the time for one leg of the light's trip as being D/(V-v) with V being the speed of light with respect to the Aether and v the speed of the interferometer with respect to the same. D is the length of the interferometer arm and (V-v) would be the speed of light wrt respect to the arm. No frame crossing here that I see.

 

You have to notice that D is the distance of the arms measured in the stationary interferometer frame, but v is the velocity of the interferometer measured in an stationary aether frame. From the stationary aether frame, the photon has a larger dustance to travel in T due to the displacement of the arms wrt the aether.

 

We can also work this out from the aether frame. Here the distance traveled by the light is D+vt and the speed of light with respect to the Aether is V,

 

Thus the time for this leg would be

 

t= (D+vt)/V

 

Nope, the time should be 2D/V (or 2D/c) for the x-axis (the entire arm moves to the right, so the distance added on the outgoing trip must be subtracted from the return trip), and the transverse beam is hard to figure out, since there should be no angle at all, remember? With that diagonal, you have to find the hypotenuse and that will yield a speed of light higher than c in the aether frame, which doesn't follow from the assumption of a fixed speed of light in the aether.

 

But if you wanted to get the time for the transverse beam in the aether frame assuming that diagonal, you must find the hypotenuse:

 

H=(D^2+(vt,)^2)^1/2

 

* Where (vt,) is the distance travelled by the interferometer in the x direction in the outgoing trip.

 

And the speed would be

 

V'=(c^2+v^2)^1/2

 

*the V is primed so we don't mistake it for Michelson's V, which is c here.

 

Then

 

t=2H/V'

 

t=2(D^2+(vt,)^2)^1/2 // (c^2+v^2)^1/2

 

t=2(D^2+v^2*t,^2) // c^2+v^2

 

t^2*c^2 + t^2*v^2 = 4D^2 + 4v^2*t,^2

 

Of course t=(2t,) which is the round trip

 

ct+2vt,=2D+2vt,

 

t=2D/c

 

 

The same as above. You see, if c is the speed that has always been measured, we know for a fact that it is the speed relative to the receptor for any two-way measurement of light, which is precisely what this math shows, and it also shows that the results are the same if you are moving relative to some background or not, which is consistent with the principle of relativity. This is the first time I analised this problem like this, since I had never noticed Michelson's inconsistency, and the math is pretty straightforward, very simple, and in agreement with experiment. I see no mistake in it.

[altergnostic]

 

The zigzag path of the transverse signal, it comes to me more and more certain, is assumed by Michelson (also by Feynmanâs explanation) to evaluate the time, only because of the width of the beam. Within the beam the light reflects at a range of angles, producing not only perfectly perpendicular signal, and the signal in consideration is the one that reflects at particular angle. The angle is such that the signalâs path aims exactly at the interferometer at the moment when the twin signal reaches the interferometer. This is why in Feynmanâs calculation of the transverse time the signal travels at speed c (!) along the arms of the triangle. Which means it was reflected into direction of the arm of the triangle and again reflected into direction along the other arm to aim the interferometer; it travels this directions versus aether frame with speed c. This seems to be the reasoning behind Michelsonâs and Feynmanâs calculation of journey time.

 

If what I said was true than the ahead signal and the transverse signal are not the same ray split on the first mirror. They are two different rays from the starting from the source of light. When the speed of the apparatus is low then this may not be a problem and the derived time delay equation is OK. What if the speed of the apparatus is much greater? The zigzag will be wide open and the rays would need to be drawn from the source on different paths â the transverse signal should start at some significant angle to the parallel ray even before reaching the splitting mirror. This is because if we want the transverse signal to reflect little bit forward it needs to hit the splitting mirror at bigger than 45 deg. to normal. What will then happen to the equation of the time delay? All SR mathematics is later based on confronting this equation with the fact of no delay.

 

I can not think otherwise than: if the signal was an Euclidean point and travelled perfectly perpendicular to the apparatus direction then the signal would not be able to meet again the interferometer (also a perfect point). The journey of that point, as seen from apparatus frame, would be a zigzag, yet the opposite direction than shown in Feynmanâs and Michelsonâs pictures â i.e. back from the apparatus. Also the predicted journey time along that zigzag would need to be calculated using speed value from parallelogram of speed c perpendicular to apparatus and the speed v of the apparatus i.e. SQRT(c2+v2). Also the predicted journey time along that zigzag would equal 2L/c.

 

 

As you see, I applied the same reasoning regarding the use of the hypotenuse and everything seems to work, and the predicted time is indeed 2L/c. The speed of light varies in the aether frame, but the observed speed remains c, which is not consistent with the expectations of MM but is consistent with the results of the experiment and both this analysis and the results of the experiment falsifies the existence of the aether, or at least the aether as hypothesized by the proposed aether theories (i.e. as a medium for the propagation of light). What amazes me is that this solution resolves the experiment's results without any new assumptions, it falsifies the aether and seems to point us in the direction of the particle nature of light.

[Lammaite2]

EUREKA! Now, I think I understand. It seems, that the explanation of the transverse journey time takes into account the result of the MM experiment; NOT JUST ONLY the pre-experiment assumed condition.

The fact that the transverse signal reached the meeting point (“Normally” it should not!), implies:
that the light MUST have travelled ALONG the arms of the isosceles triangle (Otherwise it would not be seen at meeting point.).

In addition, it travelled WITH SPEED c (as the first assumption taken from Maxwell’s theory, or as a propagation of wave as says Feynman).

From these two condition is the equation for the transverse time!

This small change in the way of explaining to indicate that the result of MM was the reason behind the transverse time made so big difference.

If this is the case the width of the beam does not matter as I thought earlier. Stil a question remains: Should not be Michelson's expectation, based on purely Euclidean geometry, law of light refraction and presence of aether as a medium for light propagation, that the light rays should never meet?

The discussion here helped a lot. I really hope to hear some conforming posts!

[altergnostic]

 

The fact that the transverse signal reached the meeting point (Normally it should not!), implies:

that the light MUST have travelled ALONG the arms of the isosceles triangle (Otherwise it would not be seen at meeting point.).

 

In addition, it travelled WITH SPEED c (as the first assumption taken from Maxwells theory, or as a propagation of wave as says Feynman).

 

One quick note, just to be annoying:

it just implies that light MUST have travelled the diaginals in the AETHER frame. Since there is no evidence of an aether, this vt/c could be (and was) questioned further.

Also, strictly speaking, the speed c was measured from the interferometer's frame, where the distance travelled is not described by the angled paths.

Hence, from this experiment alone, we can only be sure that motion through space does not affect our measurements of the speed of light. Any other assumption would require further research and confirmation. For instance, this experiment doesn't tell us at what speed light has travelled through either arm, it just tells us the speeds are equal.

[Lammaite2]

Every annoyance is welcome.
When providing the predicted time of transverse journey Feynman says: …the light travels a distance ct₃ along hypotenuse of a triangle…
If this was possible there must have been assumed some counter-intuitive relativistic effect i.e. what is seen in aether - the transverse light just bouncing up and down, is seen from apparatus frame also bouncing up and down. To achieve this, the magic invisible path is along the hypothetical hypotenuses. Next is the assumption of constant c (still like if there was aether!), hence the time equation, hence the expected delay. Then the result shows there is no delay, hence SR.
This does not make sense because the assumption of relativistic effect is used for formulating relativity. Yet this the only way I can follow up Feynman’s reasoning.

My common sense tells me: If applying classical physics, aether, speed c along the hypotenuse path as on Feynman’s diagram…then the parallel and transverse rays would have to be different just on start from the source. Some width of the light beam would need to be assumed and lights rays at different angles from the source. They would reach the first mirror already with some delay and reflect at different angles. The time equations (easy to deduce) would look entirely different from Michelson’s (and Feynman’s).
If we wanted just one ray on start then split on the first mirror then, by the classical physics and aether assumption, the transverse ray will be seen from apparatus along hypotenuse yet back to the apparatus direction and with speed SQRT(v^2+c^2) versus apparatus.

I am referring here to Feynman’s “Six not so easy pieces”.

Of course I am amateur and beginner in this area. Were there any other ways of inferring SR?

[altergnostic]

Every annoyance is welcome.

When providing the predicted time of transverse journey Feynman says: the light travels a distance ct₃ along hypotenuse of a triangle

If this was possible there must have been assumed some counter-intuitive relativistic effect i.e. what is seen in aether - the transverse light just bouncing up and down, is seen from apparatus frame also bouncing up and down. To achieve this, the magic invisible path is along the hypothetical hypotenuses. Next is the assumption of constant c (still like if there was aether!), hence the time equation, hence the expected delay. Then the result shows there is no delay, hence SR.

This does not make sense because the assumption of relativistic effect is used for formulating relativity. Yet this the only way I can follow up Feynmans reasoning.

 

My common sense tells me: If applying classical physics, aether, speed c along the hypotenuse path as on Feynmans diagramthen the parallel and transverse rays would have to be different just on start from the source. Some width of the light beam would need to be assumed and lights rays at different angles from the source. They would reach the first mirror already with some delay and reflect at different angles. The time equations (easy to deduce) would look entirely different from Michelsons (and Feynmans).

If we wanted just one ray on start then split on the first mirror then, by the classical physics and aether assumption, the transverse ray will be seen from apparatus along hypotenuse yet back to the apparatus direction and with speed SQRT(v^2+c^2) versus apparatus.

 

I am referring here to Feynmans Six not so easy pieces.

 

Of course I am amateur and beginner in this area. Were there any other ways of inferring SR?

There were many different attempts to explain data and many competing theories. Many of those applied similar math (namely, the Lorentz-Fitzergald transformations or some variation) and many worked well. SR was preferred over other theories mainly because of it's simplicity, not because it was the only one that explained data. For instance, you can work your way around many relativistic problems assuming the existance of an aether, but if you can explain data without an aether, so much better. Are you familiar with Occam's Razor? It is a criteria for choosing between competing theories that explain data just as well. It states that we should prefer the one with the less amount of assumptions. There is a theory, called Edward's Theory, that is mainly SR with a different postulate for the speed of light: c is the average speed of light for a two-way trip measurement, but the outgoing and incoming path are not necessarily crossed at c each, relative to either source or receiver.

Einstein himself was rigorous enough to remind us that this is indeed the actual evidence, because we can't measure the speed of light for a one-way trip, but he chose to define c as the speed for a one-way trip also as a convention for the sake of simplicity. Since both theories explain much data with similar accuracy, we opt for SR because it is simpler, but if someone performs an experiment that measures the one-way speed of light we can then choose the best theory based on evidence instead of for it's simplicity.

This is just to say that there is no such thing as a proven theory, only an falsified theory. The black swan's logic arguments that we can't prove that all swans are white unless we have found all the swans, and you can't prove that you've found all the swans, but it is an acceptable theory until we find a black swan, falsifying the premise that all swans are white beyond question. You can only disprove a theory because all theories are based on assumptions, and until these assumptions are tested and verified, or until some conflicting evidence comes up, we can only believe a theory is correct because it works so well. That's why a scientific theory has to be falsifiable, otherwise it can't be said to be a proper scientific theory, since you can't perform any experiment to (in)validate it.

 

That said, the current accepted model is that light moves at c in any and every frame regardless of the state of motion of the source, and we alter lengths and times to conform with c (that's what the lorentz transformations do). In GR, light is not always required to move at c, for instance, due to the fact that we are no longer dealing with inertial frames and that the laws of physics are no longer required to be the same for all observers. You may continue your analysis as far as you want, but the basic premises you already got right, and you may work on some textbook exercises to see if you have learned how to operate the equations, but it is useful to learn what is known from experimental evidence and is a model built from acceptable assumptions around some given evidence.

Currently, as an example, there are proposed theories where light moves at c only in our vicinity, but bot at distant parts of the universe, and also theories that model a universe where the laws of nature are not constants but change over time. This is all admitedly very speculative, but it is important to know where there's room to play and what is concluded from direct evidence. On the other hand, you can't ever say that the speed of light for a two-way trip is not c and expect to be taken seriously, since it contradicts direct evidence.

As for the angled transverse paths specifically, assuming they move at c is consistent with the lorentz transforms, which were first derived assuming the existence of the aether. Einstein ditched the aether and proposed that light moves at c for any inertial reference frame, which is also an assumption (remember, we can only measure the speed of light in a two-way trip experiment, where light is emitted and detected at the origin of a single frame of reference), but it is an assumption with less premises than the aether, which had to be explained with many counter-intuitive physical properties.

So it isn't a verified fact that light travels those diagonals at c, it is a working convention, and one that conforms with a very succesfull Special Theory of Relativity. To verify it, we would need to place an observer some distance away in relative motion and this observer would have to measure the speed of the transverse beam, but of course he can't do that since that beam is not reaching him and would not be detected. Any light that reaches this distant observer will have described a different path than the hypotenuse, since it has to travel directly towards him, and the speed of this beam can't be measured, since you need two consecutive detections of the same beam to plot x against t, and a single detection is not enough. Two consecutive detections of different beams or wave fronts or photons are also no good, so we are stuck until someone comes up with a cleverer experiment (if that's at all possible). This is where we decide to stop speculating and effectively work with the tools we already have, which is more than enough for most practical problems, or choose to take a chance into the theoretical realm and hope not to starve to death.