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

[altergnostic]

But he can't use the primed times along with unprimed distances to calculate anything meaningful. To correctly calculate a speed you must use the distance and time from the same frame. The "distance AB" that he is given is in the unprimed frame. That distance is *not* perpendicular to the relative motion of the light clock and the observer in the unprimed frame (it is in the *primed* frame, but it is *not* in the unprimed frame--perhaps this is one of the points where you are confused). So the observer *can't* combine it with a time in the primed frame to get a meaningful answer, because of length contraction.

That's right! But you are confusing the coordinates. He uses the primed distances and unprimed times. And that is allowed because he is directly seeing the km marks, you see. I have determined that in the setup, all distances have been previously walked and each km marked with a big sign. The observer can see those directly. The primed distances are given, and the speed of the beam is not. Without knowing the speed of the beam nor the distances, we would be short on variables. Since I am looking for the velocity of the beam, I have given the distances.

Nope, this mixes numbers from different frames again.

But we have to! Just like in a projectile analysis, the velocities are different for each frame. If I am correct and this is the case here too, we have to be given some set of coordinates. If you think it is necessary (and I don't, but I may be wrong) you can transform the distance AB into unprimed distances and do your analysis. Still you should find different observed velocities for the beam.

The velocity of an object moving at less than c *does* change when you change frames. The velocity of light does not. Please review a good basic relativity textbook, paying particular attention to the formula for relativistic velocity addition; you will note that that formula always gives c for the velocity of a light beam.

I am very aware of this, and this is where the problem lies. It is an assumption, I repeat. This beam cannot go at c on both frames just as a projectile on that same path can't have the same velocity in both frames. If you disagree, please explain HOW one setup differs from the other, and why should the beam's apparent velocity not change just like the projectile's. Remember, this beam is not seen directly, you have to find its velocity from the signals. This is a premise from the setup - how does the observer at A determines the speed of the beam given primed distances, primed times and observed (unprimed) times?

Read my latest analysis again, carefully; also read my comments above, carefully. Your error is that you are mixing numbers from different frames and expecting them to give you meaningful answers. Also, you may be mistakenly assuming that the distance AB is the same in both frames; it's not, for the reason I gave above.

Done. The distances are not the same for both frames, but the observer at A doesn't have a number for the (observed) velocity to determine the distance, he only has the observed time and the given distance (and also the primed time). Also, he has direct visual information of the primed distances, so he doesn't even need to transform h into h' - it is a given. That's enough variables to solve.

Then you don't understand SR very well.
This is all confused. All light travels at c in SR, in any inertial frame. If you read my latest analysis, you will see that the light beam traveling from B1a to D2 travels at c, even though it is not "observed" (both events are "at a distance" from the observer).

Notice we only need to focus on AB to reach our conclusions, since the question in simply if the speed of the beam is the same for both frames. You can end your analysis before taking D into consideration.

But you only get c for the beam because you are inserting the speed of the light clock into gamma, and you are not calculating the speed of the beam from the signals at all. Forget the light clock's velocity. Take it that the observer has no velocity information whatsoever. He has only primed and unprimed times and unprimed distances.
What you are actually finding is the speed of the beam relative to the light clock! Ditch the light clock's velocity, assume the observer has no acces to it. That velocity was there only to determine the hypotenuse, nothing more, but that can be a given (h').

Just as in standard light clock diagrams, you don't even need to use gamma - the time dilation transforms can actually derived from these diagrams. But they can only do it just like you did it: by inserting the speed of the light clock and with no consideration of how can the observer possibly determine the path of the beam, since it is not really observed.

The only data available to the observer is:
t'A = tA = 0s
t'B = 1s
AB = h' = 1.12 lightseconds
AC = y'= 1 lightsecond

From this, we conclude that the observed time for the event at B is simply:
tB = t'B + the time it takes for the signal from B to reach A (which is the time it takes light to cross AB).

Now find the speed of the beam as seen from A.

Conversely, replace the beam with a projectile, with this set of givens:
t'A = tA = 0s
t'B = 2s (this is the moment the projectile impacts B as seen from the primed frame)
AB = h' = 1.12 lightseconds
AC = y'= 1 lightsecond

Now find the speed of the projectile as seen from A.

The setup is the same and the operations should be equivalent. If both operations are not equivalent, tell my why. If the reason is something like "because the beam can't vary its speed" then you are assuming what you are trying to prove.

I can't settle with your analysis while you are using the speed of the light clock in the equations. You have to find the speed of the beam from the given times and distances alone, than we will have analysed the problem with the same set of givens and see in what ways we disagree.

 

[altergnostic]

No, it isn't. The invariance of the speed of light is a postulate of SR. If that is precisely your point then your point is WRONG.

Then prove it! Find the speed of the beam with my set of givens. The speed of the beam can be directly calculated with the givens.

I state again that the speed of light is c for directly observed light. It is c independent of the motion of the source or the receiver. But that says nothing about a non-receiver. The light signals arriving at A are traveling at c, as the postulate demands, but since the observer can't see the beam directly, he MUST calculate its speed based on signals from reflection events (in this setup).

I want you to calculate the speed of the beam. It is an inverse operation, to check if the beam is really observed to move at c. The givens are clear:

t'A = tA = 0s
t'B = 1s
h'= 1.12 lightseconds

you can find the time the observer sees the signal from B like so:
tB = t'B + the time it takes the signal from B to reach A

Calculate the speed of the beam.

For good measure, replace the beam with a projectile, with the givens:
t'A = tA = 0s
t'B = 2s
h'= 1.12 lightseconds

Find the speed of the projectile.

If the operations are not equivalent, please explain me why. If the reason is something like "because the beam must travel at c" than you are assuming what the setup tries to prove.

 

[altergnostic]

You're going to have to explain how this is "not hard to see", because I don't see it. The only thing that changes about the velocity from frame to frame is the sign: in the ground frame, the train moves at +v, and in the train frame, the ground moves at -v.

I suspect you are confused because you think length contraction affects the velocity from frame to frame. It doesn't, because length contraction and time dilation *both* come into play, and their effects cancel when applied to the relative velocity. If you don't see how that works, read through my analysis again, carefully; for example, look at how the coordinates of events A2 and D2 transform between the primed and the unprimed frame, and note that the time and space coordinates change in concert to keep the relative velocity the same.

[Edit: To expand on this a little more, events D0 and D2 both lie on the worldline of the light clock source/detector, which is equivalent to the "train"; and events A0 and A2 both lie on the worldline of the observer, which is equivalent to the "ground". Events A0 and D0 are co-located at the origin, so their coordinates drop out of the analysis, and we can use the coordinates of event D2 in the unprimed frame, and A2 in the primed frame, to compute the relative velocity. In the unprimed frame, event D2 has coordinates (t2, x2, y2), and the relative velocity is given by sqrt(x2^2 + y2^2) / t2, which works out to 0.5. In the primed frame, event A2 has coordinates (t2', x2', y2'), and the relative velocity is given by sqrt(x'2^2 + y2'^2) / t2', which works out to 0.5. Then we just have to remember to set the sign appropriately, based on the sign of (x2, y2) or (x2', y2'), which gives v = +0.5 in the unprimed frame and v = -0.5 in the primed frame.]



I don't see how starting another thread will help. This is an absolutely fundamental point that underlies the "problem" you posed in the OP, and all of my analysis, and indeed all of SR. If you think it's wrong, you basically think SR is wrong; and if you can't back your claim up cogently (which you haven't so far) and you won't abandon it, any further discussion of your claim is basically out of bounds here on PF, since we don't discuss personal speculations that contradict SR. Moving the discussion to another thread won't change that.

Well... I'm sorry to hear that. But I'll try again, anyway, since this is not a personal opinion, it is a direct mathematical and logical operation:

A train moves along the tracks with km markers on it, placed at every lightsecond distance. The train moves along the tracks at 0.5c, as seen on an onboard speedometer (conversely, the train observers the lightsecond markers moving at 0.5c, which amounts to the same thing).
As each lightsecond sign passes the train by, the train emits a signal to the observer at the origin. This signal acts like a clock, blinking once every 2 lightseconds.
At the origin, the observer's watch has been synchronized with the blinker, ticking once every two seconds.

By the time the train reaches the first lightsecond sign, it marks 2s and blinks. The signal has to cross the lightsecond distance back to the origin, which takes one second, so it reaches the origin 2 seconds after the blink event, so the observer will see the train (the blink) at the first lightsecond mark when his watch marks 3 seconds.

By direct observation, he will see the train moving at 1/3c, in disagreement with the train's onboard speedometer. The velocities are different for each frame.

The velocities are symmetric only if you transform from the velocity of the train as seen from the embankment to the velocity of the embankment as seen from the train, which is not what we are looking for. We are comparing the speed of the train relative to the tracks as observed from the embankment to the speed of the train relativ to the tracks as observed from the train.

 

[Dale]

Then prove it!

Sure: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

Find the speed of the beam with my set of givens.

The speed of any beam of light is c. Your givens may not be self-consistent or consistent with relativity so contradictory calculations from your givens don't prove anything. [EDIT: I checked and your "givens" are in fact wrong. My calculations agree with PeterDonis except for some rounding differences where I got e.g. 1.15 and he got 1.16. See details below]

 

[harrylin]

[..] A bullet doesn't have the same speed in both frames; I'll stick to the topic instead, which allows much better clarity. [..]

This is precisely my point! The same is true for the beam! [..] the bullet has a different velocity in each frame, and so must this beam. [..]

That is the exact contrary of the premises of SR. So, you are stuck at the starting point, as you still think that the following are irreconcilable :

We will raise this conjecture (the purport of which will hereafter be called the “Principle of Relativity”) to the status of a postulate, and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body.

- http://www.fourmilab.ch/etexts/einstein/specrel/www/

Now the calculation, which isn't anything really - it only gets interesting when you consider mutual measurements, but you didn't yet reach that point.
Anyway:

[..]
Also (from earlier):

"Just notice that by taking notes of the times of each x,y position, and the length of the line imprinted on the plate, you will directly calculate light to travel faster than c."

Perhaps it's time for you to present a calculation, based on either that figure or the photographic plates, and which you think would produce a speed of light greater than c. Probably an interpretation error will show up (but perhaps a calculation error).

[..] The setup should have the observer fixed at some origin. Consider this:

All lengths and distances are primed because they are all locally measured numbers. If you like, imagine that before the experiment we have walked all the distances and measured them with a measuring rod. You can even imagine that we have marked each km with a big sign with the km distance painted on it, so even in the moving system S we know the primed spatial coordinates from direct observation.

Fix an obsever at A, his frame denoted by S.

V=c/2= the velocity of the light clock in the x direction, as shown on the onboard speedometer. Also, this velocity has been previously agreed upon, so there's no doubt that's the speed of the light clock.

BC=y'=1 lightsecond = the distance between mirrors (measured locally, remember) and this is a given.

It's a bit unusual to call the moving frame S', but never mind, Still, I will modify your symbols a little: it's much less confusing if we give everything that is measured with S' an ' and no ' for everything that is measured with S. Thus:

BC=y=1 lightsecond = the distance between mirrors (measured with S) and this is a given.
In addition, according to the second postulate (see above) the same must be true as measured with S'. There is no disagreement about the height of the train: y=y'= 1 lightsecond according to both S and S'. I hope that you see that. A light ray that is sent from C to B will take 1 s according to S'.

AC=x't'=0.5 lightseconds = distance traveled by the light clock in the x direction in 1 s.

T_A=T'_A= 0 = time of first recorded reflection, at both S and S' origins, which are both at A.

T'_B≠1s. It is the time of the second reflection event as calculated in the rest frame S', based on the fact that CB= 1 light second. As a matter of fact, if according to S', two light flashes are sent simultaneously from A and C, the flash from C will arrive after 1s but the path length AB is longer, so it must take longer. Of course, it is assumed that the speed of light is the same in all directions; else no prediction is possible at all.

The second postulate effectively says that you may set up an inertial reference system, pretend that it is at rest in the ether, and everything will work as if your assumption was correct - just as in classical mechanics one may pretend that the system is "truly" in rest. And the error that you made here is literally a "classic": the same error was made by Michelson in 1886. As he admitted in 1887:

In deducing the formula for the quantity to be measured, the effect of the motion of the Earth through the ether on the path of the ray at right angles to this motion was overlooked.

- https://en.wikisource.org/wiki/On_the_Relative_Motion_of_the_Earth_and_the_Luminiferous_Ether

I think that you should read that, or look at an animation such as here:
http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/mmexpt6.htm
(note: ignore the word in red - hopefully that bug will be fixed in the coming week!)

Anyway, let's continue the -now trivial- analysis:

T'_B2=T'_B + T'_BA = the moment a reflection signal (i.e.: scatter event) from B is seen at A in S, which is the time light takes to travel from the reflection event T'_B to the observer at A. Of course, the trajectories AB and BA are the same, along h', and are assumed to take the same time.

This triangle represents the first two reflection events as seen in the moving frame S.

The light clock is going at 0.5c, as given by an onboard speedometer, useful in this setup to determine AC. [EDIT: correct one more "given" error: also AC was wrong!]
After 1 s in S' the light from A has not yet arrived at the top mirror and the train moves beyond x=0.5; x_C=0.5h'. The angle is 60° as you can easily verify. [STRIKE]Pythagoras[/STRIKE]Trigonometry: Light path in S' AB = h'=1/sin(60)= 1.155.
Or, still with Pythagoras: 1²-0.5²=1/k² =>k=1.155
h'=k*1=1.155 lightseconds.
Thus one mirror scatter is seen at A when T'=T=0 and the other at B when T=1, and T'=1.155

Note that you did not draw the figure for S, which is simply vertical with y=y'.

AB = h' = 1.155
T'_B2 = 2T'_B = 2.31 s
V_AB = 2h'/T'_B2 = 1 lightsecond/s

Does that help? Note that this is just the beginning...

In fact, it is quite useless to proceed if you don't master for example a classical analysis of the Michelson-Morley experiment.

 

[pervect]

Well... I'm sorry to hear that. But I'll try again, anyway, since this is not a personal opinion, it is a direct mathematical and logical operation:

A train moves along the tracks with km markers on it, placed at every lightsecond distance. The train moves along the tracks at 0.5c, as seen on an onboard speedometer (conversely, the train observers the lightsecond markers moving at 0.5c, which amounts to the same thing).

If I'm understanding you correctly, you've neglected to include the effects of length contraction. If the light second markers are placed every 1 second apart in the rest frame of the tracks, in the moving frame of the train they'll be closer together.

This is why "celerity" akak "prover velocity" is a different concept than velocity in SR.

See for instance http://arxiv.org/abs/physics/0608040 by Thomas Greber.

Intergalactic spaceflight: an uncommon way to relativistic kinematics and dynamics

In the Special Theory of Relativity space and time intervals are different in different frames of reference. As a consequence, the quantity 'velocity' of classical mechanics splits into different quantities in Special Relativity, coordinate velocity, proper velocity and rapidity. The introduction and clear distinction of these quantities provides a basis to introduce the kinematics of uniform and accelerated motion in an elementary and intuitive way.

...

Sometimes, one single quantity in a theory splits into several distinct quantities in the generalised theory. One example is the velocity in classical mechanics. In the widest sense, velocity or speed means the covered distance divided by the time needed to cover it. This is uncritical in classical physics, where time and distance are well defined operational concepts that are independent of the frame of reference, in which they are measured. In Special Relativity, these concepts depend on the frame of reference in which they are defined, if the frames are not at rest with respect to each other. This makes it necessary to distinguish between the different possibilities regarding the frame of reference in which the spatial and the temporal intervals are measured. This is best illustrated with a common situation of measuring the velocity of a rolling car. Firstly, the velocity of the car can be measured by driving past kilometer posts and reading the time at the moment of passing the post on synchronized watches mounted on the posts. Secondly, the driver can also measure the velocity by reading the corresponding times on a clock which is traveling with the car. Thirdly, a person with clock standing beside the street can measure the times on his clock at the moments, when the front and the rear ends of the car are passing him. The traveled distance is then taken from a measurement of the length of the car in the frame of reference of the car. A fourth possibility measures the velocity of the car up to an arbitrary constant by measuring its acceleration using an accelerometer traveling with the car, e.g. by measuring a force and using Newtons Second Law, and integrates the measured acceleration over the time measured with a clock, also traveling with the car.

In classical mechanics, all four measurements are equivalent and give the same value for the velocity. In Special Relativity, the first possiblility gives the coordinate velocity, which is often referred to as the genuine velocity. The second and third possibilities are equivalent, but are hybrid definitions of the speed. The temporal and spatial intervals are measured in different frames of reference. This speed is sometimes called celerity [1, 2], or proper velocity [3, 4]. In addition, proper velocity is the spatial part of the vector of four-velocity [5]. The fourth definition of a speed, sometimes called rapidity [1, 2], is somewhat distinct from the other concepts of speed in so far as it can only be determined as a change of speed. The need to measure an acceleration in the moving frame by means of measuring a force qualifies rapidity to bridge kinematics and dynamics. This seems to be not critical in classical mechanics, if the concept of force is accepted as an operational quantity. However, it can also be used to determine the relativistic version of Newtons Second Law if viewed from the accelerated frame of reference.

 

[PeterDonis]

That's right! But you are confusing the coordinates. He uses the primed distances and unprimed times. And that is allowed because he is directly seeing the km marks, you see. I have determined that in the setup, all distances have been previously walked and each km marked with a big sign. The observer can see those directly.

No, he can't. He has to get light signals from them, just like from everything else that's not at his spatial location. Putting km markers everywhere doesn't magically transmit information instantaneously.

In any case, no matter how he gets the numbers, you can't mix numbers from different frames and expect to get meaningful answers. Any valid calculation of the relative velocity has to use either an unprimed distance and corresponding unprimed time, or a primed distance and a corresponding primed time. Using a primed distance and unprimed time is just as meaningless as using an unprimed distance and a primed time.

But we have to! Just like in a projectile analysis, the velocities are different for each frame.

For an object that moves at less than the speed of light, the velocity of that same object is different in different frames, yes. But that doesn't apply to light; the velocity of a light beam is c in every frame (assuming inertial frames and flat spacetime). I've explicitly proven that to be the case in your scenarios.

Also, you have claimed that the *relative* velocity of two frames is different in one frame than in the other; this is the basis of your claim that you don't know which gamma factor to use. The fact that the velocity of the same object (if it moves slower than light) is different in different frames is irrelevant to that claim, because here you are not comparing the velocity of the same object in different frames; you are comparing the velocities of *different* objects in different frames. You are comparing the velocity of the train in the observer frame, with the velocity of the observer in the train frame. Those two velocities have the same magnitude, but opposite directions; I've proven that.

I am very aware of this, and this is where the problem lies. It is an assumption, I repeat.

And I repeat, this is *not* an assumption; it is a proven fact. I've proven it in my previous posts. You can prove that all light beams travel at c using only facts about the relative motion of the light clock and the observer (or the train and the ground), without making any assumptions about the speed of the light beam.

This beam cannot go at c on both frames just as a projectile on that same path can't have the same velocity in both frames. If you disagree, please explain HOW one setup differs from the other, and why should the beam's apparent velocity not change just like the projectile's.

Because the beam is moving at c, and the projectile is moving slower than c. The two cases are different. That's just a fact about how relativistic spacetime works. The only assumption I can see that is worth mentioning here is the isotropy of space (since that's really what underlies the claim that the velocity of the observer relative to the clock is equal in magnitude and opposite in direction to the velocity of the clock relative to the observer). That assumption is supported by a massive amount of evidence, so I don't see the point of questioning it here.

Remember, this beam is not seen directly, you have to find its velocity from the signals. This is a premise from the setup - how does the observer at A determines the speed of the beam given primed distances, primed times and observed (unprimed) times?

By using known facts about the speed of the light clock relative to him to establish the coordinates of the relevant events in the unprimed frame, independently of the speed of the light signals that travel between those events.

But you only get c for the beam because you are inserting the speed of the light clock into gamma

Well, of course. Why not?

and you are not calculating the speed of the beam from the signals at all.

I don't know what "calculating the speed of the beam from the signals" means. I'm calculating the speed of the beam from knowledge of the distance and time intervals between two known events on the beam's worldline.

Forget the light clock's velocity. Take it that the observer has no velocity information whatsoever. He has only primed and unprimed times and unprimed distances.

Then I can re-write all the formulas in terms of the unknown gamma factor for the relative velocity, and I will *still* get the same answer. All the factors of gamma end up canceling out. You should be able to see that from my analysis, but if not, I can re-write it this way (although you really should be able to do that yourself). In fact, I actually do re-write the formula for the velocity of the light beam in the light clock this way (for comparison with the velocity of a projectile in a "projectile clock") below.

The only data available to the observer is:
t'A = tA = 0s

Yes.

t'B = 1s

Yes. But you are aware, aren't you, that this implies tB = gamma, right?

AB = h' = 1.12 lightseconds

Is this supposed to be a distance in the unprimed frame? If so, how do you know it is consistent with the values for tA, t'A, and t'B that you just gave? If you start out with an inconsistent set of premises, of course you're going to get a meaningless answer.

AC = y'= 1 lightsecond

I don't see how this follows at all. AC is supposed to be the distance, in the unprimed frame, that the light clock source/detector travels during the time of flight of the beam, correct? In that case, it should be gamma, not gamma primed, because the time of flight of the beam, in the unprimed frame, is tB = gamma (see above).

So already I have spotted two wrong premises in your argument; no wonder you're getting erroneous answers.

Conversely, replace the beam with a projectile, with this set of givens:
t'A = tA = 0s

Yes.

t'B = 2s (this is the moment the projectile impacts B as seen from the primed frame)

Yes. But again, you are aware that this implies tB = 2 gamma, right?

AB = h' = 1.12 lightseconds
AC = y'= 1 lightsecond

Same comment as above. This value of AB is not consistent with the above, and AC is 2 gamma, not gamma primed, or even 2 gamma primed, which is what you should have written by the same reasoning as the above--t'B is twice as large, so AB should be twice as large as well since the light clock spends twice as much time traveling as before. You are aware, aren't you, that using a projectile traveling at 0.5c, instead of a light beam traveling at c, changes the geometry of your triangle diagram, right? It's *not* the same triangle.

The setup is the same and the operations should be equivalent. If both operations are not equivalent, tell my why. If the reason is something like "because the beam can't vary its speed" then you are assuming what you are trying to prove.

I've already done the light beam case in my previous analysis, but perhaps it would help to do a similar analysis on the "projectile clock" case; we'll assume a projectile traveling at 0.5c in the primed frame, in the direction perpendicular to the relative motion of the light clock and the observer in that frame. I'll orient the x-axis in the direction the projectile travels, and the y-axis in the direction of the relative motion of the "projectile clock" and the observer. (Yes, I know you think that the x-axis has to point directly along the path to where the mirror will be in the unprimed frame when the projectile hits it. I've already shown that that doesn't matter; I don't think it needs to be shown again. But rotating the spatial axes will be easy enough after the analysis I'm about to show, if you really insist on seeing it done that way.)

We are given the following event coordinates:

A0 = D0 = A0' = D0' = (0, 0, 0)
B1a' = B1b' = (2, 1, 0)
D2' = (4, 0, 0)

This assumes a "projectile" traveling at 0.5c in the primed frame, in the x-direction.

The transformation equations are (I'm writing them now in terms of an unknown gamma factor, since you are now saying we don't know the relative velocity of the light clock and the observer):

Unprimed to Primed:

t' = \gamma ( t - v y )
x' = x
y' = \gamma ( y - v t )

Primed to Unprimed:

t = \gamma ( t' + v y' )
x = x'
y = \gamma ( y' + v t' )

This yields the following coordinates for events in the unprimed frame:

B1a = B1b = ( 2 \gamma, 1, 2 \gamma v )
D2 = ( 4 \gamma, 0, 4 \gamma v )

The velocity of the projectile in the unprimed frame on each leg can then be computed like this:

v_{projectile} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + 4 \gamma^2 v^2}}{2 \gamma} = \frac{1}{2} \sqrt{(1 - v^2) \left( 1 + \frac{4 v^2}{1 - v^2} \right)} = \frac{1}{2} \sqrt{\frac{(1 - v^2)(1 - v^2 + 4 v^2)}{1 - v^2}} = \frac{1}{2} \sqrt{1 + 3 v^2}

This will be somewhere between 1/2, which is the velocity of the projectile in the primed frame, and 1 (but always less than 1 for v < 1); but it is only *equal* to 1/2 for v = 0, so you are correct that the velocity of the projectile in a "projectile clock" *does* change if you change frames.

I could leave it to you to spot the difference between this and the case of the light clock, but I suppose I'll go ahead and give it; the corresponding formula from my previous analysis would be (written with an unknown gamma factor to make it clear how it drops out of the analysis):

v_{beam} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + \gamma^2 v^2}}{\gamma} = \sqrt{(1 - v^2) \left( 1 + \frac{v^2}{1 - v^2} \right)} = \sqrt{\frac{(1 - v^2)(1 - v^2 + v^2)}{1 - v^2}} = 1

As you can see, this formula always gives 1, regardless of v (as long as v < 1). So unlike the projectile case, the speed of a light beam in a light clock does *not* change when you change frames.

 

[PeterDonis]

Just to put some icing on the cake, here's an even more general formula which works for arbitrary "projectile" velocities (thus including the case of the light beam as the "projectile" as well).

We have a clock using a "projectile" that travels between a source/detector and a reflector. The projectile's velocity is v_p&#039; in the rest frame of the clock. We are given the following event coordinates (the projectile travels in the x direction):

A0 = D0 = A0&#039; = D0&#039; = (0, 0, 0)
B1a&#039; = B1b&#039; = (1 / v_p&#039;, 1, 0)
D2&#039; = (2 / v_p&#039;, 0, 0)

The transformation equations are as before, and they yield the following coordinates for events in the unprimed frame:

B1a = B1b = ( \gamma / v_p&#039;, 1, \gamma v / v_p&#039;)
D2 = ( 2 \gamma / v_p&#039;, 0, 2 \gamma v / v_p&#039;)

The velocity of the projectile in the unprimed frame on each leg is then:

v_p = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + \gamma^2 v^2 / v_p&#039;^2}}{\gamma / v_p&#039; } = v_p&#039; \sqrt{(1 - v^2) \left( 1 + \frac{v^2}{v_p&#039;^2 (1 - v^2 )} \right)} = v_p&#039; \sqrt{\frac{(1 - v^2)(1 - v^2 + v^2 / v_p&#039;^2)}{1 - v^2}} = v_p&#039; \sqrt{1 + v^2 \left( \frac{1}{v_p&#039;^2} - 1 \right)}

For 0 &lt; v_p&#039; &lt; 1, this gives v_p&#039; &lt; v_p &lt; 1 for any 0 &lt; v &lt; 1. For v_p&#039; = 1, this gives v_p = 1 for any 0 &lt; v &lt; 1. QED.

 

[Dale]

The givens are clear:

t'A = tA = 0s
t'B = 1s
h'= 1.12 lightseconds

Your givens are wrong. h' is 1.15, not 1.12. In the frame where the clock is moving the clock ticks slowly by a factor of 1.15, so in 1.15 s moving at 0.5 c it travels a distance of 0.577 ls, not 0.5 ls. The Pythagorean theorem gives \sqrt{1^2+0.577^2}=1.15 ls, which contradicts your supposed "givens". The light travels 1.15 ls in 1.15 s which is c.

 

[harrylin]

Your givens are wrong. h' is 1.15, not 1.12. In the frame where the clock is moving the clock ticks slowly by a factor of 1.15, so in 1.15 s moving at 0.5 c it travels a distance of 0.577 ls, not 0.5 ls. The Pythagorean theorem gives \sqrt{1^2+0.577^2}=1.15 ls, which contradicts your supposed "givens". The light travels 1.15 ls in 1.15 s which is c.

Oops I had overlooked that error in his "givens"! Thanks I'll also correct that for consistency.

 

[altergnostic]

BC=y=1 lightsecond = the distance between mirrors (measured with S) and this is a given.*

In addition, according to the second postulate (see above) the same must be true as measured with S'. There is no disagreement about the height of the train: y=y'= 1 lightsecond according to both S and S'. I hope that you see that. A light ray that is sent from C to B will take 1 s according to S'.*

Agreed, but noticed that this y is measured at the origin (it may not seem important, but it is, because this y will not be the y after the spatial rotation - where we must line up the x-axis of the observer to AB)



AC=x't'=0.5 lightseconds = distance traveled by the light clock in the x direction in 1 s.

*

T_A=T'_A= 0 = time of first recorded reflection, at both S and S' origins, which are both at A.

*

T'_B≠1s. It is the time of the second reflection event as calculated in the rest frame S', based on the fact that CB= 1 light second. As a matter of fact, if according to S', two light flashes are sent simultaneously from A and C, the flash from C will arrive after 1s but the path length AB is longer, so it must take longer. Of course, it is assumed that the speed of light is the same in all directions; else no prediction is possible at all.*

The S' frame is the rest frame (not the moving frame), it is the measurements done from inside the light clock itself. So T'_B = 1s

^RED mine.

The second postulate effectively says that you may set up an inertial reference system, pretend that it is at rest in the ether, and everything will work as if your assumption was correct...

No problem here, actually, if that isn't true my analysis fails. I don't know where is the misunderstanding coming from, though... I may need to clarify the problem further, since you have already shown that you misunderstood what I have been calling the primed frame (it is the rest frame).

Anyway, let's continue the -now trivial- analysis:

Yes!

T'_B2=T'_B + T'_{BA[/SUB = the moment a reflection signal (i.e.: scatter event) from B is seen at A in S, which is the time light takes to travel from the reflection event T'B to the observer at A. Of course, the trajectories AB and BA are the same, along h', and are assumed to take the same time.}

Ah, the point of the setup in the first place - to check if this remains consistent.

If the observer at rest in A sent a flash of light from A to B, it would take the same time to reach B as the time the signal takes to reach A from B. BUT, we aren't dealing with that here. What we are seeking is not how fast light travels from A to B as seen from B or B to A from A. From A to B, it would seem to travel at c for an observer at B (and it would take 1.12 seconds to do so if h'=1.12 lightseconds). From B to A, it will seem to travel at c for the observer at A. But we are seeking how fast would the beam would seem to travel from A to B as seen from A. That's why we need the signal from B in the first place. As a matter of fact, our givens show that light takes 1s to go from A to B as seen from inside the lightclock (the primed frame), where the distance is 1 lightsecond.

This triangle represents the first two reflection events as seen in the moving frame S.

Pythagoras: Light path in S' AB = √(1² + 0.5²)=1.12.

Thus one mirror is seen at A when T'=T=0 and the other at B when T=1, and T'=1.12[/QUOTE]

Actually, its the other way around: T' is the time in the rest frame - the time seen from the lightclock itself - and that is 1s.*

The light clock is going at 0.5c, as given by an onboard speedometer, useful in this setup to determine AC.*

Note that you did not draw the figure for S, which is simply vertical with y=y'.

*I didn't draw it because it is not necessary. Again, you may take that he only knows the primed times, sees the marked distance signs and receives the signal from B at a time TB.

AB = h' = 1.12

T'_B2 = 2T'_B = 2.24 s

V_AB = 2h'/T'_B2 = 1 lightsecond/s

You see right there! You just assumed the beam takes 1.12 seconds to go fram A to B as seen from A! You just assumed what had to be proven here. What you need to take into consideration is the primed time of event B and calculate the observed time for that event as seen from A, and then calculate the observed time from A to B.

In fact, it is quite useless to proceed if you don't master for example a classical analysis of the Michelson-Morley experiment.

But I do, and as I keep saying, this a different case. In MM we were always measuring the speed of received light, or light detected directly, which applies to the received signal from B in this setup. The postulate of the speed of light strictly states that c is absolute for the source and the observer regardlesd of motion, but that is clearly light that is in direct contact, in direct line, with either source or observer. This setup presents a novel configuration, which is light that doesn't reach the observer, and we have never performed any such experiment seeking the speed of distant light, as far as I know, so we have to be very careful with our assumptions and with our interpretation of the postulate of the speed of light. The goal of this setup is to calculate the speed of the beam, and clearly assumed this value so your analysis can't be proof that the speed of the beam as seen from A would actually be c.

 

[harrylin]

It seems that we were talking past each other, so I'll first only address the first two or three points:

^RED mine.

Once more: there is no spatial rotation here. Nothing rotates at all.
And once more, that was already explained in the link I gave, and of which you claimed to understand it (but evidently you don't):
https://www.physicsforums.com/showthread.php?t=574757

Please explain it in your own words before discussing further.

[..] I may need to clarify the problem further, since you have already shown that you misunderstood what I have been calling the primed frame (it is the rest frame).

Yes that is what I wrote: you clarified that S' is the rest frame, in which the light clock is moving; and S is the moving frame, in which the light clock is in rest. If you agree, then there is no misunderstanding about this.

Still it may be that the problem comes from confusion between frames, as you next write that "This setup presents a novel configuration, which is light that doesn't reach the observer" - however light always only reaches light detectors!

It may also come from a misunderstanding of the light postulate, or a combination of both, as you next write:

"The postulate of the speed of light strictly states that c is absolute for the source and the observer regardlesd of motion".

That is certainly not the second postulate. Did you study the Michelson-Morley calculation? And if so, do you understand it? Then please explain it.

PS. This forum is meant to explain how SR works. It is not meant to "prove" a theory. As a matter of fact, such a thing is impossible!

 

[Dale]

The goal of this setup is to calculate the speed of the beam

This stated goal is mathematically impossible. By the Pythagorean theorem for any signal speed C we have the geometrical relationship C^2 t^2 = y^2 + v^2 t^2. If you fix y and v then assuming a value for t is the same as assuming a value for C. Here you are incorrectly assuming that t=0.5, which is equivalent to assuming that C≠c.

and clearly assumed this value so your analysis can't be proof that the speed of the beam as seen from A would actually be c.

Yes. We ASSUME that the speed of the beam is c. That is what a postulate is, a fundamental assumption of the theory. If you are assuming anything contrary (which you are) then you are not doing SR. From that assumption we can then calculate the value of t using the relationship above.

There is no mathematical proof of the validity of the assumption (you can't mathematically prove postulates or axioms, by definition), however an enormous body of experimental evidence supports the assumption and contradicts your alternative assumption.

 

[PeterDonis]

The goal of this setup is to calculate the speed of the beam

In my analysis, I *calculated* (*not* assumed) the speed of the light pulse traveling between events B1a and D2, neither of which are co-located with the observer, so this is a "distant light" beam by your definition. Its speed is *calculated* to be c. So my analysis meets your goal. What's the problem?

 

[altergnostic]

No, he can't. He has to get light signals from them, just like from everything else that's not at his spatial location. Putting km markers everywhere doesn't magically transmit information instantaneously.

I don't mean that he sees anything instantaneously, this is just how he will measure anything. Length contraction applies to vt, but the markers are just distances and they are at rest relative to the observer. These markers belong to both frames, and they are not being measured by any of them. You could call them x'' or something. They are given and are not being measured in the process. Usually, we are given V and find the distance from that, but since we are seeking V, I have to give L somehow. A given V contains a given L which is the same for both frames, you see. This is the given distance. The time it would take for light to go from A to B, as seen from B, is the same as from B to A seen from A, so this distance must be the same and it is not being measured by any of the frames. It is a given just like the distance implied in the given V from most SR thought problems. If you can have the same velocity between two frames, you already have an equivalent distance between two frames, over the standard unit of time.

In any case, no matter how he gets the numbers, you can't mix numbers from different frames and expect to get meaningful answers. Any valid calculation of the relative velocity has to use either an unprimed distance and corresponding unprimed time, or a primed distance and a corresponding primed time. Using a primed distance and unprimed time is just as meaningless as using an unprimed distance and a primed time.

See above.

For an object that moves at less than the speed of light, the velocity of that same object is different in different frames, yes. But that doesn't apply to light; the velocity of a light beam is c in every frame (assuming inertial frames and flat spacetime). I've explicitly proven that to be the case in your scenarios.

You have not proven it, you have assumed it, since the operation you developed is not the same as you would if the beam was a projectile. And you simply stated that it must be different because we are supposed to know that the beam is moving at c from A to B as seen from A, and that's not a proof, that is an assumption. You claim it is an assumption backed up by numerous experiments, but not a single experiment in history has measured the speed of light that does not reach a detector directly, you have to aknowledge this. I state again that c applies to light directly detected, light that reaches and "touches" the observer.

Also, you have claimed that the *relative* velocity of two frames is different in one frame than in the other; this is the basis of your claim that you don't know which gamma factor to use. The fact that the velocity of the same object (if it moves slower than light) is different in different frames is irrelevant to that claim, because here you are not comparing the velocity of the same object in different frames; you are comparing the velocities of *different* objects in different frames. You are comparing the velocity of the train in the observer frame, with the velocity of the observer in the train frame. Those two velocities have the same magnitude, but opposite directions; I've proven that.

That is correct, and it is obvious anyway. But in this case we ARE measuring the speed of the same "object" in different frames: the setup is meant to give us the oportunity to measure the speed of the beam from both frames.

And I repeat, this is *not* an assumption; it is a proven fact. I've proven it in my previous posts. You can prove that all light beams travel at c using only facts about the relative motion of the light clock and the observer (or the train and the ground), without making any assumptions about the speed of the light beam.

You haven't shown this at all, Peter. If you carefully check your analysis you will see that you assumed the observed speed of the beam seen from A is c, not shown it.

[/QUOTE]Because the beam is moving at c, and the projectile is moving slower than c. The two cases are different. That's just a fact about how relativistic spacetime works. The only assumption I can see that is worth mentioning here is the isotropy of space (since that's really what underlies the claim that the velocity of the observer relative to the clock is equal in magnitude and opposite in direction to the velocity of the clock relative to the observer). That assumption is supported by a massive amount of evidence, so I don't see the point of questioning it here.[/QUOTE]

Now you do assume it is an assumption. The evidence shows the speed of light is c for directly detected light, as I said above. I am not aware of any experiments that allow us to measure the speed of non-observed light, in the sense that this is light that doesn't reach a detector directly.

By using known facts about the speed of the light clock relative to him to establish the coordinates of the relevant events in the unprimed frame, independently of the speed of the light signals that travel between those events.

Again, assumptions. And not backed up by evidence at all - evidence shows that detected light moves at c. No one can propose that the signal from B to A moves at any other speed than c, since that would contradict every single experiment ever done to measure the speed of light. Since no experiment has ever been performed with distant and indirectly detected light, the assumption that the beam moves at c is not supported by any evidence whatsoever. Not for, nor against.

Well, of course. Why not?

Because you must calculate the speed of the beam before assuming it is c, since that is the point of the setup in the first place: the variables given allows us to measure the speed of the beam.

I don't know what "calculating the speed of the beam from the signals" means. I'm calculating the speed of the beam from knowledge of the distance and time intervals between two known events on the beam's worldline.

It means giving c to the signal and plotting the observed times if events A and B against the given L.

Then I can re-write all the formulas in terms of the unknown gamma factor for the relative velocity, and I will *still* get the same answer. All the factors of gamma end up canceling out. You should be able to see that from my analysis, but if not, I can re-write it this way (although you really should be able to do that yourself). In fact, I actually do re-write the formula for the velocity of the light beam in the light clock this way (for comparison with the velocity of a projectile in a "projectile clock") below.

But you didn't do it relative to A, or "as the observer at A would see it", as I said above.

Yes. But you are aware, aren't you, that this implies tB = gamma, right?

You can't assume that since you aren't given the speed of the beam. What V do you insert on gamma?

Is this supposed to be a distance in the unprimed frame? If so, how do you know it is consistent with the values for tA, t'A, and t'B that you just gave? If you start out with an inconsistent set of premises, of course you're going to get a meaningless answer.

It is a distance measured locally prior to the experiment. None of the frames are measuring it, they are given it. Just like the given distance implied if you were given V.

I don't see how this follows at all. AC is supposed to be the distance, in the unprimed frame, that the light clock source/detector travels during the time of flight of the beam, correct? In that case, it should be gamma, not gamma primed, because the time of flight of the beam, in the unprimed frame, is tB = gamma (see above).

This is a typo, it was meant to be BC, sorry for this. But I guess you did notice this honest mistake, didn't you? The diagram shows clearly where the y' distance stands for.

So already I have spotted two wrong premises in your argument; no wonder you're getting erroneous answers.

You spotted a typo and a given that is also given if you were given V, so no wrong premises here.

Yes. But again, you are aware that this implies tB = 2 gamma, right?

Not really. First you have to find the velocity, otherwise what V do you plug into gamma? It may be, it may not be, you have to show that it is, the setup is supposed to allow us to calculate the speed. Than you can show what is the value for gamma.

Same comment as above. This value of AB is not consistent with the above, and AC is 2 gamma, not gamma primed, or even 2 gamma primed, which is what you should have written by the same reasoning as the above--t'B is twice as large, so AB should be twice as large as well since the light clock spends twice as much time traveling as before. You are aware, aren't you, that using a projectile traveling at 0.5c, instead of a light beam traveling at c, changes the geometry of your triangle diagram, right? It's *not* the same triangle.

The same typo. Also, although you are right and the triangle does change, the hypotenuse is still AB, the distances are still given and the setup is the same. I haven't done the calculations to find the values, though, but it should be clear that the setup is equivalent, and so the operation should be, else you lose consistency and have to prove why this should be so. Since your reason is experimental evidence, I remind you again that no experiment has measured the speed of undetected or indirectly observed light or distant light or light that does not reach the observer/detector.

I've already done the light beam case in my previous analysis, but perhaps it would help to do a similar analysis on the "projectile clock" case; we'll assume a projectile traveling at 0.5c in the primed frame, in the direction perpendicular to the relative motion of the light clock and the observer in that frame. I'll orient the x-axis in the direction the projectile travels, and the y-axis in the direction of the relative motion of the "projectile clock" and the observer. (Yes, I know you think that the x-axis has to point directly along the path to where the mirror will be in the unprimed frame when the projectile hits it. I've already shown that that doesn't matter; I don't think it needs to be shown again. But rotating the spatial axes will be easy enough after the analysis I'm about to show, if you really insist on seeing it done that way.)

We are given the following event coordinates:

A0 = D0 = A0&#039; = D0&#039; = (0, 0, 0)

I assume that D0 here stands for A0 right? You are specifying that they are the same point in the projectile-clock, right? I'm asking this because D is really unecessary after all, we can focus on A, B and C and find what we seek from that.

B1a&#039; = B1b&#039; = (2, 1, 0)
D2&#039; = (4, 0, 0)

This assumes a "projectile" traveling at 0.5c in the primed frame, in the x-direction.

The transformation equations are (I'm writing them now in terms of an unknown gamma factor, since you are now saying we don't know the relative velocity of the light clock and the observer):

I'm saying we don't know the PROJECTILE velocity. We don't need the clock velocity at all, since we are given the basic coordinates that must compose the relative velocity - which is the unknown we seek.

Unprimed to Primed:

t&#039; = \gamma ( t - v y )
x&#039; = x

The vy term should be vAB, since the speed from A to B is not the same from A to C - that would be the speed of the clock, but we are transforming the projectile, not the clock.

y&#039; = \gamma ( y - v t )

Primed to Unprimed:

t = \gamma ( t&#039; + v y&#039; )
x = x&#039;

See above

y = \gamma ( y&#039; + v t&#039; )

This yields the following coordinates for events in the unprimed frame:

B1a = B1b = ( 2 \gamma, 1, 2 \gamma v )
D2 = ( 4 \gamma, 0, 4 \gamma v )

The velocity of the projectile in the unprimed frame on each leg can then be computed like this:

v_{projectile} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + 4 \gamma^2 v^2}}{2 \gamma} = \frac{1}{2} \sqrt{(1 - v^2) \left( 1 + \frac{4 v^2}{1 - v^2} \right)} = \frac{1}{2} \sqrt{\frac{(1 - v^2)(1 - v^2 + 4 v^2)}{1 - v^2}} = \frac{1}{2} \sqrt{1 + 3 v^2}

This will be somewhere between 1/2, which is the velocity of the projectile in the primed frame, and 1 (but always less than 1 for v < 1); but it is only *equal* to 1/2 for v = 0, so you are correct that the velocity of the projectile in a "projectile clock" *does* change if you change frames.

Yes, but that is the velocity you should plug into gamma (except in the case where the projectile and the clock motion were in the same line). You must do a vector addition and find the speed of the projectile from the unprimed frame at some point, if you want to use a velocity value, but you don't a variable v at all to solve. So gamma is compromised.

I could leave it to you to spot the difference between this and the case of the light clock, but I suppose I'll go ahead and give it; the corresponding formula from my previous analysis would be (written with an unknown gamma factor to make it clear how it drops out of the analysis):

v_{beam} = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + \gamma^2 v^2}}{\gamma} = \sqrt{(1 - v^2) \left( 1 + \frac{v^2}{1 - v^2} \right)} = \sqrt{\frac{(1 - v^2)(1 - v^2 + v^2)}{1 - v^2}} = 1

Again, how do you know what V to use, since you aren't given any? You must find a way to solve without it, since I have only given the L component of the relative velocity. You have to find the observed velocity, since that's what we are seeking. As it is, you have two different velicities in your equations, and you are given none, so the only thing I can conclude from your analysis (if it is correct, and I don't think it is for the reasons above) is that the setup doesn't give enough information to solve. But I know this isn't your conclusion.

As you can see, this formula always gives 1, regardless of v (as long as v < 1). So unlike the projectile case, the speed of a light beam in a light clock does *not* change when you change frames.

This is only so because you assume Vbeam = c in both frames. You assume what has to be proven.

Just to put some icing on the cake, here's an even more general formula which works for arbitrary "projectile" velocities (thus including the case of the light beam as the "projectile" as well).

We have a clock using a "projectile" that travels between a source/detector and a reflector. The projectile's velocity is v_p&#039; in the rest frame of the clock. We are given the following event coordinates (the projectile travels in the x direction):

A0 = D0 = A0&#039; = D0&#039; = (0, 0, 0)
B1a&#039; = B1b&#039; = (1 / v_p&#039;, 1, 0)
D2&#039; = (2 / v_p&#039;, 0, 0)

The transformation equations are as before, and they yield the following coordinates for events in the unprimed frame:

B1a = B1b = ( \gamma / v_p&#039;, 1, \gamma v / v_p&#039;)
D2 = ( 2 \gamma / v_p&#039;, 0, 2 \gamma v / v_p&#039;)

The velocity of the projectile in the unprimed frame on each leg is then:

v_p = \frac{\sqrt{\Delta x^2 + \Delta y^2}}{\Delta t} = \frac{\sqrt{1 + \gamma^2 v^2 / v_p&#039;^2}}{\gamma / v_p&#039; } = v_p&#039; \sqrt{(1 - v^2) \left( 1 + \frac{v^2}{v_p&#039;^2 (1 - v^2 )} \right)} = v_p&#039; \sqrt{\frac{(1 - v^2)(1 - v^2 + v^2 / v_p&#039;^2)}{1 - v^2}} = v_p&#039; \sqrt{1 + v^2 \left( \frac{1}{v_p&#039;^2} - 1 \right)}

For 0 &lt; v_p&#039; &lt; 1, this gives v_p&#039; &lt; v_p &lt; 1 for any 0 &lt; v &lt; 1. For v_p&#039; = 1, this gives v_p = 1 for any 0 &lt; v &lt; 1. QED.

I prefer this analysis over the previous one, but the problem remains: you are not given any v, you must find another way to solve (which I believe is only possible if you can find vp - the velocity observed from A, in my diagram), and that was the purpose of my setup, and the reason I gave the L component for the relative velocity(s).

If you still have trouble with my given x' and y' - maybe a better form would be X and Y since they are given for both frames - ask yourself what L is implied in the relative V.

 

[altergnostic]

Your givens are wrong. h' is 1.15, not 1.12. In the frame where the clock is moving the clock ticks slowly by a factor of 1.15, so in 1.15 s moving at 0.5 c it travels a distance of 0.577 ls, not 0.5 ls. The Pythagorean theorem gives \sqrt{1^2+0.577^2}=1.15 ls, which contradicts your supposed "givens". The light travels 1.15 ls in 1.15 s which is c.

h' is not being measured, it is given, it is supposed to be the constituent L of the relative V that is usually given in SR problems, which we are not given in this setup and is the unknown we seek.

 

[altergnostic]

Oops I had overlooked that error in his "givens"! Thanks I'll also correct that for consistency.

Please see my answer above and also my reply to Peter.

 

[harrylin]

Please see my answer above and also my reply to Peter.

Your "givens" are according to SR selfcontradictory; when I corrected it, I showed this by two methods of calculation and I explained it with words. Note that both my approaches differed a little from that of Dalespam and PeterDonis; I only used your input.

 

[altergnostic]

It seems that we were talking past each other, so I'll first only address the first two or three points:

Yes, it is getting harder by the relative minute to keep up with so many posts.

Once more: there is no spatial rotation here. Nothing rotates at all.
And once more, that was already explained in the link I gave, and of which you claimed to understand it (but evidently you don't):
https://www.physicsforums.com/showthread.php?t=574757

Please explain it in your own words before discussing further.

The spatial rotation refers to the observer line of sight. I propose that he places his x-axis in line with AB so he can give all the motion to beam and solve without knowledge of the clock's velocity.

From the link, I will comment only upon the same issue I have been discussing here: the beam leaves A and reaches B in the rest frame (the frame that is at rest relative to the clock) and moves at c as seen from a detector at B. From the perspective of an observer that sees the clock in relative motion, the brings no data and he can't say anything about its observed velocity, since it is not observed. By including a signal sent from B to A at the moment the beam reaches B, the observer will have the observed time for that event and may determine time of the event as recorded on the detector by taking into consideration the observed time and the predetermined distance between him and the detector at B, just like the standard relative V implies a given distance/unit of time.

Yes that is what I wrote: you clarified that S' is the rest frame, in which the light clock is moving; and S is the moving frame, in which the light clock is in rest. If you agree, then there is no misunderstanding about this.*

Maybe my terminology is incorrect? The rest frame is meant to be the frame at rest relative to the clock. The moving frame is supposed to be the observer who sees the moving clock.

Still it may be that the problem comes from confusion between frames, as you next write that "This setup presents a novel configuration, which is light that doesn't reach the observer" - however light always only reaches light detectors!*

The beam never reaches the observer at A. Only the signal from B does. He then marks the observed time of event B and plots it against the given L. He can then find the time of event B as marked on the detector's clock by subtracting the time it takes light to cross the distance between the coordinates of the event at B from the observed time. From that, he can calculate the distance from source to receiver as seen from the clock itself by transforming the calculated speed of the beam into c (the speed of light as seen locally, which is the speed supported by a huge amount of evidence). So you see, we only need to know the predetermined distance from A to B, the given time of detection as seen from the clock and the observed time of event B, to figure everything else

It may also come from a misunderstanding of the light postulate, or a combination of both, as you next write:

"The postulate of the speed of light strictly states that c is absolute for the source and the observer regardlesd of motion".*

That is certainly not the second postulate. Did you study the Michelson-Morley calculation? And if so, do you understand it? Then please explain it.

The interferometer was supposed to test the existence of the aether and find a variance of the speed of light relative to the ether. The setup was built so we had perdicular light paths. As the apparatus (and the Earth) revolved, no fringe displacement was detected (actally, no significant displacement). So it was concluded that the speed of light was constant regardless of the orientation of the beams or the detectors relative to an absolute frame. Notice that in the equations applied to this experiment, V was the relative velocity wrt the aether, so there was only one possible V, as the aether was absolute. Since then, the speed if light was measured with ever increasing accuracy, always in the same manner: by noting the return time if light as it went forth and back a specified distance. The time of detection after emission is always proportional to c. This is exactly the premise of my setup: light can't be detected to move at any V other than c. Just notice that the beam is not detected by A, only the signal from B is. There's no return, no detection, nothing that relates the path of the beam with the experiments that tested the speed of light.

PS. This forum is meant to explain how SR works. It is not meant to "prove" a theory. As a matter of fact, such a thing is impossible!

This is no theory other than SR itself, since the same postulates apply to detected light, consistent with every experiment ever done. A correction to an incorrect or incomplete diagram or a new thought problem that does not contradict but expands theory is not a new theory, since it is built upon the same postulates. This is merely the outcome of a realisation that this thought problem has never been done before and that the postulate of the speed of light applies to source and receiver, but not necessarily to a non-receiver. This setup is meant to analyse this realisation, keeping all aspects of Einstein's original theory intact and remaining consistent with all available empirical data.

It is the preconception that the postulate of the speed of light also applies to undetected light that keeps you from aknowledging the possibility I intend to discuss here.
Under close inspection, you may realize that neither the postulate nor the evidence are in contradiction with my conclusions, but that it is a necessary outcome of relativity. It is only ligical, if light reaches us at a constant speed, distance events will be seen at a later time than the time of the event determined locally (for an observer in close proximity and at rest relative to the event). Hence, if the observed time is delayed and distances are given, the observed speed must be smaller.

To realize this is consistent with SR may be difficult, but it must be, since the conclusion is strictly dependent upon the acceptance of the constancy of the speed of any received or detected light.

 

[altergnostic]

Your "givens" are according to SR selfcontradictory; when I corrected it, I showed this by two methods of calculation and I explained it with words. Note that both my approaches differed a little from that of Dalespam and PeterDonis; I only used your input.

Why are they self contradictory? If I walked from A to B and measured the distance, wouldn't this distance be the same you would measure by walking from B to A? This is as symmetrical as any relative velocity, as it must be. If there is no symmetric distance between frames, there can be no symmetric speed. Do you disagree? If so, please explain why and show me how to determine relative velocity without a equally symmetric relative distance.

Are there other givens you still disagree with?

 

[altergnostic]

A footnote: length contraction usually applies to the distance between two bodies in relative motion, meaning, it usually applies to a length that is changing with time. In the proposed setup, the distance is fixed and has been previously marked. It is the symmetric distance between A and B.

 

[Dale]

h' is not being measured, it is given

I realize that, which is why I said that your givens are wrong. Givens can be wrong either by being inconsistent with themselves (e.g. given a right triangle with three equal sides) or by being inconsistent with the laws of physics (e.g. given a moving mass 2m which comes to rest after colliding elastically with a mass m initially at rest).

it is supposed to be the constituent L of the relative V that is usually given in SR problems, which we are not given in this setup and is the unknown we seek.

I thought that v=0.5c was also a given, is it not? If v is not given then you can simply plug your other givens into c^2 t^2 = y^2 + v^2 t^2 where t=h'/c and then solve for v.

 

[Dale]

A footnote: length contraction usually applies to the distance between two bodies in relative motion, meaning, it usually applies to a length that is changing with time.

This is an incorrect understanding of length contraction. I recommend starting with the Wikipedia article on length contraction. Length contraction has nothing to do with the distance between two bodies in motion. It also does not refer to a change in length over time, it refers to a difference in the same length as measured by two different frames.

 

[altergnostic]

I realize that, which is why I said that your givens are wrong. *Givens can be wrong either by being inconsistent with themselves (e.g. given a right triangle with three equal sides) or by being inconsistent with the laws of physics (e.g. given a moving mass 2m which comes to rest after colliding elastically with a mass m initially at rest).

I thought that v=0.5c was also a given, is it not? *If v is not given then you can simply plug your other givens into c^2 t^2 = y^2 + v^2 t^2 where t=h'/c and then solve for v.

I have reestaded many times that you could ditch the speed of the light clock, since it is useless to this problem. What speed would you insert into gamma in the train and embankment problem? We would have to insert the speed of the projectile/beam and align our x-axis with AB in my setup, do you see? If you were given the projectile speed and the times you could calculate the distance AB, this is simply the other way around: given the distance and times you can calculate the speed. The speed of the light clock is completely trivial and you can solve even if the observer at A doesn't know it, since he is given distances and knows all times. From my analysis you can actually derive both the speed of the beam and of the clock, since after finding the total speed for the beam, you can subtract c from that and find the speed of the clock.

Quote altergnostic
A footnote: length contraction usually applies to the distance between two bodies in relative motion, meaning, it usually applies to a length that is changing with time.

This is an incorrect understanding of length contraction. *I recommend starting with the Wikipedia article on length contraction. *Length contraction has nothing to do with the distance between two bodies in motion. *It also does not refer to a change in length over time, it refers to a difference in the same length as measured by two different frames.

Yes, but it is usually the difference measured by two frames in relative motion (you could have length contraction between two observers at rest also). But the main point is your final sentence: it is a difference between two distances being measured by different frames. The given distance h' is not being measured at all in my setup. It has been previously marked with lightsecond signs, remember? This distance is visual data, not vt (how could it be measured with no v, after all?).

 

[PeterDonis]

I prefer this analysis over the previous one, but the problem remains: you are not given any v

You are missing the point of this version of my analysis; I let v be unknown. It is the speed of the light clock relative to the observer, but I assumed no value for it. My whole point was to show that, even if you do not know the speed of the light clock relative to the observer, you can still show that the velocity of the "projectile" inside the clock is different relative to the observer than it is relative to the clock, *if* the "projectile" moves slower than light. But if the "projectile" is a light beam, then its speed is 1 (i.e., c) in both the clock frame and the observer frame. My understanding was that that was your point of confusion: you couldn't understand how a projectile's velocity could change from frame to frame if it moves slower than light, but yet the velocity of a light beam does *not* change from frame to frame. I've now given you an explicit formula that shows why that's true for the clock scenario.

You have said several times now that you can "ditch the speed of the light clock" in the analysis. I don't understand what this means. The light clock itself is part of the scenario, so you have to model its motion to correctly analyze the scenario. If you just mean that you can't assume you *know* what its speed is, that's fine; as I said above, I let that speed be unknown in my latest analysis, and showed how it doesn't matter for the question I thought you were interested in--why the projectile's velocity changes from frame to frame while the light beam's does not.

If you mean that we can somehow model the scenario without including the light clock at all, I don't see how. The motion of the parts of the light clock gives a critical constraint on the motion of the projectile/light beam inside the clock. Also, the light clock does not change direction in the scenario, so you can define a single inertial "clock frame"; but the projectile *does* change direction, so there's no way to define a single inertial "projectile frame". That means we can't just focus on the "velocity of the projectile", because that velocity *changes* during the scenario; it's not a "fixed point" that we can use as a reference.

As for "assuming" that the light beam travels at c, I have not assumed that. The only assumptions I have made are translation and rotation invariance, plus the principle of relativity, plus an assumption about how the light clock's "projectile" reflects off the mirror. It's probably futile at this point to walk through the chain of reasoning again, but I'll do it once more anyway. I'll focus on your "triangle diagram" since it's a good illustration of the spatial geometry in the unprimed frame.

We have a "projectile clock" consisting of a source/detector, which moves from A to C to D in the triangle diagram, and a mirror/reflector, which moves on a line parallel to the source/detector that passes through B at the same time (in the unprimed frame) that the source/detector passes through C. The clock as a whole moves at speed v relative to the observer who remains motionless at A (in the unprimed frame).

The "projectile" within the clock moves at some speed v_p in the unprimed frame (which we take to be unknown at this point), along the line A to B, then B to D. At the instant that this "projectile" (call it P1) reaches B, a second "projectile" (call it P2), moving at the *same* speed v_p, is emitted back towards A, to carry the information to the observer at A that the first projectile has reached B.

Here are some key facts about the geometry that follow from the above:

- Angle ABC equals angle CBD.
- Distance AB equals distance BD.
- Distance AC equals distance CD.
- Line BC is perpendicular to line AD.

We can also define the following times of interest (in the unprimed frame): T_AB = the time for P1 to travel from A to B; T_BD = the time for P1 to travel from B to D; T_BA = the time for P2 to travel from B to A. It is then easy to show from the above that all three of these times are equal: T_AB = T_BD = T_BA. We also have T_AC = the time for the light clock source/detector to travel from A to C, and T_CD = the time for the source/detector to travel from C to D. And we have T_AC = T_CD = T_AB = T_BD, because projectile P1 and the source/detector are co-located at A and D and both of their speeds are constant.

Thus, we have the following spacetime events:

A0 = D0 = the spacetime origin; the light clock source/detector passes the observer at A at the instant projectile P1 is emitted from the source/detector.

B1 = P1 reaches the mirror/reflector and bounces off; P2 is emitted back towards A.

C1 = the light clock source/detector passes point C.

D2 = P1 reaches the light clock source/detector and is detected.

A2 = P2 reaches the observer at A.

The above facts about the times and the motion of the light clock imply that B1 and C1 are simultaneous (in the unprimed frame), and D2 and A2 are simultaneous (in the primed frame).

Finally, we have formulas about the speeds:

v_p = \frac{AB}{T_{AB}} = \frac{BD}{T_{BD}} = \frac{AB}{T_{BA}}

v = \frac{AC}{T_{AC}} = \frac{CD}{T_{CD}} = \frac{AC}{AB} v_p

The last equality follows from the fact that T_AC = T_AB. Do you see what it says? It says that, if you know v_p, AC, and AB, you know v, the "light clock velocity". You stipulated v_p, AC, and AB in your statement of the problem; therefore you implicitly gave a value for v as well. But the value for v that you stated was a *different* v; it did not satisfy the above equation, which is enforced by the geometry that you yourself gave for the problem; that's why your "givens" were inconsistent.

I'll stop at this point since this post is getting long. What I'm trying to say is that the scenario you proposed has more constraints in it than you think it does; it already *contains* the information about how the projectile/light beam's velocity has to transform between frames.

 

[Dale]

I have reestaded many times that you could ditch the speed of the light clock, since it is useless to this problem.

Yes, you have restated many times many incorrect things in this thread. Obviously, whenever you are interested in the effect of speed on the operation of a clock then the speed of the clock is important.

However, if it is not a given in the problem then everything becomes easier. Simply plug the givens into the equation I posted above and we find that v = 0.45 c. At 0.45 c the time dilation factor is 1.12, so in 1.12 s the clock travels .5 ls and the light travels a distance of 1.12 ls. 1.12 ls / 1.12 s = c.

given the distance and times you can calculate the speed.

Certainly. And as long as your givens are both self-consistent and consistent with physics then you will always get that light travels at c.

(you could have length contraction between two observers at rest also).

No, you cannot.

The given distance h' is not being measured at all in my setup. It has been previously marked with lightsecond signs

This is yet another self-contradictory statement. The lightsecond signs are a measuring device, and comparing anything to them is performing a measurement.

 

[harrylin]

Why are they self contradictory? [..]

I trust that that has been solved now; so we simply ditch v and find v=0.45 from your 1.12.

[..] server line of sight. I propose that he places his x-axis in line with AB so he can give all the motion to beam and solve without knowledge of the clock's velocity.

- spatial rotation is useless. Standard is to orient the system speed along X (don't you know that the Lorentz transformation uses that convention?)
- he can give no motion to the beam; and how he places his coordinates can have no effect on the beam! Seriously, nothing of that makes any sense to me.

From the link, [https://www.physicsforums.com/showthread.php?t=574757] I will comment only upon the same issue I have been discussing here: the beam leaves A and reaches B in the rest frame [..]

Sorry, the first thing you discussed here and which was not solved were the speed and direction - and those also appear in your last example, so you must be sure to get it right...

Maybe my terminology is incorrect? The rest frame is meant to be the frame at rest relative to the clock. The moving frame is supposed to be the observer who sees the moving clock.

That is exactly how I understand it; thus you disagree about the calculation method, which is identical to the method to explain the Michelson-Morley experiment.

The beam never reaches the observer at A. Only the signal from B does. He then marks the observed time of event B and plots it against the given L. He can then find the time of event B as marked on the detector's clock by subtracting the time it takes light to cross the distance between the coordinates of the event at B from the observed time. From that, he can calculate the distance from source to receiver as seen from the clock itself by transforming the calculated speed of the beam into c (the speed of light as seen locally, which is the speed supported by a huge amount of evidence). [..]

That is wrong: it appears that you heard some stuff about SR and some stuff about GR and mixed them up. There is nothing to "transform"; as you earlier claimed there is no transformation to make, as all distance and time measurements are made with the single reference system S', the "rest" system.

[MMX:] The interferometer was supposed to test the existence of the aether and find a variance of the speed of light relative to the ether. The setup was built so we had perdicular light paths.

Note: the essential point of the experiment is that the light paths in the ether are not perpendicular; I hope that that is clear.

As the apparatus (and the Earth) revolved, no fringe displacement was detected (actally, no significant displacement). So it was concluded that the speed of light was constant regardless of the orientation of the beams or the detectors relative to an absolute frame.

Not exactly, no... There was only one detector. And no effect was detected from changing orientation regardless of the velocity (although apparently Michelson only measured it at one velocity; others repeated it at other velocities).

Notice that in the equations applied to this experiment, V was the relative velocity wrt the aether, so there was only one possible V, as the aether was absolute. Since then, the speed if light was measured with ever increasing accuracy, always in the same manner: by noting the return time if light as it went forth and back a specified distance. The time of detection after emission is always proportional to c. This is exactly the premise of my setup: light can't be detected to move at any V other than c. Just notice that the beam is not detected by A, only the signal from B is. There's no return, no detection, nothing that relates the path of the beam with the experiments that tested the speed of light. [..]

I still can't make any sense of that last part. Sorry. But there are sufficient glitches in your descripton, and sufficent lack of equations, to make me confident that before anything else it will be better to go through a special relativity calculation example of MMX. And this thread is already too long. Please start a new one on MMX. I do think it likely that this whole topic here will disappear after that exercise. Especially because:

This is merely the outcome of a realisation that this thought problem has never been done before and that the postulate of the speed of light applies to source and receiver, but not necessarily to a non-receiver. It is the preconception that the postulate of the speed of light also applies to undetected light that keeps you from aknowledging the possibility I intend to discuss here. [..] the conclusion is strictly dependent upon the acceptance of the constancy of the speed of any received or detected light.

No. Once more: there are detectors where you like, even non-detected light is assumed to go at c, and such experiments as the one you describe have been done, both in theory and in practice. As a matter of fact, radar uses that same set-up. And the basic calculations are similar as with MMX.

if light reaches us at a constant speed, distance events will be seen at a later time than the time of the event determined locally (for an observer in close proximity and at rest relative to the event). Hence, if the observed time is delayed and distances are given, the observed speed must be smaller.

Do you think that the traffic police will observe speeding cars further away as going slower - or that GPS will tell that you are going slower as the satellite is further away??

 

[altergnostic]

Peter, thanks for your patiance and effort. I'll make a couple observations regarding your reply.
You misinterpret my assertion that we can ditch the velocity of the light clock to mean that it is an unknown that must enter the equations nonetheless. No, I mean that we don't need it to solve at all. To determine the speed of the beam all the observer needs to know is the distance teavelled over the observed time - this will be the apparent velocity. Conversely, if you were given the relative velocity of the beam/projectilr from A to B (the vector addition of the upward speed of the projectile and the perpendicular speed of the projectile-clock) you could find the distance AB. Given the distance AB you can directly calculate the observed velocity as seen from A, just plot it over the observed time. You have to put yourself in the observer shoes and really imagine how he calculate the speed of the beam.
Since we are given the time of event B as measured by the clock itself, we can easily find the time that event is observed at a distance AB. Note this is the case for any scenario, we wouldn't need a beam traveling fron A to B at all. If the blinker at B simply turns on at t=1s (as measured with the blinker's own clock), you just have to consider the spatial separation from the event and the observer to get the observed time.

Assuming the event is caused by a beam going from A to B, he only has to plot the given distance over the observed time.

The mistake I keep pointing you to is the fact that you assumed the velocity of the clock enters the equations. But it would enter only if we were doing some kind if vector addition, but since the clock is going from A to D and the observer sees the event from B (at a time T), there's no reason to take that velocity into consideration, you are only looking for the time separation between events A and B to figure out at what speed something has to travel this distance to cause the observed times.

Later you state that T_AB = T_BA, but that is an assumption. The observer has to calculate T_AB from the observed time of event B, which is the local or proper time of event B plus the time it takes for that event to be seen at A. We know T_BA as seen from A from experiment, the speed of any directly detected light must be c. We don't know T_AB as seen from A because T_AB is only observed after T_BA, orherwise the observer doesn't see event B at all. T_AB must be calculated in this setup. Of course, if the observer at rest were to send a beam of light towards B, that would take the same time as a beam would need to come back from B to A. But notice the very important fact here that, if that was the case, the time of event B as seen from B would NOT be 1s (the time light takes to cross y'), but actually 1.12s (the time it would take for light to cross h').

Your last remark reassures me that you are assuming what you are trying to prove. The setup has no information on the beam, only on the events, which are good to calculate the speed of the beam in each frame.

 

[harrylin]

[..] To determine the speed of the beam all the observer needs to know is the distance teavelled over the observed time - this will be the apparent velocity. [...]

Just one more remark: you are mixing up reference systems, just as we already discovered - in fact you try to use t instead of t' for S'. That doesn't work. This will become clearer when you discuss MMX, which is the "mother" of all such calculations.

 

[PeterDonis]

You misinterpret my assertion that we can ditch the velocity of the light clock to mean that it is an unknown that must enter the equations nonetheless. No, I mean that we don't need it to solve at all. To determine the speed of the beam all the observer needs to know is the distance teavelled over the observed time - this will be the apparent velocity.

You are correct that you could set up the scenario so that the distances AB, AC, and BC were pre-determined; then you would have to control the speed of the light clock so that the beam actually hit the light clock's mirror at point B, instead of at some other point along the mirror's trajectory. That's fine, but it doesn't change anything about my analysis; my analysis is still correct, because even if you don't need the light clock's speed to do the analysis (which you actually do--see below--but for the moment I'll assume for the sake of argument that you don't), that is not the same as saying that an analysis which does use the light clock speed (even if it is left unknown) is incorrect.

Conversely, if you were given the relative velocity of the beam/projectilr from A to B (the vector addition of the upward speed of the projectile and the perpendicular speed of the projectile-clock) you could find the distance AB.

Yes, that's true, but you did not specify the upward speed of the projectile in the unprimed (observer) frame. You specified it in the primed (clock) frame. (Btw, I mis-stated this somewhat in my previous post; I said that you specified v_p, but I should have said that you specified v&#039;_p. I can go back and continue the analysis I was doing in my last post with that corrected, but it may not be worth bothering.)

So before you do this vector addition, you have to first transform the upward speed of the projectile from the primed to the unprimed frame. The upward distance (AB in the primed frame; BC in the unprimed frame, since the clock is moving in that frame) does not change when you change frames, but the *time* does, because of time dilation, so the upward speed of the projectile (i.e., the upward *component* of its velocity) is different in the unprimed frame than in the primed frame. So you do need to know the relative velocity of the light clock and the observer; without that you can't transform the upward velocity in the primed frame to the upward velocity component in the unprimed frame.

Given the distance AB you can directly calculate the observed velocity as seen from A, just plot it over the observed time. You have to put yourself in the observer shoes and really imagine how he calculate the speed of the beam.

Yes, let's do that. We have a light beam traveling from A to B, and a second light beam (the one that is emitted at the instant the first one strikes the mirror) traveling from B to A. The round-trip travel time is measured by the observer at A, and he already knows the distance AB because he measured it beforehand (and then controlled the speed of the light clock to ensure that the mirror was just passing B at the instant the first beam hit it). So we have two light beams each covering the same distance; if we assume that both beams travel at the same speed in the unprimed frame (even if we don't assume that that speed is c), then we can simply divide the round-trip time by the round-trip distance (2 * AB) to get the beam speed. Fine. See below for further comment.

Later you state that T_AB = T_BA, but that is an assumption.

Only in the sense that we assume that both light beams (the one from A to B and the one from B to A) travel at the same speed. Do you challenge that assumption? Both beams are "observed" in your sense--one endpoint of each beam is directly observed by the observer at A. It's impossible for *both* endpoints of either beam to be directly observed by the same observer, so if that's your criterion for a beam being "directly observed", then no beam is ever directly observed. But if you accept that *receiving* a beam counts as directly observing it, then *emitting* a beam should also count as directly observing it; either one gives the observer direct knowledge of one endpoint of the beam.

The observer has to calculate T_AB from the observed time of event B, which is the local or proper time of event B plus the time it takes for that event to be seen at A.

But how do we know the time it takes for that event to be seen at A? Are you assuming that the beam from B to A travels at c? If so, then why not also assume that the beam from A to B travels at c? What makes a received beam any different from an emitted beam?

By contrast, I am only assuming that the two beams (A to B and B to A) travel at the *same* speed, *without* assuming what that speed is (we *calculate* that by dividing round-trip distance by round-trip time, as above). That seems like a much more reasonable approach, since it does not require assuming that there is any difference between an emitted beam and a received beam.

We know T_BA as seen from A from experiment, the speed of any directly detected light must be c.

But only one endpoint of the light is directly detected. Why should received light count as directly detected but not emitted light?

We don't know T_AB as seen from A because T_AB is only observed after T_BA

No, they are both "observed" (by any reasonable definition of "observed") at the same time, when the beam from B to A is received and its time of reception is observed. At that point the observer knows the round-trip travel time and the round-trip distance and can calculate the beam speed.

T_AB must be calculated in this setup.

So must T_BA. The observer doesn't directly observe the emission of the beam from B to A, any more than he directly observes the reception of the beam from A to B. He has to calculate the times of both those events. The way he does that is to use the fact that both events occur at the same instant, by construction.

Of course, if the observer at rest were to send a beam of light towards B, that would take the same time as a beam would need to come back from B to A.

Unbelievable; you now *admit* this, yet you were arguing that we could *not* assume this before.

But notice the very important fact here that, if that was the case, the time of event B as seen from B would NOT be 1s (the time light takes to cross y'), but actually 1.12s (the time it would take for light to cross h').

Which it is; the time of event B, *in the unprimed frame*, *is* 1.12s (if we allow v, the velocity of the light clock relative to the observer, to be set appropriately to 0.45 instead of 0.5, per the comments of DaleSpam, harrylin, and myself). The time of event B, in the *primed* frame, is 1s; but that's not what the observer at A is interested in. He's interested in the time of event B in his frame, the unprimed frame, and that time is different from the time of event B in the primed frame because of time dilation. Which, of course, requires you to know the velocity of the light clock relative to the observer, contrary to your repeated erroneous claim that you don't.