altergnostic said:
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.
altergnostic said:
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.
altergnostic said:
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.
altergnostic said:
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.
altergnostic said:
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.
altergnostic said:
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?
altergnostic said:
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.
altergnostic said:
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.
altergnostic said:
The only data available to the observer is:
t'A = tA = 0s
Yes.
altergnostic said:
Yes. But you are aware, aren't you, that this implies tB = gamma, right?
altergnostic said:
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.
altergnostic said:
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.
altergnostic said:
Conversely, replace the beam with a projectile, with this set of givens:
t'A = tA = 0s
Yes.
altergnostic said:
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?
altergnostic said:
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.
altergnostic said:
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.
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??
