JesseM said:
Why would symmetry indicate that?
arrell said:
My understanding is that
frame 2 is moving at v with respect to frame 1 (and points stationary in frame 1)
and symmetrically
frame 1 is moving at v with respect to frame 2 (and points stationary in frame 2)
OK, I think I see the problem. It's true that if the two points on the source which shoot the photons have a separation of L in frame 1, then they will have a separation of \sqrt{1 - v^2/c^2}L in frame 2. But what you have to realize is that this does
not mean frame 2 will see events c and d (or g and h) happen at a distance of \sqrt{1 - v^2/c^2}L apart; the reason is that if events c and d (or g and h) happen at the same time in frame 1, then they happen at
different times in frame 2, so in the time between them the source will have moved somewhat in frame 2. If g happens at coordinates x=0, t=0 in frame 1, and h happens at coordinates x=L, t=0 in frame 1, then using the Lorentz transform we can show that in frame 2, g happens at x'=0, t'=0 while h happens at x'=\frac{L}{\sqrt{1 - v^2/c^2}}, t'=\frac{-vL}{c^2 \sqrt{1 - v^2/c^2}}.
Note that in frame 2, at time t'=0 the first laser gun was at position x'=0 and the second laser gun was at position x'=\sqrt{1 - v^2/c^2}L, so since both guns move a distance vt' in any time interval t', at the earlier time t'=\frac{-vL}{c^2 \sqrt{1 - v^2/c^2}} both laser guns must have been a distance \frac{v^2 L}{c^2 \sqrt{1 - v^2/c^2}} to the right, meaning the first laser gun was at position x'=\frac{v^2 L}{c^2 \sqrt{1 - v^2/c^2}} and the second was at position x'=\sqrt{1 - v^2/c^2}L \, + \, \frac{v^2 L}{c^2 \sqrt{1 - v^2/c^2}} = L(\frac{1 - v^2/c^2}{\sqrt{1 - v^2/c^2}} \, + \, \frac{v^2/c^2}{\sqrt{1 - v^2/c^2}}) = \frac{L}{\sqrt{1 - v^2/c^2}}, just where h must have happened in frame 2 according to the Lorentz transformation.
JesseM said:
Well, as I said before, your language is confusing, since photons don't "originate" in one frame vs. another, I guess what you're saying is that the source of the photons is at rest in a given frame.
arrell said:
Photons have to emit from somewhere. In this case, I chose that photons emit originally from points a, b, c, and d which are stationary in frame 1.
I understand that, I'm just saying it initially confused me when you said that the photons "originate from frame 1" when what you really meant was that they originate from sources which are stationary in frame 1; this is not a typical shorthand used by physicists.
JesseM said:
You also have a profusion of points which are infinitesimally close to each other--if a and c and g and e are all infinitesimally close to one another, what is the point of giving them different labels, for example? Both frames will assigning a and c and g and e the same coordinates anyway.
arrell said:
Only
a and c
b and d
are infinitesimally close.
While
a and b
c and d
are separated by significant distance L.
Yes, I understand that--but I am correct that a, c, g, and e are all infinitesimally close, right? Likewise, aren't b, d, f, and h infinitesimally close? If so, you have eight different labels but only two distinct points in spacetime. What do all these points add to the thought-experiment? Why not just say something like "one photon is emitted at a, another is emitted at b at the same time in their rest frame, the distance between a and b is L in their rest frame, what is the distance in another frame moving at v relative to their rest frame?"
arrell said:
When the photons originally from a and b are reflected from frame 2, the points at which they are reflected from frame 2 are stationary in frame 2
so now
the photons travel back towards frame 1 which is moving at v with respect to the reflection points and strike at points e and f in frame 1.
Again, you're talking about "frames" like they're things which can emit and reflect photons, and which occupy some particular position in space, while the standard meaning of a "frame" is just a coordinate system which fills all of space. I think what you really mean is something like "When the photons originally from a and b are reflected
from a mirror at rest in frame 2, the
points on the mirror where they are reflected are stationary in frame 2 so now the photons travel back
towards the source which is at rest in frame 1 and which is moving at
v with respect to the reflection points and strike at points e and f
on the source which is at rest in frame 1". But again, I don't see what the mirror-reflection adds to the thought-experiment. For one thing, since you're placing the mirror infinitesimally close to the source, then each photon will be emitted, reflected and reabsorbed by the source at the same spacetime coordinates, regardless of which frame you're using. Second of all, even if the mirror wasn't infinitesimally close so these events happened at different locations, the question of how the mirror is moving is irrelevant--the photons will only make contact with the mirror for a single moment, so it doesn't make any difference whether the mirror is at rest in frame 1, at rest in frame 2, or moving in both frames; this will have no effect whatsoever on the coordinates of the three events as seen in either frame.