JesseM said:
We seem to be going in circles here. You've been saying that the magic word "interchangeable" somehow justifies the claim that the factors in your two equations should be the same, but then you said "By interchangeable, I mean that I could assign either of two observers, one in each frame, with the label A, and the other B. It won't affect the end result in any meaningful way (we might as a result prime A values)." So here again it sounds like you're asserting what you're supposed to be proving, that "interchangeability" somehow will lead to the same "end result" (keeping in mind that what you got for the 'end result' itself depended on the earlier step where you were supposed to show that the factors would be the same in the two equations you wrote). Can you define "interchangeability" in a way that doesn't make reference to any steps in your derivation that happened after the step where you assume the factors are in fact the same in both equations?
Ok, you'll probably not like it, but I am going to have to jump past the factors stage to the end result which, for me, is:
[tex]x_b'=\gamma . (x_a - vt_a)[/tex]
and
[tex]x_a=\gamma . (x_b' + vt_b')[/tex]
(There is a similar pair for time, but the same argument will apply to that pair as to this pair.)
In this pair, can you see that all
b terms are also primed and all
a terms are unprimed. So we could, if we wanted to, drop the subscripts leaving us with:
[tex]x'=\gamma . (x - vt)[/tex]
and
[tex]x=\gamma . (x' + vt')[/tex]
Alternatively, we could swap the A and B terms (what we called A before is now B and what we called B before is now A). Our v was (implicitly) defined as positive in the direction that B moves away from A, and negative in the direction that A moves away from B. We keep the same priming notation, since we have shifted focus from the erstwhile A to the new A.
This gives us:
[tex]x_a=\gamma . (x_b' - vt_b')[/tex]
and
[tex]x_b'=\gamma . (x_a + vt_a)[/tex]
Again all our
b terms are primed and all our
a are unprimed. So we could drop our subscripts, giving:
[tex]x'=\gamma . (x - vt)[/tex]
and
[tex]x=\gamma . (x' + vt')[/tex]
A and B are interchangeable labels. The possible confusion here is that in my idiosyncratic way, I have considered A and B to be labels. You seem to want to have more concrete (and thus less general) determinations of what A and B are.
JesseM said:
Huh? The event (0,0) is not referred to in the two equations where you assert the factors will be the same. The two events referred to in those equations of yours are the colocation of A and the photon (at x=0,t=ta in A's frame) and the colocation of B and the photon (at x'=0, t=t'b in B's frame). Of course the time coordinate of either event in a given frame can be understood as the time interval between that event and (0,0), but then exactly is true about the time coordinate of either event of the photon crossing the x=5 axis of one of the frames. Perhaps there is something in what you mean by "interchangeable" that equations involving the time coordinates of a pair of events in both frames can only be considered interchangeable if the two events happened on the worldlines of observers at rest in each frame who crossed paths at (0,0), but if so nothing you have written about interchangeability so far even hinted at such a requirement. And what about observers who crossed paths at (0,0) but who are not at rest? What if we considered two observers who did cross paths there, with the first observer traveling at velocity V in the A frame and the second traveling at the same velocity V in the B frame? Then if we defined our two events in terms of where the light crossed each of these observer's paths, then without doing any numerical calculations do you think "interchangeability" means the factors in the equations relating the time coordinates of these two events in each frame would be the same?
I've mentioned a few times that there are only three colocation events. I honestly thought that I didn't need to make explicit that the colocation of A and B was at (0,0). I do think I have done it anyway - posts
https://www.physicsforums.com/showpost.php?p=2232756&postcount=397" - sadly I don't have time to go on but I think there are more references where I have stated that A and B are colocated at t=0,t'=0. In at least one of those I have even made explicit that that event is (0,0).
As for the tangent, your asking questings for clarifications doesn't constitute a tangent, but your assertion that events associated with photons crossing the x=5 axis can be profitably used in my scenario indicates that you have gone off on a tangent.
The difference is like between these:
"I don't understand what you are saying, what are you saying"
and
"You are claiming this wrong thing, you are wrong"
when I never actually claimed the wrong thing you asserted that I claimed. In the real life example where I said it seems you were going off on a tangent, it might not have been so clear that you were asserting that I claimed something that I had not claimed. That's why I said you were going off on a tangent rather than saying you were making unfounded assertions.
But yes, we are going around in circles.
cheers,
neopolitan