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J. Murray
J. Murray
11:47
In the following, I'll be using ChatJax, which you can enable by following the instructions here: math.ucla.edu/~robjohn/math/mathjax.html
At $t=t'=0$, two light rays are emitted from the point $x=x'=0$ traveling in opposite directions. In the unprimed frame, the trajectories of the two light rays are $x_1(t)=ct$ and $x_2(t) = -ct$. In the primed frame, they are $x_1'(t')= ct'$ and $x_2'(t') = -ct'$.
At a given value of $t$, $x_1(t)=-x_2(t)$. Similarly, for a given value of $t'$ we have that $x_1'(t')=-x_2'(t')$. However, it is incorrect to say that $x_2=-x_1$ and $x_2'=-x_1'$ without referencing a time. What would that even mean? $x=x_1$ is an entire worldline, not an event. You need to also specify a time.
For this reason, it is incorrect to blindly apply the Lorentz transformation to $x_1$ in order to obtain $x_1'$. Lorentz transformations map events to events, not trajectories to trajectories.
This can be fixed by applying the Lorentz transformation properly as follows:
Given the event $(x_1,t)$, we obtain $x_1' = \gamma(x_1-vt)$ and $t' = \gamma(t - x_1 v/c^2)$.
Applying this to the entire trajectory $x_1=ct$ yields $x_1'=\gamma(c-v)t$ and $t'=\gamma\left(1-\frac{v}{c}\right)t$
Solving the last equation for $t$ and plugging it into the first yields the expected trajectory equation $x_1'(t') = ct'$.
Exactly the same thing holds true for the other light ray:
$x_2=-ct \implies x_2' = -(c+v)t$ and $t'= \left(1+\frac{v}{c}\right)t$, yielding $x_2'(t') = -ct'$.
No contradiction.
The contradiction appears to arise when you pretend that $x_2(t_A)=-x_1(t_B)$ and $x_2'(t'_A) = -x_1'(t'_B)$ hold for arbitrary times. That's obviously not true - the positions of the rays are only negatives of one another if we evaluate them simultaneously - but simultaneity is lost when we boost to a new frame!
In other words, for an event where $x_1=-x_2$ in the unprimed frame, the transformed coordinates are such that $x_1'\neq -x_2'$ in the primed frame because those two events are no longer simultaneous, i.e. you're talking about the positions of the light rays at different times.
Explicitly, $t_1' = \gamma\left(t - \frac{x_1 v}{c^2}\right)$ and $t_2' = \gamma\left(t - \frac{x_2 v}{c^2}\right) = \gamma\left(t + \frac{x_1 v}{c^2}\right) \neq t_1'$.
The events in the primed frame would only be simultaneous if $x_1=x_2=0$ or if $v=0$, which are (not coincidentally) the conditions you found which would resolve the apparent contradiction in your answer.
 
9 hours later…
Thomas
Thomas
20:48
I don't know why you attach any mathematical significance to writing arguments to the variables or not (the OP or even the lecture notes you linked to don't write any arguments). We explicitly know in fact what the dependences are (as given by the speed of light postulate)
$$x_1=ct ;\:\:\: x_1'=ct'$$
$$x_2=-ct ;\:\:\: x_2'=-ct'$$
The events here are simply given by the light signal reaching the corresponding x-coordinates in the unprimed and primed frame.
From this it follows by the rules of algebra
 
3 hours later…
J. Murray
J. Murray
23:38
@Thomas You don't need to write the arguments if you have a clear sense of what the Lorentz transformations do. That is the essence of the confusion expressed at the end of your answer, so that's why I chose to write them.
$x_2=-x_1$ means that the position of the rightward moving beam is the negative of the position of the leftward moving beam at any given instant of time
It doesn't mean, for example, that the position of the rightward moving beam at some time is equal to the negative of the leftward moving beam two seconds prior. This is obvious.
However, it is the essence of your confusion. If you consider two events $(x_1,t)$ and $(x_2,t)$ (corresponding to the two light beams passing some reference points), then $x_2=-x_1$.
However, if you then apply the Lorentz transformation to those two events, the resulting time coordinates will be different.
$(x_1,t)\mapsto (x_1',t_1')$ and $(x_2,t)\mapsto (x_2',t_2')$ where $t_1'\neq t_2'$ as I explicitly showed above.
This is the confusion you're running into. You're applying the Lorentz transformation to two simultaneous events and expecting them to remain simultaneous, but that is not how Lorentz transformations work.

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