topspin1617
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Hello everyone, this is my first question here. I'm a mathematics student (actually pure math), but have recently found myself interested in learning about physics. I've started reading Introduction of Special Relativity by Rindler; I actually have no background in mechanics or basic physics, but I'm pretty good at picking up on theoretical stuff, and was just hoping I could see how far I could go.
Anyway, after reading the first section, I looked at the exercises. The third exercise says, if I may paraphrase,
"Show that two events have the same temporal order in every inertial frame if and only if there exists an inertial frame in which a signal with velocity less than or equal to that of the speed of light can connect the two events."
From what I can surmise, I can assume that I have an inertial frame S oriented so that the events both spatially occur on the x-axis, and further assume that any other inertial frame S^\prime moves uniformly through S in the direction of the x-axis with velocity v and is set up in what the book calls the "standard configuration" with S. If \Delta t,\Delta t^\prime represent the elapsed time between the two events in S,S^\prime respectively, they are related via the Lorentz transformation by
\Delta t^\prime=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right).
If there exists a signal connecting the two events in S, assuming constant velocity it would have to have velocity u=\Delta x/\Delta t. Factoring \Delta t out of the previous equation,
\Delta t^\prime=\gamma\Delta t\left(1-\frac{vu}{c^2}\right).
Now, the reverse claim of the question seems obvious, for if there exists a frame in which such a signal has velocity no more than c (which WLOG may be assumed to be S, I guess), the latter equation at once forces \Delta t and \Delta t^\prime to have the same sign.
The forward implication is confusing me a bit. I think the assumption implies that the term \left(1-\frac{vu}{c^2}\right) must be positive for all possible -c<v<c. First, if the velocity u from my initial frame is \leq c, then I'm done. If not, then \left(1-\frac{vu}{c^2}\right)>0\Rightarrow c^2/u>v; but if u>c then c>c^2/u. Therefore, not all values of v keep this term positive, contradicting the assumption.
I think this logic seems fine; if not, I'd like to be corrected. But if it is, this seems to say that not only does there exist a frame in which such a signal connects the two events, but in every frame there must exist such a signal. Though, if that were true, I'm not sure why the question wouldn't be phrased in this slightly more general sounding way. Am I correct? And if so, is the existence of the signal in either one frame or every frame an obvious equivalence that I'm missing?
Anyway, after reading the first section, I looked at the exercises. The third exercise says, if I may paraphrase,
"Show that two events have the same temporal order in every inertial frame if and only if there exists an inertial frame in which a signal with velocity less than or equal to that of the speed of light can connect the two events."
From what I can surmise, I can assume that I have an inertial frame S oriented so that the events both spatially occur on the x-axis, and further assume that any other inertial frame S^\prime moves uniformly through S in the direction of the x-axis with velocity v and is set up in what the book calls the "standard configuration" with S. If \Delta t,\Delta t^\prime represent the elapsed time between the two events in S,S^\prime respectively, they are related via the Lorentz transformation by
\Delta t^\prime=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right).
If there exists a signal connecting the two events in S, assuming constant velocity it would have to have velocity u=\Delta x/\Delta t. Factoring \Delta t out of the previous equation,
\Delta t^\prime=\gamma\Delta t\left(1-\frac{vu}{c^2}\right).
Now, the reverse claim of the question seems obvious, for if there exists a frame in which such a signal has velocity no more than c (which WLOG may be assumed to be S, I guess), the latter equation at once forces \Delta t and \Delta t^\prime to have the same sign.
The forward implication is confusing me a bit. I think the assumption implies that the term \left(1-\frac{vu}{c^2}\right) must be positive for all possible -c<v<c. First, if the velocity u from my initial frame is \leq c, then I'm done. If not, then \left(1-\frac{vu}{c^2}\right)>0\Rightarrow c^2/u>v; but if u>c then c>c^2/u. Therefore, not all values of v keep this term positive, contradicting the assumption.
I think this logic seems fine; if not, I'd like to be corrected. But if it is, this seems to say that not only does there exist a frame in which such a signal connects the two events, but in every frame there must exist such a signal. Though, if that were true, I'm not sure why the question wouldn't be phrased in this slightly more general sounding way. Am I correct? And if so, is the existence of the signal in either one frame or every frame an obvious equivalence that I'm missing?