- #1
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."[itex][/itex]
From what I can surmise, I can assume that I have an inertial frame [itex]S[/itex] oriented so that the events both spatially occur on the [itex]x[/itex]-axis, and further assume that any other inertial frame [itex]S^\prime[/itex] moves uniformly through [itex]S[/itex] in the direction of the [itex]x[/itex]-axis with velocity [itex]v[/itex] and is set up in what the book calls the "standard configuration" with [itex]S[/itex]. If [itex]\Delta t,\Delta t^\prime[/itex] represent the elapsed time between the two events in [itex]S,S^\prime[/itex] respectively, they are related via the Lorentz transformation by
[itex]\Delta t^\prime=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)[/itex].
If there exists a signal connecting the two events in [itex]S[/itex], assuming constant velocity it would have to have velocity [itex]u=\Delta x/\Delta t[/itex]. Factoring [itex]\Delta t[/itex] out of the previous equation,
[itex]\Delta t^\prime=\gamma\Delta t\left(1-\frac{vu}{c^2}\right)[/itex].
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 [itex]c[/itex] (which WLOG may be assumed to be [itex]S[/itex], I guess), the latter equation at once forces [itex]\Delta t[/itex] and [itex]\Delta t^\prime[/itex] to have the same sign.
The forward implication is confusing me a bit. I think the assumption implies that the term [itex]\left(1-\frac{vu}{c^2}\right)[/itex] must be positive for all possible [itex]-c<v<c[/itex]. First, if the velocity [itex]u[/itex] from my initial frame is [itex]\leq c[/itex], then I'm done. If not, then [itex]\left(1-\frac{vu}{c^2}\right)>0\Rightarrow c^2/u>v[/itex]; but if [itex]u>c[/itex] then [itex]c>c^2/u[/itex]. Therefore, not all values of [itex]v[/itex] 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."[itex][/itex]
From what I can surmise, I can assume that I have an inertial frame [itex]S[/itex] oriented so that the events both spatially occur on the [itex]x[/itex]-axis, and further assume that any other inertial frame [itex]S^\prime[/itex] moves uniformly through [itex]S[/itex] in the direction of the [itex]x[/itex]-axis with velocity [itex]v[/itex] and is set up in what the book calls the "standard configuration" with [itex]S[/itex]. If [itex]\Delta t,\Delta t^\prime[/itex] represent the elapsed time between the two events in [itex]S,S^\prime[/itex] respectively, they are related via the Lorentz transformation by
[itex]\Delta t^\prime=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)[/itex].
If there exists a signal connecting the two events in [itex]S[/itex], assuming constant velocity it would have to have velocity [itex]u=\Delta x/\Delta t[/itex]. Factoring [itex]\Delta t[/itex] out of the previous equation,
[itex]\Delta t^\prime=\gamma\Delta t\left(1-\frac{vu}{c^2}\right)[/itex].
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 [itex]c[/itex] (which WLOG may be assumed to be [itex]S[/itex], I guess), the latter equation at once forces [itex]\Delta t[/itex] and [itex]\Delta t^\prime[/itex] to have the same sign.
The forward implication is confusing me a bit. I think the assumption implies that the term [itex]\left(1-\frac{vu}{c^2}\right)[/itex] must be positive for all possible [itex]-c<v<c[/itex]. First, if the velocity [itex]u[/itex] from my initial frame is [itex]\leq c[/itex], then I'm done. If not, then [itex]\left(1-\frac{vu}{c^2}\right)>0\Rightarrow c^2/u>v[/itex]; but if [itex]u>c[/itex] then [itex]c>c^2/u[/itex]. Therefore, not all values of [itex]v[/itex] 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?