Same Temporal Order in all Inertial Frames

[topspin1617]

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 $S$ 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?

 

[PAllen]

Welcome to Physicsforums!

The exercise wording is open to the confusion you raise. It is true that in every frame there exists such a signal (possibility). I would have worded the exercise:

Either:

Show that two events have the same temporal order in every inertial frame if and only if there exists an inertial frame in which the events can represent two times for a stationary object.

or:

Show that two events have the same temporal order in every inertial frame if and only if a light signal can connect them (in any inertial frame; also in every every inertial frame).From a physical point of view, going from any to every is sort of obvious. If a signal does go from one to another (physical reality), then it is considered unacceptable for a physical theory to posit that this fact observer dependent.

 

[topspin1617]

Welcome to Physicsforums!

The exercise wording is open to the confusion you raise. It is true that in every frame there exists such a signal (possibility). I would have worded the exercise:

Either:

Show that two events have the same temporal order in every inertial frame if and only if there exists an inertial frame in which events represent two times for a stationary object.

or:

Show that two events have the same temporal order in every inertial frame if and only if a light signal can connect them (in any inertial frame; also in every every inertial frame).


From a physical point of view, going from any to every is sort of obvious. If a signal does go from one to another (physical reality), then it is considered unacceptable for a physical theory to posit that this fact observer dependent.

Hello PAllen, thanks for your reply!

Regarding your final statement, I'm not questioning the existence of a signal; I was just wondering if it was possible for one signal to be perceived as traveling at less than the speed of light in one frame and faster in another. Because, even though the axioms say that the speed of light is the "speed limit", if a photon was moving in one frame in the positive x-direction, and then a second frame was moving near the speed of light in the negative x-direction of the original frame, would that signal not be perceived as traveling faster than the speed of light in the second frame? This would seem to suggest that the existence of a signal connecting two events moving at a speed less than that of light in one frame does not necessarily force such a signal to exist in all frames. However, my solution to this question DOES seem to suggest that this SHOULD be true. I guess this is where my confusion is coming from.

 

[PAllen]

Hello PAllen, thanks for your reply!

Regarding your final statement, I'm not questioning the existence of a signal; I was just wondering if it was possible for one signal to be perceived as traveling at less than the speed of light in one frame and faster in another. Because, even though the axioms say that the speed of light is the "speed limit", if a photon was moving in one frame in the positive x-direction, and then a second frame was moving near the speed of light in the negative x-direction of the original frame, would that signal not be perceived as traveling faster than the speed of light in the second frame? This would seem to suggest that the existence of a signal connecting two events moving at a speed less than that of light in one frame does not necessarily force such a signal to exist in all frames. However, my solution to this question DOES seem to suggest that this SHOULD be true. I guess this is where my confusion is coming from.

I thought you were using the velocity addition rule in your first post. This answers your conundrum:

If two signals are moving (along x axis, say) -c and +c in one frame, they would still be seen as moving -c and +c in every frame moving relative to the first, including .99c in +x direction. This can be seen by:

(c + .99c)/ (1 + .99c^2/c^2) = c

and:

(c-.99c)/(1 - .99c^2/c^2) = c

 

[PAllen]

Show that two events have the same temporal order in every inertial frame if and only if there exists an inertial frame in which the events can represent two times for a stationary object.

or:

Show that two events have the same temporal order in every inertial frame if and only if a light signal can connect them (in any inertial frame; also in every every inertial frame).

Both of these formulations need tightening. The first is not true if the events can only be connected by a light signal (rather than a signal going <c). So you would need an inelegant statement:

Show that two events have the same temporal order in every inertial frame if and only if there exists an inertial frame in which the events can represent two times for a stationary object, or they can only be connected by a light signal.

The second formulation conversely leaves out signals less than c. Corrected:

Show that two events have the same temporal order in every inertial frame if and only if a signal traveling ≤ c can connect them (in any inertial frame; also in every every inertial frame).

 

[topspin1617]

I thought you were using the velocity addition rule in your first post. This answers your conundrum:

If two signals are moving (along x axis, say) -c and +c in one frame, they would still be seen as moving -c and +c in every frame moving relative to the first, including .99c in +x direction. This can be seen by:

(c + .99c)/ (1 + .99c^2/c^2) = c

and:

(c-.99c)/(1 - .99c^2/c^2) = c

Ah.

Yea, I see what's happening. I actually did a correct calculation for myself earlier, and just forgot to cancel a 3/2, so I was getting a 3c/2 instead of c haha. I guess this is the part of relativity that is a bit odd, because all intuition would tell me that if I am moving in the opposite direction of a photon at a rate of c/2, then it should appear to be moving at a rate of 3c/2. But I can see directly that that does not happen, and that c really IS the speed limit in this system.

So I guess this does show me that if a signal I was talking about earlier exists in one frame, the same signal again has velocity ≤ c in any other frame. I still wonder, though, why the question wasn't phrased as "every frame"... I was confused at first because I thought there was something subtle about constructing a particular frame where a signal existed.

 

[topspin1617]

Both of these formulations need tightening. The first is not true if the events can only be connected by a light signal (rather than a signal going <c). So you would need an inelegant statement:

Show that two events have the same temporal order in every inertial frame if and only if there exists an inertial frame in which the events can represent two times for a stationary object, or they can only be connected by a light signal.

The second formulation conversely leaves out signals less than c. Corrected:

Show that two events have the same temporal order in every inertial frame if and only if a signal traveling ≤ c can connect them (in any inertial frame; also in every every inertial frame).

Yeah, the second one, as you rephrased, is pretty much the question in the book, only stating in one frame. As my last post said, I thought it odd, after writing my solution, that the question didn't just say every inertial frame. In my experience with math, they always ask for the most general solution if it's easily equivalent to a weaker version haha.