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From an older exercise sheet in my relativity class which I completed, but didn't annotate enough to remember it everything was correct:

Let $X$ be an inertial observer noticing that a spacetime event $B$ was caused by another event $A$. Show that no inertial observer $X'$ exists for whom the event $A$ is seen to be caused by $B$.

I chose the following ansatz:

Let $t_B>t_A$ in $X$. Then the Lorentz transformation of $t_A$ and $t_B$ gives us $$ t_A' = \frac{ct_A - \beta x_A}{\sqrt{1-\beta^2}} \quad t_B' = \frac{ct_B - \beta x_B}{\sqrt{1-\beta^2}} $$ Now, for causality to be given, we demand $0 < t_B - t_A$, i.e., the event $B$ happens after $A$. $$0 < t_B - t_A = \gamma (ct_B - \beta x_B - ct_A + \beta x_A)\\ =\gamma [c(t_B - t_A) - \beta (x_B - x_A)]$$ Losing the gamma and rearranging gives us $$ \beta (x_B - x_A) < c(t_B - t_A) \iff v/c^2 (x_B - x_A) < t_B - t_A $$ Now, since $v<c$, the expression $t_B - t_A \geq \frac{x_B - x_A}{c}$ holds.

I haven't written anything else after this, and I'm not quite sure whether a) this proves the causality-independence of the observer or b) my way of calculating is correct.

Looking forward to answers.

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As you correctly note, you need to prove $t_B > t_A$ is preserved by Lorentz transformations. It's not clear to me whether your final answer demonstrates this (since the sign of $x_b - x_a$ isn't totally obvious), but I think you're on the right track.

I might have responded with the following argument. The invariant interval $$ds^2 = -dt^2 + dx^2 \tag1$$ is preserved by Lorentz transformations. Since event $B$ is "caused" by event $A$, $A$ must be in $B$'s past light cone, so the finite interval $s$ between them satisfies $$s^2 < 0 \tag2 .$$

Now (1) with $s$ held constant is the equation of a hyperbola. Thus, considering WLOG a reference frame centred on $B$, Lorentz transformations will shift $A$ along a hyperbola. From (2), that hyperbola is confined everywhere to $B$'s forward light cone, and so no observer exists for whom $t_A > t_B$. This completes the argument.

Note an almost identical argument can be constructed in in general relativistic spacetimes. It depends only on the fact that $t$ and $x$ have different signs in the interval.

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  • $\begingroup$ I like your answer a lot, given it's argument on a geometric basis. $\endgroup$ – John W. Nov 21 '15 at 13:48

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