Proving Preserved Temporal Order of Cause and Effect

[omoplata]

Homework Statement

Because there is no such thing as absolute simultaneity, two observers in relative motion may disagree on which of two events A and B occurred first. Suppose, however, that an observer in reference frame S measures that event A occurred first and caused event B. For example, event A might be pushing a light switch, and event B might be a light bulb turning on. Prove that an observer in another frame S' cannot measure event B (the effect) occurring before event A (the cause). The temporal order of cause and effect is preserved by the Lorentz transformation equations. Hint: For event A to cause event B, information must have traveled from A to B, and the fastest that anything can travel is the speed of light.

Homework Equations

Suppose (x_{A},y_{A},z_{A},t_{A}),(x_{B},y_{B},z_{B},t_{B}) are the coordinates in the coordinate system S, and (x'_{A},y'_{A},z'_{A},t'_{A}),(x'_{B},y'_{B},z'_{B},t'_{B}) are the coordinates in the coordinate system S', which is traveling at speed u<c in the positive x direction. Then from the Lorentz transformation equations,
t'_{A}=\frac{t_{A}-(u x_{A}/c^{2})}{\sqrt{1-(u^{2}/c^{2})}}
t'_{B}=\frac{t_{B}-(u x_{B}/c^{2})}{\sqrt{1-(u^{2}/c^{2})}}
x'_{A}=\frac{x_{A}-u t_{A}}{\sqrt{1-(u^{2}/c^{2})}}
x'_{B}=\frac{x_{B}-u t_{B}}{\sqrt{1-(u^{2}/c^{2})}}
y'_{A}=y_{A}
y'_{B}=y_{B}
z'_{A}=z_{A}
z'_{B}=z_{B}

Because B happened after A,
t_{B} > t_{A}

Because event A caused event B,
\frac{\sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2} + (z_{B}-z_{A})^{2}}}{t_{B}-t_{A}} \leq c

The Attempt at a Solution

I need to prove that t'_{B} - t'_{A} > 0

t'_{B}-t'_{A} = \frac{(t_{B} - t_{A})-(u/c^{2})(x_{B}-x_{A})}{\sqrt{1-(u^{2}/c^{2})}}

What do I do now?

 

[vela]

Since the denominator is positive, you just need to show the numerator is positive. That is, prove that
t_B-t_A > (u/c^2)(x_B-x_A)

 

[omoplata]

I can't think of any way of proving that. Which equations should I use?

 

[vela]

The one you already wrote:
\frac{\sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2} + (z_{B}-z_{A})^{2}}}{t_{B}-t_{A}} \leq c

 

[omoplata]

But those are squared terms.

It implies that,

(x_{B} - x_{A})^{2} \leq c^{2} (t_{B} - t_{A})^{2}

So,

\lvert x_{B} - x_{A} \rvert \leq c (t_{B} - t_{A})
\frac{u}{c^{2}} \lvert x_{B} - x_{A} \rvert < (t_{B} - t_{A})

This could be either \frac{u}{c^{2}} ( x_{B} - x_{A} ) < (t_{B} - t_{A}) or \frac{u}{c^{2}} ( x_{A} - x_{B} ) < (t_{B} - t_{A}) depending on whether x_{B} > x_{A} or not.

 

[omoplata]

Oh, OK. If x_{A} > x_{B}, then,
t_{B} - t_{A} > \frac{u}{c^{2}} (x_{A} - x_{B}) > \frac{u}{c^{2}} (x_{B} - x_{A}), since then(x_{B} - x_{A}) is negative and (x_{A} - x_{B}) is positive.

Thanks!