Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Ostlie

1. Jun 23, 2011

omoplata

1. The problem statement, all variables and given/known data

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.

2. Relevant 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,
$$\frac{\sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2} + (z_{B}-z_{A})^{2}}}{t_{B}-t_{A}} \leq c$$

3. 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?

2. Jun 23, 2011

vela

Staff Emeritus
Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

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

3. Jun 23, 2011

omoplata

Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

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

4. Jun 23, 2011

vela

Staff Emeritus
Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

$$\frac{\sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2} + (z_{B}-z_{A})^{2}}}{t_{B}-t_{A}} \leq c$$

5. Jun 23, 2011

omoplata

Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

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.

6. Jun 23, 2011

omoplata

Re: Special Relativity: From an Introduction to Modern Astrophysics by Carroll and Os

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!