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omoplata

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## 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 [itex](x_{A},y_{A},z_{A},t_{A}),(x_{B},y_{B},z_{B},t_{B})[/itex] are the coordinates in the coordinate system S, and [itex](x'_{A},y'_{A},z'_{A},t'_{A}),(x'_{B},y'_{B},z'_{B},t'_{B})[/itex] are the coordinates in the coordinate system S', which is traveling at speed [itex]u<c[/itex] in the positive [itex]x[/itex] direction. Then from the Lorentz transformation equations,

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

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

[tex]x'_{A}=\frac{x_{A}-u t_{A}}{\sqrt{1-(u^{2}/c^{2})}}[/tex]

[tex]x'_{B}=\frac{x_{B}-u t_{B}}{\sqrt{1-(u^{2}/c^{2})}}[/tex]

[tex]y'_{A}=y_{A}[/tex]

[tex]y'_{B}=y_{B}[/tex]

[tex]z'_{A}=z_{A}[/tex]

[tex]z'_{B}=z_{B}[/tex]

Because B happened after A,

[tex]t_{B} > t_{A}[/tex]

Because event A caused event B,

[tex]\frac{\sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2} + (z_{B}-z_{A})^{2}}}{t_{B}-t_{A}} \leq c[/tex]

## The Attempt at a Solution

I need to prove that [itex]t'_{B} - t'_{A} > 0[/itex]

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

What do I do now?