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Text of the exercise:

Event B is a consequence of event A. Using Lorentz transformations, prove that there is no inertial frame for which Event B happens before event A.

I wanted for someone to say how much the approach I use is correct and if there is any simpler one.

Let A have coordinates $(x_1,y_1,z_1,t_1)$ and B is $(x_2,y_2,z_2,t_2)$, where $t_1<t_2$ in some inertial reference frame S.

Then there is another (moving with constant speed $v$ in x-direction of motion) inertial reference frame, where event A is $(x_1',y_1',z_1',t_1')$ and B $(x_2',y_2',z_2',t_2')$

We have to prove that $t_1'<t_2'$. Using LT:

$$ t_1' = \gamma(t_1-vx_1/c^2) $$ $$ t_2' = \gamma(t_2-vx_2/c^2) $$

Now: $x_2 = x_1 + v_{propagation}(t_2-t_1)$ (so $x_2$ will happen certain distance from $x_1$, which is speed of propagation of the "effect" (hereafter $v_p$) * time interval in $S$ frame).

Then we have: $t_1' = \gamma(t_1-vx_1/c^2)$

$t_2'=\gamma(t_2-\frac{v*x_1}{c^2}-\frac{v*v_p*(t_2 - t_1)}{c^2})$

If we now subtract $t_1'$ from $t_2'$ we have (after factoring): $t_2' - t_1' = \gamma(t_2-t_1)(1-\frac{v*v_p}{c^2})$

Since $t_2>t_1$ and $\frac{v*v_p}{c^2}\leq1$, then $t_1'\leq t_2'$

Kyrylo Lyskov
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2 Answers2

1

Text of the exercise:

Event B is a consequence of event A. Using Lorentz transformations, prove that there is no inertial frame for which Event B happens before event A.

If event B is a consequence of event A, then the space-time distance between $B$ and $A$ is "time like."

In other words, the spatial distance from $\vec x_A$ to $\vec x_B$ has to be less than or equal to the distance light would travel in time $t_B - t_A$ (otherwise A couldn't cause B, almost by definition): $$ |\vec x_B - \vec x_A| \le c(t_B - t_A)\;. \tag{1} $$

Lorentz transformation do not change the sign of the time difference for time-like separations. But be careful, because this preservation of time order is not necessarily the case for space-like separations. See the answer here.

hft
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With a time-like separation, we can define $S$:

$$E_A = (0, 0)_S$$

and

$$ E_B = (T, 0)_S$$

so

$$ \Delta t = T $$

Boosting that to $S'$ by $\beta$:

$$ E_A = (0, 0)_{S'} $$ $$ E_B = (\gamma T, [{\rm something}])_{S'}$$

so

$$ \Delta t' = \gamma T > 0 $$

That's a more specific implementation of hft's answer, which I think is a more specific implementation of yours?

JEB
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