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I can understand why the $x=ct$ line is needed as follows:

If $c \Delta t \geq \Delta x$ then the event can be causally related to your current position, so the event could not appear in reverse order for any observer; therefore that event must be in the absolute future.

Similarly, if $c \Delta t < \Delta x$ then the event cannot be causally related, and it is outside the absolute future — some observers might find the order of the event switched.

Hence clearly the $x = ct$ line plays a role in delineating absolute vs relative future.

What I don't understand is where the $x = -ct$ line comes in. What role does that play? Why are events in the absolute future also under the constraint that $\Delta x \geq -ct$?

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If $x=ct$, you have a light ray moving in the positive $x$ direction. If $x=-ct$, you have a light ray moving in the negative $x$ direction. You don't care which way the light goes, only its speed.

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  • $\begingroup$ How would you derive it from the lorentz transform for time dilation? For instance, for showing that $x=ct$ is one boundary, you can start from $$\text{$\Delta $t}'=\frac{\text{$\Delta $t}-\frac{\text{u$\Delta $x}}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$$ and if $\Delta x > 0$ in order for $\Delta t'$ to be of different sign than $\Delta t$ you need $$\frac{\text{u$\Delta $x}}{c^2}>\text{$\Delta $t}$$ which means $$\frac{u}{c}>\frac{\text{c$\Delta $t}}{\text{$\Delta $x}}$$ leading to $$\text{c$\Delta $t}>\text{$\Delta $x}$$ as the absolute future. How would it work when $\Delta x$ is negative? $\endgroup$ – 1110101001 Sep 7 '15 at 2:50
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    $\begingroup$ @1110101001: Your derivation would almost work. You would have to change your second inequality to $u/c < c\Delta t / \Delta x$ on account of $\Delta x < 0$. If $u > 0$ this is impossible, but if $u<0$ then $u/c > -1$ and you get $-\Delta x > c\Delta t$. $\endgroup$ – Javier Sep 7 '15 at 3:02

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