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 $\Delta x \geq -ct$?


1 Answer 1


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.

  • $\begingroup$ 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, 2015 at 2:50
  • 1
    $\begingroup$ @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, 2015 at 3:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.