Signal can't travel at speed greater than light speed in vacuum which is a assumption of special relativity. But if a signal travels at speed greater than $c$ then it will violate causality. I tried to prove this statement using Lorentz transformation equations. But in the denominator and imaginary number will arise. I need a mathematical proof.

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    $\begingroup$ What is the statement you are trying to prove? Signal can't travel faster than light, or "if it does it would violate causality"? Can you show your work so far - what do you think an imaginary number tells you? $\endgroup$
    – Floris
    Feb 25, 2016 at 13:59
  • $\begingroup$ I assume a particle in lab frame is at point (x_0;t_0) and after some time it will be at (x;t). Here I assume that (x-x_0)/(t-t_0) is greater than c.Now my aim is to deduce the space time co-ordinates of two events in particle rest frame.For that I use Lorentz transformation equations for time and there I show that in particles own rest frame (x,t) event happens earlier than (x_0;t_0).I think this must be violation of causality. Because in Lab frame (x_0;t_0) event can be taken as cause (reason for fact) and (x,t) as fact.But this sequence is violated in particles own rest frame. $\endgroup$ Feb 25, 2016 at 14:12
  • $\begingroup$ The only thing that bothers me is that the factor root over (1-v^2/c^2) will give imaginary number.And for that I have no justification. $\endgroup$ Feb 25, 2016 at 14:14
  • $\begingroup$ @Beman: It should not be necessary for you to translate between two frames with a greater mutual velocity than $c$ -- just assume that the signal can be made to leave one event and be received at another end such that the ratio between (difference in space coordinates) and (difference in time coordinates) is greater than $c$. You don't need to set up a frame where the signal doesn't move. $\endgroup$ Feb 25, 2016 at 14:22
  • $\begingroup$ en.wikipedia.org/wiki/Tachyonic_antitelephone "Alice will receive the message back from Bob before she sends her message to him in the first place." $\endgroup$ Feb 25, 2016 at 18:06

1 Answer 1


Suppose I have a gun that fires bullets with a superluminal velocity V, and I'm going to use it to shoot you. In our rest frame I'm standing at the origin, $x=0$ and you're standing at some distance $x = D$. I fire the gun at $t=0$ and you die at $t=D/V$. So cause an effect are pretty clear - I fire my gun and as a result of this you die a short time later. (Apologies if this seems unnecessarily bloody :-)

Now suppose my friend Fred is in a spaceship flying past at a (subluminal) velocity $v$. We can use the Lorentz transformations to find out what happens in Fred's rest frame. We'll assume Fred passes me just as I fire, so the gun is fired at the spacetime point $(0,0)$ in both our frames. It just remains to find where in Fred's frame the bullet hits you.

In my frame the bullet hits you at $(t=D/V, x=D)$ so let's use the Lorentz transformations to calculate when the bullet hits you in Fred's frame:

$$\begin{align} t' &= \gamma \left( t - \frac{vx}{c^2} \right) \\ &= \gamma \left( \frac{D}{V} - \frac{vD}{c^2} \right) \\ &= \gamma \frac{D}{V}\left( 1 - \frac{vV}{c^2} \right) \end{align}$$

But if we make the bullet velocity $V \gt c^2/v$ that means $1 - vV/c^2$ is negative, so $t' \lt 0$, and this would mean that in Fred's rest frame you died before I fired the gun.

This is where we have a problem with causality. For any superluminal bullet velocity there is a frame where you died before I fired at you. The only way to avoid this is for the bullet velocity to never exceed $c$.

  • $\begingroup$ Nice explanation indeed.I am not that familiar with SR ,but the derivation of the above equations don't assume that $C$ is constant regardless the observer and cannot have greater value? $\endgroup$
    – user98038
    Feb 25, 2016 at 18:03
  • $\begingroup$ @aK1974: Yes, $c$ is assumed constant. This is required to derive the Lorentz transformations. $\endgroup$ Feb 25, 2016 at 18:38
  • $\begingroup$ If so, the use of Lorentz transformations to prove $T$<0 implies that 'tachyons' -if they exist- would 'see' the speed of photons as $C$ also .How can be sure for that? $\endgroup$
    – user98038
    Feb 25, 2016 at 19:05
  • $\begingroup$ @aK1974: I'm not sure what you're asking. Maybe you could work it up into a question and post it here. $\endgroup$ Feb 25, 2016 at 19:26
  • $\begingroup$ @ John Rennie:In an hypothetical situation that there's indeed a value of speed V greater than $C$, there's no way to check that the constancy of $C$ is valid for an observer that moves with speed $V$. So how can someone use the Lorentz transformations to show that $T<0$? $\endgroup$
    – user98038
    Mar 30, 2016 at 20:45

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