I know about this theory but I cannot get the picture that how moving faster than light even 100.01% of light-speed would make universe run backward.

  • $\begingroup$ Good, you should not. $\endgroup$ – m4r35n357 Jan 7 '17 at 12:35
  • 1
    $\begingroup$ Special Relativity forbids faster-then-light travel. Arguing about something that isn't allowed by the laws of nature is pointless. $\endgroup$ – Prof. Legolasov Jan 7 '17 at 12:37

So, rather than make some general argument, I will show how, if you can send things faster than light between two points, you can send things backwards in time. To do this, I will need to assume two things:

  • the Lorentz transformations are correct;
  • all inertial frames are equivalent & in particular if you can do something in one frame you can do it in another.

These two assumptions really add up to the statement that special relativity is correct (in flat space). This is very well-tested, of course.

We'll just worry about one space dimension: all motion is along $x$. When I write coordinates I will write the time coordinate first: $(a,b)$ means $t=a$ and $x=b$.

The Lorentz transformations for an observer moving in the $x$ direction with velocity $v$ are:

$$ \begin{align} x' &= \gamma(x - vt)\\ t' &= \gamma\left(t - \frac{vx}{c^2}\right) \end{align}$$


$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

To make life easier we'll work in units where $c = 1$: so if our time units are seconds, our length units will be light-metres. I'm just doing this so I don't need factors of $c$ all over the place.

In these units the Lorentz transformations are

$$ \begin{align} x' &= \gamma(x - vt)\\ t' &= \gamma(t - vx) \end{align}$$


$$\gamma = \frac{1}{\sqrt{1 - v^2}}$$

(Remember that $v$ is now the velocity as a fraction of $c$).

So, OK, let's draw a picture:


In this picture we are communicating from $A$ to $B$ somehow, and as you can see, $B$ is outside the light cone originating at $A$ (drawn in grey). Drawn in blue is the world line of someone who passes through the event at $A$ at some high velocity $v$, who I have called the 'primed observer' (because their frame of reference will have primes on its coordinates). In our units, $v < 1$ (the primed observer can only move slower than light), but $v$ can be very close to $1$.

In this frame the coordinates of $A$ are $(0,0)$ and of $B$ are $(t, x)$.

Now the statement that $B$ is outside the light cone of $A$ is equivalent to saying that $x/t > 1$: the light cone of $A$ is where $x=t$ in our units.

So, now, let's calculate the coordinates of $B$ in the primed frame. They are

$$ \begin{align} t' &= \gamma(t - vx)\\ x' &= \gamma(x - vt) \end{align}$$

With $\gamma = 1/\sqrt{1-v^2}$.

Well, I don't want to think about $\gamma$ all the time, so let's just look at $x'/t'$, and we can cancel the $\gamma$s:

$$\frac{x'}{t'} = \frac{x - vt}{t - vx}$$

So, what can we say about this quantity? Let's think about the case when $v$ is very close to $1$: the primed observer is moving very fast compared to the unprimed observer. So we'll write $v = 1 - \epsilon$ where $\epsilon > 0$ but $\epsilon$ can be as small as we like. So

$$ \begin{align} \frac{x'}{t'} &= \frac{x - vt}{t - vx}\\ &= \frac{x - (1 - \epsilon)t}{t - (1 - \epsilon)x}\\ &= \frac{x - t + \epsilon t}{t - x + \epsilon x}\\ &\approx \frac{x - t}{t - x} + \frac{\epsilon t}{t - x}\\ &= -1 + \delta \end{align}$$


$$\delta = \frac{\epsilon t}{t - x}$$

and I have dropped the $\epsilon x$ in the denominator of the second-last line as it can be made as small as you like.

So, well, look at $\delta$: By construction $\epsilon > 0$, and looking at the picture above, $t > 0$ & $x > 0$, but $x > t$ (because $B$ is outside the light cone of $A$), so $t - x < 0$. So $\delta < 0$, but as $\epsilon \to 0^+$, $\delta \to 0^-$.

In other words we can make $\delta$ very small but it is always negative.

So what this means is that $x'/t' < -1$ but we can make it as close to $-1$ as we like by letting $v$ become very close to $1$.

In other words, in the frame of the primed observer, $B$ can be as close as you like to its past light cone, while always being slightly outside it.

Here is a picture in the primed frame:


As you can see, $A$ is still at the origin but $B$ now appears just outside the past light cone of $A$ from the point of view of the primed observer.

(At this point you should start to hear the slithering of tentacles as some great creature of the deep rises to devour you.)

Consider yet another inertial observer, comoving with the primed observer, whose world-line passes through $B$. This observer's world-line is in blue in the second diagram (and not shown in the first). This observer receives the message sent from $A$ to $B$ as their world-line passes through $B$.

But whatever magic we originally used to get the message from $A$ to $B$, they can use too, since all inertial frames are equivalent. In particular, they can send a message from $B$ to $C$, and if $v$ is close enough to $1$ (how close it needs to be depends on how far $B$ is outside the light cone of $A$ in the original frame), then $C$ is within the past light-cone of $A$, and so it is now trivial to send a message from $C$ to $A$ (indeed we can pick $C$ to be $A$ if we like!).

And that's time travel: I have bargained being able to send a message faster than light into sending a message into my own past.


If you exceed the speed of light the traditional gamma factor in special relativity gets an imaginary value for the ratio of times (and also the ratio of lengths) from the viewpoint of observers which is meaningless unless you want to consider Godel metric according to which time travel is allowed in the realm of general relativity.

  • $\begingroup$ This doesn't really make sense. The Godel metric doesn't have anything to do with the fact that the equation for $\gamma$ gives an imaginary result when $v>c$. $\endgroup$ – Ben Crowell Mar 4 '18 at 19:37
  • $\begingroup$ It seems that the conjunction "unless" caused a misapprehension! I just wanted to say that Godel metric allows time travel. $\endgroup$ – Mohammad Javanshiry Mar 4 '18 at 20:27

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.