# Causality and speed of light

It is accepted that the speed of light is the speed of causality. If we exceed the speed of light, the order of cause and effect breaks down. This happens as we see our surroundings moving backward in time. Right?

However, how do we know they move back in time once we move faster than light? If $v>c$, then in the formula $$t'=\frac{t}{(1-v^2/c^2)^{1/2}}$$ we get an answer that is not defined, so how do we know time moves backwards?

• Speeds faster than light are impossible in the geometry of our spacetime. In relativity, "speed" is not simply distance by time, as distance and time are no longer independent. Instead, they are two sides of the same spacetime. In this geometry, light moves with the speed of time and therefore time stops for the photons while the distance shrinks down to zero. You cannot exceed the speed of light not because time would go backwards (you pointed this out correctly in the last sentence). You simply cannot move slower in time than not moving at all. Not a technical limitation, but pure geometry. – safesphere Sep 13 '17 at 5:36
• Like you can't get any more Northern that the North Pole on the globe, because your distance to the true North is already zero and can't get any smaller than that. The hyperbolic geometry of the Minkowski spacetime is a lot less intuitive than the globe, but the idea is the same. When time stops, it can't move any slower than the zero speed. This translates into the speed of time being the fastest speed possible in the hyperbolic spacetime. – safesphere Sep 13 '17 at 5:40

The issue isn't whether time will move backwards at the speed of light, it's that having stuff sending signals around faster than light causes problems even for us nonrelativistic slower-than-light beings.

Consider a particle moving at speed $v$. It traces out a worldline $(t,v t)$ in some frame of reference, which I'll call frame 1. That is, at time "t" it will have a time coordinate of $t$ and a space coordinate of $vt$. You can also say that this traces out a "series of events", each event is at the coordinate $(t,vt)$. If it moves faster than light, there is no problem. As the time coordinate increases when $v>c$, we still are looking at a point more in the future. So there's no paradox yet and no movement backwards in time.

To another observer with speed $|u|<c$, however, this gets Lorentz transformed. To him, in frame 2, the coordinates appear at locations $(\gamma t-\gamma \frac{u v}{c^2} t,-\gamma u t+\gamma v t)$. $\gamma$ is well-defined, because here $\gamma=\frac{1}{\sqrt{1-u^2/c^2}}$. But the problem is that in this observer's frame, if $1-uv/c^2<0$, the particle moves backwards in time as $t$ increases! That is, in frame 1 the particle evolves forward in time, in frame 2 it evolves backwards in time.

If you follow this line of logic, you find that if you are allowed faster-than-light travel in arbitrary reference frames, you can cause paradoxes. eg, you could kill yourself before you can go back in time to kill yourself! I outline exactly how you can do that in this linked answer. That's why things like the Alcubierre drive are still safely in the realm of science fiction. Even though it's consistent with general relativity, if it was possible to create and destroy FTL drives arbitrarily, you would still get those paradoxes.

# More intuitive approach

In response to the comment asking for more intuition. I can't give an intuitive "why" because special relativity totally breaks people's intuitions! But I can give an intuitive "how".

Say I have a special faster-than-light bomb. In my frame, causality makes sense: I launch it at a planet one light-year away, and in half a year (faster than light) the planet blows up. My description of the universe makes sense, because the planet blew up after I launched the bomb.

In your frame, if you're moving fast enough relative to me, you observe the planet explode, then the faster-than-light bomb travels back to my planet, then I press the button to launch it.

You can see how this causes problems, and yet no observer travels faster than light, so $\gamma$ is always real in any Lorentz transformations. The proper time of the bomb won't be real, but we don't need to take that into account.

• can you please give a more intuitive explanation. i am not comfortable with the formulas ,so a simpler explanation would be really appreciated.thanks – spatialdelusion Sep 13 '17 at 4:59
• @daboss I added some exposition! Let me know if that helps. – user12029 Sep 13 '17 at 5:07
• @daboss You're asking a deep question about the symmetries of the universe, which relate to phenomena which are beyond the scope of everyday human intuition. It doesn't get much simpler than this already very simplified explanation; working your way through the math is the only way to understand it better. – J. Murray Sep 13 '17 at 5:07
• @NeuroFuzzy"In your frame, if you're moving fast enough relative to me, you observe the planet explode, then the faster-than-light bomb travels back to my planet, then I press the button to launch it."Why will a person moving relative to you first see the planet explode?can you please explain that part a little elaborately .thanks a lot for giving the intuitive explanation. – spatialdelusion Sep 13 '17 at 5:14
• @daboss The idea that for distant enough events (for which $\Delta t^2 c^2-\Delta x^2<0$) you can't determine which occurred before the other one is called the relativity of simultaneity, so that's a good place to start! I'm not a master of gedankenexperiments :) – user12029 Sep 13 '17 at 5:20