I've been wondering about this for a long time. Given that special relativity is normally true, can we still prove that causality violation must occur if FTL is found? All proofs I can find, depend on it being absolutely true.

What occurs to me, is engaging an FTL drive might do something like prove the aether really does exist, but normal physics is disconnected from it. This model has a few problems, like for example anything transitioning to FTL ends up with an instantly high velocity relative to the aether.

Question almost matches Can FTL-Communication between two points in the same frame of reference break causality?, but the answers are opposed. The first (and most likely correct) answer doesn't apply; however the second one seems to apply but contradicts the first for that question's scenario.

I can prove the answer is NO by geometry if I make an assumption that we believe is false: attempting to cross a black hole event horizon is instantly fatal. A large enough (and galactic core black holes are large enough) black hole gives an absolutely backwards-pointing light cone that might be exploitable here (FTL drive usually means /can/ escape from black hole event horizon).

EDIT: Darn. Question appears to be meaningless without a particular theory to discuss. I was trying to ask if something similar to Noether's theorem for energy and momentum conservation is known under arbitrary laws of physics, but if I try to harden up the question to allow talking about drawing the parallel I get a contradiction in the question itself.

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    $\begingroup$ The proofs that FTL is equivalent to causality violation are exceedingly general. The short answer is "yes, and it was done more than 100 years ago." $\endgroup$ Nov 1, 2015 at 21:34
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    $\begingroup$ I am aware of the 100 year old proof and it has an assumption I explicitly excluded in the first paragraph. $\endgroup$
    – Joshua
    Nov 1, 2015 at 21:38
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    $\begingroup$ What do you mean by "normally" vs. "absolutely" true? If you assume that near and above SOL special relativity has to be modified to avoid causality violations than you trivially get your FTL without causality violations. Special relativity does not apply to black holes though. $\endgroup$
    – Conifold
    Nov 1, 2015 at 21:53
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    $\begingroup$ FTL is exactly like energy conservation. All you need is one experimental example and you are done. It's really that simple and it has absolutely nothing to do with theory. Theory simply describes what has and has not been observed. Relativity describes that FTL has not been observed, so you can't get FTL out of it, no matter how hard you try on the theory side. But if you find one experiment that produces an FTL effect, then we have to modify the theory and we will. So all you really have to do is to build that one experiment... $\endgroup$
    – CuriousOne
    Nov 1, 2015 at 21:56

1 Answer 1


This is a perfectly good question; don't feel discouraged.

Coming up with a notion of 'causality' that doesn't refer to any physical theory is pretty hard. I'll discuss it here in the context of general relativity (GR). The basic entities in GR are events; i.e. spacetime points $(t, x, y, z)$.

Suppose we have two events $A$ and $B$ and that, according to some observer $\mathcal{O}$, $t_A > t_B$ - that is, $A$ occurs after $B$. We seek a necessary (not sufficient) condition under which $B$ can be said to have 'caused' $A$.

That condition is the following: $t_A > t_B$ must hold for all observers. If my turning on a stove ($B$) caused water to boil ($A$), there had better not be anyone for whom the water boiling happened before the stove turned on.

Because of the signature of the metric, temporal ordering is preserved under coordinate changes only for timelike separated events; i.e. only for events connected by a geodesic that goes "slower than the speed of light". See my answer here for an elaboration: What spacelike, timelike and lightlike spacetime interval really mean?.

  • $\begingroup$ You went down the same path I did, only I found a weaker metric that suffices under one additional assumption (the quantum black hole phase change). But if I allowed crossing a black hole event horizon, my metric broke. The problem is I understand the phase change theory fell a few years ago (perhaps with Hawking's $ matrix). $\endgroup$
    – Joshua
    Nov 20, 2015 at 17:00
  • $\begingroup$ @Joshua The ordering property depends only on the signature of the metric: that the timelike coordinate has a different sign than the spacelike ones. This is true of all GR metrics by assumption. $\endgroup$
    – AGML
    Nov 20, 2015 at 17:01
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    $\begingroup$ But we don't care if a distant observer sees them in the wrong order so long as the information can never get back to the proper time of the events before all of them happen. $\endgroup$
    – Joshua
    Nov 20, 2015 at 17:03

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