# Why is causality preserved in special relativity?

PART 1:
I was reading the article Relativity of simultaneity Wikipedia. I couldn't understand this line:

"if the two events are causally connected ("event A causes event B"), the causal order is preserved (i.e., "event A precedes event B") in all frames of reference."

Is this an assumption or a consequence of STR? Please explain.

Note: My question consists of 2 parts this is the 2nd part.
Below is a genral version of my previous question question:Breaking the simultaneity.

PART 2:
Let there be three events $A$,$B$ and $C$ s.t: $C$ is the result of Simultaneous occurrence of $A$ and $B$. In other words $C$ occurs iff $A$ and $B$ are simultaneous.
Now as we know in STR any two events separated in space are not simultaneous in different frames. So In some frames $C$ will occur and in some $C$ will not occur which will cause paradox.
I tried many thought experiments to make such a paradox but i failed. In all the experiments that i thought i could not break the causality even by breaking the simultaneity because everytime the fact: "all signals move slower than light" preserved the causality.

So why causlity remains preserved always? Is it due to the fact that nothing can move faster than light?

• In special relativity, if two events are causally connected, this means that both events lay in each other's light cone. As light cones are absolute (defined by the event and observer independent), this can happen in only one way: one event ($A$) is located in the future light cone of the other ($B$), and the event $B$ is located in the past light cone of $A$. Feb 23, 2014 at 16:51
• @Johannes I haven't read about light cone. I've just started STR. Why light cones are absolute? Is this an assumption or a consequence of the two postulates of relativity? Feb 24, 2014 at 12:17

Causality is preserved, unless Tachyons exist.

Part 1: STR doesn't assume causality. Causality is violated when you have a flow of information that goes back to the same place in space AND time, creating a contradiction. Both newtonian and STR guarantee causality. STR is more complex, but it still prevents anything from going back in time with respect to another's frame, as long as nothing starts out faster than light in the first place (i.e. Tachyons). The policeman work by creating a "light barrier" that prevents anything from being accelerated past the speed of light.

Part 2:

"C is the result of Simultaneous occurrence of A and B. In other words C occurs iff A and B are simultaneous."

You can't do this. Lets try: A and B are excited atoms that emit flashes of light. Suppose a detector halfway in-between explodes because both atoms flashed at once. The atoms and detector are stationary. Sounds like a simultaneity detector? Not in a frame moving with respect to the set up! In a moving frame, the atoms emit light at different times (light still goes the same speed in any frame), and the detector is still halfway in-between the atoms. But the set-up moves just the right amount in the time it takes for the flashes to converge that the detector will hit them at thier meeting point, and still explode.

• Let the exited atoms emit techyons. The detector will not be hit simultaneous in the moving frame. So in rest frame detector will explode but in the moving frame it will not hence creating the paradox, not to mention-STR doesn't assert the non existence of techyons.second You didn't answer the first part "Is this an assumption or a consequence of STR? Please explain." Feb 24, 2014 at 11:54
• No, it's even weirder. Causality is gone. If the Tachyons are fast enough, the detector will simultaneously be hit with a single Tachyon from atom A, emit another toward B, and explode. Feb 25, 2014 at 14:53
• Why the detector will explode? The only condition was the simultaneous hit. If simultaneity breaks( Events are not spatially separated though) the detector will not explode. There should be the paradox. Feb 25, 2014 at 15:04
• Bieng "hit" by a Tachyon in one frame is "emitting" the Tachyon in another frame. Did the detector absorb to or create the Tachyon? No one knows, causality is no longer defined! This sounds absurd, and it is. This is why Tachyons probably don't exist. Feb 25, 2014 at 15:09

Special relativity is (usually) defined as that which is invariant under the action of $SO^+(1,3)$ on Minkowski space (i.e. real affine 4-space). $SO^+(1,3)$ preserves time orientation by definition, so the preservation of causality is an immediate consequence of that definition. The answer to your first question, then, is that this is either an assumption or a conclusion depending on your wording: We have two obviously equivalent statements ("SR is $SO^+(1,3)$-invariant" and "SR preserves causality"), you can take whichever you like as an assumption, and the other immediately follows as a conclusion.

But you could equally well take SR to be defined by the action of $SO(1,3)$, i.e. without requiring the preservation of time orientation. The corresponding theory has objects traveling backward in time, but is observationally equivalent to the $SO^+(1,3)$ theory. After all, if you see an object that appears to be traveling rightward and forward in time, how do you know it's not really traveling leftward and backward in time? An $SO(1,3)$-invariant theory allows both possibilities (i.e. it allows you to replace the parameter $s$ with the parameter $-s$ when you parameterize a worldline) but makes all the same predictions about what you'll actually observe.

(Note that these backward-in-time particles are not tachyons; their worldlines are timelike.)

So: Is preservation of causality an assumption or a conclusion of SR? Answer:

• If you define SR by $SO(1,3)$ invariance, it is neither a conclusion nor an assumption. In this case, SR allows backward time travel, but still makes all the same predictions as if it didn't.
• If you define SR by $SO^+(1,3)$ invariance, it's a conclusion, but the conclusion follows immediately from the assumption; in fact it's just a restatement of the assumpton.
• If you define SR by $SO(1,3)$ invariance plus the preservation of causality, it's an assumption (once again, immediately equivalent to the assumption of $SO^+(1,3)$ invariance).

Which of these is the "real" SR? It doesn't make a bit of difference. They're all slightly different ways of rewording the same thing.

I would have liked to comment but you need 50 reputation.

I think that your thought experiment is set up incorrectly. Yes in other inertial reference frames two events will not be simultaneous. However, $C$ is stationary within its own reference frame including the simultaneous events. Thus, $C$ will occur regardless of what other reference frames see.

• Since there cannot be any privileged frame of reference you cannot assert "What happen in one frame will happen in other". You can only show by calculation taken from the other frame. Feb 24, 2014 at 11:57
• The reference frames are not symmetric. So there is Feb 24, 2014 at 15:10
• All the inertial reference frames are symmetric. So there is not any. Feb 24, 2014 at 15:51
• You have made the frames asymmetric but adding the event C Feb 24, 2014 at 17:56
• I cannot change the laws of nature so all the frames have to be symmetric acc. to STR and the experimental prove of the deficiency to find "Aether". To show $C$ will not occur in the moving(w.r.t. train) frame modify Kevin's answer by replacing "flashes of light" with techyons. $C$ will occur in one frame and not the other. Feb 25, 2014 at 11:13

The statement is a consequence rather than an assumption. It is limiting the domain of event pairs to those which can be considered causally connected. In a sense it is requiring that there be the possibility that the event A be detectable in the future of event B if it is to be considered causal. The light-cone from event A defines the boundary of that possibility.
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