1
$\begingroup$

I know that instantaneous signalling can result in different observers not agreeing on the order of events, but how can that result in causality violation in practice? In other words, if one had two devices that could communicate instantaneously in their own reference frame how could it be set up to violate causality?

$\endgroup$
5
  • 1
    $\begingroup$ What do you mean by causality? It is quite a murky concept. $\endgroup$ Nov 17, 2014 at 13:08
  • $\begingroup$ Sending a signal backwards in time between two versions of the same machine at different times. $\endgroup$
    – user56903
    Nov 17, 2014 at 13:23
  • $\begingroup$ note that any superluminal signal comes with a critical frame where is appears instantaneous; for an explicit example of how superluminal signals violate causality, see en.wikipedia.org/wiki/… $\endgroup$
    – Christoph
    Nov 17, 2014 at 13:47
  • $\begingroup$ Yes, but assuming I don't have access to ships passing at 0.8c, but I do have access to various communications media that carry info at that speed, can causality violation still be determined? $\endgroup$
    – user56903
    Nov 17, 2014 at 13:52
  • $\begingroup$ It would help if you defined what frame the signaling was going to be instantaneous in. The frame of the transmitter? Of the receiver? $\endgroup$
    – user4552
    Nov 17, 2014 at 16:04

4 Answers 4

2
$\begingroup$

A very simple example would be the following:

Pair up device A and B for instantaneous communication. Keep Device A with you and send device B on a journey at a speed close to $c$ (e.g. $v=\frac{\sqrt{3}}{2}c$), programmed to repeat any message back to you.

World line of A: $x(t) = 0$, world line of B: $x(t) = vt$

If, at an event $(0, T_0)$, you send a message to device B, it will receive it at $(vT_0, T_0)$. To determine which event is simultaneous to B at your location, we need to enter B's frame of reference via Lorentz transformation:

$$ x\rightarrow x' = \gamma (x-vt),\qquad t\rightarrow t'= \gamma(t-\frac{v}{c^2}x) $$

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is $2$ in our case and the event where B receives the message is $(0, 2(T_0 - \frac{v^2}{c^2}T_0))' = (0, \frac{T_0}{2})'$, so the time in B's frame of reference is $T_0'=\frac{T_0}{2}$.

The motion of A relative to B is $x_A'(t') = -vt'$, so, if B instantaneously replies, the message will be sent to $x_A'(T_0') = -v\frac{T_0}{2}$ in B's reference frame. This event $(-v\frac{T_0}{2}, \frac{T_0}{2})'$ then translates to

$$ \left(\gamma\left(-v\frac{T_0}{2}+v\frac{T_0}{2}\right), \gamma\left(\frac{T_0}{2} + \frac{v}{c^2}\cdot \left(-v\frac{T_0}{2}\right)\right)\right) = \left(0, \frac{T_0}{4}\right), $$

which clearly happens before $(0, T_0)$, since $T_0>0$.

$\endgroup$
1
  • $\begingroup$ If it helps Dirk or others to follow the physical meaning of this math, there's an illustration involving spacetime diagrams on this page, and there's a numerical example on the tachyonic antitelephone wiki page. $\endgroup$
    – Hypnosifl
    Nov 17, 2014 at 17:48
2
$\begingroup$

Yes, it would violate causality because of the following reasons.

(1) "instantaneous" is a reference-depending notion. A pair of events are instantaneous in a reference frame if and only if they are spacelike separated.

(2) Using a chain of spacelike segments you can connect an event $q$ on a future oriented timelike curve $\gamma$ (your worldline) with a point with an event $p$ on the same $\gamma$, in the past of $q$.

In other words, with this chain of events, you can communicate with your-self in your own past. This could generate causal paradoxes even in the absence of closed or almost closed timelike worldlines.

What I wrote is independent on the notion of synchronization you adopt and only relies on the causal structure of the spacetime, which is independent from the synchronization procedure you choose. The synchronization "à la Enstein" (which produces Lorentz transformations) is partially conventional because it uses the value of $c$ on an open path. Conversely, the causal structure can be constructed using only the experimental fact that the speed of light is constant, $c$, when measured along closed paths at rest in an inertial reference frame.

$\endgroup$
7
  • $\begingroup$ "This could generate causal paradoxes even in the absence of closed or almost closed timelike worldlines." If all events (including the emission and reception of communication) are fixed, what does term "causal paradox" even mean? It would help if you could add some example of such a paradox. $\endgroup$ Nov 17, 2014 at 15:53
  • 1
    $\begingroup$ I send a signal from the future to explode a bomb under my bed today...This causes my death so I cannot send the signal tomorrow... $\endgroup$ Nov 17, 2014 at 15:59
  • $\begingroup$ As a matter of fact similar spacelike "correlations" already exist in view of EPR phenomena. However, due to the fact that QM is intrinsically stochastic they cannot be used to produce events in the pasts controlled in the future...They just are "correlations" not causal relations. $\endgroup$ Nov 17, 2014 at 16:05
  • $\begingroup$ You assume that the reception of communication in the past will change what already happened then (past gets rewritten). When this assumption is dropped, the communication can still happen and the paradox vanishes. $\endgroup$ Nov 17, 2014 at 16:19
  • $\begingroup$ Yes you are right. Your viewpoint is consistent with the idea that spacetime is nothing but the list of what happened, and this list is filled in "at the end of time". $\endgroup$ Nov 17, 2014 at 16:29
1
$\begingroup$

if one had two devices that could communicate instantaneously [...]

Two (or several) devices that communicate instantaneously with each other (or at least nearly instantaneously among each other, in comparison with their communications with other devices),
i.e. devices which find zero ping durations between each other (or who determine at least that their ping durations between each other were much shorter than their ping durations with respect to other devices),
do therefore have a very specific geometric relation to each other: they are co-located
(or they are at least much closer to each other than to other devices; they are "practically co-located", "as good as co-located").

They took part in the same events together. Therefore they will have no difficulty to agree on the sequence in which either of them took part in these events. Any "violation" or "contradiction" concerning this sequence does not obtain.

$\endgroup$
3
  • $\begingroup$ Could you clarify your answer? Do you mean to say, that there is no violation of causality, if you allow for FTL information exchange? Or are you simply saying, that devices which can talk to each other very fast must be close together? $\endgroup$
    – M.Herzkamp
    Nov 18, 2014 at 7:53
  • $\begingroup$ @M.Herzkamp: "are you simply saying, that devices which can talk to each other very fast must be close together?" -- Exactly. Comparison of pairwise "signal round trips" (a.k.a. "pings") between "devices" under consideration is how we determine in the first place geometric relations between them. From (necessarily mutual) observation of "faster exchange of signals", i.e. "shorter ping durations" between given participants (always referring to the signal front, of course) it follows that they had been "closer together" than others. $\endgroup$
    – user12262
    Nov 18, 2014 at 17:04
  • $\begingroup$ Hm. The question does not mention anything about FTL information transfer. I like your spin on the question. +1 $\endgroup$
    – M.Herzkamp
    Nov 19, 2014 at 9:31
0
$\begingroup$

Imagine you are standing next to an evil person you is trying to shoot your friend, who is also standing still a ways away. You will tell your friend to duck when the evil person shoots his gun. Then in your frame, your telling your friend to duck and his ducking occur at the same time.

In a boosted frame, these events will not be simultaneous. By boosting the appropriate way, you can make it so that your friend ducks (because of your message telling him to duck) before you have told him to duck in the first place. Here the "effect" happens before the "cause", which violates causality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.