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According to special relativity (as I understand it) simultaneity is no longer a universal concept in special relativity. Consider two events A and B which are simultaneous in reference frame 1; in reference frame 2, A might precede B timewise; in reference frame 3, B might precede A.

Suppose in reference 2 we might consider scrambling an egg: event A, egg is dropped into the frying pan; event B, egg is cooked. It seems counter to common sense that there would be a reference frame in which the egg is cooked before being heated in the pan.

In other words, does entropy, $S$, "Time's Arrow," trump relativity?

Qmechanic
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duhem7
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    Here is your answer https://physics.stackexchange.com/questions/75763/can-special-relativity-distort-the-relative-order-in-which-events-occur –  Feb 26 '20 at 03:56

3 Answers3

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Consider two events A and B which are simultaneous in reference frame 1; in reference frame 2, A might precede B timewise; in reference frame 3, B might precede A.

This is true, if the events happen at spatially separate locations in reference frame 1. In technical language, the events must have a "spacelike" separation between them.

(Moreover, we assume that the events are true 'pointlike' events, i.e., that they have zero duration. If not, then the spatial separation between the two, in reference frame 1, must be longer than $c$ times their duration, in order for this property to hold.)

Your example (scrambling an egg) necessarily happens in the same location in a given reference frame (that of the egg) with a nonzero time between the start and the end, which means that it has a "timelike" separation between events. That means that those two events will never appear simultaneous, or in reversed order, in any other reference frame.

In other words,

does entropy, $S$, "Time's Arrow," trump relativity?

No.

Emilio Pisanty
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  • suppose in reference frame 2 you move the oven and stove during the frying process. Will your reasoning still hold? – duhem7 Feb 26 '20 at 00:13
  • Yes. If there is a material particle whose trajectory goes through the two events, then their separation is timelike, and it cannot be spacelike. Relativistic frame transformations cannot alter the spacelike or timelike character of the separation between events. – Emilio Pisanty Feb 26 '20 at 00:15
  • Hmmm. I don't understand that. I'd be grateful if you refer me to some website or reference that gives exact definitions of spacelike and timelike character of events. – duhem7 Feb 26 '20 at 00:26
  • The separation between two events is spacelike if there is a reference frame where the two are simultaneous (and at a nonzero distance); it is timelike if there is a reference frame where the two happen in the same spot, at different times (i.e. if a particle in uniform motion can pass through both events). SR tells us that the two categories are incompatible (i.e. it can't be spacelike if it's timelike, and vice versa), and that simultaneity is only relative for spacelike-separated pairs of events. – Emilio Pisanty Feb 26 '20 at 00:37
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Consider two events A and B which are simultaneous in reference frame 1;

These events must be "spacelike". They are separated in space by a sufficient degree that light cannot start from one event and reach the other. Yes we can find reference frames where one event precedes the other, but in none of the frames can light start at one event and reach the other.

Suppose in reference 2 we might consider scrambling an egg: event A, egg is dropped into the frying pan; event B, egg is cooked.

These events are not spacelike, but "timelike". It is possible for light (or in some cases, slower objects) to travel from event A (the time and place where the egg is scrambled) to event B (the time and place where the egg is cooked). Because these events are close enough to each other in space, but not time, there are no reference frames in which they are simultaneous. Another way of saying this is that the timewise ordering of the events is consistent in all reference frames (A always precedes B).

The ambiguous ordering of events is only true for certain sets of events, not all. If you take a look at a light cone diagram, you will see that for a cone centered at the breaking of the egg, the event of scrambling that egg lies in that event's future. Only events in the "elsewhere" outside of the cone can be considered simultaneous in some reference frame.

Changing reference frames can shift events around inside the cone and outside the cone, but can never move an event across the cone's boundary. Future events remain in the future in all frames.

BowlOfRed
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I don't think this is possible. Just because simultaneity is no longer universal in special relativity does not mean that you can find a frame in which causality is violated.

Your situation is a bit tricky, because in order for the pan to heat the eggs, it needs to be (roughly) at the same position. But if two events take place at the same place and time in a reference frame, they will also take place at the same place and time in any reference frame.

In order to see the situation a bit more clearly, imagine now that $A$ is an atomic bomb exploding at $t = 0$ and $x= 0$, and suppose that the radiation from the explosion (propagating at $c$, i.e the speed of light) is enough to cook instantly an egg at $x = L$. In the reference frame $R$ where both are at rest, this event takes place at a time $t=L/c$ (more generally $t=L/v$ where $v \leq c$), which is the time needed for the energy to propagate from the explosion site to the egg.

We say that the separation between $A \, (t=0, x=0)$ and $B \, (t=L/v, x=L)$ is timelike (or lightlike if $v=c$) in $R$, because one event, $A$, can be the cause of the other, $B$.

You can show that in special relativity, a timelike (resp. lightlike) interval remains timelike (resp. lightlike) in any frame of reference, with the order of causality unchanged.

If we boost to a reference frame $R'$ moving at $V$ with respect to $R$, $A$ will be unchanged in the new coordinates $(t'=0, x'=0)$, while $B$ can be written as $B(t' = \gamma (t - Vx/c^2), x' = \gamma(x - Vt))= B(t' = \gamma L/v(1-vV/c^2) > 0, x' = \gamma L (1 - V/v))$. You can see immediately that $t' > 0$ (meaning that $B$ happens after $A$ in any reference frame), and, with a bit more calculations, that $c t' \geq x'$ (meaning that the signal of the explosion is able to propagate from $A$ to $B$ in the time $t'$ between the two events.

QuantumApple
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  • I see... vV/c^2 is always < 1 and if V<0 then it's still the case that t'>0. – duhem7 Feb 26 '20 at 02:29
  • Yes, of course you have to be careful about the sign of $V$ and showing that $c t' \geq x'$ is not obvious but it can be shown by comparing carefully the two expressions. – QuantumApple Feb 26 '20 at 02:32
  • If you want a bit more formalism, you can show that for any Lorentz transform, the quantity $ds^2 \equiv \Delta x^2 - c^2 \Delta t^2$ between two events $A$ and $B$ remains unchanged. You can easily convince yourself that if $ds^2 > 0$, then the distance between $A$ and $B$ is too large for any signal coming from $A$ to have an influence on $B$ or reciprocally (the signal would need to travel faster than light). In that case, you can show that there exists a frame of reference in which $t'_A = t'_B$, as well as frames where $t'_A > t'_B$ or $t'_B> t'_A$. This is called a spacelike interval. – QuantumApple Feb 26 '20 at 02:47
  • If $ds^2<0$, and $t_A<t_B$, it is possible for a signal from $A$ to influence $B$ (said signal does not need to travel faster than light). In this case, you will always have $t'_A<t'_B$ in any frame of reference, thus preserving causality. This is called a timelike interval. Finally, in between, when $ds^2=0$, the two intervals are separated by a distance $\Delta_x$ and a time $\Delta_t$ such that it would take a photon exactly $\Delta_t$ to propagate from $A$ to $B$. This is called lightlike interval. The nature of these intervals is preserved through any change of reference frame. – QuantumApple Feb 26 '20 at 02:56