1
$\begingroup$

In the book Explorations in Mathematical Physics (p. 262) by Don Koks, he discusses a traveler who has constant proper acceleration and therefore a Rindler horizon behind her, about which he says:

On this mysterious plane, she observes that time itself has stopped, and although she cannot see what lies beyond, she knows from our previous discussion that behind the horizon time is flowing in reverse!

This makes sense to me given that an object that is stationary in her frame has a rate of time in her frame (in Rindler coordinates) given by $\frac{d\tau}{dt} = \alpha x$ where $\alpha$ is her acceleration. Since the horizon is at $x = 0$, the rate of time is $0$ there. And since $x$ is negative on the other side of the horizon, the rate of time is negative there (although technically, the Rindler coordinate system is only defined for $x \gt 0$).

Now I realize that she won't actually see anything move backwards through time, and the same is true of time dilation—it can't be directly observed in real time; it can only be inferred after the fact. But here's the main thing I'm wondering about this reversal of time in her frame, which I'll call "time inversion" (rather than "time reversal" to avoid confusing it with the unrelated topic of time reversal symmetry/invariance): Is time inversion every bit as physically real as time dilation? If so, why do we only ever hear that time's speed is relative and not that its direction is relative as well?

If your answer is that it's not as physically real (e.g., that it's the result of coordinate changes), please read on so that I can explain why it seems to me that it must be, and please refer to this spacetime diagram:

As you've probably noticed, this basically depicts the twin paradox with an additional stationary twin/triplet (A) who is just watching the action from afar—3 light years away in fact. C travels away from B at $v = 0.6c$ $(\gamma = \frac{1}{\sqrt{1-0.6^2}} = 1.25)$ until reaching a destination $3$ ly away before returning. So he travels for a total of $\frac{2 \times 3}{0.6} = 10$ years of Earth time and $\frac{10}{1.25} = 8$ years of proper time. The blue and red lines are of course the planes of simultaneity. The blue and red numbers show C's proper time $(\tau)$. The red numbers are times when he's at rest relative to A and B, and the blue numbers are times when he's moving.

Using these numbers, you can track C's accounting of A's time, and you can see that, according to it, A's time starts at $t = 0$, then quickly drops well below $0\ ($to $-1.8)$ during C's acceleration—i.e., A's time seems to move backward—and it stays in that past for more than $2$ years of C's time while it moves forward at $80\%$ of the proper rate for the first half of the journey. Then, at the turnaround, it leaps forward by several $(7.2)$ years, and then resumes its slow advance until reaching a value well above $10\ (11.8)$. It then quickly falls backward again during C's deceleration to settle at $t = 10$. Note that for simplicity, we're assuming that all acceleration is effectively infinite (hence the backward time jumps and the forward time jump at the turnaround are instantaneous).

Here's my main argument: We know that 1) C's journey is physically $2$ years shorter in his frame than it is in A and B's frame. And we know that 2) during the inertial legs of that journey, A's time and B's time physically run at $80\%$ of the speed of C's time in C's frame. Because of these two things, we also know that 3) B's time has to physically leap forward $3.6$ years (since $10 - 80\% \times 8 = 3.6$) in C's frame during the turnaround (which can be explained either in terms of rotating planes of simultaneity or the equivalence principle of general relativity and gravitational time dilation). And the same things that require that to happen make it so that 4) A's time must physically leap forward $7.2$ years in C's frame during that same turnaround (twice as many years as B's time since A is twice as far away as B (see this page from Michael Weiss and John Baez on "distance dependence"); equivalently, A is half as deep in the apparent uniform gravitational field that we can say causes A and B to "fall" towards C). And in order for this to all be true, 5) A's time must physically move backwards by a total of $3.6$ years in C's frame at some point(s) during the journey (since $80\% \times 8 + 7.2 - 10 = 3.6$), as it seems to do. These five statements are all interdependent, so if any one of them isn't valid (i.e., the effect it describes is not physically real), the others—including (1)—are all called into question until at least one other of them is modified to compensate for it. So is (5) valid? If not, which of the other statements need to be modified to preserve (1)?

The Lorentz transformations show these backward time jumps as well. E.g., the first one happens at $t' = 0$ when $v = 0.6c$, $x = -3$, and $x' = \frac{x}{\gamma} = \frac{-3}{1.25} = -2.4$ (per the length contraction formula—if there's a better way to do this, let me know), so

\begin{align} t & = \gamma \left(t' + \frac{vx'}{c^2}\right) = 1.25 \left(0 + 0.6(-2.4)\right) \\ & = -1.8. \end{align}

(Note that since all of our units are light years (ly), years (y), and light years per year (ly/y) and since $c = 1$ ly/y, we can omit units and $c$ from our calculations, as I've done.)

So while C speeds up, he goes from knowing that A's $t = 0$ is happening currently (i.e., simultaneously with his present time) to knowing that A's $t = -1.8$ is happening currently. In other words, he believes that A's time rewinds $1.8$ years, and in his reference frame, he's correct, is he not?

While I haven't shown that these backward time jumps are due to A being behind the Rindler horizon of C, it seems safe to say that the same effect is at play. Since C's acceleration is effectively infinite, the horizon's distance from him is $\frac{c^2}{\alpha} = \frac{c^2}{\infty} = 0$, so virtually everything in the other direction is behind it.

So should we say that the relativity of time applies not just to the speed of time but to its direction as well? If time dilation is the result of one observer rotating in spacetime relative to another, causing their time to advance in a somewhat different direction, could the combination of acceleration and distance (i.e., "gravitational" potential) allow that rotation to simply continue until it has a component in the opposite direction?

Notes

This all raises questions about whether an analogous effect might exist beyond the event horizons of black holes and the cosmological horizon of the universe but those are probably topics for another day. For now, I'll just say that seeing a number of answers on this site that claim that time beyond a black hole's event horizon does not run backwards relative to ours is part of what sparked my interest in this topic (and interestingly, Koks doesn't touch this question in his chapter on black holes and GR even though he discusses horizons extensively in it).

Before writing this question, I found two similar ones (here and here) but I don't think my questions and points are addressed in either of them. I've also done some pretty deep searching elsewhere and, to my surprise, have found very little on this topic, so if you happen to know of any more resources on it, please share. All I've found so far, aside from Koks's book, is a mention in a class handout called "Space-Time Diagrams, Relativity of Simultaneity, and Relative Time Reversal" that "relative time reversal can only occur between events that cannot communicate by any physically realistic signal." While I'm sure this is true, the statement seems intended to mean that said reversal would require superluminal signals or travel and is therefore impossible. Apparently, it hadn't occurred to the author that an event horizon might fulfill the criteria.

Thanks and sorry for the long question!

$\endgroup$
2

2 Answers 2

2
$\begingroup$

Is time inversion every bit as physically real as time dilation? …

If your answer is that it's not as physically real (e.g., that it's the result of coordinate changes)

Time inversion is every bit as physically real as time dilation. They are both the result of coordinate changes. Time dilation is the ratio of the change in coordinate time and the change of proper time. As such, it is no more “physically real” than the coordinates themselves.

What is indisputably physically real is the comparison of clocks that are co-located. The twin paradox can be resolved in a frame invariant way with that alone. As such, the question you have asked is inextricably tied to coordinates.

So, the first thing is to understand is what coordinates are and what restrictions they have. Coordinates are a mapping between an open set of events in spacetime and an open set of points in R4. This map has only two requirements. It must be smooth and one-to-one. This issue is described in detail in the first two chapters of Carroll’s Lecture Notes: https://arxiv.org/abs/gr-qc/9712019

Explicitly, there is no requirement that any of the coordinates need to map to time and if there is such a coordinate there is no requirement that the “future” must coincide with the direction of increasing time coordinate.

In fact, there is no need to use the Rindler transform. We can simply make the transform $t \rightarrow -t’$ This is a perfectly valid transform where time flows backwards.

Now, there are some important points to keep in mind. With respect to your idea that time flows backwards and forwards during the twin’s scenario, that cannot happen. As you described it, the coordinate system was not one to one since it mapped different coordinates to the same event. That is not a valid coordinate system, and if you try to split it into several systems then it can no longer be said to represent “C’s frame” or “C’s time” in any meaningful way. This issue is further discussed by Dolby and Gull: https://arxiv.org/abs/gr-qc/0104077

In Rindler coordinates the $x=0$ has a similar issue. The coordinates are not one-to-one there. That means that Koks’ statement is incorrect, the $x=0$ plane is not part of the Rindler coordinate chart and No claims about time on that plane can be made based on it.

So the negative time region is a disjoint open set from the positive time region. This is a general fact. You cannot smoothly transform a timelike coordinate from negative to positive. There will always be a boundary where either the coordinate stops being timelike or where the coordinate system becomes non-one-to-one. So the argument you made regarding the twins paradox unfortunately does not work since the coordinates described are invalid. Your statement 1 is correct but all of the subsequent statements are wrong both because they assert that coordinate-based statements are physical and because the coordinate system used to make those statements is not valid.

A correct resolution that is purely physical is that the invariant proper time on C’s clock is given by $\tau = \int_C \sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$ This corresponds to your first point because this holds in all reference frames, and nothing further is either needed or “physical”.

why do we only ever hear that time's speed is relative and not that its direction is relative as well?

Most likely because introductory courses are so focused on inertial frames. In inertial frames there is only time dilation, not time reversal.

$\endgroup$
4
  • $\begingroup$ "That is not a valid coordinate system." I'm not saying we should only use a single coordinate system for C's frame throughout his entire journey. I figure that each period of inertial motion and each period of acceleration needs its own coordinate system, each of which should be valid. And in two of those systems, A's time only runs backwards. Does that resolve the issue? "In Rindler coordinates the $x = 0$ has a similar issue." Do you think that Don Koks's point about time being frozen on the horizon and reversed beyond it (which he reiterates throughout the chapter) is wrong then? $\endgroup$ Apr 28, 2022 at 8:18
  • $\begingroup$ "the argument you made regarding the twins paradox unfortunately does not work" Can we zoom in on "my main argument" then and pin point exactly where it goes wrong (this is why I broke it into numbered statements)? Note that I've rewritten that paragraph (mostly at the end) to make the argument more explicit. $\endgroup$ Apr 28, 2022 at 8:18
  • $\begingroup$ @GumbyTheGreen I have made edits to the answer to address your follow up. However, I do think you would be better served at a discussion site like physicsforums.com than at a Q&A site like here. There is nothing wrong with your desire for follow up, but it just doesn’t fit well with the Q&A format of SE $\endgroup$
    – Dale
    Apr 28, 2022 at 11:58
  • 1
    $\begingroup$ Thanks for the additions. Ok, I've made a post there. I've upvoted your answer but I'll hold off on accepting it until I get some clarity. I need to understand how there could possibly be a sensible way for C to make an accounting of A's time that differs from the one I've given... $\endgroup$ Apr 29, 2022 at 5:44
1
$\begingroup$

When you say that A's time must physically move backwards at the start of C's journey, you are either wrong or you are using the word 'physically' in some narrow sense. Certainly for A, the physical reality is that no such jumping occurs. If I were A sitting three light years away from B and C at the start, with my clock synchronised with theirs, and if I were to start drinking coffee when C suddenly accelerates to 0.6c, my physical reality is that I continue drinking my coffee and living my life as before- time for me does not jump backwards. Certainly from C's perspective, having just started their journey, the event of my drinking my coffee has a time coordinate that is 3.6 years in the future. But what does that mean? The event is outside of C's light cone, and in an important sense the words 'past' and 'future' lose their meaning when talking about spacetime outside of an observers forward or backward light cone.

Specifically, you will get into all kinds of logical difficulties if you assume that the consequence of C's acceleration is that my coffee drinking has not actually happened in C's frame owing to the fact that it is 3.6 years into C's future. To see this, imagine a quadruplet, D, who is physically present with A at the outset, and that D sets off on a journey which is the mirror image of C's (ie D accelerates instantly to 0.6c in the opposite direction to C). If you assume that my drinking coffee hasn't 'physically' happened in C's frame at the start of C's journey, then you must also conclude that D has not yet taken off either. However, the arrangement is entirely symmetric, so from D's perspective it is C who has 'physically' not taken off.

The point is that events in spacetime are physical. You cannot make an event unhappen by switching to another reference frame in which that event is has a future t coordinate.

$\endgroup$
2
  • $\begingroup$ By "physical", I think I just mean that it's not conventional—it's true in a given frame regardless of coordinate system choices, etc. It can still be relative, i.e., frame dependent. So C and D can disagree on who takes off first (especially since they're behind each other's Rindler horizons)—the exact same kind of thing happens with reciprocal time dilation, so I don't see any logical difficulties in your thought experiment. No particular event is being made to unhappen since no information about that event can exist in that frame yet. Btw, I think you mean 1.8 years, not 3.6. $\endgroup$ Apr 30, 2022 at 4:39
  • 1
    $\begingroup$ I think your question is a very interesting one, but I don't think it is quite analogous to reciprocal time dilation- I think it is more fundamental than that. I have thought of some related questions, which I might try to formulate into a question on SE if I get the time. And yes, I should have said 1.8 years, so apologies for that. $\endgroup$ Apr 30, 2022 at 5:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.