This started with wondering about the nature of certain physical quantities under time-reversal - chiefly, that acceleration retains its magnitude and direction at a given time regardless of the 'direction of time' used; this is different to say momentum, which may be reversed under time-reversal (in the sense that we are integrating forward in time from our perspective, along the backward direction in time).

There may be already agreed-upon conventions I'm not aware of regarding which quantities would be reversed, but the essence of the idea is that some quantities or some combinations of quantities need to be reversed, while others (such as acceleration) don't.

With respect to the twin paradox, this got me wondering if it works under non-trivial time reversal regarding the acceleration periods of the paradox. That is, suppose it isn't enough to simply reverse the order of events; if we assume everything runs in reverse, could we start off with two people of different ages, and due to relative reverse motion cause one twin to 'catch up' in 'getting younger' such that at some future (ie their past) they are of the same age?

-PS

To put the question more simply: Given that acceleration retains direction going forward and backward in time while velocity (or momentum) may not, do the relativistic effects remain the same (and do they make physical sense) from the perspective of a forward-time observer watching events unfold in reverse?

More generally: By virtue of acceleration having the same magnitude in both directions of time, does it mean it is impossible to discern a direction of time from any measurable acceleration alone? (further questions tempt relevance)

## closed as unclear what you're asking by ACuriousMind♦, honeste_vivere, user36790, CuriousOne, GertJul 10 '16 at 1:52

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• It's not a matter of conventions but simple math. If time appears to first order in a quantity (as in $dx/dt$), then the quantity reverses its sign under time-reversal. If time appears in second order as in ($d^2x/dt^2), then there is no sign reversal. You don't have to reverse time to get people to catch up to the same age. And old man takes off for a relativistic journey and comes back as an old man eighty years from now. A baby is born now and is an old man the same eighty years from now, when the old man returns. – CuriousOne Dec 28 '14 at 23:42 • Traversing world lines in reverse is no more physical than running a movie backward. The positron as an electron running backward in time is as bad a physical proposition today as it was seventy years ago. See, if I set the positrons in my accelerator free, will they be able to go back in time to tell my dead grandma that I still love her? If not, what meaning does it have that the positron is supposedly an electron running backward in time? A positron is simply the same quantum state as the electron but with charge reversed. It's a difference of one quantum number. – CuriousOne Dec 29 '14 at 1:16 • Time reversal invariance is simply a local symmetry. One can not extend it to a global one. Time does not run backwards, not even for a femtosecond. One can prepare systems into forward running states that look like backward running ones, but that's just for looks. – CuriousOne Dec 29 '14 at 1:36 • If time ran backwards, my grandma wouldn't be dead, anymore, and I would be five years old. That's not how time works. Reversing sign on a variable called$t$on paper does just that: you add a minus sign on a piece of paper. The universe doesn't care what we do in our descriptions of it. Feynman diagrams are a naive attempt at perturbation theory that comes with a number of very nasty mathematical problems. Who said that? Feynman himself. Personally I wouldn't make him responsible for scores of people misunderstanding what he gave them. – CuriousOne Dec 29 '14 at 3:05 • I'm not sure I understand the question but I think you are asking this: If I start with a spacetime$M$, and then construct a new spacetime$M'$with the same topology and the same metric but the opposite time orientation, does the twin paradox work the same way in$M'$as it does in$M$? If that's the question, then the answer is obviously yes, because you could just as easily have started with$M'$, i.e. the general twin paradox story works as it does in an arbitrary spacetime. – WillO Oct 21 '15 at 1:36 ## 1 Answer You can have a situation where two people start out next to each other with different ages, then move apart and reunite later so that they are the same age when they meet, with the one that started out younger moving inertially between meetings and the one that started out older accelerating during the journey. If they carried clocks that showed, say "100 years minus my current age", then if you played the movie of this backwards, you would have a situation where their clocks initially show the same reading, then both clocks tick "forward" as the movie continues to play backwards, with the inertial twin's clock showing a greater reading when the twins are again next to each other at the end of the time-reversed movie. So just looking at their clocks, this would look just like a movie of the standard twin paradox scenario, and if the movie showed the rate each clock was ticking in some inertial frame at each point in the journey, this rate would be$\sqrt{1 - v^2/c^2}\$ slower than a clock at rest in this inertial frame, just like in the standard twin paradox. If the movie didn't show any thermodynamically irreversible processes (or any of the rare particle interactions that violate T-symmetry), there would be no way of knowing you were watching a movie being played backwards.

• So in general any situation where time (age) is gained / lost is a symmetrical rate of change (with respect to time and the direction of time), aside 'thermodynamically irreversible' processes (ignoring the T-symmetry-violating interactions for simplicity)? If that's so, then it begs the question - is the twin paradox a thermodynamically reversible process in general or are there situations where it isn't? I would think it is reversible in all cases though not sure... – Xeren Narcy Dec 29 '14 at 0:46
• As long as you're only interested in the proper time interval between two events on a given worldline (like departing from and reuniting with a twin), thermodynamics is not an issue. It's only relevant to the fact that certain processes are only likely to be observed in one direction of time but not the other, but calculating proper time intervals is a purely geometric matter, like calculating the lengths of paths on a surface. – Hypnosifl Dec 29 '14 at 1:07
• Yes, I'm interested in how the proper time interval would be calculated depending on your choice of temporal vantage point, paying attention to acceleration being a time-symmetrical quantity and any resulting implications (if any). What throws me off is that the twin paradox requires a background to define proper time - be it a reference object, or the 'relative background' of stars really far away or the CMB etc... – Xeren Narcy Dec 29 '14 at 1:27
• The proper time on a worldline is a frame-independent quantity that can be measured just by having an ideal clock moving along that worldline, no external reference is necessary. – Hypnosifl Dec 29 '14 at 3:52
• That's true, but it rests on the argument of an ideal clock being possible - I don't see how you can avoid treating an ideal clock as the background / external frame by it's definition as an 'ideal' clock. – Xeren Narcy Dec 29 '14 at 4:19