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I've been reading a lot about the twin paradox recently. I understand that there must be some kind of asymmetry between the twins' experiences in order for them to measure different amounts of time, and the asymmetry comes from the fact that one twin accelerates while turning around and heading back towards Earth. However, this is only an asymmetry if the twin in the rocket can't just as easily say that they're stationary and the twin on Earth is accelerating. As I understand it, the rocket twin can't do this because they are experiencing proper acceleration - something they can measure - whereas the twin on Earth would be undergoing only coordinate acceleration, not proper acceleration, from the travelling twin's frame. Another way of putting it is that the twin on Earth is in an inertial frame, while the other twin is in a non-inertial frame, so they can't use the same rules to calculate time dilation for the earthbound twin.

I have one issue with this though. What if the travelling twin's rocket and everything in it were accelerated by a force that acted equally on all points within every object. There would be no relative acceleration between parts of the ship, so the twin couldn't detect any acceleration, right? For instance, if they let go of a ball in the ship, it wouldn't move towards the back wall because the force would be accelerating all points within it along with the ship. If there is no experiment an observer can do to tell whether they're stationary or if all points in their body, and in everything at rest relative to them, are being accelerated by the same force, isn't proper acceleration meaningless? The rocket twin could now claim to be stationary and say the other twin was accelerating in the exact same way, with a force acting on every point in that twin's body. The situation between the two twins would seem to be totally symmetrical. How can this issue be resolved?

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closed as off-topic by WillO, ZeroTheHero, Jon Custer, Kyle Kanos, AccidentalFourierTransform Oct 3 '18 at 2:54

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  • $\begingroup$ Alice travels from Chicago to Detroit in a straight line. Her twin Bob travels from Chicago to Detroit along a curvy path. Their odometers record different distances for the trip. I understand this must be because Bob's path is not straight. But what if we blindfold him and apply all sorts of forces so that he never feels his car turning around or going around any curves? Does this make the situation between Alice and Bob "totally symmetrical"? Should I now expect Bob's and Alice's odometers to agree? $\endgroup$ – WillO Oct 1 '18 at 1:55
  • $\begingroup$ I don't really understand how you'd tell which path, if any, was straight in the twin paradox. If the rocket is accelerated away from Earth in the way I described, and then accelerated back, this would seem like the normal twin paradox to the twin on Earth, so they should expect to be older. However, how would they know that the Earth and spaceship weren't both being accelerated at equal rates (half the rate they observe for the ship) at all points simultaneously, so neither measured any proper acceleration? If both accelerated equally, they should be the same age. $\endgroup$ – CyborgOctopus Oct 1 '18 at 2:11
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    $\begingroup$ Welcome New contributor CyborgOctopus! Do you have a mathematical model for this conjectured force that acts (I presume) only on the rocket and all within? $\endgroup$ – Alfred Centauri Oct 1 '18 at 2:26
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    $\begingroup$ I'm saying that unless you have a mathematical model, your description is hardly different from stipulating that 'magic' is the cause of the twin's undetectable acceleration. $\endgroup$ – Alfred Centauri Oct 1 '18 at 2:35
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    $\begingroup$ This discussion is pointless, because acceleration has nothing to do with the twin paradox and cannot account for the time difference. $\endgroup$ – safesphere Oct 1 '18 at 6:11
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In special relativity, time and velocity are not absolute, but acceleration is. So you cannot say that the traveling twin is at rest, and the Earth accelerates instead. Proper acceleration, $\vec a$, is defined such that after a time $\text d t$, the object is moving with a velocity $\vec a \; \text d t$ relative to the frame it was in before the time $\text d t$. Even if your proposed force exists, it doesn't change the fact that the traveling twin is not in an inertial frame because objects in an inertial frame do not accelerate when there is zero net force on them.

The force that you propose can be modeled as geodesics in general relativity, but objects moving along such paths only undergo inertial motion. They are still in a non-inertial frame.

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  • $\begingroup$ Are you sure a free fall is not an inertial frame? See this: physics.stackexchange.com/questions/265913/… $\endgroup$ – safesphere Oct 1 '18 at 6:06
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    $\begingroup$ @safesphere: It isn't. Only on a small region of spacetime where curvature is insignificant, it can be treated locally as an inertial frame. $\endgroup$ – user7777777 Oct 1 '18 at 6:41
  • $\begingroup$ Per David McKee's answer, "one falling freely [...] is an inertial frame". $\endgroup$ – safesphere Oct 1 '18 at 6:48
  • $\begingroup$ @safesphere: Plenty of sources suggest otherwise. Inertial frames can only be defined locally. Gravity is a fictitious force, and inertial frames cannot have such forces. $\endgroup$ – user7777777 Oct 1 '18 at 6:50
  • $\begingroup$ @safesphere: I'm also quite sure that he meant to say "only locally" in his answer. $\endgroup$ – user7777777 Oct 1 '18 at 6:57
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I would like to rule out all this talking about accelerations, because it is not the real issue. Sure, it is necessary that one of the twins accelerates, but the difference in proper time when they re-encounter is not due to acceleration.

Let us be definite. Assume twin A rests in a spaceship with its motors turned off, far away from all sources of gravitational field (it's better than letting him stay on Earth, which has a gravitational field, is moving around the sun, is accelerating, and so on). Twin B's spaceship stays initially aside A's. At time 0 of its proper time B fires its rear motors, accelerating the spaceship at an acceleration $g$ (as measured by his accelerometer). After a time of - say - one hour he turns off his motors and rides for a whole month. After that, he fires his front motors for the same acceleration $g$, and keeps them burning for 2 hours. (This will invert his speed wrt A.) After one month B will arrive not far from A, will fire the rear motors for one hour, and will find himself just aside A, resting. At this point A's and B's clocks are compared, and found as expected: B's is behind - I leave for you to compute how much.

Now the experiment is repeated according the same program, apart for a variation: the time of unaccelerated journey is doubled, two months instead of one. I want to emphasize that in both experiments B's acceleration phases were the same: $g$ for one hour at start, $-g$ for two hours at mid-journey, $g$ for one hour at the end.

Do you think the clocks' differences will be the same or not? What do you believe SR says?

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  • $\begingroup$ +1. I was about to write something along these lines as a comment, but you said it much better in your answer! $\endgroup$ – D. Halsey Oct 1 '18 at 21:55
  • $\begingroup$ I would say the difference in the clocks will be the same, as when not accelerating there is no way to distinguish any of the two reference frames associated to the twins. $\endgroup$ – thermomagnetic condensed boson Oct 2 '18 at 14:28
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There is a force which has exactly the characteristics you describe: it accelerates everything equally so that no (local) measurement can be done to detect the acceleration.

That 'force' is gravity, and what you are describing is one of the twins moving in a gravitational field.

This motion can no longer be explained by special relativity: you need general relativity. But the explanation is conceptually quite simple.

In SR, if the twins are to meet more than once at least one of them must travel along a path which is not a straight line (this follows immediately from the parallel postulate in a flat space), and it is this non-straightness which accounts both for the acceleration measured by one (or both) of the twins and for the difference in the lengths of the paths between the two points at which the twins meet. As in conventional euclidean geometry straight lines are extrema of length, but unlike in euclidean geometry they are the maxima of length (between timelike-separated events). And length, of course, is proper time along the curve. Also as in euclidean geometry there is exactly one straight line between any two events (assuming no global topological oddness). So the twin who travels on this unique straight line between the two meetings experiences no acceleration and always experiences more proper time (so, is older than) a twin who follows any other curve at all. A twin who follows any other curve always experiences acceleration, which they can measure locally.

In GR everything is the same: straight lines (which are now called geodesics) are the maxima of proper time between timelike-separated events, and are also the curves on which no acceleration is felt. But because spacetime is no longer flat the parallel postulate is no longer true: two distinct geodesics can intersect more than once and, equivalently, there can be more than one geodesic connecting a pair of events. And these distinct geodesics can have different lengths: they are now only local maxima of proper time between the events.

So what you are describing, according to GR, is that the twins are now living in a gravitational field and, equivalently, in a non-flat spacetime. To know the actual lengths of the two geodesics -- the proper time experienced by each twin -- you would now have to crank up the field equations of GR and everything would become much more complicated. But in general they would not experience the same proper time (there are symmetrical cases where they would, of course).


So, what I've described above is how one 'force' which has the characteristics you describe in fact turns out (according to GR) to be not a force at all, but a manifestation of geometry. Could there be another such force?

Well, there are two options.

  • That 'force' is actually geometry as well. This would mean that the field equations of GR are wring in detail but some modified version of them is true. There have been plenty of such theories: I think from the point of view of your question they all amount to the same thing, which is that 'forces' like this are actually geometry.

  • There is some other such force which works like this even in flat spacetime. I hesitate to say such a theory could not work but it would be at least extremely problematic. To see why think about inertial frames: SR says that inertial frames exist and talks about the relationships between them. In a theory with this additional force there would be a whole other class of frames which were locally indistinguishable from inertial frames but were not actually inertial. That's eating away at the foundations of SR (and newtonian physics, too) in a bad way. Perhaps someone more up on this could say whether such a theory is even possible while being compatible with SR: I suspect not but I am not sure.

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  • $\begingroup$ In flat space time, what about distinguishing acceleration by this force from inertial frames by shining a light beam? If this force was accelerating everything else equally, it still couldn’t accelerate light, which is already going at the speed of light after all, so a light beam would bend. If this is true, would it resolve the issue? $\endgroup$ – CyborgOctopus Oct 1 '18 at 12:37
  • $\begingroup$ gravity accelerates light just fine. I recommend mastering the twin paradox without the acceleration-crutch before diving into GR. It's worth the effort. $\endgroup$ – JEB Oct 1 '18 at 17:40
  • $\begingroup$ Alright, I think I mostly understand now after looking into it more. The situation can be thought of geometrically. The non-accelerating twin is taking a straight path through spacetime, but the other one is taking a curved path. In spacetime geometry, the straight path is longer. Obviously, the paths are not the same length and the other twin can't just say "actually, my line is straight, so yours is shorter." This seems to make sense but then why would the force I'm talking about not be consistent with SR? $\endgroup$ – CyborgOctopus Oct 1 '18 at 23:32
  • $\begingroup$ @CyborgOctopus: You've understood the curve stuff: the twin paradox is all about geometry & the length of curves. I'm not sure that your proposed force is incompatible with SR but it makes me pretty queasy. $\endgroup$ – tfb Oct 3 '18 at 7:39
  • $\begingroup$ tfb, I realized the force would be a problem for Newtonian physics as well like you said. I don't think it causes paradoxes in either necessarily, it just might make it impossible to tell what a straight line is anymore. You would never know if you were going straight or if the force was manipulating things in some way. $\endgroup$ – CyborgOctopus Oct 4 '18 at 1:33
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First, you are right, speed is symmetrically relative, so as long as the twins only move with constant speeds, you will have a paradox (SR).

You need GR (time dilation).

Both twins move in the time dimension with speed c (the magnitude of the four vector). This will not change until one of them accelerates (the turning point of the spaceship).

But acceleration is absolute (GR). If the twin in the spaceship turns around, its spaceship accelerates and (just like in a gravitational field because of the equivalence principle) its speed in the time dimension changes. It will slow down in the time dimension. this is because the universe is set up so, that the magnitude of the four vector has to stay c. If the spaceship's speed in the spatial dimensions changes, its speed in the time dimension will have to compensate, its speed in the time dimension will slow down.

Now this is when the twin in the spaceship will move in the time dimension slower then the twin on earth. There will be a gap between them from now on in the time dimension, which will only be visible to them, when they meet in space again and compare each other and realize one is younger (the one from thespaceship).

This gap will not close ever, if the spaceship moves on the way back to earth with a constant speed again.

The point with this acceleration of the spaceship is not that it just accelerates at the point of return. The point is that the acceleration (because of the equivalence principle), creates an effect (same as gravity) that is stronger then earth's gravitational effects. This difference in gravitational potential (source of GR time dilation) is what causes the spaceship to move slower in the time dimension then the twin on earth at the point if return.

Otherwise there is a negligible effect that the spaceship is in outer space, where there is even less gravitational potential then on earth, so the twin on the spaceship should age more. But this is negligible, and we do not take it into account.

So it is the difference in gravitational potential that counts. The acceleration of the spaceship at the point of return creates a gravitational potential greater then that of earth's.

Now you ask how you can measure that. this acceleration of the spaceship is absolute. The way to measure that, is to release beacons from the spaceship at the same time rate. So release a beacon, and then another one a second later, again a second later. Then you have to use radio signals to measure the distance between the beacons. You will realize, that the distance between following beacons is increasing (acceleration) or is decreasing (deceleration). This is absolute, and the twin on earth using radio signals will measure the same for the spaceship. Both twins will agree that the spaceship is accelerating.

But again, you do not need a spaceship to understand GR time dilation. You do not need a spaceship to turn around and accelerate. Just put a twin close to the Sun (with extra suncream), and a twin on earh. Let's use skype or facetime. They will literally see on the screen as the one near the Sun will age slowly, and the one on earth will age faster. If they meet again one day, they will see the same thing, the one on earth aged much more then the one near the Sun. Why? Because the Sun has stronger gravitational effects, and more gravitational potential, so clocks tick at the Sun slower compared to clocks on earth (Shapiro effect). The earth has less gravitational potential, so clocks there will tick faster compared to clocks at the Sun. Same thing with chemistry. Our body is full of atomic clocks, the twin near the Sun will tick (age) slowe.

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and the asymmetry comes from the fact that one twin accelerates while turning around and heading back towards Earth.

The fundamental asymmetry is the fact that all inertial observers agree that one twin remains in one inertial frame of reference (IFR) while the other twin does not.

First, and to be sure, the context here is the flat spacetime of special relativity (SR) and I set $c = 1$.

There are just two events to consider:

  • Event A - the two twins are initially co-located
  • Event B - the two twins are finally co-located

Associated with these two events is a time-like interval $\Delta s^2$. According to SR, there is just one world line with an elapsed time between A & B equal to $\sqrt{\Delta s^2}$. This world line is the geodesic through the two events.

In the twin 'paradox', the world line of the "stay at home" twin is the geodesic through the events A & B. All inertial observers agree that the (ordinary) velocity of the "stay at home" twin is constant between events A & B. That is, there is no change of reference frame.

Since the other twin does not have the same world line as the "stay at home" twin, it follows that the elapsed time along the world line of the other twin is less and further, that all inertial observers agree that the velocity of the other twin is not constant between events A & B; all inertial observers agree that the outgoing inertial reference frame is different from the incoming inertial reference frame.

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