I have seen the classical twin paradox before. It uses a twin stationary on Earth and the other traveling away and back. I have seen many contradictory solutions for it, some use general relativity, others use special relativity, either way, I am still troubled by it. They always try to break the symmetry through the traveling twin's acceleration and deceleration, but never quite succeed.

So, let's do away with the classical twin paradox and let's explain a much simpler, perfectly symmetrical version of it where both twins are moving towards each other.

So imagine we have Twin A in a spaceship, and Twin B in another, a light year apart from each other at the beginning of the experiment. They both start traveling at the same speed towards each other when the first light of one twin since the beginning of the experiment reaches the other, so they effectively start moving one year into the experiment.

If I understand relativity properly:

  • From Twin A's frame of reference, he's stationary and Twin B is moving at a constant speed towards him, therefore, because of time dilation, Twin B's clock is ticking slower.
  • From Twin B's frame of reference, he's stationary and Twin A is moving at a constant speed towards him, therefore, because of time dilation, Twin A's clock is ticking slower.

Regardless of what their observations might be because of the Kepler effect and what not, time dilation dictates that a moving clock will absolutely tick slower than a stationary one. So, because Twin B is moving relative to Twin A, Twin B's clock is absolutely ticking slower than Twin A's. The same is supposed to be true the other way around. This is obviously a contradiction.

  • 6
    $\begingroup$ This doesn't fix the problem because they won't agree on their relative ages at the start of the exercise. You have to specify how their original positioning was set up to get an answer and then it becomes clear that this is the original twin paradox all over again. The relativity of simultaneity is a tyrannical master. $\endgroup$ Feb 17 '16 at 18:51
  • $\begingroup$ @dmckee is right: in order for them to agree on their relative ages at the beginning they need to be in the same place at the start. So now to make the thing symmetrical both twins need to follow accelerated paths if they are to meet again. $\endgroup$
    – user107153
    Feb 17 '16 at 19:59
  • $\begingroup$ @dmckee OK. Then let's assume that their initial positions are a light year apart, so each perceives the other to be one year younger than the other at the start of the experiment. I edited my question with this added. $\endgroup$ Feb 17 '16 at 20:10
  • 1
    $\begingroup$ @AxiomaticNexus they start a light year apart in which reference frame? And how are the twins agreeing on when the experiment starts? $\endgroup$ Feb 17 '16 at 20:24
  • 1
    $\begingroup$ @BySymmetry I think he means they are initially a light year apart in their common initial reference frame. They can agree on when the experiment starts, say by having some one go to the midpoint between them and flash a light. $\endgroup$ Feb 17 '16 at 21:03

It is easy to see what is going on if you make a minkowski diagram. At first, when the two observers are separated and at rest, they both agree that they are the same age. This is intuitive.

Next, they begin to move toward each other. This part is basically shown in the lower half of this figure

enter image description here

Above, the green and red are the frames of the two different people. The axis labeled $ct$ (or $ct'$) is the world line of the red (or green) person. The axes label x (or x') give the direction of the planes of simultaneity. When the poeple are at rest, the plane of simultaneity is horizontal, but when they begin to move, the plane of simultaneity tilts.

So we can see that as soon as the green guy, who starts out on the left, starts to move, his plane of simultaneity tilts up, and he sees the age of the red guy suddenly increase (the age being given by where the green plane of simultaneity intersects the red world line). So as soon as one person begins to move, they see the other person's age jump. But then as they continue to move, the other person ages more slowly, so that when they meet they have the same age.

You may want to read more about minkowski diagrams for this explanation to make sense. If someone has a better resource than wikipedia, feel free to edit it into the answer or leave a comment.

  • $\begingroup$ Suppose they both have a stopwatch and are standing still. A sends a signal to B and accellerates for t seconds according to his watch. When B sees the signal from A he also accellerates for t seconds with the same thrust. They now have the same speed v towards each other and will meet and compare watch readings while passing. Assuming they are x meters apart at the time (on A's clock) when A sends his signal and they both accellerate with g meters per second, what would the readings of A and B be when they meet. $\endgroup$
    – Jens
    Feb 17 '16 at 21:38
  • $\begingroup$ This is it! For OP, note this is exactly the same idea as the explanation for the usual twin paradox (i.e. the traveling twin sees the twin at home age really rapidly while turning around). The fact that so many twin paradox questions are asked, even when the explanations are always basically the same for every variation, proves that acceleration in SR is really, really unintuitive. $\endgroup$
    – knzhou
    Feb 18 '16 at 4:56
  • 1
    $\begingroup$ Also note that, since acceleration is finite, what is actually seen is rapid ageing during acceleration, not a sudden jump in age. (This is not disagreeing with the answer, just clarifying that everything is actually nice and continuous.) $\endgroup$
    – user107153
    Feb 22 '16 at 22:46
  • $\begingroup$ This is very nice, but the Minkowski diagram depicts what happens when A and B leave their common origin. $\endgroup$ Jan 10 at 14:45

The way to think about problems in relativity is to avoid reference frames and think in terms of invariants, in this case proper time. Proper time is just path length in Minkowski space. So if the twins travel equal path lengths their clocks will measure equal durations. It's not acceleration per se that produces the paradox, it's the fact that a straight path is longest in a Minkowski metric. But if you want to see what different frames look like, just draw the spacetime diagram in one frame, like this showing AC and BD as the initial twins (quadrulets?). Intially they are stationary with respect to each other so with light signals they can synchronize their clocks. The pips on these lines mark equal proper time intervals. A and B frame

And then Lorentz transform the whole thing to another frame. In this case, the reference frame of C after he leaves A and runs toward B.C's frame

Note that in C's frame (when he's running), he did not start at the same time as D even though their starts are at the same time in the initial AB frame...the relativity of simultaneity.

  • $\begingroup$ Very nice. My only quibble is that the OP already named A and B as the traveling twins. You've renamed A to C and B to D and introduced a new A to mean A's stationary twin clock and a new B to mean B's stationary twin clock. Overall, it still has the same meaning, but the renames might confuse a reader. It'd also be nice to concretely describe (e.g. - with an actual example + numbers) what C infers about D, A, and B when C and D coincide. $\endgroup$ Jan 10 at 17:30

They are both standing still at the beginning of the experiment, so they're both in the same reference frame when they're one light year apart. Aren't they? For your second question, they can start when the first light of their twin since the experiment started reaches the other, so they both start moving exactly one year into the experiment.

There are three inertial reference frames to consider

(1) The unprimed 'lab' frame in which both twins are initially at rest. Alice is located at $x_A = -0.5\:\mathrm{ly}$ and Bob is located at $x_B = 0.5\: \mathrm{ly}$. The lab's clocks, Alice's clock, and Bob's clock are synchronized.

(2) The primed frame with (positive) velocity $u'$ in the lab frame. The origin of the primed frame is located at $x_A$ when $t' = t = 0$.

(3) The double primed frame with velocity $u'' = -u'$ in the lab frame. The origin of the double primed frame is located at $x_B$ when $t'' = t = 0$.

From this setup, it is clear that, in the lab frame, the clock at the origin of the primed frame and the clock at the origin of the double primed frame run at the same rate and are synchronized. Thus, when these clocks meet at $x=0$, they will read the same time.

Now, we simply stipulate that Alice 'jumps' (instantaneously accelerates) into the primed frame when $t_A = t = t' = 0$. Likewise, Bob 'jumps' into the double primed frame when $t_B = t = t'' = 0$.

The instant before the jump, Alice and Bob agree on what events are simultaneous.

But the instant after, they don't since they are now in relatively moving reference frames and so they do not agree on what events are simultaneous.

According to the primed frame, Bob jumps into the double primed frame before Alice jumps into the primed frame. Likewise, according to the double primed frame, Alice jumps first.

Thus, according to Alice, the instant after her jump, Bob's clock is ahead of hers and Bob is located closer to $x = 0$ than her. Likewise for Bob.

So according to Alice less time elapses on Bob's clock than hers from the moment Alice jumps until they meet. Likewise for Bob.

  • $\begingroup$ For now, I am less concerned about what their observations are, and more interested in the actual, absolute effects of time dilation. If I understand time dilation properly, because Bob is moving relative to Alice, Bob's clock will absolutely tick slower that Alice's, and the same is supposed to be true the other way around. Clearly this is a contradiction. $\endgroup$ Feb 25 '16 at 18:30
  • 1
    $\begingroup$ @AxiomaticNexus Bob's clock runs slow in Alice's frame, and Alice's clock runs slow in Bob's frame. In the lab frame, they both run a little slow, but they run at the same speed as each other. Can you elaborate more on where the contradiction is? I am not able to see one. You say Bob's clock will absolutely tick slower than Alice's. I am not sure what you mean by "absolute" here. Could you please explain that more? I would say neither one "absolutely" runs slower, because the one that is slower depends on reference frame. $\endgroup$ Feb 25 '16 at 19:09
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs By 'absolute', I mean the state of the clocks when they finally join the same frame. We are both moving rapidly towards each other. What I see, is that I'm stationary and I see someone moving rapidly in my direction. When we both finally meet, because of time dilation, his clock is behind mine. Time ticked absolutely slower for him. What he sees, is that he's stationary and he sees me moving rapidly in his direction. When we both finally meet, because of time dilation, my clock is behind his. Time ticked absolutely slower for me. This is a contradiction. $\endgroup$ Feb 25 '16 at 19:37
  • $\begingroup$ @AxiomaticNexus Your description of what happens is wrong. You incorrectly assumed both clocks show the same time immediately after they start to move. What actually happens is you see the other person's clock ahead of yours, but running slow, so when you meet up you have the same time. There is no contradiction. $\endgroup$ Feb 25 '16 at 21:01
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Forget about what I see, imagine for a second that I closed my eyes until the end of the experiment. What I'm trying to say is that time is supposed to tick slower for him because he's moving relative to me. His clock should be behind mine when we finally meet. $\endgroup$ Feb 25 '16 at 21:25

The more questions I see about the twin paradox, the more convinced I am that relativity is badly explained in textbooks.

Relative movement causes two effects relating to time. One is that time seems to run more slowly in a moving reference frame when viewed from a stationary one-that's the effect called time dilation. The other is that clocks that are spatially separated in a moving reference frame seem out of sync with each other when viewed from a stationary frame.

Most of the questions I have seen where posters have been confused about relativity arise because they have considered only one of these effects or because they have mixed them up in some way.

So what happens when you move through a reference frame is two-fold- the individual clocks in that frame will seem to be running slowly to you, but as you move from clock to clock they get more and more out of sync too.

There are two distinct ways to compare times between moving reference frames, and they are not equivalent. One is for a single observer in one frame to compare their time with the time at multiple points in the other frame as those points pass by. The other is for multiple observers spaced in one frame to compare the time at a single point in the other frame as it passes by each of them in turn. If you think about this carefully, you will realise that the first method measures clock synchronisation, while the second measure time dilation. They are two quite separate effects.

So the answer to your so called paradox is that it ignore the fact that the clocks in the reference frames of Twin A are out of synch with the clocks in the reference frame of Twin B. Each twin will think that the other set-off later, and the combination of the later take-off and the slower running clocks will cancel, meaning their watches will agree when they meet.


I am standing on the earth, halfway between Twins A and B. Each twin has a clock.

According to me: At noon, both clocks read "noon". At 6:00, the twins pass each other, right in front of me, and both their clocks, which have been running slowly, read 4:00.

According to Twin A: At noon, my clock reads "noon", and Twin B's clock reads "3:00". At 4:00, we pass each other. At that time, my clock, which is accurate, reads "4:00". Twin B's clock, which said 3:00 at noon and has been running at one-quarter speed ever since, also reads 4:00.

According to Twin B: At noon, my clock reads "noon", and Twin A's clock reads "3:00". At 4:00, we pass each other. At that time, my clock, which is accurate, reads "4:00". Twin A's clock, which said 3:00 at noon and has been running at one-quarter speed ever since, also reads 4:00.

Note that all three observers agree that at the time when the Twins cross paths, their clocks both read 4:00.

(The actual times in this story are not perfectly accurate; perhaps I should have written 4:32 instead of 4:00 --- it would be easy to calculate the right numbers. But the basic idea is the same.)

Now what's the problem?

  • $\begingroup$ At noon, my clock reads "noon", and Twin B's clock reads "3:00". Why? $\endgroup$ Feb 26 '16 at 19:31
  • $\begingroup$ What does "why" mean? Do you mean "Why later than noon?" or do yuo mean "Why 3:00 exactly, and not 2:30 or 3:30?" For the first, relativity tells you that if Twin A's and Twin B's clocks are synchronized according to the observer on the ground, then Twin B's clock must be set later than Twin A's clock according to Twin A. For the second, I used 3:00 as an example; the exact time of course depends on how fast the twins are moving and how far apart they are. (As always, this will all be crystal clear if you just stare at the Minkowski diagram.) $\endgroup$
    – WillO
    Feb 26 '16 at 19:55
  • $\begingroup$ I was asking why to the first. But forget about that, I think I can make my confusion a lot clearer by simply asking my question this way: time dilation causes a moving clock to tick slower than a stationary clock. If clock A is moving relative to clock B, then clock A ticks slower. However, relatively speaking, clock B is also moving relative to clock A, then clock B ticks slower. These two obviously cannot be true together, so what's the resolution? $\endgroup$ Feb 26 '16 at 20:21
  • $\begingroup$ The thing you said to forget about is the resolution. $\endgroup$
    – WillO
    Feb 27 '16 at 0:34
  • $\begingroup$ Are you talking about the delay light takes to reach you? If so, shouldn't it be the other way around? When your clock reads "3:00", Twin B's clock reads "noon"? If the clocks are synchronized according to the observer on the ground, why would you be reading a future time from the other clock at noon? $\endgroup$ Feb 27 '16 at 0:40

I've yet to see an explanation of the twin paradox that makes sense to me. At its core, the question is whether, and in what circumstances, does time dilation occur when comparing inertial frames of reference. Acceleration is a red herring, as the same results can be obtained if clock#2 passes clock#1at constant velocity. And asserting that the change in direction is the cause of differences in aging implies no time dilation if there is only an outbound journey. I think time dilation has been confirmed experimentally, though I am unaware of such proof for the twin paradox's round trip. The sole explanation for the assymetry, and consequent resolution of the twin paradox, is the fact that the origin and the destination are in the same inertial frame. Round trip or not, accelerating or not, the traveler does not travel as far as observed from the origin. Imagine trying to resolve this where there is no fixed destination, where both twins are traveling (different directions &/or velocities), or where the destination is in a frame of reference distinct from those of the twins.

  • $\begingroup$ If you have constant velocities, time of one twin will always occur to be slower than the time of the other twin - and since this is symmetric, we would have a problem - but we don't, because that way, we can never be in the same frame of reference. Only there can we actually compare age. This is the key to the solution and it does mean that acceleration is NOT a red herring: it is needed to get one twin into the frame of the other, thereby breaking the symmetry. $\endgroup$
    – Martin
    Feb 22 '16 at 23:47
  • $\begingroup$ But it's "acceleration" only in the sense of a non-straight path thru spacetime, as illustrated by the triplet version of the paradox in which C hands-off his clock time to D in my above diagrams. When D arrives at A, D's clock is behind just as if a single clock had traveled the AC->CD->DA path. Yet none of the clocks were accelerated. Conceptually this is important because people start to wonder if it's some effect of stress on the clocks; but it's not. It's a purely geometric effect: This path is a different length (duration) than that path. $\endgroup$ Aug 25 '20 at 19:54
  • $\begingroup$ If the traveling twin stops at some distance and becomes stationary relative to Earth again, then the twins really will agree that less time has passed for the traveler in Earth's frame. You can even eliminate one end or both ends of the trip for the traveller (i.e. - eternal traveler whose clock happens to read 0 when they are coincident). The traveling clock and Earth's clock will always agree that more time has passed on Earth in Earth's frame. They will also always agree that more time has passed for the traveler in the traveler's frame. Transforms between the frames show this. $\endgroup$ Jan 10 at 17:09
  • $\begingroup$ Actually, it's not necessary that "the origin and the destination are in the same inertial frame", only that the two twins end up in a common frame (which might differ from their starting frame on Earth). Let's say "there is only an outbound journey" (like one scenario you hinted at). At the end of that journey, either the traveler could "stop" or the twin on earth could start traveling with the same velocity. Either way, they'll end up in a common frame and agree on the same result. $\endgroup$ Jan 15 at 9:52

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