I have known about the twin paradox for some time and as far as I know that it's not a paradox because one of the twins accelerate in his return journey.

But if this is not the case, then what will they see when they meet each other? If by some means the twins synchronize their clocks when the space twin reaches his ending point and then on his return journey both of the twins' clocks start at $0$ from the beginning of the return journey, then the effect of acceleration will be removed from the time taken for the return journey.

So they when they meet each other and can compare each others' clocks. What will they see then? A will see B's clock to run slow and B will see A's clock to run slow. Will they see that both their clocks has measured the same time for the return journey?

Here A is the twin who goes from point X to point Y at uniform velocity and B is the twin who stays at point X.

Edit: Suppose A sends light signal from Y to X and the light signal reaches X and is immediately reflected by a mirror. The it will go back to Y where it will once again be reflected back to X. In the above case, the light will take 2 years to go from X to Y and back to X. So A will figure out that light takes 1 year to cover the distance XY. When light reaches B for the second time (3 years after the first signal), he will start his clock from $0$.

And also after 3 years A will start his return journey with starting his clock from $0$. So both of the clocks will start from $0$ exactly 3 years after the first signal was sent.

  • 1
    $\begingroup$ What does "synchronize their clocks" mean? It turns out that the answer to your question depends entirely on the implementation of this process, as it means sending a signal (typically via light) from one to the other at a long distance. The nature of that signalling will affect how they set their clocks, and thus the numbers they read back home. $\endgroup$
    – Cort Ammon
    Jul 7, 2020 at 6:14
  • $\begingroup$ @CortAmmon I have edited the answer to include the process. $\endgroup$ Jul 7, 2020 at 6:42
  • $\begingroup$ The usual way to synchronize seoarated clocks that are at rest relative to each other is Einstein synchronization, but there are other options. $\endgroup$
    – PM 2Ring
    Jul 7, 2020 at 6:42
  • 1
    $\begingroup$ You haven't defined properly who is A and who is B. Also you haven't properly defined the points X and Y. No answer can be given unless you are precise on the definitions. I suspect that once you try to be precise you will indeed discover your mistake. $\endgroup$
    – Noumeno
    Jul 7, 2020 at 10:00
  • 1
    $\begingroup$ If you want to take out acceleration, just bring in a clone of the traveling twin that is on another rocket coming towards the Earth, and represent the turn around as the clone passing the traveling twin where they synchronize watches. Would this be better for what you are trying to ask? $\endgroup$ Jul 7, 2020 at 12:04

3 Answers 3


$B$ is the twin who remains on earth. $A$ is the twin who starts at $Y$ and travels to earth.

There are two frames here: The $B$-frame (which is $B$'s frame throughout and also $A$'s frame before he starts his journey) and the $A$-frame (which is $A$'s frame once the journey starts).

Your procedure synchronizes the clocks in the $B$ frame. They are not synchronized in the $A$ frame.

Here is the story in the $B$-frame:

  1. When both clocks read $0$, $A$ starts his journey.

  2. The journey takes $1$ year. During that year, $A$'s clock runs at 1/2 speed, but $B$'s runs normally.

  3. Therefore when $A$ arrives, $A$'s clock says $1/2$, and $B$'s clock says $1$.

Here is the story in the $A$ frame (where it's $B$ who does all the traveling):

  1. When $A$'s clock says $0$, $B$'s clock says $3/4$, and he's already partway along his journey.

  2. The remainder of the journey takes half a year. During that time, $B$'s clock runs at half speed (so that it advances by 1/4 of a year) while $A$'s run normally.

  3. Therefore when $B$ arrives, $A$'s clock says $1/2$ and $B$'s clock says $1$.

I arrived at all this not by thinking about time dilation or clocks running slow, but by meditating on the geometry of the spacetime diagram. Only after I understood it did I translate it into the language of "clocks running slow". In other words, m4r35n357 basically gave you good advice.

  • 1
    $\begingroup$ I couldn't agree with your penultimate sentence more! Folks: the spacetime diagram (and spacetime interval, effectively the same thing) are the ultimate truth, they should be your first port of call for all SR problems. Worry about the buzzwords after you have calculated the answer ;) $\endgroup$
    – m4r35n357
    Jul 7, 2020 at 14:36
  • 4
    $\begingroup$ Since your spacetime diagram was useful in resolving the issue, can you update your answer with it? It could help others. $\endgroup$
    – robphy
    Jul 7, 2020 at 14:41
  • 2
    $\begingroup$ @Theoretical: Draw the picture. $\endgroup$
    – WillO
    Jul 7, 2020 at 15:28
  • 2
    $\begingroup$ @PM2Ring : I think the OP would be much better served by doing this on his/her own. $\endgroup$
    – WillO
    Jul 8, 2020 at 1:23
  • 2
    $\begingroup$ @Theoretical: here's the algebra your picture should reflect: 1) By calculating in the B frame, we know that when the two come together, A's clock must say 1/2 and B's must say 1. 2) By calculating in the A frame, we know that between time 0 and time 1/2, the time elapsed on B's clock must be 1/4. 3) From 1) and 2) we know that in the A frame, when A's clock reads 0, B's must read 3/4. $\endgroup$
    – WillO
    Jul 8, 2020 at 1:27

The problem is they can't synchronize their clocks unless they are in the same place. In relativity, you need to keep in mind that there is no "universal now".

  • $\begingroup$ A long process had come up in my mind. So I did not want to include it in my question. I think I have to put it now. $\endgroup$ Jul 7, 2020 at 6:25
  • 1
    $\begingroup$ @Ken Absolutely true of course, but hardly an answer! $\endgroup$
    – m4r35n357
    Jul 7, 2020 at 10:29
  • $\begingroup$ @Ken Why can't they snchronize their clocks? I would be great if you would elaborare your answer. $\endgroup$ Jul 7, 2020 at 14:14
  • $\begingroup$ @Theoretical hope you didn't find my answer too patronizing, but if you look at my comment below it I address the reading of a remote clock ;) $\endgroup$
    – m4r35n357
    Jul 7, 2020 at 14:30

Acceleration is absolutely not necessary to understand the "Twin Paradox". Furthermore, it is a significant impediment to understanding the "Twin Paradox". The "Twin Paradox" is a great example of a really basic concept badly taught over many decades.

If you learn about the spacetime interval, $\delta \tau^2 = \delta t^2 - \delta x^2$, you will know what all the participants' clocks say, at every stage of the process. Specifically, here is the time shown on the traveler's clock:

$$\delta \tau^2 = (1 / v^2 - 1) \delta x^2$$

Anyone with basic algebra skills will be able to derive this in two lines starting from the Spacetime Interval together with $v = \delta x / \delta t$ (taking $c = 1$). So why make them worry about acceleration on top of that?

If you get hung up on distractions like "acceleration", "clocks running slow", "time dilation" and puzzle over things happening "at the same time" you will never gain an understanding.

[EDIT] following @Not_Einstein's comment, perhaps I should add that for the non-traveler, the time elapsed is simply $\delta \tau = \delta t$, the coordinate time, since $\delta x = 0$. I had not considered this useful (visual clutter), but I am adding it now anyway.

  • 4
    $\begingroup$ The tone of this post is patronizing, and it is not an answer. It is only a hint that you know the answer. Someone with the apparent background of the OP will not be able derive the solution in two lines. Please provide a complete answer. I'll also note that the usual statement of the twin paradox has the twins meeting at some future time after they leave each other so that they can compare their clocks. To explain your version of the twin paradox you will have to explain how the twins compare their clocks. (downvoted) $\endgroup$
    – garyp
    Jul 7, 2020 at 11:11
  • $\begingroup$ While acceleration may not be necessary to explain the twin paradox to someone versed in the mathematics of special relativity, it is useful to differentiate the traveling twin from the earthbound one to someone not so versed. $\endgroup$ Jul 7, 2020 at 12:58
  • $\begingroup$ @garyp I know you understand this, but for others my point is that acceleration is an unnecessary distraction, since if the traveler never turned around and just went twice as far, his age would be the same as if he had turned round. The non-traveler would of course need to read the traveler's clock remotely (or the traveler could transmit it), but this can be done in principle. Thus the turn-around (and therefore the "acceleration" phase) is ripe for Occam's Razor. en.wikipedia.org/wiki/Occam%27s_razor $\endgroup$
    – m4r35n357
    Jul 7, 2020 at 14:21
  • 1
    $\begingroup$ Please refer again to the OP title and first sentence. It is your question that has not been asked. The OP is asking specifically and explicitly about the role of acceleration in the TP, and that is precisely the aspect I have focused on. If I thought the poster was merely asking for yet another description of the TP I would not have bothered answering at all. $\endgroup$
    – m4r35n357
    Jul 7, 2020 at 17:37
  • 1
    $\begingroup$ And my point is that the traveler is the same age when he has gone the given distance whether or not he accelerates half way along. Acceleration has no effect on the ages of the traveler or the non traveler. To get me to agree with you, show me a mathematical expression (relevant to the TP) containing the acceleration, $a$. $\endgroup$
    – m4r35n357
    Jul 7, 2020 at 18:22

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.