I think one catch in Twin Paradox was about the big acceleration that can turn back the traveling twin from light speed outward bound, to become light speed inward bound.

What if there is strictly no acceleration?

  1. Peter is on a space ship, traveling 99% of light speed. He is exactly 20 years old.
  2. Michael is on Earth (or a planet similar to Earth, but with a radius so small that any centripetal acceleration is negligible... or consider him standing just on a piece of concrete in space with an oxygen supply)
  3. Michael is also exactly 20 years old.
  4. According to time dilation, Peter's clock in the spaceship is slower than Michael's clock.
  5. According to time dilation, Michael's clock on Earth is slower than Peter's clock. (since motion is relative, if we consider Peter to be stationary, and Michael is traveling)
  6. Peter's spaceship is traveling towards Michael.
  7. After 30 years on Earth, Peter's spaceship went past Michael's face, so Peter and Michael are 1 cm apart, face to face and eye to eye.
  8. Now, would Peter see Michael quite older than him, and also, Michael sees Peter quite older than him?

4 Answers 4


You say that both twins are "exactly 20 years old." I assume you mean that they are both 20 years old at the same time. But part of the point of special relativity is that a phrase like "at the same time" means different things in different reference frames.

To be specific, suppose that these two moments (Peter's birthday party and Michael's birthday party) are simultaneous in a reference frame in which the Earth is at rest. Then, performing the entire analysis in that same frame, we would say the following:

  • Peter's clock ticks slower.
  • Therefore, at the moment the two pass each other, Peter is younger than Michael.

Now let's look at things as measured in Peter's reference frame. In his frame, Michael's clock ticks slower. Therefore, from the time of his 20th birthday until the time the two of them meet, the amount of time as measured by Michael's clock is less than the amount of time as measured by Peter's clock. If we then concluded that Michael would be younger than Peter, we would indeed have a paradox. But that conclusion doesn't follow, because, in this reference frame, the two of them didn't start out the same age. To be specific, the event "Michael's 20th birthday" and the event "Peter's 20th birthday" were not simultaneous. Michael's birthday happened earlier. So in this reference frame, Michael started out older, and even though his clock ticked slower, he was still older when the two met.

All of that is the way things play out if the two birthdays were simultaneous in Michael's reference frame. If on the other hand the two events were simultaneous in Peter's reference frame, then you can just switch the names "Michael" and "Peter" throughout, and everything will work the same way.

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    $\begingroup$ What if Michael and Peter synchronize their clocks at the moment Peter passes by. Then Peter goes on traveling. There would be no problem if they were never to meet up again, that's why I'll add this assumption (which really makes the question academic): assume that the universe has no boundary and that by traveling in one direction, you can go back to your starting point. (Kind of like in Asteroids) Whill they agree on the anniversary date of their meeting? $\endgroup$ Commented May 1, 2011 at 19:58
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    $\begingroup$ That's a good twist on the twin paradox! The answer is that such a universe does have a preferred "rest frame", and the one who's sitting still in that frame ages fastest. Why doesn't this violate special relativity? I guess the best way to think about it is that, in any spacetime other than true Minkowski spacetime, the large-scale properties are the province of general relativity. In GR, the laws of physics are Lorentz-invariant, but particular solutions to those laws need not be. $\endgroup$
    – Ted Bunn
    Commented May 1, 2011 at 21:46
  • $\begingroup$ Thanks Ted, I suspected the answer would be one in GR. Maybe I should make this a question of its own so that you can elaborate on that? $\endgroup$ Commented May 2, 2011 at 8:55
  • $\begingroup$ @Ted Bunn : Einstein relativity forbids a preferred "rest frame" as you said ? or ... do not need it, under the terms choosen to do the analisys ? $\endgroup$ Commented May 3, 2011 at 17:36
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    $\begingroup$ Oh, now I understand. Special relativity says that the laws of physics are the same in all inertial reference frames. You are perfectly free to pick one frame and call it "absolute," and sometimes that's a salutary thing to do, but it's you choosing it, not the physics. In the multiply-connected universe Raskolnikov asks about, there is a frame that's actually physically different from the others. That's what special relativity says shouldn't happen. $\endgroup$
    – Ted Bunn
    Commented May 4, 2011 at 13:02

I'm rusty, but will have a stab at this.

Let's make it easy and say that Peter starts off 30LY away from Michael. And P is going .99c towards Michael.

If P beams a picture of a clock to M at the start of this 30 light-year sprint, when will Michael first see it? Well, the light takes 30 years to get there and P arrives about .3 years afterwards. So, M sees Peter's clock go from 0 to x in .3 years...

What is x? x is the time that the clock (and P) experience. It should be roughly 0.14 x 30 years or 4.2 years.

So Michael sees the clock going from 0 to 4.2 years, in just 0.3 years! The clock is spinning like crazy.

Let's say P beams the clock plus his face... The face and the clock both experience 4.2 years. So M sees P aging in fast-forward, but only to the age of 24.2 years. Still, 12 time the normal rate of aging (same as when you look after a newborn baby)

The key thing is, Michael wouldn't even know that Peter is on his way until the signal reaches him from 30 years away. If Michael can work out that Peter started 30LY away, then Michael knows that Peter has experienced time dilation.

What could Peter see? Well, again the distance comes into play. Peter's first picture of Michael (at age 20) would take many years to reach Peter. let's say, at the half-way mark (15 LY) Peter sees the video of Michael aged about 20. Meanwhile Peter is now 22.1. Peter sees a live feed of Michael aging from 20->50 in just 2 short years (depressing).

The key thing is to give up on the idea of 'the same time'. There is only time as observed by Michael or Peter.

Extend this - what does Michael see after another ten years. After Peter whizzes by?


As with most of the so called paradoxes of special relativity, the resolution involves the relativity of simultaneity.

You have assumed explicitly at the outset that Peter and Michael are 20 years old, and implicitly you assume they are twenty years old simultaneously. That cannot be a reciprocal assumption. Either they are both 20 in Michael's reference frame, or they are both 20 in Peter's reference frame, but they cannot both be 20 years old in both frames.

Whether one or other appears older when they meet will depend upon which of the two starting positions you choose.

You can if you wish pick a third starting position, in which Peter and Michael are moving toward each other at near light speed in the frame of an observer stationary at a point mid-way between them. If they were both twenty years old simultaneously in the frame of that observer, they will have aged the same amount when they meet.


The question in the twins' experiment is how to get rid of acceleration. This can be done as follows:

Consider three observers, the Earth (E), a very distant planet (P) at a constant distance from Earth, and a spaceship (S). Clocks E and P are synchronized with the Einstein method and always remain synchronized.

S starts its journey from Earth accelerating for a short time and then stabilizes its speed. It is at this point that clock S synchronizes with clock E, marking the start of the experiment. It does not matter what the S did before this synchronization, so its acceleration is not included in the experiment.

Frame E&P: When S gets too close to P, it photographs the S and P clocks at the same time and then slows down and stops very quickly. At that time the difference between S and P watches will not differ significantly from the difference shown in their photos. Note the difference of the photographed watches as D1=S-P.

In Frame E&P we examined the history in the common frame of reference of E and P. We will now look at the same story in the reference frame of S:

Frame S: When P gets too close to S, S photographs the clocks S and P at the same time, and then accelerates to immediately stabilize its speed relative to the speed of P. Note the difference of the photographed watches as D2=S-P.

After doing the above calculations or your assumptions, you will have already found that D1=D2. The reason is obvious, since both measurements are different aspects of the same event so it is not possible for the measurements to be different.

The possible differences of the measurements D1 and D2 (D1=D2=D) are D<0, D>0 and D=0. If your calculations or estimates show that it is D<0 or D>0, then you should satisfactorily explain how this happens, since the alternative stories (of the same event) are equivalent, as there is no acceleration. The only justified measurement is D=0, which means that the photographed clocks S and P show the same time. So since it is S=P, it will also be S=E, that is, at the end of the experiment all three clocks are in sync.


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