I want to simply eliminate the effect of acceleration, which is the cornerstone in resolving the traditional twin paradox, through the following thought experiment:

Assume twins $A$ & $B$ are located in the same orbital height, far away from the surface of a massive planet, on a long concrete column at rest WRT the planet. A rapid explosion that happens between the twins makes them orbit the earth with a high velocity, say, $0.6 \ c$, in the same orbital path but in different directions. According to relativity, since the orbiting satellites are completely inertial frames, each twin attributes a variable time dilation to the other one, and each twin expects the other one to be younger. Nonetheless, these twins are automatically reunited after traveling half of the orbit's circumference without being affected by additional accelerations. Which twin is younger in this case? Indeed, If each twin sees the other's clock as running slow all along the way, how can the clocks agree when they meet?

For this configuration:

  1. The acceleration due to the explosion does not affect the twins' clock rates since it occurs abruptly in the same place on the column where the twins are located back-to-back. The clocks have been synchronized just before the explosion.

  2. The traditional gamma factor, attributed to one twin from the viewpoint of the other, is variable because the resultant velocity of one twin complying with the relativistic velocity addition is a variable measured from the viewpoint of the other one. However, whether or not the corresponding time dilation is varying, each twin anticipates the other one to be younger. There is no breaking of symmetry!

  3. The $g$-field of the planet has nothing to do with orbiting satellites since they are inertial, and thus GR corrections are not inserted. (Remember the GPS settings!)

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    $\begingroup$ I’m voting to close this question because this is yet another attempt to simplify the twin paradox. What alll of these supposed simplifications inevitably do is to add unneeded complexity. The twin paradox was intentionally formed in an easily understood manner to illustrate an apparent paradox of special relativity. $\endgroup$ Jul 28 '20 at 11:11
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Jul 31 '20 at 23:47

I just want to point out, that the time passed for any object traveling from event $A$ to event $B$ on a curve $\gamma(t)$ (with tangent vector $\dot{\gamma}(t)$) in spacetime with metric $g$ is: $$\tau=\int_A^B \sqrt{-g(\dot{\gamma},\dot{\gamma})}dt,$$ or when you put it in a more intuitive language, it is given by time-like "length" of the curve.

Due to symmetry in the case of negligible rotation, the length of the curve is same wheter it goes clockwise or counterclockwise and the twin will thus age the same.

In general relativity you must be aware that there is no global lorentz frame and thus in general you do not have natural procedure for comparing two (random) clocks. So the twin paradox cannot be even formulated the way you do it, because the sentence "twin attributes a variable time dilation to the other one" makes no sense.


All the issue about time dilation in the twin paradox comes from the expression (using $c = 1$) for $$\frac{dt}{d\tau} = \gamma = (1-v^2)^{-1} $$ valid for calculations in inertial frames.

In the case of a circular orbit, using the Schwarzschild solution: $$d\tau^2 = \left(1 - \frac{2GM}{r}\right)dt^2 - \left(1 - \frac{2GM}{r}\right)^{-1}dr^2 - r^2d\theta^2 - r^2sin^2(\theta)d\phi^2$$

If we assume a circular orbit on the equator to simplify the calculations, ($dr = d\theta = 0$, and $sin(\theta) = 1$), and dividing by $dt^2$ we get:

$$\frac{dt}{d\tau} = \left(1-\frac{2GM}{r} - \omega^2r^2\right)^{-1/2}$$

The time dilation is the same for both twins, because they have the same values for $\omega$ and $r$.

Of course, it is also necessary to calculate the metric in coordinates where each twin are at rest, to confirm that the values are the same. But I can not see how it would be different due to the symmetry of the situation.

  • $\begingroup$ in twin paradox, the supposed paradox arises from symmetry that the two twins both see each others time dilated (and resolution of the paradox is due to fact that the situation is not in fact symmetric). Here we do have symmetry, but we do not have a physically meaningful way to compare clocks of the two twins directly. So they in fact cannot make an expectation that the other one is to be younger unlike in STR. There is simply no time dilation formula, because you cannot even say how the two frames move relative to each other, as parallel transport depends on the path. $\endgroup$
    – Umaxo
    Jul 29 '20 at 3:48
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    $\begingroup$ "Of course, it is also necessary to calculate the metric in coordinates where each twin are at rest" What does this solve? coordinates do not need to have physical meaning and even if they do in the neighborhood of the event in question, they do not need to retain the meaning far away from the event. In twin paradox you need global lorentz frame, which you do not have in GR. $\endgroup$
    – Umaxo
    Jul 29 '20 at 3:50
  • $\begingroup$ @ClaudioSaspinski It seems that your calculations somehow show that the paradox can be valid. Am I right? If so, please tell me if each of the twins measures the orbit to be circular or elliptical? Because I think, due to the relative motion of the orbit, it is measured to be an ellipse rather than a circle unless you want to consider the viewpoint of the observer at rest WRT the earth. $\endgroup$ Jul 29 '20 at 6:31
  • $\begingroup$ @MohammadJavanshiry As I wrote in my answer, the time dilation of each of them can be calculated from the Schwarzschild metric. But you can calculate the metric from the free fall frame of each of them. It is a good exercise of GR at least. $\endgroup$ Jul 29 '20 at 15:38
  • $\begingroup$ @umaxo Don't you think that it is possible for each orbiting twin keep track of the clock of the other one during the time between meetings? I think that is the issue here. $\endgroup$ Jul 29 '20 at 15:43

TLDR: the situation by design is completely symmetric, so both observers are the same age when they reunite. Furthermore, if they correctly apply relativity then they correctly predict that there is no time dilation between them.

The g-field of the planet has nothing to do with orbiting satellites since they are inertial, and thus GR corrections are not inserted.

This is not correct. This scenario involves curved spacetime so it requires general relativity. Specifically, if two objects move inertially (no acceleration) with different velocities in flat spacetime then they can meet at one single event at most. The only way to have them reunite after travelling without any acceleration is for the spacetime to be curved. GR is therefore unavoidable for this question.

The traditional gamma factor, attributed to one twin from the viewpoint of the other, is variable because the resultant velocity of one twin complying with the relativistic velocity addition is a variable measured from the viewpoint of the other one. However, whether or not the corresponding time dilation is varying, each twin anticipates the other one to be younger.

The traditional gamma factor is for special relativity only, and is not relevant here. To derive the relevant time dilation expression we can start with the Schwarzschild metric. Since we are working in a single orbital plane we can, without loss of generality, set $\sin(\theta)=1$ and $d\theta = 0$ and for convenience $R=2GM/c^2$. And assuming that the orbits are circular we can (with loss of generality) set $dr=0$ also. Thus we arrive at a simplified Schwarzschild metric $$ds^2 = -c^2 d\tau^2= -\left(1-\frac{R}{r}\right)c^2 dt^2+r^2 d\phi^2 \tag{1}$$

Now, for a spacecraft moving in a circular path in the plane we have $$\frac{d\phi}{dt}=\pm\frac{v}{r} \tag{2}$$ respectively for the two spacecraft and the spacecraft are inertial for $v^2=GM/r$.

Substituting $(2)$ into $(1)$ and solving for the time dilation we get $$\frac{dt}{d\tau}=\frac{1}{\sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}} \tag{3}$$ Because one spacecraft moves at $v$ and the other moves at $-v$ we see from $(3)$ that the time dilation is equal for both spacecraft in the Schwarzschild coordinates. This means that at all times the spacecraft show the same time including at the moment of their reunion. This is as expected from the symmetry of the scenario.

Now, we would like to transform to the rest frame of one of the spacecraft and calculate the time dilation in the spacecraft's frame. We will use the coordinate transform $$t'=\sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}} t \\ \phi' = \phi - \frac{v}{r} t \tag{4}$$ and differentiating $(4)$ gives $$ dt=\frac{1}{\sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}} dt' \\ d\phi = d\phi' + \frac{v}{r\sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}} \ dt' \tag{5}$$ Note that the expression for $\phi'$ involves $t$. So the two terms will no longer be independent.

Substituting this $(5)$ into $(1)$ gives $$ds^2 =-c^2 dt'^2 + r^2 d\phi'^2 + \frac{2 r v}{\sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}} dt' d\phi' \tag{6}$$

Because of the cross term in $(6)$ we cannot simply solve for $\frac{dt'}{d\tau}$. However, if we substitute $(5)$ into $(2)$ we get $$\frac{d\phi'}{dt'}=0 \\ \frac{d\phi'}{dt'}=-\frac{2v}{r\sqrt{1-\frac{R}{r}-\frac{v^2}{c^2}}} \tag{7}$$ respectively for the two spacecraft. Note that one twin is at rest, so this represents that twin’s rest frame. However, note that although this is the twin’s rest frame the rest twin is not at the center of the coordinate system. $\phi$ is measured with respect to the planet not the twin.

Substituting $(7)$ into $(6)$ and simplifying we get $$ ds^2=-c^2 dt'^2 \tag{8}$$ for both spacecraft. This indicates that in one spacecraft's reference frame the other spacecraft is never time dilated. Therefore the observers agree and expect that the clocks will read the same when they pass the other spacecraft. Because this is curved spacetime the gravitational time dilation and the velocity time dilation exactly cancel for these symmetric twins in the coordinates above representing one spacecraft's reference frame.

  • $\begingroup$ Hi Dale. How did you get $\phi^\prime=\phi-vt/r$? Are you sure about this equation? $\endgroup$ Jul 29 '20 at 20:05
  • $\begingroup$ Hi @Mohammad Javanshiry we want one of the astronauts to be at rest in the primed coordinates. So that implies $d\phi’/dt’=0$ along the spaceship’s path in the primed coordinates. Equation 4 is chosen so that equation 7 will have the desired form. So, yes, I am sure about equation 4 precisely because it results in the correct equation 7. $\endgroup$
    – Dale
    Jul 29 '20 at 21:29
  • $\begingroup$ I think that you have misunderstood the physical meaning of Eq. (7). Indeed, Eq. (6) shows how one of the twins observes the motion of the planet as well as the Schwarzschild observer. Eq. (7) shows that there is only one point (center of the planet) which is at rest WRT one of the twins from the viewpoint of himself, therefore, this twin measures no angular velocity for this point. ... $\endgroup$ Jul 30 '20 at 6:46
  • $\begingroup$ Other points at rest WRT the center of the planet as seen by the Schwarzschild observer, have a constant angular velocity as shown below Eq. (7) from the viewpoint of each twin. I think your equations, at most, show that each twin measures the spacetime (attached to him) in which the planet and the Schwarzschild observer are orbiting constantly, is in a way that each twin measures the time of the Schwarzschild observer not dilated compared to his own time. Nonetheless, to me, your calculations do not relate the viewpoint of one twin to the other one at all. ... $\endgroup$ Jul 30 '20 at 6:47
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    $\begingroup$ +1 for the comprehensive answer! I think I am a bit confused about your coordinate transformations. Sure, it is the rest frame of the orbiting twin, but, it is not its local inertial frame as evident by the presence of the $dt'd\phi'$ term, one would have to do some $\tilde{t} = f(t',\phi')$ to arrive at local inertial coordinates I think. [...] $\endgroup$
    – Dvij D.C.
    Mar 8 '21 at 7:34

The answer to the question you've asked is this: When the twins come back together (half way around the world from where they started), it's clear from symmetry that their clocks must agree. But that's so obvious that I have to suspect you meant to ask something else.

Perhaps this is a question about what happens at some other point along the journey, before the reunification? If so, the answer is that the twins are in motion with respect to each other and therefore each one's clock runs slow in the other's (instantaneous) frame. So maybe you meant to ask this: If each twin sees the other's clock as running slow all along the way, how can the clocks agree when they meet?

The answer to that is: Each twin's instantaneous frame keeps changing, so he keeps changing his description of where the other twin is "right now". Those changing descriptions end up canceling the effects of the slow-running clock. Again, it's obvious from symmetry that this cancellation must occur, though it can be instructive to confirm this with a calculation. If you have any remaining confusion, that calculation should resolve it.

  • $\begingroup$ how do you extend "right now" in case of curved spacetime - or in this case Schwarzschild geometry? How do you compare the clocks? $\endgroup$
    – Umaxo
    Jul 28 '20 at 17:10
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    $\begingroup$ "So maybe you meant to ask this: If each twin sees the other's clock as running slow all along the way, how can the clocks agree when they meet?" Yes, this is my question. I added it to my text. Thanks. $\endgroup$ Jul 29 '20 at 5:08
  • $\begingroup$ "Those changing descriptions end up canceling the effects of the slow-running clock." I am doubtful that this notion can resolve the paradox. I think If the observers were non-inertial, we could say that the changing descriptions may cancel the effects of the time dilation. However, in this example, each observer can claim that he is inertial. Remember that, in the original TP, the twin standing on the (massless) earth, who is believed to remain inertial all during the experiment, always asserts that there is a time dilation complying with the gamma factor for the other traveling twin ... $\endgroup$ Jul 29 '20 at 5:26
  • $\begingroup$ though there is a changing description for the traveling twin as he accelerates to return to the earth. This change of description is effective only for the non-inertial traveling twin when he wants to judge the clock(s) of the inertial twin. $\endgroup$ Jul 29 '20 at 5:31
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    $\begingroup$ the non-inertial one detects an additional change in the clock rate of the inertial twin at the instant of acceleration which makes the inertial clock to run faster in the period of acceleration I think "detect" is a very misleading word here. With this usage, you'd have to say that when I make a U-turn in New York, I can "detect" a 6000 mile shift in the position of Los Angeles. Nobody uses the word "detect" that way. $\endgroup$
    – WillO
    Jul 29 '20 at 6:45

Initially I did not understand your question, but the interpretation of your question by WillO makes sense, so I will go with the same interpretation.

while you do not state so, it appears what you have in mind is the case of orbital motion at constant altitude. I will discuss that case.

In your description you added a starting point in time, an instanteous launch, horizontally, at orbital level. However, such elaborate history obscures the point you are trying to make. It is not necessary to describe how the orbital motion started. It is sufficient to state that it is a case of orbital motion.

So, we have a celestial body, with two satellites in orbit, counter-orbiting. That is, the two satellites are in the same orbital plane, very close in orbital altitude. The two satellites pass each other twice every orbit. We can designate one such pass as $t = 0$

This case does not change when each orbit is filled with many satellites. I increase the number of satellites to help with a comparison that will come later.

The two fleets of satellites can set up a procedure using only the exact moment of passing each other to iterate to a synchronized fleet time that is common to all the satellites.

Let me use an arbitrary number of satellites in each orbit, let's say 12 in each orbit. Let the clocks initially be unsynchronized. Each time two satellites pass each other by they transmit the time readout of their own clock to the other satellite. On reception each satellite adjusts its own time towards the middle of the two readouts. Iterated over multiple orbits that procedure will bring all the clocks of that fleet to a synchronized fleet time. That is, at every passing the time readout received will be the same as the time on their own clock.

The above case is one where the motion is arranged in a loop. In special relativity, when you close a loop things get very interesting.

It's time now to contrast the above case - motion along a loop - with the case where the motion is linear.

Let there be two fleets of spaceships, I will arbitrarily call them the 'red fleet' and the 'green fleet'. Each fleet is arranged in a single line. The two fleets pass by each other, like trains moving in opposite direction over parallel tracks.

In this case, the linear motion case, there are far less means available to arrive at a synchonized fleet time. The ships of the red fleet can use pulses of light to establish a synchronized fleet time for the ships of the red fleet (using Einstein synchronization procedure ), and similarly the ships of the green fleet can established a synchronized fleet time.

However, the synchronized red fleet time will not have the same simultaneity as the synchronized green fleet time. That is: in the case of linear motion there is no escape from relativity of simultaneity.

That raises the question: how does the escape from relativity of simultaneity come about in the case of closing-a-loop?

To see that we look at what we need to remove to prevent the escape from relativity of simultaneity.

Consider a fleet of satellites, all in the same orbital plane, at the same orbital altitude. Let these satellites arrive at a synchonized fleet time (using pulses of light), but disallow them from closing the loop. Let's say there are 12 satellites, numbered 1 to 12. Each satellite exchanges pulses with its orbital neighbour only. Satellite 12 and satellite 1 are neighbours, but now we disallow sending or receiving between 12 and 1. Without closing the loop the procedure is once again the same as Einstein synchronization procedure.

This is a profound difference between Newtonian space and time on one hand and Minkowski spacetime on the other hand. In Newtonian space and time: being able to close a loop makes no difference. But in Minkowski spacetime it does make a difference.


I don't see how adding gravity to the Twin Paradox will make it simpler. If you take it out, and just have the twins move in circle, obviously their motion is symmetric.

To simplify it even more, you could remove the transverse motion, leaving 2 twins to execute one cycle of simple harmonic motion from the same starting point, but it on opposite directions.

If we call them the plus twin and the minus twin, then their speeds in the initial rest frame are:

$$ v_{\pm} = v\sin{\omega t} $$

for $0 \le t \le \frac{2\pi}{\omega}$.

At any time, their position is:

$$ x_{\pm} = \frac v {\omega}(1 -\cos\omega )$$

Each sees the other moving at:

$$ v^{\mp}_{\pm} = \frac{2v\sin\omega t}{1+\frac{v^2}{c^2}\sin^2\omega t} $$

leading to a Lorentz factor:

$$ \gamma = \frac 1 {\sqrt{1-\frac{v^2}{c^2}}}= \sqrt{\frac{1+\frac{v^2}{c^2}\sin^2\omega t}{1-\frac{v^2}{c^2}\sin^2\omega t}}$$

by which each sees the others' clock ticking more slowly.

However, thought experiment twins don't travel in a vacuum. The have a lattice of comoving observers. When the twins are separated by a distance $x$, the right twin's comoving observer at the same position as the left twin (and vice versa) sees the clock's offset shift as the velocity changes. As he is moving away, the other twin's bias is moving forward. By the time they reach maximum separation, and both have $v=0$: the clock bias has exactly canceled the time dilation.

Now they begin coming home: as the move towards each other, $\gamma$ causes the other's clock to tick more slowly. Meanwhile the large separation causes the other's clock bias to move forward in time. Hence, the twin's clock is ahead, but ticking more slowly.

When they rejoin, the clock rate and clock bias have exactly canceled, leaving the twins at the same age.

  • $\begingroup$ In your example, the twins are no longer inertial but in mine they are. Indeed, you try to return the acceleration to this modified version and make it similar to the original TP. $\endgroup$ Jul 30 '20 at 7:00
  • $\begingroup$ It's not about acceleration, that's the whole point. $\endgroup$
    – JEB
    Jul 30 '20 at 22:26

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