Special Relativity: the symmetric twin paradox

This post was prompted by this question on Reddit.

Most people have heard of Einstein’s theory of Special Relativity (SR), the 1905 theory that introduced the idea that time and space are affected by the state of relative motion, and the famous equation


And most people have heard of the famous “twin paradox”: Two twins separate, one of them traveling away from earth at close to the speed of light. The traveling twin reaches a distant star, turns around and then heads back to earth, again close to the speed of light. When the twins are reunited on Earth, the traveling twin is younger. The traveling twin experienced less time than the twin on Earth.

The reason it’s a paradox (an apparent contradiction) is that one of the central tenets of SR is that velocity is relative. Twin A perceives Twin B as moving forward away from earth and analyzes things from that standpoint. But it’s equally valid for Twin B to say Twin A and the Earth are moving backward, and analyze things from that standpoint.

So if they have equally valid viewpoints about the state of motion, how can one end up younger? Shouldn’t they both do the same analysis, that the other is younger?

The standard resolution of the paradox is that the twins are NOT in equivalent frames of reference. They are not symmetric. The fact that Twin B turns around turns out to make all the difference. The details of that turning around are irrelevant, it’s just the fact that it happens at all that breaks the symmetry.

Here’s a good article on the standard Twin Paradox. It includes Minkowski diagrams or spacetime diagrams that, if you know how to read them, show you why the difference in aging happens. It’s well worth learning how to draw and read these diagrams if you’re seriously interested in fiddling around with SR.

But what if we take out that turn-around so the situation is more symmetric? Now which one describes the other as younger? Or do they both? How can that be?

Let’s use some of the numbers from the original Reddit post, but some will need some adjusting. We have a rocket traveling to the nearby star system Paradoxum Geminorum (PG), which is stationary with respect to Earth. The ship is traveling at 0.8c according to Earth, and according to Earth the trip takes 25 years. Τhen the distance to the star is 0.8 * 25 = 20 light years (ly). Light signals will take 20 years to travel between the two points (according to Earth).

According to SR, the relativistic factor \gamma is equal to

\gamma = \frac{1}{\sqrt{1 - \left (\frac{v}{c} \right )^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{0.6} = 1.67

That means that according to the ship, they are traveling to a star which is approaching them at 0.8c and is a distance of 20/\gamma = 12 ly away. The trip according to the ship takes 12/0.8 = 15 years. Note that this is also 25 / \gamma years, the expected time dilation of 25 years Earth time to 15 years ship time.

We can assume that Earth has pre-placed a clock on PG which is synchronized to Earth time. For instance we could send a signal to the clock with the time t right now on Earth. Knowing that the light-speed delay is 20 years, the clock sets its local time to t + 20 years. Because the clock is in the rest frame of the earth, it stays synchronized once the synchronization procedure is complete.

Let’s say we did that procedure before our ship left on its trip so there is a local clock already ticking away in the PG system.. The ship now leaves at Earth time 0, and sets its onboard clock to also read 0.

When the ship arrives at PG, the local clock will record t = 25 Earth/PG time. Earth doesn’t know that yet, but eventually (20 years from now) we’ll hear the report.

Because we’re removing acceleration from the scenario and keeping things symmetric, the ship is going to go right past the PG system at 0.8c. But as it passes the local clock, it sends a signal back to earth that says “we got here at t = 15”. Years later (20 to be precise), Earth simultaneously receives that signal and the signal from the clock and says, “Aha! As expected, it took 25 years our time, and 15 years their time.” Earth concludes that time passes slower for the moving ship passengers.

What does the ship say? According to the onboard clocks, 15 years have passed in their own frame. They know it was 0 Earth time when they left (and were distance 0 from Earth, an important point). The question they want to ask is, “what time is it right now on Earth?”

And that’s the key. They can see that the PG clock reads t = 25. But that’s a moving clock according to them. Being well trained in Special Relativity, they know that moving clocks do not stay synchronized. That is not the time on Earth.

To answer this question, they apply the full Lorentz Transform. The fact that the Lorentz Transform for time includes an x term expresses the relativity of simultaneity, the fact that events simultaneous in one frame but physically distant are not simultaneous in other frames.

t' = \gamma \left (t - \frac{vx}{c^2} \right )

They say, “In our frame the time is t = 15, and the position of Earth is x = -12 ly. So the time right now on Earth is \gamma \left [15 - (-0.8) * (-12) \right ] = 1.67 (15 - 9.6) = 9. So Earth experienced only 9 years while we experienced 15.”

In other words, the ship passengers say “clocks on Earth are running slow compared to our clocks”.

The confusion arises from the fact that the local Earth-time clock on PG says 25, so if you’re not thinking relativistically, you might say “25 years have passed on Earth”. But from the ship’s point of view, they would not agree that the PG clock is keeping Earth time, and as the Lorentz transform shows, only 9 years have actually passed on earth.

You could work out an experiment to verify that, for instance a signal sent from earth at earth t = 9. Of course they’re still in motion so it will take time for that signal to catch up with them. When it does, they can backtrack and figure out that signal was sent at the exact moment (according to their clocks) when they passed PG.

Conclusion: Due to the relativity of simultaneity, both frames of reference can indeed conclude that the other’s clocks are running slow without contradiction.

One final note: If the ship puts on the brakes and stops on Paradoxum Geminorum, the Earth is no longer a moving frame. They are in the rest frame of the Earth. They will agree with Earth that t = 25 in Earth time and that they are younger than their contemporaries on Earth. As with the round-trip twin paradox, that change of frames in the traveling ship is what breaks the symmetry and causes one frame to experience less total time than the other.


One response to “Special Relativity: the symmetric twin paradox”

  1. Pam Your Wife Avatar
    Pam Your Wife

    Not that I understand most of this but it looks great. You are too creative!! 😘. I’m already fond of “Terra-phy” as it fits how I feel about Physics…🤫


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