# Twin Paradox

*Most recent answer: 10/22/2007*

- Pat Dolan (age 43)

Seattle, WA

It is most convenient to analyze situations like these in just one nonaccelerating frame of reference. Only one twin, the stay-at-home one, is stationary in a single frame of reference. In this frame, one can add up all the time on his watch, and compare it with the time that accumulates on the traveling twin’s watch, and compare it when they meet up. You can also do the calculation in any other nonaccelerating frame of reference, and you will always get the same answer, that the twin that accelerates will have less time on his watch when he meets his brother.

Special relativity also gives you the tools to analyze the situation from the traveling twin’s perspective.

The traveling twin does not stay put in a single nonaccelerating frame of reference during his journey, however, so life is more complicated for him. Instead, at each moment of his journey, we can define a frame of reference that is moving along with the traveling twin. In this frame of reference, we can ask the question what time is read on his brother’s watch. Then a moment later, we have to pick another frame of reference moving along with the traveling twin’s new speed, and ask what time is read on his brother’s watch. If we make a graph of the time read on the stay-at-home twin’s watch in all of these different frames of reference as a function of the traveling twin’s watch’s reading, we will find it’s not a straight line. On out outgoing part, the traveling twin thinks his stay-at-home twin’s watch is running slow, just like his stay-at-home twin thinks the traveling twin’s watch is runnning slow. On the turnaround part, the traveling twin thinks his stay-at-home brother’s watch speeds up, and then on the return trip, he thinks the stay-at-home twin’s watch runs slow again. It’s a mess, but it gives the same answer as doing everything in just one frame.

In General Relativity. a broader class of reference frames is considered. These accelerating frames, like that of the traveler, have additional peculiar effects on time, but must agree with the Special Relativity answers when they are applicable (for example, in cases with no gravitational fields).

Your argument about the symmetry between the two twins assumes that only their relative motion, including acceleration, matters. It turns out that acceleration is not only a relative quantity -- you can devise experiments which give you different outcomes depending on the acceleration of the apparatus, without regard to the acceleration of anything else.

The idea that only relative motion can matter is known as Mach’s principle. Einstein tried to follow it for a while, but ultimately dropped it in General Relativity.

Mike W. and Tom J.

*(published on 10/22/2007)*

## Follow-Up #1: twin paradox

- Neil (age 44)

UK

We've moved your question to be a follow-up to one that comes closer to it.

Your assumptions are right, but you're missing one key step. When the traveler accelerates back toward earth, he isn't using an inertial reference frame. The speed-up of the earth's clocks that he sees due to accelerating toward them is just enough to make him agree with the stay-at-home about what their age ratio is when he returns.

Mike W.

*(published on 11/21/2013)*

## Follow-Up #2: twin paradox again

- Subra (age 42)

Chennai, India

No, it's not correct. You've assumed that the correct description of how things behave is what we call Galilean Relativity, but that turns out not to be true.

In the real world, the twin B who is leaving will *not* reach the speed of light, as seen by the stay-at-home, A. No matter how long B's rocket engine fires, she will only get close to the speed of light. This behavior is directly confirmed by all sorts of particle accelerators. In this flat-spacetime picture, where we ignore cosmological effects, the signals from B will never cease reaching A.

The rate at which the signals from B reach A will be not far from what you estimate when B isn't moving too fast relative to A. When B is moving at a significant fraction of c, the type of calculation you made just doesn't work, it doesn't describe the real world.

The simple effects of the sort you're describing were known since Romer used them to find the speed of light from the observations of Jupiter's moons, whose rotations make a sort of clock. That was before 1700 AD. The world just turns out to be different from that.

Mike W.

*(published on 11/19/2014)*

## Follow-Up #3: acceleration and twin paradox

- Ben Steemers (age 29)

Nethetlands

If the acceleration is caused by anything other than gravity (e.g. rocket engines) it is indeed directly measurable with accelerometers of all types, without requiring any chosen reference frame. So for that type of problem the usual story is ok. If, on the other hand, the important acceleration (where the spaceship quits leaving and starts to return) is due to slinging around a star or planet, via gravity, then that story about feeling accelerations breaks down. The effects of uniform gravity cannot be detected in any local experiment, and the gravity here can be very nearly uniform. So the local quantity that has direct physical meaning is not acceleration but only acceleration with respect to a locally free-falling object. That has some sort of cause, like the rocket engines, so it's not surprising that it has measurable effects.

Yet the results still work. The twin who goes out and slings around the heavy planet does come back less aged than the stay-at home. The full description requires general relativity. The traveling twin spends time "down" near the heavy planet, and that slows her clocks, as seen by the "up" stay-at-home.

Mike W.

p.s. Please follow up if this doesn't quite answer your question.

*(published on 09/06/2016)*

## Follow-Up #4: local free-fall

- Ben Steemers (age 29)

Netherlands

"Could the twin back on earth not see the twin in the space-ship as his locally free-falling object? "

No, because the other twin is not local. She's far away.

We cannot in general say who is accelerating. We have a choice of coordinates that allows any object to be described as accelerating or not. What we can say is whether any object is accelerating with respect to what it would do if it were in free-fall.

So yes, change in velocity with respect to free-fall has some specific cause. Change in velocity in other frames does not have to have a cause, since you can pick different frames accelerating with respect to each other. For example on Earth, we usually pick the Earth-frame rather than the free-fall frame. We say that somebody falling is accelerating, even though they feel nothing different thatn they would in a much bigger field or no field. We say that we aren't accelerating, even though we feel forces (on my feet, I'm using a standing desk) and can measure effects of those forces like different clock rates in the basement and attic. Such effects are alway present for paths other than free-fall and never for free-fall.

In the standard version of the twin story, gravity is left out, i.e. the gravitational field is assumed to be uniform, which is treated most simply as uniformly zero. That means we call call anybody's local free-fall "stationary" and it doesn't matter who we pick. In my slingshot version, however, the gravity isn't uniform so we need a more complicated analysis.

Mike W.

*(published on 09/12/2016)*

## Follow-Up #5: twin

- Larry Elterman (age 64)

02138

You can try our old answers and follow up if you need more.

Mike W.

*(published on 08/06/2017)*