# Momentum conservation in the twin paradox

1. Oct 27, 2013

I was thinking about one of the many resolutions of the so-called 'twin paradox' within SR framework. I realize many methods of resolving this have been proposed, but I am curious about the particular solution that uses 'acceleration' of the traveling twin, to create the required asymmetry for differential clock rates (time dilation) between the twins.

As I understand it, the traveling twin's clock is slowed because (s)he is in a non-inertial frame at some point(s) during the trip because of the acceleration experienced, while the stay-at-home twin always remains in an inertial frame.

However, given momentum conservation, even the Earth along with the stay-at-home twin would have to have an acceleration in the opposite direction of the traveling twin, even if by an insignificant amount. This would put the Earth and stay-at-home twin also in a non-inertial frame, however briefly.

So, what is the logic behind this particular resolution of the twin paradox that produces the asymmetry between the twins?

2. Oct 27, 2013

### WannabeNewton

Well the twin paradox can be done with two accelerating observers as well, in the obvious way, so the dichotomy that you mentioned is not pivotal in getting a working scenario for the paradox. The solution is simply that proper times integrate to different values along different paths.

Anyways in the scenario you're talking about we are assuming gravity has no effect/presence for the purposes of the problem hence momentum conservation associated with the gravitational interaction doesn't come into play.

Last edited: Oct 27, 2013
3. Oct 27, 2013

### Staff: Mentor

Although many explanations of the twin paradox will leave you thinking this, it's not right - the acceleration does not cause the differential aging, and it's possible to set up a twin paradox in which neither twin ever experiences any acceleration and both remain in (locally) inertial frames for the duration.

The acceleration matters because, in the standard version of the paradox, it's different for the two twins. Because it's different, we can say that there is something different about the experiences of the two twins and therefore it's at least possible that the different experience includes different measurements of the passage of time. When you read a description of the standard version of the paradox, you'll always find some words along the lines of "But for the traveling twin, it's just as if he were at rest while the earth moves away" - and because of the acceleration, it's not.

A better way of thinking about it is to say that the two twins' clocks disagree at the reunion because they took different paths of different length through space-time. The acceleration is just the mechanism which sent them on their different paths.

4. Oct 27, 2013

### Staff: Mentor

This is not correct. In the standard twin paradox scenario, the traveling twin fires rockets to turn around; momentum exchange with the rocket exhaust is what changes his trajectory. There is no momentum exchange with Earth or the stay-at-home twin, so the latter can be inertial the whole time.

As other posters have noted, there are other possible twin paradox-type scenarios where both twins move non-inertially; but there is no requirement that one must just because the other does.

5. Oct 27, 2013

### Staff: Mentor

Huh? Why is that?

6. Oct 27, 2013

### yuiop

Here is one example of differential ageing with no acceleration. One twin is fired upwards from a cannon. On leaving the atmosphere he is travelling inertially and passes his twin who happens to be orbiting the Earth. If things are set up correctly the launched twin has less than escape velocity and falls back to meet his twin who has just completed a full orbit.
I was trying to think of other examples. An elliptical orbit versus a circular orbit is problematic because of precession of the elliptical orbit. Any others?

7. Oct 27, 2013

### WannabeNewton

Take Minkowski space-time and roll it up into a cylinder; one can find two points between which there are two time-like geodesics, one that goes straight up the cylinder and another which coils around it, and the former will record more proper time than the latter.

8. Oct 27, 2013

### A.T.

Similar to yours: One clock resting at the Earth's center. Other clock oscilating in a tunnel through the center.

Here we had some debates, if they actualy will age differently, if the oscilating clock stays within the Earth. But if it oscilates higher than the surface, then I'm pretty sure they will age differently.

9. Oct 27, 2013

### Staff: Mentor

Do the turnaround in free fall by executing a hyperbolic orbit around a massive object...
This is the case that I was thinking about when I added the "locally inertial" qualifier in my previous post.

And WbN, clever dude that he is , points out a solution that doesn't need that qualifier, although it's not one that describes our universe:
At this point, we've moved well beyond OP's original question, although this discussion does strongly reinforce the point made above, that the acceleration is just a convenient way of introducing some asymmetry between the two travelers and sending them on different paths through spacetime, not a fundamental requirement for differential aging.

Last edited: Oct 27, 2013
10. Oct 27, 2013

Instead of replying separately to everyone's answers, I am picking the above two as representative. The concensus seems to be that acceleration or inertial frames are not really necessary for the resolution.

True, but I think if we are going to consider non-inertial frames, the traveling twin is non-inertial three times. Is there anything specific about the 'turn around' non-inertial frame that explains his slower clock rate?

I was trying to think of it in a real life scenario. Momentum conservation would have to be considered when the traveling twin departs from Earth, briefly putting even Earth in a non-inertial frame, however insignificant the acceleration (practically zero, theoretically not zero). However, as it turns out from all the answers, consideration of who is inertial throughout and who is not is unnecessary.

11. Oct 27, 2013

### Staff: Mentor

It's easy to get rid of the acceleration and deceleration at the two ends of the journey: Start the traveling clock in motion towards the stationary clock, set the two clocks to the same time as the traveler passes the stationary clock on the outbound leg, compare the clock readings as the traveler zooms by on his return trip. That leaves just the one acceleration at the turnaround, and we've already discussed ways of getting rid of that one as well.

12. Oct 27, 2013

### Staff: Mentor

Yes. Have you read the Usenet Physics FAQ on the twin paradox? And in particular, the Doppler Shift Analysis? That should make clear what the fundamental asymmetry is in the standard scenario, and why it only comes into play when the traveling twin turns around, not at the start and end of his trip, even though he is also, as you correctly note, non-inertial at the start and end of his trip.

Not necessarily. The traveling twin could do his initial acceleration in a way that did not transfer any momentum to Earth (for example, by firing his rocket perfectly horizontally and achieving escape velocity in that direction).

Correct; the Doppler Shift Analysis that I linked to above bears that out.

13. Oct 27, 2013

### Staff: Mentor

There are certainly scenarios that you could envision where there is an initial acceleration by the stay at home twin, but there are others without such acceleration. It certainly is not necessary.

14. Oct 27, 2013