# The twins in a big circle

1. Jan 29, 2008

### jdavel

I haven't seen the twin problem configured this way before. But it seems to me it should lay to rest the argument that SR can't account for the age difference.

Instead of going from the earth to the star and back, the traveler goes from the earth in a big circle around the star and back. The star is the center of the circle and the earth is a point on the circle. The traveler then experiences a continuous acceleration throughout his trip. If the star is 100LY from earth and the traveler's speed is .99c then his acceleration (v^2/r) works out to about 8m/sec/sec. That's less than the 9.8m/sec/sec that his twin experiences staying on the earth! So, when he returns (2pi*100 years later), if he has aged any less than his twin, it can't be because of acceleration.

Now all you have to do is a relativistic doppler calculation for an object moving in a big circle. I haven't done it, but it shouldn't be hard. The distance between the twins is r*sin(theta/2) where theta is the angle at the star. So their relative speed is just 1/2*r*cos(theta/2), or about 50*cos(t/200), where t is in years. The total number of ticks on the traveler's clock should be an integral of his tick rate doppler adjusted for this speed.

2. Jan 29, 2008

### JesseM

The twin paradox ordinarily assumes the flat spacetime of SR--so although you can refer to one twin being "on Earth", the curvature of spacetime due to gravity is generally ignored. In any case, when people say that the twin paradox is explained by acceleration, no one says the magnitude of the acceleration determines the age difference--for example, in the ordinary twin paradox where the traveling twin moves out inertially, turns around, and comes back inertially, you can have the same acceleration in the turnaround but vary the length of time spent moving out and coming back and the age difference will change significantly.
The twin at the center won't see the traveling twin's clock rate affected by the doppler effect, because the distance isn't changing. The twin at the center will see the traveling twin's clock slowed down by the time dilation factor of $$\sqrt{1 - 0.99^2}$$, while the traveling twin will see the clock of the twin at the center sped up by the same amount.

3. Jan 29, 2008

### jdavel

JesseM,

"The twin at the center won't see the traveling twin's clock rate affected by the doppler effect...."

I didn't explain clearly enough. The star is at the center of the big circle; the earth is on the circle. So neither twin sits at the center of the circle.

In the interem an even simpler configuration has occurred to me. You don't need the circle. Just do the whole thing the regular way (out to the star and back) but keep the traveler's deceleration and receleration (when he's turning around) right at 1g. Since the earth twin experiences the 1g during the whole trip, if acceleration is the cause of the age difference, he should be the younger one. Right?

4. Jan 29, 2008

### 1effect

Incorrect.
You can see the correct calculations here

5. Jan 29, 2008

### JesseM

My fault, you did explain that in your original post, I just got confused. In this case the doppler shift as seen on Earth will depend on the angle of the ship's motion relative to the axis between it and the Earth at that moment.
No, what I said previously still stands, the difference in ages is not a function of the magnitude of acceleration, it depends on the entire path. As an analogy, we know the shortest distance between two points in a plane is a straight line, which in any x-y coordinate system would mean a path of constant slope (where slope at a point on the path is delta-y/delta-x for an arbitrarily small interval around that point). So, if we have two points with a path of constant slope between them, and another path where the slope is constant for a while going out from the first point, then there's a change in slope, then it's constant for a while again until the path reaches the second point, we know this second path will have a greater length (the path will look sort of like one of those drinking straws with a bend in it). But is the difference in length between the straight-line path and the bent path just a function of the rate the slope was changing during the (brief) non-constant section? Of course not, the total length of the non-constant section may be very small, with almost all of the length of the bent path being made up of the two straight segments. The key is that each straight segment was not parallel to the straight line joining the two points, but was going off at a different angle.

6. Jan 29, 2008

### geoffc

Here's a great paper dealing specifically with the twins on a ring that appeared in AJP

Last edited by a moderator: May 3, 2017
7. Jan 30, 2008

### smallphi

8. Jan 31, 2008

### yogi

djavel: "I haven't seen the twin problem configured this way before. But it seems to me it should lay to rest the argument that SR can't account for the age difference."

SR always accounts for the age difference - the issues arise in connection with whether nature works the way SR demands in order to arrive at the age difference, ...or whether the interpretations of different individuals are functionally equivalent.