# Mach and the Twin Paradox

1. Jan 24, 2010

### GRDixon

If the fixed stars/galaxies did not exist, would there be a twin paradox?

2. Jan 24, 2010

### atyy

What do you mean by fixed stars?

3. Jan 24, 2010

### GRDixon

All things physical, except for the twins and the rocketship.

4. Jan 24, 2010

### atyy

The twin paradox would not exist without the Minkowski metric.

5. Jan 24, 2010

### GRDixon

This is true.

6. Jan 24, 2010

### Mentz114

What paradox is this ? Can we call this scenario 'differential ageing' because there's no paradox.

Why not ? as atyy has said it's inherent in SR.

[snap ] I didn't see GR Dixon's post.

7. Jan 24, 2010

### bcrowell

Staff Emeritus
In GR, there is a twin paradox in an empty universe. In a purely Machian theory, there clearly couldn't be one, since there would be no way to define which twin was accelerating. However, we don't have any viable theory of gravity that is purely Machian. The next best thing is Brans-Dicke gravity with $\omega$ on the order of unity. If you look at the original Brans-Dicke paper, they do frame the paper's storyline in terms of one of these artificial thought experiments involving an empty universe, of the type often used in discussing Mach's principle. However, they can't actually calculate the limit of a zero-density limit. Similar considerations might apply to this example. My off-the-cuff guess about applying Brans-Dicke gravity to the twin paradox, without a real calculation, is that G would be very large (in BD gravity, G varies, not inertia), and therefore the twins would first recede from one another, but then be pulled back together by their mutual gravitational interaction. The situation would be symmetric, so there would be no twin paradox. If this analysis holds water, then it seems to me that it would be quite a nice elementary example of the Machian properties of the theory -- simpler than the one they actually give in their paper.

8. Jan 24, 2010

### atyy

Do you think that would be true if we took pure GR with just the metric and matter field equations, and we're not allowed to put twins who don't contribute to spacetime curvature on top of a fixed curved spacetime?

9. Jan 24, 2010

### bcrowell

Staff Emeritus
I'm not quite clear on what you're saying here. When you refer to a fixed curved spacetime, do you mean a prior geometry (which GR assumes does not exist), or a cosmological solution (but one with no matter or radiation in it, as required by the thought experiment)? In standard GR, the twins don't have any significant gravitational effect on one another; they're test particles, to a very good approximation.

10. Jan 24, 2010

### atyy

I meant no prior geometry, which means no test particles.

11. Jan 24, 2010

### bcrowell

Staff Emeritus
Sorry, I still don't follow you. Can you explain more what you had in mind?

12. Jan 24, 2010

### atyy

I'm thinking that GR consists of the Einstein field equation and the equation of state of matter, say EFE and Maxwell's equations. The universe is a solution to these equations alone, and there are no equations of motion such as the geodesic equation of motion or the Lorentz force law. Such a universe has no prior geometry, since matter and metric determine each other. A universe with test particles has prior geometry, since the test particles propagate on a fixed background. So I think in full GR with no test particles, it's not so easy to set up a twin paradox, since the clocks would have to be in a solution of the field and matter equations. I guess that could be done, along the lines of http://arxiv.org/abs/gr-qc/9912051, so the twin paradox would survive ... but would this viewpoint change the Machianess of GR?

13. Jan 24, 2010

### bcrowell

Staff Emeritus
I don't think this is right. A test particle just means a particle that has a very small mass-energy, so it has a negligible effect on the spacetime curvature, and follows geodesics that would have been the same if the particle hadn't been there. A prior geometry means one that is set by the theory itself, rather than in a self-consistent way because it just happens to be a solution to the field equations. Newtonian physics has a prior geometry, which is Galilean.

14. Jan 24, 2010

### atyy

But if the mass of the test particle has been neglected, then the geometry was not determined self consistently.

15. Jan 24, 2010

### bcrowell

Staff Emeritus
You could say it's self-consistent in the limit of small mass.

16. Jan 24, 2010

### atyy

Hmmm, wouldn't you get a black hole with a point mass?

17. Jan 25, 2010

### Staff: Mentor

Yes, and the Minkowski metric is a valid solution of the EFE for no mass. Although what you would really want is to solve a GR 3 body problem (home twin, rocket twin, rocket exhaust).

18. Jan 25, 2010

### atyy

Yes, I'd say it a bit differently, but those are my sentiments too. The alternative to treat the twins as test particles. Or are these alternatives not really that different?

19. Jan 25, 2010

### bcrowell

Staff Emeritus
I don't really understand this subthread. In standard relativity, this is completely unnecessary; SR is an excellent approximation.

In Brans-Dicke gravity, you would presumably need to include gravitational interactions in order to describe the twin paradox in an otherwise empty universe.

Which are we talking about in the last half-dozen posts, GR or Brans-Dicke?

20. Jan 26, 2010