# Twin paradox in an empty universe

1. Apr 11, 2008

### Wiemster

The obvious objection to the famous thought experiment with one of a twins moving away from the earth is, couldn't you have moved the twin together with the earth in the opposite direction to yield the opposite conclusion. Thes standard answer is that the symmetry is broken when the returning twin accelerates by returning. Moving the other twin together with the earth and the rest of the universe is not the same thing as moving only the leaving twin.

Does this imply there is some connection between all the mass in the universe and inertia a la Mach?

2. Apr 11, 2008

### michael879

what exactly are you asking? How do mass and inertia have anything to do with the twin paradox (qualitatively at least)?

3. Apr 11, 2008

### Wiemster

Well, without any other mass in the universe, the situation would be entirely symmetric and no twin can come back older than the other. Now add some mass to this universe and an asymmetry arises and there is such a thing as accelerating with respect to the background of mass in the universe. Doesn't this thought experiment clearly show the relation between inertia (as is overcome in accelerating with respect to the universe, back to earth) and the mass of the universe?

4. Apr 11, 2008

### pam

The twin that stays home goes straight up the time axis. The twin that travels and returns
moves out along the x axis and then returns to the other twin.
Since d\tau^2=dt^2-c^2 dx^2, the twin that travels has a smaller $$\int d\tau$$.
It has nothing to do with the rest of the universe.

5. Apr 13, 2008

### Wiemster

How can you say that and still talk about a 'traveling twin' and one that 'stays home'. These statements make an indirect reference to the rest of the universe. Without the rest of the universe there can be made no distinction between the twins and neither will come back younger or older than the other. Actually by conservation of momentum in order to return they must both change their momentum in an equal but opposite way from the center of mass.

Now add some mass and the 'traveling twin' can push against some massive object to change its direction back home. The symmetry breaking can only occur in a universe with mass where an acceleration is possible. Don't you think this clearly shows a connection between inertia of an object and the mass in the universe a la Mach?

6. Apr 13, 2008

### Ken G

We actually don't know that, you are in effect assuming something and then reasoning from it to create a paradox if what you assumed isn't true. All we can can say is that (in the absence of real gravity), if each twin carries an accelerometer with them, the one carried by the "younger" twin on return will have to have spiked through the roof during the turnaround. If that reading of that instrument somehow depended on the rest of the universe, then your symmetry assumption would be right, but in your empty universe inertia would not exist and physics as we know it would fail, so we really have no idea what would happen in that universe. However, if the rest of the universe has nothing to do with the reading of that accelerometer, then it is also not the thing that breaks the symmetry in your thought experiment, and the twin paradox could unfold normally even in an empty universe. I tend to think in terms of the latter picture, but it's pretty hard to suggest an observational test, given the limitations in how many universes we get to try.

Last edited: Apr 13, 2008
7. Apr 13, 2008

### luben

so is it fair to say, the difference in age is due to the asymmetry in acceleration history?

8. Apr 13, 2008

### HallsofIvy

Yes, it is. Also note that the "twin-earth" and "twin" are NOT symmetric. The one that stays on the earth is subject to a gravitational force that the other is not.

9. Apr 13, 2008

### phyti

The motion is symmetrical, but the casuse is not.
The energy to launch the spacecan, if applied to the earth would not produce the same motion, but would be dispersed as heat. The earth would not move, just get warmer.
Conservation of momentum.

10. Apr 14, 2008

### DrGreg

Nobody seems to have noticed reference to Mach in the original question, which I assume means Mach's principle, the idea that what disitinguishes inertial particles from other particles is somehow determined by all the matter in the Universe. The questioner's suggestion is that if there were only two twins in the Universe and nothing else, we wouldn't be able to say which of the two accelerated.

To mind my this is nonsense. One of the twins decides to turn round and actually does something (e.g. fire a rocket) in order to turn around. That is the twin who accelerates, and who experiences the "g-force" of acceleration. Would that g-force be absent in an empty universe? I find that hard to swallow.

11. Apr 14, 2008

### phyti

Time dilation is caused by motion relative to light. They would age differently if they moved differently. More mass is irrelevant.

12. Apr 14, 2008

### pervect

Staff Emeritus

In particular there have been several papers published on the topic that are mentioned in the thread:

http://arxiv.org/abs/physics/0006039
http://arxiv.org/abs/gr-qc/0101014 ( http://dx.doi.org/10.1103/PhysRevA.63.044104 )

http://www.maa.org/pubs/monthly_aug_sep01_toc.html
http://arxiv.org/abs/gr-qc/0101014

Talking about "Mach's principle" is a bit of a dead end, unless you happen to believe in it. Unfortunatlely, it turns out that "Mach's principle" does not actually have an operationally well-defined meaning, so it tends to cause long and pointless arguments by people who may not even realize they aren't arguing about the same thing (but are using the same words, an example of the equivocaton fallacy).

Note that GR may or may not be compatible with "Mach's principle" depending on exactly what one thinks the principle is.

Last edited by a moderator: Apr 23, 2017
13. Apr 14, 2008

### Mentz114

phyti,
What ? I think not.

14. Apr 15, 2008

### phyti

Check the equation, it's a function of v and c!

Check the light clock, it's the same!

15. Apr 15, 2008

### matheinste

Hello phyti.

c is constsnt so if you regard it as a function It is a constant function of c.

Matheinste.

16. Apr 15, 2008

### Mentz114

'Motion relative to light' is not a definable thing. What you mean is that the magnitude of the time dilation includes c as a constant.