# Twin Paradox In Empty Space

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1. Nov 12, 2014

### dman12

Whenever the twin paradox in GR seems to be discussed, it always seems to be done in the presence of a large mass such that the twins can be considered as test particles moving in some metric.

I was wondering whether the same problem could be generalised and be proposed in completely empty space in which the twins are the only things distorting the MInkowski metric? Say the twins started together at the same point in space time and then one of them did some arbitrary motion is there a way to analyse their relative proper times? It seems that the complication comes in as for one of the twins to move, they must emit matter of some kind which then distorts the metric itself.

I thought perhaps you could assume that around each of the twins was a Schwarzschild metric and that you can treat the moving twin as a Kinnersley photon rocket and then stitch together the Kinnersley solution with the two Schwarzschilds to give the overall metric. Then integrate this to give proper times for both twins. Would this work and does anyone know of any work that has been done similar to this?

2. Nov 12, 2014

### phinds

You have a misunderstanding. Gravitational fields have absolutely nothing to do with the Twin Paradox and if two twins were to start off stationary relative to each other and one were to not accelerate and the other were to accelerate and then return, you would have exactly the twin paradox you read about. Inertial motion is relative, acceleration is not.

3. Nov 12, 2014

### Staff: Mentor

I've seen very few discussions of "the twin paradox in GR"; virtually all of the discussions I've seen are set in flat spacetime. The twins themselves are assumed to have negligible gravity.

I have seen some discussion of "twin paradox" scenarios in curved spacetime; for example, one twin stays at the starting point, floating in empty space, and the other travels inertially (i.e., zero proper acceleration) to a star, "slingshots" around the star (still in free fall, zero proper acceleration), and comes back to the starting point. The presence of the star allows the traveling twin to reverse his direction of motion without any proper acceleration. But the final result is similar to the flat spacetime case: the traveling twin ends up younger when the two meet up again. The twins are also assumed to have negligible gravity themselves in this scenario.

I am not aware of any work that assigns non-negligible gravity to the twins themselves. I'm not sure why you would want to (not that it is physically unreasonable, just that I don't see what additional enlightenment you would gain from doing it, over and above analyzing the scenarios I described above).

4. Nov 12, 2014

### pervect

Staff Emeritus
As others have mentioned, it's not necessary to consider the presence of some large mass, and I don't recall offhand many treatments that do consider a large mass to be present.

Yes, you do need a metric in GR, but you can have a metric for empty space, This flat, empty-space metric is in fact the Minkowski metric you refer to later. I am interpreting your post as you thinking you need to have a large mass to have a metric - this is not the case.

5. Nov 13, 2014

### jerromyjon

Let's forget the people since observation and memory are not reliable. Instead we have video recorders with clocks labeled A and B which everyone interested can view and analyze after the experiment. I would also like to eliminate the "thought" and make this a truly realistic scenario if at all possible. I don't think this setup can get any simpler:

The distance from A to B never changes. The distance from A to P=B to P and we assume the pulse reaches A and B simultaneously. A's velocity through space at .6c causes A to slow down in time, literally everything in this non-inertial frame evolves slower than B and everything around B is normal except for A. These cameras record the video of the pulsar providing a moment of simultaneity for both to sync at :00 and for the next pulse which video B stamps the time :05 and video A stamps the time :04.

This seems like a logical setup and we all can agree there is no paradox, so the paradox occurs when A records B's time and B records A's time?

6. Nov 13, 2014

### JesseM

A is not moving inertially so relativity does not predict symmetric time dilation with B in this scenario--when A gets clock signals from B, he sees B's clock running faster than his own, not slower, whereas B sees A's clock running slower. A also sees the pulses from the pulsar happening at a faster rate than B does.

7. Nov 13, 2014

### Staff: Mentor

This is impossible by simple geometry. (I'm assuming that you intend A to be moving in a circle around B.)

Simultaneously according to whom? Simultaneity is relative.

No, it doesn't. Only A "evolves slower" than B, because only A is moving. A's motion can't make anything else evolve slower, nor can A make anything else evolve slower by adopting a particular set of coordinates.

No, they don't, because the cameras are never co-located in space; they are always spatially separated. That means relativity of simultaneity is always present; the two cameras will never agree on a notion of simultaneity, at least not if they both adopt the "natural" notion of simultaneity for their state of motion, which is what you appear to be assuming.

(Note: in the other thread where we are discussing a similar scenario, I had assumed you intended A to be moving in a straight line back and forth, passing B during each straight-line segment of motion. In that case, when A and B pass each other and are co-located, they can use the pulsar signals as a common reference to compare their elapsed times. But this only works when A and B are at the same point in space.)

8. Nov 13, 2014

### jerromyjon

So now we have an omni-directional equidistant pulse where A and B sync at t=0. A's trip around P is 1000 pulses in which t=4000 for A and t=5000 for B. This obviously complicates it slightly because A's spacetime velocity is not identical to B's relative motion away from A's unique frame of rest. Would it be acceptable to break this trip into 3 phases depicted by red, green and blue?

My only goal here is to isolate and comprehend where the paradox occurs and how relativity explains it.

9. Nov 13, 2014

### jerromyjon

B is obviously not in the plane of A's circular path but they can share a spherical distance from the center of P.

10. Nov 13, 2014

### jerromyjon

I'm gripping very tightly to this as a given.

How close do they need to get and how slow do they have to be in relative velocity?

11. Nov 13, 2014

### phinds

Speed is irrelevant. They have to be at exactly at the same place at exactly the same time but can be speeding past each other. This is a single "event" in space-time.

12. Nov 13, 2014

### jerromyjon

So my latest diagram satisfies these requirements so far? A passes as close as technologically possible to B at which point it just so happens P pulses at that instant and everyone is satisfied that t=0 at this coincidental moment in time. A completes one orbit around P at .6c and again passes very close to B. In the second coincidental event both A and B agree that P pulsed 1000 times. A reports t=4000 and B reports t=5000 and there is no paradox because whatever occurred on the departure between A and B was counteracted on the approach between A and B. Does this make sense?

13. Nov 13, 2014

### Staff: Mentor

Ok, that clarifies what you meant. But my remarks still apply (at least, I think they do--see below), because A and B are never spatially co-located in this scenario.

However, I'm still confused because this does not look like what you described in post #8. Was that supposed to be a different scenario? If so, it would really help to stick to one scenario at a time.

I'll assume this is a different scenario from the previous one (see above), because here it looks like A and B are spatially co-located, every time A returns around to B's position on the circle. If so, you are correct that both A and B will count the same number of pulses (1000) between two successive times that they meet, and that B's clock will record more elapsed time than A's clock (5000 vs. 4000) for that number of pulses. This is the invariant sense in which A's clock "runs slower" than B's.

The general explanation in relativity is that different paths through spacetime between the same two events can have different lengths. In the second example above, A and B take different paths through spacetime between the same two events (two successive instants at which they are spatially co-located), and A's path is shorter than B's, so A records less elapsed time. It's just geometry.

There is no single place "where the paradox occurs", because the two different path lengths between the two events are global properties of the paths, not local properties of some particular point on the paths. Asking "where the paradox occurs" is like asking "where" the difference in distance is between two different routes from New York to Los Angeles. It's not at any particular point on the routes; it's a global property of the two routes as a whole.

It depends on how accurate the measurements are. Atomic clocks today can detect time dilation factors on the order of one part in $10^{14}$, which corresponds to relative velocities on the order of tens of meters per second (about the speed of a car on a freeway). How close together the objects would need to be depends on how short the smallest time interval of interest is.

14. Nov 13, 2014

### jerromyjon

PeterDonis, I'm good with all that. I'm trying to back the paradox into a round corner and the latest scenario I believe eliminates irrelevant factors.

What happens from t=0 to t=5 (B's clock). A records the next pulse at t=4. B records the next pulse at t=5. B accurately predicts that A received the next pulse at A's time of t=4, but as the paradox goes, A should have some indicator that B's t=3.2?

15. Nov 13, 2014

### jerromyjon

Just on a side note, if both A and B are green (or whatever you want to call the center of optical frequency range) B should see A as red shifted to red at high enough relative velocity and A should see B disappear into infrared with relativistic doppler added to that, opposite upon approach into violet. A reaches halfway around and B sees no shift at t=2500 because relative v=0 (remove pulsar for line of sight) but A still sees some redshift because the relativistic doppler light cone moves forward where perpendicular to travel is in the redder zone "behind" A. I also realize A is not where B sees it by the time the light travels A is much further along its course.

16. Nov 13, 2014

### Staff: Mentor

I strongly advise you to take my repeated hints and think about this in terms of spacetime geometry. Imagine a "stack" of circles like the one you drew; each one represents the space you are describing at an instant of Bob's time. (Why Bob's time? See below.) Stacking all of them together creates a diagram representing the geometry of spacetime, and the worldlines of Alice and Bob as curves in that geometry. It should be easy to see that Bob's worldline is just a vertical line, while Alice's worldline is a helix, winding around the cylinder, and crossing Bob's every time Alice passes Bob. The pulsar is at the center of the circle, and its worldline is just a vertical line as well (or a vertical tube, if you want to take into account the pulsar's size, but that's not needed for this discussion).

Now, consider two events, E1 and E2, representing two successive meetings of Alice and Bob. Each of these events is a point in spacetime, where the two worldlines cross. In between the two, Bob and Alice each follow a curve--again, Bob's is just a vertical line segment between E1 and E2, while Alice's is a helical segment that winds once around the cylinder, starting at E1 and coming back around to E2. Then the length of Bob's segment is 5, and the length of Alice's segment is 4. These are invariant geometric facts; they don't depend on whose frame we pick to describe them.

Now, what about Alice thinking that Bob's clock should be running slower, because in her (non-inertial) rest frame, he is moving and she is not? The key point is this: in order to draw this conclusion, Alice has to look at the length of Bob's worldline between a different pair of events, not between E1 and E2. It's harder to visualize which pair of events this would be, because Alice's motion is circular instead of in a straight line, but the basic reason is, again, geometry: if Alice thinks that Bob's clock advanced by only 3.2 "at the same time" as her clock advanced by 4, she must be comparing the length of her worldline with the length of his worldline between some different pair of events, because we know that the length of Bob's worldline between E1 and E2 is 5, not 3.2. (That "at the same time" should also be a cue that relativity of simultaneity is involved.)

So the resolution of the "paradox" is, again, geometry: if Alice thinks Bob's clock is "running slower", it's because she is comparing lengths of worldlines between different pairs of events than the pair (E1, E2), which clearly shows Alice's clock running slower. But the events E1 and E2 are not picked out arbitrarily: they are picked out by the geometric fact that the two worldlines cross at those events. (This is another way of saying that Alice and Bob meet--are co-located--at those events; it should be clearer now why I was focusing on that as being important.) So there is a good reason why we use these events to judge whose clock is running slower.

Picking out the worldline crossings also helps to explain why we used instants of Bob's time to define the "slices" of space at an instant that we then "stacked up" to form our diagram of spacetime: this is what makes the scenario look the simplest, because only one object (Alice) moves. Any other "slicing" would make things look more complicated. Other slicings are still valid, and you could try using them to draw a diagram, for example, in which Alice's worldline was a vertical line. But the geometric facts would remain the same, just as the geometric facts about distances on the Earth's surface remain the same whether we look at a Mercator projection or a stereographic projection.

17. Nov 13, 2014

### Staff: Mentor

No, this is not correct. In relativity, there is a transverse Doppler effect as well as a Doppler due to motion along the line of sight. So B will still see a shift in A's light signals when A is on the opposite side of the pulsar.

18. Nov 13, 2014

### jerromyjon

Is that analogous to frame dragging?

19. Nov 13, 2014

### Staff: Mentor

No. Frame dragging occurs in the gravitational field of a rotating body. The transverse Doppler effect is part of the relativistic Doppler effect, and occurs whenever a light source has a component of motion perpendicular to the line of sight of the observer. In Newtonian physics, this component does not produce any Doppler effect, but in relativity, it does.

20. Nov 14, 2014

### jerromyjon

Gravitoelectromagnetism is the word of the day.

That seems like it has to be what I'm missing.

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