# Can you demonstrate the Twin Paradox using a Schwarzschild spacetime?

Short Version:

If we want to solve a problem where a particle is forced off a geodesic, can we obtain identical results if we instead solve the problem where the particle is in a different spacetime but stays on a geodesic?

Long Version (The Thought Experiment):

Twin A is floating stationary in space. Twin B is in a high-speed rocket which passes by twin A. At that moment, twin A and twin B synchronise their clocks.

Then, consider two different situations:

1. Space is globally flat Minkowski. Twin B coasts through flat space, using their thrusters to slowly accelerate before using them to turn around and travel back home to twin A.

2. Space is globally Schwarzschild, but twin A is situated very far away from the central black hole where space is, to good approximation, flat. Twin B immediately turns off their thrusters and coasts along their geodesic with no noticable acceleration in their frame. The Geodesic happens to take them close to the black hole, completely curving around the black hole and turning back on itself, back towards twin A (eventually passing by them again).

Figure 1: a suggestion of a geodesic around a Schwarzschild black hole which turns back on itself (not to scale).

When the twins pass by the second time, we find that twin A's clock has avanced more time than twin B's clock. That is, twin A has aged more than twin B.

I know this is true in case (1) because it is the typical description of the twin paradox, which has a well-known answer (although I don't know how to explicitly calculate it).

I know this is true in case (2) because twin A and twin B are both just geodesics in a schwarzchild spacetime, so I was able to numerically integrate them and sure enough, I found that $$\tau_B < \tau_A$$.

My question is: Are situation (1) and (2) equivalent when viewed from twin A's frame?

Why I think they are equivalent:

• The Einstein Equivalence Principle states that acceleration and curvature are equivalent. A description that uses curvature (geodesics) to describe a particle's trajectory should be the same as a description that uses acceleration.

Why I'm not sure:

• Twin B feels a force of acceleration in case (1) but doesn't feel anything in case (2).
• In case (2), both twins A and B stay on geodesics forever. This is more akin to time dialation in Special Relativity which is happens when two particles are on different geodesics - Perhaps the time dialation we're seeing in (2) is more related to the SR type of time dialation which is presumably different from the type in (1)?
• If twin A is in geodesic motion, he'd not be at a fixed coordinate, say, with respect to an asymptomatic observer. From your description, it looks like you're assuming otherwise. – Dvij D.C. May 28 '20 at 13:39
• @DvijD.C. I'm taking A to be so far away that their geodesic essentially stays fixed in space. Of course, it isn't literally fixed, and will drift very very slightly but this will be so negligable that I can consider the approximation "coordinate time = observer A's proper time". You could imagine taking the limit where A is infinitely far away and has 0 motion. – Adam Kiddle May 28 '20 at 13:55
• Yes, that's what I thought but then it'd take infinitely large amount of proper time of $A$ until $B$ comes back. – Dvij D.C. May 28 '20 at 14:00
• The Einstein Equivalence Principle does not say that acceleration and curvature are equivalent. In curved spacetime, you can get a twin paradox even if both twins remain in free-fall between meetings (as in your scenario 2, if I understood it correctly), which cannot happen in Minkowski spacetime. Are you really asking if they're equivalent, or just asking if a twin paradox is possible in curved spacetime if both twins remain in free-fall? If the latter, then this is related: Twin paradox with one twin in orbit, one in radial free-fall – Chiral Anomaly May 29 '20 at 2:01

Are situation (1) and (2) equivalent when viewed from twin A's frame?

No. The equivalence principle says that the outcome of any local experiment is the same if you are at rest in a gravitational field or accelerating uniformly. Situation 1 and 2 are therefore not equivalent.

First, the experiment is not a local one per the meaning used in the equivalence principle. In the equivalence principle “local” means that it is conducted over a small enough region of spacetime that tidal gravity is negligible. In other words, the gravitational field should be uniform and the spacetime should be flat to within the precision of all measurement devices. In this scenario the gravitational field changes substantially during the slingshot. So it violates the locality condition of the equivalence principle.

Second, since the equivalence principle requires that all experimental outcomes are the same. This includes measuring the readings on accelerometers. As you said:

Twin B feels a force of acceleration in case (1) but doesn't feel anything in case (2).

Which means that an accelerometer reading will be non-zero in case (1) and will be zero in case (2). This is an experimental difference which shows the non-equivalence. Personally, I find thinking about accelerometers to be very useful in figuring out what should be equivalent to what in the equivalence principle.

Suppose that we modify case 2 as follows: the twins are in a small laboratory (small enough that there are no tidal effects) where the whole laboratory is near the gravitating body and is kept stationary with respect to standard Schwarzschild coordinates by means of thrusters. Twin B sits in a chair in the lab while twin A jumps. The twins synchronize their clocks at the moment that A jumps off the floor and compare them when A lands.

Case (1) would be modified as follows. In Case (1) we will locate the lab in flat spacetime far from any gravitating source, but we will use the same lab and the same twins and the same chair etc. In particular, the thrusters will be firing at the exact same rate as before and A will jump with the exact same force as before.

All experimental results will be the same in the modified cases. Twin B will measure the same non-zero accelerometer reading in both cases and twin A’s accelerometer will read 0 during the jump. Also, the clocks carried by A and B will read the same. Since we can easily calculate the clocks in A’s frame in case (1) we can use those results to exactly predict the clocks in case (2)

• I think I follow what you're saying. An experiment that observer A could do tell the difference between case (1) and (2) would be strapping an accelerometer to twin B and watching the reading throughout twin B's flight (or asking twin B when they got back home if they felt any acceleration). In that case, situation (1) and (2) truly are inequivalent. But is the time dilation the same between the two situations? If it is, there is some value in treating situation (1) as situation (2) since it gives us a way to explicitly calculate proper time of twins A and B. – Adam Kiddle May 29 '20 at 14:27
• @AdamKiddle The amount of time dilation would depend on the mass of the gravitating object. I suspect that for any given flat spacetime scenario you could find a mass that would work, but I haven’t tried to prove it – Dale May 29 '20 at 22:03