Twin Paradox: Orbital Geodesics, Gravity, & Time Dilation

[D.S.Beyer]

TL;DR Summary

What is the result of the Twin Paradox if the round trip of one twin is entirely on a geodesic?

~ Shower Thoughts ~

Twin A is in a spaceship, Twin B is in a spaceship. Both in 'deep space'.
B follows a highly elliptical geodesic which goes around a planet (or black hole) with strong gravity, very far away.

When they meet again, who is younger and why?

I genuinely don't know what this answer will be.

( In an ideal case A stays in Flat Minkowski Spacetime during B's trip. However, A will need to accelerate a little bit away from the distant planet to maintain position. Does this small bit of acceleration may make the difference in the end..? dunno)

I'm going through Eigenchris's youtube videos, and he has an amazing explanation of the Twin Paradox, essentially saying the solution is "The reversed triangle inequality (assuming timeline vectors). In which geodesics have the longest proper time, and non-geodesics have shorter proper time."

That made sense to me in SR terms, so I thought I'd see if I could convolute it with some GR curvature. Since Twin B follows geodesics, and A must slightly accelerate... is A younger? Or is B effected by gravitational time dilation as it goes around the planet and thus B is younger?

Very curious. Thoughts?

 

[PAllen]

At first I thought you were referring to a different case, where e.g. A is hovering at a distance from the sun and B is on circular orbit meeting A every orbit. In this case, A will elapse more time between meeting even though B is geodesic and A is not. The seeming 'paradox' is resolved by noting there is a different geodesic between the meeting points that elapses more time than A: the geodesic that moves radially outward from A at an initial velocity such that it 'falls back' in exactly the time it takes B to orbit. This is the global maximum geodesic between these points.

Your case is different. If you mean for both to be geodesics, with A being very far a way in 'asymptotically flat' spacetime, while B does an eccentric orbit, then A would elapse the maximal time.

In general, with gravity, there may be two or more geodesics between a pair of events. One of them will be the global maximal elapsed time. The others will elapse less time than many non-geodesic world lines.

 

[Dale]

Summary:: What is the result of the Twin Paradox if the round trip of one twin is entirely on a geodesic?

B follows a highly elliptical geodesic

Any particular reason to choose an elliptical rather than circular orbit? It makes the math harder for no benefit that I can see.

 

[D.S.Beyer]

Any particular reason to choose an elliptical rather than circular orbit? It makes the math harder for no benefit that I can see.

I guess I was trying to get the meeting point even as far out into 'Minkowski' flat spacetime as I could, such that Twin A could basically be stationary. As PAllen says an 'asymptotically flat' spacetime.

But, I guess it really doesn't change the outcome in the end.

 

[D.S.Beyer]

In general, with gravity, there may be two or more geodesics between a pair of events. One of them will be the global maximal elapsed time. The others will elapse less time than many non-geodesic world lines.

This is great. Thank you.
How does one determine the maximal geodesic?
I'm not to keen on the hard math, so if you have a geometric example, I'd love to see it.

 

[PeterDonis]

Any particular reason to choose an elliptical rather than circular orbit?

Because that's what most naturally matches the standard twin paradox, where the traveling twin goes out for some distance, makes a 180 degree turn, and then comes back. For a massive body to produce that 180 degree turn, the traveling twin has to have a very low altitude of closest approach, compared with the total spatial distance traveled (where all the distances here are in the common rest frame of the stay at home twin and the massive body). In other words, the traveling twin's orbit has to be highly elliptical.

 

[PeterDonis]

Any particular reason to choose an elliptical rather than circular orbit?

Another reason is that if the orbit is circular, the stay at home twin won't be inertial--if the massive body's gravity is strong enough to produce a circular orbit at that distance for the traveling twin, it's strong enough to force the stay at home twin to "hover", with nonzero proper acceleration. That complicates the scenario.

 

[PAllen]

This is great. Thank you.
How does one determine the maximal geodesic?
I'm not to keen on the hard math, so if you have a geometric example, I'd love to see it.

I don't know of any general rule. In the case of Schwarzschild metric, you can just 'read' from the metric that a circular orbit will elapse less time than a non-inertial stationary observer at the same radius. Just a little harder is to demonstrate that the pure radial geodesic (out and back) is the global maximum between two events at the same radius and angular coordinates, separated by some time. However, even in this very simple metric, other cases of multiple geodesics (e.g. elliptical orbits of different shapes and positions that have a pair of event intersections) just have to be calculated. All you know is that the maximal geodesic will be a maximal proper time among all possible world lines.

 

[vanhees71]

Here is a concrete example in Schwarzschild spacetime:

https://www.physicsforums.com/insights/the-twin-paradox-for-freely-falling-observers/

 

[Dale]

Another reason is that if the orbit is circular, the stay at home twin won't be inertial--if the massive body's gravity is strong enough to produce a circular orbit at that distance for the traveling twin, it's strong enough to force the stay at home twin to "hover", with nonzero proper acceleration. That complicates the scenario.

A hovering clock is easy since all of the spatial derivatives are 0.

 

[PeterDonis]

A hovering clock is easy since all of the spatial derivatives are 0.

It's easy to compute its proper time, but it's at the same distance from the massive body as the traveling twin. I'm not sure that's what the OP intended. Although perhaps at this point we are considering several possible scenarios, not just one.

 

[PeterDonis]

Any particular reason to choose an elliptical rather than circular orbit? It makes the math harder

Actually, for the limiting case of eccentricity approaching 1, the math is easier, because for most of the traveling twin's trajectory, spacetime can be assumed to be flat and you just have the standard twin paradox. The only segment of the trajectory where spacetime curvature is significant is the turnaround, and for eccentricity approaching 1, the turnaround time is short enough compared to the total travel time that it can be ignored.

 

[Dale]

It's easy to compute its proper time, but it's at the same distance from the massive body as the traveling twin. I'm not sure that's what the OP intended. Although perhaps at this point we are considering several possible scenarios, not just one.

I would prefer to analyze the simplest possible scenario that captures the essence of the query. A circular orbit is simpler than an elliptical one. That is why I asked if the OP had a specific interest in the elliptical one. It doesn’t look like he had a specific reason for that, so I would rather use a circular orbit.

 

[PeterDonis]

I would prefer to analyze the simplest possible scenario that captures the essence of the query.

There can be different criteria of simplicity. In one sense, yes, a circular orbit is simpler because, if you're going to try and compute orbits at all, circular orbits are the simplest ones to compute. But in another sense, my suggestion of an elliptical trajectory in the limit as eccentricity approaches 1 is simpler, because it avoids the need to compute the orbit at all; you just have the standard twin paradox with a massive body instead of a rocket engine as the means of making the turnaround.

 

[Dale]

t avoids the need to compute the orbit at all; you just have the standard twin paradox with a massive body instead of a rocket engine as the means of making the turnaround.

Well, I agree in general, however, in this thread specifically I understood that the OP was interested in the fact that in curved spacetime there can be multiple geodesics of different lengths. (i.e. maximal geodesic query in post 5) If we simplify it as far as you suggest then there is only one geodesic for any worldline not including the very center. So I am not sure that answer's the OP's question. Again, I am looking for simplest possible that captures the essence of the query.

 

[PAllen]

Actually, for the limiting case of eccentricity approaching 1, the math is easier, because for most of the traveling twin's trajectory, spacetime can be assumed to be flat and you just have the standard twin paradox. The only segment of the trajectory where spacetime curvature is significant is the turnaround, and for eccentricity approaching 1, the turnaround time is short enough compared to the total travel time that it can be ignored.

Is this really true? The potential is undergoing a large change along the eccentric path, an effect not present in the SR twin version. Also, the speed is not constant along the path, unlike the SR twin case.

 

[PeterDonis]

The potential is undergoing a large change along the eccentric path, an effect not present in the SR twin version. Also, the speed is not constant along the path, unlike the SR twin case.

Hm, yes, if we insist on an elliptical trajectory, meaning a bound trajectory, you're right, it won't be the same as the SR case.

The case I was actually thinking of would be a hyperbolic trajectory with orbital energy much greater than escape energy.

 

[PeterDonis]

I understood that the OP was interested in the fact that in curved spacetime there can be multiple geodesics of different lengths. (i.e. maximal geodesic query in post 5) If we simplify it as far as you suggest then there is only one geodesic for any worldline not including the very center. So I am not sure that answer's the OP's question.

Yes, this is a fair point; the limiting case I was suggesting does not address this issue.

 

[PAllen]

Hm, yes, if we insist on an elliptical trajectory, meaning a bound trajectory, you're right, it won't be the same as the SR case.

The case I was actually thinking of would be a hyperbolic trajectory with orbital energy much greater than escape energy.

But, for an inertial slingshot trajectory, you still have both differences: potential change, and variable speed. You can simplify using constants of motion, but the calculation looks nothing like the SR case. Also, the hyperbolic path would never be closed except in a very artificial limit, so it is a stretch to call it a twin scenario. The highly eccentric ellipse participates in a true inertial twin scenario.

 

[PeterDonis]

for an inertial slingshot trajectory, you still have both differences: potential change, and variable speed

But for the limiting case I described (hyperbolic trajectory with orbital energy much greater than escape energy), both of these are negligible except during the turnaround, which is a short enough time relative to the total travel time that it can be ignored.

the hyperbolic path would never be closed except in a very artificial limit

Yes, agreed.

The highly eccentric ellipse participates in a true inertial twin scenario.

Not quite, since the stay at home twin still has some nonzero proper acceleration. But I agree that it can be made very small.

 

[PAllen]

But for the limiting case I described (hyperbolic trajectory with orbital energy much greater than escape energy), both of these are negligible except during the turnaround, which is a short enough time relative to the total travel time that it can be ignored.

I am still not convinced. The turnaround may be brief, getting very close to e.g. a BH or neutron star, but the change in potential cannot be made negligible, nor can the change in velocity. If the incoming velocity is huge, you will just have a small angle deflection rather than a turnaround. For the turnaround to occur at all, the change in velocity during approach and recession must be significant.

Not quite, since the stay at home twin still has some nonzero proper acceleration. But I agree that it can be made very small.

The home twin can do a radial out and back trajectory, which, very far from the massive body, can hardly be distinguished from stationary.

 

[PeterDonis]

If the incoming velocity is huge, you will just have a small angle deflection rather than a turnaround.

Yes, I had assumed that was what you were referring to when you said this:

the hyperbolic path would never be closed except in a very artificial limit

For the actual case I described, the path would be very, very far from closed, since, as you note, the angle deflection will be small, nowhere near 180 degrees.

 

[PeterDonis]

The home twin can do a radial out and back trajectory, which, very far from the massive body, can hardly be distinguished from stationary.

Yes, I agree this would be a reasonable approximation to stationary and would make the home twin inertial.

 

[meekerdb]

Any particular reason to choose an elliptical rather than circular orbit? It makes the math harder for no benefit that I can see.

It's so A can be in almost flat spacetime.

 

[D.S.Beyer]

Thanks for all the wonderful responses to this question.

It sounds like, no matter if the geodesic is hyperbolic, highly elliptical, or circular, Twin A will elapse more time.

One of the things that I wanted to get out of this question was if 'local acceleration' matters. Such that Twin B, being on a geodesic, should never feel like they are accelerating. Twin B is "falling" the whole time during the round trip.

@PAllen point to different geodesics between two events is something that I had not really considered before, and is extremely interesting. If there is no easy way to find the maximal geodesic, is there a way to find the minimal geodesic?

Could we say that the minimal geodesic between two points is the one which comes closest to a body (or bodies) of mass?

 

[Dale]

If there is no easy way to find the maximal geodesic, is there a way to find the minimal geodesic?

You just have to find all geodesics and then see which is the longest and which is the shortest. I don’t know of any shortcuts