The twin paradox and black holes

[BoraxZ]

Does the twin paradox hold around a black hole (or maybe less extreme gravitational fields)?

In a gravitational fields like that of the Earth it seems to apply. If two particles fall together, with synchronized clocks, and one of them rests on a platform for a while, after which it accelerates back to join the other, non-accelerated particle, that other particle, which travels on a geodesic, has traveled a small amount further in time, so upon meeting the first particle it is a bit younger than its twin particle.

But what if the particles fall together in a black hole? Will it apply as long as they are capable of meeting again?

 

[PeterDonis]

Does the twin paradox hold around a black hole (or maybe less extreme gravitational fields)?

I don't know what you mean. If you mean, is it possible to find pairs of timelike worldlines between the same two points that have different elapsed times along them, of course it is; you can do that in any spacetime.

 

[vanhees71]

One example is here:

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

 

[BoraxZ]

I don't know what you mean. If you mean, is it possible to find pairs of timelike worldlines between the same two points that have different elapsed times along them, of course it is; you can do that in any spacetime.

I meant what vanhees71 showed in his comment. Two particles falling together freely, after which one of them (or both) accelerates away from the other to meet up again later in. Can they meet up again behind the horizon?

 

[PeterDonis]

Two particles falling together freely, after which one of them (or both) accelerates away from the other to meet up again later in. Can they meet up again behind the horizon?

It depends on the specifics of the scenario. There is not one general rule that can be used.

 

[BoraxZ]

One example is here:

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

In the thread you link to, an observer (1) in free-fall on a circle meets an observer (2) moving freely radially upwards, after which they meet again when 1 has completed a circle. But in the twin paradox, two observers, initially at rest, accelerate away from each other and meet again in rest. The famous example is the observer returning to Earth, to meet an older twin (which in fact is an example in a curved spacetime, but that's ignored).

Is the general case both starting out in an arbitrary state of motion and meeting up in an arbitrary state? Can they compare clocks in that case?

 

[PeterDonis]

Can they compare clocks in that case?

Two observers who meet at an event in spacetime can compare clock readings at that event regardless of their relative velocity.

 

[BoraxZ]

Two observers who meet at an event in spacetime can compare clock readings at that event regardless of their relative velocity.

So they can see each other's proper time? Don't they see the clock of the other ticking slower?

Or are they that close so they can read each other's clocks?

Can they synchronized their clocks too when passing one another at a common event?

 

[PeterDonis]

So they can see each other's proper time?

They can see each other's clock readings when they meet (i.e., when they are at the same event in spacetime). That's true regardless of how they are moving.

Don't they see the clock of the other ticking slower?

I didn't say anything about the rates of their clocks, only about the readings of their clocks at the instant they meet. The latter is all you need to evaluate elapsed time for each of them between two successive meetings.

Or are they that close so they can read each other's clocks?

They're at the same event in spacetime, so yes. That's what "the same event" means.

Can they synchronized their clocks too when passing one another at a common event?

They could, yes, but that wouldn't change the elapsed time on each clock since their previous meeting.

 

[Ibix]

So they can see each other's proper time? Don't they see the clock of the other ticking slower?

Your proper time is the elapsed time on your clock. I just need a camera to record the value as you pass - rate is irrelevant.

 

[Ibix]

Can they synchronized their clocks too when passing one another at a common event?

They can reset their clocks to show the same time, but they won't keep time after that unless they come to rest with respect to each other.

 

[jbriggs444]

They can reset their clocks to show the same time, but they won't keep time after that unless they come to rest with respect to each other.

Whether they do or do not match rates after a meeting is a coordinate-relative statement. The rates may match or may fail to match depending on the selected standard of comparison.

Unless, as you point out, the two clocks remain stuck to one another.

 

[BoraxZ]

They can reset their clocks to show the same time, but they won't keep time after that unless they come to rest with respect to each other.

But when they return to the same event with a relative velocity, what time they see with a camera then?

 

[jbriggs444]

But when they return to the same event with a relative velocity, what time they see with a camera then?

Return to the same event?! Does this space-time include closed timelike curves?

 

[BoraxZ]

Return to the same event?! Does this space-time include closed timelike curves?

Ah yes. I should have said move on to a common event.

 

[Ibix]

But when they return to the same event with a relative velocity, what time they see with a camera then?

Depends on the paths taken.

You can do the twin paradox in a spaceship falling in to a black hole. Just have two clocks, one bolted to the floor and one on a rail so you can slide it back and forwards. You'll see less time elapsed on the rail clok each time it returns to the bolted down clock.

For a general path, you have some potentially nasty integration to do to predict the clock readings.

 

[PeterDonis]

when they return to the same event with a relative velocity, what time they see with a camera then?

The relative velocity will not affect the clock readings they see through the cameras.

 

[BoraxZ]

Depends on the paths taken.

You can do the twin paradox in a spaceship falling in to a black hole. Just have two clocks, one bolted to the floor and one on a rail so you can slide it back and forwards. You'll see less time elapsed on the rail clok each time it returns to the bolted down clock.

For a general path, you have some potentially nasty integration to do to predict the clock readings.

All right But you wrote:

"They can reset their clocks to show the same time, but they won't keep time after that unless they come to rest with respect to each other."

But if they pass each other at a common event with a relative velocity (after having left from another common event), they see each other's proper time that has elapsed since they departed. Which indeed is no problem...

So when they come to a full stop wrt each other at the second common point, the clocks just tick on at the same pace from the point in time they see at that moment on each other's clocks?

 

[Ibix]

So when they come to a full stop wrt each other at the second common point, the clocks just tick on at the same pace from the point in time they see at that moment on each other's clocks?

Yes, if I'm understanding your question correctly.

 

[BoraxZ]

The relative velocity will not affect the clock readings they see through the cameras.

Indeed. I confused pace (not space) with elapsed time.

 

[BoraxZ]

I read Feynman telling about the time passed for a freely falling object and an object brought to the same event in a forced way. Like letting it stay at some height for a while and then bring it down fast. He was mentally performing twin paradox experiments. The Hafele-Keating flight (costed them 7900$) with an atomic clock was a twin experiment too.

 

[BoraxZ]

I don't know what you mean. If you mean, is it possible to find pairs of timelike worldlines between the same two points that have different elapsed times along them, of course it is; you can do that in any spacetime.

When one twin stays behind, in the flat space scenario, and one twin departs for a trip through spacetime, then the twin that stays behind moves on a straight line connecting the two events. The other twin when they meet again, had traveled on a longer path.

Now if the length of the path rails the proper time, how is it possible that the twin that returns is younger that the other? Or is part of the tripping twin's path through space?

 

[jbriggs444]

When one twin stays behind, in the flat space scenario, and one twin departs for a trip through spacetime, then the twin that stays behind moves on a straight line connecting the two events. The other twin when they meet again, had traveled on a longer path.

To be clear, the twin that "stays behind" is coasting inertially the whole time, right? While the twin that "departs for a trip through spacetime" is thrusting hither and yon and eventually rejoining the sibling.

Now if the length of the path rails the proper time, how is it possible that the twin that returns is younger that the other? Or is part of the tripping twin's path through space?

One trajectory is "longer" than the other. That length is the elapsed proper time along the trajectory. In flat spacetime, the coasting trajectory maximizes elapsed time -- it is the "longer" trajectory.

You measure elapsed time on a path in four dimensional spacetime the same way you measure accumulated distance on a road in three dimensional space -- you take a path integral of the incremental displacement. You get different elapsed times for the same reason that the odometer reading when you arrive in Chicago depends on the road you took from New York.

 

[PeterDonis]

When one twin stays behind, in the flat space scenario, and one twin departs for a trip through spacetime, then the twin that stays behind moves on a straight line connecting the two events. The other twin when they meet again, had traveled on a longer path.

No, the other twin has traveled on a shorter path. A timelike straight line in Minkowski spacetime is the longest path between two points, not the shortest. Minkowski geometry is not the same as Euclidean geometry.

 

[Nugatory]

Now if the length of the path rails the proper time, how is it possible that the twin that returns is younger that the other?

Spacetime is non-Euclidean - the squared “distance” between two points is $\Delta T^2-\Delta X^2$, not the $\Delta T^2+\Delta X^2$ that we get by applying the Pythagorean theorem to the spacetime diagram we’ve drawn on a sheet of graph paper.

Say the journey lasts two earth years, traveller flies out at .8c, turns around and returns at .8c, gets back two earth years later. Working with coordinates in which the earth is at rest, for each leg $\Delta X$ is .8 light years, $\Delta T$ is one year, the proper time along each leg is .6 years, total proper time along the round trip path is 1.2 years. But for stay at home two years have passed.

 

[BoraxZ]

No, the other twin has traveled on a shorter path. A timelike straight line in Minkowski spacetime is the longest path between two points, not the shortest. Minkowski geometry is not the same as Euclidean geometry.

So a geodesic can actually be shorter than a non-geodesic?

 

[PeterDonis]

So a geodesic can actually be shorter than a non-geodesic?

I said no such thing. The stay at home twin's path, the straight line, is a timelike geodesic. The traveling twin's path, the shorter path, is not a geodesic.

 

[BoraxZ]

You measure elapsed time on a path in four dimensional spacetime the same way you measure accumulated distance on a road in three dimensional space -- you take a path integral of the incremental displacement. You get different elapsed times for the same reason that the odometer reading when you arrive in Chicago depends on the road you took from New York.

Yes. But if the elapsed time on the path of the twin that goes on a trip is more than the twin that stays behind, isn't that twin older than the twin that stays behind, hile it should be younger upon return?

 

[BoraxZ]

I said no such thing. The stay at home twin's path, the straight line, is a timelike geodesic. The traveling twin's path, the shorter path, is not a geodesic.

But isn't a non-geodesic path longer than a geodesic path? Or is that only in Euclidesn space?

If the path of the traveling twin has a longer proper time than the twin staying behind, why then is he younger when he returns?

 

[PeterDonis]

if the elapsed time on the path of the twin that goes on a trip is more than the twin that stays behind

It isn't. It's smaller.

isn't a non-geodesic path longer than a geodesic path?

No.

Or is that only in Euclidesn space?

Yes. Go read post #25 by @Nugatory.

If the path of the traveling twin has a longer proper time than the twin staying behind

It doesn't. It's shorter. I've already said that in post #24. You need to pay attention.

 

[PeterDonis]

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[PeterDonis]

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