Einstein's space elevator and the twins paradox

[Enuma_Elish]

A PBS program said that the theory of general relativity holds acceleration and gravity observationally equivalent. (Einstein's space elevator example was given.) Here is a thought experiment based on that mental picture.

A "gravity switch" will make the following thought experiment easier to relate, but is not essential for it.

A set of twins is brought together at a place. One of the them is set on a spaceship which starts to accelerate at a constant rate. No sooner than the ship takes off, a gravity switch is thrown on and a gravity field is created in a way that it affects only the "stationary" twin.

If the strength of the spaceship's acceleration is identical to the gravitational field, then the twins will not only be speeding away at the same relative speed but also at the same "relative acceleration." That is because nobody can tell the difference between being on a spaceship accelerating away and being in a gravity field of equal strength.

As the spaceship traverses the (closed) universe and ends up near where it started from (place where the other twin has been living under the gravitational field), it decelerates. As it does so, the gravity field is turned off gradually. The twins are united once again.

Would they have aged differently?

 

[pervect]

A PBS program said that the theory of general relativity holds acceleration and gravity observationally equivalent. (Einstein's space elevator example was given.) Here is a thought experiment based on that mental picture.

A "gravity switch" will make the following thought experiment easier to relate, but is not essential for it.

A set of twins is brought together at a place. One of the them is set on a spaceship which starts to accelerate at a constant rate. No sooner than the ship takes off, a gravity switch is thrown on and a gravity field is created in a way that it affects only the "stationary" twin.

If the strength of the spaceship's acceleration is identical to the gravitational field, then the twins will not only be speeding away at the same relative speed but also at the same "relative acceleration." That is because nobody can tell the difference between being on a spaceship accelerating away and being in a gravity field of equal strength.

As the spaceship traverses the (closed) universe and ends up near where it started from (place where the other twin has been living under the gravitational field), it decelerates. As it does so, the gravity field is turned off gradually. The twins are united once again.

Would they have aged differently?

I'll offer a different example with a similar idea without a "gravity switch" or a closed universe which I hope will answer your underlying question.

Suppose we have someone called Able, reasonably far away from any large masses, in free fall.

Passing by Able at a good fraction of C is his friend, Bonny. Bonny is also in free fall. Bonnie's course takes her near a couple of black holes (the only object with a strong enough gravity to deflect her course at relativistic velocites), and after a long journey, winds up right next to her friend Able again, without ever turning on her rocket or accelerating, by traveling in a triangular path.

Both Able and Bonny have traveled geodesic paths, in the language of GR.

Able will still be older when they re-unite. To a reasonably good approximation if her trip is long compared to the time she spent near a black hole, Bonnies age will be Able's age divided by gamma, where

gamma= 1/sqrt(1-(v/c)^2).

The moral of the story is that it is the geometry of the path that's important, not the magnitude of any felt acceleration.

 

[Enuma_Elish]

How did Able know to age, and Bonnie knew not to, given that their relative speeds were identical and neither experienced any accelerational or gravitational interference (leaving out the angular momentum parts when Bonnie was making a turn)?

 

[Wizardsblade]

I am not an expert in this but from what I understand. In Pervect story, if you think about it for a bit, you will expect Bonnie to be younger bc without the black holes she would have beat able to the ground (given that her innitial velocity was greater.) The reason they meet each other is because the black holes curved space/time.

In Enuma's original example I believe that as long as the acceloration, and therefore geodesic paths, are the same they will both be the same age in the end.

 

[Enuma_Elish]

I think my question can be stated more concisely as: "with reference to which frame are the geodesics defined?"

Does either special or general relativity allow for a preferred frame of reference?

 

[JesseM]

I think my question can be stated more concisely as: "with reference to which frame are the geodesics defined?"

Does either special or general relativity allow for a preferred frame of reference?

I think general relativity does allow for some sort of global preferred reference frames, although no locally preferred frames (since locally GR must reduce to SR). That seems to be the resolution to the standard cosmological twin paradox where the twins are both moving inertially through a closed universe--this paper on the cosmological twin paradox says in the abstract that "The resolution relies on the selection of a preferred frame singled out by the topology of the space."

 

[Wizardsblade]

Is this thought of GR and SR correct? Think of the universe (space/time) as a deformed piece of glass with buldges, curves, and dips in it. Looking at and studying the shape of the glass is GR, but SR is like looking at an object through the glass. The object will look differnt (from distortions) if you look at the object from different parts of the glass.

Does this make since/work?

 

[JesseM]

Looking at and studying the shape of the glass is GR, but SR is like looking at an object through the glass. The object will look differnt (from distortions) if you look at the object from different parts of the glass.

It makes sense to say GR is like looking at the shape of the glass, but I don't understand how "looking at an object through the glass" makes sense as a metaphor for SR--how do we "look through" spacetime? I think a better metaphor would be that SR is like zooming in on a small region of the glass with a microscope, so that if the curvature of the glass is fairly smooth, any small region of it will look approximately flat when you zoom in on it. You could also say that SR could deal with a large sheet of glass that was completely flat--basically, SR deals with any situation in which you're looking at a region of spacetime where you can treat spacetime as flat within that region.

 

[pervect]

Geodesics are not dependent on the coordinates used to describe them - there is no "preferred frame" or coordinate system.

One of the defintions of the geodesic with a certain choice of sign conventions is the path which minimizes the Lorentz interval. This is the path that maximizes proper time (i.e. wristwatch time, the time an observer experiences).

This problem is interesting because both Able and Bonnie are following geodesics, yet they have different elapsed times. Yet we have just said that a geodesic path "minimizes L and maximizes time". How do we resolve this?

This appears at first glance to be a contradiction, but it is not. What is glossed over is the process of how one goes about minimizing the interval, and the initial conditions. Both Able and Bonnie are following a path that *locally* minimizes their Lorentz interval. The minimization process is only a local minimization (in calculus, it is like a point with zero slope on a curve) - it is not necessarily a global minimum.

A few examples using geometry would help explain this point. In geometry, a geodesic path is always the shortest distance between two points. Let's look at a geodesic on the Earth's surface - it would be a great circle.

Suppose I have an object 1 meter east of me. The great circle connecting me to that object is a geodesic, so the path that minimizes the distance is a path that starts out east, and follows a great circle. So far, so good.

But suppose I go west instead of going east? I am still following a geodesic (because a great circle is a geodesic) - but I have to walk all the way around the Earth to get to the point that is one meter east.

The moral of the story is that geodesics do not necessarily have the same length, even when they connect the same two points, in curved geometries.

Therfore, if one wants to get from point A to point B in a a manner which globally (as well as locally) minimizes the distance, one has to not only chose a geodesic, but chose the "right" geodesic. In other words, the shortest global path from A to B will always be a geodesic, but the fact that a specific geodesic connects A to B does not necessarily mean it has the shortest _global_ length.

Note that if we were on a flat plane, this ambiguity does not arise. The ambiguity arises only for geometries that are curved.

Thus we see that in flat space-time, the fact that a path is a geodesic is sufficient to guarantee that it is globally the shortest Lorentz interval (and longest elapsed proper time) - in curved space, time, we not only have to pick a geodesic path, but we have to pick the correct geodesic path to ensure that we have minimized the lorentz interval and maximzed the proper time (i.e to ensure we have aged the most).

We can illustrate the situation of Able and Bonnie with a Euclidean "distance-minimizing" problem. Imagine a huge sheet of mostly flat paper.

Able always stays in flat space-time, and goes directly from point A to point B.

A-------------------------B
|
|
|
|
|
C--------------------------D

Bonnie, however goes from A to C to D to B.

Around points C and D there are "pits" in the plane, which deflect Bonnies trajectory (as seen by us, viewing from above) - though to Bonny, she is always following a "distance-minimizing" path.

Bonnies trajectory is a "straight line" at all points, it is a geodesic, but like the person who went west to go east, circumnavigating the Earth, it is a longer geodesic than the direct route that Able took.

So the longest elapsed time is always achieved by taking the "most direct route". In the presence of curvature, taking the 'most direct route' is a non-trival problem.