The Twin Paradox and the Equivalence Principle

haroldholt

I'm having a little trouble understanding the equivalence principle explanation of the twin paradox.

I understand that the resolution to the paradox according to the equivalence principle is that the non-traveling twin has a higher gravitational potential energy in the pseudo-gravitational field created when treating the traveling twin's reference frame as an inertial one, but I'm not sure that I understand why this is so.

Doesn't gravitational potential energy at a particular point depend on the strength of the gravitational field at that point? If the twins have different gravitational potential energies, does this mean that the gravitational field is a different strength for each of them? And if this is the case, where is the pseudo-gravitational field at its strongest? For a massive body, the gravitational field is the strongest at the centre of the body. But being a flat pseudo-gravitational field in the case of the twin paradox, there obviously is no centre.

Simon Bridge

The resolution of the paradox is that one twin is not in an inertial frame - so there is an asymmetry between them that has the travelling twin end up younger.

An accelerating frame is indistinguishable from gravity.
If you treat the traveler's frame as inertial, then you get the appearance of a gravitational field much as you get a centrifugal force when you go around a corner. This is not a "real" gravitational field like you are used to thinking about. Its there to make the math come out in agreement with observation and the assumption of which frame is inertial.

Imagine you are in a closed room that is lit by a bulb hanging by a cord from the ceiling. You observe that the bulb is hanging at an angle so you conclude that there must be some sort of force deflecting the bulb right?

This is provided the reference frame of the room is inertial ... the room could be accelerating.

haroldholt

I could probably simplify my question by asking the following: if the traveling observer has a lower gravitational potential energy and is therefore further down the "gravitational well", then where is the bottom of the well?

Taking the gravitational field of the earth, for example, the bottom of the gravitational well is the point at the centre of the mass.

Or is my question irrelevant since the gravitational field being created in the case of the twin paradox isn't real?

Mentz114

The twin scenario in flat spacetime can be resolved without gravity being introduced. Are you talking about twins in the earth's field or in flat spacetime ?

HallsofIvy

The equivalence principle says that acceleration is locally the same as an inertial frame with gravity. Since that is local the questions "where is the center" or "where is the bottom of the gravity well" do not arise.

haroldholt

If the equivalence principle only applies locally, how do we conclude that the non-traveling twin has a higher gravitational potential energy?

harrylin

A lower gravitational potential just corresponds to a lower height; gravitational potential energy at a particular point does not depend on the strength of the gravitational field at that point. But the equivalence principle explanation (at least the strong one, pretending a "real" field) isn't widely accepted.

- Original (and "strong") equivalence principle explanation:
https://en.wikisource.org/wiki/Dialo..._of_Relativity

- Physics FAQ commentary:
http://math.ucr.edu/home/baez/physic...x/twin_gr.html

Harald

haroldholt

Which explanation, from the point of view of the traveling twin, is the most widely accepted?

Simon Bridge

Just a side note:
You are aware that the choice of a zero potential is arbitrary?
If you put zero at infinity - then all potentials are negative and some more than others. OPs question then becomes - where is the smallest potential to be found?

One would expect to find this at the center of mass for the Universe right?
But the Universe does not have a center...

But this sort of question does not depend on the equivalence principle.

The starting point to understanding this would be that the infinite distance = zero potential is a convention only - not actually real. The question is actually about the structure of space-time.

Assuming that's where he was headed.

Mentz114

The Einstein argument is rather dated and the problem is best analysed ( as the Baez page suggests) using the proper length argument

http://math.ucr.edu/home/baez/physic...spacetime.html

harrylin

I think that the special relativistic explanation is the most widely accepted - thus perhaps not really "from the point of view of the traveling twin".

PeterDonis

Check out the Usenet Physics FAQ entry on the Twin Paradox:

http://math.ucr.edu/home/baez/physic...n_paradox.html

In particular, the "Equivalence Principle Analysis" page:

http://math.ucr.edu/home/baez/physic...x/twin_gr.html

PeterDonis

There is no center, but there is still a definite "up-down" direction, defined by the direction of the "force" you have to exert to stay stationary in the field. (Which is ultimately derived from the direction the traveling twin's rockets push him when he turns around.) The traveling twin is "below" the stay-at-home twin when he turns around because the direction from the traveling twin to the stay-at-home twin is "up".

haroldholt

When you say that the Einstein argument is rather dated, do you mean that the idea of treating the traveling twin's frame of reference as an inertial one is dated in general?

stevendaryl

I've always been slightly annoyed by the "gravitational time dilation" explanation for the twin paradox for two reasons.

First, it's kind of circular. You say that the accelerating twin is equivalent to a stationary twin in a gravitational field, and then you apply the formula for gravitational time dilation. But how do you know that clocks in a gravitational field experience gravitational time dilation? By transforming the problem to an equivalent case involving acceleration in flat spacetime, and using SR. You can skip the "gravitation" step and just use SR, plus calculus to transform to a noninertial coordinate system, if that's more convenient.

Second, it has bizarre consequences. Let me describe a variant of the twin paradox: There are two planets many light-years apart. Two twins were separated at birth and one went to live on one planet, and the other went to live on the other planet. The two planets are at rest relative to each other, and the twins are the same age, according to the reference frame of the planets. When one twin reaches age 20, he accelerates to nearly the speed of light. and travels to the other twin. When he arrives, he has only 21 years old, but the other twin is 40 years old. The traveling twin explains this using gravitational time dilation: during the time of acceleration, the distant twin aged rapidly, gaining at least 20 years during the brief time of acceleration.

What's wrong with that explanation? Well, imagine that traveling twin, after accelerating, changes his mind; he decelerates and returns to his own planet. If the whole process of accelerating and decelerating was quick (much less than one year) then the twins will still be approximately the same age. But what that means is that the distant twin gained 20 years during acceleration, and then lost 20 years during deceleration. Gravitational time dilation can work to youthen a distant twin as well as age him prematurely.

DaleSpam

The problem mentioned here isn't a problem of the gravitational explanation, but one of poorly specified non inertial coordinate systems. There is no one unique meaning to "the reference frame" of a non inertial observer. The most naive frame that you describe simply doesn't cover regions of spacetime sufficiently "low" for time to run backwards. You cannot use a chart to make claims about a region of spacetime that it doesn't cover. In any valid coordinate chart which does cover the other twin, his age will never go backwards.

lalbatros

It is possible to script another scenario where the "second twin" doesn't return home and just sends "somebody else" back and in such a way that there is no acceleration anywhere in the scenario.

In this way, the geometry appears to contain the full effect.