1. Dec 2, 2014

### Fantasist

I am surprised this has apparently not been discussed here as such in this forum (nor anywhere else on the web). The point is that in the classical space-ship twin paradox, not only the traveling twin will be subject to a proper acceleration, but, according to Einstein's equivalence principle, also the twin staying on the earth, because of the gravitational field of the latter. If the space traveler is accelerated with 1g (assume he starts from an earth orbit), it should not be too difficult to construct a scenario where both the accelerometers on the space ship and the earth will measure exactly the same time dependence of the proper acceleration according to their respective clocks. Even the turnaround of the traveler could be replicated by the earth twin (by moving to the other side of the earth). On the basis of the accelerometer measurements, it should thus not be possible to single out on twin from the other because of the equivalence principle. Does this imply that for this scenario there would be no age difference when the traveling twin returns?

2. Dec 2, 2014

### A.T.

You can create a scenario with symmetrical proper acceleration profiles in flat space-time . No need to involve gravity for this. In flat space-time symmetrical proper acceleration profiles will result in identical proper-times between the meetings.

In curved space-time symmetrical proper acceleration does not imply identical proper-times, because you have gravitational time
dilation too. You can even construct the opposite outcome of the classical flat space-time twin paradox, where the accelerated twin ages more than the inertial twin.

3. Dec 2, 2014

### Fantasist

That would be missing the point here, namely that only one twin is traveling whilst the other stays at home.

How does this go together with the equivalence principle? Should a proper acceleration of 1g not have the same effect on the proper time regardless of the nature of the proper acceleration?

4. Dec 2, 2014

### Staff: Mentor

In general relativity the amount of time on a clock is given by $\int_P d\tau$ where the P denotes integration along the worldline, P, of the clock. This formula works regardless of the spacetime curvature and regardless of the proper acceleration.

5. Dec 2, 2014

### PAllen

Gravitational time dilation is not proportional to acceleration. To first order, it is proportional to potential difference, for which distance is a factor. To apply the principle of equivalence, you need to set up equivalent situations (and the principle of equivalence is primarily local, and not exact). Thus you can say that the gravitational time dilation between the front and back of a rocket of height h, whose back is accelerating 1 g, is the same as the gravitational time dilation between static objects that same distance apart in a gravitational field that such that the acceleration of the bottom object is 1 g and the gradient of acceleration is the same as for the rocket's back to front.

Both special and general relativity state that acceleration per se has no affect at all on the passage of proper time.

6. Dec 2, 2014

### pervect

Staff Emeritus
The issue of the basics of how to handle gravity is an interesting one, but I would say that the best time of presentation comes well after learning and appreciating the solution to the twin paradox.

Formally, one introduces the concept of the Lorentz interval and how to use it to calculate proper time. If one know the particles position as a function of time, i.e one has x(t) y(t) and z(t), by using the formula $d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$ , solving for $d\tau$ (which is just $d\tau = \sqrt{1 - (dx/dt)^2 - (dy/dt)^2 - (dz/dt)^2} dt = \sqrt{1 - v^2} dt$ and integrating, one can come up with an expression for the amount of elapsed proper time.

One will certainly find discussions on the internet if one looks for this very well-known formula.

To handle GR, one points out that in General Relativity we replace the simple Minkowskii metric $d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$ with some other metric, with a single large mass we can use the Schwarzschild metric, for instance.

This provides an answer to the problem, for instance on the Earth we can hold all the position variables constant (dr, $d\theta$, $d\phi$ if we use Schwarzschild coordinates) and focus on the coefficient for $dt$. By this means we can put an upper bound on the diffference between the observer who stays in space well away from any gravitational fields, and one who spends his whole time on Earth, and demonstrate that it's tiny.

There's a bit more insight to be gained if one asks the question "What is the curve of longest proper time that starts on the surface of the Earth, and ends there" as opposed to the easier question of the observer "at infinity" wel away from any gravitational fields. Feynman in particular , has essays which ask(and answer) this question - probably his published lectures, though I can't say which lecture. I believe I actually read a not-very complete version in another popular work of his. Anyway - I digress.

The results are interesting and can provide some important physical insights. That solution is just the natural motion of a body that you throw up into the air. If the period of elapsed time is short enough, this solution is unique, and any body that moves naturally will maximize the proper time. If the period of elapsed time is long enough to allow a orbit, the solution to natural body motion is not unique (you have orbiting and non-orbiting solutions), and only the non-orbiting solutions maximize the proper time. The physical insights are put to best use when one has some (possibly minor) familiarity with Lagrangian theory and/or Hamilton's principle, both of which exist for Newtonian mechanics as well as relativit.

7. Dec 2, 2014

8. Dec 2, 2014

### Fantasist

OK, but then the equivalence principle would be violated: you would be able to tell from a purely local measurement (just assume the accelerometer/clock as small enough) whether the proper acceleration measured is due to gravity or a real (coordinate) acceleration.

Just to give Einstein's statement of the equivalence principle again:
"we [...] assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system." (Einstein 1907).

Last edited: Dec 2, 2014
9. Dec 2, 2014

### A.T.

Gravity doesn't cause any proper acceleration

The equivalence applies only locally. It says nothing about proper times accumulated along non-local worldliness in curved space time.

10. Dec 2, 2014

### Staff: Mentor

Don't forget that local in spacetime includes small distances in space as well as small durations in time.

Here is how the equivalence principle works in the twins scenario. Suppose that you have two pairs of clocks. One pair is on a rocket accelerating at 1 g in flat spacetime and the other is at rest on the earth. For each pair, one is thrown a small distance "up" and the other remains fixed to the rocket or ground. The time on each clock is measured from when the thrown clock leaves to when it returns. In both cases the thrown clock reads more time than the fixed clock. This is just as required by the equivalence principle.

Last edited: Dec 2, 2014
11. Dec 2, 2014

### PAllen

Note some implicit caveats. "corresponding acceleration of the reference system". This hides a lot, because except for tiny region of space, and a very short time, a gravitational field has no corresponding acceleration of reference frame, and vice versa (an accelerated reference system of any generality, duration, and 'size' has no corresponding gravitational field).

You also didn't take note of the fact that even in a gravitational field, it isn't the gee force of static observers (to put it in Newtonian terms) that produces different accumulation of time, it is change in potential. You could have two observers each experiencing exactly 1 g in, say, the +x direction. The a clock on the one further in the -x direction would be seen to be running slower than the other, and both observers would agree on this. The greater the distance between them, the greater the clock rate difference would be. [Note, if they both had exactly one g acceleration, each would also find the distance between them growing over time. See the Bell Spaceship paradox in our FAQ for more information.]

Basically, you need a better understanding of the equivalence principle, and also correct your misunderstanding that acceleration per se has anything to do with time dilation (in gravity or rockets).

12. Dec 4, 2014

### Fantasist

Thanks for your replies. There appears to have been a bit of a misunderstanding though: I was not trying to involve GR here, but merely the effect of the gravitational field on the usual twin paradox in SR. As has been frequently stated in other threads here before, the clock difference after the return is due to the fact that the traveling twin is not all the time in an inertial frame (i.e. an accelerometer would measure a proper acceleration different from zero), whereas the stay-at-home twin is in an inertial frame. However, if the stay-at-home twin is actually on the earth's surface, he is in reality not in an inertial frame either (as an accelerometer would confirm). So my question was, if both twins subject to the same proper acceleration over the whole period of time in this sense, would there then still be a clock difference on return (ignoring the GR effect), or would it be equivalent to the case of both twins actually traveling in a symmetrical manner (where no clock difference would be predicted)?

13. Dec 4, 2014

### PAllen

It seems to me you are contradicting yourself. If you ask about the principle of equivalence you are asking about GR, not SR, period. In SR, there is no principle of equivalence, and to the (limited) extent you can treat gravity as an ordinary force in SR, it is ambiguous what the 'charge' should be (energy? invariant mass?).

Note, that in SR, an accelerometer reading correlates to what is an inertial frame only in the absence of gravity. In SR, a free fall frame is an accelerating frame (but there is no really good way to treat gravity in SR), while an earth surface observer is an inertial frame. When an earth observer is treated as inertial SR, you are assuming that gravity can be completely ignored - it's effects are insignificant compared to pure SR effects.

You seem to have completely ignored my post which gave a pure SR example of why acceleration has nothing to do (per se) with time dilation. We have two rockets acceleration 1 g in the same direction, yet SR alone (and GR doesn't change this) predicts that both rockets agree that the forward rocket's clock is going faster than the rearward rocket's clock.

Nutshell: neither SR nor GR, at all, says there is contribution to time dilation directly related to acceleration.

14. Dec 4, 2014

### pervect

Staff Emeritus
Hopefully you now understand that if you don't want to involve GR, you need to use some force other than gravity? I'm not sure if we've made that point clear enough yet from your question.

The idea that gravity is "just a force" is a Newtonian idea. But in GR, gravity causes time dilation. No other "force" does this. To properly model gravity, one needs to consider curved space-time. One can't just "pick and choose" to include parts of GR, and ignore other parts, if one sets up a problem involving curved space time, and expects consistent results, one needs the necessary math. Fortunately, the absolute minimal amount of math that one needs to calculate the time dilation with GR isn't too onerous, it is just the equation I presented in post #6 in this thread.

To answer the question in detail, one needs to have the metric of space-time, then apply the formula I gave in post #6. If you just want the answer, we can tell you that having the same magnitude of proper acceleration isn't any guarantee that the proper times will be close. There will be three paths with three proper times (well, in the scneario I envision at least). The first is a true maximum of proper time. It is necessary, but not sufficient, that the path that this observer follows to get the true maximum experiences no proper acceleration. The second path is proper time of someone sitting on the planet, and the proper time of someone who lifts off from the planet, flies around, and returns. The first proper time will be the true maximum by definition. The second proper time will be close to the first for a typical planet). The third proper time can be very much shorter than the second.

I guess there is one more point to be made. That is that curved space-time is introduced when the non-flat metric is introduced. If you have an observer accelerating in flat space-time, you can still use the flat space-time metric form of the equation I gave in post #6 if you use inertial coordinates. You can choose to use non-inertial coordinates if you wish. You still don't have actual curvature in this case, but the mathematical form of the equations is closer to the form you'll need for full GR when you use the non-inertial coordinates. If you have actual curvature though , due to the presence of a massive body, you have no choice but to use the metric of said space-time, which cannot be reduced to a flat metric, similar to the way it's impoosible to make an accurate flat map of the curved surface of the Earth.

Last edited: Dec 4, 2014
15. Dec 4, 2014

### Staff: Mentor

You cannot do that. If you have a non-uniform gravitational field (curved spacetime) then you need to use GR.

In post 10 I gave a variant of the twins paradox in a uniform gravitational field (flat spacetime) where you could use SR. That is going to be the most you could do with SR.

16. Dec 4, 2014

### Fantasist

Sorry, but I still don't feel I received a straight answer to my question:
with no information available to each twin apart from their respective accelerometer reading curves, and with the latter turning out to be exactly identical, will they have to conclude that their clocks will show the same time or a different time in the end (or can it not be decided without further information)?
This is a question that can be answered in one sentence or even one word. There is no need to extend this to an essay.

17. Dec 4, 2014

### PAllen

AT's answer in #2 would seem to be complete for this formulation of the question. But note, that anything involving gravity falls in the curved spacetime part of his answer. And any attempt to ascribe significance to accelerometer readings in SR with gravity treated as a force (in some adhoc, approximate manner), must now account that accelerometer reading no longer relate to proper acceleration (as used in SR).

18. Dec 4, 2014

### Staff: Mentor

If the accelerometer curves are identical then obviously the clocks will show the same time since the clock reading is the horizontal axis of the curve.

19. Dec 4, 2014

### A.T.

I would think that accelerometer readings always relate to proper acceleration. But in curved space (GR) time they don't relate a coordinate acceleration in an global inertial frame, because there is none. That is the difference to SR.

20. Dec 4, 2014

### A.T.

Sneaky!

21. Dec 4, 2014

### A.T.

Do you actually mean identical or symmetrical (opposite directions)?

22. Dec 4, 2014

### PAllen

In SR, proper acceleration is deviation from geodesics of flat spacetime. If gravity is treated as force that happens to act on all parts of an accelerometer simultaneously, then a free fall accelerometer has proper acceleration but measures zero. [I am well aware of the fundamental difficulties of treating gravity in SR, but for weak field, slow motion, reasonable precision, you can get away with it.]

Last edited: Dec 4, 2014
23. Dec 4, 2014

### pervect

Staff Emeritus
If you have only the numerical reading of the accelerometer (and no information on the direction of the acceleration), the answer is "it can not be decided without further information".