Twin paradox in GR - negative time?

[Ibix]

Here's derivation of the exact formula: In my approximate derivation the Lorentz boost is wrong by gamma, and the ticking rate of the distant clock is wrong by gamma. Those two errors cancel, so the exact formula is the same. (Unless I'm missing some gammas) (Oh yes the acceleration-velocity relation is more complicated - well I have think some more about this)

Also the proper time for the accelerating observer isn't the same as coordinate time. And x isn't a constant over the acceleration period.

 

[jartsa]

So I said I would derive a formula for rate of a distant clock. Seems I haven't done so.


But isn't Schwarzschild metric the correct metric when there's gravity, and Rindler metric is a weak field approximation in that case?

And Rindler metric is the correct metric for an accelerating observer to use?

So the approximation in the case of gravity is the exact thing in case of acceleration?

 

[Ibix]

I think it's correct for the time dilation between Rindler observers, those undergoing permanent constant equal proper acceleration. But that's not appropriate for a twin paradox.

If you want to do what you seem to be trying to do - patch together two inertial frames with a Rindler frame in between - the way I did it was to note that in the stay at home's frame the traveller has coordinate acceleration $dv/dt=a_0/\gamma^3$ and integrate to get $v (t)$, $x (t)$ and $\tau (t)$. Or you can look up the results - they're fairly standard. Then you can work out the elapsed time for both in what this coordinate system calls "during the turnaround".

Radar coordinates are better because they're physically motivated, guaranteed to be continuous, and smooth if your path is smooth. This system is neither, I think.

 

[m4r35n357]

Radar coordinates are better because they're physically motivated, guaranteed to be continuous, and smooth if your path is smooth. This system is neither, I think.

I'm off on another tangent ATM, but i did stumble across this paper whilst taking a brief diversion into this thread. From the abstract: "Our results provide a critical assessment of the physical significance of radar coordinates".

 

[PAllen]

I'm off on another tangent ATM, but i did stumble across this paper whilst taking a brief diversion into this thread. From the abstract: "Our results provide a critical assessment of the physical significance of radar coordinates".

Right, in that radar coordinates, in general, do not provide globally consistent coordinates any more than Fermi coordinates do. Even for as simple a case as an eternally uniformly accelerating observer, their coverage is identical to Fermi coordinates.

However, they have a quite interesting special case. For any non rotating observer, no matter what the trajectory, as long as it is inertial for all time before some point, and again after some point, no matter what happens in between, they provide a globally consistent coordinate system, while Fermi coordinates do not.

 

[pervect]

I was reading the wikipedia page on the twin paradox (https://en.m.wikipedia.org/wiki/Twin_paradox). It says:

The mechanism for the advancing of the stay-at-home twin's clock is gravitational time dilation. When an observer finds that inertially moving objects are being accelerated with respect to themselves, those objects are in a gravitational field insofar as relativity is concerned. For the traveling twin at turnaround, this gravitational field fills the universe. In a weak field approximation, clocks tick at a rate of t' = t (1 + Φ / c^2) where Φ is the difference in gravitational potential. In this case, Φ = gh where g is the acceleration of the traveling observer during turnaround and h is the distance to the stay-at-home twin.

My question is, if the traveling twin were to accelerate away from the stay-at-home twin, would Φ be negative? If yes, then if Φ < c^2 that would mean as time goes forward for the traveling twin, time would go backward for the stay-at-home twin (negative time).

In addition to the other reference given, (Doby & Gull), I would recommend Misner, Thorne, Wheeler's book "Gravitation", section $6.3, "Constraints on the Size of an accelerated frame". It's likely you'll be able to find the appropriate section of the book with a Google search for these words (but perhaps not guaranteed, though it worked for me).

A short quote from the relevant section:

It is very easy to put together the words "the coordinate system of an accelerated observer," but it is much harder to find a concept these words might refer to. The most useful first remark one can make about these words is that, if taken seriously, they are self-contradictory.

It's a bit long to cut and paste the rest of MTW's text, but I can summarize the general theme of their answer to your question. This answer is "no". Their answer (which seems to match that of other papers and texts such as the Dolby & Gull reference) is that the "frame" of an acceleated observer (which would better be termed "the" coordinate system of an accelerated observer) does not naturally cover all of space-time.

You are apparently assuming that a coordinate system does (and must) exist for an accelerated observer that covers all of space-time. The textbook answer is that this idea is highly suspect.

A more sophisticated version of your argument might be to argue that one of the forms of the Rindler coordinate chart , $-(1+gz)dt^2 + dx^2 + dy^2 + dz^2$ covers some nearby region of the space-time of an accelerated observer. Here the "z" coordinate is what you call h, the "height". Recognizing that many coordinate systems exist, one might suggest "what happens if we extend this coordinate system to the region 1+gh < 0? What issues arise?

One obvious issue is that the metric signature becomes ++++ in this region, which breaks a lot of the mathematics. So the standard approach doesn't do this, and you'd be wise not to, either.