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Would the observer moving along the circle in this Godel space-time diagram feel fictitious forces as though he is accelerating along a circular path or would he simply arrive at an earlier point in time after heading outward in a straight path. I believe he is moving in a straight path but arrives at the earlier point due to the weirdness of the godel model, but someone online disagrees. What mathematics should I use to answer such questions? Is there an intuitive non-mathematical answer based on the the diagram alone? I know GR up to the three early classic predictions, but that is basically all I know.

I would prefer to use the diagram from Hawking/Ellis, but I cannot find it online. This one should suffice if you are familiar with the diagram,

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2 Answers 2

Why are you reading Hawking and Ellis if you want to avoid math?

I don't remember enough of the details of the Gödel universe to give you a definite answer, but you would want to check to see if the circular path is a geodesic of the spacetime. If this is the case, then your test observer will feel no force as she travels around the circle. If not, some force has to be exerted to keep her moving on her closed timelike curve.

The underlying math is differential geometry, and this matter is definitely not decidable from the diagram alone. If I have some time later, I'll go and look up the metric for the Gödel universe and do the necessary computation. It'd be good to get some practice with an unfamiliar metric.

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Thank you. I believe it is a geodesic. I read, or tried to read Hawking, years ago. I took a course that used D'Invernio. It introduces mathematical tools without really explaining in detail why they work. I've never understood killing vectors. I think this question shouldn't require advanced math. –  MadScientist Dec 13 '12 at 16:10
@Jerry Schirmer It looks as though I won't be able to accept your answer because I was't logged in. –  MadScientist Dec 13 '12 at 16:25
@Jerry Schirmer Unless I'm reading Hawking/Ellis incorrectly, the circle isn't a geodesic, but I have to review spacetime diagrams. If that is true, then I would like to calculate the acceleration of an object moving along that circle -but I'm not sure if that is possible from the metric and the fact that r>log(1+sqrt[2]) which is all the information I have about the path. –  MadScientist Dec 14 '12 at 12:14
@Jerry Schirmer If you are still interested in this and happen to have a copy of Flanders. Differential Forms with Applications, I believe it might be possible to do this calculation using the technique on page 134. That is what I'm trying now. It uses "moving frames." –  MadScientist Dec 14 '12 at 13:25
I gave up on that. Now I'm dividing the first term of upload.wikimedia.org/math/2/6/5/…, from the wiki on Godel's metric, by the second term. Hopefully, this gives the velocity. But this is a complete stab in the dark. I'm going to check whether this satisfies v^2/r =a –  MadScientist Dec 14 '12 at 14:50

The CTCs in the Gödel universe are not straight lines. I'm even hesitant to call them circles, because they have a non-zero energy balance. In fact, the energy balance is not only slightly non-zero, but many times the rest mass of the particle moving on such a CTC. The following text is from a question I asked some time ago, and (as far as I remember) it was quite easy to get access to the original publications:

Gödel published his discovery of closed timelike curves in 1949. Many years later (in 1961), S. Chandrasekhar and James P. Wright pointed out in "The geodesic in Gödel's universe" that these curves are not geodesics, and hence Gödel's philosophical conclusions might be questionable. Again some years later, the philosopher Howard Stein pointed out that Gödel never claimed that these curves are geodesics, which Gödel confirmed immediately. Again much later other physicists have computed that these closed timelike curve must be so strongly accelerated that the energy for a particle with a finite rest mass required to run through such a curve is many times its rest mass. (I admit that I may have misunderstood/misinterpreted this last part.)

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