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In this article the Ehrenhaft paradox is described. You can read in it that, according to Einstein's General Relativity (to which this paradox contributed), the spacetime around a rotating disc is non-Euclidean, i.e. the spacetime is curved.

Now if we put two rotating identical discs (infinitely thin, but with a mass) on top of each other (somewhere in an empty region of intergalactic space and with their axes perpendicular to their surfaces aligned) and let them rotate contrary to one another with equal but opposite angular velocities (and zero friction) will this cause spacetime around them to be Euclidean again? In other words, will the curvatures of spacetime caused by the two discs "cancel each other out", resulting in a Euclidean geometry of spacetime "around" them? Or, on the contrary, will the curvature of spacetime increase? In other words, do the two curvatures of spacetime just "add up", because (maybe) there is no difference in the curvature of spacetime surrounding a disc if the direction of the angular velocity of a disc changes into the opposite direction?

I think I am right by saying that the space part of spacetime (in two dimensions) of one rotating disc looks like a cone with the tip of the cone at the center of the disc. Maybe the cones of the discs are oppositely directed (also in three dimensions where there are 3-d cones for this case).

EDIT:

Thanks to a comment by Constantine Black I make this edit. The spacetime around a rotating disc is of course only curved for persons standing stationary on the disc.It seems to them that there is a gravitation field (and thus curved spacetime) which grows proportionally with $r$, the distance to the center of the disc. The question is: does the counterrotating disc contribute to the (artificial) gravity they experience. It seems to me it doesn't but I could be wrong.

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  • $\begingroup$ Why the downvote? I like to know what's the reason for it. Maybe it gives me some information. Doesn't anybody have to say anything about it? Or has the downvoter just downvoted for no reason except for the fact that it was me who asked the question? $\endgroup$ – Deschele Schilder Dec 15 '17 at 19:16
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    $\begingroup$ I didn't downvote; it seems like a pretty good question to me. $\endgroup$ – Mozibur Ullah Dec 15 '17 at 23:46
  • $\begingroup$ Alright! Finally a reaction. It doesn't answer my question though, but who cares... $\endgroup$ – Deschele Schilder Dec 16 '17 at 6:54
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    $\begingroup$ Hi. A question: doesn't the sentence:"spacetime is non-euclidean around the rotating disc" mean it isn't euclidean for an observer on it; that is a rotating observer would observe a curvature in the universe. Or have I misread the article? Thank you,+1. $\endgroup$ – Constantine Black Dec 17 '17 at 9:21
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    $\begingroup$ There are "true" (i.e. present even for stationary observers) effects of the angular momentum (cf. Kerr vs. Schwarzschild metrics or Lense-Thirring effect). Furthermore, two couterrotating discs will have no oveeall angular momentum, but mass and energy, so at least that will warp spacetime. $\endgroup$ – Toffomat Dec 19 '17 at 14:32
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In this article the Ehrenhaft paradox is described. You can read in it that, according to Einstein's General Relativity (to which this paradox contributed), the spacetime around a rotating disc is non-Euclidean, i.e. the spacetime is curved.

You've misunderstood. If a spacetime is flat, then it's flat in all coordinate systems and frames of reference. The rotating observer perceives the space as curved.

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  • $\begingroup$ But If space is curved isn't time automatically curved too? I was taught that curved space implies curved time (and vice-versa) and that you can't consider space on its own. So if a rotating observer perceives the space as curved, he perceives time as curved too. If the person on the rotating disc stands on a wall on the disc, at the right distance from the center and assuming the true gravity of the disc is canceled by the form of the wall, he can feel no difference between gravity on Earth and the artificial gravity he experiences on the disc while standing on the wall. $\endgroup$ – Deschele Schilder Jul 26 '19 at 14:55
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The comments of Constantine Black and Toffomat were very useful in trying to answer my own question.

There is no true warping (due to the gravitomegnetic effect) because both discs are rotating counterclockwise. And of course, because of the mass of the discs, there is a true curvature.

But as we ignore these real effects, the non-true curvature as "seen" by a stationary observer on one of both discs will always be present on both discs, so the answer is simply NO. The rotation of one disc doesn't have any influence on the non-true curvature of the other disc. On both discs, the same pull of gravity is felt (for people that are trying to take a walk on them), so the non-true curvatures of the discs don't cancel.

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