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Minkowski spacetime the has a flat metric of Lorentzian signature (-1,1,1,1).

It is well known (c.f. Geroch 1967 & citation there) that whether a manifold admits a metric of Lorentzian signature is equivalent to the question of whether it admits a nowhere vanishing timelike vector field. By Geroch's theorem the spacelike hypersurfaces of Minkowski spacetime are diffeomorphic to each other, the diffeomorphism being produced by the integral curves of such a timelike vector field.

Since the metric on each spacelike slice of Minkowski spacetime is the (flat) Euclidean metric, the diffeomorphism must be an isometry, but among the possible isometries are rotations.

Question: is there any sense to the idea any each slice may be related another by a rotation? What physical or mathematical justification would there be for [disallowing] allowing such diffeomorphisms? (If the timelike vector field had curl, could one have a corkscrew hole in ~Minkowski space?)

The independent rotation of spatial hypersurfaces being something like this...

enter image description here

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  • $\begingroup$ Perhaps, MacDowell-Mansouri Gravity is what you're looking for? arxiv.org/abs/gr-qc/0611154 $\endgroup$
    – Kosm
    Mar 7, 2017 at 8:01
  • $\begingroup$ They consider translations as approximated SO(1,4) rotations $\endgroup$
    – Kosm
    Mar 7, 2017 at 8:05
  • $\begingroup$ @Kosm Thank you greatly, but I'm afraid you credit me with greater expertise than I possess: that paper is way beyond me. I was thinking rather: whilst a spacelike hypersurface might be considered rotated wrt another, would consideration of Eulerian geodesics (is than an extra constraint or...?) make rotated indistinguishable from rotated, and thus a vacuous distinction?... $\endgroup$
    – Julian Moore
    Mar 8, 2017 at 6:58
  • $\begingroup$ @Kosm Transformations $\in$ ?Poincare? group applied to the whole manifold preserve inertial frames, but what happens to inertial frames if the spatial hypersurfaces are rotated as described? I'm not doing well rephrasing this, I know, but I hope your clearly superior expertise will allow you to make better sense of it! $\endgroup$
    – Julian Moore
    Mar 8, 2017 at 6:58
  • $\begingroup$ Spacial rotations are part of the Lorentz group. Lorentz group is a subgroup of Poincare. I don't understand the problem $\endgroup$
    – Kosm
    Mar 8, 2017 at 7:03

1 Answer 1

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Let me try answering this (assuming I understood the question). In 4 dimensions you do not have a unique axis of rotation. Instead, there are two of them (so-called stationary plane which is fixed under a rotation). So in Minkowski space, when you rotate something around a time axis, you also rotate it around one of spatial axes (say $z$), and it will be the same as rotation in 3D around $z$. If you instead fix stationary plane to be purely spatial, then you get a Lorentz boost.

Feel free to correct me if anything.

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  • $\begingroup$ I'll working on a question update over the weekend and get back to you. Thanks $\endgroup$
    – Julian Moore
    Mar 9, 2017 at 17:12

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