In General Relativity, we generally assume that the derivative operator is torsion-free, i.e., second covariant derivatives commute on functions.

However, in Kerr black holes, spacetime is dragged (especially in the ergosphere), so it seems that there is torsion in this case.

Geometrically, why there isn't torsion in the Kerr spacetime? What is the geometric interpretation of torsion if Kerr spacetime doesn't have it?


1 Answer 1


Torsion is not frame dragging. Torsion is having an anti-symmetric spacetime connection. As you do parallel transport in general relativity (GR) you drag frames the frames roll as they move. With torsion they would twist. The connection is GR is the Christopher symbols, symmetric in the two bottom indices. The torsion is an anti-symmetric tensor. It will give you an alternate theory to GR, not in any way part of GR. So Kerr and any other GR metric have no torsion. See the Wikipedia article on it at https://en.m.wikipedia.org/wiki/Torsion_tensor

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    $\begingroup$ Also even in a case where torsion would be present, the Kerr solution is a vacuum solution, and vacuums do not carry any torsion. $\endgroup$
    – Slereah
    Commented May 20, 2016 at 6:19
  • $\begingroup$ Re Slereah's comment, the OP may want to consider kerr-newman BH's, which are electrically charged. $\endgroup$
    – Edouard
    Commented Mar 5, 2021 at 18:23
  • $\begingroup$ As applications of torsion that may relate to the OP's question are extensively discussed by Nikodem Poplawski, in 2010-2020 papers whose preprints are freely available on Cornell University's Arxiv site, I'm suggesting an edit to add a "cosmology" tag, given that most of Poplawski's preprints deal with a torsion-based cosmological model. They mainly use Einstein-Cartan Theory (which I believe is the alternate to which Bob Bee referred), which was worked out in 1929 (after the discovery of particulate spin) thru conversations between Einstein and the mathematician Elie Cartan. $\endgroup$
    – Edouard
    Commented Mar 5, 2021 at 18:49
  • $\begingroup$ I would also add that the electromagnetic field of the Kerr-Newman solution (not a pure vacuum solution) doesn't generate any torsion, even in the Einstein-Cartan-Sciama-Kibble theory. In that theory, only matter with spin $\frac{1}{2}$ could generate torsion. Yet, there is none in the Kerr and Kerr-Newman spacetimes. $\endgroup$
    – Cham
    Commented Mar 5, 2021 at 20:37

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