I have a very simple question:

In the phrase "gravity is the manifestation of curvature", I want to know its analogue in Einstein Cartan theory where torsion is non-vanishing.

So basically I want to fill the gap in the following phrase: $\dots$ is the manifestation of Torsion.

And could someone, please, explain to me why Torsion has the same statue as: 1) Curvature when we consider the field equations of E.C.T. $$ R_{ab}-\frac{1}{2} R = \kappa P_{a b} $$ $$T_{a b}{ }^{c}+g_{a}{ }^{c} T_{b d}{ }^{d}-g_{b}{ }^{c} T_{a d}{ }^{d}=\kappa \sigma_{a b}{ }^{c} $$ where $P_{ab}$ is the energy momentum density, $ \sigma_{a b}{ }^{c}$ is what we call a spin tensor and $g_{a}{ }^{c}$ is the metric of a Riemann-Cartan space. 2) And a connection when we consider the affine connection $$ \Gamma= \tilde{\Gamma}+ K(T) $$ Where $\tilde{\Gamma}$ is the Levi-Civita connection and $K$ is the contorsion.

Am asking this question because there is no such thing in GR.

N.B.: I know the geometrical meaning of Torsion and that the so-called spin current is related to it.

  • $\begingroup$ we have the word "gravity" because it is real and we can see its effects. Spacetime "Curvature" is just reformulating gravity in terms of spacetime geometry. In contrast we don't see torsion in the real world, so there is no analogous phrase to what your wrote for gravity. In any case, "gravity is manifistation of (spacetime) curvature" is essentially tautology in terms of GR. $\endgroup$
    – Kosm
    Commented Aug 6, 2021 at 10:00
  • $\begingroup$ @Kosm I understand what you are saying, but Torsion can be seen if we have the right conditions it's like curvature when you can see its effect just next to big objects (macroscopic objects in scientific terms) $\endgroup$ Commented Aug 6, 2021 at 14:42

1 Answer 1


I would be very careful with the phrase "gravity is the manifestation of curvature". Because if one falls towards a massive body --- this observation is typically associated (by laymans) with gravity --- it is actually not a symptome of curvature, it simply might mean that one is in a "wrong" coordinate/reference system that provides one this impression.

However, if one includes in the concept of gravity the occurrence of tidal forces, then one could indeed attribute that to curvature. In this sense one could say: "gravity is the manifestation of curvature".

Actually I would prefer to say:

Curvature is generated by momentum & energy (density) whereas

Torsion is generated by spin (density).

The effect of torsion is actually very weak. As the 2. equation (also 2. equation of your post) which links torsion with spin is an algebraic equation torsion cannot propagate. One would find torsion only at places with significant spin density, for instance inside exotic astronomical objects like a neutron star or in the first instants after the big-bang.

Under the assumption that the torsion tensor is totally antisymmetric, autoparallelism equation and geodesic equation are the same. Therefore the torsion would not influence the motion of a particle in a place of high spin density. The only effect torsion produces is spin precession. So one could tentatively say: Spin precession is a manifestation of torsion. But as there are other effects that can produce spin precession --- and mostly probably stronger than the one generated by the torsion --- one should be extremely careful in the expression of such an assertion. Because the effect (spin precession) and the concept (torsion) are not uniquely related.

The most pursued approach in handling with torsion is to try to eliminate it from the equations, because torsion makes the math quite complicated. In this respect there would be no manifestation at all.

  • $\begingroup$ So we can say, roughly speaking, that torsion has the same effect of curvature in the sense that it makes objects deviate from geodesics. $\endgroup$ Commented Aug 6, 2021 at 14:29
  • $\begingroup$ Actually I don't understand your question. With curvature or without curvature or with (completely antisymmetric) torsion or not point particles will always move on a geodesic. BTW, there is no known particle found yet that would produce a non-completely-antisymmetric torsion. $\endgroup$ Commented Aug 6, 2021 at 15:55
  • $\begingroup$ Oh, you say that point particles will always move on geodesic because the existence of Torsion is not proved yet, or in other terms, there is no known particle found yet that would produce a non-completely-antisymmetric torsion and that will not move on a geodesic. $\endgroup$ Commented Aug 6, 2021 at 18:08
  • $\begingroup$ Because as far as I know, the introduction of torsion in spacetime produce a difference between geodesics and autoparallel paths $\endgroup$ Commented Aug 6, 2021 at 18:09
  • $\begingroup$ If the torsion tensor is completely antisymmetrical, its contribution cancels out in the geodesic equation, because an antisymmetric tensor is multiplied with a symmetric product yields 0. A difference between autoparallel paths and geodesic paths only occurs if the torsion is not completely antisymmetric. However, the torsion tensor computed from particles described by a Dirac spinor is always completely antisymmetric. For bosons the torsion is zero. Therefore for all known particles the torsion is either completely antisymmetric or 0 In particular there is no effect on the geodesic equation. $\endgroup$ Commented Aug 6, 2021 at 20:57

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