5
$\begingroup$

In GR we use the Riemannian manifold without any torsion to describe the theory. Hence, the geodesic equation can be interpreted as "a trajectory of a free falling particle" or "equation of motion". But in Einstein-Cartan manifold the torsion prevents us to use the geodesic equation as an equation of motion. So, my question is that, what is the meaning of a geodesic equation in this theory?

$\endgroup$
5
$\begingroup$

Torsion doesn't prevent us to use the geodesic equation as an equation of motion. When torsion is present, we need to distinguish two types of "motion": The geodesic is a curve of extremal proper time. It comes from a lagrangian formulation. The teleparallel curve is the one for which the tangent is parallely transported from one point to the next. Both are equivalent in the absence of torsion, but they're not the same when a general torsion is present. Currently, we don't know which one a massive particle should follow. The inertia principle suggest that it should be a teleparallel curve, not a geodesic curve.

Take note that, given some initial conditions, both curves are the same when torsion is completely antisymetric, which is the case when only Dirac spinors are coupled to torsion.

$\endgroup$
  • $\begingroup$ As expressed by Sabatta and Gasperini in "Introduction to Gravitation" (page 244), neither geodesic equation nor the auto-parallel curves could be considered as an equation of motion in Einstein-Cartan theory, since none of them describes the effects of torsion correctly. The former does not considered torsion and the latter effects on all particle in the same way, regardless of their spin. $\endgroup$ – Astrolabe Sep 18 '19 at 20:41
  • 1
    $\begingroup$ I agree with @Cham and I disagree with Astrolabe. See arxiv.org/abs/gr-qc/0606062, for instance $\endgroup$ – Cinaed Simson Sep 19 '19 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.