Does a metric theory of gravity plus torsion violate the equivalence principle?

Question

I have read that the Einstein-Cartan theory introduces torsion into general relativity in a way that produces coupling between gravity and the spin of particles. Then, the gravitational field is able cause the precession of this spin axis.

But what about spinless particles? Wouldn't there be a difference between the way gravity acts on particles with spin vs no spin? I guess that even for particles with nonequal nonzero spin there would be a difference

If gravity acts differently on the varied kinds of particles, the equivalence principle goes out the window. Plus, I don't know how to make the spin coupling disappear with a simple coordinate change

Answer 1

The strong equivalence principle can be expressed as the claim that the physics of special relativity applies in sufficiently small regions of spacetime. If we introduce a metric with torsion, but it is done in such a way that spacetime is still locally flat, then this version of the equivalence principle still holds.

The weak equivalence principle can be expressed as the claim that bodies close to one another have the same acceleration under gravity. When allowing for spin we usually modify this to a statement about spinless bodies. Such a version of the weak equivalence principle can still hold in a theory with torsion.

Ultimately it's the natural world that tells us whether bodies with spin fall with the same acceleration as those without. Whether in GR or another theory, there is no particular problem with the notion that a body with instrinsic angular momentum will not follow a geodesic (in the absence of forces other than gravity). If we treat the angular momentum as a classical property then the body is extended so there are various geodesics passing through the body and it will have to make a compromise of some sort. If we treat it as a quantum property then in any case we have gone beyond pure GR.