Torsion in GR and Einstein-Cartan theory

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The discussion centers on the relationship between torsion in General Relativity (GR) and Einstein-Cartan theory, specifically how both theories yield identical results in vacuum conditions. It is established that the torsion tensor's equation of motion in Einstein-Cartan theory is algebraic, meaning torsion does not propagate and is only present in the presence of matter with spin. In vacuum, where there is no spin, torsion vanishes, leading to the same geometric framework as pure GR. This highlights that GR does not account for spin, as it operates within a classical framework.

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  • Understanding of General Relativity (GR)
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  • Knowledge of differential geometry
  • Basic concepts of spin in quantum mechanics
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This discussion is beneficial for theoretical physicists, researchers in gravitational theories, and students of advanced physics looking to deepen their understanding of the interplay between torsion, spin, and the geometrical structure of spacetime.

TrickyDicky
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I was reading something about the Cartan (vierbein) formalism in GR, in which the connection is allowed to have torsion, and it got me interested in the Einstein-Cartan theory.
Apparently both GR and Cartan theory with torsion should give the same experimental results in vacuum, what I would like to understand better from the differential geometry point of view is why exactly this happens.
Why doesn't torsion make any difference in vacuum?
 
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Mathematically, the answer is that in Einstein-Cartan theory, the equation of motion for the torsion tensor is purely algebraic (i.e., without derivatives). Hence the torsion does not propagate, and is nonzero only in the presence of matter with spin.

In the vacuum, there is no spin, and so the torsion vanishes. So in that case the geometry is identical to the pure GR case. The torsion makes no difference because there is no torsion.
 
Ben Niehoff said:
Mathematically, the answer is that in Einstein-Cartan theory, the equation of motion for the torsion tensor is purely algebraic (i.e., without derivatives). Hence the torsion does not propagate, and is nonzero only in the presence of matter with spin.

In the vacuum, there is no spin, and so the torsion vanishes. So in that case the geometry is identical to the pure GR case. The torsion makes no difference because there is no torsion.

Thanks, does this mean GR assumes matter has no spin? what would be the justification to eliminate that degree of freedom in GR?
 
Because spin is quantum mechanical in nature and GR is purely classical.
 

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