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I do not understand why the torsion is set equal to zero in the general theory of relativity. The geodesics would be the same. Is there even a way to test it?

Pg 250 from the 2017 edition of MTW says

$\nabla$ is said to be a "symmetric" or "torsion-free" covariant derivative when $\nabla_{\mathbf{u}}\mathbf{v}-\nabla_{\mathbf{v}}\mathbf{u} = [\mathbf{u}, \mathbf{v}]$. Other types of covariant derivatives, as studied by mathematicians, have no relevance for any gravitation theory based on the equivalence principle.

but I don't know what does this have to do with the equivalence principle.

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    $\begingroup$ Autoparallel geodesics would not be the same though. Moreover, allowing for torsion opens up for more general theories of gravity beyond GR (such as Einstein-Cartan). It is a choice at the end of the day, but the dynamics of the theories are different. $\endgroup$
    – Eletie
    Commented Oct 2, 2023 at 21:57
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    $\begingroup$ There is no experimental evidence so far that including torsion is necessary. Therefore, by Occam’s Razor, it is excluded. There is no point in making a theory more complicated than it needs to be. $\endgroup$
    – Ghoster
    Commented Oct 2, 2023 at 23:24
  • $\begingroup$ Related: physics.stackexchange.com/q/27746/2451 , physics.stackexchange.com/q/103576/2451 , physics.stackexchange.com/a/362319/2451 $\endgroup$
    – Qmechanic ♦
    Commented Oct 3, 2023 at 3:59

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