How to measure Torsion and Non-metricity?

Question

In General Relativity, we most often work with the Levi-Civita connection (metric and torsion-free). What kind of experiment can we make to be sure that our physical space-time indeed is torsion-free and uses a metric connection?

Answer 1

Torsion affects the transport of vectors along a path. More physically speaking, it affects the propagation of spinor fields (EM fields are not affected since exterior derivatives are independent from the connection). Since the torsion tensor is directly equal to the spin tensor, this means that we can ignore pretty much all geometric interpretation and treat it as a simple interaction.

This is manifested in the Hehl-Datta equation :

\begin{equation} i\gamma^\mu \nabla_\mu \psi + \frac{3\kappa}{8}(\overline{\psi}\gamma_\mu\gamma^5\psi)\gamma^\mu\gamma^5\psi - m\psi = 0, \end{equation}

With $\kappa$ the usual coupling to gravity. This corresponds to an axial current-axial current interaction.

If you impose the torsion by hand to be something simple (the simplest torsion tensor is $T_{abc} = \varepsilon_{abc}$), this will make the spin vector rotate around some axis. Generally, the motion of the spin vector parallel propagated along a geodesic of tangent vector $u$ is

\begin{equation} S^\mu_{;\nu} u^{\nu} = 3 K^{[\alpha \beta \mu]} S_{\beta} u_{\alpha} + \mathcal{O}(\hbar) \end{equation}

With $K$ the contortion tensor.

You may know that generally, the current of a spinor field that obeys the Dirac equation can be decomposed in two parts, the so called Gordon decomposition :

\begin{equation} j_\mu = \frac{i}{2m} [\bar \psi (\partial_\mu \psi) - (\partial_\mu \bar \psi) \psi] + \frac{1}{2m} \partial_\nu (\bar \psi \sigma^{\mu\nu}\psi) \end{equation}

The orbital current $j^c$ and the spin current $j^M$ (the spin current roughly corresponds to the magnetization and polarization in classical EM). If a connection with torsion is added into it, you still get two independently conserved currents, but this time of the form

\begin{equation} j_\mu^c = \frac{i}{2m} [\bar \psi (\nabla_\mu \psi) - (\nabla_\mu \bar \psi) \psi] - \frac{1}{2m} \bar{\psi} \sigma^{\alpha\beta} \psi K_{\alpha\beta\mu} \end{equation}

\begin{equation} j^M_\mu = \frac{1}{2m} (\bar \psi \sigma^{\mu\nu}\psi)_{;\nu} \end{equation}

(I think that the dirac bilinear of the spin current here is a 2-form hence it does not depend on the connection either so it is unaffected by torsion)

So fermions in a spacetime including torsion will generate a different EM field, and test particles send in such conditions will deviate from the trajectory we would expect without torsion.

As for the non-metricity tensor :

\begin{equation} \nabla_\alpha g_{\mu\nu} = N_{\alpha\mu\nu} \end{equation}

it will affect the conservation of scalar quantities along geodesics, among other things. For instance, in the case of the mass

\begin{eqnarray} \frac{dp^2}{d\lambda}&=& \frac{d(g_{\mu\nu} p^\mu p^\nu)}{d\lambda} \\ &=& m^2 u^\alpha \nabla_\alpha (g_{\mu\nu} u^\mu u^\nu)\\ &=& m^2 u^\alpha (N_{\alpha\mu\nu} u^\mu u^\nu + g_{\mu\nu} u^\nu \nabla_\alpha u^\mu + g_{\mu\nu} u^\mu \nabla_\alpha u^\nu) \end{eqnarray}

And as you know, for geodesics, $u^\alpha \nabla_\alpha u^\mu = 0$, leaving

\begin{eqnarray} \frac{dp^2}{d\lambda}&=& m^2 N_{\alpha\mu\nu} u^\alpha u^\mu u^\nu \end{eqnarray}

Meaning a free particle would change mass along its trajectory.

In addition, consider two vectors along a curve. For instance, pick the case of the momentum vector and the spin vector of the particle. The helicity of this particle will be $h = g(p, s)$. Its evolution along the curve is

\begin{eqnarray} \frac{dh}{d\lambda}&=& \frac{d(g_{\mu\nu} p^\mu s^\nu)}{d\lambda} \\ &=& m^2 u^\alpha \nabla_\alpha (g_{\mu\nu} u^\mu s^\nu)\\ &=& m^2 u^\alpha (N_{\alpha\mu\nu} u^\mu s^\nu + g_{\mu\nu} s^\nu \nabla_\alpha u^\mu + g_{\mu\nu} u^\mu \nabla_\alpha s^\nu)\\ &=& m^2 N_{\alpha\mu\nu} u^\mu s^\nu u^\alpha + g_{\mu\nu} u^\mu u^\alpha \nabla_\alpha s^\nu\\ \end{eqnarray}

The time derivative for a free particle (outside of a magnetic field and such) of the spin should be zero, so that $g_{\mu\nu} u^\mu u^\alpha \nabla_\alpha s^\nu = 0$, but even in that case, your helicity does not stay constant either.