10
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Experimental evidence for "uses a metric connexion" is the evidence for the equivalence principle: Galileo, cannonballs, feathers, Eötvös experiment and all the rest of it. It is this that motivates a manifold with metric theory. As for the rest: Google seems to want to search on "constraining torsion with gravity probe b" for me and this does seem to throw up some interesting leads which I don't have time to track down. Try this and see how you go. $\endgroup$ – WetSavannaAnimal Oct 13 '15 at 9:56
  • $\begingroup$ Related: physics.stackexchange.com/q/27746/2451 , physics.stackexchange.com/q/192230/2451 and links therein. $\endgroup$ – Qmechanic Oct 13 '15 at 13:23
13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This is a great summary, but the OP was asking for experiments to find evidence of nonzero torsion. Can you give any comment here. Have you for example read arxiv.org/abs/gr-qc/0608121 or even recenter relevant papers, and can you comment? $\endgroup$ – WetSavannaAnimal Oct 13 '15 at 23:04
  • $\begingroup$ As this paper mentions, torsion is hard to test for, because the basic theory (Einstein Cartan) doesn't propagate in the vacuum (this is related to the torsion tensor $\propto$ the spin tensor), and the coupling constant is roughly on the Planck scale, and only affects objects with (half) spin. As a result only specific theories with torsions can be tested in the solar system, for this paper Hayashi-Shirafu and Weitzenböck theory (on which I don't know too much). At most torsion has some rather small effect on neutron stars (although still small) $\endgroup$ – Slereah Oct 14 '15 at 7:07
  • $\begingroup$ For some details on neutron stars try adsabs.harvard.edu/cgi-bin/… $\endgroup$ – Slereah Oct 14 '15 at 7:07

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.