# Why spin should influence the geometry of the space-time?

I have been studying the inclusion of (intrinsic) spin of particles in a geometrical theory of gravity. I understand how that can be done using the Einstein-Cartan theory and how spin is related with the torsion tensor and how that coupling allows for spin to contribute to the space-time geometry (or to the gravitational field). What I can't seem to understand is why should we expect (intrinsic) spin to contribute to the gravitational field? Is it because, for instance, for a Dirac field, spin contributes to the corresponding canonical energy-momentum tensor? Can someone elaborate on this?

Also, if related, why doesn't the stress-energy tensor that appears in the Einstein field equations (in standard GR) can not account for (intrinsic) spin?

At the intuitive level, intrinsic spin and orbital spin are not that different. For example, you can imagine that protons have internal spin. Or that they are made of 3 quarks and their orbital spin could look like intrinsic spin of the proton. You can actually have excited states of proton with spin higher than 1/2: they come from exciting the orbital angular momentum of the quarks that make up the proton. This is not a stable configuration and requires lots of energy. So if you have no problem assuming that orbital angular momentum curves spacetime you should also find intuitive the idea that intrinsic spin curves spacetime.

If you consider the connection in general relativity simply as a general connection on the principal bundle $OM$, the bundle of orthonormal frames, then it is neither metric nor symmetric (such a general condition is called the metric-affine gravitation theory).

I don't know the experimental status of the non-metric part of the connection (it's related to dilation and shear currents of the matter fields), but if we only account for the torsion part of the connection, then the Euler-Lagrange equations for gravity give us

$${R^a}_{\mu} - \frac12 R {e^a}_\mu + \Lambda {e^a}_\mu = \kappa \frac{\delta \mathscr L_M}{\delta {e^\mu}_a}$$ $${S^\mu}_{ab} - S_a {e^\mu}_b + S_b {e^\mu}_a = 2 \frac{\delta \mathscr L_M}{\delta {\omega^{ab}}_\mu}$$

The second equation relating the torsion tensor to the spin density tensor. By defining connections on spacetimes in the most general way possible (but still enforcing its metricity), this is the field equation we obtain from the Einstein-Hilbert action, the Sciama-Kibble equations.

Einstein's theory of relativity is defined to be torsion free not because it's the most natural condition to impose, but because it simplifies problems a lot, and, for all classical fields, it is also true. For scalar fields, EM fields and general fluids in general relativity, we have

$$\frac{\delta \mathscr L_M}{\delta {\omega^{ab}}_\mu} = 0$$

This is due to the fact that the connection on scalar fields is trivially just $\partial_\mu \phi$, that EM fields use differential forms $F = \mathrm{d}A$ which are independant of the connection, and fluids are just defined with respect to the metric tensor. In those circumstances, the Einstein-Cartan and Einstein theory of gravitation both predict the same events.

It is only for spinor fields that the spin density tensor is non-zero,

$$\frac{\delta \mathscr L_M}{\delta {\omega^{ab}}_\mu} = \frac i2 \bar \psi \gamma^{[\mu} \gamma^\nu \gamma^{\rho]} \psi e_{a\nu} e_{b\rho}$$

So in other words, the Einstein-Cartan theory is simpler in its assumptions, and the Einstein theory is simpler for calculations. For all current experimental tests, both are equivalent, but will predict different results for fermions.

• Dear Slereah, thank your for your time. Your exposure is indeed correct but it doesn't answer my question. Why do we expect spin to contribute to the gravitational field? Sure it can be coupled to space-time torsion assuming the Einstein-Cartan theory but in GR it does not couple with the metric or the (symmetric) connection. – PML May 23 '17 at 11:23
• Also, from your answer I got the idea (maybe wrong) that the way the electromagnetic interaction is included in GR is incomplete (putting a contribution to the energy-momentum tensor) and, perhaps, there could be other couplings between the Faraday tensor and the connection? – PML May 23 '17 at 11:25
• It depends what you mean by spin exactly then. The spin density tensor will indeed not contribute to the curvature, but then the stress energy tensor will be $${T^a}_\mu = \frac i2 (\bar \psi \gamma^a \psi_{;\mu} + {e^a}_\mu \bar\psi_{;\mu} + h.c.)$$The state of spin of this field will certainly affect the curvature since it will affect the angular momentum, and for instance for the Kerr metric of a spinor field will depend on its spin. – Slereah May 23 '17 at 11:32
• Also generally people do not assume that the EM field in GR as it is is wrong, although there are certainly other theories in which the EM field carries a torsion charge. – Slereah May 23 '17 at 11:33
• I've been trying to understand what you said but I'm stuck with the same doubt I had initially. We can obviously input any Lagrangian that we feel like in the RHS of the Einstein field equation in GR. For instance, we can use the Dirac Lagrangian and spin, as you said, will contribute to the space-time geometry, but then why every article I see on the inclusion of spin in GR starts with the Einstein-Cartan theory and the addition of extra degrees of freedom? So why isn't just the inclusion of a Dirac Lagrangian in GR expected to give the correct results (at some classical limit at least)? – PML May 23 '17 at 13:11