# Dirac equation, curved space time

1. Feb 16, 2013

Hi when trying to derive this equation, i am stuck on:
$[\Gamma_{\mu}(x),\gamma^{\nu}(x)]=\frac{\partial \gamma^{\nu}(x)}{\partial x^{\mu}} + \Gamma^{\nu}_{\mu p}\gamma^{p}$.

This $[\Gamma_{\mu}(x)$ term is the spin connection, if this is an ordinary commutator:
a) is it a fermionic so + commutator
b) how can one solve to find the Gamma term whilst cancelling away the
c) can anyone give a qualitative description of what the spin connection is?
thanks

2. Feb 16, 2013

### Bill_K

In curved space the gamma matrices are position-dependent, since they obey γμγν + γνγμ = 2 gμν. But we can and do take them to be covariantly constant.

A gamma matrix is a hybrid quantity, having one vector index, and two matrix indices referred to a tetrad basis. The covariant derivative contains a correction term for each index - a Christoffel symbol for the vector index and a spin connection (Ricci rotation coefficient) for each tetrad index.

The equation you wrote is the statement that the covariant derivative vanishes. You can take a look at this paper for more details.

3. Feb 16, 2013

Hi
I have reread the paper: http://arxiv.org/abs/gr-qc/0501077v1 (eq 8 and 10) and i see what you mean. How else then can one find $\Gamma_\mu$?

4. Feb 16, 2013

### Bill_K

Eqs. (8) and (9) in the paper I referred you to shows how to solve for Γμ.

5. Feb 16, 2013

thx,

am i right in thinking $\bar{S_{ab}}$ is always equal to zero when b =0 because all my answers involved $\bar{S_{a0}}$ terms?

6. Feb 16, 2013

sorry i should explain more,

I assumed that $t^{a}_{b}$ is only non zero if a = b (from 0 to 3). and the same for $t_{ab}$.

If this is correct then all my answers for $w_{abj}$ contain either a =0 and b=1,2,3 or vice versa meaning that when multiplied with $S_{ab}$ then will surely go to zero if $s^{0b} = 0$.
Are any of these assumptions incorrect?

7. Mar 13, 2014

### FunkyDwarf

Sorry for the thread necro, but could someone please provide more details regarding eqn (8) in that reference, specifically why the commutator of the spin connection with the curved space gamma matrix is defined that way.

8. Mar 14, 2014

### Bill_K

The covariant derivative of gamma contains a correction term for each index: a Christoffel symbol for the vector index, and a spin connection for each of the two spinor indices.

∂γν/∂xμ + Γνμρ γρ + Γμ γν - γν Γμ

Eq.(8) says that this quantity vanishes (we choose to define our γ's that way) and then the terms are rearranged slightly to make them look like a commutator.

9. Mar 14, 2014