A small (and perhaps trivial :-() calculation is troubling me somewhat.
Consider the action of a right-handed Dirac fermion coupled to external gravity (a background gravitational field). The action can be written as
$$S = \int\sqrt{-\text{det}(g)}\, i\overline{\psi}_{R}\gamma^\mu\left(\partial_\mu + \frac{1}{2}\omega_{\mu}\right)\psi_{R} $$
where $\omega_\mu = {\omega_\mu}^{ab}\Sigma_{ab}$ and $\Sigma_{ab} = \frac{1}{4}[\gamma_a, \gamma_b]$. Here $\mu, \nu$ denote curved indices and $a, b$ denote tangent space (flat) indices. Now, the energy-momentum tensor is given by
$$T^{\mu\nu} = \frac{2}{\sqrt{-\text{det}(g)}} \frac{\delta S}{\delta g_{\mu\nu}}$$
To vary the action with respect to the metric, there are three pieces as I see it:
- Variation of the $\sqrt{-\text{det}(g)}$ piece. This is easy.
- Variation of the pieces with Lorentz indices: this is the part that I seem to have gotten a bit confused about.
Indeed $\gamma^\mu = \gamma^a e_a^\mu$ and so $\delta \gamma^\mu = \gamma^a \delta e^\mu_a$, and
$$\delta g_{\mu\nu} = \delta(\eta_{ab}e_{\mu}^a e_{\nu}^b) = \eta_{ab}[(\delta e^a_\mu)e_{\nu}^b + e^a_\mu \delta e_{\nu}^b]$$
But what happens on varying the operator $\gamma^\mu(\partial_\mu + \frac{1}{2}\omega_\mu)$ with respect to the metric?
One should ideally get
$$T^{\mu\nu} = \frac{i}{4}\overline{\psi}_{R}\gamma_\mu\overleftrightarrow{\nabla}_\nu\psi_R + (\mu \leftrightarrow \nu)$$
from this calculation. Here, $\nabla$ denotes the spinorial covariant derivative.
