Energy Momentum Tensor for a Dirac Fermion coupled to external gravity

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:

1. Variation of the $\sqrt{-\text{det}(g)}$ piece. This is easy.
2. 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.

You can write

$\gamma^\mu(\partial_\mu + \frac{1}{2} \omega_\mu) = \gamma_\nu g^{\nu \mu}(\partial_\mu + \frac{1}{2} \omega_\mu) = \gamma_\nu g^{\mu \nu}(\partial_\mu + \frac{1}{2} \omega_\mu)$. (A)

Furthermore

$\gamma^\mu(\partial_\mu + \frac{1}{2} \omega_\mu) = \gamma_\mu g^{\mu \nu}(\partial_\nu + \frac{1}{2} \omega_\nu)$. (B)

If you add equations A and B you obtain

$2 \gamma^\mu(\partial_\mu + \frac{1}{2} \omega_\mu) = \gamma_\nu g^{\mu \nu}(\partial_\mu + \frac{1}{2} \omega_\mu)+\gamma_\mu g^{\mu \nu}(\partial_\nu + \frac{1}{2} \omega_\nu)$

and then differentiate by $g^{\mu \nu}$ to obtain the identity.

• Thanks! That is very neat. But don’t you also generate $-g_{\mu\nu}i\overline{\psi}_R\nabla\psi_R$ in the functional differentiation wrt the background metric? – leastaction Jul 23 '17 at 22:18
• The product $(\omega_\mu \psi_R)^a$ with spinor indices $a,b$ is equivalent to $\Omega_\mu^{ab} \psi_{Rb}$; this can also be transformed to with Greek letters denoted Lorentzian indices (with tetrads) to $\Omega_\mu^{\alpha \beta} \psi_{R \beta}$. Now use the same trick as showed above with representing contraction of indices with metric Tensor and derive with respect to it. – kryomaxim Jul 24 '17 at 14:10