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I am trying to derive the energy-momentum tensor of a Dirac field defined on a Riemann-Cartan background which is a space with a metric-compatible connection and non-zero torsion.

The action is

$$ S = \int_M \mathrm{d}^{3+1}x |e| \frac{i}{2} ( \bar{\psi} \gamma^\mu D_\mu \psi - \overline{D_\mu \psi} \gamma^\mu \psi ) $$

where $D_\mu \psi= \partial_\mu \psi - \frac{1}{8} \omega_{\mu a b} [ \gamma^a, \gamma^b ] \psi$. The energy-momentum tensor is defined as

$$ T^a_\mu \propto \frac{1}{|e|} \frac{\delta S}{\delta e^\mu_a}$$

Essentially, I need to evaluate the variation of $S$ with respect to $e$. Now I am unsure how the spin connection $\omega_{\mu a b}$ varies under a variation of the tetrad $e$. I am aware of this answer here, but it is for a space with zero torsion. We can split the spin-connection up as

$$ \omega_{\mu a b} = \tilde{\omega}_{\mu a b} + K_{\mu a b}$$

where $\tilde{\omega}_{\mu a b}$ the Levi-Civita connection and $K_{\mu a b}$ is the contortion tensor. $\tilde{\omega}_{\mu a b}$ depends on the dreibein and I know how this varies under $e_a^\mu$ but how does $K_{\mu a b}$ vary with respect to $e^\mu_a$?

Qmechanic
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Hermitian_hermit
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1 Answers1

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First of all, your action $$ S = \int_M \mathrm{d}^{3+1}x |e| \frac{i}{2} ( \bar{\psi} \gamma^\mu D_\mu \psi - \overline{D_\mu \psi} \gamma^\mu \psi ) $$ is wrong. It should read $$ S = \int_M \mathrm{d}^{3+1}x |e| \frac{i}{2} ( \bar{\psi} e^\mu_a\gamma^a D_\mu \psi - \overline{D_\mu \psi} e^\mu_a\gamma^a \psi ). $$

And as for deriving the energy-momentum tensor, you should variate against spin connection $\omega_{\mu a b}$ and the tetrad $e$, independently. Follow these steps:

  • Use the variation against $\omega_{\mu a b}$ to yield the torsion tensor. Use this non-zero torsion tensor equations to calculate $\omega_{\mu a b}$ as a function of $e$.
  • Use the variation against $e$ to arrive at the the energy-momentum tensor. And substitute the $\omega_{\mu a b}$ in the energy-momentum tensor equation with the corresponding function of $e$ obtained from the first step.
MadMax
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