0
$\begingroup$

I want to derive the Dirac massless equation in curved spacetime from the action. I have the symmetric form of the Dirac action: $$S = \frac{1}{2} \int \bigg[i\bar{\psi} \gamma^\mu D_\mu \psi - i D_\mu \bar{\psi} \gamma^\mu \psi \bigg] \sqrt{-g} d^4 x$$ where $D_\mu \bar{\psi}= \partial_\mu + \frac{i}{4} \omega_{\mu a b} \sigma^{ab} \bar{\psi}$ ($\omega$ is the spin connection and $\sigma^{ab} = \frac{i}{2} [\gamma^a, \gamma^b]$) and I proceed to do $$\frac{\delta S}{\delta \bar{\psi}} = \frac{1}{2} \int \bigg[i \gamma^\mu D_\mu \psi - i \frac{\delta }{\delta \bar{\psi}} (D_\mu \bar{\psi}) \gamma^\mu \psi \bigg] \sqrt{-g} d^4 x$$ If I'm not mistaken, $\delta$ and $D_\mu$ commute since I'm varying the matter fields over the background of a fixed metric and so $$\frac{\delta S}{\delta \bar{\psi}} = \frac{1}{2} \int \bigg[i \gamma^\mu D_\mu \psi - i D_\mu \frac{\delta \bar{\psi}}{\delta \bar{\psi}} \gamma^\mu \psi \bigg] \sqrt{-g} \, d^4 x \overbrace{=}^{\color{red}{!}} \frac{1}{2} \int \bigg[i \gamma^\mu D_\mu \psi - i D_\mu \gamma^\mu \psi \bigg] \sqrt{-g} d^4 x $$ If this is correct, I'm not sure how I can proceed. The Dirac equation is supposed to be $$i \gamma^\mu D_\mu \psi = 0$$ and thus the only way for that to happen is IF $$- i D_\mu \gamma^\mu \psi = i \gamma^\mu D_\mu \psi$$ but this doesn't seem to be correct. What am I misunderstanding here?

EDIT: $\frac{\delta \bar{\psi}}{\delta \bar{\psi}} \neq 1$, as mentioned in the comments. Therefore, we have $D_\mu (something) * \gamma^\mu \psi$. This means that the next term (with the red exclamation mark) in my computations is wrong. By doing partial integration, we get rid of a surface term and we change the derivative to $\gamma^\mu \psi$. Therefore, the correct relation is

$$\frac{\delta S}{\delta \bar{\psi}} = \frac{1}{2} \int \bigg[i \gamma^\mu D_\mu \psi + i D_\mu (\gamma^\mu \psi) \bigg] \sqrt{-g} d^4 x $$

where indeed the sign has changed. Now, as mentioned in the answer, $D_\mu \gamma^\nu = 0$ and thus

$$D_\mu (\gamma^\mu \psi) = (D_\mu \gamma^\mu) \psi + \gamma^\mu D_\mu \psi = \gamma^\mu D_\mu \psi$$

effectively changing the order of $D_\mu$ and $\gamma^\mu$. This means that

$$\frac{\delta S}{\delta \bar{\psi}} = \int \bigg[i \gamma^\mu D_\mu \psi \bigg] \sqrt{-g} d^4 x $$

and thus we get the desired Dirac equation in curved spacetime.

Improve this question
$\endgroup$
2
  • $\begingroup$ You can't replace $ \frac{ \delta {\bar \psi} }{ \delta {\bar \psi} } = 1$. The correct thing is $ \frac{ \delta {\bar \psi}(x) }{ \delta {\bar \psi}(x') } = \delta^4(x-x')$. $\endgroup$
    – Prahar
    Commented Feb 22 at 19:01
  • $\begingroup$ Have you tried to get the standard non-symmetric version first? Because it is difficult to disentangle the complications coming from the symmetric version from the complications coming from Dirac action, so it is much more common for standard texts to cover the non-symmetric version. Once you have the standard version done correctly, then you would be in a much better position to do other stuff. $\endgroup$
    – naturallyInconsistent
    Commented Feb 23 at 4:11

1 Answer 1

Reset to default
1
$\begingroup$

The metricity condition $\omega_{ab\mu}= - \omega_{ba\mu}$, can be used to show that
$$ \partial_\mu \gamma^\nu +\frac 12 [{\sigma^{ab}}, \gamma^\nu] \omega_{ab\mu}+ \gamma^\lambda {\Gamma^\nu}_{\lambda\mu}=0. $$ This equation can can be interpreted as ``$\nabla_\mu \gamma^\nu=0$'' or as the freedom to pass gamma matrices though covariant derivatives provided that suitable connection forms are tacitly or explicitly included.

I think you also have forgotten the minus sign from integrating by parts to take the derivative off $\bar \psi$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ You are correct, there's a missing minus sign from integrating by parts. Could you give some sources for the $\nabla_\mu \gamma^\nu = 0$ relation or explain a bit how it means that gamma matrices can pass through covariant derivatives? $\endgroup$
    – Yaezir
    Commented Feb 22 at 20:37
  • $\begingroup$ Look at my online notes: people.physics.illinois.edu/stone/torsion_review.pdf pages 5 and 8. $\endgroup$
    – mike stone
    Commented Feb 23 at 0:52