# Varying the metric with respect to the Veirbein

1. Jun 25, 2015

### Physicist97

Hello! Given a metric in terms of the Veirbein, $g_{\mu\nu}=e^{a}_{\mu}e^{b}_{\nu}{\eta}_{ab}$ , how would you go about varying it with respect to $e^{a}_{\mu}$ ? I know that ${\delta}g_{\mu\nu}={\delta}e^{a}_{\mu}e^{b}_{\nu}{\eta}_{ab}+e^{a}_{\mu}{\delta}e^{b}_{\nu}{\eta}_{ab}$ , with ${\delta}{\eta}_{ab}=0$ . Then I would divide both sides by ${\delta}e^{a}_{\mu}$ , but that leaves me with the term ${\frac{{\delta}e^{b}_{\nu}}{{\delta}e^{a}_{\mu}}}$ . What would I do with that? Thanks for any help :)

2. Jun 25, 2015

### fzero

Typically you'd assume that the components of the vierbein are independent variables, so that $\delta e^b_\nu / \delta e^a_\mu = \delta^b_a \delta^\mu_\nu$.

3. Jun 25, 2015

### Physicist97

Thank you! I have another question, assuming there is torsion, how would you go about taking the variation of the christoffel symbol with respect to the veirbein, ${\delta}{\Gamma}^{\sigma}_{\mu\nu}/{\delta}e^{a}_{\tau}$ ? If it was torsion-free I would just take the christoffel symbols in terms of the metric, and do a very long, algebraically tedious calculation. But I can't seem to find an equation for the christoffel symbols when there is torsion.

4. Jun 25, 2015

### fzero

Are you working from a particular reference? In Einstein-Cartan theory, I believe that the metric and torsion tensor are taken to be independent variables. In this theory, the covariant derivative is

$$\nabla_\mu V_\nu = \partial_\mu V_\nu - ({\Gamma^\rho}_{\mu\nu} + {K^\rho}_{\mu\nu} )V_\rho,~~~(*)$$

where ${\gamma^\rho}_{\mu\nu}$ is the usual Christoffel symbol expressed in terms of the metric and ${K^\rho}_{\mu\nu}$ is the contorsion tensor, related to the torsion tensor ${T^\rho}_{\mu\nu}$ by

$$K_{\rho\mu\nu} = \frac{1}{2} ( T_{\rho\mu\nu} - T_{\mu\nu\rho} +T_{\nu\rho\mu} ).$$

In this way, ${\Gamma^\rho}_{\mu\nu}= {\Gamma^\rho}_{\nu\mu}$ and all of the torsion is contained in ${T^\rho}_{\mu\nu}$. Because of the added term in (*) we say that the connection is no longer compatible with the metric.

I believe that you can also write this theory in terms of a vierbein and a spin connection ${\omega_\mu}^{ab}+{\gamma_\mu}^{ab}$. Here ${\omega_\mu}^{ab}$ is the parti of the spin connection that can be related to the Christoffel symbol (as in the theory without torsion), while ${\gamma_\mu}^{ab}$ contains the torsion via something like

$${\gamma_\mu}^{ab} = K_{\rho\sigma \mu} ( e^{a\rho} e^{b\sigma} - e^{a\sigma} e^{b\rho}).$$

Presumably you can take $e^a_\mu, {\omega_\mu}^{ab}$, and ${\gamma_\mu}^{ab}$ as independent variables.

I referred to this review by Shapiro to gather the formulas together, but I don't think he addresses the variational principle directly in this formalism.

5. Jun 25, 2015

### Physicist97

Thank you again. No I'm not working from any reference. I'm trying to self study GR from Sean Carroll's online notes, and am working with Veirbeins, Spin Connection, etc. because i find them interesting, and to get practice with these things. Your explanation cleared a lot of things up, I appreciate the help.

6. Jun 26, 2015

### naima

You wrote twice veirbein instead of vierbein. The word comes from Germany where vier = 4 (ein, zwei, drei,vier). Tetrad comes Greece (tetra = 4)
Greetings from France.