# Going from the Vielbein- to the metrical/Belifante-energy-momentum tensor

I was calculating the energy-momentum tensor for a spinor field by varying its action with respect to the background metric. I looked on StackExchange whether there was something that addresses the following issue, but most other questions that mention the energy-momentum tensor of a spinor field or calculating the Belifante energy-momentum tensor skip over the following thing in the notation that confuses me:

When calculating the energy-momentum tensor one can take the variation of the action with respect to the metric. If we want to know the action of e.g. a spinor field this can be done by taking the variation with respect to the vielbein. $$T^{\mu\nu} = \frac{2}{\sqrt{|g|}} \frac{\delta S}{\delta g_{\mu\nu}} = \frac{e^{\nu a}}{e} \frac{\delta S}{\delta e_\mu^a} + \frac{e^{\mu b}}{e} \frac{\delta S}{\delta e_\nu^b}$$ Here I used that $\delta(\eta_{ab} e^a_\mu e^b_\nu) = \eta_{ab} e_{( \nu}^b \delta e^a_{\mu)} = e_{( \nu a} \delta e^a_{\mu)}$. Often you see people calculate $T_a^\mu$ instead of $T^{\mu\nu}$, where $T_a^\mu$ is defined as $$T_a^\mu = \frac{1}{e} \frac{\delta S}{\delta e_\mu^a}$$ What confuses me, is how to connect these two expressions. Why can I not go from this expression for $T_a^\mu$, back to the earlier expression for $T^{\mu\nu}$? $$T^{\mu\nu} \neq \eta^{ab} e^\mu_b T_a^\nu = \eta^{ab} e^\mu_b \frac{1}{e} \frac{\delta S}{\delta e_\nu^a} = \frac{e^{\mu a}}{e} \frac{\delta S}{\delta e_\nu^a}$$ I now end up missing the second term where $\mu$ and $\nu$ are interchanged.

In the most general case, the energy-momentum tensor is not given by $$T^{\mu\nu} = \frac{2}{\sqrt{|g|}} \frac{\delta S}{\delta g_{\mu\nu}}\tag{A}$$ because in general $S$ might depend on the vielbein $e^a{}_{\mu}$ by itself, and not just through the combination $g_{\mu\nu}=\eta_{ab}e^a{}_{\mu}e^b{}_{\nu}$. Therefore, $\frac{\delta S}{\delta g_{\mu\nu}}$ need not be well-defined.
The correct, general expression for the energy-momentum tensor is $$T^{\mu\nu}=\eta^{ab} e^\mu_b T_a^\nu,\qquad\text{with}\qquad T_a^\mu \equiv \frac{1}{e} \frac{\delta S}{\delta e_\mu^a} \tag{B}$$ which is more general than the previous one, and need not be symmetric in $\mu\leftrightarrow\nu$. If and only if the action depends on $e^a{}_{\mu}$ through the combination $g_{\mu\nu}=\eta_{ab}e^a{}_{\mu}e^b{}_{\nu}$ then the energy-momentum tensor is symmetric, $$T^{\mu\nu}=\frac12\left(\eta^{ab} e^\mu_b T_a^\nu+\eta^{ab} e^\nu_b T_a^\mu\right)\tag{C}$$ and it agrees with the standard expression $(\mathrm A)$.