# Trace of the energy-momentum tensor

I had to find the canonical energy-momentum tensor defined by this Lagrangian density $$\mathcal{L} = - {1 \over 4} F_{\mu \nu} F^{\mu \nu}$$ and I got the result of $$T^{\mu \nu} = - F^{\mu \lambda} \partial^{\nu} A_{\lambda} + {1 \over 4} \eta^{\mu \nu} F^{\rho \sigma} F_{\rho \sigma}$$ In order to make it a symmetric tensor, I had to add total derivative term and knowing from equation of motion that $$\partial_{\lambda} F^{\mu \lambda}=0$$, I got new tensor $$\hat{T}^{\mu \nu} = F^{\mu \lambda}F^{\nu}_{\lambda} + {1 \over 4} \eta^{\mu \nu} F^{\rho \sigma} F_{\rho \sigma}$$ which is symmetric. The question is - what is the trace of $$\hat{T}^{\mu \nu}$$?

I know that, as far as it comes to the stress-energy tensor $$T^{\mu \nu}$$, we can write the trace as $$T^{\mu}_{\mu}$$ and using the formula for the energy-momentum tensor, do we get sth like that $$T^{\mu}_{\mu} = {{\partial \mathcal{L}} \over {\partial (\partial_{\mu} A_{\lambda})}} \partial_{\nu} A_{\lambda} - \delta^{\mu}_{\nu} \mathcal{L}$$? Honestly, I can't see what would the result for $$\hat{T}^{\mu \nu}$$ be. Would it be that the canonical stress-energy tensor is traceless? But how actually prove it?

• $T^{\mu}_{\mu} = {{\partial \mathcal{L}} \over {\partial (\partial_{\mu} A_{\lambda})}} \partial_{\nu} A_{\lambda} - \delta^{\mu}_{\nu} \mathcal{L}$ This equation doesn’t make sense because the indices are inconsistent. You have no free (i.e., uncontracted) indices on the left, but two free indices ($\mu$ and $\nu$) on the right. Dec 20, 2020 at 18:18

You can find the trance of $$T^{\mu \nu}$$ like this,
$$T^\mu_{\;\; \mu} = \eta_{\mu \nu} T^{\mu \nu} = \eta_ {\mu \nu} \Big(F^{\mu \rho}F^{\nu}_{\rho} + \frac{1}{4} \eta^{\mu \nu} F^{\alpha \beta}F_{\alpha \beta} \Big)$$
• I actually meant the trace of this improved tensor, namely $\hat{T}^{\mu \nu}$, not this $T^{\mu \nu}$. Perhaps it wasn't clear. Dec 20, 2020 at 21:05
• @Chakalaka Please observe that Kian did write the expression for your $\hat T^{\mu\nu}$, they simply did not include the hats. Besides this, the definition of the trace of a tensor is independent of the tensor's definition. Dec 21, 2020 at 1:46