In problem 3.3 of Schwartz's QFT, the first two questions ask us to prove that if we add a total derivative to the Lagrangian: $$ \mathcal{L}\mapsto\mathcal{L}+\partial_\mu X^\mu\tag{1} $$ then $$ \int\mathrm{d}^3\mathbf{x}~T^{00}\tag{2} $$ does not change.
I successfully proved it, essentially the same as that in this post, assuming $X$ does not depend on derivatives of $\phi_\alpha$.
The third question, then, asks us to find a $X^\mu$ to be added to $$ \mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu} $$ such that the resulting energy-momentum tensor $T^{\mu\nu}$ is symmetric, where $$ F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu $$
I figured out that $$ X^\mu=\frac{1}{2}A_\nu F^{\mu\nu}\tag{3} $$ does the trick.
However, here comes my problem.
I tried to show that $(2)$ is invariant under such an explicit $X^\mu$. Since now $X^\mu$ depends on $\partial A$, I do the calculation from scratch: $$ \begin{align*} \Delta T^{\mu\nu}&=\frac{\partial\left(\partial_\lambda X^\lambda\right)}{\partial\left(\partial_\mu\phi_\alpha\right)}\partial^\nu\phi_\alpha-g^{\mu\nu}\partial_\lambda X^\lambda\\ &=\frac{1}{2}\frac{\partial\left(\frac{\partial X^\lambda}{\partial\phi_\beta}\partial_\lambda\phi_\beta\right)}{\partial\left(\partial_\mu\phi_\alpha\right)}\partial^\nu\phi_\alpha-g^{\mu\nu}\partial_\lambda X^\lambda\\ &=\frac{1}{2}\left(\frac{\partial X^\lambda}{\partial\phi_\beta}\delta^\mu_\lambda\delta^\alpha_\beta+\frac{\partial\left(\frac{\partial X^\lambda}{\partial\phi_\beta}\right)}{\partial\left(\partial_\mu\phi_\alpha\right)}\partial_\lambda\phi_\beta\right)\partial^\nu\phi_\alpha-g^{\mu\nu}\partial_\lambda X^\lambda\\ &=\frac{1}{2}\left(\frac{\partial X^\mu}{\partial\phi_\alpha}+\frac{\partial\left(\frac{\partial X^\lambda}{\partial\phi_\beta}\right)}{\partial\left(\partial_\mu\phi_\alpha\right)}\partial_\lambda\phi_\beta\right)\partial^\nu\phi_\alpha-g^{\mu\nu}\partial_\lambda X^\lambda \end{align*} $$ Note that I deliberately kept $$ \frac{\partial\left(\frac{\partial X^\lambda}{\partial\phi_\beta}\right)}{\partial\left(\partial_\mu\phi_\alpha\right)} $$ because it does not vanish now.
Put $(3)$ in, replacing $\phi_\alpha=A_\alpha$, we have $$ \begin{align*} \frac{\partial X^\mu}{\partial\phi_\alpha}&=\frac{\partial \left(A_\nu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)\right)}{\partial A_\alpha}=\partial^\mu A^\alpha-\partial^\alpha A^\mu\\ \frac{\partial\left(\frac{\partial X^\lambda}{\partial\phi_\beta}\right)}{\partial\left(\partial_\mu\phi_\alpha\right)}&=\frac{\partial\left(\partial^\lambda A^\beta-\partial^\beta A^\lambda\right)}{\partial\left(\partial_\mu A_\alpha\right)}=\delta^{\lambda\mu}\delta^{\beta\alpha}-\delta^{\beta\mu}\delta^{\alpha\lambda} \end{align*} $$ thus $$ \begin{align*} \Delta T^{\mu\nu}&=\frac{1}{2}\left(\partial^\mu A^\alpha-\partial^\alpha A^\mu+\left(\delta^{\lambda\mu}\delta^{\beta\alpha}-\delta^{\beta\mu}\delta^{\alpha\lambda}\right)\partial_\lambda A_\beta\right)\partial^\nu A_\alpha-g^{\mu\nu}\partial_\lambda X^\lambda\\ &=\frac{1}{2}\left(\partial^\mu A^\alpha-\partial^\alpha A^\mu+\partial^\mu A^\alpha-\partial^\alpha A^\mu\right)\partial^\nu A_\alpha-g^{\mu\nu}\partial_\lambda X^\lambda\\ &=\left(\partial^\mu A^\alpha-\partial^\alpha A^\mu\right)\partial^\nu A_\alpha-g^{\mu\nu}\partial_\lambda X^\lambda\\ &=F^{\mu\alpha}\partial^\nu A_\alpha-g^{\mu\nu}\partial_\lambda\left(A_\gamma F^{\lambda\gamma}\right)\tag{4}\\ \end{align*} $$ Therefore, $$ \begin{align*} \int\mathrm{d}\mathbf{x}^3 \Delta T^{00}&=\int\mathrm{d}\mathbf{x}^3\left(2F^{0\alpha}\partial^0 A_\alpha-\partial_\lambda\left(A_\gamma F^{\lambda\gamma}\right)\right)\\ &=\int\mathrm{d}\mathbf{x}^3\left(2F^{0\alpha}\partial^0 A_\alpha-\partial_0\left(A_\gamma F^{0\gamma}\right)\right)\\ &=\int\mathrm{d}\mathbf{x}^3\left(2F^{0\alpha}\partial^0 A_\alpha-\left(\partial_0A_\gamma \right)F^{0\gamma}-A_\gamma\partial_0F^{0\gamma}\right)\\ &=\int\mathrm{d}\mathbf{x}^3\left(F^{0\alpha}\partial^0 A_\alpha-A_\gamma\partial_0F^{0\gamma}\right)\\ &=\int\mathrm{d}\mathbf{x}^3\left(\left(\partial^0 A^\alpha-\partial^\alpha A^0\right)\partial^0 A_\alpha-A_\alpha\partial_0\left(\partial^0 A^\alpha-\partial^\alpha A^0\right)\right)\\ \end{align*} $$ which I failed to see why it should vanish.
In summary, my question is:
- Does transformation $(1)$ preserve $(2)$ for arbitrary $X^\mu$? From the question of Schwartz's book, it does. If so, where did I go wrong in my calculation?
- this post shows that only derivatives to first order should appear in the Lagrangian. Then, is adding the total derivative of $(3)$, after which the Lagrangian will contain second order derivatives, valid?
- Furthermore, after integrating by parts, the whole transformation seems vanishing at all: $$ \begin{align*} \partial_\mu X^\mu &=\frac{1}{2}\partial_\mu\left(A_\nu\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)\right)\\ &=\frac{1}{2}\left(\partial_\mu A_\nu\right)\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)-\frac{1}{2} \left(\partial_\mu A_\nu\right)\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)\\ &=0 \end{align*} $$ from which it seems the physics should remain the same, and in particular $(2)$ should be preserved, right?
Update
This post argues it is impossible to find a $X^\mu$ such that the energy-momentum tensor of $$ \mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\partial_\mu X^\mu $$ is symmetric. However, I believe my $(3)$ is a solution. We have $$ \begin{align*} T^{\mu\nu}&=\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu A_\alpha\right)}\partial^\nu A_\alpha-g^{\mu\nu}\mathcal{L}\\ &=F^{\alpha\mu}\partial^\nu A_\alpha+\frac{1}{4}g^{\mu\nu}F^2+\Delta T^{\mu\nu}\\ \end{align*} $$ Put in $(4)$ we get $$ \begin{align*} T^{\mu\nu}&=F^{\alpha\mu}\partial^\nu A_\alpha+\frac{1}{4}g^{\mu\nu}F^2+F^{\mu\alpha}\partial^\nu A_\alpha-g^{\mu\nu}\partial_\lambda\left(A_\gamma F^{\lambda\gamma}\right)\\ &=g^{\mu\nu}\left(\frac{1}{4}F^2-\partial_\lambda\left(A_\gamma F^{\lambda\gamma}\right)\right) \end{align*} $$ which is symmetric, right?
