There are two versions when deriving Noether's theorem, see (581172). One is where both the coordinates and fields are varied, leading to a total variation of a field $\phi'(x') = \phi(x) + \Delta \phi(x)$. The other is where we are only concerned with field variations $\phi'(x) = \phi(x) + \delta \phi(x)$. In the latter we can always reformulate coordinate variations as field variations, see (359199, 759877). For example, for an infinitesimal translation $x'^\mu = x^\mu - \epsilon^\mu$ of scalar fields $\psi(x)$, we can always utilize $\psi'(x') = \psi(x)$ to find the form of the field variations as $\psi'(x - \epsilon) = \psi(x) \rightarrow \psi'(x) = \psi(x + \epsilon) = \psi(x) + \partial_\lambda \psi(x) \epsilon^\lambda$.
If we start with field variations only (see Peskin&Schroeder), the conserved current is,
\begin{equation} j^\mu = \frac{\partial \mathcal{L}}{\partial \partial_\mu \phi} \delta \phi - K^\mu \end{equation}
for some field $K^\mu$ that satisfies $\mathcal{L}' = \mathcal{L} + \delta \mathcal{L} = \mathcal{L} + \partial_\mu K^\mu$.
Let us determine the conserved current of the electromagnetic field under Lorentz transformation. Let $x'^\mu = \Lambda^\mu_{\; \nu} x^\nu$ be the Lorentz transformation of the coordinates and the infinitesimal version is $x'^\mu = x^\mu + \omega^\mu_{\;\nu} x^\nu$ where $\omega^\mu_{\;\nu}$ is an infinitesimal antisymmetric operator. A vector field $A_\nu(x)$ transforms as,
\begin{align} & A'_\nu(x') = \frac{\partial x^\mu}{\partial x'^\nu} A_\mu(x)\\ & A'_\nu(\Lambda x) = (\delta^\mu_{\; \nu} - \omega^\mu_{\;\nu}) A_\mu(x)\\ & A'_\nu(x) = (\delta^\mu_{\; \nu} - \omega^\mu_{\;\nu}) A_\mu(\Lambda^{-1} x)\\ & A'_\nu(x) = (\delta^\mu_{\; \nu} - \omega^\mu_{\;\nu}) A_\mu(x - \omega x)\\ & A'_\nu(x) = (\delta^\mu_{\; \nu} - \omega^\mu_{\;\nu}) (A_\mu(x) - \partial_\alpha A_\mu(x) \omega^\alpha_{\;\beta} x^\beta)\\ & A'_\nu(x) = A_\nu(x) - A_\mu(x) \omega^\mu_{\;\nu} - \partial_\alpha A_\nu(x) \omega^\alpha_{\;\beta} x^\beta, \qquad \text{up to order}\; \omega\\ & A'_\nu(x) = A_\nu(x) - A_\alpha(x) \omega^\alpha_{\;\nu} - \partial_\alpha A_\nu(x) \omega^\alpha_{\;\beta} x^\beta, \qquad \text{relabel}\; \mu \;\text{to}\; \alpha \end{align}
Thus, the variation of the field is $\delta A_\nu(x) = - A_\alpha(x) \omega^\alpha_{\;\nu} - \partial_\alpha A_\nu(x) \omega^\alpha_{\;\beta} x^\beta$. The variation of the electromagnetic field action is,
\begin{align} \delta S & = -\frac{1}{4} \delta \int d^4x F^{\mu \nu} F_{\mu \nu}\\ & = -\frac{1}{2} \int d^4x F^{\mu \nu} \delta F_{\mu \nu}\\ & = - \int d^4x F^{\mu \nu} \partial_\mu \delta A_\nu\\ & = \int d^4x F^{\mu \nu} \partial_\mu (\partial_\alpha A_\nu \omega^\alpha_{\;\beta} x^\beta + A_\alpha \omega^\alpha_{\;\nu})\\ & = \int d^4x (F^{\mu \nu} \partial_\mu \partial_\alpha A_\nu \omega^\alpha_{\;\beta} x^\beta + F^{\mu \nu} \partial_\alpha A_\nu \omega^\alpha_{\;\mu} + F^{\mu \nu} \partial_\mu A_\alpha \omega^\alpha_{\;\nu})\\ & = \int d^4x (F^{\mu \nu} \partial_\mu \partial_\alpha A_\nu \omega^\alpha_{\;\beta} x^\beta - F^{\mu \nu} \partial_\alpha A_\mu \omega^\alpha_{\;\nu} + F^{\mu \nu} \partial_\mu A_\alpha \omega^\alpha_{\;\nu})\\ & = \int d^4x (F^{\mu \nu} \partial_\mu \partial_\alpha A_\nu \omega^\alpha_{\;\beta} x^\beta + F^{\mu \nu} F_{\mu \alpha} \omega^\alpha_{\;\nu}) \end{align}
In the second term of the third to the last line, the indices $\mu,\nu$ were switched then applied the antisymmetry of $F^{\mu\nu}$. So, $\delta \mathcal{L} = F^{\mu \nu} \partial_\mu \partial_\alpha A_\nu \omega^\alpha_{\;\beta} x^\beta + F^{\mu \nu} F_{\mu \alpha} \omega^\alpha_{\;\nu}$. I have to find a way to express this as $\partial_\alpha K^\alpha$. One can guess by trying to let $K^\alpha = -\mathcal{L} \omega^\alpha_{\;\beta} x^\beta$ and checking if $\partial_\alpha (-\mathcal{L} \omega^\alpha_{\;\beta} x^\beta) = F^{\mu \nu} \partial_\mu \partial_\alpha A_\nu \omega^\alpha_{\;\beta} x^\beta + F^{\mu \nu} F_{\mu \alpha} \omega^\alpha_{\;\nu}$.
\begin{align} \partial_\alpha (-\mathcal{L} \omega^\alpha_{\;\beta} x^\beta) & = \frac{1}{4} \partial_\alpha (F^{\mu \nu} F_{\mu \nu})\omega^\alpha_{\;\beta} x^\beta - \mathcal{L} \omega^\alpha_{\;\alpha}, \qquad \omega^\alpha_{\;\alpha} = 0\\ & = \frac{1}{2} F^{\mu \nu} \partial_\alpha F_{\mu \nu} \omega^\alpha_{\;\beta} x^\beta\\ & = F^{\mu \nu} \partial_\alpha \partial_\mu A_\nu \omega^\alpha_{\;\beta} x^\beta \end{align}
This is equal to the first term of $\delta \mathcal{L}$, the only difference is the second term $F^{\mu \nu} F_{\mu \alpha} \omega^\alpha_{\;\nu}$. I'm not sure what to do with this or maybe the guess for $K^\alpha$ is incorrect. Any help?
If $K^\mu = -\eta^\mu_{\;\alpha} \mathcal{L} \omega^\alpha_{\;\beta} x^\beta$ is correct then,
\begin{align} j^\mu & = \frac{\partial \mathcal{L}}{\partial \partial_\mu A_\nu} \delta A_\nu - K^\mu\\ & = -F^{\mu\nu} \delta A_\nu - K^\mu\\ & = F^{\mu\nu} \partial_\alpha A_\nu \omega^\alpha_{\;\beta} x^\beta + F^{\mu\nu} A_\alpha \omega^\alpha_{\;\nu} + \eta^\mu_{\;\alpha} \mathcal{L} \omega^\alpha_{\;\beta} x^\beta \end{align}
Due to the second term, I cannot write the energy-momentum tensor $T^\mu_{\; \alpha}$ as $j^\mu = T^\mu_{\; \alpha} \omega^\alpha_{\;\beta} x^\beta$. Any hints?
