The variation of the Lagrangian density under an infinitesimal Lorentz transformation I'm trying to introduce myself to QFT following these lectures by David Tong. I've started with lecture 1 (Classical Field Theory) and I'm trying to prove that under an infinitesimal Lorentz transformation of the form
$$\tag{1.49} {\Lambda^\mu}_\nu={\delta^\mu}_\nu+{\omega^\mu}_\nu,$$
where $\omega$ is antisymmetric, the variation of the Lagrangian density $\mathcal{L}$ is
$$\tag{1.53} \delta\mathcal{L}=-\partial_\mu({\omega^\mu}_\nu{x}^\nu\mathcal{L}).$$
Using $\mathcal{L}=\mathcal{L}(\phi,\partial_\mu\phi)$, I've tried computing $\delta\mathcal{L}$ directly using
$$\tag{1.52} \delta\phi=-{\omega^\mu}_\nu{x}^\nu\partial_\mu\phi$$
[which I obtained earlier computing explicitly $\phi(x)\to\phi(\Lambda^{-1}x)$], however, I get
$$\delta\mathcal{L}=-\partial_\mu({\omega^\mu}_\nu{x}^\nu\mathcal{L})-\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}{\omega^\sigma}_\mu\partial_\sigma\phi$$
The extra term arises when I compute
$$\partial_\mu(\delta\phi)=-{\omega^\sigma}_\nu\left[{\delta^\nu}_\mu\partial_\sigma\phi+x^\nu\partial_{\mu\sigma}\phi\right]=-{\omega^\sigma}_\mu\partial_{\sigma}\phi-{\omega^\sigma}_\nu{x}^\nu\partial_{\mu\sigma}\phi$$
[because I'm assuming $\partial_\mu(\delta\phi)=\delta(\partial_\mu\phi)$]; I thought I'd get rid of it just replacing $\phi$ with $\partial_\sigma\phi$ in $(1.52)$, however $\partial_\mu(\delta\phi)=\delta(\partial_\mu\phi)$ should still hold, ain't it? I also tried using (the previous expression to) 1.27 in the lectures, namely that the derivatives of the field transform as
$$\tag{1.26b} \partial_\mu\phi(x)\to{(\Lambda^{-1})^\nu}_\mu\partial_\nu\phi(\Lambda^{-1}x),$$
but I still get (to the first order in $\omega$),
\begin{align}{(\Lambda^{-1})^\nu}_\mu\partial_\nu\phi(\Lambda^{-1}x)&=({\delta^\nu}_\mu-{\omega^\nu}_\mu)\partial_\nu\phi(x^\sigma-{\omega^\sigma}_\rho{x}^\rho)\\&=({\delta^\nu}_\mu-{\omega^\nu}_\mu)\left[\partial_\nu\phi(x)-{\omega^\sigma}_\rho{x}^\rho\partial_{\sigma\nu}\phi(x)\right]\\&=\partial_\mu\phi-{\omega^\sigma}_\rho{x}^\rho\partial_{\sigma\mu}\phi-{\omega^\nu}_\mu\partial_\nu\phi\end{align} 
I'm resisting the idea that ${\omega^\nu}_\mu\partial_\nu\phi=0$, but I don't understand what I'm doing wrong.
 A: Provided that $\mathcal{L}$ is a Lorentz scalar, the quantity $\partial\mathcal{L}/\partial(\partial_{\mu}\phi)$ has to carry an upper index. Since $\mathcal{L}$ is a function of $\phi$ and $\partial_{\mu}\phi$, the only object that can give such an index is $\partial^{\mu}\phi$. Hence
\begin{equation}
\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\phi)} \propto \partial^{\mu}\phi.
\end{equation}
Then,
\begin{equation}
\begin{split}
\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\phi)} \omega^{\sigma}{}_{\mu}\partial_{\sigma}\phi \,&\propto \,\omega^{\sigma}{}_{\mu}\,\partial_{\sigma}\phi \,\partial^{\mu}\phi\\
&=\omega^{\sigma\mu} \partial_{\sigma}\phi\,\partial_{\mu}\phi\\
&=0.
\end{split}
\end{equation}
The last expression vanishes because $\partial_{\sigma}\phi\,\partial_{\mu}\phi$ is symmetric under the interchange of indices while $\omega^{\sigma\mu}$ is antisymmetric.
I actually don't understand why Tong didn't simply write
\begin{equation}
\delta \mathcal{L} = -\omega^{\mu}{}_{\nu} x^{\nu}\partial_{\mu}\mathcal{L}.
\end{equation}
After all, $\mathcal{L}$ should have the same transformation rule as $\phi$ because they are both Lorentz scalars. One can verify the above equation by noting that
\begin{equation}
\delta \mathcal{L} = - \partial_{\mu}(\omega^{\mu}{}_{\nu}x^{\nu}\mathcal{L}) = -\omega^{\mu}{}_{\nu} x^{\nu}\partial_{\mu}\mathcal{L} - \omega^{\mu}{}_{\mu}\mathcal{L},
\end{equation}
and that
\begin{equation}
\omega^{\mu}{}_{\mu} = \eta_{\mu\rho}\omega^{\mu\rho} = 0
\end{equation}
because $\eta_{\mu\rho}$ is symmetric and $\omega^{\mu\rho}$ is antisymmetric.
A: Cool! I'm working on the exact same thing. 
The way I proved this was since $\mathcal{L}$ and $\phi$ are both Lorentz scalars they must have the same transformation law. Therefore
$$\delta \mathcal{L} = - \omega^{\mu}_{\phantom{\mu}\nu}x^{\nu}\partial_{\mu}\mathcal{L}.$$
However, note that
$$\partial_{\mu} \left( -\omega^{\mu}_{\phantom{\mu}\nu}x^{\nu} \mathcal{L} \right) = -\omega^{\mu}_{\phantom{\mu}\nu} \partial_\mu x^\nu\mathcal{L} - \omega^{\mu}_{\phantom{\mu}\nu}x^{\nu}\partial_{\mu}\mathcal{L}.$$
The first term on the right hand side of the equation is 0:
$$-\omega^{\mu}_{\phantom{\mu}\nu} \partial_\mu x^\nu\mathcal{L} = -\omega^{\mu}_{\phantom{\mu}\nu} \delta_\mu^\nu \mathcal{L}.$$
The expression $\omega^{\mu}_{\phantom{\mu}\nu} \delta_\mu^\nu \mathcal{L}$ is the trace of $\omega$, which is an antisymmetric matrix which is 0. Therefore
$$\partial_{\mu} \left( -\omega^{\mu}_{\phantom{\mu}\nu}x^{\nu} \mathcal{L} \right) = -\omega^{\mu}_{\phantom{\mu}\nu}x^{\nu}\partial_{\mu}\mathcal{L}=\delta \mathcal{L}.$$
So the variation in the Lagrangian density is equal to a total derivative as we set out to prove with
$$F^\mu = -\omega^{\mu}_{\phantom{\mu}\nu} x^{\nu} \mathcal{L}.$$
$$\partial_\mu F^\mu = \delta \mathcal{L}.$$
I'm fairly new to this myself, especially this nasty index manipulation, so if you go over my logic and find it sound please let me know.
Thanks, and cheers!
