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 $\omega$),


I'm resisting the idea that ${\omega^\nu}_\mu\partial_\nu\phi=0$, but I don't understand what I'm doing wrong.


2 Answers 2


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.

  • $\begingroup$ Okay, this seems to do it, but then is the correct way that $\partial_\mu\phi$ transforms (under an infinitesimal Lorentz transformation) still as $\partial_\mu\phi-{\omega^\sigma}_\rho{x}^\rho\partial_{\sigma\mu}\phi-{\omega^{\nu}}_\mu\partial_\nu\phi$ ? $\endgroup$
    – user24999
    Sep 14, 2014 at 0:07
  • $\begingroup$ @PedroFigueroa Right. $\endgroup$
    – higgsss
    Sep 14, 2014 at 0:13
  • $\begingroup$ Will this fail to hold if $\mathcal{L}$ is not a Lorentz scalar, then? Lorentz-invariance means the action is a Lorentz scalar, but the Lagrangian density $\mathcal{L}$ might not be, correct? $\endgroup$
    – Sam Jaques
    Aug 11, 2020 at 17:42

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!

  • $\begingroup$ Sure, I know what the correct result is, however my question is about the extra term that arises when computing directly $\delta{L}=\frac{\partial{L}}{\partial\phi}\delta\phi+\frac{\partial{L}}{\partial{(\partial_\mu\phi)}}\delta(\partial_\mu\phi)$ $\endgroup$
    – user24999
    Sep 13, 2014 at 23:56

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