# Problem 6 of Sheet 1 - Quantum field theory David Tong - Variation of Lagrangian density

Consider the infinitesimal form of the Lorentz transformation derived in the previous question: $$x^\mu \rightarrow x^\mu +\omega^{\mu}_\nu x^\nu$$. Show that the scalar field transforms as $$\phi(x)\rightarrow \phi'(x)=\phi(x)-\omega^{\mu}_\nu x^\nu \partial_\mu \phi(x)$$ and hence show that the variation of the Lagrangian density is a total derivative $$\delta \mathcal{L}=-\partial_\mu (\omega^{\mu}_\nu x^\nu \mathcal{L})$$

We consider the trasformation as $$x^\mu \rightarrow x'^\mu=x^\mu +\delta x^\mu \\ \phi\rightarrow \phi'=\phi+\delta \phi \\ \partial_\mu \phi\rightarrow (\partial_\mu\phi)'=(\partial_\mu \phi)+\delta (\partial_\mu \phi)$$

Using $$\delta \mathcal{L}=\mathcal{L}\left(\phi(x_\mu),\partial_\mu (\phi_\mu),x_\mu\right)-\mathcal{L}\left(\phi'(x'_\mu),\partial_\mu \phi'_\mu(x'_\mu),x'_\mu\right)$$ Using the Taylor expansion, we find that ( See No-Nonsense QFT Eq. 4.32 ) $$\delta \mathcal{L}=-\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi\right)-\frac{\partial \mathcal{L}}{\partial x_\mu}\delta x_\mu$$ Putting the values, we have $$\delta \mathcal{L}=-\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \omega^\lambda_\nu x^\nu \partial_\lambda \phi\right)-\frac{\partial \mathcal{L}}{\partial x_\mu} \omega^\mu_\nu x^\nu$$

I don't find any further simplification. Can you help me on how to proceed?

In the last equation in your question, the term involving $$\partial {\mathcal L}/{\partial x^\mu}$$ is zero becuase $${\mathcal L}$$ does not depend explicitly on $$x^\mu$$. Assuming that the other term is correct, you are done.

You can use the fact that $$\mathcal{L}$$, is a scalar field. Thus $$\delta \mathcal{L}$$ is of the same form as $$\delta \phi$$ just by replacing $$\phi \rightarrow \mathcal{L}$$. We further use the fact that $$\omega$$ is antisymmetric. If you use these two hints, it should be enough to solve the problem.

The rest of the answer is about how the calculation actually proceeds, if you need it. I do however recommend trying it first using just the small hints above:

With this, you get $$\delta \mathcal{L} = - {\omega^\mu}_\nu x^\nu\partial_\mu \mathcal{L}$$. Now we note that $$\omega$$ is antisymmetric, i.e. $$\omega_{\mu\nu} = - \omega_{\nu\mu} \implies Tr(\omega) = 0$$. Then $$\delta \mathcal{L} = - {\omega^\mu}_\nu x^\nu\partial_\mu \mathcal{L} = - \partial_\mu ({\omega^\mu}_\nu x^\nu \mathcal{L}) + \delta^\nu_\mu \omega^\mu_\nu \mathcal{L}$$. Now using the fact that the trace of $$\omega$$ vanishes, you get the desired result.

I will try to complement the answer by Thomas, since some things were not that clear to me.

We start by varying the Lagrangian $$\mathcal{L}(\phi(x),\partial_{\mu}\phi(x),x^{\mu}):$$

$$\begin{split}\delta \mathcal{L} &= \frac{\partial \mathcal{L}}{\partial \phi}\delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\delta(\partial_{\mu}\phi)+\frac{\partial \mathcal{L}}{\partial x^{\mu}}\delta x^{\mu} \\ &=\partial_{\mu} \bigg{(}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta \phi\bigg{)}+ \frac{\partial \mathcal{L}}{\partial x^{\mu}}\delta x^{\mu},\end{split}\tag{1}$$

where in the second line we used the chain rule and imposed the equation of motion. Then, the first term vanishes, because the Lagrangian should be a Lorentz scalar. Generally, the Lagrangian is a Lorentz scalar if and only if $$\delta \mathcal{L}=\delta x^{\mu}\partial_{\mu}\mathcal{L}.\tag{2}$$

So in the end, we just have

$$\delta \mathcal{L}=\frac{\partial \mathcal{L}}{\partial x^{\mu}}\delta x^{\mu}=-{\omega^{\ \mu}}_{\nu}x^{\nu}\partial_{\mu} \mathcal{L},\tag{3}$$

where the minus comes from doing an active rather than a passive transformation. The result then follows trivially using the property that $${\omega^{\ \mu}}_{\mu}=0$$.

Many thanks to ACuriousMind who helped me with this in The h Bar.