# Energy-momentum Tensor for a Real Scalar Field Lagrangian

I'm currently working through Schwartz's QFT book, and I'm trying to find the energy-momentum tensor for the following Lagrangian:

$$L = -\frac{1}2\phi(\Box+m^2)\phi.$$

Am I correct in thinking that,

$$\frac{\partial L}{\partial(\partial_{\mu}\phi)} = 0$$

owing to the fact that there are no first derivatives of $$\phi$$ in the Lagrangian? This seems to disagree with a solution set that I found for the book, hence why I'm asking here, just to be sure.

The other option I see is that

$$\frac{\partial L}{\partial(\partial_{\mu}\phi)} = -\frac{1}{2}\phi\partial^{\nu}\delta_{\mu\nu}$$ but I'm not sure why this would be true and the other not.

Your operator $$\Box$$ can be expressed via $$\partial^\mu \partial_\mu$$, so that you can write the Lagranian density $$\mathcal{L}$$ as $$\mathcal{L} = \frac{1}{2} \partial_\mu \Phi \partial^\mu \Phi - \frac{1}{2} m^2 \Phi^2$$
Therefore, the answer to your question is $$\frac{\partial \mathcal{L}}{\partial(\partial_\mu \Phi)} = \partial_\mu \Phi$$
• Cool, I had also considered that. And I assume that's because we can express $-\phi\Box\phi$ as $\partial_{\mu}\phi\partial^{\mu}\phi$, assuming the total derivative term vanishes? – samcarter Sep 10 '19 at 12:53
• What do you mean? Honestly I have never seen your Lagrangian. I mean, you have $(\Box + m^2)\Phi$ which is zero due to the Klein-Gordon equation and if i remember correctly you have the choice to choose your Lagrangian to be zero. This could be a way of getting yours. – Jan2103 Sep 10 '19 at 12:59
• So, $\phi\Box\phi = \phi\partial_{\mu}^2\phi = \partial_{\mu}(\phi\partial_{\mu}\phi) - \partial^{\mu}\phi\partial_{\mu}\phi$. The total derivative $\partial_{\mu}(\phi\partial_{\mu}\phi)$ can then be dropped as per Schwartz: "our fields vanish on these asymptotic boundaries, which lets us drop [...] total derivatives from Lagrangians." – samcarter Sep 10 '19 at 13:27