# Understanding the Energy-Momentum Tensor for the Klein-Gordon Field

On Peskin & Schroder's QFT Book, page 19, they give us the conserved charge associated with spatial translations (equation 2.19): $$P^i=\int T^{0i}d^3x=-\int\pi \partial_i\phi d^3x$$ where $$T^\mu_{\ \ \ \nu}$$ is the energy momentum tensor: $$T^\mu_{\ \ \ \nu}=\frac{\partial L}{\partial(\partial_\mu\phi)}\partial_\nu \phi-L\delta^\mu_\nu$$ Where $$L$$ is the lagrangian density. Now, the book doesn't actually show the intermediate steps, so I decided to try to do them myself. And although I think I am close, there is something that eludes me.

My attempt is: Start with the Lagrangian density for the Klein-Gordon Field: $$L=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-\frac{1}{2}m^2\phi^2$$ Now I plug this into the energy-momentum tensor: $$T^\mu_{\ \ \ \nu}=\partial^\mu\phi\partial_\nu\phi-\delta^\mu_\nu(\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-\frac{1}{2}m^2\phi^2)$$ Now, I know that momentum corresponds to $$T^{0i}$$, so I raise the $$\nu$$ everywhere and set $$\mu=0$$.(Recognizing it as the time derivative) $$T^{0\nu}=\partial_t\phi\partial^\nu\phi-(\frac{1}{2}\partial_t \phi \partial_t \phi-\frac{1}{2}m^2\phi^2)$$ Now if I were to take $$\nu=i$$, I would find $$T^{0i}=\partial_t\phi\partial^i\phi-(\frac{1}{2}\partial_t \phi \partial_t \phi-\frac{1}{2}m^2\phi^2)$$ Lowering the index on the first term: $$T^{0i}=-\partial_t\phi\partial_i\phi-(\frac{1}{2}\partial_t \phi \partial_t \phi-\frac{1}{2}m^2\phi^2)$$ Now, the first term resembles what I am looking for, keeping in mind that $$\pi=\partial_t\phi$$. But I don't know how to get rid of the $$1/2$$ or the 2nd and third term. What is the way forward here?

• 1. $\partial \mathcal{L}/ \partial{\phi,_\mu}= \partial^\mu \phi$ (your 4th formula). 2. Summation indices should occur only twice (also 4th formula). Feb 27, 2023 at 21:42
• 1. I Don't understand. Is that not what I did? 2. I believe I fixed that, but it doesn't seem to have had an impact on the rest of the problem Feb 27, 2023 at 22:05
• Your factor $1/2$ is wrong. As I said, $\partial \mathcal{L} / \partial \phi,_\mu = \partial^\mu \phi$ and not $\frac{1}{2} \partial^\mu \phi$. Feb 27, 2023 at 22:10
• Ok, I understand what you meant. Sorry if this is a basic question, but I dont' understand what happened to it. since we had $\partial_\mu\partial^\mu$, Where would a factor of two come from? I would understand if it was $(\partial_\mu)^2$, but I don't see it here Feb 27, 2023 at 22:12
• $\partial (\eta^{\rho \sigma} \phi,_\rho \phi,_\sigma) / \partial \phi,_\mu = 2 \phi^{,\mu}$ Feb 27, 2023 at 22:17

$$\delta_0^{\, \, i}=0$$ for $$i=1,2,3$$ solves your problem.