2
$\begingroup$

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?

$\endgroup$
6
  • 1
    $\begingroup$ 1. $\partial \mathcal{L}/ \partial{\phi,_\mu}= \partial^\mu \phi$ (your 4th formula). 2. Summation indices should occur only twice (also 4th formula). $\endgroup$
    – Hyperon
    Feb 27, 2023 at 21:42
  • $\begingroup$ 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 $\endgroup$ Feb 27, 2023 at 22:05
  • 1
    $\begingroup$ Your factor $1/2$ is wrong. As I said, $\partial \mathcal{L} / \partial \phi,_\mu = \partial^\mu \phi$ and not $\frac{1}{2} \partial^\mu \phi$. $\endgroup$
    – Hyperon
    Feb 27, 2023 at 22:10
  • $\begingroup$ 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 $\endgroup$ Feb 27, 2023 at 22:12
  • $\begingroup$ $\partial (\eta^{\rho \sigma} \phi,_\rho \phi,_\sigma) / \partial \phi,_\mu = 2 \phi^{,\mu}$ $\endgroup$
    – Hyperon
    Feb 27, 2023 at 22:17

1 Answer 1

4
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.