I am working my way through Peskin and Schroeder section 2.2 and trying to show that $T^{00}$ is equivalent to the expression $\frac{1}{2}\pi^2-\frac{1}{2}(\nabla \phi)^2-\frac{1}{2}m^2\phi^2$ in equation (2.8) as it suggests.
From $T^\mu_{\;\;\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\partial_\nu – \mathcal{L}\delta^\mu_{\;\nu}$, I get:
$\begin{equation} T^\mu_{\;\;\nu} = \frac{1}{2}\partial^\mu\phi\partial_\nu\phi – \mathcal{L}\delta^\mu_{\;\nu} \end{equation}$
and from there:
$\begin{equation} T^{00} = T^0_{\;\;0} = \frac{1}{2}\partial^0\phi\partial_0\phi – \mathcal{L}\delta^0_{\;0} \end{equation} = \frac{1}{2}\dot{\phi}^2 – \frac{1}{2}[\partial_0\phi^2-\partial_1\phi^2-\partial_2\phi^2-\partial_3\phi^2] + \frac{1}{2}m^2\phi^2$
It looks like I have an extra $-\frac{1}{2}\dot\phi^2$ in my result. Did I make a mistake somewhere?
Best Answer
The lagrangian you're dealing with is $\mathcal{L}= \frac12 (\partial_{\mu} \phi)^2 - \frac12 m^2 \phi^2$. When you take the partial with respect to $\partial_{\mu}\phi$, you should be getting $2 * (\frac12 \partial^{\mu}\phi)$. This would make the first term in your expression $\dot{\phi}^2$ instead of $\frac12 \dot{\phi}^2$ and things would work out.
If you notice in the very next couple of lines he writes the expression for $T^{0i}$. This expression has no $\frac12$ term in front of it either.