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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?

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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.

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  • $\begingroup$ So then both factors in $\partial_\mu \phi \partial^\mu \phi$ are both treated as though they had a lower index when you do the derivative? That's fine, but it's less obvious to me now why you end up with the upper index in the end. $\endgroup$ – JohnnyMo1 Feb 23 '15 at 14:37
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    $\begingroup$ $\partial_{\mu}\phi\partial^{\mu}\phi$ is an indexless quantity. When you then divide or take the derivative with respect to a lower index, it becomes an upper index. I think what I quoted is actually nontrivial to derive, but I'm unfamiliar with the chain rule when it comes to indexes so can't say for sure one way or another. If you rewrite it as $g^{\mu v}\partial_{\mu}\phi\partial_{v}\phi$ It seems like it makes sense. $\endgroup$ – MonkeysUncle Feb 23 '15 at 15:25
  • $\begingroup$ Ah, yes. Writing it out with the metric does make it more obvious. Thank you. $\endgroup$ – JohnnyMo1 Feb 23 '15 at 15:42

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