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I am trying to answer problem 3.3 part b in Schwartz's QFT book:

(b) Show that the total energy $Q=\int \mathcal{T}^{00} d^3 x$ is invariant under the addition of a total derivative to the Lagrangian, $\mathcal{L}\rightarrow \mathcal{L}+\partial_{\mu} X^{\mu}$.

When I try to show this, I find that it is in general not invariant, meaning I am making an error somewhere. Here is my attempt:

The (canonical) energy-momentum tensor $\mathcal{T}^{\mu \nu}$ is defined by:

$$\mathcal{T}^{\mu \nu}\equiv \sum_{i} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_i)}\partial^{\nu}\phi_i - \eta^{\mu\nu}\mathcal{L}$$

Since a total derivative $\partial_{\mu}X^{\mu}$ doesn't change the action, and $\mathcal{T}^{\mu \nu}$ is defined through the action, the new energy-momentum tensor will simply be the old one with $\mathcal{L}$ replaced with $\mathcal{L}+\partial_{\mu}X^{\mu}$.

$$\begin{align} \mathcal{T'}^{\mu \nu}&=\mathcal{T}^{\mu \nu}+ \sum_{i} \frac{\partial \left(\partial_{\rho}X^{\rho}\right)}{\partial (\partial_{\mu} \phi_i)}\partial^{\nu}\phi_i - \eta^{\mu\nu}\left(\partial_{\rho}X^{\rho}\right)\\ &=\mathcal{T}^{\mu \nu}+ \sum_{i} \partial_{\rho}\left(\frac{\partial X^{\rho}}{\partial (\partial_{\mu} \phi_i)}\right)\partial^{\nu}\phi_i - \eta^{\mu\nu}\partial_{\rho}X^{\rho}\\ \end{align}$$

The new energy $Q'$ is given by:

$$\begin{align} Q'&=\int d^3 x \mathcal{T}'^{00}\\ &=Q+\int d^3 x \, \left[ \sum_{i} \partial_{\rho}\left(\frac{\partial X^{\rho}}{\partial \dot{\phi}_i}\right)\dot{\phi}_i -\partial_{\rho} X^{\rho}\right] \end{align}$$

So I need to show that the second term vanishes, but I don't know how. That's a 4-derivative sitting inside the integral, so if it was $d^4 x$ I'd just use the divergence theorem and say "oh well assuming $X^{\rho}$ vanishes at infinity we get our desired result". But it's an integral over $d^3 x$.

What do I do?


marked as duplicate by Qmechanic Feb 10 '18 at 19:20

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  • $\begingroup$ @Qmechanic♦ This is not a duplicate of that question you marked. That question does not address the more important part of my question, which is the invariance of $Q$. $\endgroup$ – Arturo don Juan Feb 10 '18 at 19:53
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/257875/2451 $\endgroup$ – Qmechanic Feb 10 '18 at 20:31
  • $\begingroup$ Ah okay, now that you've edited your answer to that question to include the answer to my question, I see that it is. Thanks. $\endgroup$ – Arturo don Juan Feb 10 '18 at 20:43