# Problem 3.3 b) of Schwartz's Quantum Field Theory

In problem 3.3 b) of Schwartz's Quantum field theory you are asked to prove that $$Q = \int T_{00} d^3x$$ is invariant under changing the Lagrangian $$\mathcal{L} \rightarrow \mathcal{L} + \partial_\mu X^\mu$$ by a total derivative. In 3.3 a) I already wrote down the transformation of the stress-energy tensor $$T_{\mu\nu} \rightarrow T_{\mu\nu} + \frac{\partial \left( \partial_\lambda X^\lambda \right)}{\partial \left( \partial_\mu \phi \right)} \partial_\nu \phi - g_{\mu\nu} \partial_\lambda X^\lambda$$

Plugging this into the formula for $$Q$$ you get $$\delta Q = \int \frac{\partial \left(\partial_\lambda X^\lambda \right)}{\partial \dot{\phi}} \dot{\phi} - \partial_\lambda X^\lambda \, d^3x$$

From here I don't know how to proceed to show that $$\delta Q = 0$$. You could use the product rule with repsect to $$\partial_t$$ to then utilize the Euler-Lagrange equations to get $$\delta Q= \int d^3x \, \partial_t \left(\frac{\partial \left(\partial_\lambda X^\lambda\right)}{\partial \dot{\phi}} \phi\right) + \phi \left(\frac{\partial \left(\partial_\lambda X^\lambda\right)}{\partial \phi}- \partial_i \frac{\partial \left(\partial_\lambda X^\lambda\right)}{\partial \left(\partial_i \phi\right)}\right) -\partial_\lambda X^\lambda,$$ but I don't see how this has helped me.

The solutions available online by To Chin Yu reads $$\delta Q = \int \frac{\partial \left(\partial_\lambda X^\lambda \right)}{\partial \dot{\phi}} \dot{\phi} - \partial_\lambda X^\lambda \, d^3x \\ = -2 \int \partial_\lambda X^\lambda d^3x \\ = - 2 \partial_0 \int X^0d^3x \\ =0.$$ I understand the step from the second to the third equation where we have used $$\int \partial_i X^i dx_i = X^i|_{\partial M} =0$$ (no Einstein summation implied). But I don't see why the first line should be equal to the second line and why the third line should be zero.

If you can help me I would be most grateful :)

$$$$A \partial_{\mu} B = - (\partial_{\mu} A) B$$$$ inside Lagrangians. You do the same here. Identify
$$$$B =\partial_{\lambda} X^{\lambda}, \, \, A = \dot{\phi},$$$$ and make the switch (remembering that you are differentiating w.r.t $$\dot{\phi}$$), giving
\begin{align} &- \frac{\partial \dot{\phi}}{\partial \dot{\phi}} \partial_{\lambda} X^{\lambda} \\ &= - \partial_{\lambda} X^{\lambda} \end{align} This obviously combines with the other term, giving the factor of $$-2$$.
• Thank you for your answer. Unfortunately, I don't see why partial integration should be allowed in the way you are describing it. The integral is with respect to $d^3x$ and the derivative with respect to $\dot{\phi}$. So one would need to show that the total derivative term $$\frac{\partial}{\partial \dot{\phi}} \left( \dot{\phi} \partial_\lambda X^\lambda \right)$$ vanishes under the integral which I don't see obviously. Could you please clarify? Commented Nov 20, 2023 at 14:16