Is there a superfluous statement in Schwartz's QFT book in deriving Euler-Lagrange equations?

Question

Please help me with the following confusion.

Yesterday I was looking at the derivation of Euler-Lagrange equation in several QFT textbooks using stationarity of the action. At the last step one needs to throw away the boundary term in the action $$\int d^4x\partial_\mu\Big[\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi\Big]$$ to get the Euler-Lagrange equation. Two different books argue differently why this term can be thrown away.

Peskin and Schroeder's QFT (Page 15) It says that this term can be turned into a surface integral over the boundary of the four-dimensional spacetime region of integration. Since the initial and final field configurations are assumed given, $\delta\phi$ is zero at the temporal beginning and end of this region.

Schwartz's QFT (Page 31) It says "We will always make the physical assumption that our fields vanish on these asymptotic boundaries, which lets us drop such derivatives from Lagrangians".

P & S argument is pretty clear, and matches with the derivation in particle mechanics. But Schwartz's argument seems to be problematic. First because one can derive the EL equations in finite spacetime volume with no need to consider spatial or temporal infinity. Second, we can clearly see that the fields also need not vanish at the boundary but only requirement is that the field configurations be fixed at the temporal beginning and end of a finite spacetime region so that $\delta\phi({\bf x}_1,t_1)=\delta\phi({\bf x}_2,t_2)=0$.

So isn't Schwartz's statement superfluous in the context of this derivation?

Answer 1

When we integrate over all of non-compact spacetime $\mathbb{R}^n$, there is, strictly speaking, no boundary term. $\mathbb{R}^n$ has no boundary, the boundary term simply doesn't show up when integrating by parts on it. (Neither does it show up when we "add infinity" to our spacetime, making it the also boundary-less sphere $S^n$)

However, we must ensure that the integrals involved converge! (E.g. integrating $f(x) = x$ over all of $\mathbb{R}$ does not converge) The easiest way to control this is to postulate that all the physical fields are either compactly supported or go zero sufficiently fast towards infinity (a common choice is to make them Schwartz functions).

So while Schwartz (the one in the question, not the one after whom the functions are named) gives the wrong reason for assuming the fields fall off towards infinity, it is nevertheless a common and reasonable assumption to make.