# How the boundary term in the variation of the action vanishes

Can someone explain a little more that why the last term in equation (1.5) vanishes?

Reference:

1. David Tong, Quantum Field Theory: University of Cambridge Part III Mathematical Tripos, Lecture notes 2007, p.8. The PDF file is available here.
-
What book are you quoting from? – Emilio Pisanty Mar 19 at 15:05
– Unlimited Dreamer Mar 19 at 15:32

The term vanished because we can translate this term to one making a statement about the fields at the boundary and assume that the fields themselves vanish in spatial and temporal infinity.

By Stoke's Theorem, we can translate volume integrals into surface integrals. More specifically Gauss' Theorem states that the integral of a divergence of a field over a volume (denoted $V$) to an integral of the field itself over the surface of that volume (denoted $\partial V$)

$$\int_V \textrm{div} \vec{A}\,\textrm{d}V= \int_{\delta V} \vec{A}\,\textrm{d}\vec{S}$$

This holds true in any dimension and metric. In Minkowski-space the divergence (called a four-divergence) is exactly $\partial_\mu\phi$

Thus, you can translate $$\int_V \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\, \textrm{d}V = \int_{\delta V} \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\, \textrm{d}\Sigma_\mu$$

i.e. if we assume that the fields (and thus the Lagrangian density) vanishes in infinity, this term vanishes.

-
 Note that it's not sufficient for the fields to simply vanish at infinity, they need to vanish sufficiently rapidly as a function of the distance $r$ to the origin that the $r$-dependence of the thing you're integrating "beats" the $r$-dependence of the volume of the sphere of radius $r$. – joshphysics Mar 19 at 16:31 yes, that's true, sorry for the sloppy wording – luksen Mar 19 at 16:38

The last term is of the form $$\int d^n x \partial_\mu X^\mu$$ and by Stoke's theorem, this equals a surface integral at $\infty$ $$\int_\infty ds^\mu X_\mu$$ which vanishes provided $X^\mu\to 0$ sufficiently rapidly at infinity.

Addendum. I realize that this is a rather terse answer, so let me add some detail. Notice that the oriented surface area element $ds^\mu$ of the $n$-sphere has the behavior $$ds^\mu = dA_{n_1} r^{n-1} \,\hat n^{\mu}$$ there $n^\mu$ is the radially outward pointing unit normal and $dA_{n-1}$ is the area element on the unit $n-1$ sphere. For example, in two spatial dimensions we have $$ds^\mu = \underbrace{dA_1}_{\text{length element on unit circle}} r\,\hat n^\mu$$ while in three spatial dimensions $$ds^\mu = \underbrace{dA_2}_{\text{area element on unit 2-sphere}} r^2\, \hat n^\mu$$ The surface integral at "infinity" can be thought of as the limit of the integral over the surface of the sphere as the sphere's radius gets large; $$\int_\infty ds^\mu X_\mu = \lim_{r\to\infty}\int_{S^{n-1}}dA_{n-1}\, r^{n-1}\hat n^\mu X_\mu(r, \vec\phi)$$ Where $\vec\phi$ are coordinates on the sphere. Now notice that provided for example that $X^\mu\sim r^{-n}$ as $r\to\infty$, this surface integral will vanish in the large $r$ limit.

-