lagrangian formalism - How the boundary term in the variation of the action vanishes

In David Tong's QFT lecture notes (Quantum Field Theory: University of Cambridge Part III Mathematical Tripos, Lecture notes 2007, p.8), he states that

We can determine the equations of motion by the principle of least action. We vary the path, keeping the end points fixed and require $\delta S=0$, $$ \begin{align} \delta S &= \int d^4x\left[\frac{\partial \mathcal L}{\partial\phi_a}\delta\phi_a+\frac{\partial \mathcal L}{\partial(\partial_\mu\phi_a)}\delta(\partial_\mu\phi_a)\right] \\&= \int d^4x\left[\frac{\partial \mathcal L}{\partial\phi_a} -\partial_\mu\left(\frac{\partial \mathcal L}{\partial(\partial_\mu\phi_a)}\right) \right]\delta\phi_a +\partial_\mu\left(\frac{\partial \mathcal L}{\partial(\partial_\mu\phi_a)}\delta\phi_a\right) \tag{1.5} \end{align} $$ The last term is a total derivative and vanishes for any $\delta\phi_a(\vec x,t)$ that decays at spatial infinity and obeys $\delta\phi_a(\vec x,t_1)=\delta\phi_a(\vec x,t_2)=0$. Requiring $\delta S=0$ for all such paths yields the Euler-Lagrange equations of motion for the fields $\phi_a$, $$ \partial_\mu\left(\frac{\partial \mathcal L}{\partial(\partial_\mu\phi_a)}\right) -\frac{\partial \mathcal L}{\partial\phi_a} =0 \tag{1.6} $$

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

Answer

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 Stokes' 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 $\partial V$)

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

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

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