# What is the “surface term”?

In Peskin's quantum field theory book, There is a sentence in page 17:

...

More generally, we can allow the action to change by a surface term, since the presence of such a term would not affect our derivation of the Euler-Lagrange equations of motion ...

...

\begin{equation} \mathcal{L}(x)\to\mathcal{L}(x)+\partial_\mu\mathcal{J}^\mu(x).\tag{2.10} \end{equation}

What is the "surface term"? Is it just a partial derivative term as $\partial_\mu\mathcal{J}^\mu(x)$?

• Just a short comment on the meaning of these terms, which is not often discussed sufficiently: While surface terms have no influence on the equations of motion, they are important and some situations and can't be simply neglected. Surface terms are important to describe different phases of a system, which is something "global", whereas the equation of motion deal with local behaviour. C.f. my recent question physics.stackexchange.com/q/369840 – jak Nov 22 '17 at 8:31

Yes, a surface term is a 4-divergence of a 4-vector. The reason is that the Action

$S= \int_\Omega d^4x \mathcal{L}$

defined in some Region of spacetime $\Omega$ corresponding to such a surface term can be converted by Gauss Theorem:

$S = \int_\Omega d^4x \partial^\mu J_\mu = \int_{\partial \Omega}d \sigma n^\mu J_\mu$.

Here, $n^\mu$ is the unit normal vector pointing out of the spacetime surface $\partial \Omega$.

1. We use the divergence theorem to connect boundary terms to divergences, as already mentioned in kryomaxim's answer.

2. It should be stressed that the relevant divergence term $d_{\mu}{\cal J}^{\mu}$ is a total spacetime derivative, not a partial/explicit spacetime derivative $\partial_{\mu}{\cal J}^{\mu}$, even if many authors use confusing notation, cf. e.g. my Phys.SE answer here.