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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 $\partial_\mu\mathcal{J}^\mu(x)$?

Qmechanic
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Orient
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2 Answers2

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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 $\partial \Omega$.

kryomaxim
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  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.

Qmechanic
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