1

People always say that boundary terms don't change the equation of motion, and some people say that boundary terms do matter in some cases. I always get confused. Here I want to consider a specific case: classical scalar field.

The action for the free scalar field is (no mass term since it's not relevant to the question) $$ S = \int \phi \partial_{\mu}\partial^{\mu}\phi \mathrm{d}^4x.\tag{1} $$ We can use Stoke's theorem to extract a so-called total derivative: $$ S = \int \phi \partial^{\mu}\phi n_{\mu} \sqrt{|\gamma|} \mathrm{d}^3x -\int\partial_{\mu}\phi \partial^{\mu}\phi \mathrm{d}^4x\tag{2} $$ where $n_{\mu}$ is the normal vector to the integral 3D hypersurface and $\gamma$ is the induced metric of the hypersurface.

The second term is another common form of scalar field. But I don't think that the first term doesn't matter at all. Variation of first term should be $$ \begin{aligned} \delta \int \phi \partial^{\mu}\phi n_{\mu} \sqrt{|\gamma|} \mathrm{d}^3x &= \int (\delta\phi) \partial^{\mu}\phi n_{\mu} \sqrt{|\gamma|} \mathrm{d}^3x \\ &+ \int \phi (\partial^{\mu} \delta\phi) n_{\mu} \sqrt{|\gamma|} \mathrm{d}^3x\\ &= \int \phi (\partial^{\mu} \delta\phi) n_{\mu} \sqrt{|\gamma|} \mathrm{d}^3x \end{aligned}\tag{3} $$ where the first term vanishes because $\delta\phi \equiv 0$ on the boundary.

But then what? There is no guarantee that $\delta \partial_{\mu} \phi$ also vanishes on the boundary. (Or there is?)

My questions are:

  1. How to deal with this term?

  2. More generally, what's the exact statement about the total derivative term in an action?

Qmechanic
  • 201,751
Xiaosheng Yang
  • 299
  • 1
  • 10

1 Answers1

1

  1. Note first of all that it is usually important to specify appropriate boundary conditions (BCs) to render a variational principle well-posed, i.e. to ensure that the functional/variational derivative (i.e. the Euler-Lagrange (EL) expression) exists.

  2. Adding boundary terms (BTs) to the action $S$ can affect the existence of the functional derivative. However if it still exists, it is unchanged, cf. e.g. this related Phys.SE post.

  3. Concerning OP's scalar example: OP imposes Dirichlet BCs. This is appropriate for the 1st-order action $S=-\int\!\mathrm{d}^4x\partial_{\mu}\phi \partial^{\mu}\phi,$ but is not enough for OP's 2nd-order action (1), which requires extra BCs. Since extra BCs are likely unphysical/overconstraining, the action principle (1) is ill-posed.

Qmechanic
  • 201,751
  • So when people say that the two actions in my question are "equivalent", they actually mean that the action $S=-\int \mathrm{d}^4x \partial_{\mu} \phi \partial^{\mu}\phi$ with Dirichlet B.C. and the action $S=\int \mathrm{d}^4 x\phi \partial_{\mu}\partial^{\mu} \phi$ with Neumann B.C. give the same equation of motion in the bulk. Is this understanding correct?
    • – Xiaosheng Yang May 09 '22 at 02:10
  • Oh, I understand another thing as well. In classical mechanics (particle dynamics) we say that the variation of the coordinates vanishes at the initial and final moment: $\delta q(t_1)=\delta q(t_2)=0$. That is in fact a Dirichlet B.C.! And the reason that we only assume this B.C. but not Neumann B.C. is Lagraigans usually don't contain derivatives higher than first order, e.g. $L=\frac{1}{2} \dot{q}^2-V(q)$.
    • – Xiaosheng Yang May 09 '22 at 02:22
  • Hi Xiaosheng Yang: 1. Usually it is assumed that the BCs are not changed, but it seems you got the main idea. 2. Yes, although BCs are often dictated by the physics situation, which could call for e.g. Neumann BCs. – Qmechanic May 09 '22 at 08:05
  • Thanks for your reply! And I would like to know more precisely about the relation between the exsistence of functional derivatives and the equation of motion. Could you please recommend some textbooks on this topic? I never knew this before. It seems that it's not in Landau's book or other classical mechanics textbook that I've read. :( – Xiaosheng Yang May 09 '22 at 13:25