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My original question is like:

Why are the Euler-Lagrange equations invariant if we add a surface term to the action?

And there is an answer by Javier:

https://physics.stackexchange.com/a/205585/

I have some questions about Javier's answer, but as a new user I can not comment it so I have to ask this question.

In Javier's answer he said:

Suppose you have a lagrangian $L_0$ and add a divergence to get $L=L_0+∂_μJ^μ$. Recall that the action is (in your favorite number of dimensions): $$S=∫dx L=S_0+∫dx ∂_μJ^μ=S0+∫dS n_μJ^μ$$ Here $S_0$ is the integral of $L_0$, and $n_μ$ the normal vector to your boundary. The equations of motion are the condition that $δS=0$ to first order whenever we make a variation in $L$. So: $$δS=δS_0+∫dS n_μδ(J^μ)$$ But $J_μ$ is constructed out of the fields for which you want the equations of motion. Since by hypothesis the variation of the fields at the boundary is zero, so is the variation of $J_μ$. The last term vanishes, and we get $δS=δS_0$.

But $$\delta J^{\mu }=\frac{\partial J^{\mu }}{\partial \phi }\delta \phi +\frac{\partial J^{\mu }}{\partial(\partial_{\nu}\phi)}\delta (\partial_{\nu}\phi)$$ and we only have $\delta \phi=0$ in the boundary, and generally we don't have $\delta (\partial_{\nu}\phi)=0$, so why the variation of $J_μ$ is zero?

When I found the answer of my original question, I found in some book the author writes $J^{\mu }$ as $J^{\mu }=J^{\mu }(\phi(x),x)$ , that is $J^{\mu }$ is only a function of $\phi(x)$ and $x$ but not of $\partial_{\nu}\phi$. When we write $J^{\mu }$ in this way we have: $$\delta J^{\mu }=\frac{\partial J^{\mu }}{\partial \phi }\delta \phi $$ and the variation of $J_μ$ is zero due to $\delta \phi=0$ in the boundary. And we can also prove that: $$\left(\frac{\partial }{\partial \phi }-\partial _{\nu }\frac{\partial }{\partial(\partial_{\nu}\phi)}\right)(∂_\mu J^\mu)=0$$ so we have: $$\frac{\partial L}{\partial \phi }-\partial _{\mu }\frac{\partial L}{\partial(\partial_{\mu}\phi)}=\frac{\partial L_0}{\partial \phi }-\partial _{\mu }\frac{\partial L_0}{\partial(\partial_{\mu}\phi)}$$ so the equation of motion is invariant.

But in some books the author dose not mention the condition $J^{\mu }=J^{\mu }(\phi(x),x)$, so is this condition a necessary condition that the variation of a surface term is zero?

Qmechanic
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Hallow_Juan
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    I don't have a full answer, but I wanted to point out that the variation of a surface term isn't necessarily zero. For a simple example where it isn't (and where this non-vanishing encodes useful physical information), see the problem of the elastic bar as described in Chapter 2.15 of Lanczos's The Variational Principles of Mechanics. – Michael Seifert Aug 16 '17 at 15:28
  • If $J^{\mu}$ contains derivatives of the fields, that means that the original action contained second derivatives of the fields. If these are not of the form $(\partial^{2}\phi)(\partial^{2}\phi)$, you should integrate these out before taking the variation. – Zo the Relativist Aug 16 '17 at 17:00
  • @JerrySchirmer The Einstein-Hilbert action contains second derivatives but we don't in general integrate them out? – R. Rankin Jan 16 '22 at 03:12
  • @R.Rankin: when you compute the equations of motion, you eventually do, though (Or you use the Palantini trick of treating $\Gamma_{ab}{}^{c}$ as independent fields, which amounts to the same thing, since it's just "hiding a derivative") – Zo the Relativist Jan 16 '22 at 21:16
  • @JerrySchirmer I see, that must also be related to the fact that in the gravitational pseudotensor (in any of it's forms) the second derivatives always cancel out. Thanks! – R. Rankin Jan 17 '22 at 22:38

1 Answers1

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  1. OP asks:

    Why the variation of a surface term is zero?

    Answer: Assume that the action is schematically of the form $$S=S_1+B_2,\tag{1}$$ where $S_1$ is a bulk term and $B_2$ is a boundary term (BT). (E.g. in GR $S_1$ is the EH action and $B_2$ is the GHY BT.) Then the variation of the bulk term is of the form $$\delta S_1~=~(\text{bulk term}) + \delta B_1,\tag{2} $$ where $\delta B_1$ is a BT.

    The variations of the BTs $\delta B_1$ and $\delta B_2$ do generally not have to vanish. The condition $$\delta B_1+\delta B_2~=~0\tag{3}$$ is not automatic but has to be imposed via appropriate choice of boundary conditions (BCs) in order to guarantee the existence of a variational/functional derivative for $S$.

  2. The original question asks:

    Why are the Euler-Lagrange equations invariant if we add a surface term to the action?

    Answer: This is e.g. explained in my Phys.SE answers here and here. The main point is that a BT can never change the EL eqs. in an interior/bulk point.

Qmechanic
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  • Isn't the only famous example of a classical field theory where the boundary terms do play an important physical role...general relativity? And that's (IIRC) because the metric on the boundary is related to the metric on the bulk? No other fields have this peculiar quality. – Alex Nelson Aug 16 '17 at 17:04
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    @AlexNelson There are other examples. Yang-Mills, Chern-Simons ,... see e.g., https://inspirehep.net/record/567304 . The boundary term are related to Charges and as such have physical interpretation. – ungerade Aug 16 '17 at 18:15
  • For a simple example, see e.g. https://physics.stackexchange.com/q/138236/2451 – Qmechanic Aug 16 '17 at 18:30
  • Thanks for you answer. Now I know the boundary term not have to vanish. But I still can not figure out why this no vanish term does not change the Euler-Lagrange equations. I have read your answer. You said the functional derivative of the boundary component of action($B_2=∫dx ∂_μJ^μ$) is zero: $\frac{\delta B_2}{\delta \phi}=0$, because " variational/functional derivative is an object living in the bulk (rather than on the boundary), can never be other than identically zero in the bulk". – Hallow_Juan Aug 17 '17 at 04:43
  • But if one calculates the functional derivative directly, one has: $\frac{\delta B_2}{\delta \phi}=\left(\frac{\partial }{\partial \phi }-\partial {\nu }\frac{\partial }{\partial(\partial{\nu}\phi)}\right)(\partial_\mu J^\mu)$, and for arbitrary $ J^\mu$ I can not prove this term equals to zero, unless $J^{\mu }=J^{\mu }(\phi(x),x)$. So why doesn't the functional derivative equal to zero if I directly calculate? – Hallow_Juan Aug 17 '17 at 04:44
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    The functional derivative does vanish. Note that the formula for a functional derivative contains more terms in the case of higher order spacetime derivatives. – Qmechanic Aug 17 '17 at 05:03
  • @Qmechanic I think I have understood. The real Euler-Lagrange equation has infinity term which is like $\frac{\partial \mathcal{L}}{\partial \phi}+\cdots +(-1)^s\partial {\mu_1}\cdots\partial _{\mu_s}\frac{\partial \mathcal{L}}{\partial \phi{;\mu_1\dots\mu_s}}+\cdots=0$, and this infinite term E-L equation is zero for an arbitrary 4-divergence $\partial _{\nu}J^{\nu}$ so the equation of motion is unchanged under this extra term. Is that right? – Hallow_Juan Aug 17 '17 at 13:04
  • $\uparrow$ Yes. – Qmechanic Aug 17 '17 at 13:23