I take this Lagrangian:

$$\mathcal{L}=\mathcal{L}_0+\partial_\alpha f(\phi, \partial_\mu \phi).$$

In this topic Does a four-divergence extra term in a Lagrangian density matter to the field equations? , it is said that any 4-divergence term added to a Lagrangian doesn't modifies the equation of motion.

In my example I add $\partial_\alpha f(\phi, \partial_\mu \phi)$ to $\mathcal{L}_0$ (it is not a 4-divergence but the mechanics behind is exactly the same). And I remark that it can modify the equation of motion if $f$ contains time derivatives of $\phi$. So I don't understand.

I Write the infinitesimal variation of action to $\mathcal{L}$:

$$ \delta S = \int d^4x ~ \delta \mathcal{L}, $$

$$ \delta S = \int d^4x ~ [ \frac{\partial \mathcal{L}_0}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}_0}{\partial (\partial_\mu \phi)} \delta(\partial_\mu \phi) + \partial_\alpha [\frac{\partial f}{\partial \phi} \delta \phi + \frac{\partial f}{\partial (\partial_\mu \phi)} \delta(\partial_\mu \phi)] ~ ].$$

As usual, I know that : $\delta(\partial_\mu \phi)=\partial_\mu \delta(\phi)$. Thus I can integrate by parts:

$$ \delta S = \int d^4x ~ [ \frac{\partial \mathcal{L}_0}{\partial \phi} - \partial_\mu \frac{\partial \mathcal{L}_0}{\partial (\partial_\mu \phi)} )\delta \phi + \int d^4x ~ \partial_\mu[\frac{\partial \mathcal{L}_0}{\partial (\partial_\mu \phi)} \delta \phi] + \int d^4x ~ \partial_\alpha [\frac{\partial f}{\partial \phi} \delta \phi + \frac{\partial f}{\partial (\partial_\mu \phi)} \delta(\partial_\mu \phi)].$$

We have:

$$ \int d^4x ~ \partial_\mu[\frac{\partial \mathcal{L}_0}{\partial (\partial_\mu \phi)} \delta \phi] = \int d^3x ~ [\frac{\partial \mathcal{L}_0}{\partial (\partial_\mu \phi)} \delta \phi]_{x_i^{-}}^{x_i^{+}}=0.$$

Indeed, $\delta \phi=0$ on the boundaries by hypothesis ($x_i^{+}=+\infty$ for spatial coordinates and $t_f$ for time).

We also have:

$$ \int d^4x ~ \partial_\alpha [\frac{\partial f}{\partial \phi} \delta \phi + \frac{\partial f}{\partial (\partial_\mu \phi)} \delta(\partial_\mu \phi)]= \int d^3x ~ [\frac{\partial f}{\partial \phi} \delta \phi + \frac{\partial f}{\partial (\partial_\mu \phi)} \delta(\partial_\mu \phi)]_{x_i^{-}}^{x_i^{+}}=\int d^3x ~ [\frac{\partial f}{\partial (\partial_\mu \phi)} \delta(\partial_\mu \phi)]_{x_i^{-}}^{x_i^{+}}.$$

** And here is my problem **.

The fact $\delta \phi(x_i^{+})=\delta \phi(x_i^{-})=0$ doesn't implicate that $\partial_\mu \delta \phi(x_i^{+})=\partial_\mu \delta \phi(x_i^{-})=0$.

To be more precise, it could be true if $x_i^{+}=-x_i^{-}=+\infty$(*) but if I take the time coordinates, I have $x_i^{+}=t_f$. So it is at least not true for $\mu=t$.

Thus the extra term $\partial_\alpha f(\phi, \partial_\mu \phi)$ modifies the extremality of the action. Thus I will not have the same equation of motion.

But in this topic : Does a four-divergence extra term in a Lagrangian density matter to the field equations? the book of the author says that any four divergence doesn't affect the equation of motions.

But we've seen here (if I made no mistake which is not sure at all) that if the extra term is a total derivative that contains time derivatives of the field it can change the equations of motion.

Where am I wrong?

(*) : it is true because we ask $\phi$ to go to zero at infinity, so we only allow variations of $\phi$ that vanish at infinity (else we would end up with $\phi+\delta \phi$ not integrable). And as $(x,y,z) \mapsto \delta \phi(x,y,z,t)$ goes to $0$ at infinity, all its derivative also.

  • $\begingroup$ Comment to the post (v2): Note that the issue with dependence on derivatives in the $f$ function is in principle the same in the point mechanics case and the field theory case. In particular, the conclusion should be the same, not opposite. $\endgroup$
    – Qmechanic
    Mar 31, 2017 at 21:17
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/87628/2451 , physics.stackexchange.com/q/112036/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 31, 2017 at 21:21
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    $\begingroup$ Minor point: we need an upper $\alpha$ index on $f$. Also, the $t$ dependence of $f$ is in general an $x$ dependence. $\endgroup$
    – J.G.
    Mar 31, 2017 at 22:39
  • $\begingroup$ For the $x$ dependance I agree $(x,t)$ in the most general case. But I don't understand why you want an upper $\alpha$ ? Btw I'm reading the related post. $\endgroup$
    – StarBucK
    Mar 31, 2017 at 22:45
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    $\begingroup$ In your field theory example, the boundary condition would simply be that the field goes to zero in the infinite. The integral of the second term gives therefore exactly zero. That's different from the classical mechanics terms where you have $f(q,t_1) - f(q,t_0)$. $\endgroup$
    – Noiralef
    Apr 1, 2017 at 10:16

2 Answers 2


The correct statement is that a boundary term (BT) in the action (or equivalently, a total divergence term in the Lagrangian density) does not change the functional/variational derivative if both the old and the new functional derivatives exist. Pay attention to the important word if in the previous sentence: This does not exclude the possibility that a functional/variational does not exist.

In order for functional derivatives to exist, it is necessary to impose adequate boundary conditions (BCs). A boundary/total divergence term may change the adequate set of BCs.

In OP's example, he has correctly observed that Dirichlet BCs are not enough to remove BTs in the variation.

To summarize: OP has not shown that 2 different sets of Euler-Lagrange equations exist, cf. the title question (v6). Only that some choices of BTs & BCs may make the variational problem ill-defined.

For the point mechanical case, see also this Phys.SE post. The field theoretical case is a straightforward generalization.

  • $\begingroup$ Sorry for my late comment but to summarize, if I understood well, to add some dependencies on the derivative of the field in $V^\mu$ implicates that I have to change my boundary conditions. Thus, when the author of the book said a 4-divergence $\partial_\mu V^\mu$ doesn't change the equation of motion, if I only have B.C on the field the function in the 4-div term must only contain $\phi$. If I now add B.C on the derivatives of $\phi$, I can have derivatives of $\phi$ in the 4-divergence term. Am I correct ? Just to be sure $\endgroup$
    – StarBucK
    Apr 20, 2017 at 12:43
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    $\begingroup$ As far as the existence of functional derivatives goes: Yes. But there may be other problems: The system with all the BCs may now be overconstrained, so that the EL eqs. have no solutions. $\endgroup$
    – Qmechanic
    Apr 20, 2017 at 12:55

On the boundary, $\delta \phi(x) = 0 \implies \delta \left(\partial_{\mu} \phi(x)\right) = 0$.

Think in terms of the one-dimensional variational principle. In this case, one finds the equivalence $\delta \phi(x) = \delta \left(\dot{\phi}(x) dt\right) = \delta \left(\dot{\phi}(x)\right) dt$. Thus when one takes $\delta \phi(x) = 0$ on the boundary, we obtain immediately $\delta \left(\dot{\phi}(x)\right) = 0$ as well.

This holds for any variational principle with the given boundary condition in any dimension. I hope this solves your confusion.

  • $\begingroup$ I'm not sure I follow this. It feels like claiming that if $\frac{df}{dx}$ is zero at some point $x=a$, then necessarily $\frac{d^2 \ f}{dx^2} = \frac{d}{dx}(\frac{df}{dx}) = \frac{d}{dx}(0) = 0 $ at that point. Which is of course not correct. $\endgroup$
    – PPenguin
    Apr 7, 2017 at 19:17

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