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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.

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StarBucK
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  • 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. – Qmechanic Mar 31 '17 at 21:17
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    Related: https://physics.stackexchange.com/q/87628/2451 , https://physics.stackexchange.com/q/112036/2451 and links therein. – Qmechanic Mar 31 '17 at 21:21
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    Minor point: we need an upper $\alpha$ index on $f$. Also, the $t$ dependence of $f$ is in general an $x$ dependence. – J.G. Mar 31 '17 at 22:39
  • 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. – StarBucK Mar 31 '17 at 22:45
  • @Qmechanic : I read the related topics. In the one dealing with scalar lagrangian, I understood that I can add a function $(d/dt)[f(q,\dot{q},t)]$ to my Lagrangian and it will not change the equations of motion if I also have boundary conditions on $\dot{q}$ (because in the proof we need to do integration by part and to avoid surface term we need boundary conditions). This same argument thus apply to field theory. I could have an extra term $\partial_\alpha f(\phi,\partial_\mu \phi,x)$, but to make my problem coming from variational principle, I would need boundary condition on... – StarBucK Mar 31 '17 at 23:17
  • ...$\partial_\alpha \phi$. Ok. But I don't understand the other topic. In this one the 4 divergence contains a current that has derivative of the field. Thus, I would need boundary condition of all terms $\partial_\mu A^\mu$ to make it coming from a variational principle, and thus be physically valid. But why would we have thoose boundary conditions on the field derivative ?? And I don't understand why the author focused on $\partial_0$, there is the same problem with all the $\partial_i$ ?? Thank you – StarBucK Mar 31 '17 at 23:20
  • I edited a lot my post to clarify my questions after my reading of the topics linked. – StarBucK Apr 01 '17 at 10:03
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    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 $f(q,t_1) - f(q,t_0)$. – Noiralef Apr 01 '17 at 10:16
  • With what you said I think I get why the author of the related post focused on derivatives on time of the field. It is because my derivative of the field could not be $0$ at $t_i$ and $t_f$ (it is different from space coordinates where we integrate on $\mathbb{R}^3)$. But thus there is still a problem in a general case with a current depending on the time derivatives of the field ? – StarBucK Apr 01 '17 at 10:20
  • For a scalar field with a Lagrangian density ${\cal L}~=~\frac{1}{2} \partial_{\mu}\phi~\partial^{\mu}\phi -V(\phi)$, it enough to impose Dirichlet BC on the boundary of (possibly compactified) spacetime. – Qmechanic Apr 01 '17 at 10:47
  • Comment to the post (v3): Consider to define $j^{\mu}$ in the title. – Qmechanic Apr 01 '17 at 10:50
  • I agree that for the Lagrangian you gave Dirichlet BC are enough. But if I have a term like $L=L_0+\partial_\mu (\phi^\nu\partial_\nu \phi^\mu)$ (for example) I'm not sure that Dirichlet BC will be enough for the invariant of action because it has a time derivative of the field and we don't integrate time on $\mathbb{R}$. The citation from the book in the post linked, allows such term. This is what I don't understand. $j^\mu$ is the 4-current inside my 4-divergence. – StarBucK Apr 01 '17 at 10:54
  • The action is also integrated over time. $j^{\mu}$ is usually called $f^{\mu}$ not to confuse with, say, a Noether current. – Qmechanic Apr 01 '17 at 11:11
  • Yes but the derivatives of field are not necesseraly $0$ at $t_i$ and $t_f$ at the opposite of they are 0 at $x$ goes to infinity for example. – StarBucK Apr 01 '17 at 11:15
  • @user3183950 can you link to "the post" and "the other topic" etc? It's difficult to follow without context. – ZeroTheHero Apr 01 '17 at 12:37
  • I totally edited the post again. If you read it I think it is a lot more clear now – StarBucK Apr 01 '17 at 14:47

2 Answers2

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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.

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Qmechanic
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  • 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 – StarBucK Apr 20 '17 at 12:43
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    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. – Qmechanic Apr 20 '17 at 12:55
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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.

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  • 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. – PPenguin Apr 07 '17 at 19:17