Greiner in his book "Field Quantization" page 173, eq.(7.11) did this calculation:

${\mathcal L}^\prime=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu A_\nu\partial^\nu A^\mu-\frac{1}{2}\partial_\mu A^\mu\partial_\nu A^\nu $
$\space\space\space\space=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$

The last term is a four-divergence which has no influence on the field equations. Thus the dynamics of the electromagnetic field (in the Lorentz gauge) can be described by the simple Lagrangian

${\mathcal L}^{\prime\prime}=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu$

Yes, if it is a four-divergence of a vector whose 0-component doesn't contain time derivatives of the field, indeed according to the variational principle this four-divergence will not influence the field equation.

And actually I calculated the time-derivative dependence of 0-component of $[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$, in which only $[A_0(\partial^0 A^0)-(\partial_0 A^0) A^0]$ could possibly contain time-derivative, which vanishes fortunately, so whatever the general case it doesn't matter in this present case.

But how can he seem to claim that it holds for a general four-divergence term,The last term is a four-divergence which has no influence on the field equations?

I only assumed the boundary condition to be $A^\mu=0$ at spatial infinity, not at time infinity. And the variation of the action $S = \int_{t_1}^{t_2}L \, dt$ is due to the variation of fields which vanish at time, $\delta A^\mu(\mathbf x,t_1)=\delta A^\mu(\mathbf x,t_2)=0$, not having the knowledge of $\delta \dot A^\mu(\mathbf x,t_1)$ and $\delta \dot A^\mu(\mathbf x,t_2)$, which don't vanish generally, so the four-divergence term will in general contribute to the action, $$\delta S_j=\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \partial_\mu j(A(x),\nabla A(x),\dot A(x))^\mu =\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \dot j^0 =\int d^3\mathbf x [\delta j(\mathbf x, t_2)^0-\delta j(\mathbf x, t_1)^0]$$ which does not vanish in general!

  • 1
    $\begingroup$ As you say, in your example the time derivatives cancel so there's no issue. An example where this subtlety is important is the einstein hilbert action. There are non vanishing time derivatives on the boundary. The fix in that case is to add boundary terms (the gibbons hawking York boundary terms) that cancel the offending pieces. $\endgroup$
    – Andrew
    Dec 24, 2013 at 15:01
  • $\begingroup$ (a) I think you mean Lorenz gauge, not Lorentz. (b) If you actually apply that gauge condition, that last term becomes $\partial_{\mu}[A_{\nu}(\partial^{\nu}A^{\mu})]$. The problem becomes a triviality after that... $\endgroup$ Apr 23, 2014 at 3:59

3 Answers 3


I) The geometric argument is clear: Consider a Lagrangian density ${\cal L}=d_{\mu}F^{\mu}$ that is a total divergence. The corresponding action

$$S[\phi] ~=~ \int_M \! d^nx~{\cal L}~=~ \int_{\partial M} \! d^{n-1}x~(\ldots)\tag{0}$$

will then be a boundary integral, due to the divergence theorem. Therefore the corresponding variational/functional derivative,

$$ \frac{\delta S}{\delta\phi^{\alpha}(x)}\tag{1}$$

which is an object living in the bulk (rather than on the boundary), can never be other than identically zero in the bulk

$$ \frac{\delta S}{\delta\phi^{\alpha}(x)}~\equiv~0,\tag{2}$$

if it exists. (Note: Even for a sufficiently smooth Lagrangian density ${\cal L}$, the existence of the functional derivative is a nontrivial issue and tied to whether or not consistent boundary conditions are assumed in the variation.)

Next, recall that (the expression for) the field equations of motion is just simply given by the functional derivative (1) of the action. Then according to eq. (2), (the expression for) the field equations of motion vanish identically.

II) Finally, extend the above argument from section I by linearity to a general Lagrangian density of the form ${\cal L}+d_{\mu}F^{\mu}$ that include an extra total divergence term. Conclude via linearity, that the latter does not contribute to the field equations of motion.

  • $\begingroup$ Could you explain a bit more how linearity leads to the conclusion in your last paragraph? $\endgroup$ Nov 15, 2020 at 15:50
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    $\begingroup$ The original question asked about a four divergence but you reference a total divergence in your answer, i e., $\partial_u F^u$ versus $d_u F^u$. Does your result only hold for the total divergence? $\endgroup$
    – Jeff
    Apr 23, 2022 at 13:43
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Apr 23, 2022 at 13:56
  • $\begingroup$ Regarding Point I), what about the case $n=1$? $\endgroup$
    – Rescy_
    Nov 10, 2022 at 22:30
  • $\begingroup$ Then the boundary $\partial M$ is 0-dimensional, e.g. an initial and a final point. $\endgroup$
    – Qmechanic
    Nov 10, 2022 at 22:37

A term $\int\!d^4x\, \operatorname{Tr}\, F \wedge F$ can be added to a non-Abelian Yang-mills theory (it vanishes trivially for the Abelian case, because of the wedge), and it is a total derivative. This term doesn't influence the equations f motion. However, this is a topological charge that counts something akin to the "winding number" of the gauge field.

Another place this enters is in string theory, where it counts the number of handles in the worldsheet, allowing second quantization to arise rather naturally.


If you have a four divergence inside an integral over all of spacetime (which is what you get when you extremize the action), the result will be a term which will be some product of the field(s) and its/their derivatives, evaluated at the boundary of spacetime. Since we assume that all fields go to zero (sufficiently quickly, so that their derivatives also go to zero) at the boundary, we have a contribution of zero, and can safely ignore the term.

However, there might be some subtleties which could prevent one from using this argument in some cases, which I am unaware of. I hope someone more knowledgeable than me can shed some light on this.

EDIT: Look at the comments below for some additional information.

  • $\begingroup$ The integral does not go to zero. Part of the boundary of the space-time region is at spatial infinity, but part is at the time slices for initial and final time. But since you don't vary the degrees of freedom (here, the field) at the initial and final time, this part of the integral does not contribute to the variation in the action. So when finding the variation of the action to get the field equations, its like these terms don't exist since they don't cause variation. $\endgroup$ Nov 23, 2013 at 16:34
  • $\begingroup$ Fair enough. I guess I assumed previous knowledge of Lagrangian methods in non-field contexts. $\endgroup$
    – Danu
    Nov 23, 2013 at 16:42

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