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?
EDIT:
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!
Best Answer
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.