 Solution

If $L^{\prime }=L+dF(q,t)/dt$, the action changes by an endpoint term:In the principle of stationary action with fixed endpoints, $\delta q(t_i)=\delta q(t_f)=0$ and the endpoint times are fixed. Therefore $\delta (S^{\prime }-S)=0$, and the two actions give the same bulk Euler-Lagrange equations.

For a Lagrangian density, adding ${\partial }_{\mu }{K}^{\mu }$ changes the action by ${\int }_{\partial \Omega }{K}^{\mu }{n}_{\mu }d\Sigma$, by the divergence theorem. Its variation vanishes for compactly supported field variations, or when the appropriate boundary data are held fixed. These total divergences in a variational principle can therefore be discarded when deriving bulk equations under those conditions.

The qualification about boundaries matters: if $K^{\mu }$ depends on field derivatives, their variations on the boundary must also be controlled. A total derivative can change the canonical momenta or the chosen representative of a Noether current, even while leaving the bulk dynamics unchanged.