Total Momentum From a Standing Electromagnetic Wave How does one show the momentum imparted to a perfect conducting resonance cavity (boundary) of any shape by a classical standing electromagnetic wave inside is zero?
It should be by conservation of momentum. But I would like to see a detailed treatment argued specifically with the property of the electromagnetic wave, say with Poynting vector or electromagnetic stress-energy tensor.
Specifically, given the boundary condition of the perfect conductor cavity, how does one derive $\frac{\partial \int\mathbf S dV}{\partial t} = 0$ or $\frac{\partial \int \mathbf<S>dV}{\partial t} = 0$ where $\mathbf S$ is the Poynting vector, the integral is over the space of the cavity, $<\cdot>$ denotes time average.
One way of doing this could be to show the temporal spatial separated form of the Poynting vector $S(t,x)=S(x)e^{i\omega t}$ inside of the cavity. That is a Poynting vector of a standing wave. That form of $S(t,x)$ leads to its time average $<S>$ being zero.
I would also suppose the average pressure on the boundary within the scale of the wave length is constant. How would one argue or describe that?
 A: 
How does one show the momentum imparted to a perfect conducting resonance cavity (boundary) of any shape by a classical standing electromagnetic wave inside is zero?

For a standing EM wave in a cavity, the Poynting energy of the EM field inside is constant. This implies no energy is being transferred to the matter of the cavity from inside so the kinetic energy of the material cavity is constant in time. Therefore momentum of the material cavity is constant in time as well (if the cavity was changing its momentum during some time interval, it would be changing its kinetic energy too).
Of course, the walls of the cavity may experience pressure forces due to the EM forces (calculable with the Maxwell tensor), but if the cavity holds its shape so no work occurs, these forces cancel out and the cavity does not move.

Why is the Poynting energy of standing EM wave constant?

Let $d\boldsymbol \Sigma$ be outward area vector of an element of the inner boundary surface of the cavity. Net flux of Poynting vector
$$
\oint_\Sigma (\mathbf E\times\mathbf B)\cdot d\boldsymbol\Sigma
$$
over the inner boundary of the conductor $\Sigma$ is zero because $\mathbf E\times \mathbf B$ near the wall is parallel to the plane of the surface element nearby; it is perpendicular to $d\boldsymbol \Sigma$. That's because the electric field of a standing wave is perpendicular to the wall or vanishes. 

Why is the electric field  perpendicular to the wall? Because the component of $\mathbf E$ parallel to the wall is continuous across the boundary and on the conductor side of it, this component vanishes.

