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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?

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  • $\begingroup$ The em field outside of the cavity is zero, right? What's the momentum of a zero em field? $\endgroup$ – CuriousOne Aug 16 '14 at 2:41
  • $\begingroup$ I am talking about the EM field INSIDE of the cavity. $\endgroup$ – Hans Aug 16 '14 at 3:05
  • $\begingroup$ And I was suggesting that momentum conservation doesn't care about arbitrary physical partitions like inside and outside of arbitrary (ideal) cavities. If there are no outgoing electromagnetic waves, where is the momentum supposed to go to? $\endgroup$ – CuriousOne Aug 16 '14 at 3:21
  • $\begingroup$ Did you read the first sentence of my second paragraph? I already know how to argue it from conservation of momentum. I would like to see an argument from the perspective of Poynting vector or electromagnetic stress-energy tensor. I have edited my question to better reflect the particular perspective from which I would to derive the result. $\endgroup$ – Hans Aug 16 '14 at 4:06
  • $\begingroup$ Apologies for upsetting you by offering the easy way out, which I understand now, you don't want to take. $\endgroup$ – CuriousOne Aug 16 '14 at 4:20
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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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Manishearth Aug 22 '14 at 20:52
  • $\begingroup$ I wonder how simple it is to derive an inertia relation from this, i.e., the force differential for an accelerating box. $\endgroup$ – lionelbrits Dec 5 '14 at 14:45
  • $\begingroup$ I am revisiting this post because of others, such as @lionelbrits, are. The assertion "no energy is being transferred to the matter of the cavity from inside so kinetic energy of the material cavity is constant in time. Therefore momentum of the material cavity is constant in time as well" is patently wrong. Consider a particle running with constant speed in a circle. Besides, I am asking for the spatial integral of the field momentum. This so called answer does not answer the question. $\endgroup$ – Hans Dec 9 '14 at 5:15

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