We know from theory and experiment that an electromagnetic wave that incides on a surface will generate a radiation pressure normal to that surface as a result of the change in momentum of the wave when being reflected on it. Evidently, if all the wave is reflected, we get the maximum pressure radiation (as the total moment transfered is two times the momentum of the wave).

However, I was wondering how could one derive this pressure radiation by considering the Lorentz forces over the induced currents in the surface of a conductor. Indeed, the presence of an electromagnetic wave (which has and $\vec{E}$ and $\vec{B}$ field) will induce a surface current density $\vec{K}$ which will in turn react with the wave and induce a pressure on the surface of the very same conductor.

I did some research but I couldn't find any source on the literature, so how could one proceed in this case? For simplicity, let's consider a semi-infinite conductor plane at $x<0$ and a plane wave with vector $\vec{E}=\hat{y}E exp[i(k(zsin\theta-xcos\theta)-\omega t)]$ that incides with some angle $\theta$.

Update: Some information has been provided on how to approach this problem, but since no canonical answer has been provided I'm opening a bounty in case someone wants to further elaborate/summarize the information or provide an example that could be useful for the community.

  • $\begingroup$ the electrons on the surface moves along the polarization of the electric field. The magnetic field of the incident EM wave is perpendicular to the electric current. v cross B is along the normal vector pointing into the plane. $\endgroup$ – wcc Mar 22 at 0:55
  • $\begingroup$ You're right, the current will be produced by the movement of electrons using the parallel component of the electric field to the surface, and then apply the magnetic force. If I understand correctly, this force should push against the surface, thus creating the radiation pressure on the conductor, am I correct? $\endgroup$ – Charlie Mar 22 at 1:18
  • $\begingroup$ Let's see, assuming $E$ is polarized along $y$, $B$ along $x$ , then electrons accelerate toward $-y$, so the Lorentz force is along $-y \times x = z$. Poynting vector is along $y \times x = -z$. So it seems the Lorentz force is out of the plane, not into the plane. So I don't think I got this right. $\endgroup$ – wcc Mar 22 at 1:35
  • $\begingroup$ Sorry, because Lorentz force is $qv \times B$, there is extra minus sign from the charge of the electron. Then the Lorentz force points into the plane. $\endgroup$ – wcc Mar 22 at 1:41
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    $\begingroup$ @AdityaGarg thanks for the link, this is better explained than other articles I've found. $\endgroup$ – Charlie Mar 24 at 23:46

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