Question

I have a question regarding electromagnetic waves hitting a conducting surface. It's too long to post here, so check the first comment. If you need clarification on any part, don't hesitate to ask.

Answer 1

Let us consider a conducting plate in the \(yz\)-plane intersecting the origin of resistivity \(\rho\) and (small) thickness \(d\). Now, let us shine plane waves of light in the positive \(x\) direction onto the plate.\[\vec E = E_0\cos (\omega t-kx) \hat y\]\[\vec B = \frac{E_0}{c}\cos (\omega t-kx) \hat z\]Now, the current density induced in the conducting plate will be \(\vec J=\frac{\vec E}{\rho}\). If we take a slice of the conducting plate down the \(y\) direction with width \(dw\), the current \(dI\) through this slice will be \[dI=\vec J \cdot d\vec A = J d\ dw=\frac{E_0d}{\rho}dw\ \cos(\omega t).\]If this slice is of length \(\ell\) (i.e., the plate being that length in the \(y\) direction), then the magnetic field will impose a force on this induced current. Note that the area in the plane of the light of this strip will be \(\ell\ dw=dA\).\[dF=B\ dI\ \ell = \frac{E_0^2 d}{\rho c}(\ell\ dw) \cos^2(\omega t)=\frac{E_0^2 d\ dA}{\rho c} \cos^2(\omega t)\]Now, dividing out \(dA\) gives us the pressure on the conducting plate.\[P=\frac{dF}{dA}=\frac{E_0^2 d}{\rho c} \cos^2(\omega t)\]For convenience, we can also take the time average, which simply takes \(\cos^2(\omega t) \mapsto \frac{1}{2}\).\[\overline P = \frac{E_0^2 d}{2\rho c}\]It is clear from this formulation that the time-averaged pressure is inversely proportional to the resistivity (i.e., \(\overline P \propto \frac{1}{\rho}\). What if the conducting surface were a superconductor (in which case we'd have \(\rho = 0\)). Effectively, we'd have an infinite pressure from the electromagnetic radiation, which obviously doesn't happen. What is incorrect about this reasoning?

Answer 2

When you consider these two relations:

\(\vec E = E_0\cos (\omega t-kx) \hat y\)
\(\vec J=\Large \frac{\vec E}{\rho}\)

the problem is that the electric field is not the same in those equations.

The first one is \(\vec E_i\) of the incident wave, whereas, the second one is \(\vec E_t\) of the transmitted wave.
\(\vec E_t\)'s amplitude decreases exponentially in the conductor, but even on the surface, its amplitude is much less than \(E_o\)

On the surface, its amplitude is \(E_o\) times a transmission factor \(|t|=2\large \sqrt{\epsilon _o\rho\omega}\).

So, when \(\rho\) decreases, the field in the conductor (where the current is) tends to zero.
So the radiative pressure of the wave remains finite.
Its mean value is \(\epsilon_o\:E\,_o^2\) for normal incidence.