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Consider the situation of a plane electromagnetic wave in vacuum incident normally on an interface with a good conductor. Within the conductor there is a small transmitted electric field (proportional to the impedance of the conductor). This electric field will drive an oscillating current density.

My question is, when we solve for the boundary conditions we end up with a reflected wave that is almost exactly the same amplitude and out of phase with the incident wave. But are we implicitly (or even explicitly by assuming plane wave vacuum solutions to Maxwell's equations) neglecting the contribution from electromagnetic fields that would result outside the conductor from the oscillating current source inside the conductor?

And if so, is that always appropriate?

Edit: My concern stems from the fact that in a good conductor, almost regardless of the size of this oscillating current (which is $\propto \sigma^{1/2}$, where $\sigma$ is the conductivity), the reflected wave has almost exactly the same amplitude as the incident wave.

If someone can show me that the oscillating current density is responsible for the reflected wave then probably the problem is solved.

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  • $\begingroup$ Contribution to what? If we are solving boundary value problem, we deal only with total field, which has all contributions by definition, including primary field and secondary field of the currents on the interface of two halfspaces. $\endgroup$ Jan 4 at 18:38
  • $\begingroup$ A radio specialist would shrug his shoulders. For him, the electrical conductor is a receiving antenna rod. The conductor will 1. have losses (disspiation of energy through the surface electrons) and 2. will in turn radiate. $\endgroup$ Jan 5 at 5:43
  • $\begingroup$ The question is even more interesting for the deflection of EM radiation at edges. If the incoming EM radiation induces phonons along the surface, how does their periodicity affect the deflection of the radiation? $\endgroup$ Jan 5 at 5:46
  • $\begingroup$ The induced currents cause the "reflected" wave $\endgroup$
    – R. Emery
    Jan 5 at 8:30
  • $\begingroup$ @R.Emery read my edit. $\endgroup$
    – ProfRob
    Jan 5 at 8:31
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For posterity, here I think is the resolution to my problem. The whole set-up is self-consistent and no assumption needs to be made about ignoring any contributions due to oscillating currents in the conductor. The key is to realise that as the conductivity gets higher, the electric field in the conductor gets lower, but the resultant current density multiplied by the skin depth is almost constant and exactly that required to produced the reflected (and transmitted) waves as suggested by R.Emery. The trick is in showing that.

Take as a set-up and incident wave of the form $\vec{E_i} = E_i \cos(\omega t-kz)\hat{i}$, which is incident normally on a conductive plane at $z=0$, with a high conductivity $\sigma$. Let us assume this leads to a transmitted wave in the conductor and a reflected wave. Let us further assume that the wavelength of the incident field is much larger than any skin depth of the conductor, so that in the region where there is a current density the value of $\vec{E_i} \simeq \vec{E_i}(0,t)$.

If the average current density in the medium is $\vec{J}(\omega t)$ and the effective width of the current carrying region is $\delta$, then the electric field generated by the oscillating current sheet will be$\dagger$ $$ \vec{E_s}(z,t) = -\frac{\mu_0 c \delta}{2} \vec{J}(\omega[t- z/c]) \ \ \ \ \ {\rm for\ \ z>0}$$ $$ \vec{E_s}(z,t) = -\frac{\mu_0 c \delta}{2} \vec{J}(\omega[t+ z/c]) \ \ \ \ \ {\rm for\ \ z<0}$$ where $t - z/c$ represents a retarded time.

The total electric fields in the regions $z>0$ and $z<0$ will be $\vec{E_T} = \vec{E_i} + \vec{E_s}$ and from Ohm's law $\vec{J}(t) = \sigma \vec{E_T}(0,t)$.

At $z=0$, we can use either expression for the field generated by the oscillating current sheet to write $$\vec{J}(t) = \sigma \left( E_i \cos(\omega t)\hat{i} - \frac{\mu_0 c \delta}{2}\vec{J}(t)\right) $$ $$ \vec{J}(t) = \frac{\sigma E_i\cos(\omega t)}{1 + \sigma \mu_0 c \delta/2}\hat{i}\ . $$

This current density can then be used to find $\vec{E_s}$ at $z>0$ and $z<0$. $$ \vec{E_s} = -\frac{\mu_0 c \delta}{2} \left( \frac{\sigma E_i\cos(\omega t - kz)}{1 + \sigma \mu_0 c \delta/2} \right)\hat{i} \ \ \ \ \ {\rm for\ \ z>0}$$ $$ \vec{E_s} = -\frac{\mu_0 c \delta}{2} \left( \frac{\sigma E_i\cos(\omega t + kz)}{1 + \sigma \mu_0 c \delta/2} \right)\hat{i} \ \ \ \ \ {\rm for\ \ z<0}$$

These fields add to the incident wave to give the total electric field on either side of the current sheet: $$ \vec{E_T} = E_i \cos(\omega t - kz) \left[ 1 - \frac{\sigma \mu_0 c \delta}{2 + \sigma \mu_0 c \delta} \right] \hat{i} \ \ \ \ \ {\rm for\ \ z>0}$$ $$ \vec{E_T} = E_i \left[ \cos(\omega t - kz) - \frac{\sigma \mu_0 c \delta}{2 + \sigma \mu_0 c \delta}\cos(\omega t + kz) \right] \hat{i} \ \ \ \ \ {\rm for\ \ z<0}$$

For a very good conductor we can require that $\sigma \mu_0 c \delta \gg 2$, these reduce to $$\vec{E_T} \simeq 0\ \ \ \ \ {\rm for\ \ z>0}$$ $$\vec{E_T} \simeq 2E_i \sin(kz)\sin(\omega t) \hat{i} \ \ \ \ \ {\rm for\ \ z<0}$$

Thus the fields produced by the oscillating current sheet are exactly those required to cancel the incident electric field on the far side of the conductive boundary (and beyond the current-carrying region) and to produce a standing wave on the incident side of the boundary.

I am still thinking about whether I can refine this treatment to deal with the spatial dependence of the electric fields and current density within the current-carrying region rather than assuming it is narrow.

The solution to the paradox I posed in my own question is that although the current density in the conductor is $\propto \sigma^{1/2}$, what matters for the fields that are produced is the product $J \delta$; and since the skin depth is $\propto \sigma^{-1/2}$, this product is almost constant. This is why the reflected wave has almost the same amplitude, independent of the exact value of $\sigma$ (as long as it is large enough to be considered a good conductor).

$\dagger$ Proving the form of the fields radiating from an oscillating current sheet.

Assume a position $\vec{r}=z\hat{k}$ relative to an origin at some point on the plane $z=0$ and that there is a time-dependent current density $J(t) = J\cos(\omega t) \hat{i}$ in a sheet of thickness $\delta$.

The general solution of the inhomogeneous wave equation is $$\vec{A}(z,t) = \frac{\mu_0}{4\pi} \int \frac{J\cos[\omega(t-|r-r'|/c)]}{|r-r'|}\ d^3r'\ \hat{i},$$ where $\vec{r'}$ is a position on the current sheet, $|r-r'| = \sqrt{z^2 + r'^2}$. If we also let $\vec{J}\ d^3r' = \delta \vec{J} 2\pi r'\ dr'$ and the solution for the vector potential becomes $$\vec{A}(z,t) = \frac{\mu_0 \delta J}{4\pi} \int^{\infty}_{r'=0} \frac{\cos[\omega(t - k\sqrt{z^2 + r'^2})]}{\sqrt{z^2 + r'^2}}\ 2\pi r'\ dr'\ \hat{i} .$$

If we let $u = k\sqrt{z^2 + r'^2}$, then $du = k r'\ dr'/\sqrt{z^2 + r'^2}$ and $$\vec{A}(z,t) = \frac{\mu_0 \delta J}{2k} \int^{\infty}_{u=k|z|} \cos(\omega t - u)\ du\ \hat{i} $$ $$\vec{A}(z,t) = -\frac{\mu_0 \delta J}{2 k} \left[\sin(\omega t-k|z|) - \sin (\infty)\right]\ \hat{i}\ . $$ Taking the time derivative $$ \vec{E} = -\frac{\partial \vec{A}}{\partial t} = -\frac{\mu_0 \delta J \omega}{2 k}\cos(\omega t - kz)\ \hat{i}\ \ \ \ {\rm for}\ z>0 $$ $$\vec{E} = -\frac{\partial \vec{A}}{\partial t} = -\frac{\mu_0 \delta J \omega}{2 k}\cos(\omega t + kz)\ \hat{i}\ \ \ \ {\rm for}\ z<0\, , $$ where $\omega/k = c$ in vacuum.

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  • $\begingroup$ > I am still thinking about whether I can refine this treatment to deal with the spatial dependence of the electric fields and current density within the current-carrying region -- Yes this should be possible, I am more familiar with dielectrics but the basic assumptions and results should be the same - we assume that $\mathbf E(\mathbf r,t) = \sigma(\omega) \mathbf j(\mathbf r,t)$ where all qs. are complex. Write down the wave equation for $E_x$ (component parallel to boundary) in both regions and assume solution (on the transmission side) of the form $Ae^{-k_d z}\cos(\Omega t - k_o z)$... $\endgroup$ Jan 5 at 17:28
  • $\begingroup$ where $\Omega$ is frequency of primary wave. Wave equation should give you $k_d$ and $k_o$ as functions of $\sigma',\sigma''$ (components of complex $\sigma$). $\endgroup$ Jan 5 at 17:29
  • $\begingroup$ As a result you should get $\mathbf j$ that diverges with $\sigma$ near the boundary, but its integral over $z$ from 0 to $\infty$ should be finite. $\endgroup$ Jan 5 at 17:33
  • $\begingroup$ So your derivation is only valid for low frequency EM waves,right? $\endgroup$
    – Kksen
    Jul 27 at 16:27
  • $\begingroup$ @KamKahSen I've assumed that $\lambda > \delta$. So, not that low a frequency for good conductors. $\endgroup$
    – ProfRob
    Jul 27 at 19:19

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