Lets say we have a wave which is linearly polarized and is incident to the surface of a imperfect conductor, which we will say is the plane $z=0$. Suppose the incident wave $E_i$ is parallel to the surface.

We know there will be a wave $E_r$ which is reflected, and because the conductor is imperfect, there will also be a transmitted wave $E_t$ which weakly penetrates weakly the conductor (skin effect).

To calculate reflection/transmission coefficients we need the boundary conditions. We have:

  1. $E_i(z=0)+E_r(z=0)=E_t(z=0)$ because the electric field is continuos across the boundary in the direction tangentiel to the surface.

  2. Assuming no surface courant, we should also have continuity of the magnetic field across the boundary: $B_i(z=0)+B_r(z=0)=B_t(z=0)$.

However, according to slide 10 of https://www2.ph.ed.ac.uk/~playfer/EMlect15.pdf, the second relation should actually be $B_i(z=0)-B_r(z=0)=B_t(z=0)$ (*).

The above link gives the correct answer for the relfection/transmission coefficients, so the equation (*) they give is not a typo.

I can't figure out for he life of me the source of the minus sign!


This is easy to explain. The boundary condition in the linked lecture slide (p. 10) for the magnetic field with the minus sign before the reflected amplitude is wrong! Your boundary conditions 1. and 2. for the electric and magnetic field amplitudes are correct.

In the considered case, there are no surface charges or surface currents and the permeabilities are assumed to be $\mu_0$ so that the necessary boundary conditions for wave reflection and transmission are the continuity of the tangential electric fields $$E_1=E_2 \tag 1$$ and the continuity of the tangential magnetic fields $$B_1=B_2 \tag 2$$ to which your correct amplitude boundary conditions 1. and 2 correspond. With the (complex) vectorial electric field amplitude $\vec E_0$, the incident electric wave is described by $$\vec E=\vec E_{0i} \exp i( \vec k \vec r-\omega t) \tag 3$$ where $\vec k=k_i \hat z$ is the wave vector pointing in positive z-direction, $\vec r = z \hat z$ is the position vector, $\omega$ is the angular frequency, and $t$ is the time. The reflected wave is $$\vec E=\vec E_{0r} \exp i(- \vec k \vec r-\omega t) \tag 4$$ Maxwell's equation $$\nabla \times \vec E=-\frac {\partial B}{\partial t} \tag 5$$ relates the vectorial magnetic field amplitude $\vec B_0$ to the electrical amplitude $\vec E_0$ by the cross product $$\vec k \times \vec E_0=\omega\vec B_0 \tag 6$$ Thus, when the electric field $\vec E_{0i}$ of the incident wave is in positive x-direction, the magnetic field $\vec B_{0i}$ is in positive y-direction giving $$B_{0i}= \frac {k_i}{\omega} E_{0i}\tag 7$$ For the transmitted wave one obtains $$B_{0t}= \frac {k_t}{\omega} E_{0t} \tag 8$$ The reflected wave propagates in negative z-direction so that the wave vector in eqs. (6) is $-\vec k=-k_i \hat z$ and the relation of the reflected magnetic and electric field amplitude has a negative sign $$B_{0r}= -\frac {k_i}{\omega} E_{0r} \tag 9$$ Thus the boundary condition for the magnetic field (2) can be expressed by the electric field amplitudes $$E_{0i}-E_{0r}=\frac {k_t}{k_{i}} E_{0t} \tag {10} $$ Note the negative sign before $E_{0r}$! The boundary condition for the electric field (1) gives the other equation for the electric field amplitudes $$E_{0i}+E_{0r}=E_{0t} \tag {11}$$ From eqs. (10) and (11) follow the (amplitude) reflection and transmission coefficients for any linear (not magnetic) materials 1 and 2. In particular, also for an imperfect conductor where $k_t$ is, in general, a complex quantity!

  • $\begingroup$ I believe this answer is false. First, the magnetic field condition does not, I believe, follow from the electric field condition, as the magnetic field condition depends on the surface courant! Second, the magnetic field of the reflected wave points in the same direction as the magnetic field of the incident wave at the boundary. Here's why... $\endgroup$ – Joshua Benabou Apr 1 '18 at 15:46
  • $\begingroup$ Say the conductor is the region $z>0$. So the incident wave is $\vec E_i=E_i \exp(i (kz-\omega t)) \vec u_x$. The reflected wave is thus $\vec E_r=E_r exp(-i (kz+\omega t)) \vec u_x$. The incident magnetic field is $\vec B_i=(1/c)E_i \exp(i (kz-\omega t))\vec u_y $ and the reflected magnetic field is $\vec B_r=(1/c)E_r \exp(-i (kz+\omega t)) \vec u_y $. $\endgroup$ – Joshua Benabou Apr 1 '18 at 15:47
  • $\begingroup$ @JoshuaBenabou - In the considered case of an imperfect conductor (i.e., with a conductivity $\sigma \neq \infty$), you have no surface current and thus , assuming $\mu_1=\mu_2$, the continuity boundary condition for the parallel magnetic fields $B1=B_2$ holds. This boundary condition doesn't follow from the continuity condition of the electric field $E_1=E_2$ equation. Thus the boundary conditions 1. and 2. stated in your question are correct! $\endgroup$ – freecharly Apr 1 '18 at 19:52
  • $\begingroup$ @JoshuaBenabou - I looked at the lecture slides given in your link and it is obvious that the magnetic field boundary condition on page (10) with the minus sign $ B_{0I}-B_{R}=B_{0T}$ is wrong, it should be a plus sign! However, the following relation for the electric field with the minus sign is correct, because again a sign error is made (in deriving $E_{0R}$ from $B_{0R}$ in not changing the sign of $k_1$ for the reflected wave) to obtain the correct result. $\endgroup$ – freecharly Apr 1 '18 at 19:56
  • $\begingroup$ @JoshuaBenabou - I am sorry for hastily concluded correctness of the wrong magnetic field condition with the minus sign. I remembered the minus sign in the equation for the electric field amplitudes derived from the magnetic field continuity, which follows, as I correctly stated from the sign change of the wave vector of the reflected wave. I will edit my answer accordingly. $\endgroup$ – freecharly Apr 1 '18 at 20:04

I admit I didn't read the link you provided, but I am going to take a wild stab:

Reversing the sign on $\vec B$ but not $\vec E$ reverses the sign on the poynting vector $\vec S$, thereby changing the direction of travel of the wave for the reflected wave.

  • $\begingroup$ I know it has to do something with the direction of the reflected wave, but I can't see exactly the problem. $\endgroup$ – Joshua Benabou Apr 1 '18 at 15:26

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