In sec. 9.4.2 Griffiths shows the well known boundary conditions for E and B fields, one of them is this:


Where $\textbf{K}_{f}$ is the free surface current. Griffiths says in this section:

"... For Ohmic conductors ($\textbf{J}_{f}=\sigma\textbf{E}$) there can be no surface current, since this would require an infinite electric field at the boundary."

I just can't understand it yet. Why is this true?

I have another question: The boundary for the normal component of E is $$\epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp}=\sigma_{f}$$ Where $\sigma_{f}$ is the free surface charge.

When we treat with incident EM waves on a conductor, is it necessary to consider $\textbf{K}_{f}$ and $\sigma_{f}$ different to zero? I am asking this because in this section of the book Griffiths made $\textbf{K}_{f}=0$ and $\sigma_{f}$ vanishes naturally because he only studies normal incidence, but my question goes to the most general case in which the normal component of E is nonzero.

  • 5
    $\begingroup$ hint: what is the 3d density of a surface current? $\endgroup$
    – hyportnex
    Jan 3, 2018 at 13:46
  • 4
    $\begingroup$ We need Dirac's Delta, don't we? $\endgroup$
    – Amadeus
    Jan 3, 2018 at 23:43
  • $\begingroup$ In AC circuits, the skin depth can be arbitrarily small. $\endgroup$
    – user4552
    Jan 28, 2019 at 15:13

4 Answers 4


While Griffiths typically writes surface currents exclusively as $\mathbf{K}_f$, one can also write them as volume currents as $\mathbf{J}_f = \delta(s) \mathbf{K}_f$, where $s = 0$ corresponds to the surface being considered. For an ohmic material, this means an electric field $\mathbf{E} = \frac{1}{\sigma} \delta(s) \mathbf{K}_f$, which blows up at the surface because the volume current blows up at the surface due to the presence of a Dirac delta at the surface.

Notice Griffiths is not forbidding you from having surface currents in general. That is certainly possible. He's just forbidding you from trying to put a surface current on an ohmic material. If there is a surface current, it can't be ohmic.

As for the second question, I see no reason for why that would be necessary, but it is convenient to assume the material is overall neutral. If there were surface charges, you could simply sum their effects later, since Electromagnetism is a linear theory.

  • $\begingroup$ The issue here is mathematical idealization versus physical reality. In practical work, one may calculate the skin depth. Then one may also calculate the electric field and/or power dissipation due to the resistance of the skin. If all are negligible, go ahead and assume a skin current. Otherwise, you have a more complicated problem. $\endgroup$
    – John Doty
    Jun 3, 2022 at 13:09
  • $\begingroup$ Even ohmic materials obeying $\mathbf j = \sigma \mathbf E$ in the volume do not obey it on the surface. For example, if there is some surface charge, total electric field has normal component on the surface, but current density does not (assuming no electron emission is happening). So the argument based on this form of Ohm's law is not valid. $\endgroup$ Jun 3, 2022 at 16:55

I think this is so because for finite conductivity and for ohmic conductors, J=$\sigma$E would require that the current density be parallel electric field. Since for conductors, electric field is perpendicular to the surface, so J (current) would also be normal to surface. But the boundary condition n $\times$ H = K requires K not to be normal to surface (as it should be perpendicular to the normal), thus there would be no surface current.

  • $\begingroup$ $\mathbf K$ can't have normal component because the mobile charge would jump out of the conductor, or go the opposite direction and charge up the surface, which will stop such process. But $\mathbf K$ can have tangential component. Electric field is perpendicular to surface only in static situations; if current is flowing, electric field does have tangential component. $\endgroup$ Jun 3, 2022 at 0:53

I think Griffiths means to say that for standard ohmic conductor, we expect the "skin" current to be distributed in non-zero thickness of the conductor in a rather smooth way, with density decreasing with depth in the conductor, but without substantial contribution to current due to the boundary itself (known to be so due to magnetization current in magnetized bodies).

His argument isn't completely convincing, since we know that generalized Ohm's law

$$ \mathbf j = \sigma \mathbf E $$

does not hold in general in air/vacuum outside the conductor and its validity on the boundary itself is dubious, since in general, electric field there has normal component, and current density does not (in normal circumstances without field emission).

Some surface current may be present, in theory, due to motion of surface charges. Griffiths allows presence of these charges and denotes their surface density as $\sigma$. If surface charge density $\sigma$ changes in time, it seems plausible that a surface current accompanying this change may be present too. But since it is "much easier" for this charge to appear via currents normal to the surface coming from the conductor depth rather than via translation of charge along the surface, there is a good reason to expect that such surface current may be negligible and thus we can put $\mathbf K = 0$. Applicability of this assumption is found and confirmed by agreement of the theory results when compared to measurements, which I think is very good in this standard scenario of wave reflection from ordinary metals.

If the metal is a superconductor, there is much smaller resistance to AC currents inside the conductor, and maybe even to surface currents. Maybe then surface current density due to impacting EM waves may not be negligible anymore, but I don't know.


It is easiest to return to his boundary conditions in 9.118, EM waves in conductors. Eq (iv) \begin{equation} \nabla \times \mathbf{B} = \mu \epsilon \frac{\partial \mathbf{E}}{\partial t} + \mu \sigma \mathbf{E}. \end{equation}

which is true when $\mathbf{J}=\sigma \mathbf{E}$. Now integrate this over an infinitessimal area perpendicular to the surface norm, spanning both media to get: \begin{eqnarray} \int d\mathbf{a} \centerdot \frac{\nabla \times \mathbf{B}}{\mu} &= & \int d\mathbf{a} \centerdot \left ( \epsilon \frac{\partial \mathbf{E}}{\partial t} + \sigma \mathbf{E} \right ) \\ \int d\mathbf{l} \centerdot\frac{\mathbf{B}}{\mu} &= & \int d\mathbf{a} \centerdot \left ( \epsilon \frac{\partial \mathbf{E}}{\partial t} + \sigma \mathbf{E} \right )\\ \frac{1}{\mu_1}\mathbf{B}_1^{\parallel}-\frac{1}{\mu_2}\mathbf{B}_2^{\parallel} &=&0. \end{eqnarray}

The second equation uses Stokes theorem to transform the curl into a closed line integral around the area. The area goes to zero so unless E goes to infinity, the integral will be zero on the right hand side. In the third line the area integral is zero, and the left hand side picks out only the parallel component of $\mathbf{B}$ in the two media. This is the standard treatment of magnetic boundary conditions. This explains Griffith's comment.

Here we have a conductor with no potential across it. The only source driving electrons would be an external electric field parallel to the surface. This will penetrate by the skin depth. To get free current as opposed to bound current, this field would need to be persist in an ever shrinking area, a true surface current. Here since the area goes to zero, the integral has to go to zero unless $\mathbf{E}\rightarrow \infty$.

For the second question Griffith treats the perpendicular component in equation 9.120. He shows a current density in a conductor is cause by Gauss Law, \begin{equation} \nabla \centerdot \mathbf{E} = \frac{\rho_f}{\epsilon}. \end{equation} which picks out the perpendicular component of $\mathbf{E}$. This component dissipates with time constant $\epsilon/\sigma$, and so he is looking for effects past this time frame. This time frame is less that $10^{-14}\ s$, so charges will move a bit, but fast.

For conductors, the dialectric constant, $\epsilon_2$ will be infinite, so $\mathbf{E}_{\perp T}$ should be zero. That gives the same perpendicular $\mathbf{E}$ component for both incoming and reflected waves. This indicates that the law of reflection should work with conductors.

  • $\begingroup$ Not sure what you mean by free/bound current. The current in the skin layer is free current in the standard sense (conduction current). Griffiths mentions bound current, I think, because if the material is magnetic, there could be magnetization current on the surface (which can be called bound current). $\endgroup$ Jun 3, 2022 at 0:58
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jun 3, 2022 at 2:25
  • $\begingroup$ rewrote this and added material to reply to your comments $\endgroup$ Jun 7, 2022 at 13:51

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