Boundary Conditions at a Conducting Surface: Skin Depth

Let us consider for a moment what time dependent EM fields look like at the surface of a ``perfect'' conductor. A perfect conductor can move as much charge instantly as is required to cancel all fields inside. The skin depth $\delta = \lim_{\sigma \to \infty} \sqrt{2/\mu\epsilon_b\sigma} = 0$ as $\alpha$ diverges - effectively all frequencies are ``static'' to a perfect conductor. This is how type I superconductors expel all field flux.

If we examine the fields in the vicinity of a boundary between a perfect conductor and a normal dielectric/diamagnetic material, we get:

$$(\mbox{\boldmath$D$} - \mbox{\boldmath$D$}_c)\cdot \hat{\mbo... ...= \hat{\mbox{\boldmath$n$}}\cdot \mbox{\boldmath$D$} = \Sigma$$ (10.1)

where $\mbox{\boldmath$D$}_c$ and $\mbox{\boldmath$E$}_c$ inside the conductor vanish. Similarly,

$$\hat{\mbox{\boldmath$n$}}\times (\mbox{\boldmath$H$} - \mbox... ...\boldmath$n$}}\times\mbox{\boldmath$H$} = \mbox{\boldmath$K$}$$ (10.2)

(where in these expressions, $\Sigma$ is the surface charge density so we don't confuse it with the conductivity $\sigma$, sigh, and similarly $\mbox{\boldmath$K$}$ is the surface current density).

In addition to these two inhomogeneous equations that normal and parallel fields at the surface to sources, we have the usual two homogeneous equations:

$$\hat{\mbox{\boldmath$n$}}\cdot(\mbox{\boldmath$B$} - \mbox{\boldmath$B$}_c) = 0$$ (10.3)

$$\hat{\mbox{\boldmath$n$}}\times(\mbox{\boldmath$E$} - \mbox{\boldmath$E$}_c) = 0$$ (10.4)

Note that these are pretty much precisely the boundary conditions for a static field and should come as no surprise. For perfect conductors, we expect the fields inside to vanish, which in turn implies that $\mbox{\boldmath$E$}$ outside must be normal to the conducting surface and $\mbox{\boldmath$B$}$ outside must lie only parallel to the conducting surface, as usual.

However, for materials that are not perfect conductors, the fields don't vanish instantly ``at'' the mathematical surface. Instead they die off exponentially within a few multiples of the skin depth $\delta$. On scales large with respect to this, they will ``look'' like the static field conditions above, but of course within this cutoff things are very different.

For one thing, Ohm's law tells us that we cannot have an actual ``surface layer of charge'' because for any finite conductivity, the resistance scales like the cross-sectional area through which charge flows. Consequently the real boundary condition on $\mbox{\boldmath$H$}$ precisely at the surface is:

$$\hat{\mbox{\boldmath$n$}}\times (\mbox{\boldmath$H$} - \mbox{\boldmath$H$}_c) = 0$$ (10.5)

$$\mbox{\boldmath$H$}_{\vert\vert} = \mbox{\boldmath$H$}_{c,\vert\vert}$$ (10.6)

where $\mbox{\boldmath$H$}_{\vert\vert} = (\hat{\mbox{\boldmath$n$}}\times\mbox{\boldmath$H$})\times \hat{\mbox{\boldmath$n$}}$. However, this creates a problem! If this field varies rapidly in some direction (and it does) it will generate an electric field according to Faraday's law! If the direction of greatest variation is ``into the conductor'' (as the field is being screened by induced surface currents) then it will generate a small electric field parallel to the surface, one which is neglected (or rather, cannot occur) in the limit that the conductivity is infinite. This electric field, in turn, generates a current, which causes the gradual cancellation of $\mbox{\boldmath$H$}_{\vert\vert}$ as less and less the total bulk current is enclosed by a decending loop boundary.

If the conductivity is large but not infinite, one way to figure out what happens is to employ a series of successive approximations starting with the assumption of perfect conductivity and using it to generate a first order correction based on the actual conductivity and wavelength. The way it works is:

1. First, we assume that outside the conductor we have only $\mbox{\boldmath$E$}_\perp$ and $\mbox{\boldmath$H$}_{\vert\vert}$ from the statement of the boundary conditions assuming that the fields are instantly cancelled at the surface.
2. Assume $\delta \ll k^{-1}$ along the surface - the skin depth is much less than a wavelength and the fields (whatever they may be) vanish across roughly this length scale, so we can neglect variation (derivatives) with respect to coordinates that lie along the surface compared to the coordinate perpendicular to the surface.
3. Use this approximation in Maxwell's Equations, along with the assumed boundary conditions for a perfect conductor, to derive relations between the fields in the transition layer.
4. These relations determine the small corrections to the presumed boundary fields both just outside and just inside the surface.

The assumption of rapid variation only as one decends into the conductor is a key step, as we shall see.

Thus (from 1):

$$\hat{\mbox{\boldmath$n$}}\times(\mbox{\boldmath$H$} - \mbox{\boldmath$H$}_c) = 0$$ (10.7)

or $\mbox{\boldmath$H$}_{\vert\vert}({\rm outside}) = \mbox{\boldmath$H$}_{\vert\vert}({\rm inside}) = \mbox{\boldmath$H$}_{\vert\vert} \ne 0$, where the latter assumption is because the result is boring if there are no fields, right?

We both Ampere's law (assuming no displacement in the conductor to leading order) and Faraday's law to obtain relations for the harmonic fields in terms of curls of each other:

$$\mbox{\boldmath$\nabla$} \times \mbox{\boldmath$H$}_c = \sigma \mbox{\boldmath$E$}_c = \mbox{\boldmath$J$}$$ (10.8)

$$\mbox{\boldmath$\nabla$} \times \mbox{\boldmath$E$}_c = -\frac{\partial \mbox{\boldmath$B$}_c}{\partial t} = i\omega \mu_c \mbox{\boldmath$H$}_c$$ (10.9)

become

$$\mbox{\boldmath$E$}_c = \frac{1}{\sigma}\mbox{\boldmath$\nabla$} \times \mbox{\boldmath$H$}_c$$ (10.10)

$$\mbox{\boldmath$H$}_c = -i\frac{1}{\mu_c\omega} \mbox{\boldmath$\nabla$} \times \mbox{\boldmath$E$}_c$$ (10.11)

As we might expect, high frequencies create relatively large induced electric fields as the magnetic fields change, but high conductivity limits the size of the supported electric field for any given magnetic field strength in a frequency independent way.

Now we need to implement assumption 2 on the $\mbox{\boldmath$\nabla$}$ operator. If we pick a coordinate $\xi$ to be perpendicular to the surface pointing into the conductor (in the $-\hat{\mbox{\boldmath$n$}}$ direction) and insist that only variations in this direction will be significant only on length scales of $\delta$:

$$\mbox{\boldmath$\nabla$} \approx -\hat{\mbox{\boldmath$n$}}\frac{\partial }{\partial \xi}$$ (10.12)

then we get:

$$\mbox{\boldmath$E$}_c \approx - \frac{1}{\sigma}\left(\hat{\mbox{\boldmath$n$}} \times \frac{\partial \mbox{\boldmath$H$}_c}{\partial \xi}\right)$$

$$\mbox{\boldmath$H$}_c \approx i\frac{1}{\mu_c\omega} \left(\hat{\mbox{\boldmath$n$}} \times \frac{\partial \mbox{\boldmath$E$}_c}{\partial \xi}\right)$$ (10.13)

(Note well the deliberate use of $approx$ to emphasize that there may well be components of the fields in the normal direction or other couplings between the components in the surface, but those components do not vary particularly rapidly along the surface and so are not large contributors to the curl.)

These two equations are very interesting. They show that while the magnitude of the fields in the vicinity of the conducting surface may be large or small (depending on the charge and currents near the surface) the curls themselves are dominated by the particular components of $\mbox{\boldmath$E$}_c$ and $\mbox{\boldmath$H$}_c$ that are in the plane perpendicular to $\hat{\mbox{\boldmath$n$}}$ (and each other) because the field strengths (whatever they are) are most rapidly varying across the surface.

What this pair of equations ultimately does is show that if there is a magnetic field just inside the conductor parallel to its surface (and hence perpendicular to $\hat{\mbox{\boldmath$n$}}$) $\mbox{\boldmath$H$}_{\vert\vert}$ that rapidly varies as one descends, then there must be an electric field $\mbox{\boldmath$E$}_{\vert\vert}$ that is its partner. Our zeroth approximation boundary condition on $\mbox{\boldmath$H$}_{\vert\vert}$ above shows that it is actually continuous across the mathematical surface of the boundary and does not have to be zero either just outside or just inside of it. However, in a good conductor the $\mbox{\boldmath$E$}_{\vert\vert}$ field it produces is small.

This gives us a bit of an intuitive foundation for the manipulations of Maxwell's equations below. They should lead us to expressions for the coupled EM fields parallel to the surface that self-consistently result from these two equations.

We start by determining the component of $\mbox{\boldmath$H$}_c$ (the total vector magnetic field just inside the conductor) in the direction perpendicular to the surface:

$$\hat{\mbox{\boldmath$n$}}\cdot\mbox{\boldmath$H$}_c = \frac... ...imes \frac{\partial \mbox{\boldmath$E$}_c}{\partial \xi}) = 0$$ (10.14)

This tells us that $\mbox{\boldmath$H$}_c = \mbox{\boldmath$H$}_{\vert\vert} = (\hat{\mbox{\boldmath$n$}}\times\mbox{\boldmath$H$}_c)\times\hat{\mbox{\boldmath$n$}}$ - the magnetic field coupled by $\mbox{\boldmath$E$}_c$ by Faraday's law lies in the plane of the conducting surface to lowest order.

Next we form a vector that lies perpendicular to both the normal and the magnetic field. We expect $\mbox{\boldmath$E$}_c$ to lie along this direction one way or the other.

$$\hat{\mbox{\boldmath$n$}} \times \mbox{\boldmath$H$}_c = \hat{\mbox{\boldmath$n$}} \times \left(i\frac{1}{\mu_c\omega} \ha... ...boldmath$n$}} \times \frac{\partial \mbox{\boldmath$E$}_c}{\partial \xi}\right)$$

$$= i\frac{1}{\mu_c\omega} \frac{\partial }{\partial \xi} \left( \hat... ...mbox{\boldmath$n$}}\cdot \mbox{\boldmath$E$}_c) - \mbox{\boldmath$E$}_c \right)$$

$$= -i\frac{1}{\mu_c\omega} \frac{\partial \mbox{\boldmath$E$}_{c,\vert\vert} } {\partial \xi}$$

(where $\mbox{\boldmath$E$}_{c,\perp} = \hat{\mbox{\boldmath$n$}}(\hat{\mbox{\boldmath$n$}}\cdot\mbox{\boldmath$E$})$ and $\mbox{\boldmath$E$}_c = \mbox{\boldmath$E$}_{c,\perp} + \mbox{\boldmath$E$}_{c,\vert\vert}$) and find that it does! The fact that the electric field varies most rapidly in the $-\hat{\mbox{\boldmath$n$}}$ ($+\xi$) direction picks out its component in the plane whatever it might be and relates it to the magnetic field direction also in the plane.

However, this does not show that the two conditions can lead to a self-sustaining solution in the absence of driving external currents (for example). To show that we have to substitute Ampere's law back into this:

$$\hat{\mbox{\boldmath$n$}} \times \mbox{\boldmath$H$}_c = -i\frac{1}{\mu_c\omega} \frac{\partial }{\partial \xi} \left( -\f... ...ldmath$n$}} \times \frac{\partial \mbox{\boldmath$H_c$}}{\partial \xi}) \right)$$

$$\hat{\mbox{\boldmath$n$}} \times \mbox{\boldmath$H$}_c = i\frac{1}{\mu_c\omega\sigma} ( \hat{\mbox{\boldmath$n$}} \times \frac{\partial^2 \mbox{\boldmath$H_c$})}{\partial \xi^2})$$

$$( \hat{\mbox{\boldmath$n$}} \times \frac{\partial^2 \mbox{\boldmath$H_c$})}{\partial \xi^2}) = -i\mu_c\omega\sigma \hat{\mbox{\boldmath$n$}} \times \mbox{\boldmath$H$}_c$$

$$\frac{\partial^2}{\partial \xi^2}\left(\hat{\mbox{\boldmath$n$}}\times \mbox{\boldmath$H_c$}\right) + i\mu_c\omega\sigma \left(\hat{\mbox{\boldmath$n$}} \times \mbox{\boldmath$H$}_c\right) = 0$$

or

$$\frac{\partial^2}{\partial \xi^2} (\hat{\mbox{\boldmath$n$}}... ...}(\hat{\mbox{\boldmath$n$}} \times \mbox{\boldmath$H$}_c) = 0$$ (10.15)

where we used the first result and substituted $\delta^2 = 2/(\mu_c\omega\sigma)$.

This is a well-known differential equation that can be written any of several ways. Let $\kappa^2 = \frac{2i}{\delta^2}$. It is equivalent to all of:

$$(\frac{\partial^2 }{\partial \xi^2} + \kappa^2) (\hat{\mbox{\boldmath$n$}}\times\mbox{\boldmath$H$}_c) = 0$$ (10.16)

$$(\frac{\partial^2 }{\partial \xi^2} + \kappa^2) (\hat{\mbox{\boldmath$n$}}\times\mbox{\boldmath$H$}_c)\times\hat{\mbox{\boldmath$n$}} = 0$$ (10.17)

$$(\frac{\partial^2 }{\partial \xi^2} + \kappa^2) \mbox{\boldmath$H$}_{\vert\vert} = 0$$ (10.18)

$$(\frac{\partial^2 }{\partial \xi^2} + \kappa^2) \mbox{\boldmath$H$}_c = 0$$ (10.19)

Where:

$$(\hat{\mbox{\boldmath$n$}}\times\mbox{\boldmath$H$}_c)\times\hat{\mbox{\boldmath$n$}} = \mbox{\boldmath$H$}_{\vert\vert}$$ (10.20)

as noted above. The solution to this form is then:

$$\mbox{\boldmath$H$}_c(\xi) = \mbox{\boldmath$H$}_0 e^{\pm \sqrt{-\kappa^2} \xi}$$ (10.21)

where $\mbox{\boldmath$H$}_0$ is the magnetic field vector in the plane of the conductor at the surface and where this equation indicates how this value is attenuated as one decends into the conductor.

As always, we have two linearly independent solutions. Either of them will work, and (given the already determined sign/branch associated with the time dependence $e^{-i\omega t}$) will ultimately have the physical interpretation of waves moving in the direction of $+\xi$ ( $-\hat{\mbox{\boldmath$n$}}$) or in the direction of $-\xi$ ( $\hat{\mbox{\boldmath$n$}}$). Let us pause for a moment to refresh our memory of taking the square root of complex numbers (use the subsection that treats this in the last chapter of these notes or visit Wikipedia of there is any problem understanding).

For this particular problem,

$$\sqrt{-\kappa^2} = \sqrt{-\frac{2i}{\delta^2}} = \pm \frac{1}{\delta}(-1+i)$$ (10.22)

(draw this out in pictures). We want the solution that propagates into the surface of the conductor, decending from the dielectric medium, which is the positive branch:

$$\mbox{\boldmath$H$}_c = \mbox{\boldmath$H$}_0 e^{\sqrt{-\kappa^2}\xi} = \mbox{\boldmath$H$}_0 e^{\frac{1}{\delta}(-1 + i)\xi}$$

$$= \mbox{\boldmath$H$}_0 e^{-\frac{\xi}{\delta}} e^{i\frac{\xi}{\delta}}$$ (10.23)

(consider $e^{i\xi/\delta - \omega t}$).

Now we need to find an expression for $\mbox{\boldmath$E$}_c$, which we do by backsubstituting into Ampere's Law:

$$\mbox{\boldmath$E$}_c = - \frac{1}{\sigma}\left(\hat{\mbox{\boldmath$n$}} \times \frac{\partial \mbox{\boldmath$H$}_c}{\partial \xi}\right)$$

$$= - \frac{1}{\delta\sigma} (-1 + i) \left(\hat{\mbox{\boldmath$n$}} \times \mbox{\boldmath$H$}_0\right) e^{\frac{1}{\delta}(-1 + i)\xi}$$

$$\mbox{\boldmath$E$}_c = \sqrt{\frac{\mu_c\omega}{2\sigma}}(1 - i) (\hat{\mbox{\boldmath$n$}} \times\mbox{\boldmath$H$}_0)e^{-\frac{\xi}{\delta}} e^{i\frac{\xi}{\delta}}$$ (10.24)

Note well the direction! Obviously $\hat{\mbox{\boldmath$n$}}\cdot\mbox{\boldmath$E$}_c = 0$, (in this approximation) so $\mbox{\boldmath$E$}_c$ must lie in the plane of the conductor surface, just like $\mbox{\boldmath$H$}_{\vert\vert}$!

As before (when we discussed fields in a good conductor):

• $\mbox{\boldmath$E$}_c,\mbox{\boldmath$H$}_c$ not in phase, but out of phase by $\pi/4$.
• Rapid decay as wave penetrates surface.
• $\mbox{\boldmath$H$}_c \gg \mbox{\boldmath$E$}_c$ ($\sigma$ ``large'', $\delta$ ``small'') so energy is primarily magnetic.
• $\hat{\mbox{\boldmath$n$}} \perp \mbox{\boldmath$E$}_c \perp \mbox{\boldmath$H$}_c \perp \hat{\mbox{\boldmath$n$}}$ - fields are predominantly parallel to the surface and mutually transverse, they propagate ``straight into'' surface, attenuating rapidly as they go.
• Recall:
  

  $$\hat{\mbox{\boldmath$n$}}\times(\mbox{\boldmath$E$} - \mbox{\boldmath$E$}_c) = 0$$ (10.25)

  
  
  at the surface. Since $\mbox{\boldmath$E$}_c$ lies approximately in the surface, this yields
  

  $$\mbox{\boldmath$E$} \approx \mbox{\boldmath$E$}_c \approx \s... ...oldmath$H$}_0)e^{-\frac{\xi}{\delta}} e^{i\frac{\xi}{\delta}}$$ (10.26)

  
  
  just outside the surface - the field is approximately continuous! At this level of approximation, $\mbox{\boldmath$\nabla$} \times \mbox{\boldmath$E$} = i\omega\mbox{\boldmath$B$}$, $\mbox{\boldmath$E$}$ is parallel to the surface, and there is a small $\mbox{\boldmath$B$}_\perp$ to the surface of the same general order of magnitude as $\mbox{\boldmath$E$}$.

Since both $\mbox{\boldmath$E$}_{\vert\vert} \ne 0$ and $\mbox{\boldmath$H$}_{\vert\vert} \ne 0$ at the surface ($\xi = 0$) there must be a power flow into the conductor!

$$\frac{dP_{\rm in}}{dA} = -\frac{1}{2} {\rm Re}\left(\mbox{\b... ... \frac{\mu_c\omega\delta}{4}\vert\mbox{\boldmath$H$}_0\vert^2$$ (10.27)

where we HOPE that it turns into heat. Let's see:

$$\mbox{\boldmath$J$} = \sigma\mbox{\boldmath$E$} = \sqrt{\fra... ...dmath$n$}} \times \mbox{\boldmath$H$}_0) e^{-\xi(1-i)/\delta}$$ (10.28)

so that the time averaged power loss is (from Ohm's Law):

$$\frac{dP}{dV} = \frac{1}{\Delta A}\frac{dP}{d\xi} = \frac{1}{2} \mbox{\boldmath$J$}\cdot\mbox{\boldmath$E$}^* = \frac{1}{2\sigma}\mbox{\boldmath$J$}\cdot\mbox{\boldmath$J$}^*$$ (10.29)

$$\Delta P = \Delta A \frac{1}{2\sigma} \int_0^\infty d\xi \mbox{\boldmath$J$}\cdot\mbox{\boldmath$J$}^*$$

$$= \Delta A \frac{\mu_c\omega}{2} \vert H_0\vert^2 \int_0^\infty d\xi e^{-2\xi/\delta}$$

$$= \Delta A \frac{\mu_c\omega}{4} \vert H_0\vert^2$$ (10.30)

which just happens to correspond to the flux of the pointing vector through a surface $\Delta A$!

Finally, we need to define the ``surface current'':

$$\mbox{\boldmath$K$}_{\rm eff} = \int_0^\infty \mbox{\boldma... ...d\xi = (\hat{\mbox{\boldmath$n$}} \times \mbox{\boldmath$H$})$$ (10.31)

where $\mbox{\boldmath$H$}$ is determined just outside(inside) of the surface of a ``perfect'' conductor in an idealized limit - note that we are just adding up the total current in the surface layer and that it all works out.

Hopefully this exposition is complete enough (and correct enough) that any bobbles from lecture are smoothed out. You can see that although Jackson blithely pops all sorts of punch lines down in the text, the actual algebra of getting them, while straightforward, is not trivial!