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 as 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: (  _c)· = · = &Sigma#Sigma; where and inside the conductor vanish. Similarly, ×(  _c) = × = (where in these expressions, is the surface charge density so we don't confuse it with the conductivity , sigh, and similarly 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:
·(

_c) & = & 0
×(

_c) & = & 0
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
outside must be normal to the conducting surface and
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 . 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 crosssectional area through which charge
flows. Consequently the real boundary condition on
precisely at the surface is:
×(

_c) & = & 0
_ & = &
_c,
where
. 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
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:
Thus (from 1): ×(  _c) = 0 or , 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:
×
_c & = & &sigma#sigma;
_c =
×
_c & = & &part#partial;
_c&part#partial;t
= i&omega#omega;&mu#mu;_c
_c
become
_c & = & 1&sigma#sigma;
×
_c
_c & = & i1&mu#mu;_c&omega#omega;
×
_c
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
operator.
If we pick a coordinate
to be perpendicular to the surface
pointing into the conductor (in the
direction) and insist
that only variations in this direction will be significant only on
length scales of
:
&ap#approx;
&part#partial;&part#partial;&xi#xi;
then we get:
_c & &ap#approx;&  1&sigma#sigma;(
×&part#partial;
_c&part#partial;&xi#xi;)
_c & &ap#approx;& i1&mu#mu;_c&omega#omega; (
×&part#partial;
_c&part#partial;&xi#xi;)
(Note well the deliberate use of
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 and that are in the plane perpendicular to (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 ) that rapidly varies as one descends, then there must be an electric field that is its partner. Our zeroth approximation boundary condition on 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 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 selfconsistently result from these two equations.
We start by determining the component of (the total vector magnetic field just inside the conductor) in the direction perpendicular to the surface: · _c = i&mu#mu;_c&omega#omega; ·( ×&part#partial; _c&part#partial;&xi#xi;) = 0 This tells us that  the magnetic field coupled by 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
to lie along this
direction one way or the other.
×
_c & = &
×(i1&mu#mu;_c&omega#omega;
×&part#partial;
_c&part#partial;&xi#xi;)
& = & i1&mu#mu;_c&omega#omega; &part#partial;&part#partial;&xi#xi;
(
(
·
_c) 
_c )
& = & i1&mu#mu;_c&omega#omega; &part#partial;
_c,
&part#partial;&xi#xi;
(where
and
) and find that it does! The fact
that the electric field varies most rapidly in the
(
)
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 selfsustaining solution in the absence of driving external currents (for example). To show that we have to substitute Ampere's law back into this:
×
_c & = & i1&mu#mu;_c&omega#omega;
&xi#xi; ( 1&sigma#sigma; (
×
&xi#xi;) )
×
_c & = & i1&mu#mu;_c&omega#omega;&sigma#sigma;
(
×&part#partial;^2
)&part#partial;&xi#xi;^2)
(
×&part#partial;^2
)&part#partial;&xi#xi;^2) & = & i&mu#mu;_c&omega#omega;&sigma#sigma;
×
_c
&part#partial;^2&part#partial;&xi#xi;^2(
×
)
& + & i&mu#mu;_c&omega#omega;&sigma#sigma;(
×
_c) = 0
or
&part#partial;^2&part#partial;&xi#xi;^2 (
×
_c) +
2i&delta#delta;^2(
×
_c) = 0
where we used the first result and substituted
.
This is a wellknown differential equation that can be written any of
several ways. Let
. It is equivalent
to all of:
(&part#partial;^2 &part#partial;&xi#xi;^2 + &kappa#kappa;^2)
(
×
_c) & = & 0
(&part#partial;^2 &part#partial;&xi#xi;^2 + &kappa#kappa;^2)
(
×
_c)×
& = & 0
(&part#partial;^2 &part#partial;&xi#xi;^2 + &kappa#kappa;^2)
_ & = & 0
(&part#partial;^2 &part#partial;&xi#xi;^2 + &kappa#kappa;^2)
_c & = & 0
Where:
(
×
_c)×
=
_
as noted above. The solution to this form is then:
_c(&xi#xi;) =
_0 e^±&kappa#kappa;^2 &xi#xi;
where
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 ) will ultimately have the physical interpretation of waves moving in the direction of ( ) or in the direction of ( ). 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,
&kappa#kappa;^2 = 2i&delta#delta;^2 = ±1&delta#delta;(1+i)
(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:
_c =
_0 e^&kappa#kappa;^2&xi#xi;
& = &
_0 e^1&delta#delta;(1 + i)&xi#xi;
& = &
_0 e^&xi#xi;&delta#delta; e^i&xi#xi;&delta#delta;
(consider
).
Now we need to find an expression for
, which we do by
backsubstituting into Ampere's Law:
_c & = &  1&sigma#sigma;(
×&part#partial;
_c&part#partial;&xi#xi;)
& = &  1&delta#delta;&sigma#sigma; (1 + i)
(
×
_0) e^1&delta#delta;(1 + i)&xi#xi;
_c & = & &mu#mu;_c&omega#omega;2&sigma#sigma;(1  i)
(
×
_0)e^&xi#xi;&delta#delta; e^i&xi#xi;&delta#delta;
Note well the direction! Obviously , (in this approximation) so must lie in the plane of the conductor surface, just like !
As before (when we discussed fields in a good conductor):
Since both
and
at the surface
(
) there must be a power flow into the conductor!
dP_indA = 12
Re(
·(
_c ×
_c^*)) =
&mu#mu;_c&omega#omega;&delta#delta;4
_0^2
where we HOPE that it turns into heat. Let's see:
= &sigma#sigma;
= &mu#mu;_c&omega#omega;&sigma#sigma;2(1 
i)(
×
_0) e^&xi#xi;(1i)/&delta#delta;
so that the time averaged power loss is (from Ohm's Law):
dPdV = 1&Delta#Delta;AdPd&xi#xi; & = & 12
·
^* =
12&sigma#sigma;
·
^*
&Delta#Delta;P & = & &Delta#Delta;A 12&sigma#sigma; &int#int;_0^&infin#infty;d&xi#xi;
·
^*
& = & &Delta#Delta;A &mu#mu;_c&omega#omega;2 H_0^2 &int#int;_0^&infin#infty;d&xi#xi;
e^2&xi#xi;/&delta#delta;
& = & &Delta#Delta;A &mu#mu;_c&omega#omega;4 H_0^2
which just happens to correspond to the flux of the pointing vector
through a surface
!
Finally, we need to define the ``surface current'': _eff = &int#int;_0^&infin#infty; d&xi#xi;= ( × ) where 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!