Nothing controversial there. At AC, the fields drop to near zero just under the surface of the conductor; there is a nonzero amount in the surface, given by the boundary condition that the fields be continuous exactly at the surface (note, the direction and magnitude of E differs, effectively by refraction; D and B are equal, E and H are refracted). Therefore the Poynting vector slopes inward slightly -- which is to say, the wire is dissipating some power. Obvious enough, right?
As frequency drops, the skin depth grows; also note that dI/dt drops, so E is smaller, and so is the vector (magnitude). The active volume is larger, so losses may go up; on the other hand, the cross section is larger, so the overall resistance and therefore losses may go down. (Basically, skin effect is a 1/sqrt(f) phenomenon, so depending on which linear proportion dominates, it can get better or worse with frequency. Inductors for example tend to have a point of maximum Q, dropping off gently either side of that, as competing effects of skin effect and core loss tend to dominate. For a terminated transmission line of given length, losses generally go as sqrt(f), but loss per wavelength goes as 1/sqrt(f). So, depends how you count it.)
All the way down at DC, the electric field fully permeates the material (infinite skin depth), however because dI/dt = 0, the EMF is zero, so E is very small now -- only the resistive dropping component remains. Which means the Poynting vector is parallel to the wire, but also inside it -- pointing at itself, as it were. So yes, it's still dissipating power, in relation to its resistance anyways.
Now, for a superconductor, skin depth effectively persists all the way down to DC (which is another way of stating the Meissner effect). The depth is mediated not by frequency, but by quantum process -- on the order of the Debye length in the material (10s of nm) I think. Or maybe it's 100s, I forget, but anyway, it's very thin. So, whether the Poynting vector is pointing at the material or not, is arguable in this case; we might prefer another approach, since after all, why do you care about integrating minuscule shell layers around macroscopic conductors* when something much more numerically stable / analytically tractable would give the same result?
*To be fair, SC cables are made of extremely fine strands, for exactly this reason, which may then be embedded within a more convenient matrix (such as copper). (It's Litz for DC!) But it works out the same whether the strands are 1mm or 1um dia., embedded or free (after all, the matrix carries zero current at DC, when the SC has infinite times lower resistance than it!). The external field is just more spread out in the finely-stranded case.
Also, superconductors might not be perfectly ideal, like how type II exhibits flux pinning (basically AC loss manifest as hysteresis -- I don't know that this is a physically relevant explanation of it, but functionally at least, it's similar to magnetic hysteresis, with internal current flows / flux paths being analogous to shifts of magnetic domains).
Also also, back to the RF model of things -- it's still perfectly consistent to represent the system as a superposition of waves propagating in both directions. If we have waves building up, then even if the dynamics decay to steady state, we can still model that as the source transmitting however many multiples of itself, and the load reflecting back one less than that, or whatever. It may not be very
practical to model an open-circuit battery as constantly driving full power into an open-circuit transmission line and having it immediately reflected back in-phase nearly 100%; but it's still
consistent to do so.
Tim
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