In magnetism, what is the difference between the B and H fields?

And besides, trying to discuss vector H on the quantum level doesn't even make any sense. H is defined as:

H = B/μ° - M

And M is by definition a macroscopic value. It is the magnetic dipole moment per unit volume. Undefined on the quantum level, and hence so too is H.

...and so they are the usual assumptions for a good reason-they are the correct assumptions for the subject matter under discussion.

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The boundary condition which states that the discontinuity in the tangential component of the H at the border between two mediums is equal to the free surface current density J follows directly from the relation

∇ x H = Jfree + ∂D/∂t

...which is model independent

DrDu
And M is by definition a macroscopic value. It is the magnetic dipole moment per unit volume. Undefined on the quantum level, and hence so too is H.

That's not correct. M and P can be defined microscopically. Only in the simplest cases, it is the dipole moment density.
However, the definition is model dependent, namely, it depends on whether one splits the electric charges into internal and external ones or into bound and free + external. Any of these groups fulfills a continuity equation
##\nabla \cdot j =\partial \rho /\partial t ##.
Introducing the four vector ##j^\mu## with ##j^0=\rho## and ##j^\mu=j_i## for ##\mu \in \{1,2,3\}## and ##i \in \{x,y,z\}##, and ##\partial_\mu## as ##\partial_0=\partial_t##, ##\partial_\mu =-\partial_i##,
we can write this as
##\partial_\mu j^\mu=0##.
This equation will be fulfilled iff ##j^\mu = \partial_\nu \Pi^{\mu\nu}## where ##\Pi## is an antisymmetric tensor, i.e. ##\Pi^{\mu\nu}=-\Pi^{\nu\mu}##.
We call it the polarisation tensor.
It may be parametrized as
##\Pi = \begin{pmatrix}
0 &P_x &P_y &P_z\\
-P_x& 0 &M_z& -M_y\\
-P_y &-M_z &0 & M_x\\
-P_z & M_y & -M_x& 0
\end{pmatrix} .
##
This tensor is not unique, as the solution of ##j^\mu = \partial_\nu \Pi^{\mu\nu}## is defined only up to a solution of the homogeneous problem ##\partial_\nu \Pi^{\mu\nu}=0 ##. I suppose this is what Hendrik meant with the possibility to shuffle terms.

DrDu
...which is model independent
As I just laid out, you can shuffle terms between M and P (at least in the time dependent case) and also the choice of J is a matter of convention. E.g. instead of J_Free you may use J_external.
In QM, a unique distinction between bound and free charges is problematic.

vanhees71
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...which is model independent
It is not! What you define as ##\rho_{\text{free}}## and ##\vec{j}_{\text{free}}## and what as polarizations ##\vec{P}## and ##\vec{M}## is more or less arbitrary. You can easily shuffle contributions from the one to the other without changing the physical relevant fields ##\vec{E}## and ##\vec{B}##. Often, of course, there's a "natural choice", but it's still model dependent.

As I just laid out, you can shuffle terms between M and P (at least in the time dependent case) and also the choice of J is a matter of convention. E.g. instead of J_Free you may use J_external.
In QM, a unique distinction between bound and free charges is problematic.
How so?

Sure, here you made the usual assumptions, treating the conductor as a continuum and map everything to boundary conditions. From a microscopic point of view things are much different, and the OP asked even on the level of in-medium quantum electrodynamics, a topic, I'd recommend only after studying the classical theory and also the vacuum-QED case in detail first.

For an excellent treatment of classical in-medium electrodynamics, see Landau&Lifshitz vol. VIII.

When the cylinder is embedded, a secondary layer of bound densities is formed around it which oppose those of the P or M whose influence would otherwise be prevalent. That is why the method will measure H or D rather than E or B

Too see what I mean you can try a thought experiment using my cylinder method between the plates of a capacitor with a dielectic medium. Density on the cylinder should always be the density as the cap's plates even though E changes when permitivity changes

DrDu
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