Why div B = 0 but div H not necessary 0?

[DrDu]

H is not the field caused by external currents (excluding special cases) and M is not the field produced by microscopic currents.
In fact, M is nothing but an alternative to considering the internal (or "microscopic) currents, as $$\nabla \times M=-j_\text{int}$$. This equation does not fix M completely and hence one usually (but not always) tries to use an M so that the relation between M and B is approximately local as described by the magnetic permeability, i.e. $$B=\mu H$$.
In simple cases, M is the density of microscopic dipoles (and not of the fields they produce).

 

[lalbatros]

H is not the field caused by external currents (excluding special cases) and M is not the field produced by microscopic currents.
In fact, M is nothing but an alternative to considering the internal (or "microscopic) currents, as $$\rot M=-j_\text{int}$$. This equation does not fix M completely and hence one usually (but not always) tries to use an M so that the relation between M and B is approximately local as described by the magnetic permeability, i.e. $$B=\mu H$$.
In simple cases, M is the density of microscopic dipoles (and not of the fields they produce).

What is then the definition of H ?
I am used to think, naïvely, that H is defined by the "Ampère's circuital law":

Similarly, I am used to think of M as the field caused by the magnetisation current:

[URL]http://upload.wikimedia.org/math/d/c/2/dc24d89951a6d45b451723626c34ba38.png[/URL]

However:

- none of these definitions / assumptions imply that H or M are divergence-free
- and none of these definitions fully determine H or M from the knowledge of Jf or Jm​

After all, we should maybe first ask ourself "why is B divergence free ?".
The geometrical interpretation is clear.
But what is the physical meaning?

It might well appear that the reason why B is divergence free simply does not apply to H or M.

Any suggestion?

____________
http://en.wikipedia.org/wiki/Maxwell's_equations
http://en.wikipedia.org/wiki/Ampère's_circuital_law

 

[DrDu]

I agree on both definitions (alternatively, as B, M and H are related as B=H+M), maybe up to the sign in the definition of M. Please note that my formula wasn't displayed correctly (the rotation was missing) and I changed it. I suppose my internal current density $$j_\text{int}$$ is the same as your $$\mathbf{J}_m$$.

Both equations define only the transversal part of H or M. So you are right that they don't imply that H or M are divergence free.
You are also correct that they fully determine H or M in terms of the currents. They fix only the transversal part, the longitudinal part (which is not divergence free) is left open. In principle, we could even assume it to be 0. However this makes things ugly:

Suppose as an example theusual simple local relation between M and B (which does not hold in all materials) $$M(r)=(\mu(r)-1)/\mu(r) B(r)=\xi(r) B(r)$$. Using Helmholtz decomposition formula, we can calculate the transversal part of M:
$$M_t=\frac{1}{4\pi} \nabla \times \nabla \times \int dr' \frac{1}{|r-r'|}\xi(r')B(r')$$
In contrast to the nice local relation between B and M (or B and H) we are used to, this is a very nasty non-local relation.

That means that the divergence of M or H results from the demand for the relation between them to be as local as possible. The weakest form I can think of is the statement that asymptotically the dependence of M(r) on B(r') decays faster than $$(r-r')^{-3}$$, the highest power occurring in $$M_t$$.