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

[dgOnPhys]

Might that not be the crux of the matter?

I think indeed that in strict classical physics: $\hbox \nabla \cdot \vec M = 0$

To add magnetisation in classical electrodynamics, you need a additional empirical model.
Based on classical charges and their motions, you can't get any magnetisation.

All it takes to get classical magnetization is a distribution of classical magnetic dipoles. If you want to make a magnetic dipole through motion just spin a charged sphere on its axis. The only problem with this classical model is the wrong gyromagnetic ratio.

All it takes to have $\hbox \nabla \cdot \vec M \neq 0$ is an inhomogeneous material e.g. think of the interface between different materials

 

[lalbatros]

All it takes to get classical magnetization is a distribution of classical magnetic dipoles. If you want to make a magnetic dipole through motion just spin a charged sphere on its axis. The only problem with this classical model is the wrong gyromagnetic ratio. ...

The magnetic field associated to a (spinning) charged particle or to a "magnetic dipole" is divergence-free.
How do you explain then that the macroscopic sum of the microscopic magnetic fields could be non-divergence-free?

 

[DrDu]

lablatros, I do not exactly understand what you want to say. Do you claim that M is somehow a macroscopic sum of microscopic magnetic fields? That would be wrong.

 

[dgOnPhys]

The magnetic field associated to a (spinning) charged particle or to a "magnetic dipole" is divergence-free.
How do you explain then that the macroscopic sum of the microscopic magnetic fields could be non-divergence-free?

The thing with point-like charged particle and point-like dipoles (magnetic or not) is that they are described by a source distribution which is zero almost everywhere:
$\rho(\vec r)=q \delta(\vec r - \vec r_0)$
$\rho(\vec r)=- \vec p \cdot \nabla \delta(\vec r - \vec r_0)$

still when you are allowed to integrate on them (rather than summing) you can create any CONTINUOUS distribution of charge or dipole density.

If you just sum over a discrete set and measure the average resulting field, I am not sure things would be as peachy, perhaps this is what DrDu is saying in the post right before mine... if one tries to make a classical model out of a finite number of charges and dipoles not all average magnetization fields can be represented...

DrDu?

 

[lalbatros]

lablatros, I do not exactly understand what you want to say. Do you claim that M is somehow a macroscopic sum of microscopic magnetic fields? That would be wrong.

This is indeed my claim, and I don't think it wrong.
How else would you define the magnetisation?

By definition, B = H + M ,
where H is the magnetic field caused by "external currents"
end wher M is the magnetic field caused by "microscopic currents".

Actually, the decomposition between H and M is arbirary in principle.
It is the purpose of the "theory of magnetism" to separate "external currents" as cause and "microscopic currents" as effects and to analyze the physics behind it.

One could imagine other decompositions for other purposes.
Coupled coils could be an example.
The H field would be the field caused by the "controlled loop" and M the field from the other loop.
In this case, div H = 0 as well as div M = 0.

The question remains then:

when could we have div M <> 0 and what is the physical meaning ?

 

[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 $$M_t$$.