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

1. Nov 2, 2017

### mairzydoats

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

2. Nov 2, 2017

### mairzydoats

...which is model independent

3. Nov 3, 2017

### DrDu

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.

4. Nov 3, 2017

### DrDu

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.

5. Nov 3, 2017

### vanhees71

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.

6. Nov 3, 2017

How so?

7. Nov 3, 2017

### mairzydoats

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

8. Nov 3, 2017

### DrDu

A classical example is the Lindhard dielectric function of the free electron gas, where a gas of electrons is described by a dielectric function, although one would be tempted to treat it as free charges.
https://en.wikipedia.org/wiki/Lindhard_theory

9. Nov 3, 2017

Okay