# Magnetization density

1. Aug 3, 2014

### newbee

Can a magnetization density that is compactly supported be the gradient of a scalar field?

This question relates to magnetic resonance imaging where the signal depends upon the curl of the magnetization density.

2. Aug 5, 2014

### Jano L.

Only inside the support, for example constant magnetization in a cylinder magnet is gradient of linear function of coordinates. On the surface, however, the magnetization jumps to zero which allows forming a closed contour $\partial \Sigma$ for which the circulation of magnetization does not vanish:

$$\oint_{\partial \Sigma} \mathbf M \cdot d\mathbf r \neq 0.$$

By Stokes theorem, this implies

$$\int_\Sigma \nabla\times \mathbf M \cdot d\boldsymbol \Sigma \neq 0,$$
so curl of $\mathbf M$ does not vanish everywhere and so magnetization is not gradient field everywhere. Since outside the magnet magnetization vanishes, all the nongradiency is happening on the surface of the magnet.

3. Aug 5, 2014

### newbee

Jano

Thank you for the reply. Your proof depends upon the circulation of an arbitrary magnetization (your first integral) always being non zero for at least one contour. I do not see why that would be the case. Am I missing something?

For a cylinder if the magnetization is spatially homogeneous and directed along the axis of a cylinder then wouldn't all circulation integrals on it surface be zero?

4. Aug 5, 2014

### newbee

OK. I think that it can be shown that if the contour includes a part that is outside the region of support and a part that is along the surface of the region of support then there will always be such a circulation that is nonzero.

5. Aug 6, 2014

### vanhees71

Magnetization density can be thought of as the description of the microscopic picture of elementary magnetic dipole moment of the electrons leading to permanent magnetization of a ferromagnet. This phenomenon can be fully understood from microscopic principles only using quantum theory, but the macroscopic description in phenomenological classical electrodynamics boils down to the idea that you have a continuous distribution of magnetic dipoles within the ferromagnet. This leads to the introduction of the magnetization denisty $\vec{M}$, which gives the magnetic dipole moment per unit time.

Let's restrict ourselves to the static (time-independent) case. For simplicity, I use Heaviside-Lorentz units. Then the vector potential of the magnetic field $\vec{B}$ is given by
$$\vec{A}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{\vec{M}(\vec{x}') \times (\vec{x}-\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|^3}.$$
Now we can bring this into a form for the vector potential of the magnetic field from a current distribution by noticing that
$$\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}=\vec{\nabla}' \frac{1}{|\vec{x}-\vec{x}'|}.$$
Plugging this in the above integral and integrating by parts, leads to the Biot-Savart Law like expression
$$\vec{A}(\vec{x})=\frac{1}{4 \pi c} \int_{\mathbb{R}^3} \frac{c \vec{\nabla}' \times \vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
This leads to the conclusion that a magnetization distribution is equivalent to a magnetization current
$$\vec{j}_M=c \vec{\nabla} \times \vec{M}.$$
Together with the usual electric current density the static Ampere Law thus reads
$$\vec{\nabla} \times \vec{B}=\frac{1}{c} (\vec{j}+\vec{j}_M).$$
Now we can lump the magnetization current to the left-hand side and introduce the auxilliary field
$$\vec{H}=\vec{B}-\vec{M},$$
so that the Ampere Law becomes
$$\vec{\nabla} \times \vec{H}=\frac{1}{c} \vec{j}.$$
For a homogeneous magnetization in a finitely extended body, the curl leads to $\delta$ functions, and the magnetization current density becomes effective a surface-current density.