Gradient of Scalar Field: Compact Magnetization Density

[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.

 

[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.

 

[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?

 

[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.

 

[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.