Relation between magnetic moment and angular momentum -- classic theory How do I prove the relation between the vectors of magnetic moment $\vec\mu$ and angular momentum $\vec L$, $$\vec\mu=\gamma\vec L$$  ?
Many text books and lecture notes about the principles of magnetism show the relation of $\mu$ and $L$ as scalars only and then just state that the relation holds also for the vectors.
An example: http://folk.ntnu.no/ioverbo/TFY4250/til12eng.pdf
$$\mu=I\cdot A = \frac{q}{t}\pi r^2 = \frac{qv}{2\pi rm}m\pi r^2$$  ($I=q/t$: current, $A$: area of a loop, $q$ charge, $t=2\pi r/v$: time of 1 rotation, $v$: velocity of particle, $m$: mass )
The angular momentum is $\vec L = \vec r\times\vec p$ or $L=mrv$ and therefore
$$\Rightarrow \mu = {\frac{q}{2m}} L = \gamma L.$$
Why is this also true for the vectors? Is there a general explication by classical physics without the need of quantum theory?
 A: The direction of the magnetic moment is perpendicular to the plane of the loop. Seeing that the angular momentum is also perpendicular to that plane, and having shown that their magnitudes are proportional, is all it takes to show that two vectors are proportional.
If you insist, we can still formally go through all that. One way of enforcing an area vector perpendicular to the plane containing the area is to define it as follows:
$$\vec{A} = \vec{a}\times\vec{b}$$
for a parallelogram defined by the vectors $\vec{a}$ and $\vec{b}$.
We'll consider a planar loop carrying a current $I$. It's magnetic moment is:
$$\vec{\mu} = I\vec{A}$$
The angular momentum for a particle of mass $m$ is:
$$\vec{L} = m\vec{r}\times\frac{d\vec{r}}{dt}$$
$$\vec{L} = \frac{2m}{q}\frac{1}{2}\vec{r}\times q\frac{d\vec{r}}{dt}$$
Now, we need to observe a few things. $\frac{1}{2}\vec{r}\times d\vec{r}$ is the area of a sector of a circle, with radius given by $\vec{r}$ and a small circumference of $d\vec{r}$. Given that a charge $q$ crossed through $d\vec{r}$ in time $dt$, the current through a surface at that point is $I = \frac{q}{dt}$.
There's a little bit of a qualifier here. The current due to a point particle is non-uniform, and our expression for $\vec{\mu}$ should ideally be an integral around the loop (i.e. with $d\vec{r}$). But the current vanishes everywhere else, except where the particle is, anyway, and we can use the above shortcut to get the magnetic moment:
$$\vec{\mu} = I\vec{A} = \frac{1}{2}q\vec{r}\times\frac{d\vec{r}}{dt}$$
Our expression then becomes:
$$\vec{L} = \frac{2m}{q} \vec{\mu}$$
or
$$\vec{\mu} = \gamma\vec{L}$$
A: The easiest way to see the equality is to use a more general formula for the magnetic dipole moment of a particle.  For a flat planar loop of current, it's true that $\mu = IA$, with the direction of the dipole normal to the loop.  However, the more general case is that of a a volume current $\vec{J}$ in some finite region of space.  In this case, the the general formula for the magnetic dipole moment of the configuration is
$$
\vec{\mu} = \frac{1}{2} \int \vec{r} \times \vec{J} \,d^3r.
$$
(Showing that this reduces to the above formula for a flat planar loop is left as an exercise to the reader.)  If we further assume that the current density is due to a number of particles with number density $n$, charge $q$, velocity $\vec{v}$, and mass $m$, then we have current density $\vec{J} = nq\vec{v}$;  thus,
$$
\vec{\mu} = \frac{1}{2} \int \vec{r} \times (n q \vec{v}) \, d^3 r = \frac{q}{2 m} \int \vec{r} \times (n m \vec{v}) \, d^3 r.
$$
But $n m \vec{v} = \rho \vec{v}$, where $\rho$ is the mass density of the cloud;  thus, the above integral can be rewritten as 
$$
\vec{\mu} = \frac{q}{2 m} \int \rho \vec{r} \times \vec{v} \, d^3 r = \frac{q}{2m} \vec{L}.
$$
QED.  
The above relationship will hold so long as we can model the object as made out of particles with a definite charge-to-mass ratio, or (which is equivalent) so long as the ratio of charge density to mass density is constant throughout the body.  The easiest way for this to happen is, of course, for both densities to be constant.
