# Total angular momentum of electron in a magnetic field

In this question: Electron in the proximity of a magnetic monopole

It is stated that for an electron in the magnetic field of a monopole,

$\vec{B}(\vec{r}) = \frac{g}{r^3}\vec{r}$

, that the quantity

$\vec{J} = \vec{r} \times \vec{p} + eg\frac{\vec{r}}{r}$

is constant. It appears that $\vec{J}$ is a form of the total angular moment (is that correct?), which should indeed be conserved here since magnetic fields do no work, but I do not understand what the $eg\frac{\vec{r}}{r}$ term in $\vec{J}$ represents or where it comes from.

Can anyone elucidate where this contribution to the total angular momentum is coming from?

• I'm guessing the term $eg \vec{r}/r$ is a quantity referring the total angular momentum of the $\vec{B}$ field itself. Page 2-3 of this document has something similar physics.usu.edu/Wheeler/EM/Monopoles.pdf – eepperly16 Jun 3 '15 at 20:58
• I'm not sure why you bring up magnetic fields found no work, work is about energy and angular momentum us about momentum. I can't even figure what the $\vec p$ is in your equation, if it is just the mechanical momentum of the electron then you need to integrate $\vec r \times \vec P$ over all space where $\vec P$ is the field momentum density. And that's ignoring all the monopole issues where you need corrections to Maxwell and the force law hence also to field energy and field momentum expressions. – Timaeus Jun 3 '15 at 21:56

It's not quite clear how you jump from step 3 to step 4...

$\vec{r}\cdot \dot{\vec{r}} = r\dot{r}$

This identity does not always hold.

@Loonuh

After further examining the original question, and the source for the question, which was in the book "Electromagnetic Theory" by Ferraro on p.543, I was able to understand the conserved quantity $\vec{J}$ as thus.

Considering that with $\vec{B}(\vec{r}) = \frac{g\vec{r}}{r^3}$ the Lorentz force yields:

\begin{align} m\ddot{\vec{r}} &= q\dot{\vec{r}} \times \vec{B}\\ \vec{r}\times m\ddot{\vec{r}} &= q[\vec{r}\times(\dot{\vec{r}} \times \vec{B})]\\ \vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\vec{r}\cdot\vec{B}) - \vec{B}(\vec{r}\cdot \dot{\vec{r}})]\\ \vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\frac{g}{r}) - \vec{B}(r\dot{r})]\\ \vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\frac{g}{r}) - \vec{r}(\frac{\dot{r}g}{r^2})]\\ \vec{r}\times m\ddot{\vec{r}} &= qg\Big[\frac{\dot{\vec{r}}}{r} - \vec{r}\frac{\dot{r}}{r^2}\Big]\\ \frac{d}{dt}\Big(\vec{r}\times m\dot{\vec{r}}\Big) &= qg\frac{d}{dt}\Big(\frac{\vec{r}}{r}\Big)\\\\ \therefore \vec{r} \times \vec{p} = qg\frac{\vec{r}}{r} + \vec{J}\\ \end{align}

Where $\vec{J}$ is an arbitrary constant vector. Thus $\vec{J}$ is conserved in both magnitude and direction, so $J$ is constant.

\begin{align} \therefore \vec{J} =\vec{r} \times \vec{p} - eg\frac{\vec{r}}{r} \\ \end{align}

Thus it follows that

\begin{align} \vec{J} \cdot \vec{r} =(\vec{r} \times \vec{p} - eg\frac{\vec{r}}{r} ) \cdot \vec{r} &= (0) - egr\\ \therefore \vec{J} \cdot \vec{r} &= Jr\cos\theta = -egr\\ \therefore Jr\cos\theta &= -egr\\ \therefore \cos\theta &= -\frac{eg}{J}\\ \end{align}

Thus $\theta$ is constant, and $\dot{\theta}$ = 0