1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – eepperly16 Jun 3 '15 at 20:58
  • $\begingroup$ 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. $\endgroup$ – Timaeus Jun 3 '15 at 21:56
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.