The Angular Momentum of an Electric and Magnetic Charge

  • #1
38
0
Relevant Equations:

Angular momentum density stored in an electromagnetic field: $$\vec{l}_{em} = \epsilon_0[\vec{r} \times (\vec{E} \times \vec{B})]$$

Electric field of an electric charge: $$\frac{q_e}{4\pi\epsilon_0}\frac{r - r'}{|r - r'|^3}$$

Magnetic field of a magnetic charge: $$\frac{\mu_0q_m}{4\pi}\frac{r - r'}{|r - r'|^3}$$

Question Statement:

Suppose we have an electric change and a hypothetical magnetic charge with the fields as given in the relevant equations. Now, let's calculate the total angular momentum stored in the electromagnetic field generated by an electric and magnetic charge separated by a distance d. Suppose we orient our coordinate system so that the electric charge is located at the origin (0, 0, 0) and the magnetic charge is located at (0, 0, d). Now, we are finding the angular momentum relative to the location of the electric charge since we have put it at the origin. If you do the calculation, you will end up with the following easily computable integral. $$\vec{L}_{em} = \frac{q_eq_m\mu_0d}{16\pi^2}\int_V\frac{rsin^3{\theta}drd{\theta}d\phi}{[r^2 -2rcos{\theta}d + d^2]^{3/2}}\hat{z} = \frac{q_eq_m\mu_0}{4\pi}\hat{z}$$
Now, suppose that we placed the electric charge at (0, 0, -d/2) and the magnetic charge at (0, 0, d/2), and we still compute the angular momentum relative to the location of the electric charge, so in the angular momentum density formula $$\vec{r} = x\hat{x} + y\hat{y} + (z + \frac{d}{2})\hat{z}$$

I won't post it here, but if you do the same calculation you will get an integral that is divergent. In both cases, we computed the angular momentum relative the location of the electric charge. The only difference is that in the first example, we placed the electric charge at the origin, and in the second example, we placed the electric charge at (0, 0, -d/2). We have changed nothing physical about the situation, but we get the widely agreed upon finite answer in the former example, and a divergent integral in the latter example.

Does angular momentum depend on the coordinate system, and if so does that mean it's not an inherent physical property of the system?
 

Answers and Replies

  • #2
764
71
Yes, in general angular momentum can depend on the location of the origin. This could be seen as well just by considering what happens to the angular momentum of a point mass when you change the origin.

Consider a point mass moving at the origin, then angular momentum would be zero, right? Because the r vector is zero we get ##\mathbf{L}= \mathbf{r} \times \mathbf{p} = \mathbf{0} \times \mathbf{p} = \mathbf{0}##.
However with a different origin, we'd still have the same ##\mathbf{p}## but a different ##\mathbf{r}##. So ##\mathbf{L}## could be non-zero with a different origin.

That is also the case for the orbital angular momentum of EM field. It has possible origin dependence from the ##\mathbf{r} \times ## EM momentum density.

In physics some properties of the system can depend on the origin or the reference frame velocity. A "conserved" quantity usually means conserved over time, but still with a fixed origin/reference frame. They aren't always fixed when the origin is changed. There may even be convergence issues for some cases.

As long as that isn't an issue, the differences in these properties are just different ways of viewing the same fundamental reality upon which all the coordinate systems will agree.

It turns out there is a part of the field's angular momentum that doesn't depend on the origin. We still can get overall origin dependence from the other part. But we call the part that isn't origin dependent the spin angular momentum of the EM field. The origin dependent part is called the orbital angular momentum.

Spin angular momentum = ##\mathbf{S}_{\text{EM}}## and orbital = ##\mathbf{L}_{\text{EM}} ##
Here are expressions for the angular momentum in terms of the electric field and the vector potential. ## \mathbf{L}## has ##\mathbf{r}## in the integrand whereas ## \mathbf{S}## doesn't.

$$\mathbf{S}_{\text{EM}} = \epsilon_0\int d^{3}\mathbf{r}\,\mathbf{E}\times\mathbf{A}$$
$$\mathbf{L}_{\text{EM}} = \epsilon_0\int d^{3}\mathbf{r}\,\sum_{i} \left({E^i}\left(\mathbf{r}\times\boldsymbol{\nabla}\right)A^i\right)$$
 
  • Informative
  • Like
Likes Dale and unified
  • #3
38
0
Excellent answer, however in your example of the point mass, you are finding the angular momentum relative to 1. Where the particle itself is by placing it at the origin. 2. Some other location. In my example, I found the angular momentum relative to 1. The location of the electric charge. 2. The location of the electric charge. I only changed the fact that in one case I put the electric charge at the origin. So, nothing physical changed, but the answer changed. Do you find that mysterious?
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,480
10,245
It's dangerous to somehow split the total angular momentum of the electromagnetic field, which of course is a well-defined quantity, into orbital and spin parts. As the last 2 equations in #1 show, it's a gauge dependent split, and you cannot interpret gauge-dependent quantities physically. They are for sure not observable.
 
  • #5
164
63
Excellent answer, however in your example of the point mass, you are finding the angular momentum relative to 1. Where the particle itself is by placing it at the origin. 2. Some other location. In my example, I found the angular momentum relative to 1. The location of the electric charge. 2. The location of the electric charge. I only changed the fact that in one case I put the electric charge at the origin. So, nothing physical changed, but the answer changed. Do you find that mysterious?

There must be a calculation error. The angular momentum relative to the electric charge cannot depend on the chosen coordinate system.

E × B has a singularity at the electric charge as well as at the magnetic charge. In your integral formula, there is a division by zero when r = d and the cosine is 1. But there is no division by zero when r = 0. Is the formula correct?
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,480
10,245
I come to the same result, using Mathematica. Of course one has to assume that the magnetic charge sits at ##\vec{x}_m=(0,0,d)## and the electric charge in the origin, and the angular momentum is calculated relative to the origin, where the electric charge sits.

Of course the angular momentum depends on this arbitrarily chosen reference point, but that's already familiar from Newtonian mechanics.

Concerning the singularities note that the only non-trivial integral is the ##z##-component, and the integrand reads
$$j_z=\frac{\mu_0 q q_m d}{16 \pi^2} \, \frac{\sin^3 \vartheta}{\sqrt{d^2+r^2-2d r \cos \vartheta}^3}.$$
As you see, the singularity at ##r=0## (from the electric charge) is cancelled by the Jacobian of the spherical coordinates. The Singularity at ##\vec{x}=(0,0,d)##, i.e., at ##r=d##, ##\vartheta=0## is cancelled by the ##\sin^3 \vartheta## factor in the numerator.
 
  • Like
Likes Heikki Tuuri
  • #7
38
0
The formula for the integral is correct and produces the correct angular momentum. Yet, using the other coordinate system, I'm left with an integral which is divergent according to Mathematica. I mean the system with the origin as the midpoint between the charges. You can try writing down the integrals yourself, shouldn't take but a few minutes.
 

Related Threads on The Angular Momentum of an Electric and Magnetic Charge

  • Last Post
2
Replies
48
Views
5K
Replies
5
Views
5K
Replies
9
Views
807
Replies
7
Views
2K
Replies
8
Views
3K
Replies
3
Views
2K
Replies
11
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
Top