Tag Archives: angular momentum

Angular momentum – Poisson bracket to commutator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.8.

The classical angular momentum components are

\displaystyle   \ell_{x} \displaystyle  = \displaystyle  yp_{z}-zp_{y}\ \ \ \ \ (1)
\displaystyle  \ell_{y} \displaystyle  = \displaystyle  zp_{x}-xp_{z}\ \ \ \ \ (2)
\displaystyle  \ell_{z} \displaystyle  = \displaystyle  xp_{y}-yp_{x} \ \ \ \ \ (3)

In the position basis, we can replace each coordinate by its quantum operator {x\rightarrow X}, {y\rightarrow Y} and {z\rightarrow Z}, and each momentum component by the derivative {p_{i}\rightarrow-i\hbar\partial/\partial q_{i}}, where {q_{i}} is the {i}th coordinate. This gives

\displaystyle   L_{x} \displaystyle  = \displaystyle  -i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\ \ \ \ \ (4)
\displaystyle  L_{y} \displaystyle  = \displaystyle  -i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\ \ \ \ \ (5)
\displaystyle  L_{z} \displaystyle  = \displaystyle  -i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) \ \ \ \ \ (6)

Because coordinates always commute with momentum components of other coordinates ({x} commutes with {p_{y}} and {p_{z}}, etc), there is no ordering ambiguity in making the transition from classical to quantum mechanics. That is, we could place the coordinate on either side of the momentum in each term for all components {L_{i}}.

Classically, we can calculate the Poisson brackets for the angular momentum components. For example

\displaystyle   \left\{ \ell_{x},\ell_{y}\right\} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\ell_{x}}{\partial q_{i}}\frac{\partial\ell_{y}}{\partial p_{i}}-\frac{\partial\ell_{x}}{\partial p_{i}}\frac{\partial\ell_{y}}{\partial q_{i}}\right)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -p_{y}\left(-x\right)-yp_{z}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  xp_{y}-yp_{z}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \ell_{z} \ \ \ \ \ (10)

According to the rule for converting classical Poisson brackets to quantum commutators, we should get (since there is no ordering ambiguity)

\displaystyle  \left[L_{x},L_{y}\right]=i\hbar L_{z} \ \ \ \ \ (11)

As we’ve seen earlier, this is verified by direct calculation using the position-momentum commutator

\displaystyle  \left[q_{i},p_{j}\right]=i\hbar\delta_{ij} \ \ \ \ \ (12)

Angular momentum as an eigenvector problem

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Exercise 1.8.9.

The angular momentum in classical mechanics of a collection of point masses {m_{a}} located at positions {\mathbf{r}_{a}} and moving with a common angular velocity {\boldsymbol{\omega}} about a common axis is given by

\displaystyle  \mathbf{L}=\sum_{a}m_{a}\left(\mathbf{r}_{a}\times\mathbf{v}_{a}\right) \ \ \ \ \ (1)

where {\mathbf{v}_{a}=\boldsymbol{\omega}\times\mathbf{r}_{a}} is the linear velocity of {m_{a}}. We can use the vector identity

\displaystyle  \mathbf{A}\times\left(\mathbf{B}\times\mathbf{C}\right)=\mathbf{B}\left(\mathbf{A}\cdot\mathbf{C}\right)-\mathbf{C}\left(\mathbf{A}\cdot\mathbf{B}\right)

to write

\displaystyle   \mathbf{r}_{a}\times\mathbf{v}_{a} \displaystyle  = \displaystyle  \mathbf{r}_{a}\times\left(\boldsymbol{\omega}\times\mathbf{r}_{a}\right)\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  r_{a}^{2}\boldsymbol{\omega}-\mathbf{r}_{a}\left(\mathbf{r}_{a}\cdot\boldsymbol{\omega}\right) \ \ \ \ \ (3)

In terms of components, this is

\displaystyle   \left[\mathbf{r}_{a}\times\mathbf{v}_{a}\right]_{i} \displaystyle  = \displaystyle  r_{a}^{2}\omega_{i}-\left(r_{a}\right)_{i}\sum_{j}\left(r_{a}\right)_{j}\omega_{j}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \sum_{j}\left[r_{a}^{2}\omega_{j}\delta_{ij}-\left(r_{a}\right)_{i}\left(r_{a}\right)_{j}\omega_{j}\right]\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \sum_{j}\left[r_{a}^{2}\delta_{ij}-\left(r_{a}\right)_{i}\left(r_{a}\right)_{j}\right]\omega_{j} \ \ \ \ \ (6)

We can therefore write the angular momentum as

\displaystyle   \mathbf{L} \displaystyle  = \displaystyle  \sum_{j}\sum_{a}m_{a}\left[r_{a}^{2}\delta_{ij}-\left(r_{a}\right)_{i}\left(r_{a}\right)_{j}\right]\omega_{j}\ \ \ \ \ (7)
\displaystyle  \displaystyle  \equiv \displaystyle  \sum_{j}M_{ij}\omega_{j} \ \ \ \ \ (8)

where the matrix {M} is

\displaystyle  M_{ij}\equiv\sum_{a}m_{a}\left[r_{a}^{2}\delta_{ij}-\left(r_{a}\right)_{i}\left(r_{a}\right)_{j}\right] \ \ \ \ \ (9)

From the definition, we see that {M} is real and symmetric (interchanging {i} and {j} shows that {M_{ij}=M_{ji}}), so {M} is hermitian.

In Dirac’s notation, we have the matrix equation

\displaystyle  \left|L\right\rangle =M\left|\omega\right\rangle \ \ \ \ \ (10)

From this equation, we can see that {\mathbf{L}} and {\boldsymbol{\omega}} are parallel only if {\boldsymbol{\omega}} is an eigenvector of {M}. If the eigenvalues of {M} are non-degenerate, there are therefore three directions for {\boldsymbol{\omega}} such that {\mathbf{L}} and {\boldsymbol{\omega}} are parallel, and these directions can be found by solving for the eigenvectors of {M}.

If some of the eigenvalues are degenerate, then there is a range of directions over which {\mathbf{L}} and {\boldsymbol{\omega}} can be parallel. In the case of a sphere, all 3 eigenvalues of {M} must be the same, as all directions are axes of symmetry of the sphere.

As an example, suppose we have only one mass {m=1} with position

\displaystyle  \mathbf{r}=\left[1,1,0\right] \ \ \ \ \ (11)

We can work out {M} by substituting into 9:

\displaystyle  M=\left[\begin{array}{ccc} 1 & -1 & 0\\ -1 & 1 & 0\\ 0 & 0 & 2 \end{array}\right] \ \ \ \ \ (12)

The eigenvalues are 0, 2 and 2 with corresponding eigenvectors

\displaystyle   \left|\lambda=0\right\rangle \displaystyle  = \displaystyle  \left[\begin{array}{c} 1\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (13)
\displaystyle  \left|\lambda=2\right\rangle \displaystyle  = \displaystyle  \left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right],\left[\begin{array}{c} -1\\ 1\\ 0 \end{array}\right] \ \ \ \ \ (14)

Thus if {\boldsymbol{\omega}} is a linear combination of the two eigenvectors for {\lambda=2}, it will be parallel to {\mathbf{L}}. If {\boldsymbol{\omega}} is parallel to {\left[\begin{array}{c} 1\\ 1\\ 0 \end{array}\right]}, {\mathbf{L}=0}, as in this case {\boldsymbol{\omega}} is parallel to {\mathbf{r}} so {\boldsymbol{\omega}\times\mathbf{r}=0}, and the mass is located on the axis of rotation so has no angular momentum.

Noether’s theorem and conservation of angular momentum

References: W. Greiner & J. Reinhardt, Field Quantization, Springer-Verlag (1996), Chapter 2, Section 2.4.

Now that we’ve seen that a general Lorentz transformation can be represented as a product of a pure boost and a pure 3-d rotation, we can return to Noether’s theorem and see what conserved property it predicts when we require a physical system to be invariant under a Lorentz transformation. As usual, we consider an infinitesimal transformation, which we can write as

\displaystyle x^{\prime\mu}=x^{\mu}+\delta\omega^{\mu\nu}x_{\nu} \ \ \ \ \ (1)

 

where {\delta\omega^{\mu\nu}} is the infinitesimal rotation in 4-dimensional spacetime. Here, we are treating a pure boost as a rotation; for example, a boost in the {x_{1}} direction is given by the Lorentz transformation

\displaystyle x^{\prime}=\Lambda x \ \ \ \ \ (2)

 


where

\displaystyle \Lambda=\left[\begin{array}{cccc} \cosh\chi & \sinh\chi & 0 & 0\\ \sinh\chi & \cosh\chi & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (3)

 

for some ‘angle’ {\chi}. This reduces to the more familiar form found in introductory relativity courses if we set

\displaystyle \cosh\chi \displaystyle \equiv \displaystyle \gamma=\frac{1}{\sqrt{1-\beta^{2}}}\ \ \ \ \ (4)
\displaystyle \sinh\chi \displaystyle \equiv \displaystyle \beta\gamma=\frac{\beta}{\sqrt{1-\beta^{2}}} \ \ \ \ \ (5)

Returning to 1, we require, to first order in {\delta\omega_{\mu\nu}}, that the Minkowski length of the 4-vector is the same before and after the transformation. That is,

\displaystyle x^{\prime\mu}x_{\mu}^{\prime} \displaystyle = \displaystyle \left(x^{\mu}+\delta\omega^{\mu\sigma}x_{\sigma}\right)\left(x_{\mu}+\delta\omega_{\mu}^{\;\tau}x_{\tau}\right)\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle x^{\mu}x_{\mu}+\delta\omega^{\mu\sigma}x_{\sigma}x_{\mu}+\delta\omega_{\mu}^{\;\tau}x_{\tau}x^{\mu}\ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle x^{\mu}x_{\mu}+\delta\omega^{\mu\sigma}x_{\sigma}x_{\mu}+\delta\omega^{\mu\tau}x_{\tau}x_{\mu}\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle x^{\mu}x_{\mu}+2\delta\omega^{\mu\nu}x_{\mu}x_{\nu} \ \ \ \ \ (9)

In the last line, we renamed the dummy indices {\sigma} and {\tau} to {\nu}. To first order, we require the last term in the last line to be zero for all {x_{\mu}}, which means we must impose a condition on {\delta\omega^{\mu\nu}}. We can write this term as

\displaystyle 2\delta\omega^{\mu\nu}x_{\mu}x_{\nu}=x_{\mu}x_{\nu}\left(\delta\omega^{\mu\nu}+\delta\omega^{\nu\mu}\right) \ \ \ \ \ (10)

From this, we see that we must have

\displaystyle \delta\omega^{\mu\nu}=-\delta\omega^{\nu\mu} \ \ \ \ \ (11)

so {\delta\omega^{\mu\nu}} must be antisymmetric.

Incidentally, if this condition seems to be violated in the pure boost matrix 3, remember that 2 is an ordinary matrix product, while the last term in 1 is the product of a tensor and 4-vector, and thus includes the effect of the metric tensor

\displaystyle g=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (12)

 

To first order in {\chi}, 3 is

\displaystyle \Lambda=\left[\begin{array}{cccc} 1 & \chi & 0 & 0\\ \chi & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (13)

while for the infinitesimal rotation, we have

\displaystyle \delta\omega=\left[\begin{array}{cccc} 0 & -\chi & 0 & 0\\ \chi & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (14)

In matrix notation, 1 becomes

\displaystyle x^{\prime}=\delta\omega\times g\times x \ \ \ \ \ (15)

from which we can see that this gives the same result as 2.

In order to apply Noether’s theorem, we also need to know how the fields transform under a Lorentz transformation. The assumption is that, for infinitesimal transformations, the transformed field {\phi_{r}^{\prime}\left(x^{\prime}\right)} depends linearly on both the original fields {\phi_{r}\left(x\right)} and on the rotation {\delta\omega_{\mu\nu}}. That is, we assume that

\displaystyle \phi_{r}^{\prime}\left(x^{\prime}\right)=\phi_{r}\left(x\right)+\frac{1}{2}\delta\omega_{\mu\nu}\left(I^{\mu\nu}\right)_{rs}\phi_{s}\left(x\right) \ \ \ \ \ (16)

 

where {I^{\mu\nu}} are the infinitesimal generators of the Lorentz transformation. G & R don’t really explain this, apart from giving a reference to another book, but we won’t need to delve into the details to get the result needed in this post, so we’ll leave it for now.

From here, it’s a matter of plugging 1 and 16 into the equations for Noether’s theorem and seeing what comes out. Noether’s theorem says that

\displaystyle \partial^{\mu}f_{\mu}\left(x\right)=0 \ \ \ \ \ (17)

 

where

\displaystyle f_{\mu}\left(x\right) \displaystyle \equiv \displaystyle \frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{\mu}\phi_{r}\right)}\left(\delta\phi_{r}\left(x\right)-\partial^{\nu}\phi_{r}\delta x_{\nu}\right)+\mathcal{L}\left(x\right)\delta x_{\mu}\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle \frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{\mu}\phi_{r}\right)}\delta\phi_{r}\left(x\right)-\left(\frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{\mu}\phi_{r}\right)}\partial_{\nu}\phi_{r}-g_{\mu\nu}\mathcal{L}\left(x\right)\right)\delta x^{\nu} \ \ \ \ \ (19)

From 16

\displaystyle \delta\phi_{r}\left(x\right) \displaystyle = \displaystyle \phi_{r}^{\prime}\left(x^{\prime}\right)-\phi_{r}\left(x\right)\ \ \ \ \ (20)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\delta\omega_{\mu\nu}\left(I^{\mu\nu}\right)_{rs}\phi_{s}\left(x\right)\ \ \ \ \ (21)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\delta\omega_{\nu\lambda}\left(I^{\nu\lambda}\right)_{rs}\phi_{s}\left(x\right) \ \ \ \ \ (22)

In the last line, we renamed the indices {\mu} and {\nu} to {\nu} and {\lambda} respectively to avoid confusing the {\mu} in the first term of 19 with any of the indices in {\delta\phi_{r}\left(x\right)}.

and from 1

\displaystyle \delta x^{\nu} \displaystyle = \displaystyle x^{\prime\nu}-x^{\nu}\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle \delta\omega^{\nu\lambda}x_{\lambda} \ \ \ \ \ (24)

We also had the energy-momentum tensor

\displaystyle T_{\mu\nu}\equiv\frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{\mu}\phi_{r}\right)}\partial_{\nu}\phi_{r}-g_{\mu\nu}\mathcal{L}\left(x\right) \ \ \ \ \ (25)

 

Putting all this together, we have

\displaystyle f_{\mu}\left(x\right)=\frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{\mu}\phi_{r}\right)}\frac{1}{2}\delta\omega_{\nu\lambda}\left(I^{\nu\lambda}\right)_{rs}\phi_{s}\left(x\right)-T_{\mu\nu}\delta\omega^{\nu\lambda}x_{\lambda} \ \ \ \ \ (26)

Because {\delta\omega^{\nu\lambda}=-\delta\omega^{\lambda\nu}}

\displaystyle T_{\mu\nu}\delta\omega^{\nu\lambda}x_{\lambda} \displaystyle = \displaystyle \frac{1}{2}\left[T_{\mu\nu}\delta\omega^{\nu\lambda}x_{\lambda}-T_{\mu\nu}\delta\omega^{\lambda\nu}x_{\lambda}\right]\ \ \ \ \ (27)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\left[T_{\mu\nu}\delta\omega^{\nu\lambda}x_{\lambda}-T_{\mu\lambda}\delta\omega^{\nu\lambda}x_{\nu}\right]\ \ \ \ \ (28)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\delta\omega^{\nu\lambda}\left[T_{\mu\nu}x_{\lambda}-T_{\mu\lambda}x_{\nu}\right] \ \ \ \ \ (29)

In the second line, we swapped the dummy indices {\lambda} and {\nu} in the second term, which is allowed because both indices are summed. Therefore

\displaystyle f_{\mu}\left(x\right) \displaystyle = \displaystyle \frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{\mu}\phi_{r}\right)}\frac{1}{2}\delta\omega_{\nu\lambda}\left(I^{\nu\lambda}\right)_{rs}\phi_{s}\left(x\right)-\frac{1}{2}\delta\omega^{\nu\lambda}\left[T_{\mu\nu}x_{\lambda}-T_{\mu\lambda}x_{\nu}\right]\ \ \ \ \ (30)
\displaystyle \displaystyle = \displaystyle \frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{\mu}\phi_{r}\right)}\frac{1}{2}\delta\omega^{\nu\lambda}\left(I_{\nu\lambda}\right)_{rs}\phi_{s}\left(x\right)-\frac{1}{2}\delta\omega^{\nu\lambda}\left[T_{\mu\nu}x_{\lambda}-T_{\mu\lambda}x_{\nu}\right]\ \ \ \ \ (31)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\delta\omega^{\nu\lambda}\left[\frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{\mu}\phi_{r}\right)}\left(I_{\nu\lambda}\right)_{rs}\phi_{s}\left(x\right)-\left(T_{\mu\nu}x_{\lambda}-T_{\mu\lambda}x_{\nu}\right)\right]\ \ \ \ \ (32)
\displaystyle \displaystyle \equiv \displaystyle \frac{1}{2}\delta\omega^{\nu\lambda}M_{\mu\nu\lambda}\left(x\right) \ \ \ \ \ (33)

In the second line, we swapped the positions of the {\nu\lambda} indices on the terms {\delta\omega_{\nu\lambda}\left(I^{\nu\lambda}\right)_{rs}}. This is OK provided they are both summed over. The last line defines the term {M_{\mu\nu\lambda}\left(x\right)}.

Because the infinitesimal rotations are arbitrary (subject to the condition that {\delta\omega^{\nu\lambda}} is antisymmetric), we can choose all of them to be zero except for one. For each such choice, we have a different {f_{\mu}\left(x\right)}, which leads to a conservation law for each choice. From Noether’s theorem, the quantity that is conserved is the integral of {f_{0}} over 3-space, so we have the conserved quantities

\displaystyle M_{\nu\lambda} \displaystyle = \displaystyle \int d^{3}x\left[\frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{0}\phi_{r}\right)}\left(I_{\nu\lambda}\right)_{rs}\phi_{s}\left(x\right)-\left(T_{0\nu}x_{\lambda}-T_{0\lambda}x_{\nu}\right)\right]\ \ \ \ \ (34)
\displaystyle \displaystyle = \displaystyle \int d^{3}x\left[T_{0\lambda}x_{\nu}-T_{0\nu}x_{\lambda}+\frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{0}\phi_{r}\right)}\left(I_{\nu\lambda}\right)_{rs}\phi_{s}\left(x\right)\right] \ \ \ \ \ (35)

From 25, we see that the first two terms are

\displaystyle T_{0\lambda}x_{\nu}-T_{0\nu}x_{\lambda} \displaystyle = \displaystyle x_{\nu}\frac{\partial\mathcal{L}}{\partial\dot{\phi}_{r}}\partial_{\lambda}\phi_{r}-x_{\lambda}\frac{\partial\mathcal{L}}{\partial\dot{\phi}_{r}}\partial_{\nu}\phi_{r}\ \ \ \ \ (36)
\displaystyle \displaystyle = \displaystyle x_{\nu}\pi_{r}\partial_{\lambda}\phi_{r}-x_{\lambda}\pi_{r}\partial_{\nu}\phi_{r} \ \ \ \ \ (37)

where {\pi_{r}} is the conjugate momentum density, defined by

\displaystyle \pi_{r}\equiv\frac{\partial\mathcal{L}}{\partial\dot{\phi}_{r}} \ \ \ \ \ (38)

Using the physical momentum density

\displaystyle p_{\lambda}=\pi_{r}\partial_{\lambda}\phi_{r} \ \ \ \ \ (39)

we find that

\displaystyle T_{0\lambda}x_{\nu}-T_{0\nu}x_{\lambda}=x_{\nu}p_{\lambda}-x_{\lambda}p_{\nu} \ \ \ \ \ (40)

This is one component of the angular momentum density {\mathbf{r}\times\mathbf{p}}, so the integral

\displaystyle \int d^{3}x\left(T_{0\lambda}x_{\nu}-T_{0\nu}x_{\lambda}\right) \ \ \ \ \ (41)

is one component of the total angular momentum.

The other term in 35 is

\displaystyle \int d^{3}x\;\frac{\partial\mathcal{L}\left(x\right)}{\partial\left(\partial^{0}\phi_{r}\right)}\left(I_{\nu\lambda}\right)_{rs}\phi_{s}\left(x\right) \ \ \ \ \ (42)

depends on the generators {I_{\nu\lambda}}, and thus on the specific way in which the fields transform. G & R tell us that this term describes the spin angular momentum, but at this stage, we just have to accept this on faith.

In any case, the overall conservation rule 35 shows that the sum of the ‘traditional’ angular momentum from the first two terms in the integrand, together with this mysterious other term, is a conserved quantity, so interpreting it as some other form of angular momentum seems reasonable. We’ll just have to wait and see how this plays out.

Angular momentum conservation: example with a solenoid

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 8.13.

We’ll revisit the earlier problem of the two charged cylinders and the solenoid. To reiterate, we have a long solenoid with {n} turns per unit length carrying current {I_{0}} and a radius {R}, with its axis along the {z} axis. The magnetic field inside the solenoid is

\displaystyle  \mathbf{B}_{0}=\mu_{0}nI\hat{\mathbf{z}} \ \ \ \ \ (1)

The field is zero outside the solenoid.

We add two other cylinders (not solenoids), both coaxial with the solenoid. One cylinder has radius {a<R} (so it lies inside the solenoid) and carries surface charge {+Q}; the other cylinder has radius {b>R} (outside the solenoid) and carries charge {-Q}. Both cylinders have length {\ell}. From Gauss’s law, the electric field between these two cylinders is, for {a<r<b}

\displaystyle  \mathbf{E}_{0}=\frac{Q}{2\pi\epsilon_{0}\ell}\frac{\hat{\mathbf{r}}}{r} \ \ \ \ \ (2)

That is, the field points radially outward from the axis. The electric field is zero for {r<a} and {r>b}. (We’re neglecting end effects, so we’re assuming that {\ell\gg b>a}.)

In our earlier solution, we worked out the angular momentum contained in the fields and showed that it is equal to the mechanical angular momentum transferred to the two cylinders if the current in the solenoid is slowly reduced. However, there is another effect that we neglected: when the charged cylinders start to rotate, they generate a changing magnetic field inside them which in turn creates a circumferential electric field in the space between the cylinders. When the final rotation speeds of the two cylinders are reached (that is, when the current through the solenoid has been reduced to zero), the cylinders continue rotating, thus generating a static magnetic field that interacts with the electric field to produce an extra amount of angular momentum in the fields. We’ll consider this static magnetic field first.

The rotating cylinders are effectively solenoids themselves. The surface charge density on the two cylinders is

\displaystyle  \sigma_{a,b}=\begin{cases} \frac{Q}{2\pi a\ell} & \mbox{inner cylinder}\\ -\frac{Q}{2\pi b\ell} & \mbox{outer cylinder} \end{cases} \ \ \ \ \ (3)

The surface current densities are

\displaystyle  K_{a,b}=\begin{cases} \frac{Q\left(a\omega_{a}\right)}{2\pi a\ell} & \mbox{inner cylinder}\\ -\frac{Q\left(b\omega_{b}\right)}{2\pi b\ell} & \mbox{outer cylinder} \end{cases} \ \ \ \ \ (4)

so the magnetic field due to each cylinder is

\displaystyle  \mathbf{B}_{a,b}=\begin{cases} \frac{\mu_{0}Q\omega_{a}}{2\pi\ell}\hat{\mathbf{z}} & \mbox{inner cylinder}\\ \frac{\mu_{0}Q\omega_{b}}{2\pi\ell}\hat{\mathbf{z}} & \mbox{outer cylinder} \end{cases} \ \ \ \ \ (5)

To get the directions of {\mathbf{B}} we see from Griffiths’s Example 8.4 that the angular momentum of the inner cylinder is in the {+z} direction and of the outer cylinder is in the {-z} direction. Since the outer cylinder has negative charge, however, its magnetic field points in the same direction as that of the inner cylinder. Another way of putting this is to note that the angular velocities are

\displaystyle   \boldsymbol{\omega}_{a} \displaystyle  = \displaystyle  \omega_{a}\hat{\mathbf{z}}\ \ \ \ \ (6)
\displaystyle  \boldsymbol{\omega}_{b} \displaystyle  = \displaystyle  -\omega_{b}\hat{\mathbf{z}} \ \ \ \ \ (7)

so the minus sign for {\boldsymbol{\omega}_{b}} cancels the minus sign for the charge on the outer cylinder.

The linear momentum density is non-zero only in the region {a\le r\le b} and is

\displaystyle   \boldsymbol{\mathfrak{p}}_{em} \displaystyle  = \displaystyle  \epsilon_{0}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\mu_{0}Q^{2}\omega_{b}}{4\pi^{2}\ell^{2}r}\hat{\boldsymbol{\phi}} \ \ \ \ \ (9)

The angular momentum density is

\displaystyle   \boldsymbol{\mathfrak{L}}_{em} \displaystyle  = \displaystyle  \mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\mu_{0}Q^{2}\omega_{b}}{4\pi^{2}\ell^{2}}\hat{\mathbf{z}} \ \ \ \ \ (11)

which is constant, so the total angular momentum is

\displaystyle   \mathbf{L} \displaystyle  = \displaystyle  -\frac{\mu_{0}Q^{2}\omega_{b}}{4\pi^{2}\ell^{2}}\left[\pi\left(b^{2}-a^{2}\right)\ell\right]\hat{\mathbf{z}}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\mu_{0}Q^{2}\omega_{b}}{4\pi\ell}\left(b^{2}-a^{2}\right)\hat{\mathbf{z}} \ \ \ \ \ (13)

Now we can look at what’s happening as the cylinders are spinning up to their final speeds. As they speed up, the magnetic field due to each cylinder is changing according to

\displaystyle  \frac{\partial\mathbf{B}_{a,b}}{\partial t}=\begin{cases} \frac{\mu_{0}Q\dot{\omega}_{a}}{2\pi\ell}\hat{\mathbf{z}} & \mbox{inner cylinder}\\ \frac{\mu_{0}Q\dot{\omega}_{b}}{2\pi\ell}\hat{\mathbf{z}} & \mbox{outer cylinder} \end{cases} \ \ \ \ \ (14)

where the dot indicates a time derivative. According to Faraday’s law, this changing magnetic field induces a circumferential electric field:

\displaystyle  \oint\mathbf{E}\cdot d\boldsymbol{\ell}=-\int\frac{\partial\mathbf{B}_{a,b}}{\partial t}\cdot d\mathbf{a} \ \ \ \ \ (15)

This electric field will exert a torque on each cylinder, whose integral over time will give the angular momentum transferred to the cylinders. First we need to calculate the field at each cylinder. We choose a circular path of integration at the surface of each cylinder. Remember that the field of the inner cylinder is non-zero only for {r<a}, which the the field for the outer cylinder covers the entire region {r<b}.

\displaystyle   \mathbf{E}_{a} \displaystyle  = \displaystyle  -\frac{1}{2\pi a}\pi a^{2}\frac{\mu_{0}Q\left(\dot{\omega}_{a}+\dot{\omega}_{b}\right)}{2\pi\ell}\hat{\boldsymbol{\phi}}\ \ \ \ \ (16)
\displaystyle  \mathbf{E}_{b} \displaystyle  = \displaystyle  -\frac{1}{2\pi b}\frac{\mu_{0}Q\left(\pi a^{2}\dot{\omega}_{a}+\pi b^{2}\dot{\omega}_{b}\right)}{2\pi\ell}\hat{\boldsymbol{\phi}} \ \ \ \ \ (17)

To confirm the direction of {\mathbf{E}}, recall Lenz’s law, which states that the induced field opposes the change that produced it. Since the magnetic field is increasing in the {+z} direction, the induced electric field must oppose this increase so it must be in the {-\phi} direction.

The torque on the cylinders is then

\displaystyle   \mathbf{N}_{a} \displaystyle  = \displaystyle  \mathbf{r}\times\mathbf{F}_{a}\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \mathbf{r}\times Q\mathbf{E}_{a}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  a^{2}\frac{\mu_{0}Q^{2}\left(\dot{\omega}_{a}+\dot{\omega}_{b}\right)}{4\pi\ell}\left(-\hat{\mathbf{r}}\times\hat{\boldsymbol{\phi}}\right)\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  -a^{2}\frac{\mu_{0}Q^{2}\left(\dot{\omega}_{a}+\dot{\omega}_{b}\right)}{4\pi\ell}\hat{\mathbf{z}}\ \ \ \ \ (21)
\displaystyle  \mathbf{N}_{b} \displaystyle  = \displaystyle  -\mathbf{r}\times Q\mathbf{E}_{b}\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}Q^{2}\left(a^{2}\dot{\omega}_{a}+b^{2}\dot{\omega}_{b}\right)}{4\pi\ell}\hat{\mathbf{z}} \ \ \ \ \ (23)

Adding them together we get

\displaystyle   \mathbf{N} \displaystyle  = \displaystyle  \mathbf{N}_{a}+\mathbf{N}_{b}\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}Q^{2}\left(b^{2}-a^{2}\right)\dot{\omega}_{b}}{4\pi\ell}\hat{\mathbf{z}} \ \ \ \ \ (25)

Integrating over the time it takes to reach the final speed {\omega_{b}} we get

\displaystyle  \mathbf{L}=\frac{\mu_{0}Q^{2}\left(b^{2}-a^{2}\right)\omega_{b}}{4\pi\ell}\hat{\mathbf{z}} \ \ \ \ \ (26)

which is equal and opposite to 13. Thus the total angular momentum introduced into the system by magnetic and electric fields induced by the rotating cylinders is zero, showing that angular momentum is conserved.

Angular momentum of a charge and magnetic monopole

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 8.12.

Returning to the fantasy land where magnetic monopoles exist, suppose we have a single electric charge {e} and a single magnetic monopole (magnetic charge) {b}. Then the fields due to these charges are

\displaystyle   \mathbf{E} \displaystyle  = \displaystyle  \frac{e}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}}\ \ \ \ \ (1)
\displaystyle  \mathbf{B} \displaystyle  = \displaystyle  \frac{\mu_{0}b}{4\pi r_{b}^{2}}\hat{\mathbf{r}}_{b} \ \ \ \ \ (2)

where we’ve placed {e} at the origin. The magnetic charge {b} is not at the origin, so {\hat{\mathbf{r}}_{b}} is the vector from {b}‘s location to the point where we’re measuring the fields. We’d like to find the total angular momentum contained in these two fields.

We’ll place {b} on the {z} axis a distance {d} from {e}, so its location is {d\hat{\mathbf{z}}}. Then the vectors {d\hat{\mathbf{z}}}, {\mathbf{r}} and {\mathbf{r}_{b}} form a triangle with the spherical angle {\theta} lying between {d\hat{\mathbf{z}}} and {\mathbf{r}}, so it is the angle opposite the side {\mathbf{r}_{b}}. Therefore

\displaystyle   \mathbf{r}_{b} \displaystyle  = \displaystyle  \mathbf{r}-d\hat{\mathbf{z}}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  r\hat{\mathbf{r}}-d\left(\cos\theta\hat{\mathbf{r}}-\sin\theta\hat{\boldsymbol{\theta}}\right)\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \left(r-d\cos\theta\right)\hat{\mathbf{r}}+d\sin\theta\hat{\boldsymbol{\theta}} \ \ \ \ \ (5)

By the cosine rule for triangles

\displaystyle  r_{b}^{2}=d^{2}+r^{2}-2dr\cos\theta \ \ \ \ \ (6)

so we get

\displaystyle   \mathbf{E} \displaystyle  = \displaystyle  \frac{e}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}}\ \ \ \ \ (7)
\displaystyle  \mathbf{B} \displaystyle  = \displaystyle  \frac{\mu_{0}b}{4\pi\left(d^{2}+r^{2}-2dr\cos\theta\right)^{3/2}}\left[\left(r-d\cos\theta\right)\hat{\mathbf{r}}+d\sin\theta\hat{\boldsymbol{\theta}}\right] \ \ \ \ \ (8)

The momentum density is

\displaystyle   \boldsymbol{\mathfrak{p}}_{em} \displaystyle  = \displaystyle  \epsilon_{0}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}ebd}{16\pi^{2}}\frac{\sin\theta}{r^{2}\left(d^{2}+r^{2}-2dr\cos\theta\right)^{3/2}}\hat{\boldsymbol{\phi}} \ \ \ \ \ (10)

and the angular momentum density is

\displaystyle   \boldsymbol{\mathfrak{L}}_{em} \displaystyle  = \displaystyle  \mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\mu_{0}ebd}{16\pi^{2}}\frac{\sin\theta}{r\left(d^{2}+r^{2}-2dr\cos\theta\right)^{3/2}}\hat{\boldsymbol{\theta}} \ \ \ \ \ (12)

To find the total angular momentum, we integrate over all space

\displaystyle  \mathbf{L}=-\frac{\mu_{0}ebd}{16\pi^{2}}\int_{0}^{\pi}\int_{0}^{\infty}\int_{0}^{2\pi}\frac{\sin\theta}{r\left(d^{2}+r^{2}-2dr\cos\theta\right)^{3/2}}\hat{\boldsymbol{\theta}}r^{2}\sin\theta d\phi drd\theta \ \ \ \ \ (13)

As usual, we must first express {\hat{\boldsymbol{\theta}}} in rectangular coordinates

\displaystyle  \hat{\boldsymbol{\theta}}=\cos\theta\cos\phi\hat{\mathbf{x}}+\cos\theta\sin\phi\hat{\mathbf{y}}-\sin\theta\hat{\mathbf{z}} \ \ \ \ \ (14)

The {\phi} integral will kill off the {x} and {y} components, so we are left with

\displaystyle  \mathbf{L}=\frac{\mu_{0}ebd}{16\pi^{2}}\hat{\mathbf{z}}\left(2\pi\right)\int_{0}^{\pi}\int_{0}^{\infty}\frac{r\sin^{3}\theta}{\left(d^{2}+r^{2}-2dr\cos\theta\right)^{3/2}}drd\theta \ \ \ \ \ (15)

This is a rather unpleasant integral that can be done using Maple, with the result

\displaystyle  \int_{0}^{\pi}\int_{0}^{\infty}\frac{r\sin^{3}\theta}{\left(d^{2}+r^{2}-2dr\cos\theta\right)^{3/2}}drd\theta=\frac{2}{d} \ \ \ \ \ (16)

so the angular momentum becomes

\displaystyle  \mathbf{L}=\frac{\mu_{0}eb}{4\pi}\hat{\mathbf{z}} \ \ \ \ \ (17)

I could leave it at that, but for once I decided to solve the integral by hand, just to see how it’s done (since the difficulty of the integral would seem to be the main reason Griffiths has this marked as a ‘hard’ problem). It seems easiest to do the {r} integral first, so we can rewrite the integral as

\displaystyle  \int_{0}^{\pi}d\theta\sin^{3}\theta\int_{0}^{\infty}\frac{r}{\left(d^{2}+r^{2}-2dr\cos\theta\right)^{3/2}}dr \ \ \ \ \ (18)

If we had the derivative of the expression in parentheses in the denominator present in the numerator, the integral would be easy. So we can try something like this (I’ll call {\cos\theta\equiv c} and {\sin\theta\equiv s} to save a bit of writing; {\theta} is a constant in the {r} integral anyway so it shouldn’t cause any confusion):

\displaystyle   \int_{0}^{\infty}\frac{r}{\left(d^{2}+r^{2}-2drc\right)^{3/2}}dr \displaystyle  = \displaystyle  \frac{1}{2}\int_{0}^{\infty}\frac{2r-2dc+2dc}{\left(d^{2}+r^{2}-2drc\right)^{3/2}}dr\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  -\left.\frac{1}{\sqrt{d^{2}+r^{2}-2drc}}\right|_{0}^{\infty}+dc\int_{0}^{\infty}\frac{1}{\left(d^{2}+r^{2}-2drc\right)^{3/2}}dr\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{d}+dc\int_{0}^{\infty}\frac{1}{\left(d^{2}+r^{2}-2drc\right)^{3/2}}dr \ \ \ \ \ (21)

We can now complete the square in the remaining integral to get

\displaystyle   \int_{0}^{\infty}\frac{1}{\left(d^{2}+r^{2}-2drc\right)^{3/2}}dr \displaystyle  = \displaystyle  \int_{0}^{\infty}\frac{1}{\left(\left(r-dc\right)^{2}+\left(d^{2}-d^{2}c^{2}\right)\right)^{3/2}}dr\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  \int_{0}^{\infty}\frac{1}{\left(\left(r-dc\right)^{2}+d^{2}s^{2}\right)^{3/2}}dr \ \ \ \ \ (23)

The last integral is of the form {\int dx/\left(a^{2}+x^{2}\right)^{3/2}} which can be looked up in tables, but just to be complete, let’s work that one out too. We can use a trigonometric substitution

\displaystyle   r-dc \displaystyle  = \displaystyle  ds\tan u\ \ \ \ \ (24)
\displaystyle  dr \displaystyle  = \displaystyle  ds\sec^{2}udu\ \ \ \ \ (25)
\displaystyle  \left(r-dc\right)^{2}+d^{2}s^{2} \displaystyle  = \displaystyle  d^{2}s^{2}\left(1+\tan^{2}u\right)\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  d^{2}s^{2}\sec^{2}u \ \ \ \ \ (27)

Putting all this into 23 we get (we’ll leave the limits off for now):

\displaystyle   \int\frac{1}{\left(\left(r-dc\right)^{2}+d^{2}s^{2}\right)^{3/2}}dr \displaystyle  = \displaystyle  \int\frac{ds\sec^{2}udu}{d^{3}s^{3}\sec^{3}u}\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{d^{2}s^{2}}\int\cos u\; du\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{d^{2}s^{2}}\sin u \ \ \ \ \ (30)

To convert back to {r}, we use

\displaystyle   \cos u \displaystyle  = \displaystyle  \sqrt{\frac{1}{\sec^{2}u}}\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{1}{1+\tan^{2}u}}\ \ \ \ \ (32)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{1}{1+\frac{\left(r-dc\right)^{2}}{d^{2}s^{2}}}}\ \ \ \ \ (33)
\displaystyle  \sin u \displaystyle  = \displaystyle  \pm\sqrt{1-\cos^{2}u}\ \ \ \ \ (34)
\displaystyle  \displaystyle  = \displaystyle  \pm\frac{r-dc}{\sqrt{d^{2}s^{2}+\left(r-dc\right)^{2}}} \ \ \ \ \ (35)

so the integral comes out to (restoring the limits):

\displaystyle   \int_{0}^{\infty}\frac{1}{\left(\left(r-dc\right)^{2}+d^{2}s^{2}\right)^{3/2}}dr \displaystyle  = \displaystyle  \left.\pm\frac{1}{d^{2}s^{2}}\frac{r-dc}{\sqrt{d^{2}s^{2}+\left(r-dc\right)^{2}}}\right|_{0}^{\infty}\ \ \ \ \ (36)
\displaystyle  \displaystyle  = \displaystyle  \pm\frac{1}{d^{2}s^{2}}\left(1+c\right) \ \ \ \ \ (37)

Since the integrand is positive over its entire range, we must take the + sign in the answer. We can now plug this back into 21 to get

\displaystyle   \int_{0}^{\infty}\frac{r}{\left(d^{2}+r^{2}-2drc\right)^{3/2}}dr \displaystyle  = \displaystyle  \frac{1}{d}+\frac{dc}{d^{2}s^{2}}\left(1+c\right)\ \ \ \ \ (38)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{d}\left[1-+\frac{c\left(1+c\right)}{1-c^{2}}\right]\ \ \ \ \ (39)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{d}\left[1+\frac{c\left(1+c\right)}{\left(1-c\right)\left(1+c\right)}\right]\ \ \ \ \ (40)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\left(1-\cos\theta\right)d} \ \ \ \ \ (41)

Plugging this into 16, we now have the remaining integral over {\theta}:

\displaystyle   \int_{0}^{\pi}\int_{0}^{\infty}\frac{r\sin^{3}\theta}{\left(d^{2}+r^{2}-2dr\cos\theta\right)^{3/2}}drd\theta \displaystyle  = \displaystyle  \frac{1}{d}\int_{0}^{\pi}\frac{\sin^{3}\theta}{1-\cos\theta}d\theta\ \ \ \ \ (42)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{d}\int_{0}^{\pi}\frac{\sin\theta\left(1-\cos^{2}\theta\right)}{1-\cos\theta}d\theta\ \ \ \ \ (43)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{d}\int_{0}^{\pi}\frac{\sin\theta\left(1-\cos\theta\right)\left(1+\cos\theta\right)}{1-\cos\theta}d\theta\ \ \ \ \ (44)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{d}\int_{0}^{\pi}\left(\sin\theta+\sin\theta\cos\theta\right)d\theta\ \ \ \ \ (45)
\displaystyle  \displaystyle  = \displaystyle  \frac{2}{d} \ \ \ \ \ (46)

Q.E.D.

Now you see why I use Maple to work out integrals!

The electron as electromagnetic energy and angular momentum

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 8.11.

Early in the twentieth century, a classical model of the electron was proposed in which the electron was taken to be a spherical shell of charge of radius {R} spinning with angular speed {\omega}, and that its entire mass (strictly, mass-energy) was composed of the energy stored in its electromagnetic fields, and its angular momentum was that of the electromagentic fields as well.

First, we work out the energy, which is

\displaystyle  W=\frac{1}{2}\int_{\mathcal{V}}\left(\frac{1}{\mu_{0}}B^{2}+\epsilon_{0}E^{2}\right)d^{3}\mathbf{r} \ \ \ \ \ (1)

The electric field of a spherical shell is

\displaystyle  \mathbf{E}=\begin{cases} 0 & r<R\\ \frac{q}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}} & r>R \end{cases} \ \ \ \ \ (2)

Griffiths works out the magnetic vector potential of a spinning spherical shell of charge in his Example 5.11:

\displaystyle  \mathbf{A}=\begin{cases} \frac{1}{3}\mu_{0}R\omega\sigma r\sin\theta\hat{\boldsymbol{\phi}} & r\le R\\ \frac{1}{3}\mu_{0}R^{4}\omega\sigma\frac{\sin\theta}{r^{2}}\hat{\boldsymbol{\phi}} & r\ge R \end{cases} \ \ \ \ \ (3)

from which we can get {\mathbf{B}} from

\displaystyle  \mathbf{B}=\nabla\times\mathbf{A}=\begin{cases} \frac{2}{3}\mu_{0}R\omega\sigma\hat{\mathbf{z}} & r\le R\\ \frac{1}{3}\frac{\mu_{0}R^{4}\omega\sigma}{r^{3}}\left[2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\boldsymbol{\theta}}\right] & r\ge R \end{cases} \ \ \ \ \ (4)

The surface charge density {\sigma=q/4\pi R^{2}} so we get

\displaystyle  \mathbf{B}=\begin{cases} \frac{\mu_{0}q\omega}{6\pi R}\hat{\mathbf{z}} & r\le R\\ \frac{\mu_{0}qR^{2}\omega}{12\pi r^{3}}\left[2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\boldsymbol{\theta}}\right] & r\ge R \end{cases} \ \ \ \ \ (5)

The energy is

\displaystyle   W \displaystyle  = \displaystyle  \frac{1}{2\mu_{0}}\frac{4}{3}\pi R^{3}\left(\frac{\mu_{0}q\omega}{6\pi R}\right)^{2}+\frac{1}{2\mu_{0}}\left(\frac{\mu_{0}qR^{2}\omega}{12\pi}\right)^{2}\left(2\pi\right)\int_{R}^{\infty}\int_{0}^{\pi}\frac{4\cos^{2}\theta+\sin^{2}\theta}{r^{6}}r^{2}\sin\theta d\theta dr+\ \ \ \ \ (6)
\displaystyle  \displaystyle  \displaystyle  \frac{\epsilon_{0}}{2}\left(\frac{q}{4\pi\epsilon_{0}}\right)^{2}\left(2\pi\right)\int_{R}^{\infty}\int_{0}^{\pi}\frac{1}{r^{6}}r^{2}\sin\theta d\theta dr\nonumber
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}}{54\pi}q^{2}R\omega^{2}+\frac{\mu_{0}q^{2}R^{4}\omega^{2}}{144\pi}\int_{R}^{\infty}\int_{0}^{\pi}\frac{3\cos^{2}\theta+1}{r^{4}}\sin\theta d\theta dr+\frac{q^{2}}{16\pi\epsilon_{0}}\int_{R}^{\infty}\int_{0}^{\pi}\frac{\sin\theta}{r^{4}}d\theta dr\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}R\omega^{2}}{54\pi}+\frac{\mu_{0}q^{2}R\omega^{2}}{108\pi}+\frac{q^{2}}{8\pi\epsilon_{0}R}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \frac{3\mu_{0}q^{2}R\omega^{2}}{108\pi}+\frac{q^{2}}{8\pi\epsilon_{0}R} \ \ \ \ \ (9)

To calculate the angular momentum, we first need the linear momentum. The momentum density is

\displaystyle  \boldsymbol{\mathfrak{p}}_{em}=\epsilon_{0}\mathbf{E}\times\mathbf{B} \ \ \ \ \ (10)

Only the region outside the sphere contributes (since {\mathbf{E}=0} inside), so we get

\displaystyle   \boldsymbol{\mathfrak{p}}_{em} \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}R^{2}\omega}{48\pi^{2}}\frac{\sin\theta}{r^{5}}\hat{\boldsymbol{\phi}}\ \ \ \ \ (11)
\displaystyle  \boldsymbol{\mathfrak{L}}_{em} \displaystyle  = \displaystyle  \mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\mu_{0}q^{2}R^{2}\omega}{48\pi^{2}}\frac{\sin\theta}{r^{4}}\hat{\boldsymbol{\theta}} \ \ \ \ \ (13)

The total angular momentum is

\displaystyle  \mathbf{L}_{em}=-\frac{\mu_{0}q^{2}R^{2}\omega}{48\pi^{2}}\int_{R}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\sin\theta}{r^{4}}\hat{\boldsymbol{\theta}}r^{2}\sin\theta d\phi d\theta dr \ \ \ \ \ (14)

To do the integral, we need to express {\hat{\boldsymbol{\theta}}} in rectangular coordinates

\displaystyle  \hat{\boldsymbol{\theta}}=\cos\theta\cos\phi\hat{\mathbf{x}}+\cos\theta\sin\phi\hat{\mathbf{y}}-\sin\theta\hat{\mathbf{z}} \ \ \ \ \ (15)

The {\phi} integral will kill off the {x} and {y} components, so we are left with

\displaystyle   \mathbf{L}_{em} \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}R^{2}\omega}{24\pi}\hat{\mathbf{z}}\int_{R}^{\infty}\int_{0}^{\pi}\frac{\sin^{3}\theta}{r^{2}}d\theta dr\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}q^{2}R\omega}{18\pi}\hat{\mathbf{z}} \ \ \ \ \ (17)

The idea now is to equate the energy with relativistic rest mass and the angular momentum with the quantum mechanical spin, so we get from 9 and 17

\displaystyle   \frac{3\mu_{0}q^{2}R\omega^{2}}{108\pi}+\frac{q^{2}}{8\pi\epsilon_{0}R} \displaystyle  = \displaystyle  mc^{2}\ \ \ \ \ (18)
\displaystyle  \frac{\mu_{0}q^{2}R\omega}{18\pi} \displaystyle  = \displaystyle  \frac{\hbar}{2} \ \ \ \ \ (19)

Solving for {R} and {\omega} we get

\displaystyle   R \displaystyle  = \displaystyle  \frac{1}{8}{\frac{18\,{\hbar}^{2}{\pi}^{2}\epsilon_{{0}}+{q}^{4}\mu_{{0}}}{m{c}^{2}\pi\,\epsilon_{{0}}\mu_{{0}}{q}^{2}}}\ \ \ \ \ (20)
\displaystyle  \omega \displaystyle  = \displaystyle  72\,{\frac{\mbox{{\tt \ensuremath{\hbar}}}\,{\pi}^{2}m{c}^{2}\epsilon_{{0}}}{18\,{\mbox{{\tt \ensuremath{\hbar}}}}^{2}{\pi}^{2}\epsilon_{{0}}+{q}^{4}\mu_{{0}}}} \ \ \ \ \ (21)

We can plug in the values of the various constants (all in SI units):

\displaystyle   \hbar \displaystyle  = \displaystyle  1.0546\times10^{-34}\ \ \ \ \ (22)
\displaystyle  m \displaystyle  = \displaystyle  9.109\times10^{-31}\ \ \ \ \ (23)
\displaystyle  \epsilon_{0} \displaystyle  = \displaystyle  8.854\times10^{-12}\ \ \ \ \ (24)
\displaystyle  \mu_{0} \displaystyle  = \displaystyle  1.2566\times10^{-6}\ \ \ \ \ (25)
\displaystyle  c \displaystyle  = \displaystyle  3.0\times10^{8} \ \ \ \ \ (26)

and we get

\displaystyle   R \displaystyle  = \displaystyle  2.978\times10^{-11}\mbox{ m}\ \ \ \ \ (27)
\displaystyle  \omega \displaystyle  = \displaystyle  3.105\times10^{21}\mbox{ s}^{-1} \ \ \ \ \ (28)

This gives an equatorial speed of

\displaystyle  \omega R=9.246\times10^{10}\mbox{ m s}^{-1} \ \ \ \ \ (29)

which is more than 300 times the speed of light, so this model of the electron doesn’t work.

Angular momentum in a magnetized sphere

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 8.8.

Here’s another slightly more complex example of the conservation of angular momentum in electromagnetic fields. We have a solid iron sphere of radius {R} with a uniform magnetization {M\hat{\mathbf{z}}} and carrying a surface charge {Q}. We want to show that the angular momentem in the electric and magnetic fields is conserved as we switch off either the magnetic or electric field.

First, we need to calculate the angular momentum in the fields before anything is switched off.

The momentum density of an electromagnetic field is given by

\displaystyle  \boldsymbol{\mathfrak{p}}_{em}=\epsilon_{0}\mu_{0}\mathbf{S}=\epsilon_{0}\mathbf{E}\times\mathbf{B} \ \ \ \ \ (1)

and the angular momentum density is

\displaystyle  \boldsymbol{\mathfrak{L}}_{em}\equiv\mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}=\epsilon_{0}\mathbf{r}\times\left(\mathbf{E}\times\mathbf{B}\right) \ \ \ \ \ (2)

Griffiths shows in his Example 6.1 that the magnetic field of a uniformly magnetized sphere is

\displaystyle  \mathbf{B}=\begin{cases} \frac{2}{3}\mu_{0}M\hat{\mathbf{z}}=\frac{2}{3}\mu_{0}M\left(\cos\theta\hat{\mathbf{r}}-\sin\theta\hat{\boldsymbol{\theta}}\right) & \mbox{inside}\\ \frac{\mu_{0}MR^{3}}{3r^{3}}\left(2\cos\theta\hat{\mathbf{r}}+\sin\theta\hat{\boldsymbol{\theta}}\right) & \mbox{outside} \end{cases} \ \ \ \ \ (3)

The outside formula is just the field of a perfect magnetic dipole.

The electric field is

\displaystyle  \mathbf{E}=\begin{cases} 0 & \mbox{inside}\\ \frac{Q}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}} & \mbox{outside} \end{cases} \ \ \ \ \ (4)

The linear momentum density is then

\displaystyle  \boldsymbol{\mathfrak{p}}_{em}=\epsilon_{0}\mathbf{E}\times\mathbf{B}=\frac{\mu_{0}MQR^{3}}{12\pi r^{5}}\sin\theta\hat{\boldsymbol{\phi}} \ \ \ \ \ (5)

so the angular momentum density is

\displaystyle  \boldsymbol{\mathfrak{L}}_{em}=\mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}=-\frac{\mu_{0}MQR^{3}}{12\pi r^{4}}\sin\theta\hat{\boldsymbol{\theta}} \ \ \ \ \ (6)

The total angular momentum is thus

\displaystyle  \mathbf{L}_{em}=-\frac{\mu_{0}MQR^{3}}{12\pi}\int_{R}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\sin\theta}{r^{4}}r^{2}\sin\theta\hat{\boldsymbol{\theta}}d\phi d\theta dr \ \ \ \ \ (7)

As the unit vector {\hat{\boldsymbol{\theta}}} is not constant in direction, we need to express it in rectangular coordinates:

\displaystyle  \hat{\boldsymbol{\theta}}=\cos\theta\cos\phi\hat{\mathbf{x}}+\cos\theta\sin\phi\hat{\mathbf{y}}-\sin\theta\hat{\mathbf{z}} \ \ \ \ \ (8)

The integral of {\cos\phi} and {\sin\phi} over the interval {\left[0,2\pi\right]} gives zero, so only the {z} component survives, giving

\displaystyle   \mathbf{L}_{em} \displaystyle  = \displaystyle  \frac{\mu_{0}MQR^{3}}{12\pi}\hat{\mathbf{z}}\left(2\pi\right)\int_{R}^{\infty}\frac{dr}{r^{2}}\int_{0}^{\pi}\sin^{3}\theta d\theta\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \frac{2}{9}\mu_{0}MQR^{2}\hat{\mathbf{z}} \ \ \ \ \ (10)

Now suppose we slowly switch off the magnetic field, keeping {Q} constant. This will generate an electric field according to Faraday’s law. Consider a slice of the sphere parallel to the {xy} plane, that is, at some value of {\theta}. The electric field is circular around the axis of the sphere and satisfies the equation

\displaystyle  \oint\mathbf{E}\cdot d\boldsymbol{\ell}=-\frac{d\Phi}{dt}=-\int\frac{\partial\mathbf{B}}{\partial t}\cdot d\mathbf{a} \ \ \ \ \ (11)

The radius of the circle is {R\sin\theta} and if we switch off the magnetic field in a uniform way, then {\frac{\partial\mathbf{B}}{\partial t}} is constant over this area, so we get

\displaystyle   2\pi RE\sin\theta \displaystyle  = \displaystyle  -\pi\left(R\sin\theta\right)^{2}\frac{\partial B}{\partial t}\ \ \ \ \ (12)
\displaystyle  \mathbf{E} \displaystyle  = \displaystyle  -\frac{1}{2}R\sin\theta\frac{\partial B}{\partial t}\hat{\boldsymbol{\phi}} \ \ \ \ \ (13)

The minus sign in the last line cancels the fact that {\frac{\partial B}{\partial t}<0} and gives an electric field that opposes the decrease in magnetic field.

This field exerts a force on the strip of charge in the slice. The surface charge density {\sigma} is

\displaystyle  \sigma=\frac{Q}{4\pi R^{2}} \ \ \ \ \ (14)

so the amount of charge in the slice is

\displaystyle  dq=\frac{Q}{4\pi R^{2}}\left(2\pi R\sin\theta\right)\left(Rd\theta\right)=\frac{1}{2}Q\sin\theta d\theta \ \ \ \ \ (15)

The force on the slice is

\displaystyle  d\mathbf{F}=\mathbf{E}dq=-\frac{1}{4}RQ\sin^{2}\theta\;d\theta\frac{\partial B}{\partial t}\hat{\boldsymbol{\phi}} \ \ \ \ \ (16)

and the torque is

\displaystyle  d\mathbf{N}=\mathbf{r}\times d\mathbf{F}=-\left(R\sin\theta\right)\frac{1}{4}RQ\sin^{2}\theta\;d\theta\frac{\partial B}{\partial t}\hat{\mathbf{z}}=-\frac{1}{4}R^{2}Q\sin^{3}\theta\;d\theta\frac{\partial B}{\partial t}\hat{\mathbf{z}} \ \ \ \ \ (17)

The total torque on the sphere is

\displaystyle   \mathbf{N} \displaystyle  = \displaystyle  \int_{0}^{\pi}d\mathbf{N}\left(\theta\right)\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{4}R^{2}Q\frac{\partial B}{\partial t}\hat{\mathbf{z}}\int_{0}^{\pi}\sin^{3}\theta d\theta\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{3}QR^{2}\frac{\partial B}{\partial t}\hat{\mathbf{z}} \ \ \ \ \ (20)

The impulse is obtained by integrating this over time:

\displaystyle  \mathbf{I}=-\frac{1}{3}QR^{2}\hat{\mathbf{z}}\int_{0}^{\infty}\frac{\partial B}{\partial t}dt=-\frac{1}{3}QR^{2}\hat{\mathbf{z}}\left(-\frac{2}{3}\mu_{0}M\right)=\frac{2}{9}\mu_{0}MQR^{2}\hat{\mathbf{z}} \ \ \ \ \ (21)

which agrees with 10, showing that the angular momentum is transferred to the sphere.

Now let’s keep the magnetization constant and decrease the electric field by draining off the sphere’s surface charge. We arrange to do this by connecting a resistor {\mathcal{R}} between the north pole and ground, but we do this in a way that the surface charge density remains uniform over the sphere as the total charge decreases. Doing this creates a surface current that experiences a force from the magnetic field. Since we’re draining the charge off at the north pole, the surface current everywhere on the sphere will head towards {\theta=0}, that is, it flows in the {-\theta} direction. To work out the current at each point on the sphere, consider the same horizontal slice that we used above. The current passing through this slice must be the sum of the current passing through the slice just below it (that is, at {\theta+d\theta}) and the charge draining away from the slice itself. That is

\displaystyle   I\left(\theta\right) \displaystyle  = \displaystyle  I\left(\theta+d\theta\right)+\frac{d\sigma}{dt}\left(2\pi R\sin\theta\right)\left(Rd\theta\right)\ \ \ \ \ (22)
\displaystyle  \frac{dI}{d\theta} \displaystyle  = \displaystyle  -\frac{d\sigma}{dt}2\pi R^{2}\sin\theta\ \ \ \ \ (23)
\displaystyle  I\left(\theta\right) \displaystyle  = \displaystyle  \frac{d\sigma}{dt}2\pi R^{2}\cos\theta+\alpha \ \ \ \ \ (24)

where {\alpha} is a constant of integration. Since the charge is draining off through the north pole, we’d like the current at the south pole to be zero, so we choose

\displaystyle  \alpha=\frac{d\sigma}{dt}2\pi R^{2} \ \ \ \ \ (25)

We can treat the sphere/ground system as a capacitor {C}, so the charge density is given by

\displaystyle   \sigma\left(t\right) \displaystyle  = \displaystyle  \sigma\left(0\right)e^{-t/\mathcal{R}C}\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  \frac{Q}{4\pi R^{2}}e^{-t/\mathcal{R}C} \ \ \ \ \ (27)

so

\displaystyle  I\left(\theta\right)=-\frac{Q}{2\mathcal{R}C}e^{-t/\mathcal{R}C}\left(1+\cos\theta\right) \ \ \ \ \ (28)

This is the total current flowing across the surface of a slice at angle {\theta}. The {q\mathbf{v}} term in the force law is the current times the distance over which it flows, which in this case is {Rd\theta}. We can now apply the Lorentz force law {\mathbf{F}=q\mathbf{v}\times\mathbf{B}} to work out the force on the sphere due to the current interacting with the magnetic field. Since the current is parallel to {\hat{\boldsymbol{\theta}}}, only the {r} component of {\mathbf{B}} will survive the cross product. Note that although the {\theta} component of {\mathbf{B}} is not continuous across the surface of the sphere, the {r} component is continuous, so it doesn’t matter whether we use the inside or outside formula. Thus we get

\displaystyle   d\mathbf{F} \displaystyle  = \displaystyle  \left(-\frac{Q}{2\mathcal{R}C}e^{-t/\mathcal{R}C}\left(1+\cos\theta\right)\right)\left(Rd\theta\right)\left(\frac{2\mu_{0}M}{3}\cos\theta\right)\left(-\hat{\boldsymbol{\phi}}\right)\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}MQR}{3\mathcal{R}C}e^{-t/\mathcal{R}C}\hat{\boldsymbol{\phi}}\left(\cos\theta+\cos^{2}\theta\right) \ \ \ \ \ (30)

The torque on a slice is

\displaystyle   d\mathbf{N} \displaystyle  = \displaystyle  \mathbf{r}\times d\mathbf{F}\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}MQR^{2}}{3\mathcal{R}C}e^{-t/\mathcal{R}C}\hat{\mathbf{z}}\left(\cos\theta+\cos^{2}\theta\right)\sin\theta \ \ \ \ \ (32)

The total torque is thus

\displaystyle   \mathbf{N} \displaystyle  = \displaystyle  \frac{\mu_{0}MQR^{2}}{3\mathcal{R}C}e^{-t/\mathcal{R}C}\hat{\mathbf{z}}\int_{0}^{\pi}\left(\cos\theta+\cos^{2}\theta\right)\sin\theta d\theta\ \ \ \ \ (33)
\displaystyle  \displaystyle  = \displaystyle  \frac{2\mu_{0}MQR^{2}}{9\mathcal{R}C}e^{-t/\mathcal{R}C}\hat{\mathbf{z}} \ \ \ \ \ (34)

Finally we integrate the torque over time to get the impulse:

\displaystyle   \mathbf{I} \displaystyle  = \displaystyle  \int_{0}^{\infty}\mathbf{N}dt\ \ \ \ \ (35)
\displaystyle  \displaystyle  = \displaystyle  \frac{2\mu_{0}MQR^{2}}{9\mathcal{R}C}\hat{\mathbf{z}}\int_{0}^{\infty}e^{-t/\mathcal{R}C}dt\ \ \ \ \ (36)
\displaystyle  \displaystyle  = \displaystyle  \frac{2}{9}\mu_{0}MQR^{2}\hat{\mathbf{z}} \ \ \ \ \ (37)

This again agrees with 10 so all is well.

Angular momentum in electromagnetic fields

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 8, Post 7.

The momentum density of an electromagnetic field is given by

\displaystyle \boldsymbol{\mathfrak{p}}_{em}=\epsilon_{0}\mu_{0}\mathbf{S}=\epsilon_{0}\mathbf{E}\times\mathbf{B} \ \ \ \ \ (1)

If we have linear momentum, then we automatically have angular momentum with respect to some origin by using the classical definition of angular momentum {\mathbf{L}=\mathbf{r}\times\mathbf{p}}. We can define the angular momentum density of an electromagnetic field by

\displaystyle \boldsymbol{\mathfrak{L}}_{em}\equiv\mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}=\epsilon_{0}\mathbf{r}\times\left(\mathbf{E}\times\mathbf{B}\right) \ \ \ \ \ (2)

Just as with linear momentum, even static fields can have angular momentum. As an example, suppose we have a long solenoid with {n} turns per unit length carrying current {I_{0}} and a radius {R}, with its axis along the {z} axis. The magnetic field inside the solenoid is

\displaystyle \mathbf{B}_{0}=\mu_{0}nI\hat{\mathbf{z}} \ \ \ \ \ (3)

The field is zero outside the solenoid.

Now suppose
we add two other cylinders (not solenoids), both coaxial with
the solenoid. One cylinder has radius {a<R} (so it lies inside the solenoid) and carries surface charge {+Q}; the other cylinder has radius {b>R} (outside the solenoid) and carries charge {-Q}. Both cylinders have length {\ell}. From Gauss’s law, the electric field between these two cylinders is, for {a<r<b}

\displaystyle \mathbf{E}_{0}=\frac{Q}{2\pi\epsilon_{0}\ell}\frac{\hat{\mathbf{r}}}{r} \ \ \ \ \ (4)

That is, the field points radially outward from the axis. The electric field is zero for {r<a} and {r>b}. (We’re neglecting end effects, so we’re assuming that {\ell\gg b>a}.)

The linear momentum density is non-zero in the region {a<r<R} (where both fields are non-zero) and we have

\displaystyle \boldsymbol{\mathfrak{p}}_{em}=-\frac{\mu_{0}nIQ}{2\pi\ell}\frac{\hat{\boldsymbol{\phi}}}{r} \ \ \ \ \ (5)

so the angular momentum density is

\displaystyle \boldsymbol{\mathfrak{L}}_{em} \displaystyle = \displaystyle \mathbf{r}\times\boldsymbol{\mathfrak{p}}_{em}\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle -\frac{\mu_{0}nIQ}{2\pi\ell}\hat{\mathbf{z}} \ \ \ \ \ (7)

Conveniently, this is constant so the total angular momentum is just the density times the volume of the cylindrical tube in the region {a<r<R}

\displaystyle \mathbf{L}_{em} \displaystyle = \displaystyle -\ell\pi\left(R^{2}-a^{2}\right)\frac{\mu_{0}nIQ}{2\pi\ell}\hat{\mathbf{z}}\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle -\left(R^{2}-a^{2}\right)\frac{\mu_{0}nIQ}{2}\hat{\mathbf{z}} \ \ \ \ \ (9)

Now suppose we (quasistatically) discharge the two cylinders by connecting a resistor {\mathcal{R}} between them. We’d like to show that the angular momentum gets transferred from the fields to the physical devices in the problem. The two cylinders are effectively a capacitor with some capacitance {C}, so we know that the current in the resistor will decay exponentially

\displaystyle I\left(t\right)=\frac{V_{0}}{\mathcal{R}}e^{-t/\mathcal{R}C} \ \ \ \ \ (10)

where {V_{0}} is the potential difference between the cylinders at {t=0}. The force {d\mathbf{F}} on a segment of the resistor of length {dr} is

\displaystyle d\mathbf{F} \displaystyle = \displaystyle I\left(t\right)dr\hat{\mathbf{r}}\times\mathbf{B}_{0}\ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle -I\left(t\right)drB_{0}\hat{\boldsymbol{\phi}} \ \ \ \ \ (12)

so the torque on this segment is

\displaystyle d\mathbf{N} \displaystyle = \displaystyle \mathbf{r}\times d\mathbf{F}\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle -I\left(t\right)B_{0}rdr\hat{\mathbf{z}} \ \ \ \ \ (14)

The total torque on the resistor at time {t} is

\displaystyle \mathbf{N}\left(t\right) \displaystyle = \displaystyle -I\left(t\right)B_{0}\hat{\mathbf{z}}\int_{a}^{R}r\; dr\ \ \ \ \ (15)
\displaystyle \displaystyle = \displaystyle -\frac{1}{2}I\left(t\right)B_{0}\left(R^{2}-a^{2}\right)\hat{\mathbf{z}} \ \ \ \ \ (16)

The angular impulse is the integral of torque over time, so we get

\displaystyle \mathbf{I} \displaystyle = \displaystyle \int_{0}^{\infty}\mathbf{N}\left(t\right)dt\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle -\frac{1}{2}B_{0}\left(R^{2}-a^{2}\right)\hat{\mathbf{z}}\int_{0}^{\infty}I\left(t\right)dt\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle -\frac{1}{2}B_{0}\left(R^{2}-a^{2}\right)\hat{\mathbf{z}}\int_{0}^{\infty}\frac{V_{0}}{\mathcal{R}}e^{-t/\mathcal{R}C}dt\ \ \ \ \ (19)
\displaystyle \displaystyle = \displaystyle -\frac{1}{2}B_{0}\left(R^{2}-a^{2}\right)CV_{0}\hat{\mathbf{z}}\ \ \ \ \ (20)
\displaystyle \displaystyle = \displaystyle -\frac{1}{2}\mu_{0}nI\left(R^{2}-a^{2}\right)Q\hat{\mathbf{z}} \ \ \ \ \ (21)

where we used the relation between capacitance, charge and voltage {Q=CV}. We see that this agrees with 9, so all the angular momentum in the fields is transferred to the resistor as the electric field is reduced to zero.

Fine structure of hydrogen: spin-orbit coupling

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 6.16.

Another correction to the energy levels of hydrogen arises from spin-orbit coupling. Unlike the relativistic correction, spin-orbit coupling arises from the interaction of the magnetic moment due to the electron’s spin and the magnetic field generated by its revolution around the nucleus. The derivation using non-relatistic electromagnetic theory isn’t very satisfying since it’s fundamentally incorrect, although by a series of fortuitous coincidences, it does give the right answer. We won’t go into all the gory details here since Griffiths discusses some of the issues in his book in section 6.3.2; rather, we’ll give a sketch of the argument.

In the reference frame of the electron, the proton appears to be rotating about it and, since its speed is very fast, it can be considered to be a steady current loop, which generates a magnetic field {\mathbf{B}}, given by the Biot-Savart law:

\displaystyle  \mathbf{B}=\frac{\mu_{0}}{2r}I\hat{\mathbf{z}} \ \ \ \ \ (1)

where {I} is the effective current generated by the proton, {r} is the radius of the orbit, and we’re taking the axis of revolution to be the {z} axis.

The magnetic moment of the electron, as calculated from classical electrodynamics, is given in terms of the angular momentum {\mathbf{S}} (we’ve renamed some of the symbols in the electrodynamics formula to be consistent with Griffths’s QM book):

\displaystyle  \boldsymbol{\mu}=-\frac{e}{2m}\mathbf{S} \ \ \ \ \ (2)

where {-e} is the electron charge.

The energy of a dipole in a magnetic field is:

\displaystyle  U=-\boldsymbol{\mu}\cdot\mathbf{B} \ \ \ \ \ (3)

so the correction to the hamiltonian is

\displaystyle  H^{\prime}=\left(\frac{e}{2m}\right)\left(\frac{\mu_{0}}{2r}I\right)\mathbf{S}\cdot\hat{\mathbf{z}} \ \ \ \ \ (4)

The current {I} produced by the proton can be approximated by {I=ev/2\pi r} where {v} is the linear speed of the proton in its apparent orbit about the electron. The orbital angular momentum of the electron is {L=mvr}, so {I=eL/2\pi mr^{2}} so we get

\displaystyle  \mathbf{B}=\frac{\mu_{0}e}{4\pi mr^{3}}\mathbf{L} \ \ \ \ \ (5)

and

\displaystyle  H^{\prime}=\frac{\mu_{0}e^{2}}{8\pi m^{2}r^{3}}\mathbf{S}\cdot\mathbf{L} \ \ \ \ \ (6)

We can rewrite this using the relation {\mu_{0}=1/\epsilon_{0}c^{2}} (derived in electromagnetic theory) to get

\displaystyle  H^{\prime}=\frac{e^{2}}{8\pi\epsilon_{0}m^{2}c^{2}r^{3}}\mathbf{S}\cdot\mathbf{L} \ \ \ \ \ (7)

Since the perturbation depends on the interaction between the spin and orbital angular momentum operators, this is the spin-orbit coupling effect.

As I mentioned above, this derivation is seriously flawed because it ignores relativity but when relativistic effects are taken into consideration, they cancel each other, with the result that this equation actually turns out to be correct.

In order to apply degenerate perturbation theory to this correction to the hamiltonian, we’d like to find operators that commute with the original hamiltonian and with {H^{\prime}}, and also have eigenvectors with distinct eigenvalues. Since the unperturbed hydrogen wave functions are derived from a spherically symmetric potential, their angular factors are the spherical harmonics, which are also the eigenfunctions of the orbital angular momentum operators {L^{2}} and {L_{z}} meaning that this hamiltonian commutes with these operators, which in turn means that {L^{2}} and {L_{z}} are conserved.

The introduction of {H^{\prime}}, however, means that this is no longer true. We can see this by considering the commutation relations. The commutators for angular momentum are

\displaystyle  \left[L_{x},L_{y}\right]=i\hbar L_{z} \ \ \ \ \ (8)

with cyclic permutations for the other two relations. The same commutators apply to spin with {L} replaced by {S}. Since {L} and {S} refer to different spaces, all components of {L} commute with all components of {S}.

First, we consider the commutator with {\mathbf{L}}:

\displaystyle   \left[\mathbf{L}\cdot\mathbf{S},\mathbf{L}\right] \displaystyle  = \displaystyle  i\hbar\left(-S_{y}L_{z}+S_{z}L_{y}\right)\hat{\mathbf{x}}+i\hbar\left(S_{x}L_{z}-S_{z}L_{x}\right)\hat{\mathbf{y}}+i\hbar\left(-S_{x}L_{y}+S_{y}L_{x}\right)\hat{\mathbf{z}}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  i\hbar\mathbf{L}\times\mathbf{S} \ \ \ \ \ (10)

Thus {H^{\prime}} does not commute with {\mathbf{L}}. By exactly the same argument

\displaystyle  \left[\mathbf{L}\cdot\mathbf{S},\mathbf{S}\right]=-\hbar\mathbf{S}\times\mathbf{L}=-\left[\mathbf{L}\cdot\mathbf{S},\mathbf{L}\right] \ \ \ \ \ (11)

so {H^{\prime}} doesn’t commute with {\mathbf{S}} either. However, for the total angular momentum {\mathbf{J}=\mathbf{L}+\mathbf{S}}, we get

\displaystyle  \left[\mathbf{L}\cdot\mathbf{S},\mathbf{J}\right]=\left[\mathbf{L}\cdot\mathbf{S},\mathbf{L}\right]+\left[\mathbf{L}\cdot\mathbf{S},\mathbf{S}\right]=0 \ \ \ \ \ (12)

so the components of{\mathbf{J}} are conserved. Since {\left[\mathbf{L},L^{2}\right]=\left[\mathbf{S},S^{2}\right]=0}, we get

\displaystyle  \left[\mathbf{L}\cdot\mathbf{S},L^{2}\right]=\left[\mathbf{L}\cdot\mathbf{S},S^{2}\right]=0 \ \ \ \ \ (13)

Finally, since {J^{2}=L^{2}+S^{2}+2\mathbf{L}\cdot\mathbf{S}}

\displaystyle  \left[\mathbf{L}\cdot\mathbf{S},J^{2}\right]=0 \ \ \ \ \ (14)

Thus the conserved quantities are {J_{z}}, {J^{2}}, {L^{2}} and {S^{2}}.

Particle orbits – conserved quantities

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Boxes 10.1 – 10.2; Problem 10.2.

One form of the geodesic equation is

\displaystyle  \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{ij}}{\partial x^{a}}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (1)

A geodesic is the path followed by a free particle in a given metric, so we can apply this equation to the Schwarzschild metric to discover what orbits a particle can have around a mass. The metric is

\displaystyle  ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (2)

From the {t} component, we get from 1:

\displaystyle  \frac{d}{d\tau}\left(g_{tj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{ij}}{\partial t}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (3)

Since the metric does not depend on {t} the second term is zero. Also, since the metric is diagonal, the first term becomes

\displaystyle   \frac{d}{d\tau}\left(g_{tj}\frac{dx^{j}}{d\tau}\right) \displaystyle  = \displaystyle  \frac{d}{d\tau}\left(g_{tt}\frac{dt}{d\tau}\right)\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  -\frac{d}{d\tau}\left(\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau}\right)\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  0\ \ \ \ \ (6)
\displaystyle  \left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau} \displaystyle  = \displaystyle  e \ \ \ \ \ (7)

where {e} is a constant.

The meaning of {e} can be inferred from the following argument. At {r=\infty}, the relation becomes

\displaystyle  \frac{dt}{d\tau}=e \ \ \ \ \ (8)

We’ve seen that the Schwarzschild {t} coordinate is the same as the object’s proper time for an object at rest, when measured by an observer at infinity. Thus for an object at rest, {e=1}. However, in general {dt/d\tau=u^{t}}, the time component of the four-velocity. In this case, {t} is the time as measured by an observer at rest at infinity, and {\tau} is the proper time as measured by the object, which may be moving. The four-momentum‘s time component is the energy, and the four-velocity is the four-momentum per unit mass, so {e} is the energy per unit mass of the object which remains constant as the object moves in from infinity.

Now look at the {\phi} component of 1. Again, since the metric does not depend on {\phi} and it is diagonal, we get

\displaystyle   \frac{d}{d\tau}\left(g_{\phi j}\frac{dx^{j}}{d\tau}\right) \displaystyle  = \displaystyle  \frac{d}{d\tau}\left(g_{\phi\phi}\frac{d\phi}{d\tau}\right)\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -\frac{d}{d\tau}\left(r^{2}\sin^{2}\theta\frac{d\phi}{d\tau}\right)\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  0\ \ \ \ \ (11)
\displaystyle  r^{2}\sin^{2}\theta\frac{d\phi}{d\tau} \displaystyle  = \displaystyle  l \ \ \ \ \ (12)

where {l} is another constant. If we look at motion in the equatorial plane, {\theta=\pi/2} and {d\phi/d\tau\equiv\omega} which is the angular speed of rotation, so we get

\displaystyle  r^{2}\omega=l \ \ \ \ \ (13)

which is equivalent to the classical definition if {l} is the angular momentum.

In fact, any initial velocity lies in some equatorial plane (merely redefine the location of the {z} axis so that the velocity lies in the plane with {\theta=\pi/2}) so for any specific motion, we can assume {\theta=\pi/2} without, as they say, any loss of generality. Also, because the metric is spherically symmetric, any motion that starts in an equatorial plane must stay in the plane, since there is no asymmetry that would push the object to one side or the other of that plane. Looking at the {\theta} component of 1, we can see that an equatorial path is a geodesic:

\displaystyle   \frac{d}{d\tau}\left(g_{\theta j}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\partial_{\theta}g_{ij}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau} \displaystyle  = \displaystyle  \frac{d}{d\tau}\left(g_{\theta\theta}\frac{d\theta}{d\tau}\right)-\frac{1}{2}\left(2r^{2}\sin\theta\cos\theta\right)\left(\frac{d\phi}{d\tau}\right)^{2}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  2r\frac{dr}{d\tau}\frac{d\theta}{d\tau}+r^{2}\frac{d^{2}\theta}{d\tau^{2}}-\frac{1}{2}\left(2r^{2}\sin\theta\cos\theta\right)\left(\frac{d\phi}{d\tau}\right)^{2}\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  0 \ \ \ \ \ (16)

If {\theta=\pi/2=\mbox{constant}}, then all three terms in the second line are zero, so the geodesic equation is satisfied. Note that this last conclusion doesn’t prove that all geodesics lie in an equatorial plane; it merely verifies that a planar path is a solution. We rely on the symmetry argument to state that all geodesics must lie in an equatorial plane. Of course, in a different metric where the mass is not spherically symmetric, non-planar paths are possible.

Finally, we consider the {r} component of 1. Since all four components of the metric depend on {r}, this takes a bit more work. We can start with the universal relation for the four-velocity {\mathbf{u}\cdot\mathbf{u}=-1}, which in this metric is

\displaystyle  g_{tt}\left(\frac{dt}{d\tau}\right)^{2}+g_{rr}\left(\frac{dr}{d\tau}\right)^{2}+g_{\theta\theta}\left(\frac{d\theta}{d\tau}\right)^{2}+g_{\phi\phi}\left(\frac{d\phi}{d\tau}\right)^{2}=-1 \ \ \ \ \ (17)

From above, we know that, since {\theta=\pi/2}

\displaystyle   \frac{dt}{d\tau} \displaystyle  = \displaystyle  e\left(1-\frac{2GM}{r}\right)^{-1}\ \ \ \ \ (18)
\displaystyle  \frac{d\theta}{d\tau} \displaystyle  = \displaystyle  0\ \ \ \ \ (19)
\displaystyle  \frac{d\phi}{d\tau} \displaystyle  = \displaystyle  \frac{l}{r^{2}\sin^{2}\theta}=\frac{l}{r^{2}} \ \ \ \ \ (20)

Therefore

\displaystyle   -1 \displaystyle  = \displaystyle  -\left(1-\frac{2GM}{r}\right)\left[e\left(1-\frac{2GM}{r}\right)^{-1}\right]^{2}+\left(1-\frac{2GM}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^{2}+r^{2}\left(\frac{l}{r^{2}}\right)^{2}\ \ \ \ \ (21)
\displaystyle  -\left(1-\frac{2GM}{r}\right) \displaystyle  = \displaystyle  -e^{2}+\left(\frac{dr}{d\tau}\right)^{2}+\left(1-\frac{2GM}{r}\right)\frac{l^{2}}{r^{2}}\ \ \ \ \ (22)
\displaystyle  \frac{1}{2}\left(e^{2}-1\right) \displaystyle  = \displaystyle  \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2}+\frac{1}{2}\frac{l^{2}}{r^{2}}-GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)\ \ \ \ \ (23)
\displaystyle  \frac{dr}{d\tau} \displaystyle  = \displaystyle  \pm\left[e^{2}-1+2GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)-\frac{l^{2}}{r^{2}}\right]^{1/2}\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  \pm\sqrt{e^{2}-\left(1-\frac{2GM}{r}\right)\left(1+\frac{\ell^{2}}{r^{2}}\right)} \ \ \ \ \ (25)

Since the only dependent variable in this equation is {r} ({l} and {e} are constants), this is a first-order ODE for {r} in terms of the proper time {\tau}.

As a simple example of the use of this equation, suppose we start a particle at rest at infinity and let it fall in radially towards the mass. Since the motion is entirely radial, the angular momentum is {l=0}. Also, since the particle starts off at rest {e=1} (see above), so the equation becomes

\displaystyle   \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2} \displaystyle  = \displaystyle  \frac{GM}{r}\ \ \ \ \ (26)
\displaystyle  \pm\int\sqrt{r}dr \displaystyle  = \displaystyle  \sqrt{2GM}\int d\tau\ \ \ \ \ (27)
\displaystyle  \pm\frac{2}{3}r^{3/2} \displaystyle  = \displaystyle  \sqrt{2GM}\tau+T \ \ \ \ \ (28)

where {T} is a constant of integration. If the particle is falling inwards, then we would expect {\tau} to increase as {r} decreases, so we need to take the minus sign on the left. The proper time interval as measured by the falling object between two radii (say, {r_{A}=10GM} and {r_{B}=2GM}) is

\displaystyle   \Delta\tau \displaystyle  = \displaystyle  \frac{2}{3\sqrt{2GM}}\left(\left(10GM\right)^{3/2}-\left(2GM\right)^{3/2}\right)\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  \frac{\sqrt{2}}{3}\left(10^{3/2}-2^{3/2}\right)GM \ \ \ \ \ (30)