In Griffith's Introduction to Electrodynamics 4th edition example 5.11 the solution for the vector potential of a uniformly charged spinning spherical shell is given.

enter image description here

Now let's assume that the surface charge density is given by $\sigma(\theta)=\sigma_0sin(\theta)$ and the sphere is rotating with a constant angular velocity $\omega$ about the $\hat{z}$ axis. So, unlike the figure above we have to put the axis of rotation on $\hat{z}$. The velocity vector will be:

$\vec{v}=\vec{\omega}\times\vec{r'}=\begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ 0 & 0 & \omega\\ Rsin\theta'cos\phi' & Rsin\theta'sin\phi' & Rcos\theta' \end{vmatrix}$

After expanding we get

$\vec{v}=-R\omega sin\theta' sin\phi'\hat{x}+R\omega sin\theta' cos\phi'\hat{y}$

The surface current becomes

$\vec{K}=-R\omega\sigma (sin\theta')^2 sin\phi'\hat{x}+R\omega\sigma (sin\theta')^2 cos\phi'\hat{y}$

We can write it as $\vec{K}=-R\omega\sigma sin\theta' \hat{\phi}$

The vector potential can be written as


in which $\alpha$ is taken to be the angle between $\vec{r}$ and $\vec{r'}$ and we can express $cos\alpha$ using the polar and azimuthal angles in the problem as

$cos\alpha=cos\theta cos\theta' + sin\theta sin\theta' cos(\phi-\phi')$

Although I think the formulation of the solution is correct the integral for the vector potential is very difficult for me to solve. I also have tried to integrate using MATHEMATICA but gave up after half an hour since it didn't give me an answer. Is there a better way to solve this problem?

  • $\begingroup$ Shouldn't it be $\vec{K} = - R \omega \sigma \sin^2 \theta' \hat{\phi}$? (Since $\hat{\phi} = \sin \phi' \hat{x} + \cos \phi' \hat{y}$.) $\endgroup$ Jun 9, 2022 at 14:35

3 Answers 3


You are missing the $\hat\phi \cdot \hat \phi'$ factor in the integrand. That is, the current is along $\hat \phi'$, and the resulting vector potential is along $\hat \phi$, but these are two different directions in general. So to calculate the $\hat \phi$ component of the vector potential you need to dot the integral over $\hat \phi'$ with $\hat \phi$.

In addition, I believe there is a trivial sign error: \begin{equation} \hat z \times (x\hat x + y \hat y + z\hat z) = x \hat z \times \hat x+y\hat z \times \hat y = x\hat y-y\hat x = r\sin\theta \hat \phi \,. \end{equation} You can continue and evaluate your integral with this change.

A faster way is to realize $x'$ and $y'$ can be written as linear combinations of $r' Y_{1m}(\theta',\phi')$, writing \begin{equation} \vec A(\vec r) = \frac{\mu_0}{4\pi} \int d^3r' \frac{\vec J(\vec r')}{|\vec r-\vec r'|} \end{equation} and then expanding \begin{equation} \frac{1}{|\vec r-\vec r'|} = \sum_{\ell m} \frac{4\pi}{2\ell+1} Y_{\ell m}(\theta',\phi')^*Y_{\ell m}(\theta,\phi) \frac{r_<^\ell}{r_>^{\ell+1}} \end{equation} we see that after integrating over the solid angle, using orthogonality of the spherical harmonics, we get only $\ell=1$ terms with the same angular function of the $\theta$, $\phi$ coordinates that we had for the $\theta',\phi'$ coordinates. Therefore the integration is trivial and \begin{equation} \vec A (r,\theta,\phi) = \hat \phi \frac{\mu_0 R^3\omega\sigma_0\sin\theta}{3}\left \{ \begin{array}{cc} \frac{r}{R^2} & r< R\\ \\ \frac{R}{r^2} & r> R\\ \end{array} \right . \end{equation}

  • $\begingroup$ Thanks for the answer. As far as I can understand in order to calculate the integral using spherical harmonics we need to express $sin(\theta')$ in the basis of spherical harmonics. I think there is no close form solution for that, right? $\endgroup$
    – Ali Pedram
    Jun 8, 2022 at 7:24
  • $\begingroup$ The correct $A_\phi$ integral has $\cos(\phi-\phi') = \cos(\phi)\cos(\phi')+\sin(\phi)\sin(\phi')$ in the integrand. $\sin(\theta')\cos(\phi')$ and $\sin(\theta')\sin(\phi')$ are linear combinations of $Y_{11}$ and $Y_{1,-1}$. $\endgroup$
    – user200143
    Jun 8, 2022 at 16:25

It seems very difficult to obtain analytically integrated formulas. By the way, in Landau and Lifshitz book (See PROBLEM 2 of page 125 in "Electrodynamics of Contnuous Media"), the analytical formulas for "magnetic field of a linear current flowing in a circle" is given. In the case of axisymmetry, it would be easy to obtain a practical method of calculation if one allows for numerical integration rather than a complete analytical expression.

Define the k: $$ k^2=\frac{4(R\text{sin}\theta)r}{(R\text{sin}\theta+r)^2+(z-R\text{cos}\theta)^2}, $$ where the observation point is $(r,z)$ (Note, the definition of $r$ differs from the figure). The vector potential from the ribbon region $(\theta,\theta+d\theta)$ is expressed as $$ dA_{\phi}(r,z,\theta)=\frac{\mu_0dI}{\pi k}\sqrt{\frac{R\text{sin}\theta}{r}}\left[(1-\frac{1}{2}k^2)K(k)-E(k)\right]\;\;\;\cdots(1) $$ where $K(k)$ and $E(k)$ are complete eliptic integral of the first and second kind. Since the electric current of the ribbon is $$ dI=\sigma(\theta) R d\theta\:v =\sigma(\theta) R d\theta(\omega R\text{sin}\theta) =R^2\omega\sigma(\theta)\text{sin}\theta d\theta, \;\;\;\cdots(2) $$ substituting (2) into (1), and make integration, we get, $$ A_\phi(r,z)=\int_0^{\pi}dA_{\phi}\;\;\;\cdots(3) $$ The (3) is easy to calculate numerically provided the observation position $(r,z)$ is not on the spherical surface.


Since you ask "Is there a better way to solve this problem?", here's one:

Instead of calculating the vector potential, you can instead calculate a magnetic scalar potential $\psi$ defined such that $\vec{B} = - \vec{\nabla} \psi$. This is permissible at all points with $r \neq R$, since at these points we have $\mu_0 \vec{J} = \vec{\nabla} \times \vec{B} = 0$. Moreover, since $\vec{\nabla} \cdot \vec{B} = 0$, we have $\nabla^2 \psi = 0 $, allowing us to use the machinery (familiar to us from electrostatics) of axially symmetric solutions to Laplace's equation. I will give an outline of the procedure for this problem below, but I will skip a lot of steps, which I encourage you to fill in yourself. (Also, I did this quickly, and it's entirely possible that I've dropped a sign or a factor of 2 somewhere in here.)

The general solution to Laplace's equation for $\psi$ (assuming good behavior at the origin and at infinity) will be $$ \psi = \begin{cases} \displaystyle\sum_{\ell=0}^\infty A_\ell \left( \frac{r}{R}\right)^\ell P_\ell(\cos \theta) & r < R \\ \displaystyle \sum_{\ell=0}^\infty B_\ell \left( \frac{R}{r}\right)^{\ell+1} P_\ell(\cos \theta) & r > R \end{cases} \tag{1} $$ The boundary conditions are different from the electrostatic case, though; we must have $$ \hat{r} \cdot (\vec{\nabla} \psi_\text{int} - \vec{\nabla} \psi_\text{ext} ) = 0 \qquad \hat{r} \times (\vec{\nabla} \psi_\text{int} - \vec{\nabla} \psi_\text{ext} ) = -\mu_0 \vec{K}. \tag{2} $$ The first boundary condition (it can be shown) implies that $$ B_\ell = - \frac{\ell}{\ell+1} A_\ell. \tag{3} $$ The second one (combined with Eq. (3)) yields $$ \sum_{\ell = 0}^\infty \frac{A_\ell}{\ell + 1} P_\ell^1(\cos \theta) = R \mu_0 K_\phi(\theta) \tag{4} $$ where $P_\ell^1(\cos \theta)$ is an associated Legendre function. The orthogonality relations for associated Legendre functions then imply that $$ A_\ell = \frac{2\ell + 1}{2 \ell} R \mu_0 \int_0^\pi P_\ell^1(\cos \theta) K_\phi(\theta) \sin \theta \, d \theta. $$ You can then calculate as many $A_\ell$ coefficients as you like. Unfortunately, for the case you've described, with $K_\phi(\theta) \propto \sin^2 (\theta)$, there are infinitely many non-zero coefficients, so an exact closed-form solution is (probably) not possible using this technique; but you can get as close as you like.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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