Charged rotating sphere in five dimensions

I calculate the electromagnetic field of a charged rotating sphere in 4 + 1 dimensions. Griffiths calculates the magnetic field of a charged rotating sphere in 3 + 1 dimensions in example 5.11 in his book. In this post, I perform a similar calculation but in 4+1 dimensions.
The sphere in this post is the set of points $x_1^2 + x_2^2 + x_3^2 + x_4^2 =R^2$. It is called a 3-sphere, the radius is $R$. I use the following normalization for the Maxwell action $$S = - \frac{1}{4} \int d^5 x \sqrt{|g|} F_{\mu\nu}F^{\mu\nu} + \int d^5 x \sqrt{|g|} j^{\mu} A_{\mu}$$ with $g$ the determinant of the metric, $j^{\mu}$ the source and $F$ the electromagnetic field $$\begin{equation*} F = \frac{1}{2} F_{\mu\nu} dx^{\mu} dx^{\nu} = dA \end{equation*}$$ Digression about electromagnetic fields in higher dimensions. In 3+1 dimensions, electromagnetic theory is usually described with an electric field $\vec E$ and a magnetic field $\vec B$. When one does relativistic calculations, one combines $\vec E$ and $\vec B$ in an anti-symmetric $4 \times 4$ matrix $F_{\mu\nu}$, called the electromagnetic field tensor. This sounds more difficult than it is: just put the field $\vec E$ in the first column of $F_{\mu\nu}$, and sprinkle the components of $\vec B$ in the $3 \times 3$ block in the bottom right of $F_{\mu\nu}$. 
In 4 + 1 dimensions, the natural place to start building an electromagnetic theory is a $5 \times 5$ matrix $F_{\mu\nu}$; if one decomposes this in electric and magnetic components, the electric field is still a vector $E_i$ (with four components now), but the magnetic field is not a vector anymore, but a $4 \times 4$ anti-symmetric matrix $B_{ij}$. This is again easy to see: the electric field can be found in the first column of $F_{\mu\nu}$, the magnetic field $B_{ij}$ is the $4 \times 4$ matrix in the lower bottom right corner of $F_{\mu\nu}$.

I parametrize the space $\mathbb{R}^4$ as $$\begin{align*} x_1 & = r \sin\theta\ \cos\phi\\ x_2 & = r \sin\theta\ \sin\phi\\ x_3 & = r \cos\theta\ \cos\psi\\ x_4 & = r \cos\theta\ \sin\psi \end{align*}$$ In these coordinates (they are known as Hopf coordinates), the metric in $4+1$ dimensions is $$\begin{equation*} ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2 +r^2 \cos^2\theta d\psi^2 \end{equation*}$$
Digression about rotation in higher dimensions. In three (space) dimensions, a rotation is specified by an angular velocity vector; this is a (pseudo) vector which points along the axis of rotation and has length equal to the angular frequency. This does not work in four dimensions: after one fixes an axis, there would still be three dimensions left over for the body to move in; leaving too much freedom to define the rotation. Therefore, in four (and higher) dimensions, rotation is specified by an anti-symmetric matrix, or, equivalently, by specifying planes and angular frequencies in those planes. Notice that one could have chosen to specify rotations in three dimensions with an anti-symmetric $3 \times 3$ matrix as well; this is usually not done.

I let the sphere rotate in the $x_1, x_2$ plane with angular frequency $\omega_1$ and in the $x_3, x_4$ plane with angular frequency $\omega_2$. The current $j$ is thus $$\begin{equation*} j = \sigma \delta(r - R) \left( \partial_t + \omega_1 \partial_{\phi} + \omega_2 \partial_{\psi} \right) \end{equation*}$$ with $\sigma$ the charge density on the sphere. Maxwell's equations are $$\begin{equation}\label{eq:20160122a} \frac{1}{\sqrt{|g|}} \partial_{\nu} \left(\sqrt{|g|} F^{\mu\nu} \right)=j^{\mu} \end{equation}$$ I use as ansatz for the electromagnetic potential $$\begin{equation*} A = V(r) dt + h(r) \left( \omega_1 \sin^2\theta d\phi + \omega_2\cos^2\theta d\psi \right) \end{equation*}$$ The equations $\eqref{eq:20160122a}$ are then easily solved and give for $r \le R$ $$\begin{equation*} V(r) = -\frac{\sigma R}{2} \quad\text{and}\quad h(r)= \frac{\sigma R}{4} r^2 \end{equation*}$$ and for $r \ge R$ $$\begin{equation*} V(r) = -\frac{\sigma R^3}{2 r^2} \quad\text{and}\quad h(r) =\frac{\sigma R^5}{4 r^2} \end{equation*}$$

Observations

1. For $r < R$ $$\begin{equation*} A = -\frac{\sigma R}{2} dt + \frac{\sigma R}{4} r^2 \left( \omega_1 \sin^2\theta d\phi + \omega_2\cos^2\theta d\psi \right) \end{equation*}$$ and thus $$\begin{equation*} F = dA = \frac{\sigma R}{4} \left(\omega_1 d(r^2 \sin^2\theta) d\phi +\omega_2 d(r^2 \cos^2\theta) d\psi \right) \end{equation*}$$ With a bit of calculation, one sees that this is equal to $$\begin{equation*} F = \frac{\sigma R}{2} \left(\omega_1 dx_1 dx_2 +\omega_2 dx_3 dx_4 \right) \end{equation*}$$ Hence, inside the sphere the electric field is zero, and the magnetic field is homogeneous and proportional to the rotation matrix $\omega_1 dx_1 dx_2 +\omega_2 dx_3 dx_4$. Both of these properties are exactly as in 3+1 dimensions, see Griffiths, example 5.11
2. For $r > R$ the electric potential $V$ is proportional to $1 /r^2$. This is as expected: in $D$ space dimensions, the electric potential $V$ is proportional to $1 / r^{ D - 2 }$

In the next post, I will try to calculate again the electromagnetic field of a rotating sphere in five dimensions, not in Maxwell's theory, but in Maxwell-Chern-Simons theory.