Wednesday, February 17, 2016

Electromagnetic properties of a charged rotating sphere

I calculate some properties of the electromagnetic field of a charged rotating sphere.
The set up
Charge is glued uniformly on a sphere, which is then made to rotate around the \(z\)-axis. If I use spherical coordinates \begin{align*} x&= r \sin\theta\cos\phi\\ y& = r \sin\theta\sin\phi\\ z& = r \cos\theta \end{align*} then the sphere is the set of points \( x^2 + y^2 + z^2 = R^2 \) and the source is \begin{equation*} J^{\mu}\partial_{\mu} = \rho\ \delta(r - R) \left( \partial_t + \omega \partial_{\phi}\right) \end{equation*} with \( \omega \) the angular frequency of the rotation and \( \rho \) the charge density on the sphere. The total charge on the sphere is thus \( q = 4 \pi R^2 \rho \)

Magnetic moment
The magnetic moment is defined by the antisymmetric matrix
\begin{equation*} M_{ij} = \int d^3 x \ x_i J_j
A straightforward calculation using \( \partial_{\phi}  = x \partial_ y - y \partial_x \) gives
M_{xy} = - M_{yx} =  \frac{4\pi}{3} \rho \omega R^4
The other components are zero.

The electromagnetic field
I calculate the electromagnetic field by solving Maxwell's equations \begin{equation}\label{eq:20160214a} \frac{1}{\sqrt{|g|}} \partial_{\nu} \left(\sqrt{|g|} F^{\mu\nu} \right)=J^{\mu} \end{equation}

I use the ansatz
A = f(r) dt + g(r) \omega \sin^2\theta d \phi
\end{equation*} If I plug this ansatz into \eqref{eq:20160214a}, I get
\frac{(r^2 f')'}{r^2}&=  \rho \delta(r - R)\\ - \frac{g''}{r^2} + 2 \frac{g}{r^4} &=\rho \delta(r - R) \end{align*}

These equations are easily solved \begin{align*} f(r) &= \left\{ \begin{array}{ll} -R \rho & \text{if}\quad r < R \\ - \dfrac{R^2 \rho}{r} &\text{if}\quad r > R\end{array} \right.\\ g(r) &= \left\{ \begin{array}{ll} \dfrac{\rho R}{3} r^2 & \text{if}\quad r < R \\ \dfrac{\rho R^4}{3 r} &\text{if}\quad r > R\end{array} \right.\\ \end{align*}

The electromagnetic field inside the sphere
Inside the sphere, the potential is
\begin{equation*} A = - R \rho\ dt + \frac{1}{3} \omega \rho R\ r^2 \sin^2 \theta d\phi \end{equation*} thus \begin{align*} F& = \frac{1}{3} \omega \rho R\ d\left( r^2 \sin^2 \theta \right) d\phi \\ &= \frac{2}{3} \omega \rho R\ dx dy \end{align*} Inside the sphere, the electric field is thus zero, and the magnetic field is homogeneous and points in the \( z \)-direction.

The electromagnetic field outside the sphere
Outside the sphere \begin{equation*} A = - \dfrac{R^2 \rho}{r} \ dt + \dfrac{\rho R^4}{3 r}\ \omega \sin^2 \theta\ d\phi \end{equation*} The first term is the potential of a point particle with charge \( q \). The second term is the vector potential of a pure magnetic dipool with magnetic moment \( M_{ij} \), namely \begin{equation*} A_j dx^j = \frac{1}{4 \pi} \frac{1}{r^3} x_i M_{ij} dx^j \end{equation*}

The momentum of the electromagnetic field
The momentum of the electromagnetic field is defined by \begin{equation*} P^{\mu} = \int d^3 x \ T^{0 \mu} \end{equation*} Here, \( \mu \) are Euclidean indices and \( T^{\mu\nu} \) is the stress tensor \begin{equation*} T_{\mu \nu} = - \left( F_{\mu}^{\ \ \beta} F_{\beta \nu} + \frac{1}{4} g_{\mu \nu} F^2\right) \end{equation*} A straightforward integration gives \begin{equation*} P^0 = E = \frac{q^2}{8 \pi R} + \frac{q^2 R \omega^2}{36 \pi} \end{equation*} The other components \( P^i \) are zero.

The angular momentum of the electromagnetic field
The angular momentum of the electromagnetic field is defined by \begin{equation*} L^{\mu\nu} = \int d^3 x\ x^{\mu} \ T^{0\nu} - \mu\leftrightarrow\nu \end{equation*} A straightforward integration gives \begin{equation*} L_{xy} = - L_{yx} = \frac{q^2 R \omega }{36 \pi } \end{equation*} The other components are zero.


  • Griffiths calculates the electromagnetic field of a rotating charged sphere with the Biot-Savart law, see example 5.11 in his book
  • I wrote down all these formulas explicitly because I want to see what they look like in \( D = 4+1 \).

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