The action of the Maxwell-Chern-Simons theory is \begin{equation*} S = - \frac{1}{4} \int d^5 x \sqrt{|g|} F_{\mu\nu}F^{\mu\nu} + \lambda \int A \wedge F \wedge F + \int d^5 x \sqrt{|g|} j^{\mu} A_{\mu} \end{equation*} The first term is the usual Maxwell action, the second term is the Chern-Simons term, \( \lambda \) is a dimensionless constant. The rest of the notation is explained in a previous post, where I calculated the field of a charged rotating sphere in the pure Maxwell theory. The field equations are \begin{equation}\label{eq:20160123a} \frac{1}{\sqrt{|g|}} \partial_{\nu} \left(\sqrt{|g|} F^{\mu\nu} \right)=j^{\mu} - \frac{3}{4} \lambda \epsilon^{\mu\kappa\nu\alpha\beta} F_{\kappa \nu} F_{\alpha \beta} \end{equation} Here, \( \epsilon^{\mu\kappa\lambda\alpha\beta} \) is the Levi-Civita tensor; some care is needed about signs and factors of \( \sqrt{|g|} \) appearing in this tensor. The main difference between \eqref{eq:20160123a} and the usual Maxwell's equations is that Maxwell's equations are linear, but \eqref{eq:20160123a} is not linear because of the quadratic term in \( F_{\mu \nu} \). This makes it much more difficult to solve equations. The source of the charged rotating sphere is \begin{equation*} J^{\mu}\partial_{\mu} = \rho \delta(r - R) \left( \partial_t + \omega \partial_{\phi} + \omega \partial_{\psi} \right) \end{equation*} with \( \rho \) the charge density on the sphere, \( \omega \) the angular frequency of the sphere and \( R \) its radius. In general, the sphere could have two angular frequencies \( \omega_1 \) and \( \omega_2 \), but I have taken them equal to simplify the calculations. As in the previous post, I use the ansatz \begin{equation*} A = V(r) dt + g(r) \left( \omega \sin^2\theta d \phi + \omega \cos^2\theta d \psi \right) \end{equation*} If I plug this ansatz into \eqref{eq:20160123a}, I get \begin{align} \frac{(r^3 V')'}{r^3}& = \rho \delta(r - R) + \frac{6 \lambda \omega^2 ( g^2 )'}{r^3}\label{eq:20160128a}\\ r g' + r^2 g'' &= 4 g + 12 r \lambda g V' - R^4 \rho \delta(r - R)\label{eq:20160128b} \end{align} Integrating \eqref{eq:20160128a} gives \begin{equation} V'(r) = h_0(r) + \frac{6 \lambda \omega^2 g^2(r)}{r^3} \end{equation} with \begin{equation}\label{eq:20160128c} h_0(r) = \frac{R^3 \rho}{r^3} \quad\text{if}\quad r > R\quad\text{and}\quad h_0(r) = 0\quad\text{if}\quad r < R \end{equation} Plugging \eqref{eq:20160128c} in \eqref{eq:20160128b} gives \begin{equation}\label{eq:20160128d} r g' + r^2 g'' = 4 g + 12 r \lambda g h_0 + 72 \lambda^2 \omega^2 \frac{g(r)^3}{r^2}- R^4 \rho \delta(r - R) \end{equation} One can also see that \eqref{eq:20160128b} leads to the following gluing condition at \( r = R\) \begin{equation}\label{eq:20160128e} g'(R+\epsilon) -g'(R-\epsilon) = -R^2 \rho \end{equation} I could not find an analytical expression for the solution of equation \eqref{eq:20160128d}. I will therefore solve \eqref{eq:20160128d} only for small \( \lambda \). The solution for small \( \lambda \) is \begin{align*} V'(r) &= \frac{3}{8} \lambda \rho ^2 r R^2 \omega ^2-\frac{3}{8} \lambda ^2 \left(\rho ^3 r R^3 \omega ^2\right)+O\left(\lambda ^3\right)\quad\text{for}\quad{r < R}\\ V'(r) &= \frac{\rho }{r^3}+\frac{3 \lambda \rho ^2 \omega ^2}{8 r^7}+\lambda ^2 \left(\frac{3 \rho ^3 \omega ^2}{4 r^9}-\frac{9 \rho ^3 \omega ^2}{8 r^7}\right)+O\left(\lambda ^3\right)\quad\text{for}\quad{r > R}\\ g(r) &= \frac{1}{4} \rho r^2 R-\frac{1}{8} \lambda \left(\rho ^2 r^2 R^2\right)+\frac{3}{160} \lambda ^2 \left(5 \rho ^3 r^4 R^3 \omega ^2-9 \rho ^3 r^2 R^5 \omega ^2+5 \rho ^3 r^2 R^3\right)+O\left(\lambda ^3\right)\quad\text{for}\quad{r < R}\\ g(r) &= \frac{\rho R^5}{4 r^2}+\lambda \left(\frac{\rho ^2 R^8}{4 r^4}-\frac{3 \rho ^2 R^6}{8 r^2}\right)+\lambda ^2 \left(\frac{3 \rho ^3 R^{15} \omega ^2}{160 r^8}+\frac{3 \rho ^3 R^{11}}{32 r^6}-\frac{3 \rho ^3 R^9}{8 r^4}-\frac{3 \rho ^3 R^9 \omega ^2}{32 r^2}+\frac{3 \rho ^3 R^7}{8 r^2}\right)+O\left(\lambda ^3\right)\quad\text{for}\quad{r > R}\\ \end{align*} One can continue this expansion to higher order in \( \lambda \), but the expressions become more and more messy.

Here is a graph of the function \( V'(r) \) as function of \( r \). I have chosen \( R = 1 \), \( \rho =1 \), \( \lambda = 0.2 \) and \( \omega = 1 \). In black I have plotted the function in the Maxwell case (in other words if \( \lambda = 0\) ); the red dotted line is the solution up to order \( \lambda \), the red dashed line is the solution up to \( \lambda^2 \). Because the dotted and dashed lines lie close to one another, presumably the approximation up to first order in \( \lambda \) is already quite accurate.

The graph below is a graph of the function \( g(r) \) as function of \( r \). Again, the dotted and dashed lines lie close to one another, presumably the approximation up to first order in \( \lambda \) is already quite accurate.

__Further comments__

- I also tried solving the differential equation in Mathematica using NDSolve. However, I get many numerical problems related to singularities and stiff equations. I have therefore not included those calculations in this post.
- I also tried finding the electromagnetic field of a magnetic dipole field in the Maxwell-Chern-Simons theory. Again, I did not find a closed form solution.
- I expand the functions \( g(r) \) and \( V'(r) \) to higher order in \( \lambda \) in another post.

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