Tuesday, February 2, 2016

Charged rotating sphere in Maxwell-Chern-Simons theory

I calculate the electromagnetic field of a charged rotating sphere in five dimensional Maxwell-Chern-Simons theory.
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.
From the graph above, one can see that the electric field is not zero inside the sphere. This is different from the pure Maxwell case, then the electric field is exactly zero inside the sphere, as can be easily seen from Gauss's law.

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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