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__

- 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
- 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 \( 1 / r^{ D - 2 } \)

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