Thursday, January 19, 2017

A magnetostatic exercise in 10 dimensions

I calculate the electromagnetic field generated by electrical currents in 10 spacetime dimensions (9 space and 1 time). The set up is as follows: the current flows down the positive $x_1$-axis, hits the origin and then spreads out isotropically in the $x_2 x_3 x_4$ subspace, see figure 1 and 2. I wanted to calculate this because in string theory a similar calculation is needed to obtain the Kalb-Ramond field generated by a string ending on a $D3$-brane [1]
Figure 1: view of the currents in $x_1 x_2 x_3$subspace

Figure 2: view of the currents in $x_1 x_2 x_5$subspace

Semi-infinite wire
I first calculate the field generated by a semi-infinite wire. Suppose the current $I$ flows from the origin to infinity along the positive $x_1$-axis. The Biot-Savart law in 10 dimensions is \begin{equation*} F_{\mu\nu}(x) = \frac{1}{| S^8 |} \int\! d^9 y\ \frac{j_{\mu}(y) (x_{\nu} - y_{\nu}) }{| x-y|^9} - (\mu \leftrightarrow \nu) \end{equation*} Here, $|S^8|$ is the area of an 8-dimensional sphere, and $j_{\mu}$ is the current. For the case at hand, the integration reduces to a one-dimensional integration and gives \begin{equation}\label{eq:20170118b} F(x) = \frac{I}{| S^8 |} dx_1 \frac{r dr}{s^8} \left( \frac{16}{35} + \frac{x_1}{r}- \frac{x_1^3}{r^3}+ \frac{3}{5}\frac{x_1^5}{r^5} - \frac{1}{7}\frac{x_1^7}{r^7} \right) \end{equation} with $s^2 = x_2^2 + \cdots + x_9^2$ and $r^2 = x_1^2 + \cdots + x_9^2$. As a check, this field trivially satisfies $dF =0$. Taking care with delta-functions, one can also calculate that [2] \begin{equation}\label{eq:20170118c} \partial_{\nu}F^{\mu\nu} = j^{\mu} - \frac{I}{| S^8 |} \frac{x^{\mu}}{r^9} \end{equation} with \begin{equation*} j(x) = I\ 1(x_1 \ge 0) \delta^8(x_2, \ldots, x_9)dx_1 \end{equation*} If the wire points along the $\vec{n}$ direction, (with $n$ a unit vector), then equation \eqref{eq:20170118b} is generalized to \begin{equation}\label{eq:20170118d} F(x) = \frac{I}{| S^8 |} n_i dx_i \frac{r dr}{s^8} \left( \frac{16}{35} + \frac{n\cdot x}{r}- \frac{(n\cdot x)^3}{r^3}+ \frac{3}{5}\frac{(n\cdot x)^5}{r^5} - \frac{1}{7}\frac{(n\cdot x)^7}{r^7} \right) \end{equation} with now $s^2 = r^2 - (n \cdot x)^2$ and still $r^2 = x_1^2 + \cdots + x_9^2$.
Secondly, I calculate the field generated by currents that start in the origin and spread out istropically in the $x_1 x_2 x_3$ subspace. This field can be calculated by integrating \eqref{eq:20170118d} over a 2-sphere. I found the integration somewhat painful, but the result is \begin{equation*} F(x) = \frac{I}{| S^8 |} u du \ \frac{1}{r^3 v^6} r dr \left( \frac{1}{3} - \frac{2}{5} \frac{ u^2}{r^2} + \frac{1}{7}\frac{u^4}{r^4} \right) \end{equation*} with $u^2 = x_1^2 + x_2^2 + x_3^3$ and $v^2 = r^2 - u^2$. One can calculate that $dF = 0 $ and \begin{equation*} \partial_{\nu}F^{\mu\nu} = j^{\mu} - \frac{I}{| S^8 |} \frac{x^{\mu}}{r^9} \end{equation*} with \begin{equation*} j(x) = \frac{I}{4 \pi} \delta^6(x_4, \ldots, x_9) \frac{du}{u^2} \end{equation*} Together
Adding those two results together, and changing the names of some coordinates, the field of the original exercise is \begin{equation*} F = - F_1 + F_2 \end{equation*} with \begin{align*} F_1 &= \frac{I}{| S^8 |} dx_1 \frac{r dr}{s^8} \left( \frac{16}{35} + \frac{x_1}{r}- \frac{x_1^3}{r^3}+ \frac{3}{5}\frac{x_1^5}{r^5} - \frac{1}{7}\frac{x_1^7}{r^7} \right)\\ F_2 &=\frac{I}{| S^8 |} u du \ \frac{1}{r^3 v^6} r dr \left( \frac{1}{3} - \frac{2}{5} \frac{ u^2}{r^2} + \frac{1}{7}\frac{u^4}{r^4} \right) \end{align*} where $r^2 = x_1^2 + \cdots + x_9^2$, $u^2 = x_2^2 + x_3^2 + x_4^3$, $s^2 = r^2 - x_1^2$ and $v^2 = r^2 - u^2$.

References and comments

[1] Section 16.3 in A First Course in String Theory, Zwiebach, 2009

[2] The first term in equation \eqref{eq:20170118c} is as expected. The second term however makes that Maxwell's equations are not satisfied. The reason is that the current in the semi-infinite wire is not physical because the current just appears at the origin. Further on, these extra terms will cancel when I calculate the electromagnetic field of the full set-up. So for the full set-up, Maxwell's equations will be satisfied.

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