= Solution
Invert the two-dimensional <Fourier transform> from part (c):
$$
H(k)=\frac1{a_0^2n_e^2L}\int_{\mathbb R^2}C_{\mathrm{RM}}(s)e^{-i\mathbf k_\perp\cdot\mathbf s}\,d^2s.
$$
In <polar coordinates>, the angular integral is $2\pi J_0(ks)$, where $J_0$ is a <Bessel function of the first kind>. The <Fourier-Hankel normalization for an isotropic spectrum> therefore yields
$$
\boxed{H(k)=\frac{2\pi}{a_0^2n_e^2L}\int_0^\infty s\,J_0(ks)C_{\mathrm{RM}}(s)\,ds.}
$$
This <Hankel inversion of a rotation-measure correlation> recovers the scalar <spectral tensor> coefficient. To obtain the shell-integrated <magnetic energy spectrum>, take the <trace> of the three-dimensional tensor:
$$
\mathbb E|\mathbf B|^2=\int\frac{d^3k}{(2\pi)^3}\,2H(k)=\frac1{\pi^2}\int_0^\infty k^2H(k)\,dk.
$$
With <magnetic energy> density $B^2/(8\pi)$ in <Gaussian units>, the one-dimensional convention $\int_0^\infty E_B(k)\,dk=\mathbb E|\mathbf B|^2/(8\pi)$ gives $\boxed{E_B(k)=k^2H(k)/(8\pi^3)}$. A spectrum normalized instead to $\mathbb E|\mathbf B|^2/2$ has coefficient $k^2H(k)/(2\pi^2)$; the physical normalization must be stated.
Back to article page
