Clarify that the radiation pattern of a nonaxisymmetric dipole integrated in the azimuthal direction is rotationally invariant

doc/docs/Python_Tutorials/Cylindrical_Coordinates.md

@@ -867,7 +867,7 @@ A point-dipole source at $r_0 > 0$ can be represented as a Dirac delta function
 
 Simulating a point-dipole source involves two parts: (1) perform a series of simulations for $m = 0, 1, 2, ..., M$ for some cutoff $M$ of the Fourier-series expansion (the solutions for $\pm m$ are simply complex conjugates), and (2) because of power orthogonality, sum the results from each $m$-simulation in post processing, where the $m > 0$ terms are multiplied by two to account for the $-m$ solutions. This procedure is described in more detail below.
 
-Physically, the *total* field $E(x,y,z)$ is a sum of $E_m(r,z)e^{im\phi}$ terms, one for the solution at each $m$ (similarly for $H$). Computing the total Poynting flux, however, involves integrating $\Re [E \times H^*]$ over a surface that includes an integral over $\phi$ in the range $[0,2\pi]$. The key point is that the cross terms $E_mH^*_ne^{i(m-n)\phi}$ integrate to zero due to Fourier orthogonality. **The total Poynting flux is therefore a sum of the Poynting fluxes calculated separately for each $m$.**
+Physically, the *total* field $E(x,y,z)$ is a sum of $E_m(r,z)e^{im\phi}$ terms, one for the solution at each $m$ (similarly for $H$). Computing the total Poynting flux, however, involves integrating $\Re [E \times H^*]$ over a surface that includes an integral over $\phi$ in the range $[0,2\pi]$. (This is equivalent to arranging a *series* of dipoles with linear polarization and infinitesimal spacing along the circumference of a circle with fixed radius $r_0$. The radiation pattern from such a dipole arrangement is in consequence rotationally invariant.) The key point is that the cross terms $E_mH^*_ne^{i(m-n)\phi}$ integrate to zero due to Fourier orthogonality. **The total Poynting flux is therefore a sum of the Poynting fluxes calculated separately for each $m$.**
 
 A note regarding the source polarization at $r > 0$. The $\hat{x}$ polarization in 3d (the "in-plane" polarization) corresponds to the $\hat{r}$ polarization in cylindrical coordinates. An $\hat{r}$-polarized point-dipole source involves $\hat{r}$-polarized point sources in the $m$-simulations. Even though $\hat{r}$ is in fact $\phi$-dependent, $\hat{r}$ is only evaluated at $\phi = 0$ because of $\delta(\phi)$. $\hat{r}$ is therefore equivalent to $\hat{x}$. This property does not hold for an $\hat{x}$-polarized point source at $r = 0$ (where $\delta(\phi)$ is replaced by $1/2\pi$): in that case, we write $\hat{x} = \hat{r}\cos(\phi) - \hat{\phi}\sin(\phi)$, and the $\sin$ and $\cos$ terms yield simulations for $m = \pm 1$. See also [Tutorial/Scattering Cross Section of a Finite Dielectric Cylinder](#scattering-cross-section-of-a-finite-dielectric-cylinder) which demonstrates setting up a linearly polarized planewave using a similar approach. However, in practice, a single $\hat{r}$-polarized point source at $r = 0$ is necessary for $m = \pm 1$, because that gives a circularly polarized source that emits the same power as a linearly polarized source.