--- jupytext: formats: ipynb,md:myst text_representation: extension: .md format_name: myst format_version: 0.13 jupytext_version: 1.15.2 kernelspec: display_name: Python 3 language: python name: python3 --- ```{code-cell} ipython3 :tags: [hide-cell] import mmf_setup;mmf_setup.nbinit() import logging;logging.getLogger('matplotlib').setLevel(logging.CRITICAL) %matplotlib inline import numpy as np, matplotlib.pyplot as plt ``` :::{margin} **Prerequisites:** Please review {ref}`sec:ODEs`, {ref}`sec:SturmLiouville`, and {ref}`sec:OrthogonalPolynomials`. ::: (sec:LegendreFunctions)= # 15: Legendre Functions The [Legendre functions][] $P_{\lambda}(x)$ and $Q_{\lambda}(x)$ are the generalization of the [Legendre polynomials][] to non-integer $\lambda$ and the related Sturm Liouville problem. The [Legendre polynomials][] satisfy (see also {ref}`sec:OrthogonalPolynomials`) * Interval: $x \in [-1, 1]$. * Weight: $W(x) = 1$. :::{margin} Thus, we should normalize the orthonormal unit vectors $\ket{P_l}$ as: \begin{gather*} \braket{x|P_l} = \sqrt{\frac{2l+1}{2}}P_{l}(x). \end{gather*} ::: * Normalization (leading coefficient is $x^{l}$ with coefficient 1): \begin{gather*} \int_{-1}^{1} P_k(x)P_l(x)\d{x} = \frac{2}{2l+1}\delta_{kl}. \end{gather*} * Completeness: :::{margin} Note this is just $\sum_{l}\ket{P_l}\!\bra{P_l} = \mat{1}$. ::: \begin{gather*} \sum_{l=0}^{\infty} \frac{2l+1}{2}P_{l}(x) P_{l}(y) = \delta(x-y). \end{gather*} * Generating function: \begin{gather*} \frac{1}{\sqrt{1 - 2xt + t^2}} = \sum_{n=0}^{\infty}P_{l}(x)t^l. \end{gather*} :::{margin} Try to derive the following by differentiating the generating function. ::: * Recurrence Relation: \begin{align*} P_{0}(x) &= 1, \\ P_{1}(x) &= x, \\ (l+1)P_{l+1}(x) &= (2l+1)xP_{l}(x) - lP_{l-1}(x).\tag{15.18} \end{align*} * Rodrigues' formula: \begin{gather*} P_{l}(x) = \frac{1}{2^l l!}\diff[l]{}{x}(x^2-1)^l.\tag{15.30} \end{gather*} :::{margin} Derive this from Rodrigues' formula using Cauchy's integral formula for computing the derivatives. ::: * Schläfli Integral: \begin{gather*} P_{l}(z) = \frac{1}{2^{l}}\oint_C \frac{(w^2 -1)^l}{(w-z)^{l+1}}\frac{\d{w}}{2\pi \I} \end{gather*} where the contour is closed, encloses the pole $w=z$, and avoids any branch cuts (see below). * The [Legendre polynomials][] satisfy the following second-order homogeneous linear differential equation (see also {ref}`sec:ODEs`, {ref}`sec:SturmLiouville`): \begin{gather*} (1-x^2)y'' - 2xy' +l(l+1)y = 0, \qquad y(x) = P_{l}(x) \end{gather*} The [Legendre functions][] $P_{\lambda}(z)$ extend these to non-integer $l$. This presents the following issues: * The functions are no-longer orthogonal. * The generating function relation should still hold, but now with appropriately defined fractional derivatives. Similarly with Rodrigues' formula. * The recurrence relations likely hold, but need new based cases for all non-integer streams. * The Schläfli integral is promising, but we must now be careful of the contour since the non-integer power introduced branch cuts. * The differential equation remains valid. and satisfy \begin{gather*} -\nabla^2 \psi \end{gather*} ## Associated Legendre Functions The associated functions $P^{m}_{l}(x)$ add an extra piece to the differential equation: \begin{gather*} (1-x^2)y'' - 2xy' + \left(l(l+1) - \frac{m^2}{1-x^2}\right)y = 0, \qquad y(x) = P^{m}_{l}(x). \end{gather*} They can be found by differentiating: \begin{gather*} P^{m}_{l}(x) = (-1)^{m}(1-x^2)^{m/2} \diff[m]{}{x}P_{l}(x), \qquad m \geq 0,\\ P^{-m}_{l}(x) = (-1)^{m}\frac{(l-m)!}{(l+m)!}P^{m}_{l}(x). \end{gather*} :::{margin} E.g., \begin{gather*} P^{1}_{1}(x) = -\sqrt{\frac{1-x^2}},\\ P^{2}_{2}(x) = 3(1-x^2). \end{gather*} ::: When $m$ is even, the set of polynomials $\bigl\{P^{m}_{l}(x) \quad \mid\quad l\in \{m, m+1, \cdots\}\bigr\}$ forms a completed orthogonal basis. For odd $m$, we have a similar basis, but the functions are no longer polynomials. For each value of $0 \leq m\leq l$, we have a new set of orthogonal polynomials: * Interval: $x \in [-1, 1]$. * Weight: $W(x) = 1$. :::{margin} Thus, we should normalize the orthonormal unit vectors $\ket{P_l}$ as: \begin{gather*} \braket{x|P^{m}_l} = \sqrt{\frac{2(l+m)!}{(2l+1)(l-m)!}}P^{m}_{l}(x). \end{gather*} ::: * Orthogonality and Normalization (leading coefficient is $x^{n}$ with coefficient 1): \begin{gather*} \int_{-1}^{1} P^{m}_{k}(x)P^{m}_{l}(x)\d{x} = \frac{2(l+m)!}{(2l+1)(l-m)!}\delta_{kl}. \end{gather*} Note that they also satisfy the following orthogonality for fixed $l$: \begin{gather*} \int_{-1}^{1} \frac{P^{m}_{l}(x)P^{m}_{l}(x)}{1-x^2}\d{x} = \frac{(l+m)!}{m(l-m)!}\delta_{mn}. \end{gather*} * Parity: \begin{gather*} P^{m}_{l}(-x) = (-1)^{l-m}P^{m}_{l}(x). \end{gather*} * Recurrence Relation: (one of many!) \begin{align*} (l-m+1)P^{m}_{l+1}(x) &= (2l+1)xP^{m}_{l}(x) - (l+m)P^{m}_{l-1}(x).\tag{15.88} \end{align*} * Rodrigues' formula: \begin{gather*} P^{m}_{l}(x) = \frac{(-1)^{m}}{2^l l!}(1-x^2)^{m/2}\diff[l+m]{}{x}(x^2-1)^l. \end{gather*} * Schläfli Integral: \begin{gather*} P^{m}_{l}(z) = \frac{(-1)^{m}(l+m)!}{2^{l}l!}(1-z^2)^{m/2} \oint_C \frac{(w^2 - 1)^{l}}{(w-z)^{l+m+1}}\frac{\d{w}}{2\pi \I} \end{gather*} where the contour is closed, encloses the pole $w=z$, and avoids any branch cuts (see below). *(MMF: I am not certain yet that the same branch trick works here if $m$ is not integer since the numerator and denominator have different powers.)* ### Schläfli Integral Countour :::{margin} See e.g. {cite}`Szmytkowski:2006, Szmytkowski:2007, Whittaker:2021`. ::: ```{code-cell} :tags: [margin, hide-input] plt.rcParams['mathtext.fontset'] = 'cm' from matplotlib_inline.backend_inline import set_matplotlib_formats set_matplotlib_formats('svg') rng = np.random.default_rng(seed=2) th = np.linspace(0, 2*np.pi, 500) n = np.arange(-5, 6) dr = (np.exp(1j * n[:, None] * th[None, :]) * (rng.random(size=len(n)*2)-0.5).view(dtype=complex)[:, None]).sum(axis=0) z = 1.5 + 1j w0 = (z + 1)/2 r = 2*abs(z-w0) C = ((r + 0.1*dr.real)*np.exp(1j*th) + w0) eta = 1/np.linspace(1, 0, 200)[:-1] br = (1+eta*z)/(z+eta) fig, ax = plt.subplots(figsize=(4, 2)) ax.plot(C.real, C.imag, lw=2) ax.plot([-1, 1, z.real], [0, 0, z.imag], 'ko', ms=10) ax.plot(br.real, br.imag, '-k', lw=3) ax.plot([-3, -1], [0, 0], '-k', lw=3) ax.text(z.real-0.3, z.imag + 0.1, "$z$", size=15) ax.grid(True) ax.set(xlim=(-1.5, 2.5), aspect=1, xticks=[-1, 0, 1, 2], yticks=[-1, 0, 1], xlabel=r"$\mathrm{Re}(w)$", ylabel=r"$\mathrm{Im}(w)$"); ``` The Legendre function $P_{l}(z)$ has the following Schläfli contour integral representation: \begin{gather*} P_{l}(z) = \frac{1}{2^{l}}\oint_{C} \frac{(w^2-1)^l}{(w-z)^{l+1}}\frac{\d{w}}{2\pi \I} \end{gather*} with the contour shown on the right after choosing the branch cuts as shown. [Legendre functions]: [Legendre polynomials]: (sec:MultipoleExpansion)= ## 15.3: Multipole Expansion