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diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex index 6b23ce5..bff91ef 100644 --- a/buch/papers/kugel/spherical-harmonics.tex +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -1,13 +1,724 @@ -% vim:ts=2 sw=2 et spell: +% vim:ts=2 sw=2 et spell tw=80: -\section{Spherical Harmonics} +\section{Construction of the Spherical Harmonics} -\subsection{Eigenvalue Problem in Spherical Coordinates} +\kugeltodo{Review text, or rewrite if preliminaries becomes an addendum} -\subsection{Properties} +We finally arrived at the main section, which gives our chapter its name. The +idea is to discuss spherical harmonics, their mathematical derivation and some +of their properties and applications. -\subsection{Recurrence Relations} +The subsection \ref{} \kugeltodo{Fix references} will be devoted to the +Eigenvalue problem of the Laplace operator. Through the latter we will derive +the set of Eigenfunctions that obey the equation presented in \ref{} +\kugeltodo{reference to eigenvalue equation}, which will be defined as +\emph{Spherical Harmonics}. In fact, this subsection will present their +mathematical derivation. + +In the subsection \ref{}, on the other hand, some interesting properties +related to them will be discussed. Some of these will come back to help us +understand in more detail why they are useful in various real-world +applications, which will be presented in the section \ref{}. + +One specific property will be studied in more detail in the subsection \ref{}, +namely the recursive property. The last subsection is devoted to one of the +most beautiful applications (In our humble opinion), namely the derivation of a +Fourier-style series expansion but defined on the sphere instead of a plane. +More importantly, this subsection will allow us to connect all the dots we have +created with the previous sections, concluding that Fourier is just a specific +case of the application of the concept of orthogonality. Our hope is that after +reading this section you will appreciate the beauty and power of generalization +that mathematics offers us. + +\subsection{Eigenvalue Problem} +\label{kugel:sec:construction:eigenvalue} + +\begin{figure} + \centering + \includegraphics{papers/kugel/figures/tikz/spherical-coordinates} + \caption{ + Spherical coordinate system. Space is described with the free variables $r + \in \mathbb{R}_0^+$, $\vartheta \in [0; \pi]$ and $\varphi \in [0; 2\pi)$. + \label{kugel:fig:spherical-coordinates} + } +\end{figure} + +From Section \ref{buch:pde:section:kugel}, we know that the spherical Laplacian +in the spherical coordinate system (shown in Figure +\ref{kugel:fig:spherical-coordinates}) is is defined as +\begin{equation*} + \sphlaplacian := + \frac{1}{r^2} \frac{\partial}{\partial r} \left( + r^2 \frac{\partial}{\partial r} + \right) + + \frac{1}{r^2} \left[ + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( + \sin\vartheta \frac{\partial}{\partial\vartheta} + \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2} + \right]. +\end{equation*} +But we will not consider this algebraic monstrosity in its entirety. As the +title suggests, we will only care about the \emph{surface} of the sphere. This +is for many reasons, but mainly to simplify reduce the already broad scope of +this text. Concretely, we will always work on the unit sphere, which just means +that we set $r = 1$ and keep only $\vartheta$ and $\varphi$ as free variables. +Now, since the variable $r$ became a constant, we can leave out all derivatives +with respect to $r$ and substitute all $r$'s with 1's to obtain a new operator +that deserves its own name. + +\begin{definition}[Surface spherical Laplacian] + \label{kugel:def:surface-laplacian} + The operator + \begin{equation*} + \surflaplacian := + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( + \sin\vartheta \frac{\partial}{\partial\vartheta} + \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}, + \end{equation*} + is called the surface spherical Laplacian. +\end{definition} -\section{Series Expansions in \(C(S^2)\)} +In the definition, the subscript ``$\partial S$'' was used to emphasize the +fact that we are on the spherical surface, which can be understood as being the +boundary of the sphere. But what does it actually do? To get an intuition, +first of all, notice the fact that $\surflaplacian$ have second derivatives, +which means that this a measure of \emph{curvature}; But curvature of what? To +get an even stronger intuition we will go into geometry, were curvature can be +grasped very well visually. Consider figure \ref{kugel:fig:curvature} where the +curvature is shown using colors. First we have the curvature of a curve in 1D, +then the curvature of a surface (2D), and finally the curvature of a function on +the surface of the unit sphere. +\begin{figure} + \centering + \includegraphics[width=.3\linewidth]{papers/kugel/figures/tikz/curvature-1d} + \hskip 5mm + \includegraphics[width=.3\linewidth]{papers/kugel/figures/povray/curvature} + \hskip 5mm + \includegraphics[width=.3\linewidth]{papers/kugel/figures/povray/spherecurve} + \caption{ + \kugeltodo{Fix alignment / size, add caption. Would be nice to match colors.} + \label{kugel:fig:curvature} + } +\end{figure} + +Now that we have defined an operator, we can go and study its eigenfunctions, +which means that we would like to find the functions $f(\vartheta, \varphi)$ +that satisfy the equation +\begin{equation} \label{kugel:eqn:eigen} + \surflaplacian f = -\lambda f. +\end{equation} +Perhaps it may not be obvious at first glance, but we are in fact dealing with a +partial differential equation (PDE) \kugeltodo{Boundary conditions?}. If we +unpack the notation of the operator $\nabla^2_{\partial S}$ according to +definition +\ref{kugel:def:surface-laplacian}, we get: +\begin{equation} \label{kugel:eqn:eigen-pde} + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( + \sin\vartheta \frac{\partial f}{\partial\vartheta} + \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 f}{\partial\varphi^2} + + \lambda f = 0. +\end{equation} +Since all functions satisfying \eqref{kugel:eqn:eigen-pde} are the +\emph{eigenfunctions} of $\surflaplacian$, our new goal is to solve this PDE. +The task may seem very difficult but we can simplify it with a well-known +technique: \emph{the separation Ansatz}. It consists in assuming that the +function $f(\vartheta, \varphi)$ can be factorized in the following form: +\begin{equation} + f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi). +\end{equation} +In other words, we are saying that the effect of the two independent variables +can be described using the multiplication of two functions that describe their +effect separately. This separation process was already presented in section +\ref{buch:pde:section:kugel}, but we will briefly rehearse it here for +convenience. If we substitute this assumption in +\eqref{kugel:eqn:eigen-pde}, we have: +\begin{equation*} + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( + \sin\vartheta \frac{\partial \Theta(\vartheta)}{\partial\vartheta} + \right) \Phi(\varphi) + + \frac{1}{\sin^2 \vartheta} + \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2} + \Theta(\vartheta) + + \lambda \Theta(\vartheta)\Phi(\varphi) = 0. +\end{equation*} +Dividing by $\Theta(\vartheta)\Phi(\varphi)$ and introducing an auxiliary +variable $m^2$, the separation constant, yields: +\begin{equation*} + \frac{1}{\Theta(\vartheta)}\sin \vartheta \frac{d}{d \vartheta} \left( + \sin \vartheta \frac{d \Theta}{d \vartheta} + \right) + + \lambda \sin^2 \vartheta + = -\frac{1}{\Phi(\varphi)} \frac{d^2\Phi(\varphi)}{d\varphi^2} + = m^2, +\end{equation*} +which is equivalent to the following system of 2 first order differential +equations (ODEs): +\begin{subequations} + \begin{gather} + \frac{d^2\Phi(\varphi)}{d\varphi^2} = -m^2 \Phi(\varphi), + \label{kugel:eqn:ode-phi} \\ + \sin \vartheta \frac{d}{d \vartheta} \left( + \sin \vartheta \frac{d \Theta}{d \vartheta} + \right) + + \left( \lambda - \frac{m^2}{\sin^2 \vartheta} \right) + \Theta(\vartheta) = 0 + \label{kugel:eqn:ode-theta}. + \end{gather} +\end{subequations} +The solution of \eqref{kugel:eqn:ode-phi} is easy to find: The complex +exponential is obviously the function we are looking for. So we can directly +write the solutions +\begin{equation} \label{kugel:eqn:ode-phi-sol} + \Phi(\varphi) = e^{i m \varphi}, \quad m \in \mathbb{Z}. +\end{equation} +The restriction that the separation constant $m$ needs to be an integer arises +from the fact that we require a $2\pi$-periodicity in $\varphi$ since the +coordinate systems requires that $\Phi(\varphi + 2\pi) = \Phi(\varphi)$. +Unfortunately, solving \eqref{kugel:eqn:ode-theta} is not as straightforward, +actually, it is quite difficult, and the process is so involved that it will +require a dedicated section of its own. + +\subsection{Legendre Functions} + +\begin{figure} + \centering + \kugelplaceholderfig{.8\textwidth}{5cm} + \caption{ + \kugeltodo{Why $z = \cos \vartheta$.} + } +\end{figure} + +To solve \eqref{kugel:eqn:ode-theta} we start with the substitution $z = \cos +\vartheta$ \kugeltodo{Explain geometric origin with picture}. The operator +$\frac{d}{d \vartheta}$ becomes +\begin{equation*} + \frac{d}{d \vartheta} + = \frac{dz}{d \vartheta}\frac{d}{dz} + = -\sin \vartheta \frac{d}{dz} + = -\sqrt{1-z^2} \frac{d}{dz}, +\end{equation*} +since $\sin \vartheta = \sqrt{1 - \cos^2 \vartheta} = \sqrt{1 - z^2}$, and +then \eqref{kugel:eqn:ode-theta} becomes +\begin{align*} + \frac{-\sqrt{1-z^2}}{\sqrt{1-z^2}} \frac{d}{dz} \left[ + \left(\sqrt{1-z^2}\right) \left(-\sqrt{1-z^2}\right) \frac{d \Theta}{dz} + \right] + + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0, + \\ + \frac{d}{dz} \left[ (1-z^2) \frac{d \Theta}{dz} \right] + + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0, + \\ + (1-z^2)\frac{d^2 \Theta}{dz} - 2z\frac{d \Theta}{dz} + + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0. +\end{align*} +By making two final cosmetic substitutions, namely $Z(z) = \Theta(\cos^{-1}z)$ +and $\lambda = n(n+1)$, we obtain what is known in the literature as the +\emph{associated Legendre equation of order $m$}: \nocite{olver_introduction_2013} +\begin{equation} \label{kugel:eqn:associated-legendre} + (1 - z^2)\frac{d^2 Z}{dz^2} + - 2z\frac{d Z}{dz} + + \left( n(n + 1) - \frac{m^2}{1 - z^2} \right) Z(z) = 0, + \quad + z \in [-1; 1], m \in \mathbb{Z}. +\end{equation} + +Our new goal has therefore become to solve +\eqref{kugel:eqn:associated-legendre}, since if we find a solution for $Z(z)$ we +can perform the substitution backwards and get back to our eigenvalue problem. +However, the associated Legendre equation is not any easier, so to attack the +problem we will look for the solutions in the easier special case when $m = 0$. +This reduces the problem because it removes the double pole, which is always +tricky to deal with. In fact, the reduced problem when $m = 0$ is known as the +\emph{Legendre equation}: +\begin{equation} \label{kugel:eqn:legendre} + (1 - z^2)\frac{d^2 Z}{dz^2} + - 2z\frac{d Z}{dz} + + n(n + 1) Z(z) = 0, + \quad + z \in [-1; 1]. +\end{equation} + +The Legendre equation is a second order differential equation, and therefore it +has 2 independent solutions, which are known as \emph{Legendre functions} of the +first and second kind. For the scope of this text we will only derive a special +case of the former that is known known as the \emph{Legendre polynomials}, since +we only need a solution between $-1$ and $1$. + +\begin{lemma}[Legendre polynomials] + \label{kugel:thm:legendre-poly} + The polynomial function + \[ + P_n(z) = \sum^{\lfloor n/2 \rfloor}_{k=0} + \frac{(-1)^k}{2^n s^k!} \frac{(2n - 2k)!}{(n - k)! (n-2k)!} z^{n - 2k} + \] + is the only finite solution of the Legendre equation + \eqref{kugel:eqn:legendre} when $n \in \mathbb{Z}$ and $z \in [-1; 1]$. +\end{lemma} +\begin{proof} + This results is derived in section \ref{kugel:sec:proofs:legendre}. +\end{proof} + +Since the Legendre \emph{polynomials} are indeed polynomials, they can also be +expressed using the hypergeometric functions described in section +\ref{buch:rekursion:section:hypergeometrische-funktion}, so in fact +\begin{equation} + P_n(z) = {}_2F_1 \left( \begin{matrix} + n + 1, & -n \\ \multicolumn{2}{c}{1} + \end{matrix} ; \frac{1 - z}{2} \right). +\end{equation} +Further, there are a few more interesting but not very relevant forms to write +$P_n(z)$ such as \emph{Rodrigues' formula} and \emph{Laplace's integral +representation} which are +\begin{equation*} + P_n(z) = \frac{1}{2^n n!} \frac{d^n}{dz^n} (z^2 - 1)^n, + \qquad \text{and} \qquad + P_n(z) = \frac{1}{\pi} \int_0^\pi \left( + z + \cos\vartheta \sqrt{z^2 - 1} + \right) \, d\vartheta +\end{equation*} +respectively, both of which we will not prove (see chapter 3 of +\cite{bell_special_2004} for a proof). Now that we have a solution for the +Legendre equation, we can make use of the following lemma patch the solutions +such that they also become solutions of the associated Legendre equation +\eqref{kugel:eqn:associated-legendre}. + +\begin{lemma} \label{kugel:thm:extend-legendre} + If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre}, + then + \begin{equation*} + Z^m_n(z) = (1 - z^2)^{m/2} \frac{d^m}{dz^m}Z_n(z) + \end{equation*} + solves the associated Legendre equation \eqref{kugel:eqn:associated-legendre}. + \nocite{bell_special_2004} +\end{lemma} +\begin{proof} + See section \ref{kugel:sec:proofs:legendre}. +\end{proof} + +What is happening in lemma \ref{kugel:thm:extend-legendre}, is that we are +essentially inserting a square root function in the solution in order to be able +to reach the parts of the domain near the poles at $\pm 1$ of the associated +Legendre equation, which is not possible only using power series +\kugeltodo{Reference book theory on extended power series method.}. Now, since +we have a solution in our domain, namely $P_n(z)$, we can insert it in the lemma +obtain the \emph{associated Legendre functions}. + +\begin{definition}[Ferrers or associated Legendre functions] + \label{kugel:def:ferrers-functions} + The functions + \begin{equation} + P^m_n (z) = (1-z^2)^{\frac{m}{2}}\frac{d^{m}}{dz^{m}} P_n(z) + = \frac{1}{2^n n!}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n + \end{equation} + are known as Ferrers or associated Legendre functions. +\end{definition} + +\kugeltodo{Discuss $|m| \leq n$.} + +\if 0 +The constraint $|m|<n$, can be justified by considering Eq.\eqref{kugel:eq:associated_leg_func}, in which the derivative of degree $m+n$ is present. A derivative to be well defined must have an order that is greater than zero. Furthermore, it can be seen that this derivative is applied on a polynomial of degree $2n$. As is known from Calculus 1, if you derive a polynomial of degree $2n$ more than $2n$ times, you get zero, which is a trivial solution in which we are not interested.\newline +We can thus summarize these two conditions by writing: +\begin{equation*} + \begin{rcases} + m+n \leq 2n &\implies m \leq n \\ + m+n \geq 0 &\implies m \geq -n + \end{rcases} |m| \leq n. +\end{equation*} +The set of functions in Eq.\eqref{kugel:eq:sph_harm_0} is named \emph{Spherical Harmonics}, which are the eigenfunctions of the Laplace operator on the \emph{spherical surface domain}, which is exactly what we were looking for at the beginning of this section. +\fi + +\subsection{Spherical Harmonics} + +Finally, we can go back to solving our boundary value problem we started in +section \ref{kugel:sec:construction:eigenvalue}. We had left off in the middle +of the separation, were we had used the Ansatz $f(\vartheta, \varphi) = +\Theta(\vartheta) \Phi(\varphi)$ to find that $\Phi(\varphi) = e^{im\varphi}$, +and we were solving for $\Theta(\vartheta)$. As you may recall, previously we +performed the substitution $z = \cos \vartheta$. Now we can finally to bring back the +solution to the associated Legendre equation $P^m_n(z)$ into the $\vartheta$ +domain and combine it with $\Phi(\varphi)$ to get the full result: +\begin{equation*} + f(\vartheta, \varphi) + = \Theta(\vartheta)\Phi(\varphi) + = P^m_n (\cos \vartheta) e^{im\varphi}. +\end{equation*} +This family of functions, which recall are the solutions of the eigenvalue +problem of the surface spherical Laplacian, are the long anticipated +\emph{complex spherical harmonics}, and they are usually denoted with +$Y^m_n(\vartheta, \varphi)$. + +\begin{definition}[Spherical harmonics] + \label{kugel:def:spherical-harmonics} + The functions + \begin{equation*} + Y^m_n (\vartheta, \varphi) = P^m_n(\cos \vartheta) e^{im\varphi}, + \end{equation*} + where $m, n \in \mathbb{Z}$ and $|m| < n$ are called (unnormalized) spherical + harmonics. +\end{definition} + +\begin{figure} + \centering + \kugelplaceholderfig{\textwidth}{.8\paperheight} + \caption{ + \kugeltodo{Big picture with the first few spherical harmonics.} + \label{kugel:fig:spherical-harmonics} + } +\end{figure} + +\kugeltodo{Describe how they look like with fig. +\ref{kugel:fig:spherical-harmonics}} + +\subsection{Orthogonality of $P_n$, $P^m_n$ and $Y^m_n$} + +We shall now discuss an important property of the spherical harmonics: they form +an orthogonal system. And since the spherical harmonics contain the Ferrers or +associated Legendre functions, we need to discuss their orthogonality first. +But the Ferrers functions themselves depend on the Legendre polynomials, so that +will be our starting point. + +\begin{lemma} For the Legendre polynomials $P_n(z)$ and $P_k(z)$ it holds that + \label{kugel:thm:legendre-poly-ortho} + \begin{equation*} + \int_{-1}^1 P_n(z) P_k(z) \, dz + = \frac{2}{2n + 1} \delta_{nk} + = \begin{cases} + \frac{2}{2n + 1} & \text{if } n = k, \\ + 0 & \text{otherwise}. + \end{cases} + \end{equation*} +\end{lemma} +\begin{proof} + To start, consider the fact that the Legendre equation + \eqref{kugel:eqn:legendre}, of which two distinct Legendre polynomials + $P_n(z)$ and $P_k(z)$ are a solution ($n \neq k$), can be rewritten in the + following form: + \begin{equation} + \frac{d}{dz} \left[ + \left( 1 - z^2 \right) \frac{dZ}{dz} + \right] + n(n+1) Z(z) = 0. + \end{equation} + So we rewrite the Legendre equations for $P_n(z)$ and $P_k(z)$: + \begin{align*} + \frac{d}{dz} \left[ + \left( 1 - z^2 \right) \frac{dP_n}{dz} + \right] + n(n+1) P_n(z) &= 0, + & + \frac{d}{dz} \left[ + \left( 1 - z^2 \right) \frac{dP_k}{dz} + \right] + k(k+1) P_k(z) &= 0, + \end{align*} + then we multiply the former by $P_k(z)$ and the latter by $P_n(z)$ and + subtract the two to get + \begin{equation*} + \frac{d}{dz} \left[ + \left( 1 - z^2 \right) \frac{dP_n}{dz} + \right] P_k(z) + n(n+1) P_n(z) P_k(z) + - + \frac{d}{dz} \left[ + \left( 1 - z^2 \right) \frac{dP_k}{dz} + \right] P_n(z) - k(k+1) P_k(z) P_n(z) = 0. + \end{equation*} + By grouping terms, making order and integrating with respect to $z$ from $-1$ + to 1 we obtain + \begin{gather} + \int_{-1}^1 \left\{ + \frac{d}{dz} \left[ + \left( 1 - z^2 \right) \frac{dP_n}{dz} + \right] P_k(z) + - + \frac{d}{dz} \left[ + \left( 1 - z^2 \right) \frac{dP_k}{dz} + \right] P_n(z) - k(k+1) P_k(z) P_n(z) + \right\} \,dz \nonumber \\ + + \left[ n(n+1) - k(k+1) \right] \int_{-1}^1 P_k(z) P_n(z) \, dz = 0. + \label{kugel:thm:legendre-poly-ortho:proof:1} + \end{gather} + Since by the product rule + \begin{equation*} + \frac{d}{dz} \left[ (1 - z^2) \frac{dP_k}{dz} P_n(z) \right] + = + \frac{d}{dz} \left[ (1 - z^2) \frac{dP_n}{dz} \right] P_k(z) + + (1 - z^2) \frac{dP_n}{dz} \frac{dP_k}{dz}, + \end{equation*} + we can simplify the first term in + \eqref{kugel:thm:legendre-poly-ortho:proof:1} to get + \begin{gather*} + \int_{-1}^1 \left\{ + \frac{d}{dz} \left[ (1 - z^2) \frac{dP_k}{dz} P_n(z) \right] + - \cancel{(1 - z^2) \frac{dP_n}{dz} \frac{dP_k}{dz}} + - \frac{d}{dz} \left[ (1 - z^2) \frac{dP_n}{dz} P_k(z) \right] + + \cancel{(1 - z^2) \frac{dP_k}{dz} \frac{dP_n}{dz}} + \right\} \, dz \\ + = \int_{-1}^1 \frac{d}{dz} \left\{ (1 - z^2) \left[ + \frac{dP_k}{dz} P_n(z) - \frac{dP_n}{dz} P_k(z) + \right] \right\} \, dz + = (1 - z^2) \left[ + \frac{dP_k}{dz} P_n(z) - \frac{dP_n}{dz} P_k(z) + \right] \Bigg|_{-1}^1, + \end{gather*} + which always equals 0 because the product contains $1 - z^2$ and the bounds + are at $\pm 1$. Thus, of \eqref{kugel:thm:legendre-poly-ortho:proof:1} only + the second term remains and the equation becomes + \begin{equation*} + \left[ n(n+1) - k(k+1) \right] \int_{-1}^1 P_k(z) P_n(z) \, dz = 0. + \end{equation*} + By dividing by the constant in front of the integral we have our first result. + Now we need to show that when $n = k$ the integral equals $2 / (2n + 1)$. + % \begin{equation*} + % \end{equation*} + \kugeltodo{Finish proof. Can we do it without the generating function of + $P_n$?} +\end{proof} + +In a similarly algebraically tedious fashion, we can also continue to check for +orthogonality for the Ferrers functions $P^m_n(z)$, since they are related to +$P_n(z)$ by a $m$-th derivative, and obtain the following result. + +\begin{lemma} For the associated Legendre functions + \label{kugel:thm:associated-legendre-ortho} + \begin{equation*} + \int_{-1}^1 P^m_n(z) P^{m}_{n'}(z) \, dz + = \frac{2(m + n)!}{(2n + 1)(n - m)!} \delta_{nn'} + = \begin{cases} + \frac{2(m + n)!}{(2n + 1)(n - m)!} + & \text{if } n = n', \\ + 0 & \text{otherwise}. + \end{cases} + \end{equation*} +\end{lemma} +\begin{proof} + To show that the expression equals zero when $n \neq n'$ we can perform + exactly the same steps as in the proof of lemma + \ref{kugel:thm:legendre-poly-ortho}, so we will not repeat them here and prove + instead only the case when $n = n'$. + \kugeltodo{Finish proof, or not? I have to look and decide if it is + interesting enough.} +\end{proof} + +By having the orthogonality relations of the Legendre functions we can finally +show that spherical harmonics are also orthogonal under the following inner +product: + +\begin{definition}[Inner product in $S^2$] + \label{kugel:def:inner-product-s2} + For 2 complex valued functions $f(\vartheta, \varphi)$ and $g(\vartheta, + \varphi)$ on the surface of the sphere the inner product is defined to be + \begin{equation*} + \langle f, g \rangle + = \int_{0}^\pi \int_0^{2\pi} + f(\vartheta, \varphi) \overline{g(\vartheta, \varphi)} + \sin \vartheta \, d\varphi \, d\vartheta. + \end{equation*} +\end{definition} + + +\begin{theorem} For the (unnormalized) spherical harmonics + \label{kugel:thm:spherical-harmonics-ortho} + \begin{align*} + \langle Y^m_n, Y^{m'}_{n'} \rangle + &= \int_{0}^\pi \int_0^{2\pi} + Y^m_n(\vartheta, \varphi) \overline{Y^{m'}_{n'}(\vartheta, \varphi)} + \sin \vartheta \, d\varphi \, d\vartheta + \\ + &= \frac{4\pi}{2n + 1} \frac{(m + n)!}{(n - m)!} \delta_{nn'} \delta_{mm'} + = \begin{cases} + \frac{4\pi}{2n + 1} \frac{(m + n)!}{(n - m)!} + & \text{if } n = n' \text{ and } m = m', \\ + 0 & \text{otherwise}. + \end{cases} + \end{align*} +\end{theorem} +\begin{proof} + We will begin by doing a bit of algebraic maipulaiton: + \begin{align*} + \int_{0}^\pi \int_0^{2\pi} + Y^m_n(\vartheta, \varphi) \overline{Y^{m'}_{n'}(\vartheta, \varphi)} + \sin \vartheta \, d\varphi \, d\vartheta + &= \int_{0}^\pi \int_0^{2\pi} + e^{im\varphi} P^m_n(\cos \vartheta) + e^{-im'\varphi} P^{m'}_{n'}(\cos \vartheta) + \, d\varphi \sin \vartheta \, d\vartheta + \\ + &= \int_{0}^\pi + P^m_n(\cos \vartheta) P^{m'}_{n'}(\cos \vartheta) + \int_0^{2\pi} e^{i(m - m')\varphi} + \, d\varphi \sin \vartheta \, d\vartheta + . + \end{align*} + First, notice that the associated Legendre polynomials are assumed to be real, + and are thus unaffected by the complex conjugation. Then, we can see that when + $m = m'$ the inner integral simplifies to $\int_0^{2\pi} 1 \, d\varphi$ which + equals $2\pi$, so in this case the expression becomes + \begin{equation*} + 2\pi \int_{0}^\pi + P^m_n(\cos \vartheta) P^{m'}_{n'}(\cos \vartheta) + \sin \vartheta \, d\vartheta + = -2\pi \int_{1}^{-1} P^m_n(z) P^{m'}_{n'}(z) \, dz + = \frac{4\pi(m + n)!}{(2n + 1)(n - m)!} \delta_{nn'}, + \end{equation*} + where in the second step we performed the substitution $z = \cos\vartheta$; + $d\vartheta = \frac{d\vartheta}{dz} dz= - dz / \sin \vartheta$, and then we + used lemma \ref{kugel:thm:associated-legendre-ortho}. We are allowed to use + the lemma because $m = m'$. + + Now we just need look at the case when $m \neq m'$. Fortunately this is + easier: the inner integral is $\int_0^{2\pi} e^{i(m - m')\varphi} d\varphi$, + or in other words we are integrating a complex exponetial over the entire + period, which always results in zero. Thus, we do not need to do anything and + the proof is complete. +\end{proof} + +These proofs for the various orthogonality relations were quite long and +algebraically tedious, mainly because they are ``low level'', by which we mean +that they (arguably) do not rely on very abstract theory. However, if we allow +ourselves to use the more abstract Sturm Liouville theory discussed in chapters +\ref{buch:integrale:subsection:sturm-liouville-problem} and \kugeltodo{reference +to chapter 17 of haddouche and Löffler} the proofs can become ridiculously +short. Let's do for example lemma \ref{kugel:thm:associated-legendre-ortho}. + +\begin{proof}[ + Shorter proof of lemma \ref{kugel:thm:associated-legendre-ortho} + ] + The associated Legendre polynomials, of which we would like to prove an + orthogonality relation, are the solution to the associated Legendre equation, + which we can write as $LZ(z) = 0$, where + \begin{equation*} + L = \frac{d}{dz} (1 - z^2) \frac{d}{dz} + + n(n+1) - \frac{m^2}{1 - z^2}. + \end{equation*} + Notice that $L$ is in fact a Sturm-Liouville operator of the form + \begin{equation*} + L = \frac{1}{w(z)} \left[ + \frac{d}{dz} p(z) \frac{d}{dz} - \lambda + q(z) + \right], + \end{equation*} + if we let $w(z) = 1$, $p(z) = (1 - z^2 )$, $q(z) = -m^2 / (1 - z^2)$, and + $\lambda = -n(n+1)$. By the theory of Sturm-Liouville operators, we know that + the each solution of the problem $LZ(z) = 0$, namely $P^m_n(z)$, is orthogonal + to every other solution that has a different $\lambda$. In our case $\lambda$ + varies with $n$, so $P^m_n(z)$ with different $n$'s are orthogonal to each + other. +\end{proof} + +But that was still rather informative and had a bit of explanation, which is +terrible. Real snobs, such as Wikipedia contributors, some authors and +regrettably sometimes even ourselves, would write instead: + +\begin{proof}[ + Infuriatingly short proof of lemma \ref{kugel:thm:associated-legendre-ortho} + ] + The associated Legendre polynomials are solutions of the associated Legendre + equation which is a Sturm-Liouville problem and are thus orthogonal to each + other. The factor in front Kronecker delta is left as an exercise to the + reader. +\end{proof} + +Lemma \ref{kugel:thm:legendre-poly-ortho} has a very similar +proof, while the theorem \ref{kugel:thm:spherical-harmonics-ortho} for the +spherical harmonics is proved by the following argument. The spherical harmonics +are the solutions to the eigenvalue problem $\surflaplacian f = -\lambda f$, +which as discussed in the previous section is solved using separation. So to +prove their orthogonality using the Sturm-Liouville theory we argue that +\begin{equation*} + \surflaplacian = L_\vartheta L_\varphi \iff + \surflaplacian f(\vartheta, \varphi) + = L_\vartheta \Theta(\vartheta) L_\varphi \Phi(\varphi), +\end{equation*} +then we show that both $L_\vartheta$ and $L_\varphi$ are both Sturm-Liouville +operators (we just did the former in the shorter proof above). Since both are +Sturm-Liouville operators their combination, the surface spherical Laplacian, is +also a Sturm-Liouville operator, which then implies orthogonality. + +\subsection{Normalization and the Phase Factor} + +At this point we have shown that the spherical harmonics form an orthogonal +system, but in many applications we usually also want a normalization of some +kind. For example the most obvious desirable property could be for the spherical +harmonics to be ortho\emph{normal}, by which we mean that $\langle Y^m_n, +Y^{m'}_{n'} \rangle = \delta_{nn'}$. To obtain orthonormality, we simply add an +ugly normalization factor in front of the previous definition +\ref{kugel:def:spherical-harmonics} as follows. + +\begin{definition}[Orthonormal spherical harmonics] + \label{kugel:def:spherical-harmonics-orthonormal} + The functions + \begin{equation*} + Y^m_n(\vartheta, \varphi) + = \sqrt{\frac{2n + 1}{4\pi} \frac{(n-m)!}{(m+n)!}} + P^m_n(\cos \vartheta) e^{im\varphi} + \end{equation*} + where $m, n \in \mathbb{Z}$ and $|m| < n$ are the orthonormal spherical + harmonics. +\end{definition} + +Orthornomality is very useful, but it is not the only common normalization that +is found in the literature. In physics, geomagnetism to be more specific, it is +common to use the so called Schmidt semi-normalization (or sometimes also called +quasi-normalization). + +\begin{definition}[Schmidt semi-normalized spherical harmonics] + \label{kugel:def:spherical-harmonics-schmidt} + The Schmidt semi-normalized spherical harmonics are + \begin{equation*} + Y^m_n(\vartheta, \varphi) + = \sqrt{2 \frac{(n - m)!}{(n + m)!}} + P^m_n(\cos \vartheta) e^{im\varphi} + \end{equation*} + where $m, n \in \mathbb{Z}$ and $|m| < n$. +\end{definition} + +Additionally, there is another quirk in the literature that should be mentioned. +In some other branches of physics such as seismology and quantum mechanics there +is a so called Condon-Shortley phase factor $(-1)^m$ in front of the square root +in the definition of the normalized spherical harmonics. It is yet another +normalization that is added for physical reasons that are not very relevant to +our discussion, but we mention this potential source of confusion since many +numerical packages (such as \texttt{SHTOOLS} \kugeltodo{Reference}) offer an +option to add or remove it from the computation. + +Though, for our purposes we will mostly only need the orthonormal spherical +harmonics, so from now on, unless specified otherwise when we say spherical +harmonics or write $Y^m_n$, we mean the orthonormal spherical harmonics of +definition \ref{kugel:def:spherical-harmonics-orthonormal}. + +\subsection{Recurrence Relations} + +\section{Series Expansions in $L^2(S^2)$} + +We have now reached a point were we have all of the tools that are necessary to +build something truly amazing: a general series expansion formula for functions +on the surface of the sphere. Using the jargon: we will now see that the +spherical harmonics together with the inner product of definition +\ref{kugel:def:inner-product-s2} +\begin{equation*} + \langle f, g \rangle + = \int_{0}^\pi \int_0^{2\pi} + f(\vartheta, \varphi) \overline{g(\vartheta, \varphi)} + \sin \vartheta \, d\varphi \, d\vartheta +\end{equation*} +form a Hilbert space over the space of complex valued $L^2$ functions $S^2 \to +\mathbb{C}$. We will see later that this fact is very consequential and is +extremely useful for many types of applications. If the jargon was too much, no +need to worry, we will now go back to normal words and explain it again in more +detail. + +\subsection{Spherical Harmonics Series} + +To talk about a \emph{series expansion} we first need a series, so we shall +build one using the spherical harmonics. + +\begin{definition}[Spherical harmonic series] + \begin{equation*} + \hat{f}(\vartheta, \varphi) + = \sum_{n \in \mathbb{Z}} \sum_{m \in \mathbb{Z}} + c_{m,n} Y^m_n(\vartheta, \varphi) + \end{equation*} +\end{definition} + +\subsection{Fourier on $S^2$} |