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diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex index 6b23ce5..9349b61 100644 --- a/buch/papers/kugel/spherical-harmonics.tex +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -1,13 +1,977 @@ -% 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. -\section{Series Expansions in \(C(S^2)\)} +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} + +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)\footnote{ + Considering the fact that we are dealing with a PDE, + you may be wondering what are the boundary conditions. Well, since this eigenvalue problem is been developed on + the spherical surface (boundary of a sphere), the boundary in this case are empty, i.e no boundary condition has to be considered.}. +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 to 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, \quad |m|<n + \end{equation} + are known as Ferrers or associated Legendre functions. +\end{definition} +The constraint $|m|<n$, can be justified by considering eq.\eqref{kugel:eq:associated_leg_func}, where we differentiate $m+n$ times. We all know that a differentiation, to be well defined, must have an order that is greater than zero \kugeltodo{is that always true?}. 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, that would be a trivial solution. This is the power of zero: It is almost always a (boring) solution. + +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*} + +\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 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}, \quad |m|<n. +\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}, \quad |m|<n + \end{equation*} + where $m, n \in \mathbb{Z}$ 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 two 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 + \label{kugel:eq:spherical-harmonics-inner-prod} \\ + &= \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', \nonumber \\ + 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) \sin \vartheta \, d\vartheta + \int_0^{2\pi} e^{i(m - m')\varphi} + \, d\varphi. + \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 $m = m'$. After the just mentioned substitution we can write eq.\eqref{kugel:eq:spherical-harmonics-inner-prod} in this form + \begin{equation*} + \langle Y^m_n, Y^{m'}_{n'} \rangle_{\partial S} = \langle P^m_n, P^{m'}_{n'} \rangle_z \; \langle e^{im\varphi}, e^{-im'\varphi} \rangle_\varphi. + \end{equation*} + 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 exponential 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 the separation Ansatz. 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}\kugeltodo{replace x with z} +The idea of this subsection is to introduce first some recursive relations regarding the Associated Legendre Functions, defined in eq.\eqref{kugel:def:ferrers-functions}. Subsequently we will extend them, in order to derive recurrence formulas for the case of Spherical Harmonic functions as well. +\subsubsection{Associated Legendre Functions} +To start this journey, we can first write the following equations, which relate the Associated Legendre functions of different indeces $m$ and $n$ recursively: +\begin{subequations} + \begin{align} + P^m_n(z) &= \dfrac{1}{(2n+1)x} \left[ (m+n) P^m_{n-1}(z) + (n-m+1) P^m_{n+1}(z) \right] \label{kugel:eq:rec-leg-1} \\ + P^m_n(z) &= \dfrac{\sqrt{1-z^2}}{2mz} \left[ P^{m+1}_n(z) + [n(n+1)-m(m-1)] P^{m-1}_n(z) \right] \label{kugel:eq:rec-leg-2} \\ + P^m_n(z) &= \dfrac{1}{(2n+1)\sqrt{1-z^2}} \left[ P^{m+1}_{n+1}(z) - P^{m+1}_{n-1}(z) \right] \label{kugel:eq:rec-leg-3} \\ + P^m_n(z) &= \dfrac{1}{(2n+1)\sqrt{1-z^2}} \left[ (n+m)(n+m-1)P^{m-1}_{n-1}(z) - (n-m+1)(n-m+2)P^{m-1}_{n+1}(z) \right] \label{kugel:eq:rec-leg-4} + \end{align} +\end{subequations} +Much of the effort will be proving this bunch of equalities. Then, in the second part, where we will derive the recursion equations for $Y^m_n(\vartheta,\varphi)$, we will basically reuse the ones presented above. + +Maybe it is worth mentioning at least one use case for these relations: In some software implementations (that include lighting computations in computer graphics, antenna modelling softwares, 3-D modelling in medical applications, etc.) +they are widely used, as they lead to better numerical accuracy and computational cost lower by a factor of six\cite{usecase_recursion_paper}. +\begin{enumerate}[(i)] + \item + \begin{proof} + This is the relation that links the associated Legendre functions with the same $m$ index but different $n$. Using \ref{} \kugeltodo{search the general equation of recursion for orthogonal polynomials (is somewhere in the book)}, we have + \begin{equation*} + (n+1)P_{n+1}(z)-(2n+1)xP_n(z)+nP_{n-1}(z)=0, + \end{equation*} + that can be differentiated $m$ times, obtaining + \begin{equation}\label{kugel:eq:rec_1} + (n+1)\frac{d^mP_{n+1}}{dz^m}-(2n+1) \left[z \frac{d^m P_n}{dz^m}+ m\frac{d^{m-1}P_{n-1}}{dz^{m-1}} \right] + n\frac{d^m P_{n-1}}{dz^m}=0. + \end{equation} + To continue this derivation, we need the following relation: + \begin{equation}\label{kugel:eq:rec_2} + \frac{dP_{n+1}}{dz} - \frac{dP_{n-1}}{dz} = (2n+1)P_n. + \end{equation} + The latter will not be derived, because it suffices to use the definition of the Legendre Polynomials $P_n(x)$ to check it. + + We can now differentiate the just presented eq.\eqref{kugel:eq:rec_2} $m-1$ times, that will become + \begin{equation}\label{kugel:eq:rec_3} + \frac{d^mP_{n+1}}{dx^m} - \frac{d^mP_{n-1}}{dx^m} = (2n+1)\frac{d^{m-1}P_n}{dx^{m-1}}. + \end{equation} + Then, using eq.\eqref{kugel:eq:rec_3} in eq.\eqref{kugel:eq:rec_1}, we will have + \begin{equation}\label{kugel:eq:rec_4} + (n+1)\frac{d^mP_{n+1}}{dx^m}- (2n+1)\frac{d^mP_{n+1}}{dx^m} -m\left[\frac{d^m P_{n+1}}{dx^m}+ \frac{d^{m}P_{n-1}}{dx^m}\right] + n\frac{d^m P_{n-1}}{dx^m}=0. + \end{equation} + Finally, multiplying both sides by $(1-x^2)^{\frac{m}{2}}$ and simplifying the expression, we can rewrite eq.\eqref{kugel:eq:rec_4} in terms of $P^m_n(x)$, namely + \begin{equation*} + (n+1-m)P^m_{n+1}(x)-(2n+1)xP^m_n(x)+(m+n)P^m_{n-1}(x)=0, + \end{equation*} + that rearranged, will be + \begin{equation*} + (2n+1) x P^m_n(x)= (m+n) P^m_{n-1}(x) + (n-m+1) P^m_{n+1}(x). + \end{equation*} + \end{proof} + + \item + \begin{proof} + This relation, unlike the previous one, link three expression with the same $n$ index but different $m$. + + In the proof of Lemma \ref{kugel:lemma:sol_associated_leg_eq}, at some point we ran into this expression. + \begin{equation*} + (1-x^2)\frac{d^{m+2}P_n}{dx^{m+2}} - 2(m+1)x \frac{d^{m+1}P_n}{dx^{m+1}} + [n(n+1)-m(m+1)]\frac{d^mP_n}{dx^m} = 0, + \end{equation*} + that, if multiplied by $(1-x^2)^{\frac{m}{2}}$, will be + \begin{equation*} + (1-x^2)^{\frac{m}{2}+1}\frac{d^{m+2}P_n}{dx^{m+2}} - 2(m+1)x (1-x^2)^{\frac{m}{2}}\frac{d^{m+1}P_n}{dx^{m+1}} + [n(n+1)-m(m+1)](1-x^2)^{\frac{m}{2}}\frac{d^mP_n}{dx^m} = 0. + \end{equation*} + Therefore, as before, expressing it in terms of $P^m_n(x)$: + \begin{equation*} + P^{m+2}_n(x) - \frac{2(m+1)x}{\sqrt{1-x^2}}P^{m+1}_n(x) + [n(n+1)-m(m+1)]P^m_n(x)=0. + \end{equation*} + Further, we can adjust the indeces and terms, obtaining + \begin{equation*} + \frac{2mx}{\sqrt{(1-x^2)}} P^m_n(x) = P^{m+1}_n(x) + [n(n+1)-m(m-1)] P^{m-1}_n(x). + \end{equation*} + + \end{proof} + + \item + \begin{proof} + To derive this expression, we can multiply eq.\eqref{kugel:eq:rec_3} by $(1-x^2)^{\frac{m}{2}}$ and, as always, we could express it in terms of $P^m_n(x)$: + \begin{equation*} + P^m_{n+1}(x) - P^m_{n-1}(x) = (2n+1)\sqrt{1-x^2}P^{m-1}_n(x). + \end{equation*} + Afer that we can divide by $2n+1$ resulting in + \begin{equation}\label{kugel:eq:helper} + \frac{1}{2n+1}[P^m_{n+1}(x) - P^m_{n-1}(x)] = \sqrt{1-x^2}P^{m-1}_n(x). + \end{equation} + To conclude, we arrange the indeces differently: + \begin{equation*} + \sqrt{1-x^2}P^{m}_n(x)=\frac{1}{2n+1}[P^{m+1}_{n+1}(x) - P^{m+1}_{n-1}(x)]. + \end{equation*} + \end{proof} + + \item + \begin{proof} + For this proof we can rely on eq.\eqref{kugel:eq:rec-leg-1}, and therefore rewrite eq.\eqref{kugel:eq:rec-leg-2} as + \begin{equation*} + \frac{2m}{(2n+1)\sqrt{1-x^2}} \left[ (m+n)P^m_{n-1}(x) + (n-m+1)P^m_{n+1}(x) \right] = P^{m+1}_n(x) + [ n(n+1)-m(m-1) ]P^{m-1}_n(x). + \end{equation*} + Rewriting then $P^{m-1}_n(x)$ using eq.\eqref{kugel:eq:helper}, we will have + \begin{align*} + \frac{2m}{(2n+1)\sqrt{1-x^2}} &\left[ (m+n)P^m_{n-1}(x) + (n-m+1)P^m_{n+1}(x) \right] = P^{m+1}_n(x) \\ + &+ \frac{n(n+1)-m(m-1)}{(2n+1)\sqrt{1-x^2}} \left[ P^m_{n+1}(x)-P^m_{n-1}(x) \right]. + \end{align*} + The last equation, after some algebric rearrangements, it is easy to show that it is equivalent to + \begin{equation*} + \sqrt{1-x^2} P^m_n(x) = \dfrac{1}{2n+1} \left[ (n+m)(n+m-1)P^{m-1}_{n-1}(x) - (n-m+1)(n-m+2)P^{m-1}_{n+1}(x) \right] + \end{equation*} + \end{proof} + +\end{enumerate} + +\subsubsection{Spherical Harmonics} +The goal of this subsection's part is to apply the recurrence relations of the $P^m_n(z)$ functions to the Spherical Harmonics. +With some little adjustments we will be able to have recursion equations for them too. As previously written the most of the work is already done. Now it is only a matter of minor mathematical operations/rearrangements. + +We can start by listing all of them: +\begin{subequations} + \begin{align} + Y^m_n(\vartheta, \varphi) &= \dfrac{1}{(2n+1)\cos \vartheta} \left[ (m+n)Y^m_{n-1}(\vartheta, \varphi) + (m-n+1)Y^m_{n+1}(\vartheta, \varphi) \right] \label{kugel:eq:rec-sph_harm-1} \\ + Y^m_n(\vartheta, \varphi) &= \dfrac{\tan \vartheta}{2m}\left[ Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi} \right] \label{kugel:eq:rec-sph_harm-2} \\ + Y^m_n(\vartheta, \varphi) &= \dfrac{e^{-i\varphi}}{ (2n+1)\sin \vartheta } \left[ Y^{m+1}_{n+1}(\vartheta, \varphi) - Y^{m+1}_{n-1}(\vartheta, \varphi) \right] \label{kugel:eq:rec-sph_harm-3} \\ + Y^m_n(\vartheta, \varphi) &= \dfrac{e^{i\varphi}}{(2n+1)\sin \vartheta} \left[ (n+m)(n+m-1)Y^{m-1}_{n-1}(\vartheta, \varphi) - (n-m+1)(n-m+2)Y^{m-1}_{n+1}(\vartheta, \varphi) \right] \label{kugel:eq:rec-sph_harm-4} + \end{align} +\end{subequations} + +\begin{enumerate}[(i)] + \item + \begin{proof} + We can multiply both sides of equality in eq.\eqref{kugel:eq:rec-leg-1} by $e^{im \varphi}$ and perform the substitution $z=\cos \vartheta$. After a few simple algebraic steps, we will obtain the relation we are looking for + \end{proof} + \item + \begin{proof} + In this proof, as before, we can perform the substitution $z=\cos \vartheta$, and notice that $\sqrt{1-z^2}=\sin \vartheta$, hence, the relation in eq.\eqref{kugel:eq:rec-leg-2} will be + \begin{equation*} + \frac{2m \cos \vartheta}{\sin \vartheta} P^m_n(\cos \vartheta) = P^{m+1}_n(\cos \vartheta) + [n(n+1)-m(m-1)]P^{m-1}_n P^m_n(\cos \vartheta). + \end{equation*} + The latter, multiplied by $e^{im\varphi}$, becomes + \begin{align*} + \frac{2m \cos \vartheta}{\sin \vartheta} P^m_n(\cos \vartheta)e^{im\varphi} &= P^{m+1}_n(\cos \vartheta)e^{im\varphi} + [n(n+1)-m(m-1)]P^{m-1}_n P^m_n(\cos \vartheta)e^{im\varphi} \\ + &= P^{m+1}_n(\cos \vartheta)e^{i(m+1)\varphi}e^{-i\varphi} + [n(n+1)-m(m-1)]P^{m-1}_n (\cos \vartheta)e^{i(m-1)\varphi}e^{i\varphi} \\ + &= Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi}. + \end{align*} + Finally, after some ``cleaning'' + \begin{equation*} + Y^m_n(\vartheta, \varphi) = \frac{\tan \vartheta}{2m} \left[ Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi} \right] + \end{equation*} + \end{proof} + \item + \begin{proof} + Now we can consider eq.\eqref{kugel:eq:rec-leg-3}, and multiply it by $e^{im\varphi}$. After the usual substitution $z=\cos \vartheta$, we have + \begin{align*} + \sin \vartheta P^m_n(\cos \vartheta)e^{im\varphi} &= \dfrac{e^{im\varphi}}{2n+1}\left[ P^{m+1}_{n+1}(\cos \vartheta) - P^{m+1}_{n-1}(\cos \vartheta)\right] \\ + &= \dfrac{e^{-i\varphi}}{2n+1}\left[ P^{m+1}_{n+1}(\cos \vartheta)e^{i(m+1)\varphi} - P^{m+1}_{n-1}(\cos \vartheta)e^{i(m+1)\varphi}\right]. + \end{align*} + A few manipulations later, we will obtain + \begin{equation*} + Y^m_n(\vartheta, \varphi) = \frac{e^{-i\varphi}}{(2n+1)\sin \vartheta} \left[ Y^{m+1}_{n+1}(\vartheta, \varphi)-Y^{m+1}_{n-1}(\vartheta, \varphi) \right]. + \end{equation*} + \end{proof} + \item + \begin{proof} + This proof is very similar to the previous one. We just have to perform the substitution $z = \cos \vartheta$, as always. Secondly we can multiply the right side by $e^{im\varphi}$ and the left one too but in a different form, namely $e^{im\varphi}=e^{i(m-1)\varphi}e^{i\varphi}$. Then it is only a question of recalling the definition of $Y^m_n(\vartheta, \varphi)$. + \end{proof} +\end{enumerate} + +\section{Series Expansions in $L^2(S^2)$} +We have now reach a point where we have all the tools that are necessary to build something truly amazing: a general series expansion formula for +function on the surface of the sphere. +Before starting we want to recall the definition of the inner product on the spherical surface 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*} +To be a bit technical we can say that the set of spherical harmonic functions, toghether with the inner product just showed, +form something that we call Hilbert Space\footnote{For more details about Hilber space you can take a look in section \ref{kugel:sec:preliminaries}}. +This function space is defined over the space of ``well-behaved'' \footnote{The definitions of ``well-behaved'' is pretty ambigous, even for mathematicians. +It depends basically on the context. +You can sumarize it by saying: functions for which the theory we are considering (Fourier theorem) is always true. In our case we can say that well-behaved functions +are functions that follow some convergence contraints (pointwise, uniform, absolute, ...) that we don't want to consider further anyway.} functions. +We can say that the theory we are about to show can be applied on all twice differentiable complex valued functions, +to be more concise: complex valued $L^2$ functions $S^2 \to \mathbb{C}$. + +All these jargons are not really necessary for the practical applications of us mere mortals, namely physicists and engineers. +From now on we will therefore assume that the functions we will dealing with fulfill these ``minor'' conditions. + +The insiders could turn up their nose, but we don't want to dwell too much on the concept of Hilbert space, convergence, metric, well-behaved functions etc. +We simply think that this rigorousness could be at the expense of the possibility to appreciate the beauty and elegance of this theory. +Furthermore, the risk of writing 300+ pages to prove that $1+1=2$\cite{principia-mathematica} is just around the corner (we apologize in advance to Mr. Whitehead and Mr. Russel for using their effort with a negative connotation). + +Despite all, if you desire having definitions a bit more rigorous (as rigorous as two engineers can be), you could take a look at the chapter \ref{}. + +\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] + \label{kugel:definition:spherical-harmonics-series} + \begin{equation} + f(\vartheta, \varphi) + = \sum_{n=0}^\infty \sum_{m =-n}^n + c_{m,n} Y^m_n(\vartheta, \varphi). \label{kugel:definition:spherical-harmonics-series} + \end{equation} +\end{definition} + +With this definition we are basically saying that any function defined on the spherical surface can be represented as a linear combination of spherical harmonics. +Does eq.\eqref{kugel:definition:spherical-harmonics-series} sound familiar? Well that is prefectly normal, since this is analog to the classical Fourier theory. +In the latter is stated that ``any'' $T$-periodic function $f(x)$, on any interval $[x_0-T/2,x_0+T/2]$, can be represented as a linear combination of complex exponentials. More compactly: +\begin{equation*} + f(x) = \sum_{n \in \mathbb{Z}} c_n e^{i \omega_0 x}, \quad \omega_0=\frac{2\pi}{T} +\end{equation*} +In the case of definition \ref{kugel:definition:spherical-harmonics-series} the kernels, instead of $e^{i\omega_0x}$, have become $Y^m_n$. In addition, the sum is now over the two indices $m$ and $n$. + +\begin{lemma}[Spherical harmonic coefficients] + \label{kugel:lemma:spherical-harmonic-coefficient} + \begin{align*} + c_{m,n} + &= \langle f, Y^m_n \rangle_{\partial S} \\ + &= \int_0^\pi \int_0^{2\pi} f(\vartheta,\varphi) \overline{Y^m_n(\vartheta,\varphi)} \sin\vartheta \,d\varphi\,d\vartheta + \end{align*} +\end{lemma} +\begin{proof} + To develop this proof we will take advantage of the orthogonality property of the Spherical Harmonics. We can start and finish by applying the inner product on both sides of eq.\eqref{kugel:definition:spherical-harmonics-series}: + \begin{align*} + \langle f, Y^{m}_{n} \rangle_{\partial S} + &= \left\langle \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} + c_{m',n'} Y^{m'}_{n'}(\vartheta, \varphi) \right\rangle_{\partial S} \\ + &= \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} + \langle c_{m',n'} Y^{m'}_{n'}, Y^{m}_{n} \rangle_{\partial S} \\ + &= \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} c_{m',n'} \langle Y^{m'}_{n'}, Y^{m}_{n} \rangle_{\partial S} = c_{m,n} + \end{align*} + We omitted the $\vartheta, \varphi$ dependency to avoid overloading the notation. +\end{proof} +Thanks to Lemma \ref{kugel:lemma:spherical-harmonic-coefficient} we can now calculate the series expansion defined in \ref{kugel:definition:spherical-harmonics-series}. + +It can be shown that, for the famous ``well-behaved functions'' $f(\vartheta, \varphi)$ mentioned before, theorem \ref{fourier-theorem-spherical-surface} is true +\begin{theorem}[Fourier Theorem on $\partial S$] + \label{fourier-theorem-spherical-surface} + \begin{equation*} + \lim_{N \to \infty} + \int_0^\pi \int_0^{2\pi} \left\| f(\vartheta,\varphi) - \sum_{n=0}^N\sum_{m=-n}^n c_{m,n} Y^m_n(\vartheta,\varphi) + \right\|_2 \sin\vartheta \,d\varphi\,d\vartheta = 0 + \end{equation*} +\end{theorem} +The connection to Theorem \ref{fourier-theorem-1D} is pretty obvious. + +\subsection{Spectrum} + +\begin{figure} + \centering + \kugelplaceholderfig{.8\textwidth}{5cm} + \caption{\kugeltodo{Rectangular signal and his spectrum.}} + \label{kugel:fig:1d-fourier} +\end{figure} + +In the case of the classical one-dimensional Fourier theory, we call \emph{Spectrum} the relation between the fourier coefficients $c_n$ and the multiple +of the fundamental frequency $2\pi/T$, namely $n 2\pi/T$. In the most general case $c_n$ are complex numbers, so we divide the concept of spectrum in +\emph{Amplitude Spectrum} and \emph{Phase Spectrum}. In fig.\ref{kugel:fig:1d-fourier} a function $f(x)$ is presented along with the amplitude spectrum. + +\begin{figure} + \centering + \kugelplaceholderfig{.8\textwidth}{7cm} + \caption{\kugeltodo{Confront between image reconstructed only with phase and one only with amplitued}} + \label{kugel:fig:phase&litude-2d-fourier} +\end{figure} + +The thing that is easiest for us humans to visualize and understand is often the Amplitude Spectrum. +This is a huge limitation, since for example in Image Processing can be showed in a nice way that much more information is contained in the phase part (see fig.\ref{kugel:fig:phase-2d-fourier}). + +\begin{figure} + \centering + \kugelplaceholderfig{.8\textwidth}{9cm} + \caption{\kugeltodo{fig that show fourier style reconstruction on sphere (with increasing index)}} + \label{kugel:fig:fourier-on-sphere-increasing-index} +\end{figure} + +The same logic can be extended to the spherical harmonic coefficients $c_{m,n}$. In fig.\ref{kugel:fig:fourier-on-sphere-increasing-index} you can see the same concept as in fig.\ref{kugel:fig:1d-fourier} +but with a spherical function $f(\vartheta, \varphi)$. + +\subsection{Energy of a function $f(\vartheta, \varpi)$} + +\begin{lemma}[Energy of a spherical function (\emph{Parseval's theorem})] + \begin{equation*} + \int_0^{2\pi}\int_0^\pi |f(\vartheta, \varphi)|^2 \sin\vartheta \, d\varphi \, d\varphi = \sum_{n=0}^\infty \frac{1}{2n+1} \sum_{m=-n}^n |c_{m,n}|^2. + \end{equation*} +\end{lemma} +\begin{proof} +\end{proof} + +\subsection{Visualization}
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