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diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex index 2ded50b..5d394a9 100644 --- a/buch/papers/kugel/spherical-harmonics.tex +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -178,7 +178,7 @@ write the solutions 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 as straightforward, +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. @@ -220,7 +220,7 @@ 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} + (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 @@ -236,7 +236,7 @@ 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} + (1 - z^2)\frac{d^2 Z}{dz^2} - 2z\frac{d Z}{dz} + n(n + 1) Z(z) = 0, \quad @@ -250,7 +250,7 @@ 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:lem:legendre-poly} + \label{kugel:thm:legendre-poly} The polynomial function \[ P_n(z) = \sum^{\lfloor n/2 \rfloor}_{k=0} @@ -275,7 +275,7 @@ 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} \frac{d^n}{dz^n} (x^2 - 1)^n, + 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} @@ -287,7 +287,7 @@ 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:lem:extend-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*} @@ -300,7 +300,7 @@ such that they also become solutions of the associated Legendre equation See section \ref{kugel:sec:proofs:legendre}. \end{proof} -What is happening in lemma \ref{kugel:lem:extend-legendre}, is that we are +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 @@ -312,8 +312,8 @@ obtain the \emph{associated Legendre functions}. \label{kugel:def:ferrers-functions} The functions \begin{equation} - P^m_n (z) = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m}}{dz^{m}} P_n(z) - = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n + 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} @@ -356,9 +356,10 @@ $Y^m_n(\vartheta, \varphi)$. \label{kugel:def:spherical-harmonics} The functions \begin{equation*} - Y_{m,n}(\vartheta, \varphi) = P^m_n(\cos \vartheta) e^{im\varphi}, + 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 spherical harmonics. + where $m, n \in \mathbb{Z}$ and $|m| < n$ are called (unnormalized) spherical + harmonics. \end{definition} \begin{figure} @@ -366,12 +367,214 @@ $Y^m_n(\vartheta, \varphi)$. \kugelplaceholderfig{\textwidth}{.8\paperheight} \caption{ \kugeltodo{Big picture with the first few spherical harmonics.} + \label{kugel:fig:spherical-harmonics} } \end{figure} -\subsection{Normalization} +\kugeltodo{Describe how they look like with fig. +\ref{kugel:fig:spherical-harmonics}} -\kugeltodo{Discuss various normalizations.} +\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 $S^2$] + 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} + +\kugeltodo{Briefly mention that we could have skipped the tedious proofs by +showing that the (associated) Legendre equation is a Sturm Liouville problem.} \if 0 As explained in the chapter \ref{}, the concept of orthogonality is very important and at the practical level it is very useful, because it allows us to develop very powerful techniques at the mathematical level.\newline @@ -397,14 +600,63 @@ Inoltre, possiamo provare l'ortogonalità di $\Theta(\vartheta)$ utilizzando \eq Ora, visto che la soluzione dell'eigenfunction problem è formata dalla moltiplicazione di $\Phi_m(\varphi)$ e $P_{m,n}(x)$ \fi -\subsection{Properties} + +\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 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 reasons that are not very relevant to our +discussion, but we are mentioning its existence 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 $C(S^2)$} -\subsection{Orthogonality of $P_n$, $P^m_n$ and $Y^m_n$} - -\subsection{Series Expansion} +\subsection{Spherical Harmonics Series} \subsection{Fourier on $S^2$} |