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-rw-r--r--buch/papers/kugel/spherical-harmonics.tex45
1 files changed, 24 insertions, 21 deletions
diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex
index 4f393d4..9d055e0 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -394,7 +394,7 @@ will be our starting point.
\end{equation*}
\end{lemma}
\begin{proof}
- To start, consider the fact that that the Legendre equation
+ 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:
@@ -483,19 +483,19 @@ $P_n(z)$ 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
+ \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' \text{ and } m = m', \\
+ & \text{if } n = n', \\
0 & \text{otherwise}.
\end{cases}
\end{equation*}
\end{lemma}
\begin{proof}
- \kugeltodo{Is it worth showing? IMHO no, it is mostly the same as Lemma
- \ref{kugel:thm:legendre-poly-ortho} with the difference that the $m$-th
- derivative is a pain to deal with.}
+ \kugeltodo{Is this correct? And Is it worth showing? IMHO no, it is mostly the
+ same as Lemma \ref{kugel:thm:legendre-poly-ortho} with the difference that the
+ $m$-th derivative is a pain to deal with.}
\end{proof}
By having the orthogonality relations of the Legendre functions we can finally
@@ -507,7 +507,7 @@ product:
\varphi)$ on the surface of the sphere the inner product is defined to be
\begin{equation*}
\langle f, g \rangle
- = \int_{-\pi}^\pi \int_0^{2\pi}
+ = \int_{0}^\pi \int_0^{2\pi}
f(\vartheta, \varphi) \overline{g(\vartheta, \varphi)}
\sin \vartheta \, d\varphi \, d\vartheta.
\end{equation*}
@@ -520,12 +520,12 @@ product:
right?}
\begin{equation*}
\langle Y^m_n, Y^{m'}_{n'} \rangle
- = \int_{-\pi}^\pi \int_0^{2\pi}
+ = \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'}
+ = \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', \\
+ \frac{4\pi}{2n + 1} \frac{(m + n)!}{(n - m)!} & \text{if } n = n', \\
0 & \text{otherwise}.
\end{cases}
\end{equation*}
@@ -533,15 +533,15 @@ product:
\begin{proof}
We will begin by doing a bit of algebraic maipulaiton:
\begin{align*}
- \int_{-\pi}^\pi \int_0^{2\pi}
+ \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_{-\pi}^\pi \int_0^{2\pi}
+ &= \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_{-\pi}^\pi
+ &= \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
@@ -552,19 +552,22 @@ product:
$m = m'$ 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_{-\pi}^\pi
+ 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'},
+ = -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}. Now we just need look at
- the case when $m \neq m'$. Fortunately this is easy: 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.
+ 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 easy:
+ 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}
\subsection{Normalization}