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| -rw-r--r-- | buch/papers/kugel/spherical-harmonics.tex | 45 |
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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} |