Mtg 41: Tue, 30 Nov10

NOTE:Generating functions

Eq.(5)p.40-3 : General function for ${\displaystyle \left\{P_{n}\right\}\Rightarrow \ (1-2\mu \ \rho \ +\rho ^{2})^{-{\frac {1}{2}}}\ }$

Eq.(6)p.40-4 : General function for "r choose k" ${\displaystyle {r \choose k}\Rightarrow \ (1+x)^{r}\ }$
END NOTE

NOTE: Inverse square law, pg.40-1

Newton’s law is a claim—that could have been wrong—about the actual relation between the force F on a particle with mass m and its acceleration a. One tests it by calculating the acceleration with a presumed force and comparing it to the measured value. The test will fail if either Newton’s law or the presumed force is wrong. Could F =ma be tested more generally, without recourse to positing forces and looking at actual solutions? It seems not. Kane, String Theory, Physics Today, November 2010



Where ${\displaystyle F=kr\ }$ and ${\displaystyle d(t)=r\ }$



Where ${\displaystyle \mathbf {f} ={\frac {-GMm\mathbf {{\hat {r}}\ } }{r^{2}}}\ }$
END NOTE

2 recurrence relationships:

${\displaystyle \displaystyle {\begin{aligned}\mu \ P_{n}'-P_{n-1}'=nP_{n}\end{aligned}}}$

(RR1)

NOTE: not useful to generate ${\displaystyle P_{n}\ }$ from previously known ${\displaystyle P_{k},k=0,1,...,n-1\ }$ but useful to obtain Legendre differential equation together with recurrence relationship 2.
END NOTE

${\displaystyle \displaystyle {\begin{aligned}(n+1)P_{n+1}-(2n+1)\mu \ P_{n}+nP_{n-1}'=0\end{aligned}}}$

(RR2)

NOTE: RR2 useful to generate ${\displaystyle \left\{P_{n}\right\}\ }$ knowing ${\displaystyle P_{0}(x)=1,P_{1}(x)=x\ }$
END NOTE

HW7.9 Generate ${\displaystyle \left\{P_{2},...,P_{6}\right\}\ }$ using RR2 starting from ${\displaystyle P_{0},P_{1}\ }$ cf Eq.(4) - Eq.(6) p.36-2 END HW7.9

Derive RR1: ${\displaystyle A(\mu \ ,\rho \ ):=1-2\mu \ \rho \ +\rho ^{2}\ }$

Where ${\displaystyle -x=-2\mu \ \rho \ +\rho ^{2}\ }$

Then from Eq.(6) and Eq.(7) p.40-3: ${\displaystyle {\frac {1}{\sqrt {A(\mu \ ,\rho \ )}}}=\alpha _{0}+\alpha _{1}(2\mu \ \rho \ -\rho ^{2})+\alpha _{2}(2\mu \ \rho \ -\rho ^{2})^{2}+...\ }$

Where ${\displaystyle (2\mu \ \rho \ -\rho ^{2})^{2}=4\mu ^{2}\rho ^{2}-4\mu \ \rho ^{3}+\rho ^{4}\ }$

${\displaystyle \displaystyle {\begin{aligned}{\frac {1}{\sqrt {A}}}=\alpha _{0}+(2\mu \ \alpha _{1})\rho \ +(-\alpha _{1}+4\mu ^{2}\alpha _{2})\rho ^{2}+...\end{aligned}}}$

(1)

${\displaystyle \displaystyle {\begin{aligned}P_{0}(\mu \ )=\alpha _{0}=1\end{aligned}}}$

(2)

${\displaystyle \displaystyle {\begin{aligned}P_{1}(\mu \ )=2\mu \ \alpha _{1}=\mu \ \end{aligned}}}$

(3)

HW7.10 Continue power series expansion to find ${\displaystyle \left\{P_{3},...,P_{6}\right\}\ }$ and compute results to those obtained by (a) Eq.(7) and Eq.(8) p.36-2 and (b) HW7.9 END HW7.10

Plan: Find 1. ${\displaystyle {\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}\ }$ and 2. ${\displaystyle {\frac {d}{d\rho \ }}{\frac {1}{\sqrt {A}}}\ }$

${\displaystyle \displaystyle {\begin{aligned}{\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}=-{\frac {1}{2}}A^{-{\frac {3}{2}}}{\frac {dA}{d\mu \ }}={\frac {\rho \ }{A^{\frac {3}{2}}}}\end{aligned}}}$

(5)

Where: ${\displaystyle {\frac {dA}{d\mu \ }}=-2\rho \ \ }$

From Eq.(5) p.40-3 :

${\displaystyle \displaystyle {\begin{aligned}{\frac {d}{d\mu \ }}{\frac {1}{\sqrt {A}}}={\frac {\rho \ }{A^{\frac {3}{2}}}}=\sum _{n=1}^{\infty }P_{n}'(\mu \ )\rho ^{n}\end{aligned}}}$

(6)

${\displaystyle P_{n}'(\mu \ ):={\frac {d}{d\mu \ }}P_{n}(\mu \ )\ }$

Recall ${\displaystyle P_{0}(\mu \ )=1\Rightarrow \ P_{0}'(\mu \ )=0\ }$

${\displaystyle \Rightarrow \ \ }$ n starts from 1 in Eq.(6) p.41-3

${\displaystyle \displaystyle {\begin{aligned}{\frac {d}{d\rho \ }}{\frac {1}{\sqrt {A}}}={\frac {-\rho \ +\mu \ }{A^{\frac {3}{2}}}}=\sum _{n=1}^{\infty }P_{n}(\mu \ )n\rho ^{n-1}\end{aligned}}}$

(1)