CrusaderSean
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I'm doing Griffith Problem 5.11 and I'm stuck on how to do the integration. My prof. gave hints on how to simplify the integration with the following formula:
\frac{1}{|{\vec r}_{1}-{\vec r}_{2}|} = \sum_{\ell = 0}^{\infty}<br /> P_{\ell}( \cos \theta) \frac{r_{<}^{\ell}}{r_{>}^{\ell +1}}
where
\cos \theta = \frac{{\vec r}_1 \cdot {\vec r}_2}{r_1 r_2}
P_{\ell}(x)=\frac{1}{2^{\ell}\ell !}(\frac{d}{dx})^{\ell}(x^{2}-1)^{\ell}
r_< is the smaller of |{\vec r}_1|, |{\vec r_2}|
I haven't had advanced electrodynamics yet so it's my first time seeing this type of expansion.
Since I don't really know if \vec r}_1 is bigger or smaller than \vec r}_2, do I need to do expansion for both cases and add them?...
How many terms do I keep in Legendre polynomial? I recall professor saying something about only one or two terms, but I don't see how you can truncate the series like that...
\frac{1}{|{\vec r}_{1}-{\vec r}_{2}|} = \sum_{\ell = 0}^{\infty}<br /> P_{\ell}( \cos \theta) \frac{r_{<}^{\ell}}{r_{>}^{\ell +1}}
where
\cos \theta = \frac{{\vec r}_1 \cdot {\vec r}_2}{r_1 r_2}
P_{\ell}(x)=\frac{1}{2^{\ell}\ell !}(\frac{d}{dx})^{\ell}(x^{2}-1)^{\ell}
r_< is the smaller of |{\vec r}_1|, |{\vec r_2}|
I haven't had advanced electrodynamics yet so it's my first time seeing this type of expansion.
Since I don't really know if \vec r}_1 is bigger or smaller than \vec r}_2, do I need to do expansion for both cases and add them?...
How many terms do I keep in Legendre polynomial? I recall professor saying something about only one or two terms, but I don't see how you can truncate the series like that...