Expanding potential in Legendre polynomials (or spherical harmonics)

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The discussion focuses on expanding the expression $$\frac{1}{\sqrt{(\boldsymbol{r-r'})^2+a}}$$ using Legendre polynomials or spherical harmonics. The challenge arises from incorporating the additional term "a," which complicates the extraction of factors like ##r## or ##r'## necessary for applying the generating function for Legendre polynomials. A suggested approach involves factoring out $$\sqrt{r^2+a}$$ to simplify the expression. This leads to a reformulation that allows for the use of the generating function, requiring the determination of suitable expressions for the variables ##t## and ##x##. The thread emphasizes the mathematical techniques for handling such expansions in potential theory.
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Homework Statement
Expand $$\frac{1}{\sqrt{ (\boldsymbol{r-r'})^2+a} }$$ in legendre polynomials.
Relevant Equations
$$ \sum_{n=0}^{\infty} P_{n}(x) t^{n}=\frac{1}{\sqrt{1-2 x t+t^{2}}} $$
Using the generating function for the legendre polynomial: $$ \sum_{n=0}^{\infty} P_{n}(x) t^{n}=\frac{1}{\sqrt{1-2 x t+t^{2}}} $$ It's possible to expand the coulomb potential in a basis of legendre polynomials (or even spherical harmonic ) like this: $$ \begin{aligned} &\frac{1}{\left.\mid \vec{r}-\vec{r}^{\prime}\right]}=\frac{1}{\sqrt{r^{2}+r^{\prime 2}-2 r r^{\prime}\left(\hat{r} \cdot \hat{r}^{\prime}\right)}}= \sum_{\ell=0}^{\infty} \frac{r_{<}^{\ell}}{r_{>}^{\ell+1}} P\left(\hat{r} \cdot \hat{r}^{\prime}\right) \\ &=\sum_{\ell=0}^{\infty} \frac{4 \pi}{2 \ell+1} \frac{r_{<}^{\ell}}{r_{>}^{\ell+1}} \sum_{m=-\ell}^{\ell} Y_{\ell m}^{\star}\left(\vartheta^{\prime}, \varphi^{\prime}\right) Y_{\ell m}(\vartheta, \varphi) \end{aligned} $$ Where ##r_{<}## and ##r_{>}## represent the smaller and larger of ##r## and ##r^{\prime}##.

But I need to expand
$$\frac{1}{\sqrt{ (\boldsymbol{r-r'})^2+a} }=\frac{1}{\sqrt{r^{2}+r^{\prime 2}-2 r r^{\prime}\left(\hat{r} \cdot \hat{r}^{\prime}\right)+a}}$$
in a similar way. However, I can't seem to pull out a factor of ##r## or ##r'## to get the generating function as you can above, because of the addition of ##a##.
So how do I expand this expression in legendre polynomials?
 
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Nicklas said:
But I need to expand
$$\frac{1}{\sqrt{ (\boldsymbol{r-r'})^2+a} }=\frac{1}{\sqrt{r^{2}+r^{\prime 2}-2 r r^{\prime}\left(\hat{r} \cdot \hat{r}^{\prime}\right)+a}}$$
in a similar way. However, I can't seem to pull out a factor of ##r## or ##r'## to get the generating function as you can above, because of the addition of ##a##.
You might try pulling out a factor of ##\sqrt{r^2+a}## : $$\frac{1}{\sqrt{r^{2}+r^{\prime 2}-2 r r^{\prime}\left(\hat{r} \cdot \hat{r}^{\prime}\right)+a}} = \frac{1}{\sqrt{r^2+a}} \frac{1}{\sqrt{1 -2xt+t^2}}$$ where you will need to determine expressions for ##t## and ##x##.
 
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I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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