Finding Multipole Expansion for Azimuthally Symmetric Charge Distribution

[fluidistic]

Homework Statement

I'm given a charge density rho ($\rho (r) = r^2 \sin ^2 \theta e^{-r}$) and I'm asked to find the multipole expansion of the potential as well as writing explicitely all the non vanishing terms.

Homework Equations

Not sure and this is my problem.

The Attempt at a Solution

I notice that phi doesn't appear in the expression for rho, so that there's an azimuthal symmetry and I might be lucky.
I do not know what formula to use for the multipole expansion. In wikipedia I see tons of formulae, in Jackson's book I see $\Phi (\vec x ) = \frac{1}{\varepsilon _0 } \sum _{l,m } \frac{1}{2l+1} \left [ \int Y^*_{lm (\theta ', \varphi ' ) r'^{l} \rho (\vec x ) d^3 x' \right ] \frac{Y_{lm} (\theta, \varphi)}}{r^{l+1}}$.
While in Griffith's book (he does not mention when his formula is valid as far as I know, to me it looks like it's valid only in the case of azimuthal symmetry but I may be wrong. Any comment on his formula is welcome) I see $V(\vec r ) = \frac{1}{4\pi \varepsilon _0 } \sum _{n=0}^{\infty } \frac{1}{r^{n+1 }} \int (r')^n P_ n (\cos \theta ' ) \rho (\vec r' )d \tau '$.
Anyway I've followed a bit Jackson's formula and I reached for the integral to be evaluated:
$\int Y^*_{lm} (\theta ' , \varphi ' ) r'^l r'^{l+4}e^ \sin ^3 \theta ' e^ {-r'} dr' d \theta ' d \varphi '$. (Hmm I do not see any latex error but this text won't compile for some reason.)
I realize that I almost have an integral of $Y^*_{lm}$ and $Y_{2,-2}$ over a sphere of radius 1. Would the integral be worth $\int r' ^6 dr$?
If so, then I don't really understand the mathematical justification. It's like saying that $\int gfh dV =\int h dr$ because $\int fg d \Omega=1$ where g,f and h are functions (h is function of the variable r). That's not something obvious to me. Is that true?

 

[gabbagabbahey]

I'm asked to find the multipole expansion of the potential

Okay, but where? Interior to the distribution? Exterior? Both? Along a certain axis?

While in Griffith's book (he does not mention when his formula is valid as far as I know, to me it looks like it's valid only in the case of azimuthal symmetry but I may be wrong. Any comment on his formula is welcome) I see $V(\vec r ) = \frac{1}{4\pi \varepsilon _0 } \sum _{n=0}^{\infty } \frac{1}{r^{n+1 }} \int (r')^n P_ n (\cos \theta ' ) \rho (\vec r' )d \tau '$.

If you take a look at the one page derivation of his formula, you'll see that he defines $\theta'$ as the angle between the source and field point vectors (not as the azimuthal or polar angle in spherical coordinates!), and that his formula is valid for all distributions. It is particularly useful (easy to calulate) when you are looking at problems where you can easily express the charge density in spherical coordinates with your field point on the polar axis, far outside the distribution. The expansion is valid interior to a distribution as well, but utterly useless since it is in powers of 1/r instead of powers of r.

Anyway I've followed a bit Jackson's formula and I reached for the integral to be evaluated:
$$\int Y^{*}_{lm} (\theta',\varphi') r^l' r^{l+4}'e^ \sin ^3 \theta ' e^ {-r'} dr' d \theta ' d \varphi '$$. (Hmm I do not see any latex error but this text won't compile for some reason.)
I realize that I almost have an integral of $Y^*_{lm}$ and $Y_{2,-2}$ over a sphere of radius 1. Would the integral be worth $\int r' ^6 dr$?
If so, then I don't really understand the mathematical justification. It's like saying that $\int gfh dV =\int h dr$ because $\int fg d \Omega=1$ where g,f and h are functions (h is function of the variable r). That's not something obvious to me. Is that true?

Not sure how you got here (sooooo hard to read!), but it can be readily shown that when the charge density is axially symmetric, all the $m\neq 0$ moments vanish and the expansion simplifies to the one in Griffiths' with $\theta'$ being equal to the polar angle. (At least, the exterior expansion anyway, the interior one also simplifies to something similar but in powers of r instead). See wiki page.

 

[fluidistic]

First of all, nice to see you back.
Here come the fixed latex and typos: $$\Phi (\vec x ) = \frac{1}{\varepsilon _0 } \sum _{l,m } \frac{1}{2l+1} \left [ \int Y^*_{lm} (\theta ', \varphi ' ) r'^{l} \rho (\vec x ) d^3 x' \right ] \frac{Y_{lm} (\theta, \varphi)}{r^{l+1}}$$ and $$\int Y^*_{lm} (\theta ' , \varphi ' ) r'^{l+4} \sin ^3 \theta ' e^ {-r'} dr' d \theta ' d \varphi '$$.
They do not precise the region in which they want the multipole expansion.
Apart this, they also want me to write the potential as a finite expansion of Legendre polynomials for large distances and they want me to calculate the exact potential in all the space and more questions later. Basically I seek help here only for the very first part.

I realize that $Y_ {2,-2} (\theta , \varphi ) =\frac{1}{4} \sqrt {\frac {15}{2\pi } } \sin ^2 \theta e^{-2i \varphi }$ or maybe even better $Y_{2,0}(\theta,\varphi)={1\over 4}\sqrt{5\over \pi}\, (2-3\sin^{2}\theta)$.
So that I almost have $\int Y^*_{l,m} Y_{2,0}r'^{l+4} e^{-r'} d\Omega ' dr'$.
What I'm asking is that since $\int Y^*_{l,m} Y_{2,0} d\Omega =1$ when $l=2$ and m=0 or 0 otherwise, is $\int Y^*_{l,m} Y_{2,0}r'^{l+4} e^{-r'} d\Omega ' dr' = \int r' ^{l+4}e^{-r'}dr'$? If so, then it's not obvious to me.

 

[gabbagabbahey]

First of all, nice to see you back.

Thanks, I just took a little break from PF to focus on work (new career as a software developer) and family.

I realize that $Y_ {2,-2} (\theta , \varphi ) =\frac{1}{4} \sqrt {\frac {15}{2\pi } } \sin ^2 \theta e^{-2i \varphi }$ or maybe even better $Y_{2,0}(\theta,\varphi)={1\over 4}\sqrt{5\over \pi}\, (2-3\sin^{2}\theta)$.
So that I almost have $\int Y^*_{l,m} Y_{2,0}r'^{l+4} e^{-r'} d\Omega ' dr'$.
What I'm asking is that since $\int Y^*_{l,m} Y_{2,0} d\Omega =1$ when $l=2$ and m=0 or 0 otherwise, is $\int Y^*_{l,m} Y_{2,0}r'^{l+4} e^{-r'} d\Omega ' dr' = \int r' ^{l+4}e^{-r'}dr'$? If so, then it's not obvious to me.

Yes,

$$\int\int f(\Omega')g(r') d\Omega ' dr' = \left(\int f(\Omega')d\Omega '\right)\left(\int g(r')dr'\right)$$

Provided the integral is over a surface where $r'$ and $\Omega$ are independent (a circular disk, or all space for example).

Thus you get $\int r' ^{l+4}e^{-r'}dr'$ for $m=0$ and 0 for $m\neq 0$.

 

[fluidistic]

Ok thank you very much!
I get that the integral $\int Y^*_{lm} (\theta ' , \varphi ' ) r'^{l+4} \sin ^3 \theta ' e^ {-r'} dr' d \theta ' d \varphi '= \int Y^*_{l,m} r'^{l+4} e^{-r'} \left [ \frac{2}{3} -\frac{4}{3} \sqrt {\frac{\pi}{5}} Y_{2,0} (\theta, \varphi ) \right ] d\Omega ' d r'$. So yes, the integral will split into 2 integrals in which one can be simplified a lot but at first glance the other one will still be horrible.