Point charges and multipole expansion


by Shyan
Tags: charges, expansion, multipole, point
Shyan
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#1
Feb19-14, 06:39 AM
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P: 734
Consider the following charge distribution:A positive charge of magnitude Q is at the origin and there is a charge -Q on each of the x,y and z axes a distance d from the origin.
I want to expand the potential of this charge distribution using spherical coordinates.Here's how I did it:
[itex]
\phi=\frac {Q} {4\pi \varepsilon_0} \left[ \frac{1}{r} - \frac{1}{\sqrt{r^2+d^2-2rd \cos\theta}}- \frac{1}{\sqrt{r^2+d^2-2rd \cos\gamma_1}}- \frac{1}{\sqrt{r^2+d^2-2rd \cos\gamma_2}}\right]=\\
\frac {Q} {4\pi \varepsilon_0 r} \left[ 1 - \frac{1}{\sqrt{1+(\frac d r)^2-2 \frac d r \cos\theta}}- \frac{1}{\sqrt{1+(\frac d r)^2-2\frac d r \cos\gamma_1}}- \frac{1}{\sqrt{1+(\frac d r)^2-2\frac d r \cos\gamma_2}}\right]=\\
\frac {Q} {4\pi \varepsilon_0 r} \left[
1-\sum_{n=0}^\infty P_n(\cos\theta) (\frac d r)^n-\sum_{n=0}^\infty P_n(\cos\gamma_1) (\frac d r)^n-\sum_{n=0}^\infty P_n(\cos\gamma_2) (\frac d r)^n
\right]=\\
\frac {Q} {4\pi \varepsilon_0 r} \left[
1-\sum_{n=0}^\infty P_n(\cos\theta) (\frac d r)^n-\sum_{n=0}^\infty \frac{4\pi}{2n+1} (\frac d r)^n\sum_{m=-n}^n Y^{m*}_n(\frac \pi 2,0) Y^m_n(\theta,\varphi)-\sum_{n=0}^\infty \frac{4\pi}{2n+1} (\frac d r)^n\sum_{m=-n}^n Y^{m*}_n(\frac \pi 2,\frac \pi 2) Y^m_n(\theta,\varphi)\right]
[/itex]
The monopole term(n=0) is [itex] \phi^{(1)}=-\frac{2Q}{4\pi \varepsilon_0 r } [/itex],as it should be.
My problem is, the dipole term(n=1) turns out to be complex.What's wrong?
Thanks
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Meir Achuz
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#2
Feb19-14, 05:08 PM
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P_n is clearly real. You are making some algebraic mistake with the e^{i\phi}.
Be careful about factors like (-1)^m which differ in different textbooks.
Shyan
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#3
Feb19-14, 10:57 PM
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The dipole term is:
[itex]
\phi^{(2)}=\frac{Qd}{4\pi\varepsilon_0r^2} \left\{P_1(\cos\theta)- \frac{4\pi}{ 3 } \left[\left(Y^{-1*}_1(\frac \pi 2,0)+Y^{-1*}_1(\frac \pi 2,\frac \pi 2)\right) Y^{-1}_1(\theta,\varphi)\\+\left(Y^{0*}_1(\frac \pi 2,0)+Y^{0*}_1(\frac \pi 2,\frac \pi 2)\right)Y^0_1(\theta,\varphi)\\+\left(Y^{1*}_1(\frac \pi 2,0) +Y^{1*}_1(\frac \pi 2,\frac \pi 2)\right)Y^1_1(\theta,\varphi)\right]\right\}
[/itex]
The definition I use for spherical harmonics is:
[itex]
Y^m_n(\theta,\varphi)=\sqrt{ \frac{2n+1}{4\pi} \frac{ (n-m)! }{ (n+m)! } } P^m_n(\cos\theta) e^{im\varphi}
[/itex]
So we have:
[itex]
Y^{-1}_1=\frac 1 2 \sqrt{\frac{3}{2\pi}}\sin\theta e^{-i\varphi}\\
Y^0_1=\frac 1 2 \sqrt{\frac 3 {2\pi}}\cos\theta\\
Y^1_1=-\frac 1 2 \sqrt{\frac{3}{2\pi}}\sin\theta e^{i\varphi}\\
[/itex]
And:
[itex]
Y^{-1*}_1(\frac \pi 2,0)=\frac 1 2 \sqrt{\frac3 {2\pi}}\\
Y^{-1*}_1(\frac \pi 2,\frac \pi 2)=\frac 1 2 \sqrt{\frac3 {2\pi}}e^{i \frac \pi 2}=\pm i \frac 1 2 \sqrt{\frac3 {2\pi}}\\
Y^{0*}_1(\frac \pi 2,\varphi)=0\\
Y^{1*}_1(\frac \pi 2,0)=-\frac 1 2 \sqrt{\frac 3 {2\pi}}\\
Y^{1*}_1(\frac \pi 2,\frac \pi 2)=-\frac 1 2 \sqrt{\frac 3 {2\pi}}e^{-i \frac \pi 2}=\pm i \frac 1 2 \sqrt{\frac 3 {2\pi}}
[/itex]
So we'll have:
[itex]
\phi^{(2)}=\frac{Qd}{4\pi \varepsilon_0 r^2} \left\{ \cos\theta-\left( 1\pm i \right)\sin\theta e^{-i\varphi}+\left( -1\pm i \right)\sin\theta e^{i\varphi} \right\}=\\ \frac{Qd}{4\pi \varepsilon_0 r^2} \left\{ \cos\theta-\sin\theta e^{-i\varphi}\mp i \sin\theta e^{-i\varphi} -\sin\theta e^{i\varphi}\pm i \sin\theta e^{i\varphi} \right\}=\\ \frac{Qd}{4\pi \varepsilon_0 r^2} \left\{ \cos\theta-\sin\theta (e^{i\varphi}+e^{-i\varphi}) \pm i \sin\theta ( e^{i\varphi}-e^{-i\varphi} ) \right\}=\\ \frac{Qd}{2\pi \varepsilon_0 r^2} \left\{ \frac 1 2 \cos\theta-\sin\theta \left[ \cos\varphi \mp \sin\varphi\right] \right\}
[/itex]
What's wrong?

EDIT:
I found what was wrong.
Ok,another question.How can I decide which sign is the right one for the [itex] \sin\varphi [/itex]?
Thanks

Meir Achuz
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#4
Feb20-14, 08:32 PM
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Point charges and multipole expansion


[itex]e^{i\pi/2}=+i[/itex].
Shyan
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#5
Feb20-14, 10:30 PM
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P: 734
Ok,So we have:
[itex]
\phi^{(2)}=\frac{Qd}{2\pi \varepsilon_0 r^2} \left[ \frac 1 2 \cos\theta - \sin\theta \left( \cos\varphi-\sin\varphi\right) \right]
[/itex]
But we can also use the formulas [itex]\vec{p}=\int \vec{r} \rho dV [/itex] and [itex] \phi^{(2)}=\frac{\vec{p}\cdot\vec{r}} {4\pi \varepsilon_0 r^3} [/itex] and we should arrive at the same result.But when I use the charge density [itex] \rho=Q \left[ \delta(x)\delta(y)\delta(z)-\delta(x-d)\delta(y)\delta(z)-\delta(x)\delta(y-d)\delta(z)-\delta(x)\delta(y)\delta(z-d) \right] [/itex], I'll get [itex] \vec{p}=-Qd(\hat{x}+\hat{y}+\hat{z}) [/itex] and [itex] \phi^{(2)}=-\frac{Qd(x+y+z)}{4\pi \varepsilon_0 r^3}=-\frac{Qd}{4\pi \varepsilon_0 r^2}\left[ \cos\theta+\sin\theta \left(\cos\varphi+\sin\varphi\right)\right][/itex] which differs from the result obtained above.I can't see why this happens!
Meir Achuz
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#6
Feb21-14, 10:43 AM
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The signs and algebra for Y^m_L and P^m_L are tricky. I use (-1)^m in the definition of Y^m_L.
How do you define P^{-m}_L? Your Y^0_1 n your first post seems wrong.
Check everything. Get all your signs from the same place.
Shyan
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#7
Feb21-14, 11:07 AM
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P: 734
My references are here and here.
Meir Achuz
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#8
Feb21-14, 02:24 PM
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Your Y^0_1 is wrong and you lost a factor of 1/2 elsewhere. There must also be a mistake in sign that I don't readily see.


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