# Point charges and multipole expansion

• ShayanJ
In summary: You should check your algebra and sign conventions.In summary, the conversation discusses the expansion of a charge distribution using spherical coordinates and the resulting potential. The dipole term is found to be complex, but the mistake is discovered to be in the sign conventions and algebra. The conversation also mentions the importance of checking for consistency in these conventions.
ShayanJ
Gold Member
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:
$\phi=\frac {Q} {4\pi \varepsilon_0} \left[ \frac{1}{r} - \frac{1}{\sqrt{r^2+d^2-2rd \cos\theta}}- \frac{1}{\sqrt{r^2+d^2-2rd \cos\gamma_1}}- \frac{1}{\sqrt{r^2+d^2-2rd \cos\gamma_2}}\right]=\\ \frac {Q} {4\pi \varepsilon_0 r} \left[ 1 - \frac{1}{\sqrt{1+(\frac d r)^2-2 \frac d r \cos\theta}}- \frac{1}{\sqrt{1+(\frac d r)^2-2\frac d r \cos\gamma_1}}- \frac{1}{\sqrt{1+(\frac d r)^2-2\frac d r \cos\gamma_2}}\right]=\\ \frac {Q} {4\pi \varepsilon_0 r} \left[ 1-\sum_{n=0}^\infty P_n(\cos\theta) (\frac d r)^n-\sum_{n=0}^\infty P_n(\cos\gamma_1) (\frac d r)^n-\sum_{n=0}^\infty P_n(\cos\gamma_2) (\frac d r)^n \right]=\\ \frac {Q} {4\pi \varepsilon_0 r} \left[ 1-\sum_{n=0}^\infty P_n(\cos\theta) (\frac d r)^n-\sum_{n=0}^\infty \frac{4\pi}{2n+1} (\frac d r)^n\sum_{m=-n}^n Y^{m*}_n(\frac \pi 2,0) Y^m_n(\theta,\varphi)-\sum_{n=0}^\infty \frac{4\pi}{2n+1} (\frac d r)^n\sum_{m=-n}^n Y^{m*}_n(\frac \pi 2,\frac \pi 2) Y^m_n(\theta,\varphi)\right]$
The monopole term(n=0) is $\phi^{(1)}=-\frac{2Q}{4\pi \varepsilon_0 r }$,as it should be.
My problem is, the dipole term(n=1) turns out to be complex.What's wrong?
Thanks

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.

The dipole term is:
$\phi^{(2)}=\frac{Qd}{4\pi\varepsilon_0r^2} \left\{P_1(\cos\theta)- \frac{4\pi}{ 3 } \left[\left(Y^{-1*}_1(\frac \pi 2,0)+Y^{-1*}_1(\frac \pi 2,\frac \pi 2)\right) Y^{-1}_1(\theta,\varphi)\\+\left(Y^{0*}_1(\frac \pi 2,0)+Y^{0*}_1(\frac \pi 2,\frac \pi 2)\right)Y^0_1(\theta,\varphi)\\+\left(Y^{1*}_1(\frac \pi 2,0) +Y^{1*}_1(\frac \pi 2,\frac \pi 2)\right)Y^1_1(\theta,\varphi)\right]\right\}$
The definition I use for spherical harmonics is:
$Y^m_n(\theta,\varphi)=\sqrt{ \frac{2n+1}{4\pi} \frac{ (n-m)! }{ (n+m)! } } P^m_n(\cos\theta) e^{im\varphi}$
So we have:
$Y^{-1}_1=\frac 1 2 \sqrt{\frac{3}{2\pi}}\sin\theta e^{-i\varphi}\\ Y^0_1=\frac 1 2 \sqrt{\frac 3 {2\pi}}\cos\theta\\ Y^1_1=-\frac 1 2 \sqrt{\frac{3}{2\pi}}\sin\theta e^{i\varphi}\\$
And:
$Y^{-1*}_1(\frac \pi 2,0)=\frac 1 2 \sqrt{\frac3 {2\pi}}\\ Y^{-1*}_1(\frac \pi 2,\frac \pi 2)=\frac 1 2 \sqrt{\frac3 {2\pi}}e^{i \frac \pi 2}=\pm i \frac 1 2 \sqrt{\frac3 {2\pi}}\\ Y^{0*}_1(\frac \pi 2,\varphi)=0\\ Y^{1*}_1(\frac \pi 2,0)=-\frac 1 2 \sqrt{\frac 3 {2\pi}}\\ Y^{1*}_1(\frac \pi 2,\frac \pi 2)=-\frac 1 2 \sqrt{\frac 3 {2\pi}}e^{-i \frac \pi 2}=\pm i \frac 1 2 \sqrt{\frac 3 {2\pi}}$
So we'll have:
$\phi^{(2)}=\frac{Qd}{4\pi \varepsilon_0 r^2} \left\{ \cos\theta-\left( 1\pm i \right)\sin\theta e^{-i\varphi}+\left( -1\pm i \right)\sin\theta e^{i\varphi} \right\}=\\ \frac{Qd}{4\pi \varepsilon_0 r^2} \left\{ \cos\theta-\sin\theta e^{-i\varphi}\mp i \sin\theta e^{-i\varphi} -\sin\theta e^{i\varphi}\pm i \sin\theta e^{i\varphi} \right\}=\\ \frac{Qd}{4\pi \varepsilon_0 r^2} \left\{ \cos\theta-\sin\theta (e^{i\varphi}+e^{-i\varphi}) \pm i \sin\theta ( e^{i\varphi}-e^{-i\varphi} ) \right\}=\\ \frac{Qd}{2\pi \varepsilon_0 r^2} \left\{ \frac 1 2 \cos\theta-\sin\theta \left[ \cos\varphi \mp \sin\varphi\right] \right\}$
What's wrong?

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

Last edited:
$e^{i\pi/2}=+i$.

Ok,So we have:
$\phi^{(2)}=\frac{Qd}{2\pi \varepsilon_0 r^2} \left[ \frac 1 2 \cos\theta - \sin\theta \left( \cos\varphi-\sin\varphi\right) \right]$
But we can also use the formulas $\vec{p}=\int \vec{r} \rho dV$ and $\phi^{(2)}=\frac{\vec{p}\cdot\vec{r}} {4\pi \varepsilon_0 r^3}$ and we should arrive at the same result.But when I use the charge density $\rho=Q \left[ \delta(x)\delta(y)\delta(z)-\delta(x-d)\delta(y)\delta(z)-\delta(x)\delta(y-d)\delta(z)-\delta(x)\delta(y)\delta(z-d) \right]$, I'll get $\vec{p}=-Qd(\hat{x}+\hat{y}+\hat{z})$ and $\phi^{(2)}=-\frac{Qd(x+y+z)}{4\pi \varepsilon_0 r^3}=-\frac{Qd}{4\pi \varepsilon_0 r^2}\left[ \cos\theta+\sin\theta \left(\cos\varphi+\sin\varphi\right)\right]$ which differs from the result obtained above.I can't see why this happens!

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.

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.

## 1. What is a point charge?

A point charge is a theoretical concept in electrostatics where a charged particle is considered to have no physical size, and all of its charge is concentrated at a single point in space.

## 2. How is a point charge different from a real charge?

A real charge, such as an electron or proton, has a finite size and its charge is distributed over its entire volume. This results in a non-uniform electric field around the charge, unlike a point charge where the electric field is uniform in all directions.

## 3. What is multipole expansion?

Multipole expansion is a mathematical technique used to describe the electric field of a system of point charges. It breaks down the complex electric field into simpler components, making it easier to analyze and calculate.

## 4. What are the different types of multipole expansions?

There are two main types of multipole expansions: monopole and higher-order multipole expansions. Monopole expansion describes the electric field of a single point charge, while higher-order multipole expansions take into account the electric field of multiple point charges and their relative positions.

## 5. How is multipole expansion used in real-world applications?

Multipole expansion is used in various fields, including physics, chemistry, and engineering. It is commonly used to model and analyze complex electric fields in electronic devices, such as capacitors and antennas, and in studying the behavior of molecules and atoms in chemistry.

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