Mathematical Methods of Physics/The multipole expansion

Tensors are useful in all physical situations that involve complicated dependence on directions. Here, we consider one such example, the multipole expansion of the potential of a charge distribution.

Introduction

Consider an arbitrary charge distribution $\rho ({𝐫}^{\prime })$. We wish to find the electrostatic potential due to this charge distribution at a given point $𝐫$. We assume that this point is at a large distance from the charge distribution, that is if ${𝐫}^{\prime }$ varies over the charge distribution, then $𝐫>>{𝐫}^{\prime }$

Now, the coulomb potential for a charge distribution is given by $V(𝐫)=\frac{1}{4\pi {ϵ}_0}\underset{V^{\prime }}{\int }\frac{\rho ({𝐫}^{\prime })}{|𝐫-{𝐫}^{\prime }|}d{V}^{\prime }$

Here, $|𝐫-{𝐫}^{\prime }|=|r^2-2𝐫\cdot {𝐫}^{\prime }+r{\text{'}}^2{|}^{\frac{1}{2}}=r{|1-2\frac{\widehat{𝐫}\cdot {𝐫}^{\prime }}{r}+{\left(\frac{r^{\prime }}{r}\right)}^2|}^{\frac{1}{2}}$, where $\widehat{𝐫}\triangleq 𝐫/r$

Thus, using the fact that $𝐫$ is much larger than ${𝐫}^{\prime }$, we can write $\frac{1}{|𝐫-{𝐫}^{\prime }|}=\frac{1}{r}\frac{1}{{|1-2\frac{\widehat{𝐫}\cdot {𝐫}^{\prime }}{r}+{\left(\frac{r^{\prime }}{r}\right)}^2|}^{\frac{1}{2}}}$, and using the binomial expansion,

$\frac{1}{{|1-2\frac{\widehat{𝐫}\cdot {𝐫}^{\prime }}{r}+{\left(\frac{r^{\prime }}{r}\right)}^2|}^{\frac{1}{2}}}=1+\frac{\widehat{𝐫}\cdot {𝐫}^{\prime }}{r}+\frac{1}{2r^2}{(3(\widehat{𝐫}\cdot {𝐫}^{\prime })}^2-r{\text{'}}^2)+O{\left(\frac{r^{\prime }}{r}\right)}^3$ (we neglect the third and higher order terms).

The multipole expansion

Thus, the potential can be written as $V(𝐫)=\frac{1}{4\pi {ϵ}_{0}r}\underset{V^{\prime }}{\int }\rho ({𝐫}^{\prime })(1+\frac{\widehat{𝐫}\cdot {𝐫}^{\prime }}{r}+\frac{1}{2r^2}{(3(\widehat{𝐫}\cdot {𝐫}^{\prime })}^2-r{\text{'}}^2)+O{\left(\frac{r^{\prime }}{r}\right)}^3)d{V}^{\prime }$

We write this as $V(𝐫)=V_{\text{mon}}(𝐫)+V_{\text{dip}}(𝐫)+V_{\text{quad}}(𝐫)+\dots$, where,

$V_{\text{mon}}(𝐫)=\frac{1}{4\pi {ϵ}_{0}r}\underset{V^{\prime }}{\int }\rho ({𝐫}^{\prime })d{V}^{\prime }$

$V_{\text{dip}}(𝐫)=\frac{1}{4\pi {ϵ}_0{r}^2}\underset{V^{\prime }}{\int }\rho ({𝐫}^{\prime })(\widehat{𝐫}\cdot {𝐫}^{\prime })d{V}^{\prime }$

$V_{\text{quad}}(𝐫)=\frac{1}{8\pi {ϵ}_0{r}^3}\underset{V^{\prime }}{\int }\rho ({𝐫}^{\prime })(3{(\widehat{𝐫}\cdot {𝐫}^{\prime })}^2-r{\text{'}}^2)d{V}^{\prime }$

and so on.

Monopole

Observe that $Q=\underset{V^{\prime }}{\int }\rho ({𝐫}^{\prime })d{V}^{\prime }$ is a scalar, (actually the total charge in the distribution) and is called the electric monopole. This term indicates point charge electrical potential.

Dipole

We can write $V_{\text{dip}}(𝐫)=\frac{\widehat{𝐫}}{4\pi {ϵ}_0{r}^2}\cdot \underset{V^{\prime }}{\int }\rho ({𝐫}^{\prime }){𝐫}^{\prime }d{V}^{\prime }$

The vector $𝐩=\underset{V^{\prime }}{\int }\rho ({𝐫}^{\prime }){𝐫}^{\prime }d{V}^{\prime }$ is called the electric dipole. And its magnitude is called the dipole moment of the charge distribution. This terms indicates the linear charge distribution geometry of a dipole electrical potential.

Quadrupole

Let $\widehat{𝐫}$ and ${𝐫}^{\prime }$ be expressed in Cartesian coordinates as $(r_1,r_2,r_3)$ and $(x_1,x_2,x_3)$. Then, $(\widehat{𝐫}\cdot {𝐫}^{\prime }{)}^2=(r_i{x}_i{)}^2=r_i{r}_j{x}_i{x}_j$

We define a dyad to be the tensor $\widehat{𝐫}\widehat{𝐫}$ given by ${\left(\widehat{𝐫}\widehat{𝐫}\right)}_{ij}=r_i{r}_j$

Define the Quadrupole tensor as $T=\underset{V^{\prime }}{\int }\rho ({𝐫}^{\prime })(3({𝐫}^{\prime }{𝐫}^{\prime })-𝐈r{\text{'}}^2)d{V}^{\prime }$

Then, we can write $V_{\text{qua}}$ as the tensor contraction $V_{\text{qua}}(𝐫)=-\frac{\widehat{𝐫}\widehat{𝐫}}{4\pi {ϵ}_0{r}^3}::T$

Template:BookCat this term indicates the three dimensional distribution of a quadruple electrical potential.