# 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.

## IntroductionEdit

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

Now, the coulomb potential for a charge distribution is given by $V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int_{V'}\frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r'}|}dV'$

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

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

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

## The multipole expansionEdit

Thus, the potential can be written as $V(\mathbf{r})=\frac{1}{4\pi\epsilon_0r}\int_{V'}\rho(\mathbf{r}')\left(1-\frac{\hat{\mathbf{r}}\cdot\mathbf{r}'}{r}+\frac{1}{2r^2}\left(3(\hat{\mathbf{r}}\cdot\mathbf{r}'\right)^2-r'^2)+O\left(\frac{r'}{r}\right)^3\right)dV'$

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

$V_{\text{mon}}(\mathbf{r})=\frac{1}{4\pi\epsilon_0r}\int_{V'}\rho(\mathbf{r}')dV'$

$V_{\text{dip}}(\mathbf{r})=-\frac{1}{4\pi\epsilon_0r^2}\int_{V'}\rho(\mathbf{r}')\left(\hat{\mathbf{r}}\cdot\mathbf{r}'\right)dV'$

$V_{\text{quad}}(\mathbf{r})=\frac{1}{8\pi\epsilon_0r^3}\int_{V'}\rho(\mathbf{r}')\left(3\left(\hat{\mathbf{r}}\cdot\mathbf{r}'\right)^2-r'^2\right)dV'$

and so on.

### MonopoleEdit

Observe that $Q=\int_{V'}\rho(\mathbf{r}')dV'$ is a scalar, (actually the total charge in the distribution) and is called the electric monopole

### DipoleEdit

We can write $V_{\text{dip}}(\mathbf{r})=-\frac{\hat{\mathbf{r}}}{4\pi\epsilon_0r^2}\cdot\int_{V'}\rho(\mathbf{r}')\mathbf{r}'dV'$

The vector $\mathbf{p}=\int_{V'}\rho(\mathbf{r}')\mathbf{r}'dV'$ is called the electric dipole. And its magnitude is called the dipole moment of the charge distribution.

Let $\hat{\mathbf{r}}$ and $\mathbf{r}'$ be expressed in Cartesian coordinates as $(r_1,r_2,r_3)$ and $(x_1,x_2,x_3)$. Then, $(\hat{\mathbf{r}}\cdot\mathbf{r}')^2=(r_ix_i)^2=r_ir_jx_ix_j$
We define a dyad to be the tensor $\hat{\mathbf{r}}\hat{\mathbf{r}}$ given by $\left(\hat{\mathbf{r}}\hat{\mathbf{r}}\right)_{ij}=r_ir_j$
Define the Quadrupole tensor as $T=\int_{V'}\rho(\mathbf{r}')\left(3(\mathbf{r}'\mathbf{r}')-\mathbf{I}r'^2\right)dV'$
Then, we can write $V_{\text{qua}}$ as the tensor contraction $V_{\text{qua}}(\mathbf{r})=-\frac{\hat{\mathbf{r}}\hat{\mathbf{r}}}{4\pi\epsilon_0r^3}::T$