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 $$\rho(\mathbf{r}')$$. 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 expansion
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.

Monopole
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. This term indicates point charge electrical potential.

Dipole
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. This terms indicates the linear charge distribution geometry of a dipole electrical potential.

Quadrupole
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 $$V_{\text{qua}}(\mathbf{r})=-\frac{\hat{\mathbf{r}}\hat{\mathbf{r}}}{4\pi\epsilon_0r^3}::T$$

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