Last modified on 9 February 2011, at 13:20

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.

QuadrupoleEdit

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