Multi-pole Expansion

• Multi-pole expansion is useful for molecular physics, where the size of the charge distribution is much smaller then where we are measuring the potential.

$$V(r)=\frac{1}{4\pi {ϵ}_0}\int \frac{\rho ({\vec{r}}^{\prime })}{|\vec{r}-{\vec{r}}^{\prime }|}d{V}^{\prime }=V_0+V_1+V_2+V_3\dots$$ $$\frac{1}{|\vec{r}-{\vec{r}}^{\prime }}=\sum _{n=0}^{\infty }$$

Consider:

$$|\vec{r}-\vec{r}{|}^2=r^2+r^{\prime 2}-2r{r}^{\prime }\cos \alpha$$ $$=r^2\left(1+\frac{r^{\prime }}{r}\left(\frac{r^{\prime }}{r}-2\cos \alpha \right)\right)$$

LHS Terms are small because $\frac{r^{\prime }}{r}$ is small

Expand:

$$\frac{1}{|\vec{r}-{\vec{r}}^{\prime }|}=\frac{1}{r}(1+ϵ{)}^{-1/2}$$

Result:

$$\frac{1}{|\vec{r}-{\vec{r}}^{\prime }|}=\frac{1}{r}\sum _{n=0}^{\infty }\left(\frac{r^{\prime }}{r}\right)p_n\cos \alpha =V_1+V_2+V_3+\dots$$

The first three terms of the multipole expansion

Electric Monopole:

$$V_0=\frac{1}{4\pi {ϵ}_0}\frac{1}{r}{\int }_{V^{\prime }}\rho ({\vec{r}}^{\prime })d{V}^{\prime }$$

Electric Dipole:

$$V_1=\frac{1}{4\pi {ϵ}_0}\frac{1}{r^2}{\int }_{V^{\prime }}{r}^{\prime }\cos \alpha \rho ({\vec{r}}^{\prime })d{V}^{\prime }$$

Electric Quadrapole:

$$V_2=\frac{1}{4\pi {ϵ}_0}\frac{1}{r^3}{\int }_{V^{\prime }}{r}^{\prime 2}\left(\frac{3}{2}{\cos }^2\alpha -\frac{1}{2}\right)\rho ({\vec{r}}^{\prime })d{V}^{\prime }$$

>insert diagram of mono, di, quadru and octo pole configurations