When calculating the potential outside a uniformly charged sphere with charge density $\rho$ and radius R using the multipole expansion, it makes sense that only the monopole term survives (since it looks like a point charge).

However, mathematically speaking, I am unsure why the dipole and higher-order terms vanish. Just by looking at the multipole expansion equation

$$V(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \sum_{n=0}^{\infty} \frac{1}{r^{(n+1)}} \int\left(r^{\prime}\right)^{n} P_{n}(\cos \alpha) \rho\left(\mathbf{r}^{\prime}\right) d \tau^{\prime}$$

I believe that $r'$ must be zero for this to work, where $r'$ is the magnitude of the vector from some reference point to some infinitesimal volume element. However, I cannot seem to intuit why this would be the case.


1 Answer 1


It has nothing to do with $r’$ being zero. The only place $r’$ is zero is at the origin.

When $\rho$ is spherically symmetric, it is independent of the angular coordinates, say the usual ones $\theta’$ and $\phi’$. Then, if one takes the $z$-axis along $\mathbf r$ so that the angle $\alpha$ between $\mathbf r$ and $\mathbf{r}’$ is just $\theta’$, the $\theta’$ integral is

$$\int_0^\pi P_n(\cos\theta’)\sin\theta’\,d\theta’=\int_{-1}^1 P_n(u)\,du =0$$

when $n\ne 0$, by the orthogonality of Legendre polynomials.


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