# Multipole moments of line charge

I am supposed to calculate the multipole moments of a line charge with total charge Q spread from $$z=-a$$ to $$z=a$$ on the z-axis. I know that each multipole moment is given by:

$$q_{l,m}=\int_{\mathbb{R}^3} \rho(\vec{x})|\vec{x}|^l Y^*_{l,m}(\theta,\phi)\ \ dr`d\Omega$$

I think because the charge distribution doesn’t depend on $$\phi$$ only the $$q$$ for $$m =0$$ remain. So I only have to perform the integral with the $$Y_{l,0}:=\sqrt{\frac{2l+1}{4 \pi}}P_l(\cos(\theta))$$. I have, however, not really an idea how to perform this integration (maybe because I‘m lacking knowledge about the legendre polynomials), I‘d be really happy if someone could help me out.

• Is the charge evenly distributed along the line? Commented Jul 23, 2022 at 17:52
• Are you supposed to do the multipole expansion as an expansion in spherical harmonics? (cf. Jackson) If not, I would recommend doing it in cartesian coordinates. See Panofsky&Phillips, chapter 1-7. (Or see Wikipedia.) Find an expression for the charge density involving delta functions and you will be left with a one-dimensional integral. In spherical coordinates you can also work with delta functions, but it's more difficult for the given charge distribution. Commented Jul 23, 2022 at 19:01

Take a look at the spherical harmonics in cartesian coordinates, where you'll see things like:

$$Y_4^0(x,y,z) \propto \frac{35 z^4-30z^2r^2+3r^4}{r^4}$$

since you're evaluating the integral along $$r=z$$, that reduces to:

$$Y_4^0(0,0,z) \propto \frac{35z^4-30z^2z^2+3z^4}{z^4}=35-30+3=8$$

All the integrals can be reduced to an integral over a constant. You just have to figure out the general formula.

The charge distribution lies completely on the $$z$$ axis. If you think about this statement, it is apparent that the only region of space of interest (gives a non-zero contribution) is the z axis. This is because $$\rho$$ will be 0 along anything that is not the $$z$$ axis.

Mathematically:

$$\rho = \frac{Q}{2a}\delta(\theta)\space\space\space\space\text{(from z = a to z = -a)}.$$

So:

$$q_{l,m} = \sqrt{\frac{(2l+1)}{4\pi}}\int_{0}^{\pi}d\theta\delta(\theta)P_l(\text{cos}\theta)\int_{-a}^a\frac{Q}{2a}r'^l dr'.$$

Where $$P_l$$cos$$(0) = 1 \space \forall \space l$$

So:

$$q_{l,m} = \sqrt{\frac{(2l+1)}{4\pi}}\int_{-a}^a\frac{Q}{2a}r'^l dr'$$

Which evaluates to: $$q_{l,m} = \sqrt{\frac{(2l+1)}{4\pi}}\frac{Q}{a}\frac{a^{l+1}}{l+1}.$$

For $$l=0$$ at least, this gives the correct answer, i.e., $$= \sqrt{\frac{1}{4\pi}}Q$$ (Jackson 3ed, 146).

• shouldn't the odd $l$'s be zero? Also, a $\sin\theta$ in the solid angle differential?
– JEB
Commented Jul 25, 2022 at 3:18