The Expansion of a Finite Dipole in Legendre Polynomials

There is yet another way to see Legendre Polynomials in action, through the expansion of the potential energy of point dipoles. To start, we assume that we have a dipole at the origin, with its positive charge (q) at (0,0,-a/2) and its negative charge (-q) at (0,0,+a/2), so that the ``bond length'' is ``a'', and therefore the ``dipole moment'' is ``q a''.

At some point P(x,y,z), located (also) at r,$\theta$,$\phi$, we have that the potential energy due to these two point charges is

$$U(x,y,z,a) = \frac{-q}{\sqrt{x^2+y^2 + (z-a/2)^2}} + \frac{q}{\sqrt{x^2+y^2 + (z+a/2)^2}}$$

which is just Coulomb's law.

If we expand this potential energy as a function of ``a'', the ``bond distance'', we have

$$U(x,y,z,a) = U(x,y,z,0) + \frac{1}{1!}\left . \frac{d U}{da}\... ...}{3!}\left . \frac{d^3 U}{da^3}\right \vert _{a=0} a^3+ \cdots$$

All we need do, now, is evaluate these derivatives. We have, for the first

$$\frac{1}{1!}\left . \frac{d U}{da}\right \vert _{a=0} = q \le... ...{1}{2})}{(x^2+y^2+(z+a/2)^2))^{\frac{3}{2}}} \right ] \right )$$

which is, in the limit $a\rightarrow 0$,

$$q\left ( - \frac{z}{r^3} \right )$$

so, we have, so far,

$$U(x,y,z,a) = 0 -qa \left ( \frac{z}{r^3} \right )+ \cdots = -qa \left ( \frac{\cos \theta}{r^2} \right )+ \cdots$$

i.e., to the dipolar form.

For the second derivative, we take the derivative of the first derivative:

$$\frac{1}{2!}\left . \frac{d \frac{dU}{da}}{da}\right \vert _{... ...^2+(z+a/2)^2))^{\frac{3}{2}}} \right ] \right )} {da} \right )$$

which equals zero. The next term gives

$$q \frac{\cos \theta \left (3 - 5 \cos^2\theta \right )}{8r^4}a^3$$

and so it goes.