# Choice of Origin in Dipole moment calculation

I understand (Choice of Origin in Multipole Expansion in Electrostatics) that in multipole expansion I need not choose any particular origin during calculation. But in case of, say 2 point charges $+3q$ at $(0,0,a)$ and $-q$ at $(0,0,0)$, and I am asked to calculate dipole moment, I know that I have to use the superposition principle. But am I still free to choose the origin? I believe if I choose any other origin than (0,0,0), the result would not be same. [This problem is from Griffith.] Can someone please explain why in multipole expansion choice of origin is independent and in this case it is dependent?

No. Only the lowest order multipole is position independent. All higher multipoles will have contributions that are dependent on the choice of coordinate axes. To see why this is the case, take the simplest charge distribution, a monopole with charge $q$, and put it at $z=a$. The potential produced by this charge is: $$\Phi(\mathbf{x}) = \frac{q}{\epsilon_0 4\pi \sqrt{x^2 + y^2 +(z-a)^2}}.$$ Now do a series expansion in $|\mathbf{x}| / a$: \begin{align} \Phi(\mathbf{x}) &= \frac{q}{\epsilon_0 4\pi \mathbf{|x|}}\left[\frac{1}{ \sqrt{1 + \left(\frac{a}{|\mathbf{x}|}\right)^2 - 2\frac{a z}{|\mathbf{x}|^2}}} \right] \\ & = \frac{q}{\epsilon_0 4\pi \mathbf{|x|}} \left[\sum_{n,m=0}^\infty \frac{\Gamma\left(\frac{1}{2}\right)}{n!m!\Gamma\left(\frac{1}{2} - n - m\right) } \left(\frac{a}{|\mathbf{x}|}\right)^{2n} \left(- 2\frac{a z}{|\mathbf{x}|^2}\right)^m \right] \ \mathrm{for\ } |a| < |\mathbf{x}| \\ &\approx \frac{q}{\epsilon_0 4\pi \mathbf{|x|}} \left[1 + \frac{az}{|\mathbf{x}|^2} + \mathcal{O}\left(\frac{a^2}{|\mathbf{x}|^2}\right)\right]\end{align} where the second line uses the multinomial theorem. Notice how there are terms for every multipole, with the expansion explicitly carried out to the dipole term in the last line.
You can do a similar expansion for $|a|> |\mathbf{x}|$ to get the multipole expansion for the potential nearer to the origin than $a$.
• Not wrong, jut incomplete. You can do the expansion about any origin point you like, that is true. Most of the resulting expansion will depend on that choice of origin, though. The only term invariant under translations will be the lowest non-vanishing term. So, say I have a pure dipole - $+q$ and $-q$ at the tip and tail of the vector $\mathbf{a}$ - the dipole moment, $\mathbf{p} = q\mathbf{a}$, of that charge distribution won't depend on origin. The quadrupole and higher will, though. – Sean E. Lake Sep 21 '16 at 10:42
• The dipole moment is defined by one of these equations:\begin{align}\mathbf{p} & = \sum_{i} \mathbf{x}_i q_i \\ & = \int \mathbf{x} \rho(\mathbf{x}) \operatorname{d}^3 x, \end{align} with $\rho$ the charge density. Try applying the formula at different origins and see what happens. It's also worth calculating the monopole moment (total charge). Can you find the origin where the dipole moment is $|\mathbf{p}| = qa$? – Sean E. Lake Sep 22 '16 at 9:39