Why 3 dipole terms in a multipole expansion? As can be seen on this page http://en.wikipedia.org/wiki/Multipole_expansion when we take a multipole expansion without assuming azimuthal symmetry we end up with $2l+1$ coefficients for the $l^{th}$ moment in the expansion. So the dipole moment has 3 terms, the quadrupole has 5 and so on. This is different to the case of azimuthal symmetry as each we need only one co-efficient for each term.
Interpreting 3 coefficients for the dipole moment isn't too bad. I'm guessing it represents the dipole moments along the 3 Cartesian axes? And how do we interpret having to have $2l+1$ coefficients for each term?
 A: For the $l$-th term you take totally symmetric tensors of rank $l$ which are totally traceless under contractions of each index. This comes because that is how the multipole moments, or rather the spherical harmonics, are built from the cartesian coordinates. For example $l=2$ has $5$ components that look like $x_i x_j -\frac{1}{3}\delta_{ij} x^2$ which is a symmetric traceless $3 \times 3$ matrix. Being symmetric is has $3(3+1)/2 = 6$ components, being traceless removes another component, leaving the $5$ which is $2(2)+1$
A: The spherical multipole expansion arises from the solution of the Laplace equation in spherical coordinates. We try to solve by separation of variables and a eigenvalue equation appears .
The number 3 in the second term is caused by the degeneracy of certain eigenvalue (the are 3 linearly independent solutions for the same eigenvalue). The same argument is valid for higher terms.
A happy coincidence allow us to identify the potential of the dipole (a pair of charges of opposite sign with a little separation) in the second term but for higher terms it isn't so easy.  
