Multipole Expansion and Potential in Charge-Free Regions

1. Last remark before going on: the condition we required at the beginning to deal only with azimuthally symmetric system is not necessary: the consequence of that condition is that we end up with the Legendre polynomials, which have a rather simple expression. If we deal with ”any” system, not symmetric, we will end up with more complicated angular functions (associated Legendre functions and spherical harmonics, Jackson 3.5). Recapitulate Note that an azimuthal symmetry does not necessarily mean a sphere. It could be a bottle (not a Coca Cola bottle), a lampshade, a pear,….. These are all cases of symmetry in ψ but not in . We have found a general solution for the potential in a region free of charges for cases in which the boundary condition is given as the potential function on the boundaries of the charge free region, and is, of course, azimuthally symmetric. This general solution is called expansion in series of multipoles and takes the form of where the coefficients An andBn are determined by requiring the series to satisfy the boundary condition. The polynomials Pn are functions of cosθ odd or even according to n. The radial dependence of the potential is a (sum of two) powers of r, of which one describes the potential Outside the boundary surface and the other Inside. Note that the r dependence, which used to be of the type 1/r,for a multipole of order n becomes 1/rn+1for large r. Classical EM - Master in Physics - AA2011-2012

2. Note that each Pn has exactly n zeros in the range of -1≤cosθ≤1 Classical EM - Master in Physics - AA2011-2012

3. Multipole expansion of the potential from a charge distribution with azimuthal symmetry Now consider the case in which the boundary condition is not given as the potential over a surface but as charge distributions in space (always with azimuthal symmetry). But… in such case a solution exists, yes, and.. We already know it. Why bother to try and find a different one? The solution we already know is: This is the general solution. Very general…. We still want to express the solution in the form of a multipole expansion: because there is a lot of physical meaning in it. We know what a multipole has for an angular dependence, and also for a radial dependence. The multipole expansion is valid in a region without charges. Well, given such a region, there are two cases, which we will distinguish: The charges are at smaller radius than the point where want to determine the potential; and The charges are at larger radius. Ie, internal and external charges. Of course, the superposition principle will aid in the determination of the potential in the general case. Classical EM - Master in Physics - AA2011-2012

4. We start with the case of internal charge distribution . The solution will have the form: Now the problem is to expand in series of r the general formula We expand in power series a function of type to obtain We now have to integrate over ψ and ψ’, taking advantage of the fact that ρ does not depend on ψ. The only thing that varies in the integral is the product . Therefore we can set ψ=0 (because of the symmetry of the charge distribution and average over ψ’. Classical EM - Master in Physics - AA2011-2012

5. We are now in a position to calculate the powers of averaged over ψ’. And eventually Which leads us to: Or, in a concentrated formula Comparing this formula with the previous general one we expected to get: We finally obtain the formula for the coefficients Bn which are computed from the charge distribution ρ. Classical EM - Master in Physics - AA2011-2012

6. +q +q 2d -q +q What have we obtained now? We have expanded a the potential from charge distribution constrained only to be azimuthally symmetric into a multipole series. This not only a mathematical game: now we can speak of a charge distribution having a dipole moment (already known….), quadrupole moment, mainly this or mainly that. And we will know with which power of r will the potential decay at large distances. Let us compute the first moments and see what they look like. Total charge Dipole moment (remember?) Quadrupole moment B0 = +q-q = 0 Monopole B1 = qd +(-q)(-d) = 2qd Dipole B2 = qd2 + (-q)d2 = 0 Quadrupole B0 = +q-2q+q = 0 Monopole B1 = qd +2q*0+(+q)(-d) = 0 Dipole B2 = qd2+(-2q)*(0)+(+q)d2 = 2qd2 Quadrupole 2d -2q Classical EM - Master in Physics - AA2011-2012

7. What has been obtained so far is the potential in a charge free region, generated by a charge distribution inside the potential region. What we can do now is solve the inverse problem: given a hollow charge distribution, how can we calculate the potential in the charge-free region inside it? The problem is mathematically equal to the previous one, with the only inversion of r and r’. The result will be very similar: Where the coefficients An, the multipole moments of the charge distribution are given by the formula: The two multipole expansions – for charge distributions inside and outside the sphere on whose surface lies the point where we want to know the potential- can be added to compute the potential in every point in space, even within the charge distribution. Classical EM - Master in Physics - AA2011-2012

8. Electrostatic Energy The Electrostatic Energy of a charge distribution ρis given by the formula: This formula is simply the generalization of the formula for the energy of the system of two charges: It tells us that the electrostatic energy is distributed in space and is non-zero only where charges are present! The integral is in fact equal to zero where ρ=0. We have before developed a formula which gives us the potential in all situations in which charges are present both inside and outside the point where we compute. We can even imagine a system in which two azimuthally symmetric charge distributions are at work, one inside the other; and they are separated by a more or less large charge-free region. (Btw, could you find in nature an example of such a system?) Let us examine this case, and call the two charge distributions ρ1andρ2. They will give origin in the free space in between to the potentials Φ1andΦ2, and the total potential there will be Φ = Φ1 + Φ2. The total electrostatic energy: Classical EM - Master in Physics - AA2011-2012

9. Self energies of the two charge distributions Interaction energies Let us compute a term of the interaction energy: The two interaction energies are equal!!! And now, for a system which has azimuthal symmetry and therefore has charge density and potential parametrizable as series of Legendre polynomials: The substitution of the 2 indices “1” And “2” with the indices “e” and “n” for the two charge distribution is NOT casual. Classical EM - Master in Physics - AA2011-2012

10. The density distribution of the electron cloud in atoms is usually very well known. This knowledge allows for each atom a precise determination of its multipole moments – in particular of its quadrupole moments. An accurate measurement of the interaction energy of the two charges allows a determination of the quadrupole moment of the nucleus –or at least of its sign – which indicates at the very least if the nucleus has a non-spherical shape; and which way it is non-spherical. Being it a symmetrical system around one axis, it can be cigar-like (egg-like) or pancake-like. The nucleus charge distribution being all positive, a positive moment will indicate a cigar-like shape. Classical EM - Master in Physics - AA2011-2012