Today in Physics 217: Multipole Expansion
Today in Physics 217: multipole expansion
Multipole expansions Electric multipoles and their moments • Monopole and dipole, in detail • Quadrupole, octupole, … Example use of multipole expansion as approximate solution to potential from a charge distribution (Griffiths problem 3.26)
-q q q q -q q -q q q -q -q -q q -q q
21 October 2002 Physics 217, Fall 2002 1 Solving the Laplace and Poisson equations by sleight of hand The guaranteed uniqueness of solutions has spawned several creative ways to solve the Laplace and Poisson equations for the electric potential. We will treat three of them in this class: Method of images (9 October). Very powerful technique for solving electrostatics problems involving charges and conductors. Separation of variables (11-18 October) Perhaps the most useful technique for solving partial differential equations. You’ll be using it frequently in quantum mechanics too. Multipole expansion (today) Fermi used to say, “When in doubt, expand in a power series.” This provides another fruitful way to approach problems not immediately accessible by other means.
21 October 2002 Physics 217, Fall 2002 2 Multipole expansions
Suppose we have a known charge distribution for which we want to know the potential or field outside the region where the charges are. If the distribution were symmetrical enough we could find the answer by several means: direct calculation using Gauss’ Law; direct calculation Coulomb’s Law; solution of the Laplace equation, using the charge distribution for boundary conditions. But even when ρ is symmetrical this can be a lot of work. Moreover, it may give more precise information on the potential or field than is actually needed. Consider instead a direct calculation combined with a series expansion…
21 October 2002 Physics 217, Fall 2002 3 Multipole expansions (continued)
P If the reference point for potential is (and can be) at infinity, then ρτd ′ V ()r = , ∫ r V r where r22=+rr′′ 2 − 2 rr cosθ r 2 rr′′ =+r2 12cos − θ rr dτ ′ ≡+r2 ()1.ε ρ ()r′ r′ If point P is far away from the charge θ distribution, then ε 1. V
21 October 2002 Physics 217, Fall 2002 4 Multipole expansions (continued) 11 1 So consider = . But first recall this infinite series: r r 1 + ε ∞ s! ()1 +=xxs n ∑ − () n=0 nsn!! s ss()−−−112 ss()() s =+1 xx +23 + x +… 1! 2! 3! where x < 1 and s is any real number. (This is one form of the binomial theorem.) Then
11 1 323 5 =−+−1 εε ε +… r r 28 16
21 October 2002 Physics 217, Fall 2002 5 Multipole expansions (continued)
22 11 1rr′′ 3 r ′ r ′ =−12cos2cos −θθ + − r rrr 28 rr 33 2 5 rr′′ rr′′2 −−+ 2 cosθ … =+ 4 cosθθ − 4 cos rr 16 rr 22 3 4 11333rrr′′′ 2 r ′ r ′ =+1 cosθθθ − + cos − cos + rr 22 r r 2 r 8 r 3 323 5 r′′rr4 ′′′2 r r − +−cosθθ 4 cos 16 r rr r r 2 rr′′32 −−++ 2 cosθθ 8cos 8 cos θ … rr
21 October 2002 Physics 217, Fall 2002 6 Multipole expansions (continued)
Collect terms with the same powers of rr ′ , and ignore 3 higher powers than () rr ′ , for now: 2 11rr′′ 32 1 =+1cosθθ + cos − r rr r22 3 r′ 533 +−+ cosθθ cos … r 22 2 1 rr′′ =++PP01()cosθθ() cos P 2() cos θ rrr 3 r′ ++P3 () cosθ … r 21 October 2002 Physics 217, Fall 2002 7 Multipole expansions (continued) ∞ ′ n Thus, 11 r = Pn ()cosθ r ∑ rrn=0 ∞ n () 1 r′ VPdrr= ρθτ()′′n ()cos rr∑ ∫ n=0 V 11 =+ρτ()rr′′drd ρ() ′′cos θτ ′Monopole, Dipole r ∫∫2 VVr
13122 +−ρθτ()r′′rd cos ′ Quadrupole 3 ∫ 22 r V
15333 +−+ρθθτ()r′′rd cos cos ′… Octupole 4 ∫ 22 r V
21 October 2002 Physics 217, Fall 2002 8 Electric multipoles
This is a useful approximation scheme, the more useful the further away point P is from the charges within V, because one can neglect the higher-order terms in the series after the desired accuracy is achieved. The monopole term: 1 q Vd()rr==ρτ()′′ monopole rr∫ V If a charge distribution has a net total charge, it will tend to look like a monopole (point charge) from large distances.
21 October 2002 Physics 217, Fall 2002 9 Electric multipoles
The dipole term: 1 rrˆˆ Vrdd()rr==⋅=⋅ρθτρτ()′′cos ′() rrp ′′′ , 222∫∫ rrrVV where prr = ∫ ρτ () ′′′ d is called the dipole moment. V As usual, for surface, line and point charges, we have n ′′′ ′′′ ′ prrprrpr===∫∫σλ()da,,. d () ∑ qi i SCi=1 The simplest dipole has two point charges, ±q, separated by a displacement vector d that points from –q to +q.
21 October 2002 Physics 217, Fall 2002 10 The dipole potential and field
qdcosθθ p cos P Vdipole == rr22 Edipole=−—V dipole θ r ∂∂VVdipole1 dipole ˆ ˆ =−r + θ ∂∂rrθ q d 2cosppθθ sin 1 -q =+∝rˆ θˆ rrr333 Dipole moment is defined the same way in cgs and MKS. Expressions for potential and field still need a factor of 14πε0 to convert from cgs to MKS.
21 October 2002 Physics 217, Fall 2002 11 Quadrupole, octupole,…
A simple way to envision what the higher-order multipoles “look like” is to construct them from the lower-order ones: take two of the lower-order ones, invert one, and place the two in close proximity. -q q q q -q q -q q -q -q -q q -q q Dipole Quadrupole Octupole
The monopole moment (charge) is a scalar. The dipole moment is a vector. Higher order multipole moments are represented by higher-order tensors: the quadrupole moment is a second-rank tensor, etc.
21 October 2002 Physics 217, Fall 2002 12 Example of the use of multipole expansions
Griffiths problem 3.26: A sphere of radius R, centered at the origin, carries charge density R ρθ()rkRr,2sin,=−() θ r2 where k is a constant and r and θ are the usual spherical coordinates. Find the approximate potential for points on the z axis, far from the sphere. Scheme: start by calculating the monopole term. If it’s not zero, then it’s a good approximation to the potential, since it’s larger by r/r’ than the dipole term. If it is zero, move on to the dipole term. And so on…
21 October 2002 Physics 217, Fall 2002 13 Example (continued)
Monopole moment (charge): 2ππR 1 q ==ρτ()r′′dkR() Rr −2sin θ r2 sin θθφ drdd ∫∫∫∫2 V 000r 2ππ R() =−kR∫∫ dφθθsin2 d ∫ R 2 r dr 00 0 2ππ R =−=kR dφθθsin22 d Rr r 0 ∫∫ 0 00
No net charge, so move on to the dipole term.
21 October 2002 Physics 217, Fall 2002 14 Example (continued)
Dipole moment, or lack thereof: pr= ∫ cosθρ()r′′ d τ V 2ππR 1 =−kR rcosθθθθφ() R 2 r sin r2 sin drd d ∫∫∫ 2 000 r 2ππ R 22 =−kR∫∫ dφθθθsin cos d ∫() Rr 2 r dr 00 0 π 2π 3 R sin θ 2 =−=kR dφ () Rr20 r dr ∫∫3 000
21 October 2002 Physics 217, Fall 2002 15 Example (continued)
Quadrupole “moment” (simple only because this is spherical): 2231 Qr=−cos θρτ()r′′ d ∫ 22 V 2ππR kR 221 2 =−−rRrrdrdd()3cosθθθθφ 1() 2 sin sin 2 ∫∫∫ 2 000 r 2ππ R kR 22 23 =−−ddRrrdrφθ()()3cos 1 sin θθ 2 2 ∫∫ ∫ 00 0 kRππ R44 R k 25 R =−−=≠π ()20 283248 21 October 2002 Physics 217, Fall 2002 16 Example (continued)
Thus, for a point way up the z axis, 11kRπ 25 VQ()r ≅= × for the MKS answer 33 zz48 4πε0
21 October 2002 Physics 217, Fall 2002 17