Today in Physics 217: multipole expansion

Transcription

1 Today in Physics 17: multipole expansion Multipole expansions Electic multipoles and thei moments Monopole and dipole, in detail Quadupole, octupole, Example use of multipole expansion as appoximate solution to potential fom a chage distibution (Giffiths poblem 3.6) Octobe 00 Physics 17, Fall 00 1

2 Solving the Laplace and Poisson euations by sleight of hand The guaanteed uniueness of solutions has spawned seveal ceative ways to solve the Laplace and Poisson euations fo the electic potential. We will teat thee of them in this class: Method of images (9 Octobe). ey poweful techniue fo solving electostatics poblems involving chages and conductos. Sepaation of vaiables (11-18 Octobe) Pehaps the most useful techniue fo solving patial diffeential euations. You ll be using it feuently in uantum mechanics too. Multipole expansion (today) Femi used to say, When in doubt, expand in a powe seies. This povides anothe fuitful way to appoach poblems not immediately accessible by othe means. 1 Octobe 00 Physics 17, Fall 00

3 Multipole expansions Suppose we have a known chage distibution fo which we want to know the potential o field outside the egion whee the chages ae. If the distibution wee symmetical enough we could find the answe by seveal means: diect calculation using Gauss Law; diect calculation Coulomb s Law; solution of the Laplace euation, using the chage distibution fo bounday conditions. But even when ρ is symmetical this can be a lot of wok. Moeove, it may give moe pecise infomation on the potential o field than is actually needed. Conside instead a diect calculation combined with a seies expansion 1 Octobe 00 Physics 17, Fall 00 3

4 Multipole expansions (continued) If the efeence point fo potential is (and can be) at infinity, then ρdτ ( ) =, whee = + cosθ ( ε ) 1 +. If point P is fa away fom the chage distibution, then ε 1. = 1+ cosθ ρ ( ) dτ P θ 1 Octobe 00 Physics 17, Fall 00 4

5 Multipole expansions (continued) So conside =. But fist ecall this infinite seies: 1 + ε ( 1 x) s + = n= 0 s! x n! s n! n ( ) ( ) ( )( ) s s s 1 s s 1 s 3 = 1 + x + x + x + 1!! 3! whee x < 1 and s is any eal numbe. (This is one fom of the binomial theoem.) Then = 1 ε + ε ε Octobe 00 Physics 17, Fall 00 5

6 Multipole expansions (continued) = 1 cosθ + cosθ 8 5 cosθ = 1 + cosθ + cos θ cosθ cos θ 4 cos θ 16 3 cosθ 8 cos θ + 8 cos θ + = + 4 cos θ 4 cosθ 1 Octobe 00 Physics 17, Fall 00 6

7 Multipole expansions (continued) Collect tems with the same powes of, and ignoe highe powes than ( ) 3, fo now: = 1+ cosθ + cos θ cos θ cosθ + 1 = P + P + P ( cosθ) ( cosθ) ( cosθ) P3 ( cosθ ) + 1 Octobe 00 Physics 17, Fall 00 7

8 Multipole expansions (continued) 1 1 = Thus, P ( cosθ ) n= 0 1 ( ) = ρ( ) Pn ( cosθ) dτ n= 0 n n 1 1 = ρ( ) dτ + ρ( ) cosθdτ ρ( ) cos θ dτ 3 n ρ( ) cos θ cosθ dτ 4 + Monopole, Dipole Quadupole Octupole 1 Octobe 00 Physics 17, Fall 00 8

9 Electic multipoles This is a useful appoximation scheme, the moe useful the futhe away point P is fom the chages within, because one can neglect the highe-ode tems in the seies afte the desied accuacy is achieved. The monopole tem: monopole 1 = dτ = ( ) ρ( ) If a chage distibution has a net total chage, it will tend to look like a monopole (point chage) fom lage distances. 1 Octobe 00 Physics 17, Fall 00 9

10 Electic multipoles The dipole tem: 1 ˆ ˆ ( ) = ρ( ) cos θdτ = ρ( ) dτ = p, ( ) d whee p = ρ τ is called the dipole moment. As usual, fo suface, line and point chages, we have ( ) λ( ) p = σ da, p = d, p = i. i S C The simplest dipole has two point chages, ±, sepaated by a displacement vecto d that points fom to +. n i= 1 1 Octobe 00 Physics 17, Fall 00 10

11 E dipole dipole d cosθ = = = The dipole potential and field dipole p cosθ dipole = ˆ + θ pcosθ psinθ ˆ 1 = ˆ + θ 1 dipole ˆ Dipole moment is defined the same way in cgs and MKS. Expessions fo potential and field still need a facto of 14πε 0 to convet fom cgs to MKS. θ d θ - P 1 Octobe 00 Physics 17, Fall 00 11

12 Quadupole, octupole, A simple way to envision what the highe-ode multipoles look like is to constuct them fom the lowe-ode ones: take two of the lowe-ode ones, invet one, and place the two in close poximity. - - Dipole Quadupole Octupole The monopole moment (chage) is a scala. The dipole moment is a vecto. Highe ode multipole moments ae epesented by highe-ode tensos: the uadupole moment is a second-ank tenso, etc Octobe 00 Physics 17, Fall 00 1

13 Example of the use of multipole expansions Giffiths poblem 3.6: A sphee of adius, centeed at the oigin, caies chage density ρ(, θ) = k ( ) sin θ, whee k is a constant and and θ ae the usual spheical coodinates. Find the appoximate potential fo points on the z axis, fa fom the sphee. Scheme: stat by calculating the monopole tem. If it s not zeo, then it s a good appoximation to the potential, since it s lage by / than the dipole tem. If it is zeo, move on to the dipole tem. And so on 1 Octobe 00 Physics 17, Fall 00 13

14 Monopole moment (chage): π Example (continued) ππ 1 = ρ( ) dτ = k ( ) sinθ sinθddθdφ π ( ) = k dφ sin θdθ d π π = k dφ sin θdθ = No net chage, so move on to the dipole tem. 0 1 Octobe 00 Physics 17, Fall 00 14

15 Example (continued) Dipole moment, o lack theeof: p = cosθρ dτ ππ π π ( ) 1 = k cosθ ( ) sinθ sinθddθdφ π 3 π ( ) = k dφ sin θ cosθdθ d sin θ ( = k dφ ) d = Octobe 00 Physics 17, Fall 00 15

16 Example (continued) Quadupole moment (simple only because this is spheical): 3 1 Q = cos θ ρ( ) dτ ππ k ( ) 1 = 3cos θ 1 ( ) sinθ sinθddθdφ π π k ( ) ( 3 = dφ 3cos θ 1 sin θdθ ) d k π kπ = ( π ) = Octobe 00 Physics 17, Fall 00 16

17 Example (continued) Thus, fo a point way up the z axis, ( ) 5 1 kπ 1 Q = 3 3 z 48z 4πε 0 fo the MKS answe 1 Octobe 00 Physics 17, Fall 00 17