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Multipole expansion Introductory example: the potential of two point charges q q (r ) 1 2 | r r1 | | r r2 | The exact expression is What is the potential far away from the point charges? Namely, what is the potential when r r1 , r2 ? Each of the terms is 1 | r ri | 1 r 2 2rri cos i ri 2 with i=1 or 2. Let us denote 2rri cos i ri 2 ri 2 2 i ri rˆ 2 r r2 r so that 1 r 2 2rri cos i ri 2 1 1 1 3 1 i i2 ... r 1 i r 2 8 1 rˆ ri 3(rˆ ri ) 2 ri 2 2 ... r r 2r 3 Using this expansion, the potential of the two point charges is q q q q rˆ 1 (r ) 1 2 1 2 2 q1r1 q2 r2 3 3q1 (rˆ r1 ) 2 3q2 (rˆ r2 ) 2 q1r12 q2 r22 | r r1 | | r r2 | r r 2r To see the meaning of this expansion we note that q q1 q2 is the total charge of the system generating the potential; p q1r1 q2 r2 is the total dipole moment of the system of charges. So, 1 up to order r 2 q rˆ p the potential is ( r ) 2 . The next term is the quarupole moment which r r is a tensor. To see this, we note that rˆ ri and then (note that rˆ r 1, 2 , 3 i , , 3(rˆ ri ) 2 ri 2 3 rˆ rˆm ri , ri ,m ri 2 rˆ 3ri , ri ,m ,m ri 2 rˆm rˆ 2 ,m ,m 1). The quadrupole tensor of our example is thus Qm q1 3(r1 ) (r1 ) m r12 m q2 3(r2 ) (r2 ) m r22 m The expansion of the potential to order r 3 is finally q rˆ p rˆ Q rˆ (r ) 2 ... r r 2r 3 The general formalism The potential created by a general charge density is (r ' ) (r ) dr ' | r r '| where r r , , , r ' r ' , ' , ' and the integration is dr ' (r ' dr ' )(sin ' d ' )d ' r ' dr ' d' 2 2 We look for an expression for the potential far away from the region where the charge density is, in other words, the potential where r r ' . This means that now we have to expand 1 r 2 2rr ' sin sin ' cos( ' ) cos cos ' r ' 2 It turns out that this expansion can be written using the formula 2 r ' 1 1 Y*m ( ' , ' )Ym ( , ) 4 1 | r r '| 2 1 m 0 r which is valid for side.) r r ' . ( For r ' r one interchanges the roles of r and r' on the right-hand- Using this formula, the potential becomes (r ' ) 1 1 (r ) dr ' 4 Ym ( , ) 1 dr ' (r ' )r 'Ym ( ' , ' ) | r r '| r 0 m 2 1 It is therefore convenient to define the following moments of the charge density: qm dr ' r 'Y*m ( ' , ' ) (r ' ) Then the potential becomes Y ( , ) (r ' ) 1 (r ) dr ' 4 qm m 1 | r r '| r 0 m 2 1 This is the multipole expansion. Exercise: A spherical shell of radius a is carrying a uniform surface charge density, given by q , except on a cone around the north pole, of angle . Find the potential inside the spherical 4a 2 shell. Solution: Since we need the potential inside the spherical shell, we use the formula above with r interchanged with r', that is r 1 1 Y*m ( ' , ' )Ym ( , ) 4 1 | r r '| 2 1 m 0 r ' and then (r ' ) 1 1 (r ) dr ' 4 Ym ( , )r dr ' (r ' ) 1 Ym ( ' , ' ) | r r '| r' 0 m 2 1 3 Let us consider the integral here for our case, in which (r ' ) q (r 'a) for ' 4a 2 The charge density is independent of Ym ( ' , ' ) ' . Since (2 1)( m)! m P (cos ' )e im ' 4 ( m)! this means that we are left only with the terms for which m=0, Y 0 ( ' ) (2 1) P (cos ' ) 4 The expressions for the potential inside the spherical shell is now 1 r q (r ) 4 Y 0 ( ) 1 2 d ' sin 'Y 0 ( ' ) a 4 0 2 1 0 r q P (cos ) 1 d ' sin ' P (cos ' ) a 2 4 The first moments of the multipole expansion The multipole expansion of the potential is given by Y ( , ) (r ' ) 1 (r ) dr ' 4 qm m 1 | r r '| r 0 m 2 1 with qm dr ' r 'Y*m ( ' , ' ) (r ' ) and (2 1)( m)! m P (cos )e im 4 ( m)! Ym ( , ) (Note that Y m ( , ) (1) Ym ( , ) and therefore m * The first moment comes from the term q, m (1) m q*m .) 0 which implies m=0, q00 dr 'Y00* ( ' , ' ) (r ' ) dr ' 1 1 1 P0 (cos ' ) (r ' ) dr ' (r ' ) q 4 4 4 where q is the total charge of the system. Therefore, to order 1/r, the multipole expansion is just The second moment comes from the term Y10 ( , ) (r ) q / r . 1 which implies m=1,0,-1. We have 3 3 P1 (cos ) cos 4 4 and therefore q10 3 d r ' r ' cos ' ( r ') 4 3 d r ' z ' ( r ') 4 3 pz 4 Likewise Y11 ( , ) 3 sin e i 8 and therefore 5 q11 q11 3 3 3 dr ' r ' sin ' e i ' (r ' ) dr ' ( x'iy ' ) (r ' ) ( p x ip y ) 8 8 4 3 3 i ' d r ' r ' sin ' e ( r ' ) d r ' ( x ' iy ' ) ( r ') 8 8 3 ( px ip y ) 4 so that the term gives 4 1 3 z 3 x iy 3 x iy pr pz ( p x ip y ) ( p x ip y ) 3 3 r 2 4 r 8 r 8 r r where the dipole moment of the charge density is defined by p dr ' r ' (r ' ) The next contribution, of the terms with q22 2 is a little more complicated 15 1 15 dr ' ( x'iy ' ) 2 (r ' ) (Q11 2iQ12 Q22 ) 32 12 2 q21 15 1 15 dr ' ( x'iy ' ) z ' (r ' ) (Q13 iQ23 ) 8 3 8 q20 5 1 5 2 2 d r ' ( 3 z ' r ' ) ( r ' ) Q33 16 2 4 where Q is the quadrupole moment tensor, Qij dr ' (3x'i x' j r '2 ij ) (r ' ) Collecting terms, the multipole expansion is 6 xi x j q pr 1 (r ) 3 Qij 5 ... r 2 ij r r Once we have this expansion, we can find the expansion of the electric field. For example, the electric field of the dipole term is moment and r as , we find p r Edipole(r ) 3 . Taking the angle between the dipole r p r ˆ 1 p cos p 3rˆ( p rˆ) p ˆ E dipole(r ) 3 (rˆ ) 3 3rˆ cos (rˆ cos sin ) r r r r2 r r3 Exercise: Given the following charge distribution: (r ) Cr, r a 0, r a where C is a constant (dimensions: charge over length to the fourth), find the first terms of the multipole expansion for r>a. Solution: All the required integrations are now carried only over a spherical volume of radius a. We find 1 a4 q00 dr ' (r ' ) C 4 4 4 for the first moment. q11 3 d r ' ( x ' iy ' ) ( r ' ) 0 8 q10 3 d r ' z ' ( r ' ) 0 4 so there is no dipole moment to such a charge distribution. The next terms vanish as well, q22 15 15 dr ' ( x'iy ' ) 2 (r ' ) 0 q21 dr ' ( x'iy ' ) z ' (r ' ) 0 32 8 5 2 2 q20 d r ' ( 3 z ' r ' ) ( r ' ) 0 16 There is no quadrupole moment to this charge distribution. 7 Exercise: Derive the multipole expansion of the charge distribution (dimensionless units) (r ) 1 2 r r e sin 2 64 Solution: Firstly, we need to revise a bit the formalism above. Because of the charge density is independent of , all the terms with m 0 are zero. SO, we need only the terms with m=0 and therefore we use Y 0 2 1 P (cos ) . 4 The expansion for any r is then (r ) 4 r 1 1 Y 0 1 d r ' r ' ( r ' ) Y ( ' , ' ) 4 r Y d r ' ( r ' )Y 0 ( ' , ' ) 0 0 2 1 r 1 0 r ' 1 2 1 r The Legendre polynomials have the identity 1 d (1 1 2 4 8 ) P ( ) 0 2 3 105 ( cos ) Our potential becomes r 1 4 Y 0 4 8 ( r ) ( 0 2 ) dr ' r ' 4 e r ' 1 32 2 1 r 3 105 0 1 4 4 8 r Y 0 ( 0 2 ) dr ' e r ' r '3 32 2 1 3 105 r 8 r 1 r 1 1 1 4 r ' 3 r ' 6 r ' 2 r ' dr ' r ' e dr ' r ' e P (cos ) dr ' r ' e r dr ' r ' e 2 3 24 r 0 r r 420 r 0 The terms in the brackets can The terms in the brackets can be now easily calculated. In particular, as r tends to zero, the first brackets tend to 6, while the second tend to 1. r 1 r 1 1 1 4 r ' 3 r ' 6 r ' 2 dr ' r ' e dr ' r ' e 3 dr ' r ' e r dr ' r ' e r ' P (cos ) 2 420 r 24 r 0 r r 0 The terms in the brackets can be now easily calculated. In particular, as r tends to zero, the first brackets tend to 6, while the second tend to 1. 9 10