Solutions to PS 5 in Physics 210A by Anthony Karmis: Multipole Moments and Potentials, Exams of Physics

Download Solutions to PS 5 in Physics 210A by Anthony Karmis: Multipole Moments and Potentials and more Exams Physics in PDF only on Docsity! Physics 210A PS 5 Solutions Anthony Karmis February 24, 2009 Jackson 4.1 Part (a) The multipole moments are given by: qlm = ∫ ρ (~x) rlY ∗lm (θ, φ) d 3x = qal [ Y ∗lm (π 2 , 0 ) + Y ∗lm (π 2 , π 2 ) − Y ∗lm (π 2 , π ) − Y ∗lm ( π 2 , 3π 2 )] This is given in terms of associated Legendre polynomials by: qlm = qa l √ 2l + 1 4π (l − m)! (l + m)! Pml (0) [1 + (−ı)m − (−1)m − (ı)m] The moments vanish unless m is odd. Writing m = 2k + 1 gives: ql,2k+1 = 2qa l [ 1 − ı (−1)k ] √ 2l + 1 4π (l − (2k + 1))! (l + (2k + 1))! P 2k+1l (0) = 2qal [ 1 − ı (−1)k ] Yl,2k+1 ( π 2, 0 ) Note that by parity this vanishes unless l is odd. Therefore, only the odd l and m moments are present. The lowest non-trivial ones are: q1,1 = −q∗1,−1 = −2qa (1 − ı) √ 3 8π q3,3 = −q∗3,−3 = −2qa3 (1 + ı) 1 4 √ 35 4π q3,1 = −q∗3,−1 = 2qa3 (1 − ı) 1 4 √ 21 4π 1 Part (b) In this case, we have: qlm = qa l [Y ∗lm (0, 0) + Y ∗ lm (π, 0)] for l > 0 and q00 = 0. By azimuthal symmetry, only the m = 0 moments survive. Therefore, ql0 = qa l √ 2l + 1 4π [Pl (1) + Pl (−1)] = qal [ 1 + (−1)l ] √ 2l + 1 4π for l > 0. We end up with even multipoles: ql0 = qa l √ 2l + 1 4 l = 2, 4, 6, ... Explicity: q20 = qa 2 √ 5 π q40 = qa 4 √ 9 π Part (c) The expansion of the potential is: Φ (~x) = 1 4πǫ0 ∑ l,m 4π 2l + 1 qlm Ylm (θ, φ) rl+1 = 1 ǫ0 ∑ l=2,4,... qal 2l + 1 √ 2l + 1 π Yl0 (θ, φ) rl+1 = q 2πǫ0 ∑ l=2,4,... al rl+1 Pl (cosθ) = q 4πǫ0 a2 r3 ( 3cos2θ − 1 ) + ... In the x − y plane, we have cosθ = 0, so the lowest order term is: Φ = − q 4πǫ0a (a r )3 + ... 2 Jackson 4.6 Part (a) The quadrupole interaction energy is: W = −1 6 Qij ∂Ei ∂xj Note that the cylindrical symmetry gives rise to the non-vanishing quadrupole moments: Q11 = Q22 = 1 1 2 Q33 Since the nuclear quadrupole moment Q is Q33/e, we end up with: Q11 = Q22 = − 1 2 eQ Q33 = eQ Therefore, the energy is: W = −eQ 6 ( −1 2 ∂Ex ∂x − 1 2 ∂Ey ∂y + ∂Ez ∂z ) We now make use of the fact that the external electric field is divergence free (~∇ · ~E = 0). This, as well as the cylindrical symmetry, give us: ∂Ex ∂x = ∂Ex ∂x = −1 2 ∂Ex ∂x Making this substituion above, we have: W = −eQ 4 ( ∂Ez ∂z ) 0 Part (c) Let us assume that the spheroid is obtained by a rotation about the semimajor axis. The equation for a spheroid is given by: x2 + y2 b2 + z2 a2 = 1 The volume of the spheroid is: V = ∫ 2π 0 dφ ∫ b 0 ρdρ ∫ a √ 1−ρ2/b2 −a √ 1−ρ2/b2 dz = 4π 3 ab2 5 The charge density of the nucleus is: ρc = 3Ze 4πab2 Therefore, we have: Q33 = ρc2π ∫ b 0 ρdρ ∫ a √ 1−ρ2/b2 −a √ 1−ρ2/b2 ( 2z2 − ρ2 ) dz = ρc2π ∫ b 0 ρ ( 2 3 a √ b2 − ρ2 b2 2a2b2 − 2a2ρ2 − 3ρ2b2 b2 ) dρ = 3Ze 4πab2 2π 4ab2 ( a2 − b2 ) 15 = 2 5 Ze ( a2 − b2 ) So: Q = 2 5 Z ( a2 − b2 ) = 4 5 ZR (a − b) Or: a − b R = 5Q 4ZR2 = 5 · 2.5 × 10−28m2 4 · 63 · (7 × 10−15m)2 ≈ 0.101 6 Jackson 4.7 Part (a) This charge distribution is azimuthally symmetric. As a result, only m = 0 moments will be non-vanishing. Also, noting that: sin2θ = 1 − cos2θ = 2 3 [P0 (cosθ) − P2 (cosθ)] we may write down the moments: ql0 = ∫ rlY ∗l0 (θ, φ) ρ (r, θ) r 2drdφd (cosθ) = 2π √ 2l + 1 4π ∫ rlPl (θ, φ) ρ (r, θ) r 2drd (cosθ) = 2π 64π 2 3 √ 2l + 1 4π ∫ ∞ 0 rl+4e−rdr ∫ 1 −1 Pl (cosθ) [P0 (cosθ) − P2 (cosθ)] d (cosθ) = 1 48 √ 2l + 1 4π Γ (l + 5) [ 2δl,0 − 2 5 δl,2 ] The multipole expansion then yields the large distance potential: Φ = 1 4πǫ0 ∑ l,m 4π 2l + 1 qlm Ylm (θ, φ) rl+1 = 1 4πǫ0 ∑ l 4π 2l + 1 ql0 Pl (cosθ) rl+1 = 1 4πǫ0 [ 1 r − 6 r3 P2 (cosθ) ] Part (b) We may use a Green function to obtain the potential at any point in space. In general, we have: G (~x, ~x′) = 1 |~x − ~x′| = ∑ lm 4π 2l + 1 rl< rl+1> Y ∗lm (θ ′, φ′)Ylm (θ, φ) However, for azimuthal symmetry, it is sufficient to focus on the m = 0 terms in the expansion: G (~x, ~x′) = ∑ l rl< rl+1> Pl (cosθ)Pl (cosθ ′) + ... 7