PHYS 7312 Spring 2023 - Homework #5 Problem 3

ρ[θ] = ρ0  (1 + γ Cos[θ]);

ωvec = ω{0,0,1};
rvec={r Sin[θ] Cos[ϕ],r Sin[θ] Sin[ϕ],r Cos[θ]};

qrule=ρ0->q/(4/3 π a^3);

j= ρ[θ] Cross[ωvec,rvec]

Cartesian magnetic dipole vector using shortcut: (1/2)∫(OverVector[r])×(OverVector[J])dV

mvec = 1/2 Integrate[Cross[rvec,j] r^2 Sin[θ],{r,0,a},{θ,0,π},{ϕ,0,2π}]

mvec/.qrule

Cartesian magnetic dipole vector long way: -(1/2)∫(OverVector[r])(OverVector[∇])·((OverVector[r])×(OverVector[J]))dV

crule={x->r Sin[θ] Cos[ϕ], y->r Sin[θ] Sin[ϕ], z-> r Cos[θ]};

mvec2 =- 1/2 Integrate[({x,y,z}Div[Cross[{x,y,z},ρ[θ]Cross[ωvec,{x,y,z}]],{x,y,z}] r^2 Sin[θ])/.crule,{r,0,a},{θ,0,π},{ϕ,0,2π}]

mvec2/.qrule

Spherical magnetic multipoles (had to multiply bySqrt[4 π/3] for L=1 to match Cartesian see lecture 21, page 1)

m10 = -1/2 Integrate[(r^1 Div[Cross[{x,y,z},ρ[θ]Cross[ωvec,{x,y,z}]],{x,y,z}] SphericalHarmonicY[1,0,θ,ϕ]r^2 Sin[θ])/.crule,{r,0,a},{θ,0,π},{ϕ,0,2π}]*(Sqrt[4 π/3])

m10/.qrule

m11 = -1/2 Integrate[(r^1 Div[Cross[{x,y,z},ρ[θ]Cross[ωvec,{x,y,z}]],{x,y,z}] Conjugate[SphericalHarmonicY[1,1,θ,ϕ]]r^2 Sin[θ])/.crule,{r,0,a},{θ,0,π},{ϕ,0,2π}]*(Sqrt[4 π/3])

m1n1 = -1/2 Integrate[(r^1 Div[Cross[{x,y,z},ρ[θ]Cross[ωvec,{x,y,z}]],{x,y,z}] Conjugate[SphericalHarmonicY[1,-1,θ,ϕ]]r^2 Sin[θ])/.crule,{r,0,a},{θ,0,π},{ϕ,0,2π}]*(Sqrt[4 π/3])

Cartesian quadrupole irreducible tensor

mq[i_,k_] := 1/3 Integrate[(rvec[[i]]Cross[rvec,j][[k]]+rvec[[k]]Cross[rvec,j][[i]] )r^2 Sin[θ],{r,0,a},{θ,0,π},{ϕ,0,2π}]

Table[mq[i,k]/.qrule,{i,1,3},{k,1,3}]//TableForm

Spherical magnetic multipoles (had to multiply bySqrt[4 π/5] for L=2 to match Cartesian see lecture 21, page 1)

m20 = -1/3 Integrate[(r^2 Div[Cross[{x,y,z},ρ[θ]Cross[ωvec,{x,y,z}]],{x,y,z}] Conjugate[SphericalHarmonicY[2,0,θ,ϕ]]r^2 Sin[θ])/.crule,{r,0,a},{θ,0,π},{ϕ,0,2π}]*(Sqrt[4 π/5])

m20/.qrule

m22 = -1/3 Integrate[(r^2 Div[Cross[{x,y,z},ρ[θ]Cross[ωvec,{x,y,z}]],{x,y,z}] Conjugate[SphericalHarmonicY[2,2,θ,ϕ]]r^2 Sin[θ])/.crule,{r,0,a},{θ,0,π},{ϕ,0,2π}]*(Sqrt[4 π/5])

m21 = -1/3 Integrate[(r^2 Div[Cross[{x,y,z},ρ[θ]Cross[ωvec,{x,y,z}]],{x,y,z}] Conjugate[SphericalHarmonicY[2,1,θ,ϕ]]r^2 Sin[θ])/.crule,{r,0,a},{θ,0,π},{ϕ,0,2π}]*(Sqrt[4 π/5])

See https : // farside.ph.utexas.edu/teaching/jk1/Electromagnetism/node32.html to convert spherical to Cartesian