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Lecture 8 Notes, Electromagnetic Theory I Dr Lecture 8 Notes, Electromagnetic Theory I Dr. Christopher S. Baird University of Massachusetts Lowell 1. Multipole Expansion of the Potential - Consider a localized charge density completely contained within some region R. - Very far away from the region R, the charge density behaves more and more like a sphere or a point charge. - Far away from the region R then we can make an expansion of the potential in spherical harmonics and keep only the first few terms and it will still be a valid approximation to the solution. - This is useful when the charge density is localized but too complex to be approached in an exact way. - Because we want the potential far away from the charge density, where there is no charge, we can use the spherical coordinates solution to the Laplace equation when a valid solution is required on the full azimuthal range: ∞ l Φ( θ ϕ)=∑ ∑ ( l + −l −1) (θ ϕ) r , , Alm r Blm r Y lm , l=0 m=−l − 2l 1 l m ! where Y ,= ei m Pm cos are the spherical harmonics lm 4 lm! l - The region we are interested in includes infinity, but not the origin. To ensure the solution approaches zero at infinity, we require Al = 0. The solution now becomes: ∞ l (θ ϕ) Φ( θ ϕ)=∑ ∑ Y lm , r , , B lm l +1 l=0 m=−l r π = 1 4 - For later convenience, we redefine the arbitrary constant, Blm π ϵ + ql m so that: 4 0 2 l 1 ∞ l π (θ ϕ) Φ( θ ϕ)= 1 ∑ ∑ 4 Y lm , r , , π ϵ + ql m l+1 4 0 l=0 m=−l 2l 1 r - This equation is called a multipole expansion. The l = 0 term is called the monopole term, l = 1 are the dipole terms, etc. - We must now determine the coefficients qlm to fully solve the problem. - The solution in integral form was already obtained as Coulomb's law for the potential: = 1 ∫ x' ∣ − ∣d x' 4 0 x x' - We expand the 1/∣x−x'∣ factor into spherical harmonics, remembering that we are interested in the solution far away from the charge so that we want the x > x' case. ∞ l l 1 = ∑ ∑ r ' 1 * ∣ − ∣ 4 l1 Y l m ' , ' Y lm , x x' l =0 m=−l r 2l 1 so that: ∞ l = 1 ∑ ∑ 4 [∫ * l ] Y lm , Y l m ' , ' r ' x ' d x' l 1 4 0 l=0 m=−l 2l 1 r - Comparing this solution to the one above, it becomes apparent that: =∫ * (θ ϕ ) l ρ( ) ql m Y l m ' , ' r ' x' d x' - These coefficients are called the spherical multipole moments. Their physical significance can be seen by representing the first few terms explicitly in Cartesian coordinates. - The l = 0 term is just proportional to the total charge q, which is known as the monopole moment, and has no angular dependence: 1 q = ∫ x'd x ' 00 4 1 q =√ q 00 4π - The l = 1 terms are proportional to the components of the electric dipole moment p. 3 q = ∫ cos'r 'x' d x' 10 4 3 q = ∫ z 'x' d x' 10 4 3 q =√ p 10 4 π z =∓ 3 ∫ ∓i ' q ± sin ' e r ' x' d x ' 1, 1 8 =∓ 3 ∫ ∓ q ± sin ' cos ' i sin ' r ' x ' d x' 1, 1 8 =∓ 3 ∫ ∓ q ± x ' i y ' x' d x ' 1, 1 8 =∓ 3 ∓ q ± p i p 1, 1 8 x y - The total Cartesian dipole moment is defined as p=∫ x' x' d x' - The l = 2 terms are proportional to the Cartesian quadrupole moments Qij: 5 2 2 q = ∫3 cos '−1r ' x' d x' 20 16 5 2 2 q = ∫3 z ' −r ' x' d x ' 20 16 5 q =√ Q 20 16 π 33 =∓ 15 ∫ ∓i ' 2 q ± sin ' cos ' e r ' x ' d x' 2, 1 8 =∓ 15 ∫ ∓ 2 q ± sin ' cos ' cos ' i sin ' r ' x' d x' 2, 1 8 =∓ 15 ∫ ∓ q ± z ' x ' i y ' x' d x' 2, 1 8 =∓ 15 ∫ ∓ q ± z ' x ' i y ' x' d x' 2, 1 8 =∓ 1 15 ( ∓ ) q ± √ Q i Q 2, 1 3 8π 13 23 =± 15 ∫ 2 ∓i 2' 2 q ± sin ' e r ' x' d x' 2, 2 32 =± 15 ∫ 2 ∓ 2 2 q ± sin ' cos ' i sin ' r ' x ' d x' 2, 2 32 =± 15 ∫ ∓ 2 q ± x ' i y ' x' d x' 2, 2 32 =1 15 ( ∓ − ) q ± √ Q 2i Q Q 2, 2 3 32π 11 12 22 - In general, the Cartesian quadrupole moments are defined as: =∫ − 2 Qij 3 xi ' x j ' r ' ij x ' d x' - With the first few coefficients found, we can write out the potential explicitly: ∞ l = 1 ∑ ∑ 4 Y lm , r , , ql m l1 4 0 l =0 m=−l 2l 1 r = 1 r , , q0,0 Y 0,0 0 r 1 [ ] q Y q Y q − Y − 2 1,0 1,0 1,1 1,1 1, 1 1, 1 3 0 r 1 [ ] q Y q Y q − Y − q Y q − Y − ... 3 2,0 2,0 2,1 2,1 2, 1 2, 1 2,2 2,2 2, 2 2, 2 5 0 r - The spherical harmonics for these first few terms are simple enough to be written out explicitly: = 1 1 x q0,0 0 r 4 1 3 [ − i − −i ] 2q cos q e q − e sin 2 1,0 1,1 1, 1 3 0 r 8 1 5 [ 2− − i − −i / i 2 −i 2 2] q 3cos 1 6 q e q − e sin cos 3 2 q e q − e sin ... 3 2,0 2,1 2, 1 2, 2 2, 2 5 0 r 16 - Now switch from spherical multipole moments to Cartesian multipole moments: q 1 i −i x= [2 p cos p −i p e p i p e sin] 2 z x y x y 4 0 r 8 0 r 1 2 i −i [Q 3cos −12Q −i Q e Q i Q e sin cos] 3 33 13 23 13 23 16 0 r 1 i 2 −i 2 2 [1/2 Q −2iQ −Q e Q 2i Q −Q e sin ] ... 3 11 12 22 11 12 22 16 0 r ⋅ 3 3 x x Φ( )= 1 [ q+ p x+ 1 ∑∑ i j + ] x 3 Qi j 5 ... π ϵ = = 4 0 r r 2 i 1 j 1 r =− − - The last step was obtained using the traceless nature of the quadrupoles, i.e. Q33 Q11 Q22 2. Multipole Expansion of the Electric Field - The electric field is most easily expressed in spherical coordinates. - The potential in spherical coordinates was found to be: ∞ l = 1 ∑ ∑ 4 Y lm , r , , ql m l1 4 0 l =0 m=−l 2l 1 r - The electric field is just the negative gradient: E=−∇ - In spherical coordinates: ∂ 1 ∂ 1 ∂ E=−[r ] ∂ r r ∂ r sin ∂ ∞ l q ∂ Y , ∂ Y , ∂ Y , = 1 ∑ ∑ l m [− lm − 1 lm − 1 lm ] E r l 1 l 1 l1 = =− ∂ ∂ ∂ 0 l 0 m l 2l 1 r r r r r sin r ∞ l q ∂ = 1 ∑ ∑ l m 1 [ − − i m ] E l2 r l 1 Y lm , ∂ Y lm , Y lm , 0 l=0 m=−l 2l 1 r sin - The monopole contribution to the electric field (l = 0) is then: = 1 1 = [ ] El 0 q00 2 Y 00 , r 0 r = q E = r l 0 2 4 0 r - This is, of course, the electric field due to a point charge q. This means that far enough away from a finite localized charge distribution with total charge q, the electric field is approximately equal to the field produced from a point charge q at its center. - The dipole contributions to the electric field (l = 1) evaluate to: q − ∂ = 1 1, 1 1 [ − i ] = − − − El 1 3 r 2Y 1, 1 , ∂ Y 1, 1 , Y 1, 1 , 0 3 r sin q ∂ 1 1, 0 1 [ − ] 3 r2Y 1,0 , ∂ Y 1,0 , 0 3 r q ∂ 1 1,1 1 [ − − i ] 3 r 2Y 1,1 , ∂ Y 1,1 , Y 1,1 , 0 3 r sin = 1 E = r [2 p cos 2 p sin cos 2 p sin sin ] l 1 3 z x y 4 0 r 1 [ p sin − p cos cos− p sin cos] 3 z x y 4 0 r 1 [ p sin − p cos] 3 x y 4 0 r - After transforming every part of this equation into Cartesian coordinates and collecting terms, the dipole contribution of a localized charge distribution simplifies to a coordinate-independent form: ̂ ( ⋅̂)− = 3x p x p E = =∫ ρ( ) l 1 π ϵ ∣ ∣3 where p x' x' d x ' (dipole at the origin) 4 0 x - The development thus far has assumed the multipoles are centered on the origin. If we desire to add together the effects of multiple charge distributions or multiple dipoles that are not at the same location, they can not both be at the origin. To generalize, the dipole at location x0 creates the field: ( −̂ )( ⋅( −̂ ))− = 3 x x0 p x x0 p E = (dipole at x ) l 1 π ϵ ∣ − ∣3 0 4 0 x x0 where the hat over the terms in parentheses means that the vector that results after taking the difference is normalized to be a unit vector. Sample Electric Order Name Potential Spherical Moments Cartesian Moments Point Form field Φ∝ 1 ∝ 1 l = 0 monopole + E q q r r2 00 + Φ∝ 1 ∝ 1 l = 1 dipole E q , q , q p , p , p - r 2 r3 1-1 10 11 x y z Q , Q , Q + - Φ∝ 1 ∝ 1 xx xy xz l = 2 quadrupole E q , q , q , q , q Q , Q , Q - + r 3 r4 2-2 2-1 20 21 22 yx yy yz Qzx, Qzy, Qzz - + Φ∝ 1 ∝ 1 q , q , q , Q l = 3 octupole + - E 3-3 3-2 3-1 ijk + - 4 5 - + r r q30, q31, q32, q33 i, j, and k = x, y, or z ..
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