Multipole Expansion - Electricity and Magnetism - Lecture Notes, Study notes of Electromagnetism and Electromagnetic Fields Theory

Download Multipole Expansion - Electricity and Magnetism - Lecture Notes and more Study notes Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! 1 PHY481 - Lecture 14: Multipole expansion Griffiths: Chapter 3 Expansion of 1/|~r − ~r′| (Legendre’s original derivation) Consider a charge distribution ρ(~r′) that is confined to a finite volume τ . For positions ~r that are outside the volume τ , we can find the potential using either superposition, or Laplace’s equation, i.e., V (~r) = k ∫ τ ρ(~r′)d~r′ |~r − ~r′| or ∇2V = 0 (1) In cases where there is no φ dependence the Laplace solution in polar co-ordinates is,∑ l (Alrl + Bl rl+1 )Pl(cosθ) (2) How are these two approaches related? The multipole expansion of 1/|~r − ~r′| shows the relation and demonstrates that at long distances r >> r′, we can expand the potential as a multipole, i.e. Eq. (2), with Al = 0. More than that, we can actually get general expressions for the coefficients Bl in terms of ρ(~r′). First lets see Eq. (1) and (2) are related, but doing a systematic expansion of 1/|~r − ~r′|, in the case where r′/r < 1. We write, 1 |~r − ~r′| = 1 [r2 + r′2 − 2~r · ~r′]1/2 = 1 r 1 [1 + ( r′r ) 2 − 2~r·~r′r2 ]1/2 (3) We use x = ( r ′ r ) 2 − 2~r·~r ′ r2 = a − b, where a = ( r′ r ) 2 and b = 2~r·~r ′ r2 = 2(r ′/r)cosθ, and make a Taylor expansion of 1/(1 + x)1/2, i.e. use f(y) = f(y0) + (y − y0)f ′)y0) + 1 2! (y − y0)2f ′′(y0) + 1 3! (y − y0)3f ′′′(y0) + .... (4) with f(y) = 1/(1 + y)1/2, y0 = 0 and y = x. Then f(y0) = 1, f ′(y0) = −1/2; f ′′(y0) = 3/4, f ′′′(y0) = −15/8, so that, 1 (1 + x)1/2 = 1− x 2 + 3x2 8 − 5x 3 16 + ... (5) Substituting x = a− b gives, 1 (1 + x)1/2 = 1− a− b 2 + 3(a2 − 2ab+ b2) 8 − 5(a 3 − 3a2b+ 3ab2 − b3) 16 + .. = 1− a− b 2 + 3(−2ab+ b2) 8 − 5(−b 3) 16 +O(( r r′ )4 (6) where we kept terms to octapole order (i.e. keeping terms up to (r′/r)3). Now collecting terms according to their order in the expansion we get; 1 |~r − ~r′| = 1 r [1 + b 2 + ( 3b2 8 − a 2 ) + ( 5b3 16 − 3ab 4 ) +O(( r r′ )3 (7) Finally we use b = 2~r·~r ′ r2 = 2(r ′/r)cosθ, a = (r′/r)2 to find, 1 |~r − ~r′| = 1 r [1 + r′ r cosθ + ( r′ r )2( 3cos2θ 2 − 1 2 ) + ( r′ r )3( 5cos3θ 2 − 3cosθ 2 ) +O(( r r′ )4 (8) Recall Bonnet’s recursion formula for Legendre polynomials, (l + 1)Pl+1(u) = (2l + 1)uPl(u)− lPl−1(u) (9) With P0 = 1 and P1 = u, we find P2 = (3u2 − 1)/2, P3 = (5u3 − 3u)/2, and with u = cosθ demonstrates that, 1 |~r − ~r′| = ∑ l=0 r′l rl+1 Pl(cosθ) r > r′ (10) 2 This expansion is for the case where r′/r < 1 and is called the exterior solution. A similar expansion may be carried out for r′/r > 1 and this is called the interior expansion 1 |~r − ~r′| = ∑ l=0 rl r′l+1 Pl(cosθ) r < r′ (11) Using the exterior expansion (10) for a dipole charge configuration, we have, V (r, θ) = kq r ∑ l ( d 2r )lPl(cosθ)− kq r ∑ l=0 ( −d 2r )lPl(cosθ) (12) The even terms in the sum cancel, while the odd terms add so that, V (r, θ) = kq r ∑ l odd 2( d 2r )lP1(cosθ) = kq r [ d r cosθ + 2( d 2r )3P3(cosθ) + ...]. (13) The leading term is the dipole potential, though higher order terms do exist and are important for smaller distances. Monopole and dipole terms for a general localized charge distribution First consider a discrete charge distribution consisting of charges qi at positions ~ri. The potential at position ~r is then expanded as, V (~ri) = ∑ i kqi |~r − ~ri| = ∑ i kqi r + ∑ i kqiricosθi r2 + ∑ i kqir 2 i 2r3 [3cos2θi − 1] +O( 1 r4 ) (14) It is then natural to define the quantities, Q = ∑ i qi; ~p = ∑ i qi~ri (15) which are the total charge and the total dipole moment of the charge distribution. The definition of the quadrupole term is more subtle, however a matrix form is convenient so that, we finally have,∑ i kqi |~r − ~ri| = kQ r + k~p · r̂ r2 + kr̂ · Q̃2 · r̂ r3 +O(1/r4) (16) where the quadrupole matrix is given by, Q̃2 = ∑ i 1 2 qi(3~ri ⊗ ~ri − r2i Ĩ) (17) where ⊗ is the outer product and Ĩ is a 3 × 3 identity matrix. The continuum version of the monopole and dipole terms are, Q = ∫ ρ(~r′)d~r′ and ~p = ∫ ~r′ρ(~r′)d~r′. (18) In general the dipole and higher order terms depend on the choice of origin for the co-ordinate system. However if the monopole term is zero it is easy to show that the dipole term is independent of the co-ordinate system. To prove this substitute ~r′+~a for ~r′ in the dipole expression and show that the dipole moment is unaltered provided that Q = 0. An example - Cartesian co-ordinates Consider a square region of space, centered at the origin and with dimensions a × a. The sides of the square are parallel to the x and y axes. The sides at y = ±a/2 are held at a fixed potential V0, while the sides at x = ±a/2 are grounded, ie V = 0 there. Find an expression for the potential everywhere on the interior of the square domain. Solution The first observation is that the boundary conditions in the x direction are symmetric about the origin so we choose functions of the form X(x) cos(kx). Similarly the boundary conditions in the y-direction are symmetric so we choose Y (y) cosh(ky). Since there is dependence on both x and y directions, we expect the one dimensional solutions will not be useful, so we do not include them. We then have, V (x, y) = ∑ k A(k)cos(kx)cosh(ky) (19)