Download Legendre Polynomials and Solutions to Laplace's Equation - Prof. Phillip Duxbury and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 17 Chapter 5.1-5.3 of PS - We will not cover 5.4, 5.5 A. Legendre Polynomials - Series solution ∂ ∂u [(1 − u2) ∂P (u) ∂u ] + l(l + 1)u = 0 (1) This equation is call Legendre’s equation and is solved using a series solution, P (u) = ∑ ∞ n=0 Cnu n. Substituting this series into Legendre’s equation we find that, ∂ ∂u [1 − u2] ∞∑ n=1 Cnnu n−1 + ∞∑ n=0 l(l + 1)Cnu n = 0 (2) and, ∞∑ n=2 Cnn(n − 1)u n−2 − ∞∑ n=1 Cnn(n + 1)u n + ∞∑ n=0 l(l + 1)Cnu n = 0 (3) The coefficient of un in this equation must be zero, implying that, Cn+2(n+2)(n+1)+Cn[l(l +1)−n(n+1)] = 0; hence Cn+2 = Cn n(n + 1) − l(l + 1) (n + 1)(n + 2) (4) For a given value of l, we get two series of solutions, one starting with a value of C0 and producing C2, C4... and the other starting with C1 and producing C3, C5... The key physical observation is that if l is an integer, the series terminate at n = l, leading to a finite polynomial solution. In contrast if l is not an integer, the series does not terminate and the coefficients remain finite at infinity, a solution that does not have physicial meaning. The constants C0 and C1 are fixed by requiring that the polynomials be normalized on the interval [−1, 1], which corresponds to [−π, π] in the original variable θ. The normalization condition is ∫ 1 −1 P 2l (u)du = 2 2l + 1 or Pl(1) = 1 (5) With this choice the first few Legendre polynomials are, P0(u) = 1; P1(u) = u; P2(u) = (3u 2 − 1)/2; P3(u) = (5u 3 − 3u)/2 (6) Higher order functions Pl(u), are found by choosing an arbitrary starting constant, applying the recursion relation till termination, then normalization using Eq. (5) fixes the starting constant. 1 The solution to Laplace’s equation for cases where there is no dependence on φ is then of the form, V (r, θ) = ∞∑ l=0 (Alr l + Bl rl+1 )Pl(cosθ) (7) If we set Al = 0 in this expansion, we get terms which are of the form of the general multipole expansion, with B0 the monopole term, B1 the dipole term and B2 the quadrupole term. In lecture 9, we carried out the multipole expansion to order l = 2 using the expansion, 1 |~r − ~ri| = 1 r + ricosθ r2 + r2i 2r3 [3cos2θ − 1] + .... (8) Now we recognize the angle dependent terms in this expansion as Legendre polynomials. In fact this expansion can be carried out to arbitrary order in terms of Legendre polynomials yielding, 1 |~r − ~ri| = ∞∑ l=0 rli rl+1 Pl(cosθ) r > ri, (9) which of course reproduces the multipole terms we found before. Also since this is an expansion to infinite order, we can extend it to the regime r < ri, where the expansion takes the form, 1 |~r − ~ri| = ∞∑ l=0 rl rl+1i Pl(cosθ) r < ri, (10) Of course these two terms are just the terms in the general solution to Laplace’s equation in polar co-orindates (Eq. (7)). A simple example. Consider a sphere in a constant applied electric field ~E = E0k̂. The potential is then V (z) = −E0z = −E0rcosθ. Again we tried the simplest solution first, which corresponds to l = 1, so that, V (rθ) = (A1r + B1 r2 )P1(cosθ) = (A1r + B1 r2 )cosθ (11) To satisfy the boundary condition at infinity, we set A1 = −E0, and to satisfy the boundary condition at the surface of the sphere, we require V (R, θ) = 0 = (−E0R + B1 R2 )cosθ so that B1 = E0R 3 (12) yielding the solution, V (r, θ) = (−E0r + E0R 3 r2 )cosθ (13) From this expression it is clear that the induced dipole moment can be found by comparing, k pindcosθ r2 = E0R 3 r2 cosθ so that pind = 4πǫ0E0R 3 (14) 2
