Orthogonal functions

Transcription

1 Orthogonl functions Given rel vrible over the intervl (, b nd set of rel or complex functions U n (ξ, n =, 2,..., which re squre integrble nd orthonorml b U n(ξu m (ξdξ = δ n,m ( if the set of of functions is complete n rbitrry squre integrble function f(ξ cn be expnded s where n = f(ξ = b n U n (ξ (2 n= Substituting the expressions for n bck into f we find f(ξ = b U n(ξ f(ξ dξ. (3 Un(ξ U n (ξf(ξ dξ (4 but this is the definition of the Dirc delt-function nd so we hve n= Un(ξ U n (ξ = δ(ξ ξ (5 n= which is known s the completeness reltion. Fourier Series A fmilir exmple is Fourier series, where the function is periodic function on the intervl ( L/2, L/2. The complete set of functions re lbelled by m Z with m 0 ( 2 2πmx L sin, L 2 ( 2πmx L cos L (6 nd m = 0 / L. We cn expnd n rbitrry function s f(x = ( ( 2πmx 2πmx 2 A 0 + A m cos + B m sin L L m= (7 where A m = 2 L B m = 2 L L/2 L/2 L/2 L/2 ( 2πmx dx f(x cos L ( 2πmx dx f(x sin L. (8

2 If we tke the intervl to be infinite we require continuum of functions. Tht is for f(x defined on the rel line where f(x = 2π A(k = 2π nd the orthonormlity nd completeness reltions re 2π 2π A(ke ikx dk (9 f(xe ikx dx (0 e i(k k x dx = δ(k k 2 Lplce s Eqution in two-dimensions If we consider Lplce s eqution in two-dimensions e ik(x x dk = δ(x x. ( 2 Φ x Φ y 2 = 0 (2 we cn look for solutions by seprtion of vribles i.e we tke n nstz Φ(x, y = X(xY (y nd we find tht or equivlently d 2 X(x X(x dx 2 + d 2 Y (y Y (y dy 2 = 0 (3 d 2 X(x X(x dx 2 = α 2, d 2 Y (y Y (y dy 2 = α 2, (4 for n rbitrry constnt (possibly complex α. The solutions of these equtions re X = Ae ±iαx, Y = A e ±αy. (5 Now consider the problem in Fig.. We wnt to find the potentil Φ for 0 x nd y 0 with the boundry conditions s in the figure. We cn see tht the potentil must oscillte in the x-direction nd fll off in the y-direction. Hence we tke α to be rel. We lso see tht we wnt the solution to be of the form sin(αx s Φ = 0 t x = 0. Thus Φ e αy sin αx. The boundry condition t x = sets α = πn/. Hence the potentil cn be written s where Φ(x, y = ( nπx A n sin e nπy/ (6 n= A n = 2 ( nπx Φ(x, 0 sin dx 0 = 2 ( nπx V sin dx 0 { 4V/πn n odd, = 0 n even. 2 (7

3 Figure : Potentil on two-dimensionl semi-infinite strip. Figure 2: Potentil on two-dimensionl semi-infinite strip. Hence Φ(x, y = 4V π n odd This cn ctully be resumed (see Jckson 2.0 ( nπy nπx e sin n Φ(x, y = 2V ( sin πx π tn sinh πy 3 Lplce s Eqution in Sphericl coordintes. (8. (9 Now let us consider the three-dimensionl problem however rther thn using crtesin coordintes we will consider the usul sphericl coordintes in three-dimensions (r, θ, φ, see Fig. 2, in terms of which the Crtesin coordintes (x, y, z re given s x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ. (20 3

4 In these coordintes Lplce s eqution, 2 Φ = 0, is ( r 2 r 2 Φ ( + r r r 2 sin θ Φ sin θ θ θ + 2 Φ r 2 sin 2 θ φ 2 = 0. (2 We look for solutions of the form Φ(r, θ, φ = U(r r Y (θ, φ; substituting this Anstz into (2 nd dividing by Φ/r 2 we find r 2 d 2 U(r dr 2 U(r + Y (θ, φ θ,φy (θ, φ = 0 (22 where θ,φ Y = ( sin θ Y + 2 Y sin θ θ θ sin 2 θ φ 2 (23 which depends only on coordintes (θ, φ. Thus we cn set r 2 U (r = l(l + U(r Y (θ, φ θ,φy (θ, φ = l(l +. (24 We cn mke the further nstz Y (θ, φ = P (θq(φ, which substituting into (24 nd multiplying by sin 2 θ we find sin θ d (sin θ ddθ P (θ dθ P + d 2 Q(φ dφ 2 Q = l(l + sin2 θ which in turn implies tht Q(φ dφ 2 Q = m2, nd (25 d (sin θ ddθ sin θp (θ dθ P m2 sin 2 + l(l + = 0. (26 θ d 2 We cn immeditely write down the solutions to (25: Q = e ±imφ. It is importnt to note tht becuse we require Φ(φ + 2π = Φ(φ this implies m Z. In (26 we define x = cos θ (which is conventionl nd hs nothing to do with the Crtesin coordinte x nd divide by sin 2 θ to find, ( d ( x 2 d dx dx P + (l(l + m2 x 2 P = 0. (27 Eqution (27 is the ssocited Legendre eqution. 4 Legendre Eqution When m = 0 eqution (27 simplifies to the Legendre eqution which we cn write s P Here nd below we will use Lgrnge s prime nottion f (x = df(x dx. 2x ( x 2 P l(l + + ( x 2 P = 0. (28 4

5 This is n ODE with singulr points t x = ± s the coefficients of P nd P diverge t these points. They re regulr singulr points s they diverge s (x ±. The point t x requires little more effort (see [] section 8.4 for exmple but is lso regulr singulr point. One cn find solution to such n ODE by mking series expnsion bout point s long s the behviour is no worse thn tht of regulr singulr point; this is lso known s the Frobenius method. We thus write P (x = n 0 n x n+α (29 with α rbitrry nd 0 the lowest non-vnishing coefficient of the series. Substituting into (28 (nd multiplying by ( x 2 we find x n 0( 2 n (α + n(α + n x α+n 2 2 n (α + nx α+n + l(l + n x α+n = 0 n 0 n 0 nd we cn determine the coefficients by demnding the ech term in the x n+α expnsion vnishes. Strting with while for generl n we find n = 0 0 α(α = 0 n = α(α + = 0 (30 n+2 = n (α + n(α + n + l(l + (α + n + 2(α + n +. (3 If we choose α = we must tke = 0 nd, from the generl formul (3, we find series of odd powers of x P ( = 0 x + 2 x x (32 while if we tke α = 0 nd choose = 0 then we find series with even terms only P (2 = x x (33 In both cses the coefficients 2, 4,... re relted to 0 by the generl formul (3. This gives us two linerly independent solutions ] P ( = 0 [x + 6 (2 l(l + x (2 l(l + (2 l(l + x ] P (2 = 0 [ 2 l(l + x2 24 l(l + (6 l(l + x4 +..., which s we re considering second order differentil eqution gives us complete bsis of solutions. We cn nturlly consider α = 0 nd 0 this will give us liner combintion of the bove solutions. 2 In generl (i.e. for rbitrry l these solutions re infinite series nd it is importnt to question whether they converge. In fct for x < they do, however they will not for x = ± unless the series termintes t some vlue of n; tht is to sy n = 0 for ll n > N. This will be the cse if l is some positive integer. For exmple, l+2 = 0 (nd hence ll n = 0 for ll n > l when α = 0 nd l = n s (α + n(α + n + l(l + = 0. Thus the even series termintes when l is n even integer nd the odd series termintes when l is n odd integer. For ech l we will cll this solution P l (x. We fix the overll constnt by choosing to normlise P l ( =. The first few Legendre functions re P 0 (x =, P (x = x, P 2 (x = 2 (3x2, P 3 (x = 2 (5x3 3x. (34 Note tht from the construction it is immeditely pprent tht P l ( x = ( l P l (x. 2 Jckson llows 0 = 0 nd for the α = 0 cse considers 0. This is, by simple relbelling, our second solution. 5

6 Useful Properties These functions re orthogonl nd normlised such tht P l (xp l (xdx = 2 2l + δ l l. (35 A compct nd useful expression for the Legendre functions is given by Rodrigues eqution Moreover these functions stisfy vrious recursion reltions e.g. d l P l (x = 2 l l! dx l (x2 l. (36 (l + P l+ (2l + xp l + lp l = 0. (37 See for exmple [] for more complete discussion of their mny properties. Second solutions Finlly we note tht for ech l this gives us only one solution, the other series does not terminte nd is singulr t x = ±. These solutions cn be written s 4. Generting function for Legendre functions P n (x = 2 P n(x ln + x x. (38 There is nice wy to write the Legendre functions: Consider the function It is esy to check tht Now expnd G(x, t in powers of t G(x, t = ( 2xt + t 2 /2. (39 ( x 2 2 G x 2 2x G x + t 2 (tg = 0. (40 t2 G(x, t = l 0 t l L l (x, (4 nd substituting this into (40 it esy to see tht it implies l 0 t l[ ( x 2 L l 2xL l + l(l + L l] = 0 (42 i.e. tht ech L l stisfies the Legendre eqution i.e. L l (x = P l (x, nd so G(x, t = l 0 t l P l (x. (43 This reltion cn be explicitly checked t low orders by inspection G(x, t = + xt + 2 (3x2 t + 2 (5x3 3xt (44 We in fct proved version of this in clss when considering the potentil due to point chrge in terms of Legendre functions. 6

7 5 Associted Legendre eqution We now return to the more generl cse of the ssocited Legendre eqution (27. It is in fct strightforwrd to solve this hving lredy solved the Legendre eqution. Let us write P = ( x 2 m/2 v(x, this implies tht (27 becomes ( x 2 v 2x( + mv + v(l(l + m(m + v = 0. (45 Now consider ny solution u(x of the Legendre eqution i.e. ( x 2 u 2xu + l(l + u = 0. (46 We now differentite this eqution m times w.r.t. x nd find ( x 2 u (m+2 2x(m + u (m+ + u (m (l(l + m(m + = 0, (47 where we hve used the nottion u (m = dm dx m u(x This implies tht v=u (m solves (45 nd so P m l = ( m ( x 2 m/2 dm dx m P l(x (48 solves the ssocited Legendre eqution (27 nd hence we cll these ssocited Legendre functions. Pl m(x is regulr everywhere, including x = ±, s P l is. This derivtion lso implies m l s otherwise the function would be zero given tht P l is n l-fold polynomil in x. We cn now use Rodrigues formul to find P m l ( m (x = 2 l ( x 2 m/2 dm+l l! dx m+l (x2 l. (49 This formul is, perhps surprisingly, lso vlid for m < 0 s long s l + m 0 i.e. m l. Useful Properties The ssocited Legendre functions lso hve mny useful properties, here we will mention just two: They re relted under chnge of the sign of m by P m m (l m! l (x = ( (l + m! P l m (x. (50 Secondly, for fixed m the ssocited Legendre functions form n orthogonl set on the intervl x nd they re normlised so tht 6 Sphericl Hrmonics Pl m (xp l m (xdx = 2 (l + m! 2l + (l m! δ l l. (5 We hd decomposed the function Y (θ, φ = P (θq(φ, now recombining them we cn define our sphericl hrmonics (fter choosing n overll constnt 2l + (l m! Y lm (θ, φ = 4π (l + m! P l m (cos θeimφ. (52 From (50 we cn see tht Y l( m (θ, φ = ( m Ylm (θ, φ nd, from the normlistion, tht the sphericl hrmonics re orthonorml 2π 0 dφ π 0 sin θdθ Y l m (θ, φy lm(θ, φ = δ l,l δ m,m. (53 7

8 Furthermore the sphericl hrmonics form complete set of functions nd hence stisfy the completeness reltion l l=0 m= l Some simple exmples re, Y lm (θ, φ Y lm (θ, φ = δ(φ φ δ(cos θ cos θ. (54 Y 00 = 4π, 3 Y = 8π sin θeiφ, 3 Y 0 = 4π cos θ, Y 22 = 5 4 2π sin2 θe 2iφ, 5 Y 2 = 8π sin θ cos θeiφ, 5 Y 20 = 4π ( 3 2 cos2 θ 2. Given this completeness nd orthogonlity we cn expnd n rbitrry function, g(θ, φ, in sphericl hrmonics g(θ, φ = l l=0 m= l A lm Y lm (θ, φ (55 where the coefficients re given by A lm = dω Ylm (θ, φg(θ, φ. (56 Finlly let us recll tht n rbitrry potentil which stisfies Lplce s eqution cn thus be expnded s Φ(r, θ, φ = l l=0 m= l [ Alm r l + B lm r l ] Y lm (θ, φ. (57 Addition Theorem for Sphericl Hrmonics An importnt property of sphericl hrmonics is encoded in the theorem (see Jckson sec. 3.6: Given two vectors x nd x with coordintes (r, θ, φ nd (r, θ, φ with the ngle between them being γ one cn write cos γ = cos θ cos θ + sin θ sin θ cos(φ φ (58 P l (cos γ = 4π 2l + l m= l Y lm (θ, φ Y lm (θ, φ. (59 8

9 Reclling the expnsion of / x x in terms of P l (cos γ from clss notes or Jckson eqution (3.38, this cn be used to write x x = 4π 7 Multipole Expnsion x x = l l=0 m= l l=0 2l + r< l r> l+ r< l r> l+ P l (cos γ (60 Y lm (θ, φ Y lm (θ, φ. (6 Let us consider distribution of chrge with some chrge density, ρ(x, loclised bout the origin. The potentil t position x outside of the chrge distribution is given by Φ(x = ρ(x 4πɛ 0 x x d3 x. (62 We cn now use the expnsion (6, with r < = r nd r > = r, so tht Φ(x = ɛ 0 l,m [ Ylm 2l + (θ, φ ρ(x r l d 3 x ] Y lm (θ, φ r l+. (63 Tht is to sy we cn write the potentil s n expnsion, clled the multipole expnsion, in sphericl hrmonics Φ(x = 4π 4πɛ 0 2l + q Y lm (θ, φ lm r l+, l,m where the coefficients q lm = Y lm (θ, φ ρ(x r l d 3 x re clled the multipole moments of the chrge distribution. Note tht becuse of the properties of the sphericl hrmonics q l,( m = ( m qlm. The first few terms in the expnsion re clled: l = 0 term is clled the monopole term l = term is clled the dipole term l = 2 term is clled the qudrupole term etc. Explicitly they re given by: q 00 = 4π where q is the totl chrge. The dipole terms re 3 q = 8π (p x ip y, q 0 = ρ(x d 3 x = 4π q (64 3 4π p z (65 where p x,y,z re the components of the electric dipole moment p = x ρ(x d 3 x. (66 9

10 The qudrupole terms re q 22 = 5 2 2π (Q 2iQ 2 Q 22 q 2 = 5 3 8π (Q 3 iq 23, q 20 = 5 2 4π Q 33 (67 where the qudrupole moments re given by Q ij = (3x ix j r 2 δ ij ρ(x d 3 x (68 which re trceless tensors of rnk two. In rectngulr coordintes, by directly mking Tylor expnsion of the x x potentil, one cn find Φ(x = 4πɛ 0 [ q r + p x r i,j x i x ] j Q ij r where s usul r = x. Of course one cn mke similr expnsion for the electric field. References [] G. Arfken, nd H. Weber, Mthemticl Methods for Physicists, 5th edition, Acdemic Press. (69 0