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Lecture 6 Notes, Electromagnetic Theory I Dr Lecture 6 Notes, Electromagnetic Theory I Dr. Christopher S. Baird University of Massachusetts Lowell 1. Associated Legendre Polynomials - We now return to solving the Laplace equation in spherical coordinates when there is no azimuthal symmetry by solving the full Legendre equation for m = 0 and m ≠ 0: m 2 d 2 d Pl x m m [1− x ][l l1− ]P x=0 where x=cos dx dx 1−x2 l - Once this equation is solved, the general solution for the Laplace equation in spherical coordinates will have the form: =∑ l −l−1 m=0 r , , Al r B l r Am0 Bm0 P l cos l ∑ l −l −1 i m −i m m Al r Bl r Am e B m e P l cos m≠0,l - Encouraged by the Rodrigues' form of the solution to the m = 0 Legendre equation, we try a m( )=( − 2)m/ 2 d m ( ) solution of the form P l x 1 x dx m g x and substitute in: m1 m m =− − 2m/2[ d ] 2 d 2 − 2m/2−1− d − 2m/ 2 0 m x 1 x g x m g x x 1 x m g x 1 x dxm 1 dxm dx m m2 m1 − 2m/21[ d ]− − 2m/ 2 d 1 x g x m 2 x 1 x g x dxm 2 dxm 1 m m 2 m/2 d 2 2 m/ 2−1 d [l l1]1−x g x−m 1−x g x dxm dx m - Collect all similar terms: m 2 m1 m = − 2[ d d ]− [ ] d [ − ] d 0 1 x g x 2 m 1 x g x l l 1 m m 1 g x dxm dx2 dxm 1 dxm - We want to move as much as possible inside the derivatives. We can do this by using the product rule. m m m− k k d =∑ m ! [ d ][ d ] - The product rule used m times states: m u v − m−k u k v dx k=0 k ! m k ! dx dx - Let us try to apply this to rearrange the first term in the differential equation. 2 = − 2 d =− d =− - If u 1 x then dx u 2 x , dx2 2 and all higher derivatives in the expansion are zero so that: m m m−1 m−2 d − 2 = − 2[ d ]− [ d ]− − [ d ] 1 x v 1 x v 2 m x − v m m 1 − v dxm dxm dxm 1 dxm 2 - Moving terms around: m m m−1 m−2 − 2[ d ]= d − 2 [ d ] − [ d ] 1 x v 1 x v 2m x − v m m 1 − v dxm dxm dxm 1 dxm 2 = d 2 ( ) - Now if we set v dx 2 g x this product rule expansion becomes: m 2 m 2 m1 m − 2[ d d ]= d − 2 d [ d ] − [ d ] 1 x g x 1 x g x 2 m x g x m m 1 g x dxm dx2 dxm dx2 dxm 1 dxm - We can use this product rule to rearrange the first term of the Legendre equation so it becomes: m 2 m1 m m = d [ − 2 d ]− d d − [ d ] 0 1 x g x 2 x g x l l 1 g x 2m g x dxm dx2 dxm 1 dxm dxm - Now we want to use the product rule expansion on the second term of the Legendre equation. = du = = d ( ) With u x , dx 1 , all higher derivatives are zero and v dx g x , the general product rule becomes: m m1 m d [ d ]= [ d ] [ d ] x g x x g x m g x dxm dx dxm 1 dx m - After rearranging: m1 m m [ d ]= d [ d ]− [ d ] x g x x g x m g x dxm 1 dxm dx dx m - Substitute in: m 2 d 2 d d 0= [1− x g x−2[x g x]ll1 gx] dxm dx2 dx m d d 2 d 0= [ ((1− x ) g( x))+l (l+1) g(x)] dxm dx dx - It is worth noting that if this is true for m = 0, then it will automatically be true for all m. So we can set m = 0: d 2 d 0= 1− x g xl l1g x dx dx - Now this is just the m = 0 Legendre equation, which we have already solved. We found the = solutions to be the ordinary Legendre polynomials, Pl(x), so that g x P l x . - We now have the solution to the full Legendre equation: m m = − 2m/2 d P l x 1 x Pl x dxm - There is an arbitrary phase factor that we have assumed to be one for simplicity, but is conventionally set to −1m , so that the associated Legendre functions become: m m( )=(− )m( − 2)m/2 d ( ) P l x 1 1 x P l x dxm - Or written explicitly using Rodrigues' equation: (− )m m+l m( )= 1 ( − 2 )m/2 d ( 2− )l P x 1 x + x 1 l 2l l ! dxm l - It is worth noting upon examination of the equation m d d 2 d 0= [ 1−x gxl l1 g x] dxm dx dx that the highest power of x inside the brackets is l, so that m cannot be greater than l, otherwise the equation would be trivially satisfied and thus there would be no unique solution. In summary, both l and m are integers and the possible m are -l, -(l-1), ..., 0, ..., (l-1), l. - Several useful mathematical relations involving the associated Legendre functions can be found (the derivations are left to the interested student). − − l m ! P m x=−1m Pm x l lm! l − m = m − m l m 1 Pl1 x 2l 1 x Pl x l m Pl −1 x − 2 m1 = − m − m 1 x Pl x l m x Pl x l m Pl −1 x 2− d m = m − m x 1 P x l x P x l m P − x dx l l l 1 - For fixed m, the Legendre functions form an orthogonal set: 1 ∫ m m = 2 l m ! Pl ' x P l x dx − l ' l −1 2l 1 l m ! 2. Spherical Harmonics - With the full Legendre equation now solved, the general solution of the Laplace equation in spherical coordinates has been found: ∞ =∑ l −l −1 m=0 r , , Al r B l r Am0 B m0 P l cos l =0 ∞ l ∑ ∑ l −l −1 i m −i m m Al r Bl r Am e B m e Pl cos l =0 m=1 m where Pl are the associated Legendre functions. - Typically, most problems require a valid solution on the full range of . The single-valued requirement in this case forces Bm0 = 0, and the m = 0 term can now be included in the sum with the rest of the terms: ∞ l =∑ ∑ l −l −1 i m −im m r , , Al r Bl r Am e Bm e Pl cos l =0 m=0 - An alternate way of presenting this is to let m sum from -l to l and thus combine the Am and Bm terms with the other constants: ∞ l =∑ ∑ l −l−1 i m m r , , Al , m r Bl ,m r e Pl cos l =0 m=−l - The normalizing terms are included in the undetermined constants Alm and Blm. When a boundary condition is applied, these constants become specified and the normalizing terms result naturally. However, to make the math cleaner, we can explicitly pull the normalizing terms out of the constants in advance. ∞ l − =∑ ∑ l −l−1 2l 1 l m ! i m m r , , Al , m r B l ,m r e Pl cos l =0 m=−l 4 l m ! l - It is now evident that P m(cos θ) times its normalization term forms a complete set of orthonormal functions in the variable θ, indexed by l for a fixed m, and eimφ times its normalization term forms a complete set of orthonormal functions in the variable φ, indexed by m. The product of both thus creates orthonormal functions that spans all angles of the unit circle. These are known as “spherical harmonics” Ylm: + ( − ) 2l 1 l m ! i mϕ m Y (θ ,ϕ)=√ √ e P (cosθ) so that lm 4π (l+m)! l ∞ l Φ( θ ϕ)=∑ ∑ ( l + −l −1) (θ ϕ) r , , Al , m r Bl , m r Y lm , l=0 m=−l - The orthonormality of the spherical harmonics explicitly means that: 2 1 ∫∫ * = = Y l ' m' x , Y l m x , dx d l ' l m'm where x cos 0 −1 2π π ∫∫ * (θ ϕ) (θ ϕ) θ θ ϕ=δ δ or Y l ' m' , Y l m , sin d d l 'l m'm 0 0 - The validity of this equation can be checked trivially by expressing the spherical harmonics in terms of its definition and applying the orthonormality of the associated Legendre functions and the complex exponentials. - Using the definitions of the spherical harmonics, the patient student can work out the explicit analytic form for any given l and m. The lowest-order spherical harmonics are especially simple and are typically tabulated in textbooks. - Because the spherical harmonics form an orthonormal set of equations, an arbitrary function f , can be expanded in terms of spherical harmonics: π ∞ l 2 π (θ ϕ)=∑ ∑ (θ ϕ) =∫∫ (θ ϕ) * (θ ϕ) θ θ ϕ f , Alm Y lm , where Alm f , Y lm , sin d d = =− l 1 m l 0 0 - There are several useful special cases for spherical harmonics that we should keep in mind.
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