EMLecture6 - Lecture 6 Notes, 95.657 Electromagnetic Theory...

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Unformatted text preview: Lecture 6 Notes, 95.657 Electromagnetic Theory I, Fall 2011 Dr. Christopher S. Baird, UMass 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: d dx [ 1 x 2 d P l m x dx ] [ l l 1 m 2 1 x 2 ] P l m x = where x = cos - Once this equation is solved, the general solution for the Laplace equation in spherical coordinates will have the form: r , , = l A l r l B l r l 1 A m B m P l m = cos m 0, l A l r l B l r l 1 A m e i m B m e i m P l m cos - Encouraged by the Rodrigues' form of the solution to the m = 0 Legendre equation, we try a solution of the form P l m ( x )=( 1 x 2 ) m / 2 d m dx m g ( x ) and substitute in: = m x 1 x 2 m / 2 [ d m 1 dx m 1 g x ] m 2 d m dx m g x x 2 1 x 2 m / 2 1 m d m dx m g x 1 x 2 m / 2 1 x 2 m / 2 1 [ d m 2 dx m 2 g x ] m 2 x 1 x 2 m / 2 d m 1 dx m 1 g x [ l l 1 ] 1 x 2 m / 2 d m dx m g x m 2 1 x 2 m / 2 1 d m dx m g x - Collect all similar terms: = 1 x 2 [ d m dx m d 2 dx 2 g x ] 2 [ m 1 ] x d m 1 dx m 1 g x [ l l 1 m m 1 ] d m dx m g x - We want to move as much as possible inside the derivatives. We can dot his by using the product rule.- The product rule used m times states: d m dx m u v = k = m m ! k ! m k ! [ d m k dx m k u ][ d k dx k v ]- Let us try to apply this to rearrange the first term in the differential equation.- If u = 1 x 2 then d dx u = 2 x , d 2 dx 2 = 2 and all higher derivatives in the expansion are zero so that: d m dx m 1 x 2 v = 1 x 2 [ d m dx m v ] 2 m x [ d m 1 dx m 1 v ] m m 1 [ d m 2 dx m 2 v ]- Moving terms around: 1 x 2 [ d m dx m v ] = d m dx m 1 x 2 v 2 m x [ d m 1 dx m 1 v ] m m 1 [ d m 2 dx m 2 v ]- Now if we set v = d 2 dx 2 g ( x ) this product rule expansion becomes: 1 x 2 [ d m dx m d 2 dx 2 g x ] = d m dx m 1 x 2 d 2 dx 2 g x 2 m x [ d m 1 dx m 1 g x ] m m 1 [ d m dx m g x ]- We can use this product rule to rearrange the first term of the Legendre equation so it becomes: = d m dx m [ 1 x 2 d 2 dx 2 g x ] 2 x d m 1 dx m 1 g x l l 1 d m dx m g x 2 m [ d m dx m g x ]- Now we want to use the product rule expansion on the second term of the Legendre equation. - Now we want to use the product rule expansion on the second term of the Legendre equation....
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EMLecture6 - Lecture 6 Notes, 95.657 Electromagnetic Theory...

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