PHYS 500 - Southern Illinois University October 13, 2016 PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 1 / 10
The Laplacian in Spherical Coordinates The Laplacian is given by 2 = = 2 x 2 + 2 y 2 + 2 z 2 = 1 r 2 r ( r 2 r ) + 1 r 2 sin θ ( sin θ ) + θ θ 1 r 2 sin 2 θ 2 φ 2. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 2 / 10
The Laplacian in Spherical Coordinates The Laplacian is given by 2 = = 2 x 2 + 2 y 2 + 2 z 2 Example = 1 r 2 r ( r 2 r In electrostatics, Gauss s law says ) + 1 r 2 sin θ ( sin θ ) + θ θ 2 ρ(r, θ, φ) Φ(r, θ, φ) =. ɛ 0 1 r 2 sin 2 θ 2 φ 2. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 2 / 10
The Laplacian with Azimuthal Symmetry Suppose that the potential Φ(r, θ, φ) is known to have azimuthal symmetry (i.e. Φ(r, θ) is independent of φ). PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 3 / 10
The Laplacian with Azimuthal Symmetry Suppose that the potential Φ(r, θ, φ) is known to have azimuthal symmetry (i.e. Φ(r, θ) is independent of φ). Gauss s law in free space then takes the form: [ ( 1 r 2 r 2 ) + 1 ( r r r 2 sin θ ) ] Φ(r, θ) = 0. sin θ θ θ PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 3 / 10
The Laplacian with Azimuthal Symmetry Suppose that the potential Φ(r, θ, φ) is known to have azimuthal symmetry (i.e. Φ(r, θ) is independent of φ). Gauss s law in free space then takes the form: [ ( 1 r 2 r 2 ) + 1 ( r r r 2 sin θ ) ] Φ(r, θ) = 0. sin θ θ θ Separation of Variables Assuming a solution of the form Φ(r, θ) = R(r)P(θ) leads to the pair of diff. eq.: ( 1 d r 2 dr ) ( 1 d = l(l + 1), sin θ dp ) = l(l + 1). R dr dr P sin θ dθ dθ PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 3 / 10
Radial Equation The radial equation has the general solution R(r) = Ar l + Br l 1. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 4 / 10
Radial Equation The radial equation has the general solution Polar Equation R(r) = Ar l + Br l 1. To solve the polar equation, make the change of variable x = cos θ so that d dθ = dx d dθ dx = sin θ d dx. Note this requires 1 x 1. Then 0 = 1 ( d sin θ dp ) + l(l + 1)P P sin θ dθ dθ = d [ (1 x 2 ) dp(x) ] + l(l + 1)P(x) dx dx = (1 x 2 )P 2xP + l(l + 1)P. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 4 / 10
To solve 0 = (1 x 2 )P 2xP + l(l + 1)P, try the power series solution P(x) = a k x k. k=0 PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 5 / 10
To solve 0 = (1 x 2 )P 2xP + l(l + 1)P, try the power series solution P(x) = This leads to the recursion relation a k x k. k=0 (k + 1)(k + 2)a k+2 = [(k + 1)k l(l + 1)]a k. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 5 / 10
To solve 0 = (1 x 2 )P 2xP + l(l + 1)P, try the power series solution P(x) = This leads to the recursion relation a k x k. k=0 (k + 1)(k + 2)a k+2 = [(k + 1)k l(l + 1)]a k. We pursue two types of solutions P + (x) and P (x), corresponding to even choices of k for P + (x) and odd k for P (x). PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 5 / 10
Even Solutions l(l + 1) a 2 = a 0 2 3 2 l(l + 1) (3 2 l(l + 1))( l(l + 1)) a 4 = a 2 = a 0 4 3 4! 5 4 l(l + 1) 5 4 l(l + 1) (3 2 l(l + 1))( l(l + 1)) a 6 = a 4 = a 0 6 5 6 5 4! a 2k = [(2k 1)(2k 2) l(l + 1)] [3 2 l(l + 1)][ l(l + 1)] a 0 (2k)! = a k 1 0 [(2j 1)2j l(l + 1)]. (2k)! j=1 PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 6 / 10
Even Solutions Therefore, k 1 P + (x) = a 0 [(2j + 1)2j l(l + 1)] x 2k (2k)!. Key Observation: k=0 j=1 When x = 1, the solution takes the form k 1 P + (1) = a 0 [(2j + 1)2j l(l + 1)] 1 (2k)! k=0 This infinite series diverges! j=1 PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 7 / 10
Proof: Write P + (1) = a 0 k=0 α k where Perform the ratio test: k 1 1 α k = [(2j + 1)2j l(l + 1)] (2k)!. j=1 α k+1 (2k + 1)2k l(l + 1) = α k (2k + 2)(2k + 1) 1 k 2 k + 1. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 8 / 10
Proof: Write P + (1) = a 0 k=0 α k where Perform the ratio test: k 1 1 α k = [(2j + 1)2j l(l + 1)] (2k)!. j=1 α k+1 (2k + 1)2k l(l + 1) = α k (2k + 2)(2k + 1) 1 k 2 k + 1. Compare with the divergent series k=1 1 k whose ratio test also gives k k+1. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 8 / 10
For large k, the series P + (1) = a 0 k=0 α k behaves like a divergent geometric series. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 8 / 10 Legendre s Equation Proof: Write P + (1) = a 0 k=0 α k where Perform the ratio test: k 1 1 α k = [(2j + 1)2j l(l + 1)] (2k)!. j=1 α k+1 (2k + 1)2k l(l + 1) = α k (2k + 2)(2k + 1) 1 k 2 k + 1. Compare with the divergent series k=1 1 k whose ratio test also gives k k+1.
Therefore, in order for P + (1) not to diverge, the sequence a 0 k=0 α k must terminate. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 9 / 10
Therefore, in order for P + (1) not to diverge, the sequence a 0 k=0 α k must terminate. In other words, the terms k 1 1 α k = [(2j + 1)2j l(l + 1)] (2k)! j=1 must vanish for all k N, where N is some integer. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 9 / 10
Therefore, in order for P + (1) not to diverge, the sequence a 0 k=0 α k must terminate. In other words, the terms k 1 1 α k = [(2j + 1)2j l(l + 1)] (2k)! j=1 must vanish for all k N, where N is some integer. This is only possible if l = 2N. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 9 / 10
Therefore, in order for P + (1) not to diverge, the sequence a 0 k=0 α k must terminate. In other words, the terms k 1 1 α k = [(2j + 1)2j l(l + 1)] (2k)! j=1 must vanish for all k N, where N is some integer. This is only possible if l = 2N. Summary In order for P + (x) to not diverge on the interval [ 1, 1], the separation constant l must be an even integer. PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 9 / 10
Odd solutions: A similar argument shows that for odd choices of k, the solution is given by k P (x) = a 1 [2j(2j 1) l(l + 1)] x 2k+1 (2k + 1)!, k=0 where l is some odd integer. j=1 PHYS 500 - Southern Illinois University Legendre s Equation October 13, 2016 10 / 10