F07_hw05a

# F07_hw05a - Physics 505 Homework Assignment#5 Solutions...

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Physics 505 Fall 2007 Homework Assignment #5 — Solutions Textbook problems: Ch. 3: 3.13, 3.17, 3.26, 3.27 3.13 Solve for the potential in Problem 3.1, using the appropriate Green function obtained in the text, and verify that the answer obtained in this way agrees with the direct solution from the differential equation. Recall that Problem 3.1 asks for the potential between two concentric spheres of radii a and b (with b > a ), where the upper hemisphere of the inner sphere and the lower hemisphere of the outer sphere are maintained at potential V , and where the other hemispheres are at zero potential. Since this problem involves the potential between two spheres, we use the Dirichlet Green’s function G ( ~x, ~x 0 ) = X l,m 4 π 2 l + 1 1 1 - ( a b ) 2 l +1 r l < - a 2 l +1 r l +1 < 1 r l +1 > - r l > b 2 l +1 Y m l (Ω) Y m * l 0 ) (1) Because there are no charges between the spheres, the Green’s function solution for the potential only involves the surface integral Φ( ~x ) = - 1 4 π Z S Φ( ~x 0 ) ∂G ∂n 0 da 0 Here, we note a subtlety in that the boundary surface is actually disconnected, and includes both the inner sphere of radius a and the outer sphere of radius b . This means that the potential may be written as a sum of two contributions Φ( ~x ) = - 1 4 π Z r 0 = a Φ( ~x 0 ) ∂G ∂n 0 a 2 d Ω 0 - 1 4 π Z r 0 = b Φ( ~x 0 ) ∂G ∂n 0 b 2 d Ω 0 (2) We now compute the normal derivatives of the Green’s function (1) ∂G ∂n 0 r 0 = a = - ∂G ∂r 0 r < = r 0 = a = = - 4 π a 2 X l,m 1 1 - ( a b ) 2 l +1 a r l +1 - a b l +1 r b l Y m l (Ω) Y m * l 0 ) and ∂G ∂n 0 r 0 = b = ∂G ∂r 0 r > = r 0 = b = = - 4 π b 2 X l,m 1 1 - ( a b ) 2 l +1 r b l - a b l a r l +1 Y m l (Ω) Y m * l 0 )

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Inserting these expressions into (2) yields Φ( ~x ) = X l,m Z V a 0 ) Y m * l 0 ) d Ω 0 1 1 - ( a b ) 2 l +1 a r l +1 - a b l +1 r b l Y m l (Ω) + X l,m Z V b 0 ) Y m * l 0 ) d Ω 0 1 1 - ( a b ) 2 l +1 r b l - a b l a r l +1 Y m l (Ω) This is the general expression for the solution to the boundary value problem where V a (Ω) is the potential on the inner sphere and V b (Ω) is the potential on the outer sphere. For the upper/lower hemispheres problem, we note that azimuthal symmetry allows us to restrict the m values to m = 0 only. In this case, the spherical harmonic expansion reduces to a Legendre polynomial expansion Φ( ~x ) = X l Z 1 - 1 V a ( ζ ) P l ( ζ ) 2 l + 1 2 1 - ( a b ) 2 l +1 a r l +1 - a b l +1 r b l P l (cos θ ) + X l Z V b ( ζ ) P l ( ζ ) 2 l + 1 2 1 - ( a b ) 2 l +1 r b l - a b l a r l +1 P l (cos θ ) where ζ = cos θ 0 . Since V a = V for ζ > 0 and V b = V for ζ < 0, this above expression reduces to Φ( ~x ) = X l (2 l + 1) V N l 2 1 - ( a b ) 2 l +1 a r l +1 - a b l +1 r b l P l (cos θ ) + X l (2 l + 1)( - 1) l V N l 2 1 - ( a b ) 2 l +1 r b l - a b l a r l +1 P l (cos θ ) = X l (2 l + 1) V N l 2 1 - ( a b ) 2 l +1 a r l +1 - a b l +1 r b l + ( - 1) l r b l - a b l a r l +1 P l (cos θ ) where N l = Z 1 0 P l ( ζ ) = ( 1 l = 0 ( - 1) j +1 Γ( j - 1 2 ) 2 πj ! l = 2 j - 1 odd
If desired, the potential may be rearranged to read Φ( ~x ) = X l (2 l + 1) V N l 2 1 - ( a b ) 2 l +1 1 + ( - 1) l +1 a b l a r l +1 + ( - 1) l 1 + ( - 1) l +1 a b l +1 r b l P l (cos θ ) = V 2 + V X j =1 ( - 1) j +1 (4 j - 1)Γ( j - 1 2 ) 4 πj ! 1 - ( a b ) 4 j - 1 " 1 + a b 2 j - 1 a r 2 j - 1 + a b 2 j r b 2 j - 1 # P 2 j - 1 (cos θ ) which agrees with the solution to Problem 3.1 that we have found earlier.

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