F07_hw10a - Physics 505 Homework Assignment#10 Solutions...

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Physics 505 Fall 2007 Homework Assignment #10 — Solutions Textbook problems: Ch. 6: 6.1, 6.4, 6.13, 6.18 6.1 In three dimensions the solution to the wave equation (6.32) for a point source in space and time (a light flash at t 0 = 0 , ~x 0 = 0) is a spherical shell disturbance of radius R = ct , namely the Green function G (+) (6.44). It may be initially surprising that in one or two dimensions, the disturbance possesses a “wake”, even though the source is a “point” in space and time. The solutions for fewer dimensions than three can be found by superposition in the superfluous dimension(s), to eliminate dependence on such variable(s). For example, a flashing line source of uniform amplitude is equivalent to a point source in two dimensions. a ) Starting with the retarded solution to the three-dimensional wave equation (6.47), show that the source f ( ~x 0 , t ) = δ ( x 0 ) δ ( y 0 ) δ ( t 0 ), equivalent to a t = 0 point source at the origin in two spatial dimensions, produces a two-dimensional wave Ψ( x, y, t ) = 2 c Θ( ct - ρ ) p c 2 t 2 - ρ 2 where ρ 2 = x 2 + y 2 and Θ( ξ ) is the unit step function [Θ( ξ ) = 0 (1) if ξ < ( > ) 0.] Using Ψ( ~x, t ) = Z [ f ( ~x 0 , t 0 )] ret | ~x - ~x 0 | d 3 x 0 we find Ψ( ~x, t ) = Z δ ( x 0 ) δ ( y 0 ) δ ( t - R/c ) R dx 0 dy 0 dz 0 = Z -∞ δ ( t - R/c ) R dz 0 where R = | ~x - ~x 0 | = p ρ 2 + ( z - z 0 ) 2 when x 0 = y 0 = 0 By shifting z 0 z 0 + z , we end up with the integral Ψ( ρ, t ) = Z -∞ δ ( t - p ρ 2 + z 0 2 /c ) p ρ 2 + z 0 2 dz 0 (1) Using δ ( f ( ζ )) = X i 1 | f 0 ( ζ ) | δ ( ζ - ζ i ) (2)
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where the sum is over the zeros of f ( ζ ), we see that δ ( t - p ρ 2 + z 0 2 /c ) = X i c p ρ 2 + z 0 2 | z 0 | δ ( z 0 - z 0 i ) The zeros z 0 i are given by ρ 2 + z 0 2 = c 2 t 2 z 0 = ± p c 2 t 2 - ρ 2 However it is clear that there are real zeros only if c 2 t 2 ρ 2 or ρ < ct . Going back to (1), and noting there are two zeros (one for each sign of the square root), we end up with Ψ( ρ, t ) = 2 c Θ( ct - ρ ) p c 2 t 2 - ρ 2 b ) Show that a “sheet” source, equivalent to a point pulsed source at the origin in one space dimension, produces a one-dimensional wave proportional to Ψ( x, t ) = 2 πc Θ( ct - | x | ) For the sheet source, we use f ( ~x 0 , t 0 ) = δ ( x 0 ) δ ( t 0 ) to write Ψ( ~x, t ) = Z δ ( x 0 ) δ ( t - R/c ) R dx 0 dy 0 dz 0 where R = p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 . By integrating x 0 and shifting y 0 y 0 + y and z 0 z 0 + z we end up with Ψ( x, t ) = Z δ ( t - p x 2 + y 0 2 + z 0 2 /c ) p x 2 + y 0 2 + z 0 2 dy 0 dz 0 = Z δ ( t - p ρ 0 2 + x 2 /c ) p ρ 0 2 + x 2 ρ 0 0 0 where we have gone to polar coordinates in the y 0 - z 0 plane. The φ 0 integral is now trivial. Treating the delta function as in (2) results in Ψ( x, t ) = 2 π Z 0 X i ( ρ 0 - ρ 0 i ) 0 where the zeros ρ 0 i corespond to ρ 0 2 + x 2 = c 2 t 2 ρ 0 = ± p c 2 t 2 - x 2 Since ρ 0 is non-negative, only the positive zero contributes, and we end up with Ψ( x, t ) = 2 πc Θ( ct - | z | )
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where the step function enforces the condition for a real zero to exist. 6.4 A uniformly magnetized and conducting sphere of radius R and total magnetic mo- ment m = 4 πMR 3 / 3 rotates about its magnetization axis with angular speed ω . In the steady state no current flows in the conductor. The motion is nonrelativistic; the sphere has no excess charge on it.
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