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Unformatted text preview: 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 ,~x = 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 threedimensional wave equation (6.47), show that the source f ( ~x ,t ) = δ ( x ) δ ( y ) δ ( t ), equivalent to a t = 0 point source at the origin in two spatial dimensions, produces a twodimensional 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 ,t )] ret  ~x ~x  d 3 x we find Ψ( ~x,t ) = Z δ ( x ) δ ( y ) δ ( t R/c ) R dx dy dz = Z ∞∞ δ ( t R/c ) R dz where R =  ~x ~x  = p ρ 2 + ( z z ) 2 when x = y = 0 By shifting z → z + z , we end up with the integral Ψ( ρ,t ) = Z ∞∞ δ ( t p ρ 2 + z 2 /c ) p ρ 2 + z 2 dz (1) Using δ ( f ( ζ )) = X i 1  f ( ζ )  δ ( ζ ζ i ) (2) where the sum is over the zeros of f ( ζ ), we see that δ ( t p ρ 2 + z 2 /c ) = X i c p ρ 2 + z 2  z  δ ( z z i ) The zeros z i are given by ρ 2 + z 2 = c 2 t 2 ⇒ z = ± 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 onedimensional wave proportional to Ψ( x,t ) = 2 πc Θ( ct  x  ) For the sheet source, we use f ( ~x ,t ) = δ ( x ) δ ( t ) to write Ψ( ~x,t ) = Z δ ( x ) δ ( t R/c ) R dx dy dz where R = p ( x x ) 2 + ( y y ) 2 + ( z z ) 2 . By integrating x and shifting y → y + y and z → z + z we end up with Ψ( x,t ) = Z δ ( t p x 2 + y 2 + z 2 /c ) p x 2 + y 2 + z 2 dy dz = Z δ ( t p ρ 2 + x 2 /c ) p ρ 2 + x 2 ρ dρ dφ where we have gone to polar coordinates in the y z plane. The φ integral is now trivial. Treating the delta function as in (2) results in Ψ( x,t ) = 2 π Z ∞ X i cδ ( ρ ρ i ) dρ where the zeros ρ i corespond to ρ 2 + x 2 = c 2 t 2 ⇒ ρ = ± p c 2 t 2 x 2 Since ρ is nonnegative, only the positive zero contributes, and we end up with Ψ( x,t ) = 2 πc Θ( ct  z  ) where the step function enforces the condition for a real zero to exist.where the step function enforces the condition for a real zero to exist....
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This note was uploaded on 11/21/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Stephens,P
 Magnetism, Work, Light

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