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# emsol12 - Electromagnetic Theory I Problem Set 12 Due 21...

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Electromagnetic Theory I Problem Set 12 Due: 21 November 2011 45. J, Problem 6.1 Solution : a) (2 Dimensional Space) ψ ( r , t ) = Z d 3 x 0 δ ( x 0 ) δ ( y 0 ) δ ( t - | r - r 0 | /c ) 4 π | r - r 0 | = Z dz 0 δ ( t - p x 2 + y 2 + ( z - z 0 ) 2 /c ) 4 π p x 2 + y 2 + ( z - z 0 ) 2 (1) If ρ 2 = x 2 + y 2 > c 2 t 2 the argument of the delta function never vanishes so ψ = 0 in that case. When ρ < ct , the delta function contributes at | z - z 0 | = p c 2 t 2 - ρ 2 , and the derivative of its argument w.r.t. z 0 is ( z - z 0 ) /c p ρ 2 + ( z - z 0 ) 2 Then ψ = 2 ( ct - ρ ) 4 π p c 2 t 2 - ρ 2 (2) where the factor of two because both z - z 0 = ± p c 2 t 2 - ρ 2 contribute. b)(1 Dimensional Space) ψ ( r , t ) = Z d 3 x 0 δ ( x 0 ) δ ( t - | r - r 0 | /c ) 4 π | r - r 0 | = Z dy 0 dz 0 δ ( t - p x 2 + ( y 0 ) 2 + ( z 0 ) 2 /c ) 4 π p x 2 + ( y 0 ) 2 + ( z 0 ) 2 = Z 2 π 0 Z 0 rdr δ ( t - x 2 + r 2 /c ) 4 π x 2 + r 2 = Z 0 rdr δ ( t - x 2 + r 2 /c ) 2 x 2 + r 2 (3) If | x | > ct the argument of the delta function never vanishes so ψ = 0 in that case. When | x | < ct , the delta function contributes at r = c 2 t 2 - x 2 . and the derivative of its argument w.r.t. r is r/c x 2 + r 2 Then ψ = ( ct - | x | ) 2 (4) 46. J, Problem 6.3. In part d), omit relating the discussion to Sections 5.18B and Problems 5.35 and 5.36. Solution : 1

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a) We write the Fourier transform of A as A ( r , t ) = Z d 3 ke i · a ( k , t ) (5) Then ( 2 - μσ∂ t ) A = 0 implies that a ( k , t ) = a ( k , 0) e - k 2 t/μσ = Z d 3 x 0 (2 π ) 3 A ( r 0 , 0) e - k 2 t/μσ e - i · 0 Comparing with the desired form we read off G ( r - r 0 , t ) = Z d 3 k (2 π ) 3 e i · ( - 0 ) - k 2 t/μσ , t > 0 (6) b) We include a factor of the step function θ ( t ) in the definition of G , so that it is zero for t < 0. Then -∇ 2 + μσ ∂t θ ( t ) Z d 3 k (2 π ) 3 e i · ( - 0 ) - k 2 t/μσ = μσδ ( t ) Z d 3 k (2 π ) 3 e i · ( - 0 ) = μσδ ( t ) δ ( r - r 0 )
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emsol12 - Electromagnetic Theory I Problem Set 12 Due 21...

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