W11HW1 - 6.1 Problem 6.1 6.1.1 Substituting f ( ~x ; t ) =...

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Unformatted text preview: 6.1 Problem 6.1 6.1.1 Substituting f ( ~x ; t ) = ( x ) ( y ) ( t ) into equation 6.47 in Jackson: ( ~x; t ) = Z [ f ( ~x ; t )] ret j ~x ~x j d 3 x = Z [ ( x ) ( y ) ( t )] ret j ~x ~x j dx dy dz Noting that [ t ] ret = t j ~x ~x j =c , ( ~x; t ) = ZZZ ( x ) ( y ) t p ( x x ) 2 + ( y y ) 2 + ( z z ) 2 =c p ( x x ) 2 + ( y y ) 2 + ( z z ) 2 dx dy dz = Z t p x 2 + y 2 + ( z z ) 2 =c p x 2 + y 2 + ( z z ) 2 dz Letting = p x 2 + y 2 and ~ z = z z : ( ~x; t ) = Z t p 2 + ~ z 2 =c p 2 + ~ z 2 d ~ z We will use the following identity: ( f ( z )) = X i 1 j f ( z ) j ( z z i ) (6.1) where z i are the zeroes of f ( z ): z i = p c 2 t 2 2 . Hence, the delta function our expression for ( ~x; t ) is equal to: t p 2 + z 2 =c = X i c p 2 + z 2 j z j ( z z i ) ( ~x; t ) = Z 1 p 2 + ~ z 2 " c p 2 + ~ z 2 j ~ z j ~ z p c 2 t 2 2 + c p 2 + ~ z 2 j ~ z j ~ z + p c 2 t 2 2 # d ~ z = 1 r 2 + p c 2 t 2 2 2 2 6 6 4 c r 2 + p c 2 t 2 2 2 p c 2 t 2 2 + c r 2 + p c 2 t 2 2 2 p c 2 t 2 2 3 7 7 5 = 1 j ct j 2 c j ct j p c 2 t 2 2 1 Note that this solution is imaginary for ct < as a result of the delta function we're using. However, it is important to note that we're integrating over the real number line{ therefore, the imaginary solutions are forbidden. Hence, ( ~x; t ) is zero for ct < . We will multiply it by the unit step function: ( ~x; t ) = 2 c ( ct ) p c 2 t 2 2 6.1.2 Substituting f ( ~x ; t ) = ( x ) ( t ) into equation 6.47 in Jackson: ( ~x; t ) = Z [ f ( ~x ; t )] ret j ~x ~x j d 3 x = Z [ ( x ) ( t )] ret j ~x ~x j dx dy dz Noting that [ t ] ret = t j ~x ~x j =c , ( ~x; t ) = ZZZ ( x ) t p ( x x ) 2 + ( y y ) 2 + ( z z ) 2 =c p ( x x ) 2 + ( y y ) 2 + ( z z ) 2 dx dy dz = ZZ t p x 2 + ( y y ) 2 + ( z z ) 2 =c p x 2 + ( y y ) 2 + (...
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W11HW1 - 6.1 Problem 6.1 6.1.1 Substituting f ( ~x ; t ) =...

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