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Unformatted text preview: 9.1 Problem 9.1 9.1.1 Part a The general solution for the vector potential in Lorentz gauge with source ~ J ( ~x; t ) is: ~ A ( ~x; t ) = 4 Z d 3 x Z dt ~ J ( ~x ; t ) j ~x ~x j t + j ~x ~x j x t Then if we look at a Fouriertransformed (in time) eld, ~ A ( ~x; ! ) = 1 2 Z 1 1 dt ~ A ( ~x; t ) e i!t = 1 2 4 Z 1 1 dte i!t Z d 3 x Z dt ~ J ( ~x ; t ) j ~x ~x j t + j ~x ~x j c t = 8 2 Z d 3 x e ik j ~x ~x j j ~x ~x j Z dt e i!t ~ J ( ~x ; t ) = 4 Z d 3 x e ik j ~x ~x j j ~x ~x j ~ J ( ~x ; ! ) where k = !=c and we have made no assumptions about the source. 9.1.2 Part b If we have a charge q rotating in a circle of radius R about the z axis, it is easiest to use cylindrical polar coordinates, with the charge at ~x q with coordinates = R , ' = ! t , z = 0. The current density is ~ J ( ~x ; ! ) = 1 2 qR! ( R ) ( z ) Z 1 1 dt ^ e ' ( ' ! t ) e i!t A little care is needed in converting ( ' ! t ) to a delta function in t , because it requires its argument to be zero modulo 2 , so ( ' ! t ) = 1 ! X n 2Z t ( ' + 2 n ) ! Thus, Z 1 1 dt ^ e ' ( ' ! t ) e i!t = 1 ! ^ e ' e i!'=!...
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This note was uploaded on 12/08/2011 for the course PHYSICS 505 taught by Professor Liu during the Winter '08 term at University of Michigan.
 Winter '08
 LIU

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