HW09'sol

# HW09'sol - 1 C\User\Teaching\505-506\S11\HW09'sol PHY 506...

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Unformatted text preview: 1 C:\User\Teaching\505-506\S11\HW09'sol.doc PHY 506 Classical Electrodynamics II Spring 2011 Homework 9 with solutions Problem 9.1 (10 points). Analyze the polarization of the EM radiation of a linearly accelerated relativistic particle. Solution : Let us consider the expression for the electric field of the radiation, given by the second term of Eq. (20a) of the lecture notes: cR q 3 ) 1 ( 4 ) ( n β β β ) (n n r E , (*) where all variables have to be taken in the retarded point. Using the coordinate choice shown in Fig. 10.4a (reproduced on the right), in which the observation point r = R n is within plane [ x , z ], we see that vectors ( n- ) and β are also within this plane. Hence the internal vector product in Eq. (*) is perpendicular to that plane (antiparallel to axis y ). As a result, vector E is in plane [ x , z ] of the drawing, and perpendicular to the direction n toward the radiation observer. Thus we conclude that the radiation is linearly polarized, with vector E in the common plane of the observation point and particle’s trajectory. Problem 9.2 (10 points). Find the time dependence of the kinetic energy E of a charged relativistic particle performing synchrotron motion in a constant and uniform magnetic field B , and hence emitting synchrotron radiation. Sketch particle’s trajectory. Hint : You may assume that the energy loss is relatively slow (- d E / dt << c E ), but should spell out the condition of validity of this assumption. Solution : Using Eq. (9.153) of the lecture notes for the cyclotron orbit’s radius R , , qB mc qB u m R ( * * ) we can rewrite Eq. (10.43) for radiative energy loss as 2 4 2 3 2 2 2 2 2 4 2 2 4 2 2 2 2 4 3 where , 2 1 1 6 6 B q Z c m mc mc c m B q Z m B q Z dt d E E ....
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HW09'sol - 1 C\User\Teaching\505-506\S11\HW09'sol PHY 506...

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