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Fin'sol - Classical Electrodynamics II PHY 506 Spring 2011...

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1 C:\User\Teaching\506\S11\Fin'sol.doc PHY 506 Classical Electrodynamics II Spring 2011 Final Exam with solutions Problem F.1 (to be graded of 200 points). Use the Born approximation to calculate the differential cross-section of scattering of a plane wave normally incident on a very thin, round dielectric disk of radius R . Analyze the result in the limits kR << 1 and kR >> 1. Solution : This problem is just a particular case of Optional Problem O.10, with q z L << 1, and q = k sin , where is the scattering angle between vectors k and k 0 . (Note that it is different from angle between vectors k and E !) Thus we can immediately write the differential cross-section: 2 1 2 2 4 2 4 2 / ) sin ( sin sin ) 1 ( 16 kR kR J R L k d d r . (*) If the disk radius is small ( kR << 1), this result is reduced to Eq. (8.53), with V = R 2 d . Taking into account that 3 8 sin 2 sin 0 3 4 2 d d , the full cross-section, in this limit, is 2 4 2 4 ) 1 ( 6 r R L k . For a large disk ( kR >> 1), the last factor in Eq. (*) limits scattering to small forward and back angles, with sin ~ 1/ kR << 1, and sin 1, so that may be calculated as   . ) 1 ( 2 4 ) 1 ( 16 4 2 / ) ( ) 1 ( 16 4 4 2 2 4 2 4 0 2 1 2 4 2 4 0 2 1 2 4 2 4 1 1   r r r R L k d J R L k d kR kR J R L k d d d d d d Thus, is described by essentially the same formula as at kR << 1, just with a different numerical coefficient. Problem F.2 (300 points). A relativistic particle with energy E and rest mass m collides with a similar particle, initially at rest in the laboratory reference frame. Find: (i) the velocity of the center of mass of the system, in the lab frame, (ii) the total energy of the system, in the center-of-mass frame, and (iii) final velocities of both particles (in the lab frame), if they move along the same direction. Solutions : (i) Before the collision, the total momentum p s of the system in the lab frame equals that ( p ) of the only moving particle: 2 / 1 2 2 s ) ( / mc c p p E ,
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2 C:\User\Teaching\506\S11\Fin'sol.doc while the total energy is 2 s mc E E . By definition, the center of mass of a system is an imaginary particle having the same total momentum p s and total mass M s as the system. (In relativity, the latter is the dynamic mass M s = E s /c 2 .) Applying to the c.o.m. the general formula discussed in class, p = M u , i.e. = p / Mc = p c / E , we get 1 2 / 1 2 2 2 2 / 1 2 2 s ) ( / mc mc mc c mc c E E E E , so that 2 / 1 2 2 2 / 1 2 s 2 1 1 mc mc β s E . (*) Notice that despite the particles are similar, s =
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