410ahw9soln - HW#9Solutions Phys410Fall 2011 Prof. Ted...

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Unformatted text preview: HW#9Solutions Phys410Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ jacobson@umd.edu S9.1 ( Approximate relativistic relation between energy and momentum ) (a) E = ( p 2 + m 2 ) 1 / 2 = m (1+ p 2 /m 2 ) 1 / 2 = m (1+ 1 2 p 2 /m 2 + ... ) = m + p 2 / (2 m )+ ... (b) E = ( p 2 + m 2 ) 1 / 2 = p (1 + m 2 /p 2 ) 1 / 2 = p (1 + 1 2 m 2 /p 2 + ... ) = p + m 2 / (2 p ) + ... p = ( E 2- m 2 ) 1 / 2 = E (1- m 2 /E 2 ) 1 / 2 = E (1- 1 2 m 2 /E 2 + ... ) = E- m 2 / (2 E ) + ... (c) E mc 2 + p 2 / (2 m ), E pc + m 2 c 3 / (2 p ), p E/c- m 2 c 3 / (2 E ) S9.2 ( Inverse Compton scattering ) We consider the case in which the incoming electron is highly relativistic ( E m ) and moving in the + x direction, and collides head on with a photon of frequency moving along the- x direction. Suppose that after the collision the outgoing photon is moving along the + x direction. (a) Let p , k , p , and k , be the initial and final 4-momenta of the electron and photons. Then p + k = p + k , so ( p + k ) ( p + k ) = ( p + k ) ( p + k ). Since p p = m 2 = p p , and k k = 0 = k k , this implies k p = k p . (b) We can use 2-vectors instead of 4-vectors, since there are only t and x com- ponents. The photon 2-vector is lightlike, so k k = 0, which implies k = ( ,- ), the minus sign coming form the direction of travel. Thus k p = ( ,- ) ( E,p ) = ( E + p ) and k p = ( , ) ( E,p ) = ( E- p ). [In the 2-vector components p and p refer to the spatial momentum rather than the 4-vector. Sorry for the notational ambiguity.] These are equal, hence = ( E +...
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This note was uploaded on 12/29/2011 for the course PHYSICS 410 taught by Professor Jacobson during the Fall '11 term at Maryland.

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410ahw9soln - HW#9Solutions Phys410Fall 2011 Prof. Ted...

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