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Unformatted text preview: Phys 253a 1 Problem Set 10 Solutions December 2, 2010 1. (a) The phase space integral for 1 2 is 1 2 m Z d 3 p (2 ) 3 2 E p d 4 q (2 ) 3 E q (2 ) 4 4 ( P p q ) This is easiest to evaluate in the center of mass frame, P = ( m , ). If we first do the integral over q , the spatial part of the delta function enforces p q = 0, so q = p and E q = E p . We get 1 2(2 ) 2 m Z d 3 p 4 E 2 p ( m  2 p m 2 e + p 2 ) This is spherically symmetric in p , so we can switch to spherical coordinates 4 8(2 ) 2 m Z p 2 dp p 2 + m 2 e ( m  2 p m 2 e + p 2 ) = 1 8 m p 2 p 2 + m 2 e p m 2 e + p 2 2 p m 2 e + p 2 = 1 2 m = 1 16 m s 1 4 m 2 e m (b) i. The amplitude is i M = ig S u ( p ) v ( q ), which satisfies X spins M 2 = g 2 S Tr[( 6 p + m e )( 6 q m e )] = 4 g 2 S ( p q m 2 e ) = 2 g 2 S ( m 2  4 m 2 e ) = g 2 S m 8 (1 4 x 2 ) 3 / 2 ii. The amplitude is i M = g P u ( p ) 5 v ( q ), so we get X spins M 2 = g 2 S Tr[ 5 ( 6 p + m e ) 5 ( 6 q m e )] = 4 g 2 S ( p q m 2 e ) = 2 g 2 S m 2 = g 2 S m 8 p 1 4 x 2 Phys 253a 2 iii. The amplitude is i M = ig V u ( p ) v ( q ). We get 1 3 X spins M 2 = 1 3 g 2 V Tr[( 6 p + m e ) ( 6 q m e ) ] X i i i = 4 g 2 V 3 ( p q + p q  ( p q + m 2 e ) ) + P P m 2 ! = 4 g 2 V 3 2 p q + 2 ( p P )( q P ) m 2 + 3( p q + m 2 e ) ! = 4 g 2 V 3 (2 m 2 e + m 2 ) = g 2 V m 12 (1 + 2 x 2 ) p 1 4 x 2 iv. The amplitude is g A u ( p ) 5 v ( q ), so we have 1 3 X spins M 2 = 1 3 g 2 A Tr[( 6 p + m e ) 5 ( 6 q m e ) 5 ] X i i i = 4 g 2 A 3 (( m 2 e p q ) + p q + q p ) + P P m 2 ! = 4 g 2 A 3 ( m 2  4 m 2 e ) = g 2 A m 12 (1 4 x 2 ) 3 / 2 v. The tricky part of this one is to write down the polarization sum for an antisymmetric tensor. Well assume its transverse to pick out the spin 1 part. Then we know the polarization sum = X i i i * must satisfy P = 0 (and similarly if we dot p into any index), along with = = . The only objects we have to build with are and P , and the only tensors we can build out of these with the correct index symmetries are  and P P  P P + P P  P P Phys 253a 3 The transverseness condition implies that we should take the combination X i i i =   1 m 2 ( P P  P P + P P  P P ) (Actually, this only fixes the spin sum up to an overall scale. We must additionally check a special case, say ...) Alternatively, we can write this as X i i i * = C C  C C Where C :=  P P m 2 is manifestly transverse. The amplitude isis manifestly transverse....
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 SCHWARTZ
 Center Of Mass, Mass

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