[SOLVED] kinematics

[phd15@phys.keele.ac.uk]

my question concerns two photon annihilation in the lab frame.

a positron and an electron annihilate resulting in photon1 and photon2,
such that
e+ + e- -> gamma1 + gamma2

below respectively,

p+,p- are the positron,electron momenta
k1,k2 are photon 1,2 k-vectors
omega 1,2 are the photon 1,2 frequencies
m is the mass of the electron,positron
E+,E- are the positron,electron energies
Ef is the energy of the final state (after annhiliation)
d{}/d{} represents the partial derivative wrt the contents of the
brace in the denominator
A^n denotes A to the power n
|p| denotes modulus p ie (p^2)^(1/2)

in the lab fame the momentum of the electron is given by
p- = (0,0,0,im), where i=sqrt(-1)

given the kinematical relations

|p+ - k1| = Sqrt(|p+|^2 - 2 |p+|omega1 cos theta1 + omega1^2)
= |k2|
= omega2
and
m + E+ omega1 + omega2

where
(d{Ef}/d{|k1|}) = d{omega1 + |p+ - k1|}/d{|k1|}
= 1 + (|k1| - |p+| cos theta1)/omega2
= m(m + E+)/(omega1 omega2)

Can anyone explain the final equality ? incidentally if anyone has
Sakurai Advaced QM this is on page 217 eq. (4.183) & (4.184)

Any help on this is greatly appreciated
Zak

[Arnie]

Here's one way to get it.

I'll let P stand for P+ and E stand for E+.
Also, w1 = omega1 and w2 = omega2.

>From (4.183) find

2 |P| w1 cos(theta1) = w1^2 - w2^2 + |P|^2

In the expression on the left side of the last equality in (4.184) let
|k1| = w1 and write the expression as

(w2 + w1 - |P| cos(theta1)) / w2

Multiply numerator and denominator by 2*w1:

( 2 w1*w2 + 2 w1^2 - 2 |P| w1 cos(theta1) ) / (2 w1 w2)

Substitute the first expression given above:

( 2 w1*w2 + 2 w1^2 - w1^2 + w2^2 - |P|^2) / (2 w1 w2)

= (w1 + w2)^2 - |P|^2 ) / (2 w1 w2)

Finally, substitute w1 + w2 = m + E and |P|^2 = E^2 - m^2 and
simplify.

Todd