PENELOPE 2003 - OECD Nuclear Energy Agency

210 Appendix A. Collision kinematics
is the total energy in units of the rest energy. From the relation (A.5), it follows that
√
E = (cp) 2 + m 2 c 4 − mc 2
(A.9)
and
dp
dE = 1 v = 1
cβ .
(A.10)
For a photon (and any other particle with m = 0), the energy and momentum are
related by
E = cp.
(A.11)
A.1 Two-body reactions
Consider a reaction in which a projectile “1” collides with a target “2” initially at rest in
the laboratory frame of reference. We limit our study to the important case of two-body
reactions in which the final products are two particles, “3” and “4”. The kinematics of
such reactions is governed by energy and momentum conservation.
We take the direction of movement of the projectile to be the z-axis, and set the
x-axis in such a way that the reaction plane (i.e. the plane determined by the momenta
of particles “1”, “3” and “4”) is the x-z plane. The energy-momentum 4-vectors of the
projectile, the target and the reaction products are then (see fig. A.1)
˜P 1 = (W 1 c −1 , 0, 0, p 1 )
˜P 2 = (m 2 c, 0, 0, 0)
˜P 3 = (W 3 c −1 , p 3 sin θ 3 , 0, p 3 cos θ 3 )
˜P 4 = (W 4 c −1 , −p 4 sin θ 4 , 0, p 4 cos θ 4 )
(A.12a)
(A.12b)
(A.12c)
(A.12d)
Energy and momentum conservation is expressed by the 4-vector equation
˜P 1 + ˜P 2 = ˜P 3 + ˜P 4 .
(A.13)
From this equation, the angles of emergence of the final particles, θ 3 and θ 4 , are uniquely
determined by their energies, W 3 and W 4 . Thus,
m 2 4c 2 = ( ˜P ˜P 4 4 ) = ( ˜P 1 + ˜P 2 − ˜P 3 )( ˜P 1 + ˜P 2 − ˜P 3 )
= ( ˜P ˜P 1 1 ) + ( ˜P ˜P 2 2 ) + ( ˜P ˜P 3 3 ) + 2( ˜P ˜P 1 2 ) − 2( ˜P ˜P 1 3 ) − 2( ˜P ˜P 2 3 )
= m 2 1c 2 + m 2 2c 2 + m 2 3c 2 + 2W 1 W 2 c −2
− 2 ( )
W 1 W 3 c −2 − p 1 p 3 cos θ 3 − 2W2 W 3 c −2 ,
(A.14)
and it follows that
cos θ 3 = m2 4 c4 − m 2 1 c4 − m 2 2 c4 − m 2 3 c4 + 2W 1 (W 3 − W 2 ) + 2W 2 W 3
2 (W 2 1 − m 2 1c 4 ) 1/2 (W 2 3 − m 2 3c 4 ) 1/2 . (A.15)