Consider the beta decay of and suppose **q**, , **p** and **k**
are the 4-momenta of the , , and
respectively. The relations below follow from momentum conservation:

with

where

The expressions for the energies and momenta of the final state particles are obtained by considering the 2-body processes below:

with final state masses ,
and
respectively [Byc-73]. The explicit calculations
* in the rest frame of the decaying * reveal:

and

The momenta follow from the energy-momentum relation

The kinematic constraints on the final state particle energies are derived as follow: from equation (), reaches a minimum when

To arrive at the relation (), one writes

where is the angle between the 3-momenta.
Equation () goes through a minimum at .
This
means that the 3-momenta of and are collinear. Suppose
that the minimum is some value **a** to be specified. One gets:

which leads immediately to:

where and are the magnitudes of the velocities of the neutral pion and the electron respectively. Setting

Taking the derivative of equation (), the minimum is reached at where : () is proven and in addition, the neutral pion and the positron have the same velocity as can be seen from (). From (), reaches a maximum when the neutrino is born at rest. It is straightforward to establish the kinematic constraints on and . Gathering all the information together:

The above relations together with equations (, and ) give the allowed energies of the neutral pion, the positron and the neutrino as:

and

Neglecting the neutrino mass, the explicit calculations show:

The maximum kinetic energy of the is about . Because of this recoil energy, the momenta of two gamma rays which originate from the decay of are not necessarily collinear. Their opening angle can be computed as follows:

- In a frame where the is at rest, the two gammas have opposite
momenta and same energy
- In the rest frame of the decaying :
- The Lorentz transform from the rest frame of to a frame
where the momentum
and are the polar angles of the gamma in the rest frame of where the momenta of the gammas are isotropic with and satisfying the conditions:

- To get the 4-momentum of the
gamma in the rest frame of , one performs the rotation
as follows, with and
being the polar angles of the momentum
- The energy and momentum of the
second gamma follow from space reflection:
- Finally, the opening angle of the two gammas is computed as:

Mon Jan 15 14:57:06 MET 1996