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
the relation () becomes
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:
is at rest, the two gammas have opposite
momenta and same energy
:
to a frame
where the momentum 
of the neutral pion lies
along the z-axis gives the 4-momentum of one of the gamma rays
as:
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:
of the
gamma in the rest frame of 
, one performs the rotation
as follows, with 
and 
being the polar angles of the momentum 
of 
in the rest frame of 
:
of the
second gamma follow from space reflection:
shows the distribution of the opening angle
of the two gamma rays, and the energy spectrum of one of the photons
is displed in figure .
