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## Kinematics

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:

• 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 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:

• 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 of in the rest frame of :

• The energy and momentum of the second gamma follow from space reflection:

• Finally, the opening angle of the two gammas is computed as:

The figure 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 .

Next: Pion Beta Decay Up: Pion Beta Decay Previous: Pion Beta Decay

Bernward Krause
Mon Jan 15 14:57:06 MET 1996