3. Two-body decays of unstable particles.

The simplest kind of particle reaction is the two-body decay of unstable particles. Awell

known example from nuclear physics is the alpha decayofheavy nuclei. In particle physics one

observes, for instance, decays of charged pions or kaons into muons and neutrinos, or decays of

neutral kaons into pairs of pions,

etc

. The unstable particle is the

mother particle

and its decay

products are the

daughter particles

.

Consider the decay of a particle of mass

M

which is initially at rest. Then its 4-momentum

is P = (

M;

0

;

0

;

0). This reference frame is called the

centre-of mass frame

(CMS). Denote the 4-

momenta of the two daughter particles byp

1

and p

2

:p

1

=(

E

1

;~p

1

), p

2

=(

E

2

;~p

2

). 4-momentum

conservation requires that

P=p

1

+p

2

(13)

and hence

~p

2

=

;

~p

1

.We can therefore omit the subscript on the particle momenta and hence

energy conservation takes on the form

E

1

+

E

2

=

q

m

2

1

+

p

2

+

q

m

2

2

+

p

2

=

M

(14)

Solving this equation for

p

weget

p

=

1

2

M

q

[

M

2

;

(

m

1

;

m

2

)

2

][

M

2

;

(

m

1

+

m

2

)

2

](15)

An immediate consequence of Eq. (14) is that

M



m

1

+

m

2

(16)

i.e.

a particle can decay only if its mass exceeds the sum of the masses of its decay products.

Conversely, if some particle has a mass that exceeds the masses of two other particles, then this

particle is unstable and decays, unless the decay is forbidden by some conservation law, such

as conservation of charge or of angular momentum

etc

.

Another point to note is that the momenta of the daughter particles and hence also their

energies are xed by the masses of the three particles. In the next section we shall see that

there is no equivalent statement in the case of three-bo dy decays: the momenta of the daughter

particles in three-bo dy decays can takeonanyvalue from zero to some maximum, and it is

only the maximum momentum that is xed by the masses of the particles.

Let us complete our calculation by deriving the formul for the energies of the daughter

particles. This is straightforward if we begin from the energy conservation formula (14) and

express

E

2

in terms of

E

1

,viz.

E

2

=

q

E

2

1

;

m

2

1

+

m

2

2

, and then solvefor

E

1

to get

E

1

=

1

2

M



M

2

+

m

2

1

;

m

2

2



(17)

and similarly

E

2

=

1

2

M



M

2

+

m

2

2

;

m

2

1



(18)

We also note that there is no preferred direction in which the daughter particles travel (the

decayissaidtobe

isotropic

), but if the direction of one of the particles is chosen (e.g. by

the positioning of a detector), then the direction of the second particle is xed by momentum

conservation: the daughter particles are traveling

back-to-back

in the rest frame of the mother

particle.

Frequently one observes that the masses of the twodaughter particles are equal, for instance

in the decay of a neutral kaon into a pair of pions. In this case the previous formul simplify: the

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