Question 13.3: The intermediate vector boson Z0, which has a rest mass ener...

According to Eq. (13.14), the energy and momentum of the $Z_0$ satisfy the equation

$E^{2} = p^{2}c^{2} + m^{2}c^{4}.$                                    (13.14)

$E^{2}_{Z} − p^{2}_{Z}c^{2} = M_{Z}^{2}c^{4}.$                                    (13.21)

Denoting the energy and momentum of the positron by $E_{p}$ and $p_{p}$ and the energy and momentum of the electron by $E_{e}$ and $p_{e}$, the requirements that the energy and momentum be conserved in the collision are

$E_{p} + E_{e} = E_{Z}$

$p_{p} + p_{e} = p_{Z}.$

Substituting these conservation laws into Eq. (13.21) gives

$(E_{p} + E_{e})^{2} − (p_{p} + p_{e})^{2}c^{2} = M_{Z}^{2}c^{4}.$                (13.22)

In a colliding beam experiment involving particles of equal mass, the center of mass of two colliding particles is stationary, and the momenta of the particles is equal and opposite

$p_{p} = −p_{e}.$

The energies of the two particles is equal

$E_{p} = c \sqrt{p_{p}^{2} + m_{p}^{2}c^{2} }= E_{e}.$

Eq. (13.22) can thus be written simply

$2E_{p} = M_{Z}c^{2},$

or

$E_{p}=M_{Z}c^{2}/2.$

The kinetic energy of the positron is equal to the energy $E_{p}$ minus its rest energy $m_{p}c^{2} = 0.511$ MeV. So, the kinetic energy of the positrons – and the electrons – is

$KE = M_{Z}c^{2}/2 − m_{p}c^{2} =$45.08 GeV