Question 13.3: The intermediate vector boson Z0, which has a rest mass ener...
The intermediate vector boson Z_{0}, which has a rest mass energy of 91.187 GeV, is produced in collisions of positrons and electrons
e^{+} + e^{−} → Z_{0}.
How much energy must the positrons and electrons in symmetric colliding beams have to produce the Z_{0}?
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