Question 13.4: Suppose that a beam of positrons strike electrons that are a...

As in the previous example, the energy and momentum of the  $Z_{0}$ satisfy Eq. (13.21). However, the conditions that the energy and momentum of the particles be conserved in the laboratory frame of reference are

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

$E_{p} + m_{e}c^{2} = E_{Z}$

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

Substituting these values into Eq. (13.21) gives

$(E_{p} + m_{e}c^{2})^{ 2} − p^{2}_{ p}c^{2} = M_{Z}^{2}c^{4}.$                    (13.23)

Using Eq. (13.14), the left-hand side of this equation may be written

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

$E^{2}_{ p} + 2m_{e}c^{2}E_{p} + m^{2} _{e} c^{4} − p^{2} p_{c}^{2} = 2m_{e}c^{2}(E_{p} + m_{e}c^{2})$,

and Eq. (13.23) becomes

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

Solving this equation for $E_{p}$, we get

$E_{p}=\frac{M_{Z}^{2}c^{4}}{2m_{e}c^{2}}− m_{e}c^{2},$

and the kinetic energy of a positron in the beam is

$KE =\frac{M_{Z}^{2}c^{4}}{2m_{e}c^{2}}− m_{e}c^{2}= 8.136 \times 10^{6} \ GeV.$

For an experiment in which the electrons are at rest in the laboratory, the kinetic energy of the positrons must be 180, 000 times larger than they would have to be in a colliding beam experiment. A few calculations of this kind are sufficient for one to understand the popularity of collider experiments in recent years. Much of the energy in traditional scattering experiments goes into increasing the velocity of the center of mass of the colliding particles rather than increasing the velocity with which the particles approach each other in the center of mass frame.