Question 13.5: A rho meson (ρ+) with a rest mass energy of 775.5 MeV decays...

We shall suppose the rho meson is at rest when it decays. Denoting the energy and the momentum of the rho meson by  E_{ρ} and p_{ρ}, the energy and momentum of the pion by  E_{π} and p_{π} , and the energy and momentum of the gamma ray by E_{γ} and  p_{γ} , the requirement that the energy and the momentum be conserved in the rest frame of the rho leads to the equations

m_{ρ} c^{2 }= E_{π} + E_{γ}

0 = p_{π} + p_{γ} .

Since a gamma ray is a photon with zero mass, the energy of the gamma is equal to |p_{γ} |c and the second equation for the conservation of momentum implies that  p_{π} = −p_{γ} . Using this information, the first equation for the conservation of energy may thus be written

m_{ρ} c^{2} = E_{π} + c|p_{π} |

We may now use Eq. (13.14) to rewrite this last equation

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

m_{ρ} c^{2} = E_{π} + \sqrt{E^{2}_{π} − m^{2}_{π} c^{4}}

To solve this equation we first bring E_{π} over to the left-hand side and square both sides of the resulting equation to obtain

(m_{ρ} c^{2} − E_{π} )^{2} = E^{2}_{π}−m^{2}_{π} c^{4}

Explicitly squaring the term on the left-hand side, we obtain

m^{2}_{ρ} c^{4} − 2m_{ρ} c^{2}E_{π} + E^{2}_{π} = E^{2}_{π} − m^{2}_{π} c^{4}

The two E^{2} _{π} terms cancel, and we may solve for E_{π} to obtain

E_{π}=\frac{m^{2}_{ρ} c^{4} + m^{2}_{π} c^{4}}{2m_{ρ} c^{2}}

Substituting the rest mass energies of the rho meson and pi meson into this last equation, we get

E_{π} = 400.3 MeV.

We may then solve the equation, E_{π} = γ(m_{π} c^{2}), to find γ has the value 2.867 and use Eq. (13.47) to get

\begin{matrix}γ ^{i} = β \alpha ^{i} ,& γ^{ 0} = β, \end{matrix}                         (13.47)

\frac{v}{c}=\sqrt{1 – 1/γ^{2}}= 0.937.