A rho meson (ρ+) with a rest mass energy of 775.5 MeV decays into a pion (π+) with a rest mass energy of 139.6 MeV and a gamma ray according to the reaction formula ρ+ → π+ + γ Find the velocity of the π+ produced by the reaction.

Question

A rho meson [latex](ρ^{+})[/latex] with a rest mass energy of 775.5 MeV decays into a pion [latex](π^{+})[/latex] with a rest mass energy of 139.6 MeV and a gamma ray according to the reaction formula

[latex]ρ^{+} → π^{+} + γ[/latex]

Find the velocity of the [latex]π^{+}[/latex] produced by the reaction.

Accepted Answer

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

[latex]m_{ρ} c^{2 }= E_{π} + E_{γ}[/latex]

[latex]0 = p_{π} + p_{γ} .[/latex]

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

[latex]m_{ρ} c^{2} = E_{π} + c|p_{π} | [/latex]

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

[latex]E^{2} = p^{2}c^{2} + m^{2}c^{4}.[/latex] (13.14)

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

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

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

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

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

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

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

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

[latex]E_{π} = 400.3[/latex] MeV.

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

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

[latex]\frac{v}{c}=\sqrt{1 - 1/γ^{2}}= 0.937.[/latex]

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

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