(a) Calculate the relativistic quantity \(\gamma {\rm{ = }}\frac{{\rm{1}}}{{\sqrt {{\rm{1 - }}{{\rm{v}}^{\rm{2}}}{\rm{/}}{{\rm{c}}^{\rm{2}}}} }}\) for \({\rm{1}}{\rm{.00 - TeV}}\) protons produced at Fermilab.

(b) If such a proton created a \({\pi ^{\rm{ + }}}\) having the same speed, how long would its life be in the laboratory?

(c) How far could it travel in this time?

The relativistic quantity\(\gamma \)is given by,

\(\gamma = \frac{E}{{m{c^2}}}\)

Here\(\gamma \)is relativistic quantity,\(E\) is the energy,\(m\)is the mass of proton,\(c\)is the speed of the light.

(a)

The relativistic quantity \(\gamma \) is given in terms of the proton's energy, mass and the speed of light by the relation: \(E = \gamma m{c^2}\). Solve this equation for \(\gamma \), giving that the mass of the proton is \(m = 1.67 \times {1^{ - 27}}{\rm{ kg}}\).

\(\begin{aligned}{}\gamma &= \frac{E}{{m{c^2}}}\\ &= \frac{{{\rm{1}}{\rm{.00TeV \times 1}}{\rm{.60 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;J/TeV}}}}{{{\rm{1}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{\;kg \times }}{{\left( {{\rm{3}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m/s}}} \right)}^{\rm{2}}}}}\\ &= {\rm{1}}{\rm{.06 \times 1}}{{\rm{0}}^{\rm{3}}}\end{aligned}\)

Therefore, the value for relativistic quantity is obtained as \(\gamma = 1.06 \times {10^3}\).

(b)

Given that the proper lifetime of \({\pi ^0}\) is \(\Delta {t_0} = 2.60 \times {10^{ - 8}}{\rm{\;s}}\), solve time dilation equation for the lifetime that would be in the laboratory given by: \(\Delta t = \gamma \Delta {t_0}\), where \(\gamma \) is the same as calculated earlier.

\(\begin{aligned}{}\Delta t &= \gamma \Delta {t_0}\\ &= 1.06 \times {10^3} \times 2.60 \times {10^{ - 8}}{\rm{\;s}}\\ &= 2.76 \times {10^{ - 5}}{\rm{\;s}}\end{aligned}\)

Therefore, the value for time is obtained as \(\Delta t = 2.76 \times {10^{ - 5}}{\rm{\;s}}\).

(c)

Solve the equation \(\gamma = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\), and then multiply this \({\rm{v}}\) by \({\rm{\Delta t}}\) to get the distance.\(\begin{aligned}{c}\gamma &= \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\\v &= \sqrt {{c^2}\left( {1 - \frac{1}{{{\gamma ^2}}}} \right)} \\ &= \sqrt {{{\left( {{\rm{3}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m/s}}} \right)}^{\rm{2}}}\left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{\left( {{\rm{1}}{\rm{.06 \times 1}}{{\rm{0}}^{\rm{3}}}} \right)}^{\rm{2}}}}}} \right)} \approx {\rm{c}}\end{aligned}\)

The distance is given by,

\(\begin{aligned}{}d &= c\Delta t\\ &= {\rm{3}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m/s \times 2}}{\rm{.76 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{\;s}}\\ &= {\rm{8}}{\rm{.28 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;m}}\\ &= {\rm{8}}{\rm{.28\;km}}\end{aligned}\)

Therefore, the value for distance is obtained as \(d = 8.28\;{\rm{km}}\).