Question

(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?

Short answer

(a) 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 is\(\gamma {\rm{ = 1}}{\rm{.06 \times 1}}{{\rm{0}}^{\rm{3}}}\).

(b) The life of the proton created in a laboratory will be\(\Delta t = 2.76 \times {10^{ - 5}}{\rm{\;s}}\).

(c) The proton could travel a distance of \(d = 8.28\;{\rm{km}}\) in time \(\Delta t = 2.76 \times {10^{ - 5}}{\rm{\;s}}\).

Step-by-step solution

Concept Introduction

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.

Calculation for relativistic quantity

(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 \(\gamma = 1.06 \times {10^3}\).

Calculation for time

(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 \(\Delta t = 2.76 \times {10^{ - 5}}{\rm{\;s}}\).

Calculation for distance

(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 \(d = 8.28\;{\rm{km}}\).