Question

In some proton accelerators, proton beams are directed toward each other to produce head-on collisions. Suppose that in such an accelerator, protons move with a speed relative to the lab reference frame of \(0.9972 c\) a) Calculate the speed of approach of one proton with respect to another one with which it is about to collide head on. Express your answer as a multiple of \(c\), using six significant figures. b) What is the kinetic energy of each proton (in units of MeV) in the laboratory reference frame? c) What is the kinetic energy of one of the colliding protons (in units of \(\mathrm{MeV}\) ) in the rest frame of the other proton?

Short answer

Answer: The kinetic energy of one colliding proton in the rest frame of the other proton is approximately \(104,860\text{ MeV}\).

Step-by-step solution

a) Calculate the speed of approach

To find the speed of approach of one proton with respect to another, we can use the formula for the relativistic sum of velocities: $$ V_\text{relative} = \frac{u + v}{1 + \frac{u v}{c^2}} $$ Here, \(u = 0.9972c\) and \(v = 0.9972c\) are the speeds of the two protons relative to the lab frame. Plug these values into the formula: $$ V_\text{relative} = \frac{0.9972c + 0.9972c}{1 + \frac{(0.9972c)(0.9972c)}{c^2}} $$ Solving this equation: $$ V_\text{relative} = 0.999894c $$ The speed of approach of one proton with respect to the other is \(0.999894c\).

b) Calculate the kinetic energy in laboratory reference frame

To find the kinetic energy of each proton in the laboratory reference frame, we can use the relativistic kinetic energy formula: $$ K = (\gamma - 1)mc^2 $$ Where \(K\) is the kinetic energy, \(\gamma\) is the Lorentz factor, \(m\) is the mass of the proton, and \(c\) is the speed of light. The Lorentz factor is given by: $$ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $$ For a proton, \(m = 1.67 \times 10^{-27}\) kg, and its speed \(v=0.9972c\). First, calculate the Lorentz factor: $$ \gamma = \frac{1}{\sqrt{1 - \frac{(0.9972c)^2}{c^2}}} \approx 13.032 $$ Now, calculate the kinetic energy: $$ K = (13.032 - 1)(1.67 \times 10^{-27}\text{ kg})(3 \times 10^8\text{ m/s})^2 = 6.738\times 10^{-10}\text{ J} $$ To convert this value to MeV, we use the conversion factor 1 J = \(6.242\times 10^{12}\) MeV: $$ K_\text{MeV} = 6.738 \times 10^{-10}\text{ J} \times \frac{6.242 \times 10^{12}\text{ MeV}}{\text{J}} \approx 4207.7\text{ MeV} $$ The kinetic energy of each proton in the laboratory reference frame is approximately \(4207.7\text{ MeV}\).

c) Calculate the kinetic energy in the rest frame of the other proton

To find the kinetic energy of one colliding proton in the rest frame of the other proton, we can use the Lorentz transformation given by: $$ \gamma' = \frac{\gamma}{\sqrt{1 - \frac{V_\text{relative}^2}{c^2}}} $$ Where \(\gamma'\) is the Lorentz factor in the rest frame of the other proton, and \(V_\text{relative}=0.999894c\) is the relative velocity. First, calculate the new Lorentz factor: $$ \gamma' = \frac{13.032}{\sqrt{1 - \frac{(0.999894c)^2}{c^2}}} \approx 3203.51 $$ Now, calculate the kinetic energy in the rest frame of the other proton: $$ K' = (3203.51 - 1)(1.67 \times 10^{-27}\text{ kg})(3 \times 10^8\text{ m/s})^2 = 1.678\times 10^{-8}\text{ J} $$ To convert this value to MeV, we use the conversion factor 1 J = \(6.242\times 10^{12}\) MeV: $$ K'_\text{MeV} = 1.678 \times 10^{-8}\text{ J} \times \frac{6.242 \times 10^{12}\text{ MeV}}{\text{J}} \approx 104860\text{ MeV} $$ The kinetic energy of one colliding proton in the rest frame of the other proton is approximately \(104,860\text{ MeV}\).