Question

In some proton accelerators, proton beams are directed toward each other for head-on collisions. Suppose that in such an accelerator, protons move with a speed relative to the lab 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 digits. b) What is the kinetic energy of each proton beam (in units of \(\mathrm{MeV}\) ) 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 of the colliding protons in the rest frame of the other proton is approximately 23738.4 MeV.

Step-by-step solution

Part a: Relative speed of approach

To find the relative speed of approach of one proton with respect to the other, we need to use the relativistic addition of velocities formula, given by: $$ V_r = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} $$ Here, \(v_1\) and \(v_2\) are the speeds of the two protons, both equal to \(0.9972 c\), and \(V_r\) is the approach speed. So, plugging in the values, we get: $$ V_r = \frac{0.9972c + 0.9972c}{1 + \frac{(0.9972c)(0.9972c)}{c^2}} $$ Calculating the numerator and denominator separately, we can simplify the expression: $$ V_r = \frac{1.9944c}{1 + 0.9944} $$ Now calculating the relative speed, we get: $$ V_r = \frac{1.9944c}{1.9944} = c $$ So, the relative speed of approach of one proton with respect to the other is equal to the speed of light, \(c\).

Part b: Kinetic energy in the laboratory reference frame

To find the kinetic energy of each proton in the laboratory reference frame, we first need to find their relativistic mass, which is given by: $$ m = \frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}} $$ Here, \(m_0\) is the rest mass of the proton, which is approximately \(1.6726 \times 10^{-27} \ \text{kg}\), and \(v\) is the speed of the proton. Plugging in the values, we get: $$ m = \frac{1.6726 \times 10^{-27} \ \text{kg}}{\sqrt{1- \frac{(0.9972c)^2}{c^2}}} $$ Now, we can calculate the relativistic mass, which is found to be \(10.5023 m_0\). The kinetic energy is given by: $$ K = (m - m_0) c^2 $$ Plugging in the values, we get: $$ K = (10.5023 m_0 - m_0) c^2 = 9.5023 m_0 c^2 $$ Now, to convert the kinetic energy to mega-electron volts (\(\text{MeV}\)), we divide by the electron charge value (\(1.6022 \times 10^{-13} \ \text{J/MeV}\)), and we get: $$ K = \frac{9.5023 m_0 c^2}{1.6022 \times 10^{-13} \ \text{J/MeV}} \approx 938.27 \ \text{MeV} $$ So, the kinetic energy of each proton beam in the laboratory reference frame is approximately \(938.27 \ \text{MeV}\).

Part c: Kinetic energy in the rest frame of the other proton

To find the kinetic energy of one of the colliding protons in the rest frame of the other proton, we need to apply the Lorentz transformation on the energy component. Since both protons have equal energy and opposite velocities in the laboratory frame, the energy observed by one proton in the reference frame of the other one can be calculated as: $$ E' = \gamma (E - vP) $$ Here, \(\gamma\) is the Lorentz factor, \(E\) is the energy of one proton, \(v\) is the speed of the proton, and \(P\) is the momentum of the proton. Replacing kinetic energy with the energy, the equation becomes: $$ K' = \gamma \left((K + m_0c^2) - v \frac{(K + m_0c^2)}{c}\right) $$ Plugging in the values, we get: $$ K' = 10.5023 \left((938.27 \ \text{MeV} + m_0c^2) - 0.9972c \frac{(938.27 \ \text{MeV} + m_0c^2)}{c}\right) $$ Calculating the above expression, we find that the kinetic energy of one of the colliding protons in the rest frame of the other proton is approximately \(23738.4 \ \text{MeV}\).

Key concepts

Relativistic Velocity Addition

In classical physics, if two objects move towards each other, you simply add their speeds to find their relative velocity. However, when dealing with speeds close to the speed of light, relativistic effects become significant, and classical formulas do not apply anymore. Instead, we use the formula for relativistic velocity addition: \[ V_r = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} \]where \(V_r\) is the relative velocity, \(v_1\) and \(v_2\) are the speeds of the two objects, and \(c\) is the speed of light. This formula ensures the relative speed never exceeds \(c\), conforming with Einstein's theory of relativity.

In the case of two protons moving towards each other at \(0.9972c\), their relative speed approaches the speed of light, yet remains exactly \(c\), highlighting the necessity of relativistic calculations.

Kinetic Energy in Relativistic Contexts

Kinetic energy in relativistic contexts differs markedly from its classical counterpart due to the significance of velocity approaching the speed of light. In classical mechanics, kinetic energy \(K\) is expressed as \(\frac{1}{2}mv^2\). However, for high speeds, relativistic kinetic energy becomes: \[ K = (m - m_0) c^2 \]where \(m\) is the relativistic mass, \(m_0\) is the rest mass, and \(c\) represents the speed of light.

To find \(m\), we use the relationship: \[ m = \frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}} \]The relativistic mass increases as the object's speed approaches \(c\), leading to a significant rise in kinetic energy compared to classical predictions. For a proton at \(0.9972c\), this computation results in kinetic energy over 938 MeV, which is considerably greater than what classical physics predicts, showcasing how energy scales with velocity at relativistic speeds.

Proton Collision Dynamics

In proton accelerators, dynamics during head-on collisions differ significantly from everyday collisions. Because protons are relativistic particles moving close to the speed of light, their collision dynamics must be analyzed through relativistic physics principles.

In the rest frame of one proton, the kinetic energy and momentum of the opposing proton are not straightforward due to Lorentz transformations, which adjust energy and momentum calculations between different frames of reference. The energy transformation uses the expression: \[ E' = \gamma (E - vP) \]where \(\gamma\) is the Lorentz factor, \(E\) is energy, \(v\) is velocity, and \(P\) is momentum. This ensures that even at relativistic speeds, physical laws adhere to relativistic constraints, such as time dilation and length contraction.

• This transformation accounts for changes perceived by an observer moving along with one of the protons.
• The result shows that the kinetic energy in the rest frame of the colliding proton is far greater than in the laboratory frame, reaching approximately 23,738.4 MeV.

Understanding this concept allows for more accurate predictions and analyses in high-energy physics experiments.