Question

A nucleus with rest mass \(23.94 \mathrm{GeV} / c^{2}\) is at rest in the lab. An identical nucleus is accelerated to a kinetic energy of \(10,868.96 \mathrm{GeV}\) and made to collide with the first nucleus. If instead the two nuclei were made to collide head on in a collider, what would the kinetic energy of each nucleus have to be for the collision to achieve the same center-of-mass energy?

Short answer

Question: Calculate the kinetic energy of each nucleus in the collider case to achieve the same center-of-mass energy as in the lab frame scenario, given that one nucleus is at rest and the other nucleus has been accelerated to a kinetic energy of 10,868.96 GeV. The mass of each nucleus is 23.94 GeV/c². Answer: To calculate the kinetic energy of each nucleus in the collider case, follow these steps: 1. Calculate the energy and momentum of the nuclei in the lab frame, where one nucleus is at rest and the other has a kinetic energy of 10,868.96 GeV. 2. Find the center-of-mass energy of the system, using the formula: √s = √((E1+E2)² - (p1+p2)²c²). 3. Calculate the kinetic energies of the nuclei in the collider frame, where the two nuclei have equal and opposite momenta, and the same center-of-mass energy as in the lab frame scenario: K₁' = K₂' = E' - mc². 4. Use the derived formulas to calculate the kinetic energy required for each nucleus to achieve the same center-of-mass energy as in the lab frame scenario.

Step-by-step solution

Calculating the energy and momentum of the nuclei in the lab frame

We are given the rest mass of the nucleus \(m = 23.94 \mathrm{GeV}/c^2\). The first nucleus is at rest, so it has energy \(E_1 = mc^2\) and momentum \(\vec{p_1}=0\). The second nucleus has a kinetic energy of \(K_2 = 10868.96 \mathrm{GeV}\). Its total energy is \(E_2 = mc^2 + K_2\). The momentum of the second nucleus can be found using the following relativistic equation: \(p_2 = \dfrac{mc}{\sqrt{1-\dfrac{v^2}{c^2}}}\), where \(v\) is the velocity of the nucleus. Since we have the total energy of the second nucleus, we can calculate its momentum using the energy-momentum relationship: \(E_2^2 = (mc^2)^2 + (p_2c)^2\) Rearrange to find \(p_2\): \(p_2 = \dfrac{\sqrt{E_2^2 - (mc^2)^2}}{c}\)

Finding the center-of-mass energy of the system

In the lab frame, the total energy of the system is \(E_1+E_2\) and the total momentum is \(\vec{p_1}+\vec{p_2} = \vec{p_2}\). Using the formula for the center-of-mass energy \(\sqrt{s} = \sqrt{(E_1+E_2)^2-(\vec{p_1}+\vec{p_2})^2c^2}\): \(\sqrt{s}=\sqrt{(E_1+E_2)^2 - (\vec{p_2})^2c^2}\)

Calculating the kinetic energies of the nuclei in the collider frame

In the collider frame, the two nuclei have equal and opposite momenta, so their total momentum is \(\vec{p_{1'}} + \vec{p_{2'}} = 0\). Therefore, for the same center-of-mass energy, we have: \(\sqrt{s} = E_{1'}+E_{2'}\) Since the nuclei have the same energy in the collider frame, we can assume that \(E_{1'}=E_{2'}=E'\). So, \(\sqrt{s} = 2E'\) Now, we can find the kinetic energy of each nucleus in the collider frame: \(K_{1'} = K_{2'} = E' - mc^2\)

Putting the results together

Calculate the center-of-mass energy and the kinetic energy of each nucleus for the collider case using the formulas derived in the previous steps. This will give you the kinetic energy required for each nucleus to achieve the same center-of-mass energy as in the lab frame scenario.