Relativistic energy and momentum

Problem:

Solution:

• Concepts:
  Energy and momentum conservation, relativistic dynamics
• Reasoning:
  We are instructed to use energy and momentum conservation.
• Details of the calculation:
  In a frame moving with the electron we have:
  Energy conservation: mc² = γmc² + hf.
  Momentum conservation: 0 = γmv - hf/c
  Since γ greater or equal to one, energy conservation cannot be satisfied unless hf = 0, γ = 1 and v = 0.
  No photon can be emitted in this or any other frame.

Problem:

A relativistic particle is stopped in a detector.  The momentum is determined to be 2 GeV/c, and it deposits a kinetic energy T = 1 GeV in the detector before it comes to rest.   What is its mass?

Solution:

• Concepts:
  Relativistic expression for energy and momentum
• Reasoning:
  E = mc² + T = (m²c⁴ + p²c²)^½.
• Details of the calculation:
  2Tmc² + T² = p²c². 
  mc² = (p²c² - T²)/(2T) = (4 -1)/2 GeV = (3/2) GeV.

Problem:

Cerenkov radiation is given off when a particle moves in a medium at a speed greater than the speed of light in that medium.  What is the minimum kinetic energy (in eV) that an electron
(mc² = 511 keV) must have while traveling in crown glass (n = 1.52) in order to create Cerenkov radiation?

Solution:

• Concepts:
  Relativistic expression for kinetic energy
• Reasoning:
  The speed of light in the glass is v = c/n = c/1.52.  The minimum speed the electron must have is c/1.52.
• Details of the calculation:
  β = 1/1.52 = 0.66,  γ = (1 - β²)^-½ = 1.33,  K = (γ - 1)mc² = 167.5 keV.

Problem:

An electron is accelerated from rest through a potential difference of 10⁶ V.  Find is its final energy, momentum, and speed.

Solution:

• Concepts:
  Relativistic energy
• Reasoning:
  E = mc² + T = (m²c⁴ + p²c²)^½,  p = γmv.
• Details of the calculation:
  E = (9.1*10^-31*9*10¹⁶ J) + 1.6*10^-13 J = 2.43*10^-13 J
  pc = 2.28*10^-13 J, p = 7.6*10^-22 kg m/s
  v/c = pc/E = 0.938,  v = 2.82*10⁸ m/s.

Problem:

Find the magnitude of the velocity and momentum of an electron which has kinetic energy equal to its rest mass energy.

Solution:

• Concepts:
  Relativistically correct expressions for energy and momentum
• Reasoning:
  We have an electron moving wit a relativistic speed.
• Details of the calculation:
  E = γmc² = 2mc²,   γ = 2,   v = ¾^½c = √3c/2.
  4m²c⁴ - m²c⁴ = p²c²,   p² = 3 m²c⁴,   p = √3mc.

Problem:

A fast proton is produced in an accelerator with energy 6.5 TeV and travels a distance of 10¹⁰ km before it collides with a target. 
(a)  By how much does the speed of the proton differ from the speed of light?
(b)  How much time elapses its own rest frame between the production and the collision events?

Solution:

• Concepts:
  Relativistic momentum and energy
• Reasoning:
  The proton moves with relativistic speed.
• Details of the calculation:
  (a)  p²c² = E² - m²c⁴,  v/c = pc/E,  v/c = (1 - m²c⁴/E²)^½ ≈ 1 - m²c⁴/(2E²).
  ∆v/c = m²c⁴/(2E²) = (½*938 MeV/6.5*10⁶ MeV)² = 5.2*10^-9. ∆v = 1.56 m/s.
  (b)  τ = t/γ,  t = d/v ≈ (d/c) = 3.33*10⁴ s ≈ 4 hours.
  1/γ = (1 - (v/c)²)^½ = mc²/E = 1.44*10^-4.  τ = 1.44*10^-4 * 3.33*10⁴ s = 4.81 s.

Problem:

A supernova at a distance d from Earth explodes, and photons and neutrinos are emitted.  What is the difference between the arrival time on Earth for the photons and neutrinos if neutrinos have an energy E_ν?  Assume that the neutrino mass m fulfills mc²/E_ν << 1.
Give a numerical answer for d = 10⁵ light years and mc²/E_ν = 10^-6.

Solution:

• Concepts:
  Relativistic momentum and energy
• Reasoning:
  For photons v/c = 1.
  The neutrinos are relativistic and have a small mass m.  We have for their momentum
  p²c² = E² - m²c⁴,  v/c = pc/E,  v/c = (1 - m²c⁴/E²)^½ ≈ 1 - m²c⁴/(2E²).
• Details of the calculation:
  For the photons, the travel time is T_photon = d/c.
  T_neutrino = d/v ≈ (d/c)(1 + m²c⁴/(2E²)).  T_neutrino - T_photon = T_photonm²c⁴/(2E²)). 
  If mc² = 10^-6E we have ∆T = T_photon*5*10^-13.
  T_photon =  10⁵ years= 3.15*10¹² s,  ∆T ≈ 1.5 s.

Problem:

Calculate the binding energy of the deuteron, which consists of a proton and a neutron, given that the mass of a deuteron is 2.013553 u.