Relativistic energy and momentum
Problem:
Use conservation of energy and momentum to show that a moving electron cannot
emit a photon unless there is a third body present (e.g., an atom or a nucleus).
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: mc^{2} = γmc^{2} + 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 = mc^{2} + T = (m^{2}c^{4} + p^{2}c^{2})^{½}.
 Details of the calculation:
2Tmc^{2} + T^{2} = p^{2}c^{2}.
mc^{2} = (p^{2}c^{2}  T^{2})/(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
(mc^{2} = 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  β^{2})^{½} = 1.33, K = (γ – 1)mc^{2}
= 167.5 keV.
Problem:
An electron is accelerated from rest through a potential difference of 10^{6}
V.
Find is its final energy, momentum, and speed.
Solution:
 Concepts:
Relativistic energy
 Reasoning:
E = mc^{2} + T = (m^{2}c^{4} + p^{2}c^{2})^{½},
p = γmv.
 Details of the calculation:
E = (9.1*10^{31}*9*10^{16} 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*10^{8} 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 = γmc^{2} = 2mc^{2},
γ = 2, v = ¾^{½}c = √3c/2.
4m^{2}c^{4} 
m^{2}c^{4} = p^{2}c^{2}, p^{2} = 3
m^{2}c^{4}, p = √3mc.
Problem:
A fast proton is produced in an accelerator with energy 6.5 TeV and travels a
distance of 10^{10} 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) p^{2}c^{2} = E^{2} – m^{2}c^{4},
v/c = pc/E, v/c = (1  m^{2}c^{4}/E^{2})^{½}
≈ 1  m^{2}c^{4}/(2E^{2}).
∆v/c = m^{2}c^{4}/(2E^{2}) = (½*938 MeV/6.5*10^{6}
MeV)^{2} = 5.2*10^{9}. ∆v = 1.56 m/s.
(b) τ = t/γ, t = d/v ≈ (d/c) = 3.33*10^{4} s ≈ 4 hours.
1/γ = (1 – (v/c)^{2})^{½} = mc^{2}/E^{
}=
1.44*10^{4}. τ = 1.44*10^{4} * 3.33*10^{4} 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 mc^{2}/E_{ν} << 1.
Give a numerical answer for d = 10^{5} light years and mc^{2}/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
p^{2}c^{2} = E^{2} – m^{2}c^{4},
v/c = pc/E, v/c = (1  m^{2}c^{4}/E^{2})^{½}
≈ 1  m^{2}c^{4}/(2E^{2}).
 Details of the calculation:
For the photons, the travel time is T_{photon} = d/c.
T_{neutrino} = d/v ≈ (d/c)(1 + m^{2}c^{4}/(2E^{2})).
T_{neutrino} – T_{photon} = T_{photon}m^{2}c^{4}/(2E^{2})).
If mc^{2} = 10^{6}E we have ∆T = T_{photon}*5*10^{13}.
T_{photon} = 10^{5} years= 3.15*10^{12} s, ∆T ≈
1.5 s.
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.