A photon of energy E (massless) hits a proton of mass M_{p} at rest.
After the collision the photon is converted into an e^{+}e^{-}
pair. Assuming that the electron and positron each have rest mass m_{e},
what is the largest possible value of the recoil momentum of the proton?

Solution:

- Concepts:

Relativistic collisions, energy and momentum conservation, frame transformations - Reasoning:

In relativistic collisions between free particles energy and momentum are always conserved. Often the physics is best visualized in the center of momentum frame. - Details of the calculation:

The proton will have the largest possible value of recoil momentum, if in the CM frame the electron and the positron move as a unit with minimum internal energy (positronium) in the opposite direction.

Neglect the binding energy of the positronium. Let m_{1}= m_{e}^{+}+ m_{e}^{-}, m_{2}= M_{p}.

In the laboratory frame we have:

energy conservation: E + m_{2}c^{2}= (m_{1}^{2}c^{4}+ p_{1}^{2}c^{2})^{½}+ (m_{2}^{2}c^{4}+ p_{2}^{2}c^{2})^{½},

momentum conservation: (E/c)**i**= p_{1}**i**+ p_{2}**i**.

Let c = 1. Then E + m_{2}= (m_{1}^{2}+ p_{1}^{2})^{½}+ (m_{2}^{2}+ p_{2}^{2})^{½}, E = p_{1}+ p_{2}.

We need to eliminate p_{1}and solve for p_{2}in terms of E, m_{1}, and m_{2}.

E^{2}+ m_{2}^{2}_{ }+ 2Em_{2}= m_{1}^{2}+ p_{1}^{2}+ m_{2}^{2}+ p_{2}^{2}+ 2(m_{1}^{2}+ p_{1}^{2})^{½}(m_{2}^{2}+ p_{2}^{2})^{½}.

E^{2}= p_{1}^{2}+ p_{2}^{2}+ 2p_{1}p_{2}.

(2p_{1}p_{2}+ 2Em_{2}- m_{1}^{2})^{2 }= 4(m_{1}^{2}+ p_{1}^{2})(m_{2}^{2}+ p_{2}^{2}).

Write out all the terms, and substitute p_{1}= E - p_{2}. Obtain a quadratic equation for p_{2}.

p_{2}^{2}+ Ap_{2}+ B = 0, p_{2}= -(A/2) ± [(A/2)^{2}- B]^{½}. With m_{1}<< m_{2}we have

A = E(m_{1}^{2}- 2m_{2}^{2}- 2Em_{2})/(m_{2}^{2}+ 2Em_{2}) ≈ -E(2m_{2}+ 2E)/(m_{2}+ 2E) ,

B = m_{1}^{2}(m_{2}^{2}+ Em_{2}- m_{1}^{2}/4)/(m_{2}^{2}+ 2Em_{2}) ≈ m_{1}^{2}(m_{2}+ E)/(m_{2}+ 2E)

p_{2max}= -(A/2) + [(A/2)^{2}- B]^{½}.

Assume a photon with energy hν incident along the y-direction scatters off an
electron with momentum p_{0} along the x-direction. After the
scattering the photon travels in the x-direction.

(a) Find an expression for the energy of the scattered photon hν' in terms
of hν and p_{0}.

(b) What is the energy of the scattered photon if the energy of the incident
photon is 5 eV and the kinetic energy of the incident electron is 100 eV?

(c) What is the energy of the scattered photon if the energy of the incident
photon is 5 eV and the energy of the incident electron is 1 GeV?

Solution:

- Concepts:

Energy and momentum conservation in relativistic collisions - Reasoning:

In relativistic collisions between free particles energy and momentum are always conserved. - Details of the calculation:

(a)

before: after: electron energy: E _{0}= (m^{2}c^{4 }+ p_{0}^{2}c^{2})^{½ }E = (m ^{2}c^{4 }+ p^{2}c^{2})^{½}electron momentum: p _{0 }**i**pcosθ **i**+ psinθ**j**photon energy: hν hν' photon momentum: (hν/c) **j**(hν'/c) **i**conservation of:

p _{x }p _{0 }= pcosθ + hν'/c(1) p _{y}hν/c = psinθ (2) E E _{0}+ hν = E + hν'(3) squaring (1): (p

_{0 }- hν'/c)^{2}_{ }= p^{2}cos^{2}θsquaring (2): (hν/c)

^{2}= p^{2}sin^{2}θadding: p

^{2}= p_{0}^{2}+ (hν'/c)^{2}_{ }- 2p_{0}hν'/c + (hν/c)^{2}(4)squaring (3): [E

_{0}+ h(ν - ν')]^{2}= E^{2}

h^{2}(ν - ν')^{2}+ E_{0}^{2}+ 2E_{0}h(ν - ν') = m^{2}c^{4 }+ p^{2}c^{2 }= m^{2}c^{4 }+ p_{0}^{2}c^{2}+ (hν')^{2}_{ }- 2p_{0}chν' + (hν)^{2}(from 4)

-h^{2}νν' + E_{0}hν - E_{0}hν' + p_{0}chν' = 0

hν' = E_{0}hν/(hν + E_{0}- p_{0}c)

(b) Let hν = 5 eV, E_{0}= T + mc^{2}= 100 eV + mc^{2}.

p_{0}^{2}c^{2}= E_{0}^{2}- m^{2}c^{4}= m^{2}c^{4}(1 - 100 eV/mc^{2})^{2}- m^{2}c^{4}≈ 2mc^{2 }100 eV.

hν' ≈ 5 eV mc^{2}/(mc^{2}+ 5 eV - (mc^{2 }200 eV)^{½})

≈ 5 eV mc^{2}/{(mc^{2})^{½}[(mc^{2})^{½}-(200 eV)^{½}]} ≈ 5 eV.

For a slow electron the photon energy does not increase appreciably.

(c) Let hν = 5 eV, E_{0}= T + mc^{2}= 1 GeV + mc^{2}≈ 1 GeV.

p_{0}^{2}c^{2}= E_{0}^{2}- m^{2}c^{4}≈ (1 GeV)^{2}.

hν' ≈ 5 eV 1 GeV/(5 eV + 1 GeV -1 GeV) ≈ 1 GeV.

The photon energy increases appreciably for a relativistic electron.**Link:**

An interesting experiment that uses the high-energy photons created that way:

"Turning Light into Matter"

(a) Find the photon threshold energy in MeV for the photoproduction process

γ + p --> p + π^{0}

(m_{π0}^{ }= 135 MeV/c^{2}, m_{p}^{
}= 938 MeV/c^{2}).

(b) Now consider the following problem. The isotropic CMBR
(Cosmic Microwave Background Radiation) of photons in space has a temperature of
T = 2.726 K. The most probable photon frequency (distributed according to
Plank's radiation law) is

f = c/λ = 1.6*10^{11} Hz, the
corresponding photon energy is E_{g }= hf ~ 658 μeV.

Calculate the
energy in MeV of a cosmic-ray proton that by interacting with a CMBR photon will
produce a pion (π^{0}) like in part (a).

(c) Calculate the mean free path of the proton in
space for this process if the density of CMBR photons is 411 cm^{-3} and
the cross section is 0.5 mbarn (1 barn = 10^{-24} cm^{2}).

Solution:

- Concepts:

Relativistic collisions, energy and momentum conservation, frame transformations - Reasoning:

In relativistic collisions between free particles energy and momentum are always conserved. Often the physics is best visualized in the center of momentum frame. - Details of the calculation:

(a) The photon has minimum energy when the reaction products are at rest in the CM frame. In the lab frame the proton and pion move together as one particle with mass

M = m_{π0}^{ }+ m_{p}.

In the CM frame after the collision we have for the "length"^{2}of the momentum 4-vector

P_{0}^{2}- P^{2}= P_{0}^{2}= (∑p_{0})^{2 }= (Mc)^{2}.

The "length"^{2}of the total momentum 4-vector is (Mc)^{2}before and after the collision in the lab frame.

Before the collision we have

P_{0}^{2}- P^{2}= (hf/c + m_{p}c)^{2}- (hf/c)^{2}= m_{p}^{2}c^{2}+ 2hfm_{p}= (Mc)^{2}.

hf = m_{π0}(m_{π0}+ 2m_{p})c^{2}/(2m_{p}) = 144.7 MeV is the minimum photon energy.(b) In the rest frame of the proton, the photon energy must be greater than 144.7 MeV.

How fast does the proton have to move for the energy of the CMBR photon to be Doppler shifted up to 144.7 MeV?

Doppler shift: f'/f = 658 μeV/144.7 MeV = [(1-v/c)/(1+ v/c)]^{½}.

f'/f = 4.547*10^{-12}. v/c = (1 - (f'/f)^{2})/(1 - (f'/f)^{2}) = (1 - α)/(1 + α) with α = (f'/f)^{2}.

1 - v^{2}/c^{2}= [1 + α)^{2}- (1 - α)^{2}]/(1 + α)^{2}= 4α/(1 + α)^{2}= 8.271*10^{-23}.

γ = 1.1*10^{11}.

γm_{p}c^{2}~ 1*10^{14}MeV is the energy of a cosmic-ray proton that by interacting with CMBR photon can produce a pion π^{0}.

(c) Mean free path: λ = (σn)^{-1}.

σ = cross section, n = density of targets.

λ = (411 cm^{-3}* 0.5*10^{-27}cm^{2})^{-1}= 4.87*10^{22}m ~ 5.14*10^{6}ly.

Assume a γ-ray collides with a particle at rest.

(a) What is the threshold energy for the photon (expressed in terms of the
electron mass) for the process γ + Pb --> e^{+} + e^{-} + Pb?

(c) What is the threshold energy for the photon (expressed in terms of the
electron mass) for the process γ + e^{-} --> e^{+} + e^{-}
+ e^{-} where e^{-} is a free electron?

(c) Why are the threshold energies in Parts (a) and (b) different?

Solution:

- Concepts:

Relativistic collisions, energy and momentum conservation, frame transformations - Reasoning:

In relativistic collisions between free particles energy and momentum are always conserved. Often the physics is best visualized in the center of momentum frame.

The photon has minimum energy when the reaction products are at rest in the CM frame. - Details of the calculation:

(a) In the CM frame after the collision we have for the "length"^{2}of the momentum

4-vector P_{0}^{2}- P^{2}= P_{0}^{2}= (∑p_{0})^{2 }= (Mc)^{2}, where Mc^{2}= ∑m_{i}c^{2}= (m_{Pb}+ 2m_{e})c^{2}.

The "length"^{2}of the total momentum 4-vector is (Mc)^{2}before and after the collision in the lab frame.

Before the collision we have

P_{0}^{2}- P^{2}= (hf/c + m_{Pb}c)^{2}- (hf/c)^{2}= m_{Pb}^{2}c^{2}+ 2hfm_{Pb}= (Mc)^{2}.

hf = 4m_{e}(m_{e}+ m_{Pb})c^{2}/(2m_{Pb}) ≈ 2m_{e}c^{2}= is the minimum photon energy.

(b) (hf/c + m_{e}c)^{2}- (hf/c)^{2}= m_{e}^{2}c^{2}+ 2hfm_{e}= (3m_{e}c)^{2}.

hf = 4m_{e}c^{2}= is the minimum photon energy.

(c) The collision has to conserve energy and momentum. For a given momentum transfer ∆p, the energy transfer ∆E is much smaller for a massive Pb nucleus than for an electron.