Scattering of radiation

Problem:

Consider a free electron in the field of a plane electromagnetic wave.  Calculate the ratio of the energy per unit time radiated by the electron to the light energy incident per unit area per unit time.  From this calculate the scattering cross section.  (Assume the light wave is of low frequency and neglect the effect of the B field of the wave on the electron.)

Solution:

• Concepts:
  The Poynting vector, the Larmor formula, the scattering cross section
• Reasoning:
  The electron is a driven oscillator, driven by the electric field of the incident plane wave.  The magnitude of the acceleration of the electron is proportional to the magnitude of the electric field vector.  The power radiated by the accelerating electron is given by the Larmor formula.
• Details of the calculation:
  Let E = Ek.  The force on the electron is F = -q_eE = ma.
  Let E = E₀e^iωt at the position of the electron.  Then d²z/dt² = -(q_eE₀/m)e^iωt.
  z = z₀e^iωt,  -ω²z₀ = -(q_eE₀/m),  z₀ = q_eE₀/(mω²).
  The electron oscillates with frequency ω and amplitude z₀.  The total power radiated by such an electron is
  <P> = q_e²ω⁴z₀²/(12πε₀c³) = q_e⁴E₀²/(12πε₀c³m²).
  The light energy incident per unit area per unit time is given by the magnitude of the average Poynting vector.
  <S> = ½ε₀cE₀².
  Scattering cross section:
  σ = <P>/<S> = q_e⁴/(6πε₀²c⁴m²) = (8π/3)(e⁴/(c⁴m²)) = (8π/3)r₀²,
  with r₀ = e²/(mc²).

Problem:

A plane electromagnetic (EM) wave is incident on a free particle of charge q and mass m.  The EM wave causes the particle to oscillate and hence to radiate.  The interaction can be considered as a scattering of EM radiation with cross section
σ_T = (power radiated)/(incident flux).
Assume the interaction can be treated non-relativistically.
Using Larmor's  radiation formula, show that
σ_T  = (1/(4πε₀))²(8π/3)(q_e²/(c²m))².
Evaluate σ_T for an electron.

Solution:

• Concepts:
  Radiation produced by accelerating charges, the Larmor formula
• Reasoning:
  We are asked to determine the power radiated by an accelerating electron and compare this power to the incident flux.
• Details of the calculation:
  Let E = Ek.  The force on the electron is F = -q_eE = ma.
  Let E = E₀e^iωt at the position of the electron.  Then d²z/dt² = -(q_eE₀/m) e^iωt.
  z = z₀e^iωt,  -ω²z₀ = -(q_eE₀/m),  z₀ = (q_eE₀/(mω²)).
  The electron oscillates with frequency ω and amplitude z₀.  The total power radiated by such an electron is
  <P> = q_e²ω⁴z₀²/(12πε₀c³) = q_e⁴E₀²/(12πε₀c³m²).
  The light energy incident per unit area per unit time is given by the magnitude of the average Poynting vector.
  <S> = ½ε₀cE₀².
  Scattering cross section:
  σ = <P>/<S> = q_e⁴/(6πε₀²c⁴m²) = (1/(4πε₀))²(8π/3)(q_e⁴/(c⁴m²)).
  For an electron: σ = (8π/3)r₀², where r₀ = e²/(mc²) = 2.82*10^-15 m.
  σ = 6.65*10^-29 m².

Problem:

Consider a point particle of mass m and charge q attached to a peg at the origin via a spring of spring constant k.  The particle can move in three dimensions.  What is the effective total cross section for the scattering of a linearly polarized electromagnetic plane wave with angular frequency ω by this particle?

Solution:

• Concepts:
  The Poynting vector, the Larmor formula, the scattering cross section
• Reasoning:
  The charge q on the spring  is a driven oscillator, driven by the electric field of the incident plane wave.  The power radiated by the accelerating electron is given by the Lamor formula.
• Details of the calculation:
  Let E = Ek.  The force on the charge is F = qE - kr = ma.
  Let E = E₀e^iωt at the position of the electron.  Then d²z/dt² + (k/m)z = (qE₀/m)e^iωt.
  Let ω₀² = k/m.
  z = z₀e^iωt,  -ω²z₀ + ω₀²z₀= (qE₀/m),  z₀ = qE₀/(m(ω₀² - ω²).
  The charge oscillates with frequency ω and amplitude z₀.  The total power radiated by such an electron is
  <P> = q²ω⁴z₀²/(12πε₀c³) = q⁴ω⁴E₀²/(12πε₀c³m²(ω₀² - ω²)²)
  The light energy incident per unit area per unit time is given by the magnitude of the average Poynting vector.
  <S> = ½ε₀cE₀².
  Scattering cross section:
  σ = <P>/<S> = q⁴ω⁴/(6πε₀²c⁴m²(ω₀² - ω²)²)
  = (1/(6πε₀²))(q²/(mc²))²[ω⁴/(ω₀² - ω²)²].