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**= E**k**. The force on the electron is**F**= -q_{e}**E**= m**a**.

Let E = E_{0}e^{iωt}at the position of the electron. Then d^{2}z/dt^{2}= -(q_{e}E_{0}/m)e^{iωt}.

z = z_{0}e^{iωt}, -ω^{2}z_{0}= -(q_{e}E_{0}/m), z_{0}= (q_{e}E_{0}/(mω^{2})).

The electron oscillates with frequency ω and amplitude z_{0}. The total power radiated by such an electron is

<P> = q_{e}^{2}ω^{4}z_{0}^{2}/(12πε_{0}c^{3}) = q_{e}^{4}E_{0}^{2}/(12πε_{0}c^{3}m^{2}).

The light energy incident per unit area per unit time is given by the magnitude of the average Poynting vector.

<S> = ½ε_{0}cE_{0}^{2}.

Scattering cross section:

σ = <P>/<S> = q_{e}^{4}/(6πε_{0}^{2}c^{4}m^{2}) = (8π/3)(e^{4}/(c^{4}m^{2})) = (8π/3)r_{0}^{2},

with r_{0}= e^{2}/(mc^{2}).

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/(4πε_{0}))^{2}(8π/3)(q_{e}^{2}/(c^{2}m))^{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**= E**k**. The force on the electron is**F**= -q_{e}**E**= m**a**.

Let E = E_{0}e^{iωt }at the position of the electron. Then d^{2}z/dt^{2}= -(q_{e}E_{0}/m) e^{iωt}.

z = z_{0}e^{iωt}, -ω^{2}z_{0}= -(q_{e}E_{0}/m), z_{0}= (q_{e}E_{0}/(mω^{2})).

The electron oscillates with frequency ω and amplitude z_{0}. The total power radiated by such an electron is

<P> = q_{e}^{2}ω^{4}z_{0}^{2}/(12πε_{0}c^{3}) = q_{e}^{4}E_{0}^{2}/(12πε_{0}c^{3}m^{2}).

The light energy incident per unit area per unit time is given by the magnitude of the average Poynting vector.

<S> = ½ε_{0}cE_{0}^{2}.

Scattering cross section:

σ = <P>/<S> = q_{e}^{4}/(6πε_{0}^{2}c^{4}m^{2}) = (1/(4πε_{0}))^{2}(8π/3)(q_{e}^{4}/(c^{4}m^{2})).

For an electron: σ = (8π/3)r_{0}^{2}, where r_{0}= e^{2}/(mc^{2}) = 2.82*10^{-15}m.

σ = 6.65*10^{-29}m^{2}.

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**= E**k**. The force on the charge is**F**= q**E**- k**r**= m**a**.

Let E = E_{0}e^{iωt}at the position of the electron. Then d^{2}z/dt^{2}+ (k/m)z = (qE_{0}/m)e^{iωt}.

Let ω_{0}^{2}= k/m.

z = z_{0}e^{iωt}, -ω^{2}z_{0}+ ω_{0}^{2}z_{0}= (qE_{0}/m), z_{0}= qE_{0}/(m(ω_{0}^{2}- ω^{2}).

The charge oscillates with frequency ω and amplitude z_{0}. The total power radiated by such an electron is

<P> = q^{2}ω^{4}z_{0}^{2}/(12πε_{0}c^{3}) = q^{4}ω^{4}E_{0}^{2}/(12πε_{0}c^{3}m^{2}(ω_{0}^{2}- ω^{2})^{2})

The light energy incident per unit area per unit time is given by the magnitude of the average Poynting vector.

<S> = ½ε_{0}cE_{0}^{2}.

Scattering cross section:

σ = <P>/<S> = q^{4}ω^{4}/(6πε_{0}^{2}c^{4}m^{2}(ω_{0}^{2}- ω^{2})^{2})

= (1/(6πε_{0}^{2}))(q^{2}/(mc^{2}))^{2}[ω^{4}/(ω_{0}^{2}- ω^{2})^{2}].