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Optics of Moving Sources

Problem 5.238

A gas consists of atoms of mass mm being in thermodynamic equilibrium at temperature T.T . Suppose ω0\omega_{0} is the natural frequency of light emitted by the atoms. (a) Demonstrate that the spectral distribution of the emitted light is defined by the formula Iω=I0e−a(1−ω/ω0)2 I_{\omega}=I_{0} \mathrm{e}^{-a\left(1-\omega / \omega_{0}\right)^{2}} (I0\left(I_{0}\right. is the spectral intensity corresponding to the frequency ω0\omega_{0}, a=mc2/2kT⊤)\left.a=m c^{2} / 2 k T^{\top}\right) (b) Find the relative width Δω/ω0\Delta \omega / \omega_{0} of a given spectral line, i.e. the width of the line between the frequencies at which Iω=I0/2I_{\omega}=I_{0} / 2.

Reveal Answer
5.238. (a) Let vxv_{x} be the projection of the velocity vector of the radiating atom on the observation direction. The number of atoms with projections falling within the interval vx,vx+dvxv_{x}, v_{x}+d v_{x} is n(vx)dvx∼exp⁡(−mvx2/2kT)⋅dvx n\left(v_{x}\right) d v_{x} \sim \exp \left(-m v_{x}^{2} / 2 k T\right) \cdot d v_{x} The frequency of light emitted by the atoms moving with velocity vxv_{x} is ω=ω0(1+vx/c).\omega=\omega_{0}\left(1+v_{x} / c\right) . From the expression the frequency distribution of atoms can be found: n(ω)dω=n(vx)dvxn(\omega) d \omega=n\left(v_{x}\right) d v_{x}. And finally it should be taken into account that the spectral radiation intensity Iω∼n(ω)I_{\omega} \sim n(\omega)  (b) Δω/ω0=2(2ln⁡2)kT/mc2 \text { (b) } \Delta \omega / \omega_{0}=2 \sqrt{(2 \ln 2) k T / m c^{2}}
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