The

x

⟨ v_{x} ⟩ = 0

. Therefore

⟨ ω ⟩ = ω_{0} (1 + \frac{⟨ v _{x} ⟩}{c}) = ω_{0} .

ii

Solution

The same one-dimensional normal distribution has

Var (v_{x}) = \frac{k _{B} T}{m} .

Because

ω - ω_{0} = (ω_{0} / c) v_{x}

Var (ω) = \frac{ω _{0}^{2}}{c ^{2}} \frac{k _{B} T}{m}, σ_{ω} = \frac{ω _{0}}{c} \frac{k _{B} T}{m} .

The Doppler relation is the affine change of variables

v_{x} = c (\frac{ω}{ω _{0}} - 1), \frac{d v _{x}}{d ω} = \frac{c}{ω _{0}} .

Transforming the one-dimensional Maxwell-Boltzmann velocity distribution therefore gives the normalized thermal Doppler broadening profile

I (ω) = \frac{c}{ω _{0}} \frac{m}{2 π k _{B} T} exp [- \frac{m c ^{2}}{2 k _{B} T ω _{0}^{2}} (ω - ω_{0})^{2}] .

Thus

a = \frac{m c ^{2}}{2 k _{B} T ω _{0}^{2}} .

The measured Gaussian line width satisfies

σ_{ω} / ω_{0} = k_{B} T / m / c

, so the temperature follows from

T = \frac{m c ^{2}}{k _{B}} (\frac{σ _{ω}}{ω _{0}})^{2} .

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