# Doppler broadening thermal line width

Access list of astrophysics formulas download page:

### Doppler broadening thermal line width

In thermal equilibrium the particles of a gas follow the Maxwell-Boltzmann distribution in velocity and photons emitted from such particles will have measured energies, wavelengths, and frequencies that differ from their rest-frame values because of the Doppler effect. Since only line-of-sight motion contributes to the Doppler effect, it is the one-dimensional Maxwell-Boltzmann velocity factor that is relevant: $\exp{-mv_{x}^{2}/2kT}$ (the three-dimensional distribution has an additional $v^{2}$ factor). In the case of nonrelativistic velocities, the fractional frequency shift of a photon is $\sim v_{x}/c$ so the resulting profile of an atomic line is a Gaussian:

\[

F(\nu) = \frac{1}{\sqrt{\pi}W_{D}} \exp{[-(\nu-\nu_{0})^{2}/W_{D}^{2}]}

\]

where the Doppler width, $W_{D}$ is *defined* as

\[

W_{D} \equiv \nu_{0} \left(\frac{2kT}{mc^{2}}\right)^{\frac{1}{2}}.

\]

Here $m$ is the mass of the atom or molecule and $v_{0}$ is the photon rest-frame frequency.

However, note that the equivalent $\sigma$ of the Gaussian is

\[

\sigma = \frac{1}{\sqrt{2}}W_{D} = \nu_{0} \left(\frac{kT}{mc^{2}}\right)^{\frac{1}{2}}

\]

because the standard Gaussian is $e^{-\nu^{2}/2\sigma^{2}}$ and $2\sigma^{2} = W_{D}^{2}$.