Doppler Broadening - Derivation

Derivation

When thermal motion causes a particle to move towards the observer, the emitted radiation will be shifted to a higher frequency. Likewise, when the emitter moves away, the frequency will be lowered. For non-relativistic thermal velocities, the Doppler shift in frequency will be:

where is the observed frequency, is the rest frequency, is the velocity of the emitter towards the observer, and is the speed of light.

Since there is a distribution of speeds both toward and away from the observer in any volume element of the radiating body, the net effect will be to broaden the observed line. If is the fraction of particles with velocity component to along a line of sight, then the corresponding distribution of the frequencies is

where is the velocity towards the observer corresponding to the shift of the rest frequency to . Therefore,

We can also express the broadening in terms of the wavelength . Recalling that in the non-relativistic limit, we obtain

In the case of the thermal Doppler broadening, the velocity distribution is given by the Maxwell distribution

where is the mass of the emitting particle, is the temperature and is the Boltzmann constant.

Then,

We can simplify this expression as

$P_f(f)df=\sqrt{\frac{mc^2}{2\pi kT {f_0}^2}}\, \exp\left(-\frac{mc^2\left(f-f_0\right)^2}{2kT {f_0}^2}\right)df$ ,

which we immediately recognize as a Gaussian profile with the standard deviation

and full width at half maximum (FWHM)

Similarly,

$P_\lambda(\lambda)d\lambda = \sqrt{\frac{mc^2} {2\pi kT\lambda_0^2}}\,\exp\left(-\frac{mc^2(\lambda-\lambda_0)^2}{2kT\lambda_0^2}\right)d\lambda$

with the standard deviation

and FWHM

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