# FWHM Gaussian FWHM Calculation

### FWHM Calculation for a Gaussian Line Profile

Below, the FWHM Gaussian FWHM calculation is shown with an example of how to estimate velocity broadening of emission or absorption lines. The FWHM is the full width half maximum parameter of an emission or absorption line that characterizes the width of the line in a single parameter. It depends on the shape of the line but does not capture all of the properties of the shape of the line. If the line can be approximated with a Gaussian it is straightfoward to calculate.

There are two situations: you are either required to calculate the FWHM for actual data or you want it for a theoretical curve. Either way you need the width of the Gaussian ($\sigma$). In the case that you have data, you either fit a model Gaussian to extract the best-fitting $\sigma$, or you estimate it directly by calculating the standard deviation (which is the estimate for $\sigma$), the formula for which you can easily look up in an elementary text book or the internet. In the case of a theoretical curve as the starting point, you will be given $\sigma$ so you can proceed to the next step.

Note that for a line that is thermally-broadened due to the Doppler effect, the $\sigma$ of the Gaussian line profile is

$\sigma = \left(\frac{kT}{mc^{2}}\right)^{\frac{1}{2}}$

where $T$ is the gas temperature and $m$ is the mass of the atom or molecule.

Since the FWHM does not depend on the absolute position of the centroid (peak) we can simply assume the line is centered on the origin and the choice of absolute normalization does not matter because the FWHM is calculated relative to the peak of the line. Suppose the $x$ coordinate represents whatever units we are measuring the position of the line with (e.g., energy), we obtain the half width at half maximum (HWHM) by setting

$\frac{1}{2} = \exp{[-x^{2}/(2\sigma^{2})]}$

and solving for $x$ to get

$x_{1/2} = \sigma \sqrt{2 \ln{2}}$
and the FWHM is just twice this width so

${\rm FWHM} = [2 \sqrt{2\ln{2}}]\sigma = 2.35482 \sigma .$

You can see from this last equation that it is very easy to calculate FWHM.
Example: Estimate the FWHM velocity broadening of an emission line with a peak at $E_{0}=2.0$ keV, and $\sigma=50$ eV.
Solution: Since $(0.050/2)\ll 1$, we can use the nonrelativistic relation $(v/c) \sim ({\rm FWHM}/E_{0})$ to get $v ({\rm FWHM}) \sim 17,661 \ \rm km \ s^{-1}$.