---

---

Access list of astrophysics formulas download page:

The Lorentzian FWHM calculation (or full width half maximum) is actually straightforward and can be read off from the equation. The Lorentzian function is encountered whenever a system is forced to vibrate around a resonant frequency. The peak is at the resonance frequency. In the case of emission-line profiles, the frequency at the peak (say $\nu_{0}$) usually corresponds to the energy of an atomic or ionic transition between electronic energy levels. The basic function is

\[
L(\nu) \ = \ A \frac{\gamma}{(\nu – \nu_{0})^{2} + \gamma^{2}}
\]

Here $A$ is a normalization constant and $\gamma$ is a width parameter. However, note that $\gamma$ may be given as a different constant, or with a constant multiplier. Inspect the given equation carefully. In any given situation make the left-hand side of the equation equal to half of the value that the right-hand side would have if $\nu = \nu_{0}$. Solving for $\nu$ will give you the half-width, so double it to get the FWHM. In the form above, the peak value is $A/\gamma$. Then,

\[
\frac{1}{2}\frac{A}{\gamma} = A \frac{\gamma}{(\nu – \nu_{0})^{2} + \gamma^{2}}
\]

so

\[
{\rm FWHM} \ = \ 2\gamma \ .
\]

For the Lorentzian function to have a total area under the curve of 1.0, the normalization should be $A=(1/\pi)$ for the form of the function given above.

---

Physics Formulas and Tables E-book:
Click image to see details and preview:

---

---

Area under a Lorentzian: details

To find the area under a Lorentzian (i.e. to integrate from $-\infty$ to $+\infty$), make the following substitution:
\[
(\nu – \nu_{0}) = \gamma \tan{\theta}
\]
(Obviously if the Lorentzian is given with different constants, choose appropriately: you want to end up with just $(1+\tan^{2}{\theta})$ times some constant in the denominator.) Then,
\[
d\nu = \gamma \ \sec^{2}{\theta} = \gamma \ (1+\tan^{2}{\theta}) \ d\theta
\]
(the latter following from the Pythagoras identity). With the above substitution, the new limits on the integral become $-\pi/2$ and $+\pi/2$ because that’s where $\tan{\theta}$ goes to $-\infty$ and $+\infty$ respectively. So now we have
\[
\int^{+ \frac{\pi}{2}}_{- \frac{\pi}{2}}
{\frac{\gamma \cdot \gamma (1+\tan^{2}{\theta}) \ d\theta}{\gamma^{2} (1+\tan^{2}{\theta}) }}
\]
which simplifies very nicely to
\[
\int^{+ \frac{\pi}{2}}_{- \frac{\pi}{2}}{ \ d\theta} = \left[\left(\frac{\pi}{2}\right) – \left(- \frac{\pi}{2}\right)\right] \ = \ \pi
\]