<p><span class="math-container">$\nu_0$</span> is the frequency you would expect for the absorption/emission line in the absence of any broadening. i.e. It is the centre of the line profile.</p>

<p><span class="math-container">$\delta \nu$</span> is just <span class="math-container">$\nu - \nu_0$</span>. (i.e. it is the separation in frequency from the centre of the line profile and is equivalent to <span class="math-container">$f -f_0$</span> in the second expression).</p>

<p><span class="math-container">$\Delta \nu_{th}$</span> is a measure of the broadening due to (I guess) thermal motions in whatever gas/plasma is emitting/absorbing the radiation. It is similar to the Gaussian "sigma" term in the expression
<span class="math-container">$$ f(x) = \exp(-x^2/2\sigma^2)$$</span></p>

<p>If you have a gas that follows Maxwell-Boltzmann statistics, then <span class="math-container">$\Delta \nu_{th}$</span> will be related to the speeds of atoms/ions/molecules in the gas and hence to the temperature of the gas.
<span class="math-container">$$ \Delta \nu_{th} \simeq \sqrt{\frac{2k_BT}{m}}\times \frac{\nu_0}{c} \,$$</span>
where <span class="math-container">$m$</span> is the mass of the particle emitting the light.</p>

CC BY-SA 4.0
2019-10-14T11:06:53.913Z
2019-10-14T11:06:53.913Z
ProfRob
<p>How to plot Doppler profile corresponding to the spectral line? The spectral line was plotted from data: flux with respect to wavelenght and I know the temperature. </p>

<p>These two function are Doppler profile? What should I substitue for <span class="math-container">$\Delta \nu_{th}$</span>, <span class="math-container">$\nu_0$</span> and <span class="math-container">$\Delta \nu$</span>. These values are from wavelenght according to <span class="math-container">$\lambda = \frac{c}{\nu}$</span>?</p>

<p><a href="https://i.sstatic.net/KYoGT.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/KYoGT.png" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/9sh2E.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9sh2E.png" alt="enter image description here"></a></p>

<p><a href="https://en.wikipedia.org/wiki/Doppler_broadening" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Doppler_broadening</a></p>

<p>Or could you recommend me another formula for plotting Doppler profile?</p>

CC BY-SA 4.0
2019-10-14T12:53:15.633Z
Thanks for the edit!
uhoh
<spectroscopy>
Thermal Doppler broadening
Arrara
# Thermal Doppler broadening

(Asked by: username_0 on 2019-10-14)

How to plot Doppler profile corresponding to the spectral line? The spectral line was plotted from data: flux with respect to wavelenght and I know the temperature.

These two function are Doppler profile? What should I substitue for $\Delta \nu\_{th}$, $\nu\_0$ and $\Delta \nu$. These values are from wavelenght according to $\lambda = \frac{c}{\nu}$?

[![enter image description here](https://i.sstatic.net/KYoGT.png)](https://i.sstatic.net/KYoGT.png)

[![enter image description here](https://i.sstatic.net/9sh2E.png)](https://i.sstatic.net/9sh2E.png)

<https://en.wikipedia.org/wiki/Doppler_broadening>

Or could you recommend me another formula for plotting Doppler profile?

**Tags:** spectroscopy

> **Comments:**
> **username_1**: Thanks for the edit!  

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## 1 answers

### Answer by username_2 on 2019-10-14. Score: 3 | Accepted answer

$\nu\_0$ is the frequency you would expect for the absorption/emission line in the absence of any broadening. i.e. It is the centre of the line profile.

$\delta \nu$ is just $\nu - \nu\_0$. (i.e. it is the separation in frequency from the centre of the line profile and is equivalent to $f -f\_0$ in the second expression).

$\Delta \nu\_{th}$ is a measure of the broadening due to (I guess) thermal motions in whatever gas/plasma is emitting/absorbing the radiation. It is similar to the Gaussian "sigma" term in the expression
$$ f(x) = \exp(-x^2/2\sigma^2)$$

If you have a gas that follows Maxwell-Boltzmann statistics, then $\Delta \nu\_{th}$ will be related to the speeds of atoms/ions/molecules in the gas and hence to the temperature of the gas.
$$ \Delta \nu\_{th} \simeq \sqrt{\frac{2k\_BT}{m}}\times \frac{\nu\_0}{c} \,$$
where $m$ is the mass of the particle emitting the light.


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