<ol>
<li><p><span class="math-container">$x$</span> is in terms of <span class="math-container">$\log \lambda$</span> because increments in <span class="math-container">$\log \lambda$</span> correspond to increments in line of sight velocity. e.g. see <a href="https://astronomy.stackexchange.com/questions/14518/why-do-linear-velocity-redshifts-correspond-to-linear-pixel-shifts-when-the-spec">Why do linear velocity redshifts correspond to linear pixel shifts when the spectra are binned in constant log wavelength?</a>  Thus a <span class="math-container">$\Delta v$</span> from the line centre corresponds to a <span class="math-container">$\Delta \log \lambda$</span>. There will be a constant of proportionality but that isn't relevant to the mathematics.</p>
</li>
<li><p>Most spectral lines could be treated as delta functions, but it depends on how big the resolving power of the instrument is. In particular, there are lines in stellar spectra where pressure broadening or rotational broadening could easily be resolved in high resolution spectra. If you could assume all the lines were delta functions then your expression for <span class="math-container">$S(x)$</span> looks ok.</p>
</li>
<li><p>You can't remove instrumental broadening by deconvolution in data which has noise. The usual procedure would be to do forward modelling, where a model is convolved with the intrumental profile before comparison with data. If the instrumental profiles are well-behaved normal distributions then it is of course possible to convolve the highest resolution spectra with a Gaussian to simulate what would be seen at the resolution of the lowest resolution spectra; but not vice-versa.</p>
</li>
</ol>

CC BY-SA 4.0
2022-08-16T12:50:28.143Z
2023-02-19T08:12:41.263Z
2023-02-19T08:12:41.263Z
ProfRob
<p>I am trying to understand the mathematical formulation of instrumental dispersion in galaxy and star spectra. Let <span class="math-container">$x=ln(\lambda)$</span>. Assuming that the galaxy spectra G(x) is composed of many identical stars with spectrum S(x), and let B(x) be the LOSVD broadening the spectral lines.
Then we can write the galaxy's spectrum as the convolution <span class="math-container">$G(x) = B(x) \ast S(x)$</span>.</p>
<p>1.) I am not sure about the functional argument of B. Since it is a velocity distribution, we should be in velocity space, so we need something like B(cx), where c is the speed of light. I still see many papers using B(x), what is the relationship between B(x) and the LOSVD in velocity space? Would it be correct to write the convolution as <span class="math-container">$G(x) = B(cx) \ast S(x)$</span>?</p>
<p>2.) Let's call the instrumental profile I(x), and assume that the width of the spectral lines in the template spectra is only due to instrumental broadening (is this assumption reasonable?). What is physically happening in the instrument/spectrosgraph so the spectral lines get broadened? I assume mathematically one also writes the template spectra S(x) as a convolution of the instrumental profile I(x) with the &quot;natural&quot; non broadened template spectrum S'(x) (Which are delta functions, if you ignore natural line broadening due to Heisenbergs uncertainty principle, temperature and pressure broadening ?), since the Instrument is nothing more then some sort of filter. Would it be correct in this case to write <span class="math-container">$S(x) = I(x) \ast S'(x)$</span> ? Following the assumption in the beginning the template spectra could be written as <span class="math-container">$S(x) = \sum \alpha_{m}I(x-x_{m})$</span>, where <span class="math-container">$\alpha_{m}$</span> is the strength of the m-th line and <span class="math-container">$x_m$</span> is the location of the m-th line.</p>
<p>3.) How does one remove instrumental broadening in the template and galaxy spectra in kinematical analysis if different instruments and therefore different functions I(x) have been used to observe the spectra, that don't cancel in the division G/S (f.e. like in Fourier-Quotient-Algorithm)?</p>
<p>I am happy about any response to the questions, comments and further discussion to clarify the topic for me.</p>

<observational-astronomy><galaxy><spectroscopy><spectra><kinematics>
How does instrumental dispersion work?
trynerror
# How does instrumental dispersion work?

(Asked by: username_0 on 2022-08-16)

I am trying to understand the mathematical formulation of instrumental dispersion in galaxy and star spectra. Let $x=ln(\lambda)$. Assuming that the galaxy spectra G(x) is composed of many identical stars with spectrum S(x), and let B(x) be the LOSVD broadening the spectral lines.
Then we can write the galaxy's spectrum as the convolution $G(x) = B(x) \ast S(x)$.

1.) I am not sure about the functional argument of B. Since it is a velocity distribution, we should be in velocity space, so we need something like B(cx), where c is the speed of light. I still see many papers using B(x), what is the relationship between B(x) and the LOSVD in velocity space? Would it be correct to write the convolution as $G(x) = B(cx) \ast S(x)$?

2.) Let's call the instrumental profile I(x), and assume that the width of the spectral lines in the template spectra is only due to instrumental broadening (is this assumption reasonable?). What is physically happening in the instrument/spectrosgraph so the spectral lines get broadened? I assume mathematically one also writes the template spectra S(x) as a convolution of the instrumental profile I(x) with the "natural" non broadened template spectrum S'(x) (Which are delta functions, if you ignore natural line broadening due to Heisenbergs uncertainty principle, temperature and pressure broadening ?), since the Instrument is nothing more then some sort of filter. Would it be correct in this case to write $S(x) = I(x) \ast S'(x)$ ? Following the assumption in the beginning the template spectra could be written as $S(x) = \sum \alpha\_{m}I(x-x\_{m})$, where $\alpha\_{m}$ is the strength of the m-th line and $x\_m$ is the location of the m-th line.

3.) How does one remove instrumental broadening in the template and galaxy spectra in kinematical analysis if different instruments and therefore different functions I(x) have been used to observe the spectra, that don't cancel in the division G/S (f.e. like in Fourier-Quotient-Algorithm)?

I am happy about any response to the questions, comments and further discussion to clarify the topic for me.

**Tags:** observational-astronomy, galaxy, spectroscopy, spectra, kinematics

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## 1 answers

### Answer by username_1 on 2022-08-16. Score: 2

1. $x$ is in terms of $\log \lambda$ because increments in $\log \lambda$ correspond to increments in line of sight velocity. e.g. see [Why do linear velocity redshifts correspond to linear pixel shifts when the spectra are binned in constant log wavelength?](https://astronomy.stackexchange.com/questions/14518/why-do-linear-velocity-redshifts-correspond-to-linear-pixel-shifts-when-the-spec) Thus a $\Delta v$ from the line centre corresponds to a $\Delta \log \lambda$. There will be a constant of proportionality but that isn't relevant to the mathematics.
2. Most spectral lines could be treated as delta functions, but it depends on how big the resolving power of the instrument is. In particular, there are lines in stellar spectra where pressure broadening or rotational broadening could easily be resolved in high resolution spectra. If you could assume all the lines were delta functions then your expression for $S(x)$ looks ok.
3. You can't remove instrumental broadening by deconvolution in data which has noise. The usual procedure would be to do forward modelling, where a model is convolved with the intrumental profile before comparison with data. If the instrumental profiles are well-behaved normal distributions then it is of course possible to convolve the highest resolution spectra with a Gaussian to simulate what would be seen at the resolution of the lowest resolution spectra; but not vice-versa.


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