<p>Short version: velocity resolution is the smallest velocity <em>difference</em> you can measure between two moving objects, using a given spectrum.</p>
<p>More details:</p>
<p>As you probably know (based on your implicit use of the formula in your question), we can measure velocities by using the Doppler shift.  To do that, we need to measure a feature (an absorption or emission line) in an object's spectrum, and then determine how much it is shifted from the known &quot;rest&quot; wavelength or frequency.</p>
<p>Now imagine that there are two gas clouds along the line of sight when you take a spectrum, both emitting light in some spectral line.  (This is pretty common when observing <a href="https://physicsopenlab.org/2020/09/08/milky-way-structure-detected-with-the-21-cm-neutral-hydrogen-emission/" rel="nofollow noreferrer">neutral hydrogen in the plane of the Milky Way</a>, for example.)  If they have the same velocity, then their emission spectra will lie exactly on top of each other, and you won't be able to see that there are two clouds there.  If they have slightly different velocities, then the different Doppler shifts of the two lines will give them slightly different wavelengths.  You can measure the two different velocities <em>if</em> the shift between them is larger than the resolution of your spectrograph, i.e. if the finite width of spectral features imposed by your optics and detector doesn't smear them together too much.</p>
<p>You can probably see from this that a smaller velocity resolution is better.  A resolution of 1 km/s means that you can distinguish features as close as that, while 10 km/s means that a 1 km/s difference in two features would be undetectable.  It is similar to spatial resolution in this sense.  A spatial resolution of 1 arcsec is better than a resolution of 10 arcsec, since it means you can distinguish finer details.</p>
<p>Note that a given velocity resolution does <em>not</em> mean that that is the smallest Doppler shift velocity you can measure for a single object.  That is because you can generally find the central wavelength of a given spectral feature (which is what you need for a Doppler shift measurement) very precisely, even if the feature is broad.  For example, the spectrographs used to measure radial velocities in searches for extrasolar planets (e.g. <a href="http://www.eso.org/sci/facilities/lasilla/instruments/harps/inst/description.html" rel="nofollow noreferrer">HARPS</a>) typically have spectral resolution on the order of 1 part in 100,000, i.e. the resolution is 1/100,000 of the observing wavelength.  If you put that in the Doppler shift formula, that translates to a velocity resolution of 1/100,000 of the speed of light, or about 3 km/s.  Yet those instruments routinely measure velocities of just a few <em>meters</em> per second, which is 1,000 times smaller than their nominal velocity resolution.</p>
<p>As a final technical note, in your question you calculated the velocity width of one channel in the spectrum, but the velocity resolution would be twice as large as this, because of the need to <a href="http://microscopy.berkeley.edu/courses/dib/sections/02Images/sampling.html" rel="nofollow noreferrer">Nyquist sample</a> the spectrum.  One way to think about that would be to imagine a series of spectral features that were separated by exactly the velocity width of one channel.  If that were the case, then every channel would have a feature, and your spectrum would just appear flat. (Even if you just had two features in consecutive channels, they would blend to appear as one.)  But if the features were separated by twice that velocity, then only every other channel would have a feature and you could see that there were distinct features there.  So the spectral resolution is always at least twice your channel width, but could be limited by other parts of your instrumentation to be larger than that.</p>

CC BY-SA 4.0
2021-04-14T01:09:16.983Z
2021-04-14T01:09:16.983Z
Eric Jensen
<p>I've always struggled to conceptualize the term &quot;velocity resolution&quot; in spectral observations. In principle, it is simple to calculate. For example, one can take the spectral width and divide it by the number of channels. ie. A 16 MHz wide window with 4096 channels yields a 3.906 kHz / channel velocity resolution. For an observation centered on the H1 spectral line, we get 0.82 km / s / channel utilizing the speed of light and the photon energy at the 21-cm H1 line ~ 1.42 Hz.</p>
<p>So, why is there a &quot;velocity&quot; defined in the resolution term? I understand dimensionally it comes from the speed of light, but what does the final velocity signify? Does it have anything to do with how the source moves (ie. object of interest is moving) with respect to the observer? Is it better to have a larger or smaller velocity resolution or does it depend? In general, why is the value important? Thanks in advance!</p>

<observational-astronomy><spectroscopy><radio-telescope><hydrogen><radio>
What is the significance of the velocity resolution in spectral line observations using radio interferometry?
Astroturf
# What is the significance of the velocity resolution in spectral line observations using radio interferometry?

(Asked by: username_0 on 2021-04-12)

I've always struggled to conceptualize the term "velocity resolution" in spectral observations. In principle, it is simple to calculate. For example, one can take the spectral width and divide it by the number of channels. ie. A 16 MHz wide window with 4096 channels yields a 3.906 kHz / channel velocity resolution. For an observation centered on the H1 spectral line, we get 0.82 km / s / channel utilizing the speed of light and the photon energy at the 21-cm H1 line ~ 1.42 Hz.

So, why is there a "velocity" defined in the resolution term? I understand dimensionally it comes from the speed of light, but what does the final velocity signify? Does it have anything to do with how the source moves (ie. object of interest is moving) with respect to the observer? Is it better to have a larger or smaller velocity resolution or does it depend? In general, why is the value important? Thanks in advance!

**Tags:** observational-astronomy, spectroscopy, radio-telescope, hydrogen, radio


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## 1 answers

### Answer by username_1 on 2021-04-14. Score: 4 | Accepted answer

Short version: velocity resolution is the smallest velocity *difference* you can measure between two moving objects, using a given spectrum.

More details:

As you probably know (based on your implicit use of the formula in your question), we can measure velocities by using the Doppler shift. To do that, we need to measure a feature (an absorption or emission line) in an object's spectrum, and then determine how much it is shifted from the known "rest" wavelength or frequency.

Now imagine that there are two gas clouds along the line of sight when you take a spectrum, both emitting light in some spectral line. (This is pretty common when observing [neutral hydrogen in the plane of the Milky Way](https://physicsopenlab.org/2020/09/08/milky-way-structure-detected-with-the-21-cm-neutral-hydrogen-emission/), for example.) If they have the same velocity, then their emission spectra will lie exactly on top of each other, and you won't be able to see that there are two clouds there. If they have slightly different velocities, then the different Doppler shifts of the two lines will give them slightly different wavelengths. You can measure the two different velocities *if* the shift between them is larger than the resolution of your spectrograph, i.e. if the finite width of spectral features imposed by your optics and detector doesn't smear them together too much.

You can probably see from this that a smaller velocity resolution is better. A resolution of 1 km/s means that you can distinguish features as close as that, while 10 km/s means that a 1 km/s difference in two features would be undetectable. It is similar to spatial resolution in this sense. A spatial resolution of 1 arcsec is better than a resolution of 10 arcsec, since it means you can distinguish finer details.

Note that a given velocity resolution does *not* mean that that is the smallest Doppler shift velocity you can measure for a single object. That is because you can generally find the central wavelength of a given spectral feature (which is what you need for a Doppler shift measurement) very precisely, even if the feature is broad. For example, the spectrographs used to measure radial velocities in searches for extrasolar planets (e.g. [HARPS](http://www.eso.org/sci/facilities/lasilla/instruments/harps/inst/description.html)) typically have spectral resolution on the order of 1 part in 100,000, i.e. the resolution is 1/100,000 of the observing wavelength. If you put that in the Doppler shift formula, that translates to a velocity resolution of 1/100,000 of the speed of light, or about 3 km/s. Yet those instruments routinely measure velocities of just a few *meters* per second, which is 1,000 times smaller than their nominal velocity resolution.

As a final technical note, in your question you calculated the velocity width of one channel in the spectrum, but the velocity resolution would be twice as large as this, because of the need to [Nyquist sample](http://microscopy.berkeley.edu/courses/dib/sections/02Images/sampling.html) the spectrum. One way to think about that would be to imagine a series of spectral features that were separated by exactly the velocity width of one channel. If that were the case, then every channel would have a feature, and your spectrum would just appear flat. (Even if you just had two features in consecutive channels, they would blend to appear as one.) But if the features were separated by twice that velocity, then only every other channel would have a feature and you could see that there were distinct features there. So the spectral resolution is always at least twice your channel width, but could be limited by other parts of your instrumentation to be larger than that.


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