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I have been tasked to find the resolutions of a few synthetic spectra (wavelength in A vs. flux) of different stars and degrade them to the resolution of the observed spectra. But, I am not sure how to find the resolution of a given spectrum.

I tried calculating the values for $\dfrac{\lambda}{\Delta\lambda}$ and averaging them, but it is giving me a value that is a lot more than what it should be.

Any help will be greatly appreciated!

Edit:

I have added a part of an observed spectrum below. I have calculated the dLambda value and the value $\frac{\lambda}{\Delta\lambda}$ (named Resolution here). The resolution should be around 30000 (taken using VLT UVES) but I am getting a value that is 10 times larger.

enter image description here

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  • $\begingroup$ Resolution is measured in wavelength; resolving power is unitless. @EdV $\endgroup$
    – ProfRob
    May 16 at 17:00
  • $\begingroup$ You are confused between resolution and resolving power. The resolving power might be 330,000 but that would imply your spectrsa were undersampled. $\endgroup$
    – ProfRob
    May 16 at 17:28

2 Answers 2

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You really need to find the resolution that the synthetic spectra were generated at. This isn't something you should be trying to find from the spectra themselves.

From the Table you have shown, you appear to have just divided the wavelength by the bin size to get 330,000 for the resolving power. (NB Resolving power is the wavelength divided by the FWHM of a resolution element. The resolution is the FWHM of the resolution element.)

It is highly unlikely that this gives the right result because then your synthetic spectra would be undersampled.

However, if you can assume that the line broadening in the spectra is dominated by the imposed resolution of the spectra, rather than the intrinsic broadening of the lines and turbulence present in the atmosphere itself - which might be true for relatively low resolution spectra - then a possible procedure is to take the Fourier transform.

The FT will be that of the convolution of random delta functions with the line spread function. The result will be the product of the FT of the random delta functions and that of a Gaussian (which is also a Gaussian). This should give you a total product that has a Gaussian envelope. The inverse of the frequency-space width of this Gaussian envelope should then be an estimate of the Gaussian sigma corresponding to your resolution.

Perhaps an easier rough way of doing this is to try and find what look like isolated absorption lines and fit them with Gaussian functions. The FWHM of those Gaussians is approximately your resolution (it is actually an upper limit, because it assumes the intrinsic broadening is negligible).

A final thought is that perhaps the resolution is limited by the number density of pixels in the synthetic spectra. In which case, it is possible that the spectra have been binned to have a resolution (FWHM) of 2 pixels? But from the information you have given I suspect that the resolution element is more like 10 pixels.

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  • $\begingroup$ Thank you, I'll try out the methods that you have mentioned. $\endgroup$
    – SaptarshiS
    May 16 at 19:33
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Your concept of 'resolution' seems to be incorrect here. An observed spectrum does not have a resolution as such. The spectral flux is a continuous function of wavelength. Only the instrument by which the spectrum has been obtained can be said to have a resolution. The convolution of the instrumental profile with the original spectrum yields then the observed spectrum. Of course, the original spectrum is in general not known, so the process is in practice the other way around i.e. it is derived from the observed spectrum and the known instrumental response function via 'deconvolution' techniques.

If you have a synthetic spectrum with a sufficiently fine resolution (i.e. enough data points), you convolute this then with the instrumental profile to obtain a spectrum to compare to the observed one.

In other words, there is no intrinsic information in the spectrum itself (i.e. if you measure flux per wavelength interval) that would tell you what resolution it has. You would need additional information regarding how the spectrum was produced and strictly speaking even how the radiation constituting the spectrum was produced in the first place. Hence the term 'resolution of a spectrum' is really meaningless as it can not be answered by just looking at the spectrum.

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    $\begingroup$ The phrase "[wavelength] resolution of a spectrum" is in common usage, just like "the [spatial] resolution of an image". It is obvious what is meant. $\endgroup$
    – ProfRob
    May 21 at 13:41
  • $\begingroup$ @ProfRob I don't recall having heard that phrase before, and I don't think the meaning of it is obvious here, in fact I think it is misleading. I quote from Wikipedia (Spectrum) "A spectrum is a condition that is not limited to a specific set of values but can vary, without gaps, across a continuum." I mean what resolution do you want for instance assign to a spectrum of sunlight formed by a prism? The OP is approaching the issue from the wrong angle. $\endgroup$
    – Thomas
    May 21 at 14:11
  • $\begingroup$ It's in common usage. Just google "resolution of the spectrum" for thousands of examples. In your example - the resolution that the OP would wish to know is the FWHM of the Fraunhofer lines present in the spectrum. If the spectrum were truly continuous then of course you are stuck, without further information. $\endgroup$
    – ProfRob
    May 21 at 14:25
  • $\begingroup$ @ProfRob What I am saying is, there is no intrinsic information in the spectrum itself (i.e. if you measure flux per wavelength interval) that would tell you what resolution it has. You would need additional information regarding how the spectrum was produced and strictly speaking even how the radiation constituting the spectrum was produced in the first place. Hence the term 'resolution of a spectrum' is meaningless as it can not be answered by just looking at the spectrum. $\endgroup$
    – Thomas
    May 21 at 14:36
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    $\begingroup$ In your example - the resolution that the OP would wish to know is the FWHM of the Fraunhofer lines present in the spectrum. Because we DO have additional information. $\endgroup$
    – ProfRob
    May 21 at 15:12

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