wavelength-based IFU data cubes VS velocity-based IFU data cubes

Question

I have 2 simulated IFU data cubes of the same observation, let's say A and B. The 3rd dimension of cube A is in wavelengths. In order to create cube B I just rebinned cube A in log-wavelengths and then converted that to velocity (km/s). The rebinning to log-wavelength was necessary in order to convert the 3rd dimension of cube B in km/s. For an explanation see this:

Converting ångström spectral dimension to galaxy speed (km/s)

Now I want to apply instrumental broadening on both cubes. In order to do that, for each cube, I have to create a Gaussian profile with FWHM equal to the resolution of the instrument (or Line Spread Function, or LSF) and then convolve that with every spectrum of the cube. Of course the LSF of cube A will be in wavelengths while the LSF of cube B in km/s. Here is my question now:

• Since all pixel shifts in cube A do not correspond to the same km/s value, one of the two LSFs won't be a perfect Gaussian, right? Shall I assume that would be the velocity LSF?

• What is the correct way of creating the velocity LSF? Do I just create a wavelength LSF and then rebin it to km/s, the same way I rebinned cube A to cube B?

• Does it worth the effort though? Will it make a difference? I have seen peer reviewed papers using velocity LSFs which are perfect Gaussians. Is this because the difference is negligible or my whole thinking/question is totally wrong?

Answer 1

Your logic seems entirely correct. Whatever you do is an approximation, the difference between a Gaussian profile in cube B and what it "should" look like will be small for any sensible resolution. By sensible, I mean if your wavelength/velocity axis covers many resolution elements.

However, there is a "correct" way. Generate a Gaussian in wavelength space $f(\lambda)$ (if you think that's what your LSD is, where $\lambda$ here is short hand for $\lambda - \lambda_0$, the separation from the central wavelength) and then convert to velocity space in a flux-conserving manner: $$f(\lambda)\ d\lambda = g(\log \lambda)\ d\log \lambda$$ $$ g(\log \lambda) = \lambda f(\lambda)$$

Maybe a better way to do this is to get some data, like an arc calibration and see how unresolved lines look in your observational space?