In Tonry & Davis (1979), they describe spectroscopic redshift measurement via correlating with templates at known redshift. In Section IIIa, they say "Because the spectra are binned linearly with ln$\lambda$, a velocity redshift is a uniform linear shift. Why is this the case?

Here is the relevant paragraph from the paper:

enter image description here

  • 1
    $\begingroup$ Tonry & Davis (1979) used Fast Fourier transforms to implement the cross-correlation. Wondering if we implement cross-correlation directly, do we still need the log-wavelength binning? Alex $\endgroup$
    – Alex
    Jun 13 '17 at 9:02

All this means is that you need to bin your spectra in equal intervals of log wavelength for each pixel to be a constant interval in velocity.

First consider the case whereeachpixel is worth a constant interval in linear wavelength. Here we have $$ \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c},$$ and the $\Delta v$ represented by each pixel depends on $\lambda$, which changes across the spectrum.

But now if you bin in equal increments of $$\Delta \log \lambda = \frac{\Delta \lambda}{\lambda} = \frac{\Delta v}{c}$$ and each pixel has the same fixed velocity interval, $\Delta v = c \Delta \log\lambda$, independent of $\lambda$.

This log wavelength binning is a prerequisite for cross-correlation procedures that yield velocities.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.