# Why do linear velocity redshifts correspond to linear pixel shifts when the spectra are binned in constant log wavelength?

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:

• 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
– Alex
Jun 13, 2017 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.