<p>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.</p>

<p>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 <em>depends on $\lambda$, which changes across the spectrum.</em></p>

<p>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 $\lambda$.</p>

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

CC BY-SA 3.0
2016-04-13T05:43:53.477Z
2016-04-13T05:43:53.477Z
ProfRob
<p>In Tonry &amp; 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?</p>

<p>Here is the relevant paragraph from the paper:</p>

<p><a href="https://i.sstatic.net/bvY80.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bvY80.png" alt="enter image description here"></a></p>

CC BY-SA 3.0
2017-06-13T09:02:12.873Z
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
<redshift><spectroscopy>
Why do linear velocity redshifts correspond to linear pixel shifts when the spectra are binned in constant log wavelength?
Tim
# Why do linear velocity redshifts correspond to linear pixel shifts when the spectra are binned in constant log wavelength?

(Asked by: username_0 on 2016-04-12)

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](https://i.sstatic.net/bvY80.png)](https://i.sstatic.net/bvY80.png)

**Tags:** redshift, spectroscopy
> **Comments:**
> **username_1**: 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  

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## 1 answers

### Answer by username_2 on 2016-04-13. Score: 6 | Accepted answer

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.


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