<p>The spectrum is recorded on a detector. The 30nm refers to a physical displacement on the detector.</p>
<p>I e. If you measured the centroid of a spectral line and then applied a 2 m/s shift, the line centroid would move, physically, by 30 nm on the detector.</p>
<p>Your calculation gives the wavelength shift (also in nm) of a spectral line. Combining these calculations tells us that there are roughly 0.084 nm of spectrum per mm on the detector.</p>
<p>Most CCDs have pixels of size <span class="math-container">$\sim 15$</span> <span class="math-container">$\mu$</span>m, so each such pixel would cover 0.00126 nm of wavelength, which is indeed, very high resolution.</p>

CC BY-SA 4.0
2022-03-24T23:10:37.930Z
My understanding is that the $ \Delta_s $ calculated above is the physical shift of spectral lines on detector, which is way off from the document's value.
Nirmala
CC BY-SA 4.0
2022-03-24T23:28:46.157Z
@Nirmala no,it isn't  You have calculated the shift in wavelength of spectral line. Where in your calculation did you insert the properties of the camera (e.g. the focal length)?
ProfRob
CC BY-SA 4.0
2022-03-24T23:45:06.130Z
Got it. Thank you for the explanation.
Nirmala
CC BY-SA 4.0
2022-03-25T08:32:03.317Z
@Nirmala if this answer helped you, make sure to accept it by clicking the checkmark.
Nilay Ghosh
CC BY-SA 4.0
2022-03-24T22:42:07.520Z
2022-03-25T07:53:49.270Z
2022-03-25T07:53:49.270Z
ProfRob
<p>I am engineer new to Astronomy and am trying to understand spec of an spectrograph (RV stability = 2m/s). There is a note in document saying &quot;RV shift of 2 m/s is equivalent to a shift of the spectrum on the detector of about 30 nm&quot;. The measurement range is 380nm-850nm.</p>
<p>My understanding so far is as follow: Stars moving away/towards earth causes shift in known spectrum lines towards red or blue depending on direction of movement. The magnitude of shift can be derived from Doppler effect as:</p>
<p><span class="math-container">$\Delta_\lambda = \lambda_0 \frac{V_r}{c}$</span></p>
<p>where <span class="math-container">$\Delta_\lambda$</span> is the shift in wavelength, <span class="math-container">$\lambda_0$</span> is the rest wavelength, and <span class="math-container">$V_r$</span> is the radial velocity</p>
<p>or <span class="math-container">$V_r = c\frac{\Delta_\lambda }{\lambda_{0}}$</span></p>
<p>RV stability would be difference between radial velocity of consecutive measurements.</p>
<p>Taking an example here:</p>
<p><span class="math-container">$V_{r1} = \frac{\Delta\lambda_1 c}{\lambda_0}$</span>
<span class="math-container">$V_{r2} = \frac{\Delta\lambda_2 c}{\lambda_0}$</span></p>
<p>where <span class="math-container">$V_{r1}$</span> and <span class="math-container">$V_{r2}$</span> are the two radial measurements, and <span class="math-container">$\Delta\lambda_1$</span> and <span class="math-container">$\Delta\lambda_2$</span> are the two wavelength shifts.</p>
<p><span class="math-container">$\Delta V_{r} = \frac{(\Delta\lambda_1 - \Delta\lambda_2 )*c}{\lambda_0}$</span></p>
<p>where <span class="math-container">$\Delta V_{r} $</span> is Radial velocity stability between measurements.</p>
<p><span class="math-container">$\Delta_s  = \frac{\Delta V_r \lambda_0}{c}$</span></p>
<p>Where <span class="math-container">$\Delta_s$</span> is <span class="math-container">$\Delta\lambda_1- \Delta\lambda_2$</span> = shift on ccd due to measurement</p>
<p>Plugging numbers from note.</p>
<p>Shift of spectrum on detector  = <span class="math-container">$\frac{2 \cdot 380nm}{3\cdot 10^8} \neq  30nm$</span></p>
<p>I am definitely doing something wrong but do not know what exactly.</p>

CC BY-SA 4.0
2022-03-24T22:19:15.247Z
@fasterthanlight Thanks a lot for making it legible.
Nirmala
<spectroscopy><redshift><radial-velocity><doppler-effect>
Radial velocity stability to spectrum shift calculation
Nirmala
# Radial velocity stability to spectrum shift calculation

(Asked by: username_0 on 2022-03-24)

I am engineer new to Astronomy and am trying to understand spec of an spectrograph (RV stability = 2m/s). There is a note in document saying "RV shift of 2 m/s is equivalent to a shift of the spectrum on the detector of about 30 nm". The measurement range is 380nm-850nm.

My understanding so far is as follow: Stars moving away/towards earth causes shift in known spectrum lines towards red or blue depending on direction of movement. The magnitude of shift can be derived from Doppler effect as:

$\Delta\_\lambda = \lambda\_0 \frac{V\_r}{c}$

where $\Delta\_\lambda$ is the shift in wavelength, $\lambda\_0$ is the rest wavelength, and $V\_r$ is the radial velocity

or $V\_r = c\frac{\Delta\_\lambda }{\lambda\_{0}}$

RV stability would be difference between radial velocity of consecutive measurements.

Taking an example here:

$V\_{r1} = \frac{\Delta\lambda\_1 c}{\lambda\_0}$
$V\_{r2} = \frac{\Delta\lambda\_2 c}{\lambda\_0}$

where $V\_{r1}$ and $V\_{r2}$ are the two radial measurements, and $\Delta\lambda\_1$ and $\Delta\lambda\_2$ are the two wavelength shifts.

$\Delta V\_{r} = \frac{(\Delta\lambda\_1 - \Delta\lambda\_2 )\*c}{\lambda\_0}$

where $\Delta V\_{r} $ is Radial velocity stability between measurements.

$\Delta\_s = \frac{\Delta V\_r \lambda\_0}{c}$

Where $\Delta\_s$ is $\Delta\lambda\_1- \Delta\lambda\_2$ = shift on ccd due to measurement

Plugging numbers from note.

Shift of spectrum on detector = $\frac{2 \cdot 380nm}{3\cdot 10^8} \neq 30nm$

I am definitely doing something wrong but do not know what exactly.

**Tags:** spectroscopy, redshift, radial-velocity, doppler-effect
> **Comments:**
> **username_0**: @fasterthanlight Thanks a lot for making it legible.  

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## 1 answers

### Answer by username_1 on 2022-03-24. Score: 1 | Accepted answer

The spectrum is recorded on a detector. The 30nm refers to a physical displacement on the detector.

I e. If you measured the centroid of a spectral line and then applied a 2 m/s shift, the line centroid would move, physically, by 30 nm on the detector.

Your calculation gives the wavelength shift (also in nm) of a spectral line. Combining these calculations tells us that there are roughly 0.084 nm of spectrum per mm on the detector.

Most CCDs have pixels of size $\sim 15$ $\mu$m, so each such pixel would cover 0.00126 nm of wavelength, which is indeed, very high resolution.

> **Comments:**
> **username_0**: My understanding is that the $ \Delta_s $ calculated above is the physical shift of spectral lines on detector, which is way off from the document's value.  
> **username_1**: @Nirmala no,it isn't  You have calculated the shift in wavelength of spectral line. Where in your calculation did you insert the properties of the camera (e.g. the focal length)?  
> **username_0**: Got it. Thank you for the explanation.  
> **username_2**: @Nirmala if this answer helped you, make sure to accept it by clicking the checkmark.  

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