Radial velocity stability to spectrum shift calculation

Question

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.

Answer 1

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.