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What is the spectral resolution, $\Delta z,$ when detecting damped Lyman alpha systems at different epochs, $z \sim 0-3,$ through current and/or future observational instruments? And is this redshift dependent? (e.g. the best observers can do in making precise spectroscopic redshift measurement)

My guess is $\sim 10^{-5} - 10^{-4}$ but I am not sure.

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Damped Lyman Alpha systems (DLAS) are often observed with quite high resolution spectrographs, with resolving powers of up to $\sim 50000$. i.e. $\lambda/\Delta \lambda \sim 50000$, where $\Delta \lambda$ is the wvelength resolution of the spectrograph (e.g. Noterdaeme et al. 2015).

The expression for redshift is $$ z = \frac{\lambda}{\lambda_0} - 1,$$ where $\lambda_0$ is the rest wavelength of a feature and $\lambda$ is the observed wavelength. Thus the resolution (in the sense of the finest redshift difference that can be resolved) would be $$\Delta z \simeq \frac{\Delta \lambda}{\lambda_0} = \frac{\Delta \lambda}{\lambda}(1+z) \simeq \frac{1+z}{R},$$ where $R$ is the resolving power of your spectrograph. This assumes that the DLA absorption features are narrow (which is the case for some of them).

So for $z \sim 1$, then basically it is the reciprocal of the spectrograph resolving power.

Note that this is not the same as asking to what precision can the redshift of a DLA be estimated. Providing the system is isolated the the centroid of a Gaussian absorption line can be located far more precisely than just the FWHM that defines the instrumental resolution. And of course there may be multiple absorption features that define the redshift of a single DLA. The answer then just depends on the resolving power, the signal-to-noise ratio and perhaps how accurately the spectrograph wavelength calibration is known.

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  • $\begingroup$ thanks! this means that (1+z)/R is the minimum (e.g. the best) given that everything else is just perfect. Right? I have seen quite similar precision in some observational papers for z<0.5. How does the actual precision change at high z given lower signal-to-noise ratio and the calibration? $\endgroup$ – Ash Jan 8 '20 at 0:13

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