When I measure RV in a program for more spectral lines, will it be equivalent? I mean I measure RVs of a line in blue spectral region, RVs of a line in red spectral region... Many thanks.
-
1$\begingroup$ The usual case would be to assume that they're not equal, then measure RV for as many lines as possible, that gives you a distribution which will have a mean and a sample variance. If that variance is small enough, for some measure of 'small', then you can say that there is only one RV. $\endgroup$– AtmosphericPrisonEscapeApr 3, 2020 at 18:42
-
$\begingroup$ Thank you. So, when I measured RV for He 6678 and I get a radial velocity curve (radial velocity with respect to phase) and it is worth measuring other lines? The phased RVs will be similar? $\endgroup$– AlexApr 4, 2020 at 5:46
1 Answer
It depends what lines you are measuring and in what kind of star.
When you measure the RV from a spectral line, you are measuring an intensity-weighted average RV over the region where the line is formed.
For a star like the Sun, the photospheric lines are all formed within a layer no thicker than about a 1000 km and the differences in RV with depth in the atmosphere are small (though not negligible if you are interested in finding low mass exoplanets using the RVs). On the other hand, if you were to start measuring RVs from lines originating in the chromosphere or Corona then you could find different RVs caused by mass motions of the emitting gas.
Similarly, hot stars are surrounded by fast moving winds with a temperature structure. Different lines can originate at different heights from gas with different velocity. This can lead to different RVs from different lines.
A better way to measure RV is by cross-correlation with a synthetic or observed template of a similar star with a known RV.
-
$\begingroup$ Thank you very much and when I measure RV by cross-correlation for red spectra (He 6678) it is worth measuring RV by cross-correlation for a line in blue region? $\endgroup$– AlexApr 4, 2020 at 9:12
-
1$\begingroup$ The result should be the same, if the template is perfect, but of course you improve your precision with more measurements. @Alex $\endgroup$– ProfRobApr 4, 2020 at 10:14
-