# What limits the use of the H-R diagram to measure distance (main sequence fitting), what distances is it useful for?

Is it only possible to measure objects that form around the same time? Is it possible to measure clusters from distant galaxies other than our own?

• I've just answered this on Physics SE too. Delete one or the other. – Rob Jeffries May 1 '19 at 16:37

Here is a diagram from Martignoni et al. (2014) showing the ZAMS and TAMS for stars of different masses. there is typically a factor 2-3 in luminosity between them (a larger difference at larger masses). That means whether you use the ZAMS, the TAMS, or something in between to determine the distance from a vertical displacement in the HR diagram, you could vary your answer for the distance by $$\sqrt{2}$$ to $$\sqrt{3}$$. In other words you need to know the age of a main sequence star before "main-sequence fitting" can give you an accurate distance.
If you were to try to use main-sequence fitting to estimate the distance to individual stars then there are several hazards. For one, it can be nearly impossible to estimate the age of an individual star. Therefore if it has a mass greater than 0.7 solar masses then there will be an uncertainty in its position in the absolute HR diagram that leads to an inevitable uncertainty in estimated distance. Further, the intrinsic position in the HR diagram depends on the star's chemical composition. Such ancillary information might be available, but it might not, in which case that is another source of error. A further source of systematic uncertainty is stellar rotation. Fast rotating stars have extended lifetimes on the main sequence and somewhat different intrinsic positions on the HR diagram; again a source of systematic uncertainty that is especially problematic for high mass stars. Finally, it can be that what you think is an isolated main sequence star is in fact a binary system. A companion can increase the luminosity of the system and make a star appear closer than it actually is (by up to a factor $$\sqrt{2}$$).