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Measuring gravitational field of stars


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2 hours ago, Physics_dude said:

How exactly can we measure the gravitational field of stars? I know they get redshifted, but how do we tell velocity redshift from gravity redshift? 

This is equivalent to asking for the mass. It is not possible to do this directly for an isolated star. You can get an estimate from its spectral class as this is related to (amongst other things) mass.

Regards Andrew 

Edited by andrew s
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ok thanks, I'm new to this stuff and learning a lot the last few days.

My understanding of current methods:

Any star with a clearly measurable spectrum can have it's spectral class deduced and mass / gravity approximated by fitting to a model. It might be inaccurate or wrong in certain situations, but meh it's an easy technique for an approximate answer.

- close binaries: We can use parallax + angular separation to get orbital distance and redshift to get orbital speed and hence period. From Kepler's law we then deduce the mass and then calculate gravity.

- Close lonely stars: Parallax + angular movement to deduce proper motion & tangential speed. Measurement of redshift to get velocity redshift + gavity redshift. hmmm, we can nearly solve this to get gravity field at surface. Unfortunately there's many radial velocity/gravity combinations that would give same redshift. dammit, so close! So astronomy scientists currently can't solve this case unless fitting to spectral models?

- more distant lonely stars: We know redshift, spectrum profile and intensity. Sounds tricky to accurately determine it's distance? Tricky to separate velocity redshift from gravity redshift? Fitting spectral models are the only way?

Any mistakes / major techniques I've missed? 

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In general relativity red shift is only separable into components in simple cases. You cannot in general say this bit is due to gravity, this bit is due to velocity or cosmic expansion.

The surface gravity of a star is a function of its size and mass. Giant stars have a lower surface gravity than their dwaf counterparts of the same spectral class. This shows up in the shape of the spectral lines.

Regards Andrew 

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