Stars seem to be brown, red, yellow, blue & white.<br />In the light of this how is red shift determined i.e. how is the original colour found. Is it something to do with the spectral line spacing?
yes. the shift in spectral lines tells you how redshifted or blue shifted a source is. The source's color is irrelevant, and it is a misnomer to think that redshifting means everything turns red in color. Red wavelengths shift to infrared, yellow shifts to red, green to yellow, blue to green, and ultraviolet shifts to blue. There is thus little resulting apparent change in the appearance of light sources that are redshifted.
<font color="yellow">"Is it something to do with the spectral line spacing?"</font><br /><br />Yes. The spectrum of a star includes lines produced by the common elements of that star. The spectral pattern of the elements is well known and can be picked out no matter how far they are shifted into the red. So it is the shift in the pattern that is measured, not the color of light from the star, or the star's temperature. <div class="Discussion_UserSignature"> </div>
So brown dwarfs or blue variables don't actually exist but are spectrally backtracked to give a local view reference.<br />For instance brown dwarfs may be invisible & blue variables yellow.
<blockquote><font class="small">In reply to:</font><hr /><p>So brown dwarfs or blue variables don't actually exist but are spectrally backtracked to give a local view reference. <br />For instance brown dwarfs may be invisible & blue variables yellow. <p><hr /></p></p></blockquote><br /><br />No. We've found brown dwarfs very close to us, as little as 8 ly away, and we may have a brown dwarf orbiting the Sun outside the Oort Cloud. The class of the star has absolutely nothing to do with red shift. Redshift is a measure of the velocity at which an object is receding from or advancing toward us.
But has a brown dwarf ever been <i>seen</i> or is its presence inferred from its gravitational effect on other bodies? <div class="Discussion_UserSignature"> </div>