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It is well known that the surface temperature of the sun can be determined by fitting the solar spectrum to the black body radiation spectrum.

Is this scheme feasible for other stars?

Possibly the problem is obtaining the spectrum precisely.

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  • $\begingroup$ Why would spectral analysis of other stars be more difficult than the sun? $\endgroup$ – Lewis Miller Feb 11 '16 at 2:42
  • $\begingroup$ Because it is too faint? $\endgroup$ – John Feb 11 '16 at 2:55
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    $\begingroup$ The spectrum does not depend on the intensity as much as one would think. You only need to know the wavelength dispersion for which you don't need a large number of photons coming in through your telescope/ device. $\endgroup$ – user106422 Feb 11 '16 at 3:00
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Yes of course. There are techniques which allow astronomers and astrophysicists to calculate these quantities quite efficiently and correctly and it depends on the spectrum, the luminosity and the distance of the star. In fact, from just the luminosity and the distance, one can compute almost everything important about the star including its age, composition, temperature, distance, mass and how much longer it will continue to live. A very useful book to study this at a relatively introductory level is the book by Prialnik (http://www.amazon.com/Introduction-Theory-Stellar-Structure-Evolution/dp/0521866049). A good advanced book is the book by Chandrasekhar (http://www.amazon.com/Introduction-Study-Stellar-Structure-Astronomy/dp/0486604136).

However, I don't see where there is a spectrum problem because once you know the distance to the star and measure it's spectrum, you would need to account for the correct red-shift or blue shift and that's it.

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  • $\begingroup$ The redshift is typically determined by finding intensity spikes from known atomic transitions and comparing with the spikes' wavelengths at zero relative velocity. $\endgroup$ – Carl Witthoft Feb 11 '16 at 14:51
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The question operates under a false premise. It is not the case that fitting a blackbody spectrum to the Sun gives you its "surface temperature". The Sun does not have a black body spectrum, although sometimes that approximation is made. A star also does not have a single "surface temperature".

What you can do is divide the luminosity of the Sun by $4\pi R^2 \sigma$, and this gives you $T_{\rm eff}^4$, where $T_{\rm eff}$ is known as the "effective temperature" and this is often what is being referred to when talking about the "surface temperature".

The effective temperature of a star is rarely deduced in this way because for most stars, we do not know what their radius is - this is actually the main difficulty in deducing the $T_{\rm eff}$ of other stars. This is why the idea of fitting a blackbody function might be thought attractive. Unfortunately, the photons in the real spectrum of a star arises at different depths and different temperatures. The image below shows the real spectrum of the Sun compared with a blackbody at the $T_{\rm eff}$ implied by the Sun's luminosity and radius. Though you will sometimes see it said that this is the "best fitting blackbody spectrum", it isn't, although Wien's law would give you roughly the right answer.

Solar spectrum vs blackbody In fact for most stars the $T_{\rm eff}$ is deduced by a detailed analysis of lots of spectral line features caused by the absorption due to various chemical elements in the atmosphere. For instance the ratio of line strengths due to iron atoms and singly ionised iron ions is diagnostic of the temperature.

Some other types of star are better fitted by a blackbody spectrum - very hot white dwarfs and perhaps hot neutron stars. But even there it is rarely the case that a blackbody spectrum is accurate enough to yield the correct temperature from a simple fit to the whole spectrum.

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  • $\begingroup$ Edit was just a typo fix. $\endgroup$ – Kyle Oman May 23 '16 at 8:41

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