It is well known that the surface temperature of the Sun can be determined by fitting the solar spectrum with a black body radiation spectrum.
Is this scheme feasible for other stars?
Possibly the problem is obtaining the spectrum precisely.
[Physics] Can we determine the surface temperature of stars other than the Sun by using black body radiation theory
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Related Solutions
A real object will radiate less energy than a black body at the same temperature. The ratio of the emission to the black body emission is called the emissivity. The Engineering Toolbox web site has data on emissivities for a range of materials. For example a polished silver surface has an emissivity of 0.02 i.e. it radiates only 2% of the power that a black body of the same temperature radiates.
The emissivity is related to the reflectivity by E + R = 1. A black body reflects none of the light falling on it so R = 0 and E = 1. Anything that has a non-zero reflectivity must have an emissivity of less than 1 otherwise it could be used to build a perpetual motion machine.
The total radiative power emitted by the Sun is equivalent to the total radiative power emitted by an ideal black body with a temperature of 5778 K and a surface area equal to that of the Sun. This 5778 K is the Sun's effective temperature. The spectrum of the Sun is very close to that of a 5778 K black body, but there are deviations. Some are due to absorption and emission, but others result from three key items:
There is no such thing as black body. The concept of a black body is an idealization based on some simplifying assumptions. The Sun doesn't exactly satisfy those simplifying assumptions.
That effective temperature of 5778 K is based on total radiative power, the area under the curve of the Planck distribution. If the spectrum of sunlight falls short of the 5778 K black body spectrum some wavelengths it must necessarily rise above the 5778 K black body spectrum at others.
The primary reason the Sun fails to satisfy the assumptions that underly the Planck distribution is that we are seeing light from multiple temperature sources. The rest of this answer goes into this in detail
The Sun is not a solid body. It doesn't have a surface from which the radiation originates. The radiation we see from the Sun comes primarily from the Sun's photosphere, a roughly 500 kilometer thick layer near the top of the Sun. The chromosphere, transition region, and corona are above the photosphere. While these higher layers do make solar radiation deviate from the ideal black body curve, the primary source is the photosphere itself.
The amount of light that is transmitted into empty space is a sharply increasing function of distance from the center. However, it is not a delta distribution. The light that does get through from those deeper layers has a higher temperature than the layers above it. The bulk of the radiation we see from the Sun comes from a ~500 km thick layer called the photosphere. The top of the photosphere has a temperature of about 4400 K and has a pressure of about 86.8 pascals. The bottom has a temperature of about 6000 K and a pressure of about 12500 pascals.
What we see is a blend of the radiation from throughout the photosphere. Some of the light comes from the top of the photosphere, some from the middle, some from the bottom, roughly weighted by pressure. The total spectrum looks close to that of a 5778 K black body, but the contribution from the bottommost part of the photosphere tilts the spectrum away from the ideal a bit, making the a tiny bit heavy for shorter wavelength radiation.
Best Answer
The question operates under a false premise. It isn't the case that fitting a blackbody spectrum to the Sun gives you its "surface temperature". On any detailed inspection, the Sun does not have a black body spectrum, although sometimes that approximation is made. A star cannot be characterised with 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 by dividing the luminosity by $4\pi R^2\sigma$, because for most stars we do not know their radius - this is actually the main difficulty in deducing an accurate $T_{\rm eff}$ for other stars. This is why the idea of fitting a blackbody function might be thought attractive. You can get an approximate answer in this way, but unfortunately, the photons in the real spectrum of a star come from different depths and at different temperatures depending on the wavelength considered and depending at what angle the line of sight makes to the stellar surface. 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.
There are some types of star that 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.