I Determine the temperature of a star via its spectrum

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1. Jul 12, 2016

Phys12

In the following video (, diagram also given for reference), the professor says that if the graph peaks in a short wavelength, then the star is a hot star (or galaxy) and if it does so in a long wavelength, then it's a cold one. However, I fail to understand this. How does it happen that if a star emits more energy in shorter wavelengths, then it's a hot star and a cold one if it does so in longer wavelengths?

Thank you!

2. Jul 12, 2016

Fightfish

This has to do with the property of the black-body spectrum. As a first idealisation, most stars can be well-approximated as behaving as blackbodies, and so the spectral distribution of the emitted radiation follows what is known as the black-body spectrum, which has a general shape resembling what you've drawn there: the power increases quite quickly to a peak value and then exponentially falls off as the wavelength increases.

More explicitly, it follows the celebrated Planck's law:
$$E_{\lambda}(\lambda,T) \propto {2 h c^2\over \lambda^5}{1\over e^{h c/\lambda k_{B}T}-1}$$
With some calculus, you can work out the wavelength at which the peak occurs, and this result, $\lambda_{peak} = \frac{2.898\times 10^{-3}}{T}$, is known as Wien's displacement law.

This is why the peak wavelength of the emission is in fact inversely proportional to the temperature of the star.

3. Jul 12, 2016

Phys12

Brilliant! Thank you.

Although I don't know much about the Plank's law, the Wien's displacement law helped. :)