Question

How does Planck's radiation law relate to temperature? What conclusions can be drawn from it?

Answer 1

Planck's radiation law expresses the radiant energy density `rho_(nu)` in terms of either the frequency `nu`:

`rho_(nu)(nu,T) = (2hnu^3)/(c^2) 1/(e^(hnu"/"k_BT) - 1)`

or in terms of the wavelength `lambda`:

`rho_(lambda)(lambda,T) = (2hc^2)/(lambda^5) 1/(e^(hc"/"lambdak_BT) - 1)`

For the frequency version, it is a cubic function of `nu` at small frequencies and an exponential decay at large frequencies. This means it increases quickly at first, and then drops exponentially, as a skewed Gaussian.

For the wavelength version, since `lambda prop 1/nu`, the graph of the radiant energy density has a similar shape whether graphed against the frequencies or the wavelength. The wavelength version of the graph can be seen below:

It can be seen that at higher and higher temperature `T`, the maximum of the graph is higher and further left (shorter wavelength). This means that overall, hotter blackbodies radiate more brightly, but also with progressively shorter wavelengths.

A useful relation called Wien's law is derived from this, which is:

`lambda_(max) = (2900 mu"m"cdot"K")/T`,

where `2900` `mu"m"cdot"K"` is Wien's displacement constant.

From this, if we knew the color of something we observe and approximate that as a blackbody, we could approximate the temperature.

For example, if we knew that the wavelength of the color radiated by the Beeteljeuse star was about `0.88` `mu"m"`, we could determine that its temperature is about `"3300 K"` (even without taking a thermometer up to it and trying such a stunt).