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

##### 1 Answer
May 15, 2017

Planck's radiation law expresses the radiant energy density ${\rho}_{\nu}$ in terms of either the frequency $\nu$:

${\rho}_{\nu} \left(\nu , T\right) = \frac{2 h {\nu}^{3}}{{c}^{2}} \frac{1}{{e}^{h \nu \text{/} {k}_{B} T} - 1}$

or in terms of the wavelength $\lambda$:

${\rho}_{\lambda} \left(\lambda , T\right) = \frac{2 h {c}^{2}}{{\lambda}^{5}} \frac{1}{{e}^{h c \text{/} \lambda {k}_{B} T} - 1}$

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 $\lambda \propto \frac{1}{\nu}$, 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:

${\lambda}_{\max} = \frac{2900 \mu \text{m"cdot"K}}{T}$,

where $2900$ $\mu \text{m"cdot"K}$ 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 $0.88$ $\mu \text{m}$, we could determine that its temperature is about $\text{3300 K}$ (even without taking a thermometer up to it and trying such a stunt).