# Wiens displacement law

Wien's displacement law is a law of physics that states that there is an inverse relationship between the wavelength of the peak of the emission of a black body and its temperature.

$\displaystyle \lambda_\mbox{max} = \frac{0.002898\ldots}{T}$

where $\displaystyle T$ is the temperature of the blackbody in kelvins (K) and $\displaystyle \lambda_\mbox{max}$ is the peak wavelength in meters. The 0.002898... is a proportionality constant with units meter-kelvins (m·K). It is sometimes expressed in cgs units as 0.29 cm·K.

Basically, the hotter an object is, the shorter the wavelength at which it will emit radiation. For example, the surface temperature of the sun is 5780 K, giving a peak at 500 nm. As can be seen in the article Color, this is fairly in the middle of the visual spectrum, due to the spread resulting in white light. Due to the Rayleigh scattering of blue light by the atmosphere this white light is separated somewhat, resulting in a blue sky and a yellow sun.

A lightbulb has a glowing wire with a somewhat lower temperature, resulting in yellow light, and something that is "red hot" is again a little less hot.

Although the law was first formulated by Wilhelm Wien, we now derive it from Planck's law of black body radiation.

This can be turned into an equation for the frequency of the radiation, using the speed of light to get

$\displaystyle f_{max} \approx 10^{11} T.$

## Derivation

From Planck's law of black body radiation we know that

$\displaystyle u(\lambda) = {8\pi h c\over \lambda^5}{1\over e^{h c/\lambda kT}-1}$

The value of $\displaystyle \lambda$ for which this function is maximized is sought. To find it, we differentiate $\displaystyle u(\lambda)$ with respect to $\displaystyle \lambda$ and set it equal to zero

$\displaystyle \partial_{\lambda}u(\lambda) = 8\pi h c\left( {hc\over kT \lambda^7}{e^{h c/\lambda kT}\over \left(e^{h c/\lambda kT}-1\right)^2} - {1\over\lambda^6}{5\over e^{h c/\lambda kT}-1}\right)=0$
$\displaystyle {hc\over\lambda kT }{1\over 1-e^{-h c/\lambda kT}}-5=0$

if we define $\displaystyle x\equiv{hc\over\lambda kT }$ , then

$\displaystyle {x\over 1-e^{-x}}-5=0$

This equation cannot be solved in terms of elementary functions. It can be solved in terms of Lambert's Product Log function but an exact solution is not important in this derivation. One can easily find the numerical value of $\displaystyle x$

$\displaystyle x = 4.965114231744276\ldots$

Solving for $\displaystyle \lambda$ in units of meters, and using units of kelvins for the temperature yields:

$\displaystyle \lambda = {hc\over kx }{1\over T} = {0.00289776829\ldots\over T}$

and

$\displaystyle \nu = {c\over\lambda} = {x k\over h}T = 1.03456325\ldots \times 10^{11} T.$