PlanetPhysics/Wien Displacement Law

The Wien Displacement Law can be used to find the peak wavelength of a blackbody at a given [[../BoltzmannConstant/|temperature]]. [[../PlancksRadiationLaw/|Planck's radiation law]] gives us a [[../Bijective/|function]] of $\lambda$ and temperature so we can find the maximum of this function and hence the peak wavelength emitted [1].

So for a given T we have

$f(\lambda )=\frac{2\pi c^{2}h}{{\lambda }^5}\frac{1}{e^{hc/\lambda kT}-1}$

To find the peak of this function differentiate with respect to $\lambda$ and set it equal to 0

$\frac{df(\lambda )}{d\lambda }=0$

Use the product rule to carry out this differentiation

$0=\frac{-10\pi c^{2}h}{{\lambda }^6}\frac{1}{e^{hc/\lambda kT}-1}+(\frac{2\pi c^{2}h}{{\lambda }^5})\frac{d}{d\lambda }(e^{hc/\lambda kT}-1{)}^{-1}$

Next use the chain rule to get

$0=\frac{1}{{\lambda }^6}\frac{-10\pi c^{2}h}{e^{hc/\lambda kT}-1}+(\frac{2\pi c^{2}h}{{\lambda }^5})(-(e^{hc/\lambda kT}-1{)}^{-2})\frac{d}{d\lambda }(e^{hc/\lambda kT}-1)$

Apply the chain rule again

$0=\frac{1}{{\lambda }^6}\frac{-10\pi c^{2}h}{e^{hc/\lambda kT}-1}+(\frac{2\pi c^{2}h}{{\lambda }^5})(-(e^{hc/\lambda kT}-1{)}^{-2})(-\frac{hc}{{\lambda }^{2}kT}{e}^{hc/\lambda kT})$

Multiply both sides by ${\lambda }^6(e^{hc/\lambda kT}-1)$

$0=-10\pi c^{2}h+(\frac{2\pi c^3{h}^2}{\lambda kT})\frac{e^{hc/\lambda kT}}{(e^{hc/\lambda kT}-1)}$

Pull the e term into the denominator and divide out $2\pi c^{2}h$ to get

$\frac{ch}{\lambda kT(1-e^{-hc/\lambda kT})}-5=0$

This leaves us with a transendental function, which must be solved numerically

Set $\alpha =\frac{ch}{\lambda kT}$ and substitute into above

$\frac{\alpha }{(1-e^{-\alpha })}-5=0$

After solving this equation for $\alpha$, the result yields Wien's Law

$\alpha =\frac{ch}{\lambda kT}$

rearranging

$\lambda =\frac{hc}{\alpha k}\frac{1}{T}$

A simple way to find $\alpha$ is to use Newton's Method. This can be done by hand or with your favorite numerical [[../Program3/|program]]. Some matlab routines have been attached to see how to get $\alpha$.

To use Newton's Method we need we rewrite and arrange (8) to get

$F(\alpha )=\alpha -5+5e^{-\alpha }$

We also need the first derivative of this so

$\frac{dF(\alpha )}{d\alpha }=1-5e^{-\alpha }$

Then through iteration we can converge on the solution

${\alpha }_{i+1}={\alpha }_i-\frac{F({\alpha }_i)}{dF({\alpha }_i)}$

For our accuracy needs we choose $1𝗑10^{-8}$ so we stop iterating when

$|{\alpha }_{i+1}-{\alpha }_i|<1𝗑10^{-8}$

In matlab you can run WienConstant.m which depends on fWien.m and dfWien.m and will get a value for $\alpha$. So we see

$\alpha =4.9651142$

Plugging this value into (10) and evaluating the other constants yields the Wien Displacement Law, which gives the peak wavelength for a given temperature of a blackbody.

$\lambda =\frac{2.897𝗑10^{-3}[Km]}{T}$

Note that the temperature must be in Kelvin [K] and then $\lambda$ will have units of meters [m]. At different temperatures a blackbody's peak wavelength is displaced, hence the name Wien's Displacement Law.

[1] Krane, K., "Modern Physics." Second Edition. New York, John Wiley \& Sons, 1996.