# PlanetPhysics/Wien Displacement Law

The Wien Displacement Law can be used to find the peak wavelength of a blackbody at a given temperature. Planck's radiation law gives us a 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 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 $1{\mathsf {x}}10^{-8}$ so we stop iterating when
   $|\alpha _{i+1}-\alpha _{i}|<1{\mathsf {x}}10^{-8}$
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.
   $\lambda ={\frac {2.897{\mathsf {x}}10^{-3}\,[Km]}{T}}$
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.