wien displacement law
WienDisplacementLaw
2004-11-20 00:53:11
2004-11-20 02:27:41
Definition
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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.
So for a given T we have
\begin{equation}
f(\lambda) = \frac{2 \pi c^2 h}{\lambda^5} \, \frac{1}{e^{hc/ \lambda kT} - 1}
\end{equation}
To find the peak of this function differentiate with respect to $\lambda$ and set it equal to 0
\begin{equation}
\frac{df(\lambda)}{d\lambda} = 0
\end{equation}
Use the product rule to carry out this differentiation
\begin{equation}
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}
\end{equation}
Next use the chain rule to get
\begin{equation}
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)
\end{equation}
Apply the chain rule again
\begin{equation}
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})
\end{equation}
Multiply both sides by $\lambda^6 (e^{hc/\lambda kT} - 1)$
\begin{equation}
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)}
\end{equation}
Pull the e term into the denominator and divide out $2 \pi c^2h$ to get
\begin{equation}
\frac{ch}{\lambda kT(1 - e^{-hc/\lambda kT})} - 5 = 0
\end{equation}
This leaves us with a transendental function, which must be solved numerically
Set $\alpha = \frac{ch}{\lambda kT}$ and substitute into above
\begin{equation}
\frac{\alpha}{(1 - e^{-\alpha})} - 5 = 0
\end{equation}
After solving this equation for $\alpha$, the result yields Wien's Law
\begin{equation}
\alpha = \frac{ch}{\lambda kT}
\end{equation}
rearranging
\begin{equation}
\lambda = \frac{hc}{\alpha k} \, \frac{1}{T}
\end{equation}
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
\begin{equation}
F(\alpha) = \alpha - 5 + 5e^{-\alpha}
\end{equation}
We also need the first derivative of this so
\begin{equation}
\frac{dF(\alpha)}{d\alpha} = 1 - 5e^{-\alpha}
\end{equation}
Then through iteration we can converge on the solution
\begin{equation}
\alpha_{i+1} = \alpha_i - \frac{F(\alpha_i)}{dF(\alpha_i)}
\end{equation}
For our accuracy needs we choose $1\mathsf{x}10^{-8}$ so we stop iterating when
\begin{equation}
|\alpha_{i+1} - \alpha_i| < 1\mathsf{x}10^{-8}
\end{equation}
In matlab you can run WienConstant.m and you will get a value for $\alpha$. So we see
\begin{equation}
\alpha = 4.9651142
\end{equation}
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.
\begin{equation}
\lambda = \frac{2.897 \mathsf{x} 10^{-3} \, [Km]}{T}
\end{equation}
Note that the temperature must be in Kelvin [K] and then $\lambda$ will have units of meters [m].