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 $$2 \pi c^2h$$ 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.