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

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

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

$\displaystyle \frac{df(\lambda)}{d\lambda} = 0$ (2)

Use the product rule to carry out this differentiation

$\displaystyle 0 = \frac{-10 \pi c^2 h}{\lambda^6} \, \frac{1}{e^{hc/\lambda kT}...
...+ (\frac{2 \pi c^2 h}{\lambda^5})\frac{d}{d\lambda}(e^{hc/\lambda kT} - 1)^{-1}$ (3)

Next use the chain rule to get

$\displaystyle 0 = \frac{1}{\lambda^6} \, \frac{-10 \pi c^2 h}{e^{hc/\lambda kT}...
...\, (-(e^{hc/\lambda kT} - 1)^{-2}) \, \frac{d}{d\lambda}(e^{hc/\lambda kT} - 1)$ (4)

Apply the chain rule again

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

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

$\displaystyle 0 = -10 \pi c^2 h + (\frac{2 \pi c^3 h^2}{\lambda kT}) \, \frac{e^{hc/\lambda kT}}{(e^{hc/\lambda kT} - 1)}$ (6)

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

$\displaystyle \frac{ch}{\lambda kT(1 - e^{-hc/\lambda kT})} - 5 = 0$ (7)

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

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

$\displaystyle \frac{\alpha}{(1 - e^{-\alpha})} - 5 = 0$ (8)

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

$\displaystyle \alpha = \frac{ch}{\lambda kT}$ (9)

rearranging

$\displaystyle \lambda = \frac{hc}{\alpha k} \, \frac{1}{T}$ (10)

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

$\displaystyle F(\alpha) = \alpha - 5 + 5e^{-\alpha}$ (11)

We also need the first derivative of this so

$\displaystyle \frac{dF(\alpha)}{d\alpha} = 1 - 5e^{-\alpha}$ (12)

Then through iteration we can converge on the solution

$\displaystyle \alpha_{i+1} = \alpha_i - \frac{F(\alpha_i)}{dF(\alpha_i)}$ (13)

For our accuracy needs we choose $ 1\mathsf{x}10^{-8}$ so we stop iterating when

$\displaystyle \vert\alpha_{i+1} - \alpha_i\vert < 1\mathsf{x}10^{-8}$ (14)

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

$\displaystyle \alpha = 4.9651142$ (15)

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.

$\displaystyle \lambda = \frac{2.897 \mathsf{x} 10^{-3} \, [Km]}{T}$ (16)

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.



Contributors to this entry (in most recent order):

As of this snapshot date, this entry was owned by bloftin.