# Wiens Law confuses me

1. Jan 25, 2014

### PsychonautQQ

My physics textbook is talking about blackbodies and has a graph showing Power radiated as a function or wavelength on the y axis and wavelength on the x axis. The slope grows to it's maximum value exponentially and then decays exponentially. They clarify that the peak of the graph is the wavelength that provides the maximum power radiated, and go on to talk about Wien's law.

My question is why is the most power not radiated at the shortest wavelength? E = hc/λ? Why is it dependent on the temperature?

2. Jan 25, 2014

### vanhees71

This comes from Planck's Law, i.e., the quantization of the electromagnetic field.

The shortcut derivation from semiclassical arguments, following mostly Planck's original derivation, goes as follows. First you start counting the number of modes in a frequency interval $\mathrm{d} \omega$ of electromagnetic waves. To this end think of a cubic volume $L^3$. Only waves with wave numbers $\vec{k} \in \pi/L \mathbb{N}^3$ "fit" into the volume. The corresponding modes are given by
$$\vec{E} \propto \sin(k_1 x) \sin(k_2 y) \sin(k_3 z),$$
fullfilling the boundary condition that $\vec{E}=0$ at the boundary of the volume. The number of modes in a frequency interval of width $\mathrm{d} \omega$ around $\omega=c k$ thus is
$$\mathrm{d} N=\mathrm{d}^3 \vec{k} 2 \frac{L^3}{\pi^3} = \mathrm{d} k k^2 4 \pi 2 \frac{L^3}{(2 \pi)^3}.$$
The factor 2 comes from the two polarization states of the em. waves. In the last step we have written the volume elment in spherical coordinates and integrated out the angles. We have to multiply by $1/8$ because only the octand with positive wave numbers has to be taken. Written in terms of frequency we finally get
$$\mathrm{d} N=\mathrm{d} \omega 8 \pi \frac{L^3}{(2 \pi c)^3}.$$
To get the energy spectrum we need the average energy in each mode at given temperature. According to classical physics each mode represents an oscillator, and its mean energy is thus $k T$ according to the equipartition law. This leads, however, to a wrong energy spectrum, particularly a divergent total energy in the electromagnetic field (Rayleigh-Jeans UV catastrophy), and only then your "intuitive" idea that the highest frequencies give the most energy in the radiation field.

Now, according to Planck's and Einstein's hypothesis on the quantization of radiation energy, each oscillator can take only discrete energy levels, $E_n=n \hbar \omega$ with $n \in \mathbb{N}_0$. That means that the mean energy is
$$\frac{\sum_{n=0}^{\infty} n \hbar \omega \exp(-n \hbar \omega/(k T))}{\sum_{n=0}^{\infty} \exp(-n \hbar \omega/(k T))}=\frac{\hbar \omega}{\exp(\hbar \omega/(kT))-1}.$$
The energy spectrum is thus given by
$$\mathrm{d} \epsilon=\frac{8 \pi \hbar \omega^3}{(2 \pi c)^3 [\exp(\hbar \omega/(kT))-1]} \mathrm{d} \omega,$$
where $\epsilon=E/V=E/L^3$ denotes the energy density. This can be rewritten in terms of the wavelength, and finding the value for the maximum of the corresponding spectral function leads to Wien's displacement law: The wavelength of maximal emission fulfills the rule
$$\lambda_{\text{max}} T=\text{const}.$$

3. Jan 25, 2014

### Jano L.

Could you give a reference? I do not think what you wrote is Planck's original derivation. This is more like Debye or Drude if I remember correctly. Planck quantized material harmonic oscillators, not EM modes.

4. Jan 25, 2014

### j824h

E in E=hc/λ you mentioned, seems to be the energy of a single photon with wavelength λ. What you're dealing with in Wien's law is the energy distribution for whole blackbody radiation - the number of photon is never distributed uniformly over the wavelength in blackbody radiation.

That explains why you cannot apply E=hc/λ to get the wrong Wien's law that would state λmax=0.

5. Jan 26, 2014

### vanhees71

That's true. The derivation I've given, is a pretty hand-waving way using a photon picture (free em. field as a set of harmonic oscillators; I'm not sure who did this first, perhaps Einstein?) Plancks derivation is pretty involved. In a way it's using Bose-Einstein statistics but of course that's not very explicit. I only know the German Annalen article, which can be downloaded for free here:

http://onlinelibrary.wiley.com/doi/10.1002/andp.19013090310/abstract

6. Jan 26, 2014

### SteamKing

Staff Emeritus