# Deeper underlying explanation for color-shift in Wien's Law?

Suppose we have a blackbody object, maybe a star or a metal (although I understand neither of these are actually blackbody objects, to some extent my understanding is they can approximate one). According to Wien's Law, as temperature increases, peak wavelength will decrease, so the color we observe will "blueshift." Despite having reached Wien's Law in my research, my understanding is that this is in fact not an answer to why the blueshift occurs, but just a statement that the blueshift does occur.

So, is there some deeper reason why blackbodies peak wavelength emitted decreases with temperature?

## 4 Answers

Wien's displacement is qualitatively quite easy to understand.

Consider a black body with temperature $$T$$. Its atoms are moving around chaotically with an average kinetic energy of $$\bar{E}_\text{atom}\approx kT \tag{1}$$ where $$k$$ is Boltzmann's constant.

On the other hand, you have the black-body radiation. Because the radiation is in thermal equilibrium with the black body, the radiation has the same temperature $$T$$. This means the photons have also an average energy of $$\bar{E}_\text{photon}\approx kT$$

A single photon of frequency $$\nu$$ has the energy $$E_\text{photon}=h\nu$$ where $$h$$ is Planck's constant.

You can rewrite this in terms of the photon's wavelength $$\lambda$$ $$E_\text{photon}= \frac{hc}{\lambda} \tag{2}$$

By equating (1) and (2) you get $$kT\approx \frac{hc}{\lambda}$$ or $$\lambda \approx \frac{hc}{kT}$$ which (apart from a factor $$4.97$$) is Wien's displacement law.

A quantitative derivation is much more difficult because the atoms and photons don't have all the same energy, but instead their energies vary quite a lot around their average values.

• At the same time this kind of analysis exposes that the difference can more or less only be a bounded (but not necessarily parameter-independent) factor. Unless the distribution somehow undergoes some kind of qualitative change.
– Ian
Jul 29, 2020 at 16:37

As an object gets hotter, it radiates more energy. Since light with shorter wavelengths has more energy, it shouldn’t be too surprising that the object radiates more in the shorter wavelengths in order to radiate more energy.

• Thank you for an intuitive answer! Jul 28, 2020 at 7:53
• Well there is still the possibility that the wavelengths are temperature independent, and that only the intensity increases when the temperature increases. Jul 29, 2020 at 8:22

It's just thermal equilibrium. A typical populated energy state will be at $$E\sim k_B T$$, so when a transition is made, of order $$k_BT$$ of energy goes into a photon. Thus $$h\bar{\nu}\propto k_B T$$ and $$\bar{\lambda}\propto T^{-1}$$.

Both higher and lower energy states are less likely to be populated and therefore there is a peak in the photon energy distribution.

For a true blackbody, the average photon energy is about $$2.7 k_B T$$.

The emission of radiation occurs because the electrically charged constituents within the material are vibrating randomly, causing random, "noisy" waves to be set up in the surrounding electromagnetic field, as they disturb it with their movements - think about an oar put into a pond, and you shake that oar around, and it produces waves in the surrounding water. And the faster you shake the oar, the smaller and denser (higher frequency and shorter wavelength) the waves produced get. Likewise, if charges, as "oars" for the electromagnetic field, vibrate faster, the waves they produce will get denser and higher in frequency.

And in hotter objects, those charged constituents have more kinetic energy and are thus vibrating faster. Hence, the radiation emitted will be at generally higher frequencies, or "blue shift", as you call it.