othereyeshmt4l

2022-05-15

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?

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?

lutzantsca885

Beginner2022-05-16Added 15 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

$\begin{array}{}\text{(1)}& {\overline{E}}_{\text{atom}}\approx kT\end{array}$

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

${\overline{E}}_{\text{photon}}\approx kT$

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 $

$\begin{array}{}\text{(2)}& {E}_{\text{photon}}=\frac{hc}{\lambda}\end{array}$

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.

Consider a black body with temperature $T$. Its atoms are moving around chaotically with an average kinetic energy of

$\begin{array}{}\text{(1)}& {\overline{E}}_{\text{atom}}\approx kT\end{array}$

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

${\overline{E}}_{\text{photon}}\approx kT$

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 $

$\begin{array}{}\text{(2)}& {E}_{\text{photon}}=\frac{hc}{\lambda}\end{array}$

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.

llunallenaipg5r

Beginner2022-05-17Added 5 answers

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 ${k}_{B}T$ of energy goes into a photon. Thus $h\overline{\nu}\propto {k}_{B}T$ and $\overline{\lambda}\propto {T}^{-1}$

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 $2.7{k}_{B}T$

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 $2.7{k}_{B}T$

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