Deshawn Cabrera

2022-05-09

Area under Wien's displacement graph

Why does the area under Wien's displacement graph give Stefan-Boltzmann law for a black body?

I couldn't find any proof of this. (I could just find this expression). I am not aware of the function of Wien's displacement graph as well (I just know that it is between Intensity and wavelength emitted by a black body).

Is there a mathematical way to prove this?

Why does the area under Wien's displacement graph give Stefan-Boltzmann law for a black body?

I couldn't find any proof of this. (I could just find this expression). I am not aware of the function of Wien's displacement graph as well (I just know that it is between Intensity and wavelength emitted by a black body).

Is there a mathematical way to prove this?

ele5ph1a7jl1

Beginner2022-05-10Added 21 answers

Since the question is a little terse, it is difficult to interpret. I think what must be happening is that the phrase 'Wien's displacement graph' is being used to mean the graph of $\rho (\omega )$ as a function of frequency $\omega $, where $\rho (\omega )$ is the energy density per unit frequency range in thermal or black body radiation. This graph implies Wien's displacement law if one studies it as a function of temperature. And the area under this graph is the total energy density in the radiation, which obeys the Stefan-Boltzmann law, as follows:

$\rho (\omega )=\frac{\hslash}{{\pi}^{2}{c}^{2}}\frac{{\omega}^{3}}{{e}^{\beta \hslash \omega}-1},$

$u={\int}_{0}^{\mathrm{\infty}}\rho (\omega )d\omega =\frac{{k}_{B}^{4}{T}^{4}}{{\pi}^{2}{c}^{3}{\hslash}^{3}}{\int}_{0}^{\mathrm{\infty}}\frac{{x}^{3}}{{e}^{x}-1}dx=\frac{{\pi}^{2}{k}_{B}^{4}{T}^{4}}{15{c}^{3}{\hslash}^{3}}=\frac{4\sigma}{c}{T}^{4}.$

The power per unit area emitted by the surface of a black body is related to this by

$I=\frac{1}{4}uc=\sigma {T}^{4}.$

$\rho (\omega )=\frac{\hslash}{{\pi}^{2}{c}^{2}}\frac{{\omega}^{3}}{{e}^{\beta \hslash \omega}-1},$

$u={\int}_{0}^{\mathrm{\infty}}\rho (\omega )d\omega =\frac{{k}_{B}^{4}{T}^{4}}{{\pi}^{2}{c}^{3}{\hslash}^{3}}{\int}_{0}^{\mathrm{\infty}}\frac{{x}^{3}}{{e}^{x}-1}dx=\frac{{\pi}^{2}{k}_{B}^{4}{T}^{4}}{15{c}^{3}{\hslash}^{3}}=\frac{4\sigma}{c}{T}^{4}.$

The power per unit area emitted by the surface of a black body is related to this by

$I=\frac{1}{4}uc=\sigma {T}^{4}.$

Yasmine Larson

Beginner2022-05-11Added 3 answers

Actually this question has two answers modern and classic,viz. 1- Deriving the Stefan-Boltzmann law and the Wien's displacement law from the modern Planck radiation law or 2-Deriving both laws from the classic Wien,s radiation law as Wien himself established by introducing his law u(f,T) =const. f **3 .exp (-hf/kT) ,as early as 1893 (short before Max Planck modern quantization hypothesis proposed in 1900). Personally I prefer Steane,s hyper modern answer which is perfect according to modern physics of the black body radiation .Only on the other hand I suppose that more classic answer, hopefully, can add some light to the subject at minimal from the historical point of view. Here it is obvious that integrating Wien formula w.r.t (f) from 0 to infinity gives the area under the curve or the total radiation density u(T) which if multiplied by c/4 yields the well known Stefan-Boltzmann law . On the other hand differentiating Wien radiation-distribution law w.r.t (f) and equating with zero results in the Wiens displacement law i.e. Wavelength x T=.0029 m.K. Obviously The differentiation or integration of the Wien,s form is simpler than that for Planck,s law.The superiority of Planck,s form lies not only in introducing the correction term -1 (which actually did slightly improve the calculations for low frequencies of radiation )but mostly in introducing the idea of quantization of radiation thus opening the route for Quantum Mechanics.

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