radiationthermal-radiation
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?
This is a two-part answer based on my own ideas and on the specifically referenced book. I'm quite confident about the first part, but less so about the second.
The question arises because of the refractive index dependence of the radiated power. This dependence is hardly mentioned anywhere online or in books, and it's not obvious that it's true (indeed John implied that it isn't true in his answer). So first we establish the claim: The power radiated per unit surface are of a black body is proportional to $n^2$, where $n$ is the index of refraction of the medium it is radiating into.
This follows from Stefan's law, when the speed of light $c$ in Stefan's constant is replaced by $c_0/n$, where $c_0$ is the speed of light in vacuum. The $n^2$ dependence of the radiated power can also be derived by a modification of the normal derivation of Planck's law. The Bose-Einstein factor is constant for constant $E$, but the density of states $g(E)dE$ is directly wavelength-dependent via the mode-number. The wavelength dependence of $g(E)dE$ gives an $n^2$ term for fixed $E$, which integrates to give an $n^2$ term in the total radiated power.
The $n^2$ dependence of the radiated power is also discussed explicitly in the book by R. Siegel and John Reid Howell: "Thermal Radiation Heat Transfer", 4th Ed.: "The ... and total emissive power of a blackbody into a medium with constant index of refraction $n$ are given by the Stefan-Boltzmann law: $\pi i_b = e_b = n^2 \sigma T^4$."
The above reference could be wrong, although I have only heard good things about that book, but if it is wrong then we will need a good reference or derivation.
I just found it and haven't gone through it myself, but Siegel and Howell give a possible resolution in their book: "Section 18-5 shows how a portion of blackbody radiation from within a medium with $n>1$ cannot pass through an interface into an adjacent medium with smaller $n$. When entering into vacuum, there is a $1/n^2$ interface-reflection factor that removes the $n^2$ provided by [Stefan's law], so the maximum energy passing into the vacuum is $\sigma T^4$."
The answer can be calculated by referring to band emission tables of any introductory heat transfer book. I use the heat transfer book by Incropera and DeWitt where are concept is explained and the tables given in chapter 12.
Here is how its done. The fraction of total energy emitted as wavelengths from 0 to a certain wavelength $\lambda$ can be obtained from the band emission table ($F_{(0-\lambda)}$). This way the emissions from 0 to 400 nm ($F_{(0-0.4\mu m)}$) and between 0 to 800 nm ($F_{(0-0.8\mu m)}$) can be obtained. The difference in these two quantities gives the emission from 400 nm to 800 nm and is the required answer.
The energy fractions $F_{(0-\lambda)}$ as a function of $\lambda$ and $T$ are given in the table
Here $T$ = 2,573 K, while $\lambda$ is the upper limit of wavelength.
Best Answer
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: $$ \rho(\omega) = \frac{\hbar}{\pi^2 c^2} \frac{\omega^3}{e^{\beta \hbar \omega}-1}, $$ $$ u = \int_0^\infty \rho(\omega) d\omega = \frac{k_B^4 T^4}{\pi^2 c^3 \hbar^3} \int_0^\infty \frac{x^3}{e^x - 1} dx = \frac{\pi^2 k_B^4 T^4}{15 c^3 \hbar^3} = \frac{4 \sigma}{c}T^4 . $$ The power per unit area emitted by the surface of a black body is related to this by $$ I = \frac{1}{4} u c = \sigma T^4 . $$