[Physics] How to derive Stefan constant from Planck’s Blackbody radiation

statistical mechanicsthermal-radiation

How to derive Stefan constant from Planck's Blackbody radiation?
Consider the following expression relating to blackbody radiation:
$$\phi(\lambda) d\lambda= E({\lambda}) \, f({E(\lambda}))\,D({\lambda})d{\lambda}$$
$$\phi(\lambda) d\lambda=\left( \frac{hc}{\lambda}\right) \left(\frac{1}{e^g-1}\right) \left( \frac{8\pi}{\lambda^4} \right) d\lambda \, \, ,$$ where $g = \frac{hc}{k_BT\lambda}$.

I know that $D({\lambda})d{\lambda}$ is the density of states within $d{\lambda}$.

What is $\phi(\lambda) d\lambda$? The book says radiation energy density.

What does it mean that $\phi(\lambda)$ = (energy of state) * (probability distribution) * (density of states) = energy of state distributed among the density of the states?
And then $\int\phi(\lambda) d\lambda$ is the density of energy distributed within the interval $d\lambda$?

What can I do to relate $\phi(\lambda) d\lambda$ to intensity, and then get $I=\sigma T^4$?

Best Answer

  • $E\left(\lambda\right)$ is the energy of one photon of light with wavelength $\lambda$
  • $f\left(E\left(\lambda\right)\right)$ is the number of photons in a state with wavelength $\lambda$
  • $D\left(\lambda\right)d\lambda$ is the number of states with wavelengths between $\lambda$ and $\lambda+d\lambda$. ($D\left(\lambda\right)$ is the density of states.)

Multiply those together (energy*number in each state*number of states), and you have the total energy in the light from photons with wavelengths between $\lambda$ and $\lambda+d\lambda$. That's $\phi\left(\lambda\right)d\lambda$.

If you integrate over all $\lambda$, you can get the total energy in a given volume.

That can be used to calculate the total energy in a cavity. A black body is equilivent to a cavity with a small hole that lets the light inside to escape. Find the amount of energy escaping from this cavity per unit time, and you have the Stefan–Boltzmann law.

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