# How to derive Stefan constant from Planck's Blackbody 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 $I=\sigma T^4$?

• Based on what you have posted so far, I can confirm that $\int_0^{\infty}\phi(\lambda) d\lambda$ is the energy density of radiation in thermal equilibrium at temperature T. It has units of energy per volume. Do you see how that makes sense, given the the three quantities you are multiplying together? Commented Dec 10, 2013 at 8:38
• Commented Dec 10, 2013 at 9:49
• @kleingordon_ actually I don't really understand why 3 of them are multiplying together. Can you please explain?________@Qmechanic_ thanks for the link. Commented Dec 10, 2013 at 22:59

• $E\left(\lambda\right)$ 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.