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 $I=\sigma T^4$?

  • $\begingroup$ 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? $\endgroup$ Commented Dec 10, 2013 at 8:38
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/77660/2451 $\endgroup$
    – Qmechanic
    Commented Dec 10, 2013 at 9:49
  • $\begingroup$ @kleingordon_ actually I don't really understand why 3 of them are multiplying together. Can you please explain?________@Qmechanic_ thanks for the link. $\endgroup$
    – Outrageous
    Commented Dec 10, 2013 at 22:59

1 Answer 1

  • $E\left(\lambda\right)$ 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.

  • $\begingroup$ Related to this, see this answer on the relation between energy density inside the blackbody (like with a spherical cavity where the radiation in the cavity is just thermal radiation from the inner walls of the cavity) and rate that energy is escaping through the surface of the blackbody (like if you made a small hole in the cavity). $\endgroup$
    – Hypnosifl
    Commented Oct 13, 2021 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.