Planck’s radiation law

This law tells us the energy density of the radiation of a black body, given the temperature of a black body, and usually one of the frequency of the light, or it’s wavelength (wiki page, specially this section).

When using the frequency, it can be written like:

$$ \rho(\omega, T) = \frac{1}{\pi^2 c^3}\frac{\hbar \omega^3}{e^{\frac{\hbar \omega}{\kappa_B T}} - 1} $$

Now, sometimes it’s useful to have it depending on the wavelength instead of $\omega$, but that transformation is not as simple as it would seem.

We have the equivalence:

$$ \omega = \frac{2\pi c}{\lambda} $$

Now, a blind replacement would give (using also $\hbar = \frac{h}{2\pi}$):

$$ \rho(\lambda, T) \ = \frac{1}{\pi^2 c^3}\frac{\hbar \left(\frac{2\pi c}{\lambda}\right)^3}{e^{\frac{\hbar \frac{2\pi c}{\lambda}}{\kappa_B T}} - 1} \ = \frac{8\pi\hbar}{\lambda^3}\frac{1}{e^{\frac{2\pi\hbar c}{\lambda\kappa_B T}} - 1} \ = \frac{4h}{\lambda^3}\frac{1}{e^{\frac{hc}{\lambda\kappa_B T}} - 1} \ $$

But this would be wrong! Well, kinda, let me explain.

The issue comes with what is the meaning of that formula. The full formula would include a differential:

$$ \rho(\omega, T)d\omega = \frac{1}{\pi^2 c^3}\frac{\hbar \omega^3}{e^{\frac{\hbar \omega}{\kappa_B T}} - 1} d\omega $$

So what we would want to do is to use lambda instead, that way we get:

$$ \rho(\omega, T)d\omega = \rho(\lambda, T)d\lambda $$

And it’s with that $d\lambda$ that issues come in. And there’s two things to keep into account:

  • Remember from before $\omega = \frac{2\pi c}{\lambda}$
  • These are energy distributions, both must be positive to have physical meaning

So from the first, we easily get $d\omega$:

$$ \frac{d\omega}{d\lambda} = \frac{d}{d\lambda}\frac{2\pi c}{\lambda} = - \frac{2\pi c}{\lambda^2} $$

And from the second, when isolating the formula we want:

$$ \begin{align*} \rho(\lambda, T)d\lambda &= \rho(\omega, T)d\omega \\ \rho(\lambda, T) &= \rho(\omega, T)\left|{\frac{d\omega}{d\lambda}}\right|\\ \end{align*} $$

Note that absolute value! That is due to the second restriction above, the energy has to still be positive!

With all this into account, we actually end up with the formula:

$$ \rho(\lambda, T) \ = \frac{8\pi hc}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda\kappa_B T}} - 1} \ $$

That is a factor $\frac{1}{\lambda^2}$ of difference!