# Relationship between Stefan-Boltzmann Law and Radiance

In a textbook, the equation for radiance is given

I am struggling to understand the relationship between the two equations. From what I know, multiplying the value obtained by the Stefan-Boltzmann law by area gives radiant flux. Therefore, how does this give rise to the equation seen for radiance?

It's not just about integrating over wavelengths, it's also about integrating over space.

There are quite a few steps to get from the $L$ you have to the $R$ and I'm stuck on one integral's derivation but here goes:

The radiance shown here is the $\textit{spectral radiance}$ for a $\textit{perfect}$ blackbody ($\epsilon=1$) so let's start with that.

$L(\lambda)=\frac{2hc^2}{\lambda^5}\frac{1}{\exp({\frac{hc}{\lambda k T}})-1}$

Using the definition of spectral radiance, what the equation above gives you is the amount of radiation emitted by a surface per unit wavelength, per unit projected area and per unit solid angle.

To get the 'complete' radiance, i.e. radiance across all wavelengths you need to take the integral

$L=\int_{0}^{+\infty}L(\lambda)d\lambda$

With a change of variable $x=\frac{hc}{\lambda kT}$ you get to:

$L=\frac{2k^4T^4}{h^3c^2}\int_{x=0}^{x=\infty}\frac{x^3}{\exp(x)-1}dx$

Noting that both $x$ and the integral are dimensionless, we can do a dimension check:

$[L]=\frac{J^4K^{-4}K^4}{J^3s^3m^2s^{-2}}=Jm^{-2}s^{-1}=Wm^{-2}$

(note that steradian is dimensionless so this makes sense).

In general, radiance can be angle dependent but a blackbody is a Lambertian surface, which means by definition that radiance is direction independent. The radiance itself is direction-independent. However, to relate to exitance you need to take into account the fact that radiation transfer between two surfaces is very much angle dependent.

The radiance $L$ is defined as:

$L\triangleq \frac{d^2\Phi}{\cos\theta dA d\Omega}$

Where $d \Omega$ is the solid angle subtended by the infinitesimal 'receiving' object. $dA$ is the area of the infinitesimal surface emitting the flux $\theta$ is the angle made by the normal to the emitting surface and the direction of propagation.

Note that $d\Omega$ can also be written $d\Omega=\frac{\cos(\theta_s)dA_s}{r^2}$

where $\theta_s$ is the angle made by the normal to the receiving surface and the direction of propagation.

In other words, the flux exchanged from the infinitesimal emitting surface of area $dA$ and a receiving infinitesimal surface of area $dA_s$

$d^2\phi=L\frac{\cos\theta \cos\theta_s dA dA_s}{r^2}$

This equation is symetric in angles and areas which makes sense: If what we called the receiving surface has a radiance $L_s=L$ then we should expect just as much flux going in one direction and in the other direction.

Exitance is by definition the total radiant flux coming out of the surface. To relate radiance to exitance you need to calculate how much radiant flux is transmitted from an infinitesimal surface into a hemisphere 'above' it.

Many people think this means $R=2\pi L$ since the solid angle of a hemisphere is $2\pi$. This is however incorrect because it fails to take into account the direction of propagation i.e. $\theta$.

The radiation exchange is more efficient right above the emitting surface (where $\cos\theta=1$) than it is 'on the extreme side' (where $\cos\theta =0$).

$R\triangleq\iint_{\text{hemisphere}}\frac{L}{r^2}\cos\theta dA_s$

Parametrize the infinitesimal area $dA_s$ using spherical coordinates: $dA_s=r^2\sin\theta d\theta d\phi$

$R=\int_{\theta=0}^{\frac{\pi}{2}}\int_{\phi=0}^{2\pi}L\cos\theta sin\theta d\theta d\phi$

Using

$\sin\theta\cos\theta=\frac{\sin2\theta}{2}$,

you get to

$R=L\pi$

and finally

$R=\frac{2\pi k^4T^4}{h^3c^2}\int_{x=0}^{x=\infty}\frac{x^3}{\exp(x)-1}dx$

Writing

$\sigma=\frac{2\pi k^4}{h^3c^2}\int_{x=0}^{x=\infty}\frac{x^3}{\exp(x)-1}$

You get for a perfect blackbody $R=\sigma T^4$

I'm not good enough at integration to solve the integral above but evidently the limit is $\frac{\pi^4}{15}$ (you can check that with Wolfram Alpha or Mathematica) and you end up with

$\sigma=\frac{2\pi^5 k^4}{15h^3c^2}$ which is indeed $5.67*10^{-8}Wm^{-2}K^{-4}$

Edit, bonus: you can remember the SF constant with '5-6-7-8'.. 'amaze your friends'.

Now if you have a gray body of emittance $\epsilon$ the radiance changes to:

$L(\lambda)=\epsilon \frac{2hc^2}{\lambda^5}\frac{1}{\exp({\frac{hc}{\lambda k T}})-1}$

Using the same derivation, you get:

$R=\sigma \epsilon T^4$

Note that an implicit assumption in the equations you gave is that

$\epsilon(\lambda)=\epsilon \forall \lambda$ i.e. spectral emissivity is independent of wavelength. Not anyways true..

Radiance is the energy radiated per unit wavelength. Integrating it over all wavelengths gives you the total energy radiated given by Stefan-Boltzmann law.