Blackbody radiation and emissive power

Question

According to blackbody radiation theory, and thanks to Planck, we now know that there is a energy density, $u(\lambda,T)$ [$J/m^3$], associated with a certain wavelength at a particular temperature. This is known as Planck's radiation formula:

$u(\lambda,T)= \frac{8 \pi h c}{\lambda^5} \frac{1}{e^{\frac{hc}{kT\lambda}}-1}$

What I am trying to figure out is how we can get the relationship between energy density and emissive power , $E$ in units of $[$W/m^2$]$. Serway, Modern Physics, states that they are simply off by a multpilicative factor:

$u(\frac{c}{4})=E $

and the units check out. Serway seems to shy away from the mathematical rigor, understandably it is aimed as an introductory book, and was wondering if anyone has a good reference to understand how this relationship holds?

Answer 1

For collimated light, E=cu (do you see why?). For light traveling uniformly in every direction, if you have an imaginary plane, cu/4 passes through in one direction and cu/4 in the other. This is the basis for the relationship you mention.

(The relevant mathematical fact is that if you have a sphere of radius 1, the average z-coordinate over the z>0 hemisphere is 1/2. You can prove this by spherical integration. There is an additional 1/2 factor from the fact that only half the light is traveling in the direction of one hemisphere. So that's the "4" in cu/4.)

I think kittel & kroemer is an example of a book that works through this in detail. :)

Answer 2

The emissive power of a blackbody is $\sigma T^4$ - the power per unit area from its surface.

This is derived by firstly establishing that the flux is the integral of the Planck function $B_{\lambda}$ (which is a specific intensity, in units of Watts per square metre per metre per steradian) over the solid angle subtended by radiation outwards into a hemisphere: $$ \int B_\lambda \cos \theta \ d\Omega = \int_{0}^{2\pi} \int^{\pi/2}_{0} B_\lambda \cos\theta \sin \theta \ d\theta\ d\phi = \pi B_\lambda$$ and then by integrating the Planck function over all wavelengths. I.e. What you call $E$ (I prefer $j$) is $$ E = \pi \int B_\lambda \ d\lambda .$$

The energy density that you quote is actually $$ u_\lambda = \frac{4\pi}{c} B_\lambda$$ So $$ u = \frac{4\pi}{c} \int B_\lambda \ d\lambda = \frac{4E}{c}. $$