# Blackbody radiation and emissive power

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?

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 $$u = \frac{4\pi}{c} \int B_\lambda \ d\lambda = \frac{4E}{c}.$$