Radiosity

Remarks

For an opaque, grey and diffuse surface at absolute temperature $T$ with emissivity $\varepsilon$, the reflectivity is $1 - \varepsilon$, and $$J = \varepsilon \sigma T^4 + (1 - \varepsilon) G,$$ where $\sigma$ is the Stefan–Boltzmann constant. The net radiative heat flux leaving the surface is $q'' = J - G = \dfrac{\varepsilon}{1 - \varepsilon} (\sigma T^4 - J)$ for $\varepsilon < 1$.

Example: a black body at temperature $T$ has $\varepsilon = 1$ and reflects nothing, so $J = \sigma T^4$ whatever the irradiation.

Non-example: the irradiation $G$ is not the radiosity: it is the power arriving at the surface, not leaving it. The two are equal only when the net radiative flux is zero.

Radiosity has the unit W/m$^2$, like a heat flux.