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Heat flux density is a vector $\mathbf{q}$ describing the rate of heat transfer per unit area, measured in $\mathrm{W}{\mathrm{m}}^{-2}$. Through a planar face of area $A$ with uniform normal component $q$, the signed rate is $\dot{Q}=Aq$ in $\mathrm{W}$; positive $q$ means heat travels in the chosen positive normal direction. Fourier's law determines the conductive contribution. In a bar with a constant cross-sectional area, steady thermal conduction without internal sources or side losses makes both $q$ and $\dot{Q}$ constant along the bar. Constant $\dot{Q}$ need not mean constant $q$ when the cross-sectional area varies.

The equilibrium photon gas is isotropic, so the energy density per unit solid angle in a narrow angular-frequency interval is $u(T,\omega )d\omega /(4\pi )$. For rays making angle $\theta$ with an outward surface normal, a surface element $dA$ is crossed in time $dt$ by rays initially in a volume $cdtdA\cos \theta$. Only the outward hemisphere contributes. ThusThe hemispherical restriction supplies a factor $1/2$ and the directional projection supplies another $1/2$. A small aperture in a cavity therefore emits this spectral power per unit area. A blackbody at the same temperature emits the same amount: otherwise radiation exchange between it and the aperture would violate thermal equilibrium.

Integrating Planck's law over all positive angular frequencies gives the total emitted heat flux densitySet $x=\hslash \omega /(k_{B}T)$. ThenThe remaining integral is independent of temperature, proving $M\propto T^4$. Its value follows by expanding $(e^x-1{)}^{-1}={\sum }_{m=1}^{\infty }{e}^{-mx}$ and integrating termwise:Consequently the Stefan–Boltzmann law isThis is emitted power alone. A body in a radiation environment also absorbs incoming radiation, which must be included when calculating net radiative cooling.