Energy losses (and gains) need to be accounted for during the convective element rise/fall. When these are not negligible, \(\nabla \neq \nabla\ \neq {\nabla_{\mathrm{\!ad}}}\). In the deep interiors of stars energy loss can be due to neutrinos, while gains are produced by nuclear burning. In the outer layers however radiation is usually responsible for departure from adiabaticity. These gains/losses are regarded as purely “horizontal” in the sense that one has to think as the average effect created by all the convective elements over a shell. Therefore no direct contribution to the net vertical heat transport is expected (on average there are as many hot and cool elements crossing the shell), but an indirect contribution can arise due to the modification to the convective properties (e.g. convective velocities). Focusing on horizontal radiative losses, the usual assumption is that the convective element is optically thick. The change in temperature of a rising convective element is due to both its adiabatic expansion and the horizontal radiative losses. For optically thick turbulence, from the equation of radiative transfer (\ref{eq:radtransfer}) \[F = -\frac{4ac}{3}\frac{T^3}{\kappa \rho}\frac{\Delta T}{\lambda/2},\] with \(\Delta T \) the temperature difference between the center of the turbulent element and the surrounding matter averaged over its lifetime. Given the area \(A\) of the element and its lifetime \(\lambda/\bar{v}\) the total radiated energy is \[E_{rad} = \frac{4ac}{3}\frac{T^3}{\kappa \rho}\frac{\Delta T}{\lambda/2} \frac{\lambda A}{\bar{v}}.\] The energy excess content carried by the convective element before dissolving is \({c_{\mathrm{P}}}\rho \Delta T^{*} V\) with \(\Delta T^{*}\) here being the temperature excess of the element over its surroundings at the end of its path. Obviously \(\Delta T^{*} \simeq \Delta T\), and in the classic Böhm-Vitense work the choice \(\Delta T^{*} = 2 \Delta T\) is made. Using these quantities one can define a convective efficiency \(\Gamma\) as \[\Gamma = \frac{\rm Energy Transported}{\rm Energy Lost} = \frac{2 {c_{\mathrm{P}}}\rho \Delta T V}{\frac{4ac}{3}\frac{T^3}{\kappa \rho}\frac{\Delta T}{\lambda/2} \frac{\lambda A}{\bar{v}}} = \frac{{c_{\mathrm{P}}}}{6ac}\frac{\kappa \rho^2 \bar{v} \lambda}{T^3} \sim \frac{\bar{v} \lambda}{K} \equiv {\rm Pe}\] where we used the fact that \(V/A=\lambda/6\) for a sphere of diameter \(\lambda\). Note how \(\Gamma\) is related to the Peclet number \(\rm{Pe}\), which is defined as the ratio between the thermal and the dynamical timescales. The next step is substituting the calculated average velocity from \ref{eq:velocity} to obtain \[\label{eq:gamma} \Gamma = \frac{{c_{\mathrm{P}}}}{12\sqrt{2} a c}\frac{\kappa g Q^{1/2} \rho^{5/2} \lambda^2}{{\mathrm{P}}^{1/2} T^3}(\nabla - \nabla')^{1/2} = \mathscr{A} \, (\nabla - \nabla')^{1/2},\] where \[\label{eq:aconv} \mathscr{A} \equiv \frac{Q^{1/2} {c_{\mathrm{P}}}\kappa g \rho^{5/2} \lambda^2}{12\sqrt{2}ac{\mathrm{P}}^{1/2}T^3}\] is essentially the ratio of the convective to the radiative conductivities.