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\subsection{Convective Velocities}
We can calculate the accelaration of a convective fluid element
\begin{equation}
\ddot{r} = - g - \frac{1}{\rho} \frac{\partial
\P}{\partial \p}{\partial r},
\end{equation}
which expanded at first order around the equilibrium state leads to an estimate of the total net force (buoyant minus gravitation) $f=-g \Delta \rho$. Along the distance $\Delta r$ a convective fluid element experience the force $f(\Delta r) = - g \Delta \rho (\Delta r)$. Here the variation of $g$ with height is considered negligible. The work done per unit volume over the distance $\Delta r$ is therefore
\begin{equation}
...
\begin{equation}\label{eq:velocity2}
\bar{v}^2 = - \frac{1}{8} g (\Delta \rho (\lambda)/\rho) \lambda.
\end{equation}
Relating $\Delta \rho$ and $\Delta T$ requires the equation of state for the gas $\rho = \rho (\mu, T,
\P)$, \p)$, which in differential form can be written as
\begin{equation}
\D\rho = \bigg(\frac{\partial \rho}{\partial \mu}
\bigg)_{\P,T} \bigg)_{\p,T} \D\mu + \bigg(\frac{\partial \rho}{\partial T} \bigg)_{\P,\mu} \D T + \bigg(\frac{\partial \rho}{\partial
\P} \p} \bigg)_{\mu,T}
\D\P, \D\p,
\end{equation}
,
\begin{equation}
\D\rho = \frac{\rho}{\mu}\bigg(\frac{\partial \ln \rho}{\partial \ln \mu}
\bigg)_{\P,T} \bigg)_{\p,T} \D\mu + \frac{\rho}{T} \bigg(\frac{\partial \ln \rho}{\partial \ln T}
\bigg)_{\P,\mu} \bigg)_{\p,\mu} \D T +
\frac{\rho}{\P} \frac{\rho}{\p} \bigg(\frac{\partial \ln \rho}{\partial \ln
\P} \p} \bigg)_{\mu,T}
\D\P \D\p
\end{equation}
and finally
\begin{equation}
\frac{\D\rho}{\rho} = \alpha \frac{\D\P}{\P} - \delta \frac{\D T}{T} + \phi \frac{\D\mu}{\mu}
\end{equation}
Where
$ \phi \equiv \big(\frac{\partial \ln \rho}{\partial \ln \mu}
\big)_{\P,T}, \big)_{\p,T}, \; \delta \equiv - \big(\frac{\partial \ln \rho}{\partial \ln T}
\big)_{\P,\mu}, \big)_{\p,\mu}, \; \alpha \equiv \big(\frac{\partial \ln \rho}{\partial \ln
\P} \p} \big)_{\mu,T}$
and for an ideal gas $\alpha = \delta = \phi = 1$, so that using pressure equilibrium
($\D\P= ($\D\p= 0$) we obtain $\Delta \ln \rho = \Delta \ln T$.
In the case of a mixture of perfect gas and radiation we can write
\begin{equation}\label{eq:deltarho}
\Delta \ln \rho = -Q \Delta \ln T
\end{equation}
where
\begin{equation}
Q = \frac{4-3 \beta}{\beta} - \bigg(\frac{\partial \ln \mu}{\partial \ln T}
\bigg)_{\P}, \bigg)_{\p},
\end{equation}
with $\beta =
\P_{Gas}/\P$. \P_{Gas}/\p$. Plugging Eq.~\ref{eq:deltarho} in \ref{eq:velocity2} one obtains
\begin{equation}
\bar{v}^2 = \frac{1}{8} g Q (\Delta T (\lambda) / T) \lambda .
\end{equation}
...
We can now rewrite the convective flux using the derived form of the convective velocities and Eq.~\ref{eq:convectiveflux}
\begin{displaymath}\label{eq:convectiveflux2}
F_{c} = \frac{1}{2} \rho \bar{v}\, \cp \, T \frac{\lambda}{\hp} ( \nabla - \nabla') \\
= \frac{1}{4 \sqrt{2}} g^2 \rho^{5/2} \cp Q^{1/2} T
\p^{-3/2} P^{-3/2} \lambda^2 \, (\nabla - \nabla')^{3/2}.
\end{displaymath}
