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\title{Scalings}
\author{}
\date{\today,~ $ $Revision: 1.1 $ $}
\begin{document}
\maketitle

\section{Solving the disk structure by integration}

In our approach, we prescribe a heating function,
${\cal H}(z)$, in the energy equation
\begin{equation}
{\dd F\over\dd z}={\cal H}(z)
\end{equation}
which leads to a fixed flux profile, $F(z)=F_\infty\,f(z)$,
where $f(z)$ goes from 0 (in the midplane) to 1 for $z\ge L$.
In our approach, the flux at infinity is given in terms of
$\dot{M}$ via
\EQ
F_\infty={\cal H}_0 L={\dot{\Sigma}\over3\pi}
(\threehalf\Omega)^2=\sigmaSB\Teff^4,
\EN
which thus also determines the effective temperature.
Next, we have
\begin{equation}
{\dd T\over\dd z}=-F(z)/K(\rho,T)
\end{equation}
\begin{equation}
{\dd P\over\dd z}=-\rho g
\end{equation}
where $g=\Omega^2z$ is the local gravitational acceleration.
Divide by each other, and use $P=\Rgas T\rho/\mu$, so
\begin{equation}
\nabrad\equiv{\dd\ln T\over\dd\ln P}
={\Rgas\over\mu}\,{F_\infty\over K\Omega^2}\,g(z)
\end{equation}
where
\begin{eqnarray}
g(z)=f(z)/z=\,\left\{\!
\begin{array}{cc}
1   & \mbox{(for $z\leq L$)},\\
1/z & \mbox{(for $z > L$)}.
\end{array}
\right.
\end{eqnarray}
and
\EQ
\dot{M}=3\pi\;{2F_\infty\over(\threehalf\Omega)^2}\,{H\over L}
\EN
Furthermore,
\EQ
F_\infty f(z)=-{16\sigmaSB T^{3-b}\over3\kappa_0\rho^{1+a}}\,{\dd T\over\dd z}
\EN
\footnote{The following rewrite {\it may} be useful:
\EQ
F_\infty f(z)=-{16\sigmaSB\over3\kappa_0}{\left({T^n\over\rho}\right)^{1+a}}\,{\dd T\over\dd z}
\EN
where $n=(3-b)/(1+a)$ is the polytropic index.}
Next, replacing $\dd T/\dd z\to T/H$ and $T\to T_{\max}$ and using
$(\gamma-1)\cp T_{\rm max}=\cs^2=\Omega^2 H^2$ we have
\EQ
F_\infty f(z)\propto {16\sigmaSB H^{7-2b}\over3\kappa_0\rho^{1+a}}
\EN
\EQ
F_\infty f(z)\propto {16\sigmaSB H^{7-2b}\over3\kappa_0(\rho H)^{1+a}}H^{1+a}
\EN
\EQ
F_\infty f(z)\propto {16\sigmaSB H^{7-2b}\over3\kappa_0\Sigma^{1+a}}H^{1+a}
\EN
\EQ
F_\infty f(z)\propto {16\sigmaSB H^{8-2b+a}\over3\kappa_0\Sigma^{1+a}}
\EN

\EQ
{4\sigmaSB T^{4-b}\over3\kappa_0\Sigma^{1+a}}H^{a}=\nut\Sigma (\threehalf\Omega)^2
\EN

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pstrat_compstrat0.0050_conv0a0bm3p5H8.0.ps}
\end{center}\caption[]{
conv0a0bm3p5H8.0
}\label{pstrat_compstrat0.0050_conv0a0bm3p5H8.0.ps}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pstrat_compstrat0.0050_conv0a0bm3p5H7.0.ps}
\end{center}\caption[]{
conv0a0bm3p5H7.0
}\label{pstrat_compstrat0.0050_conv0a0bm3p5H7.0.ps}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pstrat_compstrat0.0050_conv0a1bm3p5H8.0.ps}
\end{center}\caption[]{
conv0a1bm3p5H8.0
}\label{pstrat_compstrat0.0050_conv0a0bm3p5H8.0.ps}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pstrat_compstrat0.0050_conv0a1bm3p5H7.0.ps}
\end{center}\caption[]{
conv0a1bm3p5H7.0
}\label{pstrat_compstrat0.0050_conv0a0bm3p5H7.0.ps}\end{figure}

\section{Optically thin case}

In the optically thin case, we have $T=\const\propto\Mdot^{1/4}$.
Furthermore, from $\Mdot\propto\Sigma H^2$, and since $H^2\propto T$
we have $\Mdot^{3/4}\propto\Sigma$ or $\Mdot\propto\Sigma^{4/3}$.

\section{Disk parameters}

factor 2?
\EQ
\alpha_{\rm SS}=\dot{M}/3\pi\;\Sigma\Omega H^2
\EN
\EQ
\sigmaSB\Teff^4=(\Mdot/3\pi)\,	(\threehalf\Omega)^2
\EN
and

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\end{document}
