= Square-root exponential atmospheric profile
{title2=$P=P_0e^{\alpha\sqrt{T-T_0}}$}
For $P_0>0$, $\alpha\ne0$, $T\ge T_0$ and physical $T>0$, the <atmospheric pressure-temperature profile> $P=P_0\exp(\alpha\sqrt{T-T_0})$ is equivalent to
$$
T=T_0+\frac{\log^2(P/P_0)}{\alpha^2},\qquad \frac{\log(P/P_0)}{\alpha}\ge0.
$$
The branch restriction is essential: squaring does not create a second physical branch. Its <temperature gradient> is $dT/d\log P=2\sqrt{T-T_0}/\alpha$. Thus, in <hydrostatic equilibrium>, $\alpha<0$ gives an <atmospheric thermal inversion>, while $\alpha>0$ gives outward cooling. At $T=T_0$ the inverse branch has zero <derivative>; $\alpha=0$ gives constant <pressure> and cannot describe a nontrivial hydrostatic vertical interval.
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