Square-root exponential atmospheric profile ($P=P_0{e}^{\alpha \sqrt{T-T_0}}$)

For $P_0>0$, $\alpha \ne 0$, $T\ge T_0$ and physical $T>0$, the atmospheric pressure-temperature profile $P=P_0\exp (\alpha \sqrt{T-T_0})$ is equivalent toThe 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.

For a homogeneous ideal gas of constant specific-heat ratio $\gamma >1$, the Schwarzschild criterion applied to the square-root exponential atmospheric profile givesFor $T_0>0$ and $\alpha >0$, put $s=\sqrt{T-T_0}$ and $A={\nabla }_{\mathrm{ad}}$. The quadratic inequality $A\alpha s^2-2s+A\alpha T_0<0$ has solutions precisely when $A\alpha \sqrt{T_0}<1$, between $s_{\pm }=[1\pm \sqrt{1-(A\alpha {)}^2{T}_0}]/(A\alpha )$. This is a local instability of the imposed temperature gradient; actual convection adjusts that gradient and the prescribed profile need not survive.