The thermal <optical depth> increases inward, so an <atmospheric thermal inversion> requires $dT/d\tau<0$. Differentiating the <semi-grey irradiated atmosphere> profile gives
$$
\frac{dT^4}{d\tau}=\frac34\left[T_{\rm int}^4-f_rT_{\rm irr}^4(\gamma^2-1)e^{-\sqrt3\gamma\tau}\right].
$$
Therefore the <inversion criterion with intrinsic planetary flux> is
$$
\boxed{f_rT_{\rm irr}^4(\gamma^2-1)e^{-\sqrt3\gamma\tau}>T_{\rm int}^4}.
$$
An inverted layer exists at the top precisely when
$$
\boxed{\gamma^2>1+\frac{T_{\rm int}^4}{f_rT_{\rm irr}^4}}.
$$
For $T_{\rm int}>0$, it extends over
$$
0\leq\tau<\frac{1}{\sqrt3\gamma}\ln\left[\frac{f_rT_{\rm irr}^4(\gamma^2-1)}{T_{\rm int}^4}\right],
$$
provided the logarithm is positive. Otherwise the profile has no inverted interval. For a strongly irradiated <hot Jupiter>, $T_{\rm int}\ll T_{\rm irr}$ and the threshold is approximately $\gamma>1$; the intrinsic flux eventually restores an inward-increasing <temperature> at greater depth.

Large visible absorption <opacity> deposits stellar heat high in the <exoplanet atmosphere>. Gas-phase <titanium monoxide> and <vanadium monoxide> are candidate absorbers in sufficiently hot layers. An <atmospheric cold trap> or other condensate loss can remove them, while sufficiently high <temperature> and vigorous mixing can help keep them in the gas. The condition concerns absorbing <opacity>: highly reflective scattering alone does not deposit the required heat.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 315 / 2 / a / ii / Solution

For the usual <semi-grey irradiated atmosphere> with downward thermal <optical depth> $\tau$, an inversion occurs where $f_rT_{\rm irr}^4(\gamma^2-1)e^{-\sqrt3\gamma\tau}>T_{\rm int}^4$, with $\gamma$ the visible-to-thermal <opacity> ratio. At negligible intrinsic flux the threshold reduces to $\gamma>1$.

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 Incoming links: Inversion criterion with intrinsic planetary flux