Potential temperature

Remarks

For dry air $R / c_p \approx 0.286$. For a reversible adiabatic change of an ideal gas, the first law gives $c_p \, d \ln T - R \, d \ln p = 0$, that is $d \ln \theta = 0$. So $\theta$ is conserved by a parcel that moves without heating, and $c_p \ln \theta$ equals the specific entropy up to a constant.

Example: at $p = 500$ hPa and $T = 253$ K, $\theta = 253 \cdot 2^{0.286} \approx 253 \cdot 1.219 \approx 308.5$ K. At $p = p_0$ the potential temperature equals the temperature.

Non-example: moist air in which water vapour condenses receives latent heat, so its $\theta$ increases as it rises. Then $\theta$ is not conserved, and the equivalent potential temperature is used instead.

The vertical gradient of $\theta$ measures static stability: the squared Brunt–Väisälä frequency is $N^2 = (g / \theta) \, \partial \theta / \partial z$, with $g$ the gravitational acceleration. The atmosphere is statically stable for dry displacements where $\partial \theta / \partial z > 0$.