On page 142 of Lackmann's Midlatitude Synoptic Meteorology, Equation 6.7 gives the derivative of the Boussinesq buoyancy $b = \frac{g \theta'}{\theta_0 (z)}$ (perturbation potential temperature $\theta'$ and base state $\theta_0 (z)$) as

\begin{equation} \frac{db}{dt} = -w \frac{g}{\theta_{00}} \frac{d \theta_0}{dz} \end{equation}

where $\theta_{00}$ is the base state potential temperature $\theta_0 (z)$ at $z = 0$. While I understand this to be an application of the chain rule

\begin{equation} \frac{db}{dt} = \frac{\partial z}{\partial t} \frac{\partial \theta_0}{\partial z} \frac{\partial b}{\partial \theta_0}, \end{equation}

I am unable to see why

\begin{equation} \frac{\partial b}{\partial \theta_0} = - \frac{g}{\theta_{00}} \end{equation}

when I instead would expect

\begin{equation} \frac{\partial b}{\partial \theta_0} = - \frac{g \theta'}{\theta_0^2 (z)}. \end{equation}

Any help would be much appreciated!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.