In chapter 13.1 in 'Mesoscale Meteorology in Midlatitudes' (Markowski and Richardson,2011) they use a Bernoulli equation and the hydrostatic equation and a lot of assumptions to derive an equation that predicts the height $z_{crit}$ at where an air parcel with an initial height of $z_0$ that is advected towards a barrier loses all horizontal velocity. In their argumentation they state the following identity $$\dfrac{\partial p}{\partial z}(x,z) = \rho(x,z)c_p\theta(x,z)\dfrac{\partial \pi}{\partial z} (x,z)\tag{1}\label{wanted}$$ where $\pi(x,z)=\left(\frac{p(x,z)}{p_0}\right)^{\frac{R}{c_p}}$ is the Exner function with reference pressure $p_0\in\mathbb{R}_+$ constant, $p$ is pressure, $\theta$ is potential temperature, $\rho$ is density, and $c_p$ and $R$ are constant $\in \mathbb{R}_+$.

If I try to derive this equation by taking the partial derivative of $\pi$ with respect to height $z$ and use the identity $\pi(x,z)=\frac{T(x,z)}{\theta(x,z)}$ as can be found for example on wikipedia ($T$ is temperature), then I end up with (suppressing arguments $x$ and $z$ in favour of readability) $$\dfrac{\partial \pi}{\partial z} = \left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}\dfrac{\partial}{\partial z}\exp\left(\frac{R}{c_p}\log(p)\right)=\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}* \frac{T}{\theta}*{\frac{R}{c_p}}*\frac{1}{p}*\dfrac{\partial p}{\partial z}.$$ Multiplying this by $\rho c_p\theta$ we get $$\rho c_p \theta \dfrac{\partial \pi}{\partial z} = \left(\frac{1}{p_0}\right)^{\frac{R}{c_p}} \frac{T R \rho}{p}*\dfrac{\partial p}{\partial z}$$ and because $\frac{T R \rho}{p}=1$ by the ideal gas law we end up with $$\rho c_p \theta \dfrac{\partial \pi}{\partial z} = \left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}\dfrac{\partial p}{\partial z}.$$ This is almost what is stated in $\eqref{wanted}$, but not quite and I do not know how to get rid of the factor $\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}$ resp. do not see why it remains unmentioned in the textbook. Any help is appreciated.


Start with the original equation. Let's first write the Hydrostatic equation:

$$\frac{\partial p}{\partial z}=-\rho g$$

So let's prove that $$-g=c_p\theta\frac{\partial \pi}{\partial z}$$

If we use the product rule, we observe $$-g=c_p(\frac{\partial \theta \pi}{\partial z}-\pi\frac{\partial \theta}{\partial z})$$

Since $\pi=\frac{T}{\theta}$, we can say that $T=\pi\theta$, which makes the above equation

$$-g=c_p(\frac{\partial T}{\partial z}-\pi\frac{\partial \theta}{\partial z})$$

It can be shown that $$\frac{\partial \theta}{\partial z}=\frac{\theta}{T}(\frac{\partial T}{\partial z}+\frac{g}{c_p})$$

which can be rewritten as $$\frac{\partial \theta}{\partial z}=\frac{1}{\pi}(\frac{\partial T}{\partial z}+\frac{g}{c_p})$$

Substituting this into my fourth equation $$-g=c_p(\frac{\partial T}{\partial z}-\frac{\pi}{\pi}(\frac{\partial T}{\partial z}+\frac{g}{c_p}))$$

From here I think you can figure it out.

Derivation for $\frac{\partial \theta}{\partial z}$ $$\theta=T(\frac{p_0}{p})^\frac{R_d}{c_p}$$ $$log(\theta)=log(T)+\frac{R_d}{c_p}(log(p_0)-log(p))$$ $$ \frac{1}{\theta}\frac{\partial \theta}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}-\frac{R_d}{c_p P}\frac{\partial P}{\partial z}$$ Utilizing hydrostatic equation $$ \frac{1}{\theta}\frac{\partial \theta}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+\frac{R_d \rho g}{c_p P}$$ $$ \frac{1}{\theta}\frac{\partial \theta}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+\frac{g}{c_p T}$$

$$ \frac{\partial \theta}{\partial z}=\frac{\theta}{T}(\frac{\partial T}{\partial z}+\frac{g}{c_p })$$

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    $\begingroup$ Thanks already! I do not see the error. Perhaps your formula for $\dfrac{\partial \theta}{\partial z}$ has a wrong sign in the parentheses? Can you point me to a source for said formula? $\endgroup$ – chriss Mar 26 '20 at 14:24
  • $\begingroup$ For the formula for $\frac{\partial \theta}{\partial z}$? I'll add it in. $\endgroup$ – BarocliniCplusplus Mar 26 '20 at 14:32
  • $\begingroup$ I think the mistake in that argumentation is indeed that $\dfrac{\partial \theta}{\partial z} = \frac{1}{\pi}\left(\dfrac{\partial T}{\partial z} + \frac{g}{c_p}\right)$. This can be derived from $\theta = \frac{T}{\pi}$ by product rule. I will add the steps later. Thank you! $\endgroup$ – chriss Mar 26 '20 at 14:37
  • $\begingroup$ I solved my negative sign and updated my answer. You're welcome! $\endgroup$ – BarocliniCplusplus Mar 26 '20 at 14:42

After looking back into this I found my error to lie in the long equation

$\dfrac{\partial \pi}{\partial z}=\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}\dfrac{\partial}{\partial z}\exp\left(\frac{R}{c_p}\log(p)\right)=\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}* \frac{T}{\theta}*{\frac{R}{c_p}}*\frac{1}{p}*\dfrac{\partial p}{\partial z}$.

I was wrong, because $$\dfrac{\partial}{\partial z}\exp\left(\frac{R}{c_p}\log(p)\right)\neq \frac{T}{\theta}*{\frac{R}{c_p}}*\frac{1}{p}*\dfrac{\partial p}{\partial z},$$ but rather $$\dfrac{\partial}{\partial z}\exp\left(\frac{R}{c_p}\log(p)\right) =\left(p\right)^{\frac{R}{c_p}}*{\frac{R}{c_p}}*\frac{1}{p}*\dfrac{\partial p}{\partial z}.$$


The mistake is hidden in the identity ${\partial \over \partial z} \log (p) = {1\over p} {\partial p \over \partial z}$.

This formula looks harmless and would be correct if $p$ was a real valued function of $z$, but $p$ is actually a pressure value and so $\log(p)$ is undefined. Instead, let $p^*=p/p_0$ be a non-dimensional pressure, and then: $$ {\partial \over \partial z} \log(p^*) = {1\over p^*} {\partial p^*\over \partial z} = {1\over p} {\partial p\over \partial z}. $$

Put this in your derivation and the spurious $\left( 1 \over p_0 \right)^{R\over c_p}$ will disappear.

  • $\begingroup$ I do not think that $log(p)$ is undefined. After all, $p$ is a real value with physical dimension of pressure. See my answer, I pointed out my fault there. $\endgroup$ – chriss Mar 30 '20 at 6:46
  • $\begingroup$ The log of the real value is defined, but the log of the dimension is not. For example, if $p = 1hPa = 100 Pa$, what do you think $log(p)$ is? But, yes, your correction above is dealing with the critical error (which I had missed). $\endgroup$ – M Juckes Mar 30 '20 at 11:15
  • $\begingroup$ Ah yes, I get your meaning now. But then $\left(\frac{p}{p_0}\right)^{\frac{R}{c_p}}\neq p^{\frac{R}{c_p}}\cdot\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}$, because the rhs is undefined? $\endgroup$ – chriss Apr 1 '20 at 8:24
  • $\begingroup$ Raising units to a power is defined, so that is not the same problem. I suppose you could say that your approach works at a mechanistic level (with the correction you've posted in your answer), so it is OK. I'm coming from a maths background and have had it drummed into me that functions and operations which have been carefully defined to work on real values should not be applied to dimensional quantities. $\endgroup$ – M Juckes Apr 1 '20 at 10:01
  • $\begingroup$ But the standard defintion for the expression $a^b$ for $a>0, b \in \mathbb{R}$ is that $a^b=\exp(b\log(a))$, is it not? Hence my question in the previous comment. $\endgroup$ – chriss Apr 1 '20 at 11:33

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