Derivative of Exner function

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.

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

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 $$\frac{1}{\theta}\frac{\partial \theta}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+\frac{R_d \rho g}{c_p P}$$ $$\frac{1}{\theta}\frac{\partial \theta}{\partial z}=\frac{1}{T}\frac{\partial T}{\partial z}+\frac{g}{c_p T}$$

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

• 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? Mar 26 '20 at 14:24
• For the formula for $\frac{\partial \theta}{\partial z}$? I'll add it in. Mar 26 '20 at 14:32
• 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! Mar 26 '20 at 14:37
• I solved my negative sign and updated my answer. You're welcome! 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.

• 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. Mar 30 '20 at 6:46
• 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). Mar 30 '20 at 11:15
• 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? Apr 1 '20 at 8:24
• 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. Apr 1 '20 at 10:01
• 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. Apr 1 '20 at 11:33