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1

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, ...


0

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 ...


2

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 ...


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