# ThetaS derivation

## Potential Temperature and evolution equation

Beginning from the fundamental Equation of Thermodynamics for a three phase flow system (dry air, vapor and liquid)

$$dU (S, \alpha, \rho_{l}, \rho_{v}) = \frac{\partial U}{\partial S} dS + \frac{\partial U}{ \partial \alpha} d \alpha + \frac{\partial U}{\partial \rho_{v}} d \rho_{v} + \frac{\partial U}{\partial \rho_{l}} d \rho_{l}$$

with some basic definitions like

$$\frac{\partial U}{\partial \alpha} = T$$; $$\frac{\partial U}{\partial \alpha} = -p$$; $$\frac{\partial U}{\partial \rho_{v}} = \mu_{v}$$; $$\frac{\partial U}{\partial \rho_{l}}= \mu_{l}$$ and $$d \rho_{l} = -d \rho_{v}$$

we get:

$$dU (S, |alpha, \rho_{l}, \rho_{v}) = TdS – pd \alpha + (\mu_{v} – \mu_{l}) d \rho_{v}$$

with the Definition of Enthalpy

$$H = U +pV$$

we get
$$dH= TdS + \alpha dp + (\mu_{v} – \mu_{l}) d \rho_{l} = TdS – \alpha dp + L_{v}d \rho_{v}$$

Introducing the definitions of specific heat capacity at constant pressure and the ideal gas constant for a mixture and the equation of state for a moist air

$$C_{pml} = \frac{\rho_{d}C_{pd} + \rho_{v}C_{pl} + \rho_{l} C_{pl}}{\rho}$$ and $$R_{ml} = \frac{\rho_{d}R_{d}+ \rho_{v}R_{v}}{\rho}$$
$$dH = C_{pml}dT$$
$$\alpha = \frac{R_{ml}T}{p}$$

we get
$$C_{pml}dT = TdS + L_{v}d \rho_{l} + \frac{R_{ml}T}{p}dp$$

or

$$dS = \frac{C_{pml}}{T}dT – \frac{L_{v}}{T}d \rho_{l} – \frac{R_{ml}}{p}dp $$

integration gives
$$S = C_{pml} ln(T) – R_{ml} ln(p) – C_{\frac{L_{v}}{T}d \rho_{l}} + S_{0}$$

introducing the potential temperature

$$S = C_{pml}ln(T) – R_{ml} ln (p) – C_{\frac{L_{v}}{T}d \rho_{l}} + S_{0} = C_{pml} ln(\theta) – R_{ml} ln (p_{0}) – C_{\frac{L_{v}}{T}d \rho_{l}} + S_{0}$$

$$\theta = T (p_{0}/p)^{\frac{R_{ml}}{C_{pml}}}$$

This gives an equation where the the partial time derivative would give an easy to use expression for the case without any phase transition. But in the case with phase transitions we get an explicit time dependence of $$C_{mL}$$ and $$which leads to uncomfortable source terms in the evolution equation for the potential temperature. For the case without phase transitions the first law of thermodynamics: [latex]Tds = Q \delta t$$

leads to the evolution equation for Entropy

$$T(\frac{\delta S}{\delta t} + (\vec{v} \nabla)S) = Q$$

or for the potential temperature without phase transition

$$(\frac{\delta \theta}{\delta t} + (\vec{v} \nabla) \theta) = \frac{\theta}{C_{pml}T}Q$$

In the transition case we will get a lot of terms from the time dependence of the heat capacity and the gas “constant” for the mixture. But as a solution to this we could define a new quantity like this:

$$S = C_{pml}ln(T) – R_{ml}ln{p} – C_{\frac{L_{v}}{T}d \rho_{l}} + S_{0} = \theta_{new} – C_{\frac{L_{v}}{T}d \rho_{l}} + S_{0}$$

so that:
$$\theta_{new} = C_{pml}ln(T) – R_{ml} ln(p)$$

So we have only one time dependend term in the equation for entropy and so an easy evolution equation for our new quantity
$$\theta_{new}$$
$$S = \theta_{new} – C_{\frac{Lv}{T}d \rho_{l}} + S_{0}$$
$$\frac{dS}{dt}=\frac{d \theta_{new}}{dt} – \frac{L_{v}}{T} \frac{\partial \rho_{l}}{\partial t}$$

with
$$\frac{d \theta_{new}}{dt}= ln(T) \frac{C_{pl}-C_{p v}}{\rho}d \rho_{l} + (ln(p)+1) \frac{R_{v}}{\rho}d \rho_{l}+ \frac{C–{pml} – R_{ml}}{T}dT$$
$$T \frac{dS}{dt} = Q$$
$$(\frac{\partial \theta_{new}}{\delta t}+ (\vec{v} \nabla) \theta_{new}) = L_{v} \frac{\partial \rho_{l}}{\partial t}$$

One advantage of this quantity compared to the equivalent potential temperature is that we can easy obtain an explicit equation for pressure or the normal Temperature
$$p = (R_ml \rho \theta_{new}^{C_{d}/C_{pml}})^{\frac{1}{1-R_{ml}/C_{pml}}}$$
$$T = \frac{1}{R_{ml} \rho} (R_{ml} \rho \theta_{new}^{C_{d}/C{pml})})^{\frac{1}{1-R_{ml}/C_{pml}}}$$

## Logarithmic scales and Entropy

Starting at the equation for entropy

$$S = C_{pml} kn (T) – R_{ml} ln(p) + \frac{L_{v}}{\rho T} \rho_{v} + S_{0}$$

with some redefinitions
$$T = exp{\tilde{T}}$$ and $$p = exo{\tilde{p}}$$; $$\tilde{p} = \tilde{T} + ln(\rho R_{ml})$$
$$\tilde{\theta} = C_{pml}\tilde{T} – R_{ml}\tilde{p}$$

$$S = C_{pml}\tilde{T} – R_{ml} \tilde{p} + \frac{L_{v}}{\rho T}\rho–{v} + S_{0} = \tilde{\theta} + \frac{L_{v}}{\rho T} \rho_{v} + S_{0}$$
$$dS = d\tilde{\theta}+ \frac{L_{v}}{\rho T}d \rho_{v} – \frac{L_{v}}{T^{2}} \frac{\rho–{v}}{\rho}dT$$

so $$\tilde{\theta}$$ is conserved under adiabatic transformations without phase transitions. In the case of an reversible phase transition we get:

$$d \tilde{\theta} = \frac{L_{v}}{T^{2}} \frac{\rho_{v}}{\rho}dT – \frac{L_{v}}{\rho T}d \rho_{v}$$

while
$$d \tilde{\theta} = \frac{\partial \tilde{\theta}}{\partial \rho_{v}}d \rho_{v} + \frac{\partial \tilde{\theta}}{\partial \rho_{l}}d \rho_{l} + \frac{\partial \tilde{\theta}}{\partial \tilde{T}} d \tilde{T} + \frac{\partial \tilde{\theta}}{\partial \tilde{p}}d \tilde{p} + \frac{\partial \tilde{\theta}}{\partial t} + (\vec{v} \nabla) \tilde{\theta}$$
$$\frac{\partial \tilde{\theta}}{\partial \rho_{}} = \tilde{T} \frac{\partial C_{pml}}{\partial \rho_{v}} – \tilde{p} \frac{\partial R_{ml}}{\partial \rho_{v}}$$
$$\frac{\partial C_{pml}}{\partial \rho_{v}} = \frac{C_{pv}}{\rho} \frac{\partial R_{ml}}{\partial \rho_{vn}} = \frac{R_{vn}}{\rho}$$
$$\frac{\partial \tilde{\theta}}{\partial \rho_{l}} = \tilde{T} \frac{\partial C_{pml}}{\partial \rho_{l}} – \tilde{p} \frac{\partial R_{ml}}{\partial \rho_{l}}$$
$$\frac{\partial C_{pml}}{\partial \rho_{l}} = \tilde{T} \frac{\partial C_{pl}}{\rho} \frac{\partial R_{ml}}{\partial \rho_{l}} = 0$$
$$\frac{\partial \tilde{\theta}}{\partial \tilde{T}} = C_{pml}$$
$$\frac{\partial \tilde{\theta}}{\partial \tilde{p}} = -R_{ml}$$
$$d \tilde{\theta} = \tilde{T} \frac{C_{pv}}{\rho}d \rho_{v} – \tilde{p} \frac{R_{v}}{\rho}d \rho_{v} + \tilde{T} \frac{C_{pl}}{\rho}d \rho_{l} + C_{pml} d \tilde{T} – R_{ml} d \tilde{p} + \frac{\partial \tilde{\theta}}{\partial t} + (\vec{v} \nabla) \tilde{\theta} = \frac{L_{v}}{T^{2}} \frac{\rho_{v}}{\rho}dT – \frac{L_{v}}{\rho T}d \rho_{v}$$
$$\tilde{T} \frac{C_{pv}}{\rho}d \rho_{v} – \tilde{T} \frac{C_{pl}}{\rho}d \rho_{v} – \tilde{p} \frac{R_{v}}{\rho}d \rho_{v} + C_{pml}d \tilde{T} – R_{ml}d \tilde{p} + \frac{\partial \tilde{\theta}}{\partial t} + (\vec{v} \nabla) \tilde{\theta} = \frac{L_{v}}{T^{2}} \frac{\rho_{v}}{\rho}dT – \frac{L_{v}}{\rho T}d \rho_{v}$$

with
$$d\tilde{T} =\frac{1}{T}dT=-\frac{L_v}{\rho T C_{pml}}d\rho_v$$
$$C_{pml}d \tilde{T} =- \frac{L_v}{\rho T}d\rho_v$$
$$\tilde{T} (\frac{C_{pv}}{\rho} – \frac{C_{pl}}{\rho}) d \rho_{v} – \tilde{p} \frac{R_{v}}{\rho} d \rho_{v} + \frac{\partial \tilde{\theta}}{\partial t} + (\vec{v} \nabla) \tilde{\theta} = \frac{L_{v}}{T^(2)} \frac{\rho_{v}}{\rho}dT + R_{ml}d \tilde{p}$$
$$\frac{\partial \tilde{\theta}}{\partial t} + (\vec{v} \nabla) \tilde{\theta} = (\tilde{p} + 1) \frac{Rv}{\rho} d \rho_{v} – \tilde{T} ( \frac{C_{pv}}{\rho} – \frac{C_{pl}}{\rho}) d \rho_{v} + (\frac{L_{v}^{2}}{\rho^{2}T^{2}} \frac{\rho_{v}}{C_{pml}} – \frac{R_{ml}}{C_{pml}} \frac{Lv}{\rho T})d \rho_{v}$$

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