It makes sense to set the mass density $\rho$ as the first prognostic variable and the continuity equation
\[ \begin{align} \frac{\partial\rho}{\partial t} + \nabla\cdot\left(\rho\mathbf{v}\right) = 0 \end{align} \]
as the first prognostic equation. This is because mass density is a very fundamental quantity (it simply describes how much mass is in a given location) and the continuity equation is included in the derivation of many other prognostic equations. Due to the thermal equation of state of ideal gases, one must choose exactly one additional thermodynamic variable $q_1 \not= \rho$ in order to clearly determine the thermodynamic state of the ideal gas. You then record the Hamilton function of the atmosphere
\[ \begin{align} H = H\left(\rho, q_1, \mathbf{v}\right) = \int_A\frac{1}{2}\rho\mathbf{v}^2 + \rho\phi + \newtilde{I}\left(\rho, q_1\right)d^3r. \end{align} \]
$\newtilde{I}\left(\rho, q_1\right)$ is the internal energy density as a function of the mass density and the prognostic variable $q_1$. The thermal equation of state of ideal gases is implicit in this functional context.
The scalar product of two functions of scalar functions $f$, $g$ is defined
\[ \begin{align} \left\langle f \big| g\right\rangle \coloneqq \int_Af^\star\left(\mathbf{r}, t\right)g\left(\mathbf{r}, t\right)d^3r. \end{align} \]
This corresponds to the unitary product Eq. defined in quantum mechanics. (4.1). Real- and complex-valued functions are allowed for $f$ and $g$. In the case of real functions, the complex conjugation of $f$ does not matter. For vector-valued functions one defines analogously
\[ \begin{align} \left\langle\mathbf{f}\big|\mathbf{g}\right\rangle \coloneqq \int_A\mathbf{f}^\star\left(\mathbf{r}, t\right)\cdot\mathbf{g}\left(\mathbf{r}, t\right)d^3r. \end{align} \]
Let $F$ be any functional of the atmospheric state variables. The time evolution of $F$ can be calculated according to Eq. (2.74) using the Poisson bracket as
\[ \begin{align} \frac{dF}{dt} = \left\lbrace F, H\right\rbrace + \frac{\partial F}{\partial t}. \end{align} \]
note down. If $F$ does not depend explicitly on time, which is usually the case, it follows
\[ \begin{align} \frac{dF}{dt} = \left\lbrace F, H\right\rbrace.\tag{10.6}\label{eq:poisson_bracket_non_explicit_t} \end{align} \]
To derive an explicit expression for $\left\lbrace F, H\right\rbrace$, one begins by entering Eq. (2.73) $K = H$ inserts:
\[ \begin{align} \left\lbrace F, H\right\rbrace \coloneqq \sum_{i = 1}^f\left(\frac{\partial F}{\partial q_i}\frac{\partial H}{\partial p_i} - \frac{\partial F}{\partial p_i}\frac{\partial H}{\partial q_i}\right) \end{align} \]
The atmospheric state variables $\mathbf{u} = \left(\rho, q_1, \mathbf{v}\right)^T$ are now used for the generalized coordinates. However, since $F$ and $H$ are functionals and are generalized coordinate functions, the Poisson bracket must first be generalized for this case. Let $f = f\left(\mathbf{u}\right)$ be a scalar or vector valued function of the atmospheric state functions. One can now find a functional belonging to $f$
\[ \begin{align} F\left[\mathbf{u}\right] &\coloneqq \int_Af\left(\mathbf{u}\right)d^3r \end{align} \]
define. If you derive this in terms of time, you get
\[ \begin{align} \frac{dF\left[\mathbf{u}\right]}{dt} &= \int_A\frac{\partial f\left(\mathbf{u}\right)}{\partial t}d^3r = \int_A\frac{df\left(\mathbf{u}\right)}{d\mathbf{u}}\cdot\frac{\partial\mathbf{u}}{\partial t}d^3r, \end{align} \]
whereby the multidimensional chain rule was used in the last step. This can be further transformed into:
\[ \begin{align} \frac{dF\left[\mathbf{u}\right]}{dt} &= \left\lbrace F, H\right\rbrace = \int_A\frac{df\left(\mathbf{u}\right)}{d\rho}\cdot\frac{\partial\rho}{\partial t}d^3r + \int_A\frac{df\left(\mathbf{u}\right)}{dq_1}\cdot\frac{\partial q_1}{\partial t}d^3r + \int_A\frac{df\left(\mathbf{u}\right)}{d\mathbf{v}}\cdot\frac{\partial\mathbf{v}}{\partial t}d^3r. \end{align} \]
The derivatives of $f$ are the so-called functional derivatives of $F$, one defines
\[ \begin{align} \frac{\delta F}{\delta x} \coloneqq \frac{df}{dx} \end{align} \]
for $x \in \left\{\rho, q_1, \mathbf{v}\right\}$. Therefore applies
\[ \begin{align} \frac{dF\left[\mathbf{u}\right]}{dt} &= \left\lbrace F, H\right\rbrace = \int_A\frac{\delta F}{\delta \rho}\frac{\partial\rho}{\partial t}d^3r + \int_A\frac{\delta F}{\delta q_1}\frac{\partial q_1}{\partial t}d^3r + \int_A\frac{\delta F}{\delta \mathbf{v}}\cdot\frac{\partial\mathbf{v}}{\partial t}d^3r.\tag{10.12}\label{eq:poisson_bracket_atm_gen_deriv_0} \end{align} \]
When choosing $q_1$ you now limit yourself to the sizes for which
\[ \begin{align} \frac{\partial q_1}{\partial t} = -\nabla\cdot\left(q_1\mathbf{v}\right)\tag{10.13}\label{eq:deriv_q_1_entropy} \end{align} \]
applies. Define $q_1' \coloneqq \frac{q_1}{\rho}$, then Eq. (10.13)
\[ \begin{align} \frac{\partial\left(\rho q_1'\right)}{\partial t} &= -\nabla\cdot\left(\rho q_1'\mathbf{v}\right)\nonumber\\ \Leftrightarrow q_1'\frac{\partial\rho}{\partial t} + \rho\frac{\partial q_1'}{\partial t} &= -q_1'\nabla\cdot\left(\rho\mathbf{v}\right) - \rho\mathbf{v}\cdot\nabla q_1'\nonumber\\ \stackrel{\text{Kontinuitätsglg.}}{\Leftrightarrow} \rho\frac{\partial q_1'}{\partial t} &= -\rho\mathbf{v}\cdot\nabla q_1' \Leftrightarrow \md{q_1'} = 0. \end{align} \]
$q_1'$ is therefore a thermodynamic state variable that does not change during adiabatic processes. This implies
\[ \begin{align} q_1' = \frac{q_1}{\rho} = f\left(s\right),\tag{10.15}\label{eq:poisson_bracket_atm_gen_deriv_13} \end{align} \]
where $s$ is the specific entropy.
Define $p_x$ for $x \in \left\{\rho, q_1, \mathbf{v}\right\}$ as the canonical momentum belonging to the state variable $x$. Due to Eq. (2.71) applies
\[ \begin{align} \frac{\delta H}{\delta p_\rho} &= \frac{\partial\rho}{\partial t} = -\nabla\cdot\left(\rho\mathbf{v}\right),\tag{10.16}\label{eq:poisson_bracket_atm_gen_deriv_6}\\ \frac{\delta H}{\delta p_{q_1}} &= \frac{\partial q_1}{\partial t} = -\nabla\cdot\left(q_1\mathbf{v}\right) = -\nabla\cdot\left(\frac{q_1}{\rho}\rho\mathbf{v}\right),\tag{10.17}\label{eq:poisson_bracket_atm_gen_deriv_7}\\ \frac{\delta H}{\delta p_\mathbf{v}} &= \frac{\partial\mathbf{v}}{\partial t} = -\frac{1}{\rho}\nabla p - \frac{\etabi}{\rho}\times\rho\mathbf{v} - \nabla\phi - \nabla k,\tag{10.18}\label{eq:poisson_bracket_atm_gen_deriv_8} \end{align} \]
here $\phi$ is the geopotential and $k$ is the specific kinetic energy. For the globally integrated kinetic energy $K$ of the atmosphere applies
\[ \begin{align} K \coloneqq \int_A\frac{1}{2}\rho\mathbf{v}^2d^3r. \end{align} \]
The sum of globally integrated internal and potential energy $I'$ is
\[ \begin{align} I' \coloneqq \int_A\newtilde{I} + \rho\phi d^3r = \int_A\rho\phi + c_d^{(v)}\rho Td^3r \end{align} \]
This means that $H = K + I'$, since $H$ is the total energy of the atmosphere ($H$ is not explicitly time dependent). It applies regardless of the choice of $q_1$
\[ \begin{align} \frac{\delta I'}{\delta\mathbf{v}} = 0 \Rightarrow \frac{\delta H}{\delta\mathbf{v}} = \frac{\delta K}{\delta\mathbf{v}} = \rho\mathbf{v}.\tag{10.21}\label{eq:poisson_bracket_atm_gen_deriv_9} \end{align} \]
For $K$ still applies
\[ \begin{align} \frac{\delta K}{\delta q_1} &= 0,\tag{10.22}\label{eq:poisson_bracket_atm_gen_deriv_15}\\ \frac{\delta K}{\delta\rho} &= \frac{1}{2}\mathbf{v}^2 = k. \end{align} \]
If you insert this into the equations (10.16) - (10.18), you get
\[ \begin{align} \frac{\delta H}{\delta p_\rho} &= \frac{\partial\rho}{\partial t} = -\nabla\cdot\frac{\delta H}{\delta\mathbf{v}},\tag{10.24}\label{eq:poisson_bracket_atm_gen_deriv_10}\\ \frac{\delta H}{\delta p_{q_1}} &= \frac{\partial q_1}{\partial t} = -\nabla\cdot\left(q_1\mathbf{v}\right) = -\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right),\tag{10.25}\label{eq:poisson_bracket_atm_gen_deriv_11}\\ \frac{\delta H}{\delta p_\mathbf{v}} &= \frac{\partial\mathbf{v}}{\partial t} = -\frac{1}{\rho}\nabla p - \frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}} - \nabla\phi - \nabla\frac{\delta K}{\delta\rho},\tag{10.26}\label{eq:poisson_bracket_atm_gen_deriv_12} \end{align} \]
Now the pressure gradient must be expressed in terms of $H$. According to Eq. (5.91) applies
\[ \begin{align} p &= -\left(\frac{\partial E}{\partial V}\right)_{S, N}, \end{align} \]
where $E$ is the total energy of the system. The system used is an air particle, in which both the mass $m$ and the particle number $N$ are constant. The total energy here corresponds to the internal energy $I$. Therefore applies
\[ \begin{align} p &= -\left(\frac{\partial I/m}{\partial V/m}\right)_{s} = -\left(\frac{\partial i}{\partial V/m}\right)_{s}, \end{align} \]
where $i = I/m$ is the specific internal energy. $\frac{V}{m} = \alpha = \frac{1}{\rho}$ is the specific volume. Therefore applies
\[ \begin{align} p &= -\left(\frac{\partial i}{\partial\alpha}\right)_{s} = -\frac{\partial\rho}{\partial\alpha}\left(\frac{\partial i}{\partial\rho}\right)_{s} = \frac{1}{\alpha^2}\left(\frac{\partial i}{\partial\rho}\right)_{s} = \rho^2\left(\frac{\partial i}{\partial\rho}\right)_{s}. \end{align} \]
It now applies
\[ \begin{align} \left(\frac{\partial i}{\partial\rho}\right)_{s} &= \left(\frac{\partial i}{\partial\rho}\right)_{\newtilde{s}} + \left(\frac{\partial i}{\partial\newtilde{s}}\right)_{\rho}\frac{\partial\newtilde{s}}{\partial\rho}. \end{align} \]
For the internal energy density $\newtilde{I}$ applies
\[ \begin{align} \newtilde{I} = \rho i \Rightarrow i = \frac{\newtilde{I}}{\rho}, \end{align} \]
it follows from this
\[ \begin{align} p &= \rho^2\left(\frac{\partial\newtilde{I}/\rho}{\partial\rho}\right)_{\newtilde{s}} + \rho^2\left(\frac{\partial\newtilde{I}/\rho}{\partial\newtilde{s}}\right)_{\rho}\frac{\partial\newtilde{s}}{\partial\rho} = \rho\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}} - \newtilde{I} + \newtilde{s}\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}. \end{align} \]
This gives the pressure gradient
\[ \begin{align} \nabla p &= \left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}}\nabla\rho + \rho\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}} - \nabla\newtilde{I} + \left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}\nabla\newtilde{s} + \newtilde{s}\nabla\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}\nonumber\\ &= \left(\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}}\nabla\rho + \left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}\nabla\newtilde{s} - \nabla\newtilde{I}\right) + \rho\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}}\newtilde{s}\nabla\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}\nonumber\\ &= \rho\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}} + \newtilde{s}\nabla\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}. \end{align} \]
This therefore applies to the pressure gradient acceleration
\[ \begin{align} -\frac{1}{\rho}\nabla p &= -\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}} - s\nabla\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}.\tag{10.34}\label{eq:poisson_bracket_atm_gen_deriv_14} \end{align} \]
Due to Eq. (10.15) is $\frac{q_1}{\rho} = f\left(s\right)$, which
\[ \begin{align} \left(\frac{\partial i}{\partial x}\right)_s = \left(\frac{\partial i}{\partial x}\right)_{q_1/\rho} \end{align} \]
implied. Thus, Eq. (10.34) to
\[ \begin{align} -\frac{1}{\rho}\nabla p &= -\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{q_1} - \frac{q_1}{\rho}\nabla\left(\frac{\partial\newtilde{I}}{\partial q_1}\right)_{\rho}. \end{align} \]
be generalized. This can be expressed by functional derivatives of the globally integrated internal energy $I$:
\[ \begin{align} -\frac{1}{\rho}\nabla p &= -\nabla\left(\frac{\delta I}{\delta\rho}\right)_{q_1} - \frac{q_1}{\rho}\nabla\left(\frac{\delta I}{\delta q_1}\right)_{\rho}. \end{align} \]
With Eq. (10.22) follows
\[ \begin{align} -\frac{1}{\rho}\nabla p &= -\nabla\left(\frac{\delta I}{\delta\rho}\right)_{q_1} - \frac{q_1}{\rho}\nabla\left(\frac{\delta H}{\delta q_1}\right)_{\rho}. \end{align} \]
Putting this into Eq. (10.26), you get
\[ \begin{align} \frac{\delta H}{\delta p_\mathbf{v}} &= \frac{\partial\mathbf{v}}{\partial t} = -\nabla\left(\frac{\delta I}{\delta\rho}\right)_{q_1} - \frac{q_1}{\rho}\nabla\left(\frac{\delta H}{\delta q_1}\right)_{\rho} - \frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}} - \nabla\phi - \nabla\frac{\delta K}{\delta\rho}\nonumber\\ &= -\nabla\frac{\delta H}{\delta\rho} - \frac{q_1}{\rho}\nabla\frac{\delta H}{\delta q_1} - \frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}}.\tag{10.39}\label{eq:poisson_bracket_atm_gen_deriv_16} \end{align} \]
In the second line, the markings of the quantities kept constant were omitted because the definition of $H$ makes it clear how the derivatives are to be understood. If you put the equations (10.24) - (10.25), (10.39) into Eq. (10.12), you get
\[ \begin{align} \left\lbrace F, H\right\rbrace &= -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}}d^3r - \int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &-\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}} + \nabla\frac{\delta H}{\delta\rho} + \frac{q_1}{\rho}\nabla\frac{\delta H}{\delta q_1}\right)d^3r.\tag{10.40}\label{eq:poisson_bracket_atm_gen_deriv_1} \end{align} \]
Rearrangement results
\[ \begin{align} \left\lbrace F, H\right\rbrace &= -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta\rho}d^3r\nonumber\\ &- \int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right) + \frac{\delta F}{\delta\mathbf{v}}\cdot\frac{q_1}{\rho}\nabla\frac{\delta H}{\delta q_1}d^3r\nonumber\\ &-\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}}\right)d^3r.\tag{10.41}\label{eq:poisson_bracket_atm_gen_deriv_2} \end{align} \]
They apply
\[ \begin{align} \frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta\rho} &= \nabla\cdot\left(\frac{\delta H}{\delta\rho}\frac{\delta F}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta\rho}\nabla\cdot\left(\frac{\delta F}{\delta\mathbf{v}}\right),\\ \frac{\delta F}{\delta\mathbf{v}}\cdot\frac{q_1}{\rho}\nabla\frac{\delta H}{\delta q_1} &= \nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta q_1}\frac{\delta F}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\right). \end{align} \]
Putting this into Eq. (10.41), you get
\[ \begin{align} \left\lbrace F, H\right\rbrace &= -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \nabla\cdot\left(\frac{\delta H}{\delta\rho}\frac{\delta F}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta\rho}\nabla\cdot\left(\frac{\delta F}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &- \int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right) + \nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta q_1}\frac{\delta F}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &-\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}}\right)d^3r. \end{align} \]
You are now leaving
\[ \begin{align} \int_A\nabla\cdot\left(\frac{\delta H}{\delta\rho}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r = \int_A\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta q_1}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r = 0 \end{align} \]
which is satisfied by Gauss's theorem if $\frac{\delta F}{\delta \mathbf{v}}\cdot\mathbf{n} = 0$ on $\partial A$. Under these conditions, the time evolution equation applies to any functional $F$ in a reversible dry atmosphere
\[ \begin{align} \frac{dF}{dt} = \lbrace F, H\rbrace &= -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} - \frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta F}{\delta\mathbf{v}}d^3r\nonumber\\ &- \int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &- \int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}}\right)d^3r.\tag{10.46}\label{eq:poisson_bracket_atm_gen_explicit} \end{align} \]
This results as expected
\[ \begin{align} \lbrace H, H\rbrace &= 0. \end{align} \]
You now define
\[ \begin{align} \lbrace F, H\rbrace_\rho &\coloneqq -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} - \frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta F}{\delta\mathbf{v}}d^3r = -\int_A-\frac{\delta H}{\delta\mathbf{v}}\cdot\nabla\frac{\delta F}{\delta\rho} + \frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta\rho}d^3r\nonumber\\ &= -\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta\rho} - \frac{\delta H}{\delta\mathbf{v}}\cdot\nabla\frac{\delta F}{\delta\rho}d^3r,\\ \lbrace F, H\rbrace_{q_1} &\coloneqq -\int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r = -\int_A-\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\cdot\nabla\frac{\delta F}{\delta q_1} + \frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta q_1}d^3r\nonumber\\ &= -\int_A\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta q_1} - \frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\cdot\nabla\frac{\delta F}{\delta q_1}d^3r \end{align} \]
It becomes clear that each bracket can be formulated in a „ scalar field x divergence“ and in a „ vector field x gradient“ notation. Each of the two brackets is antisymmetric:
\[ \begin{align} \lbrace F, H\rbrace_\rho &= -\lbrace H, F\rbrace_\rho\tag{10.50}\label{eq:poisson_bracket_atm_gen_deriv_3}\\ \lbrace F, H\rbrace_{q_1} &= -\lbrace H, F\rbrace_{q_1}\tag{10.51}\label{eq:poisson_bracket_atm_gen_deriv_4} \end{align} \]
Now we define so-called Nambu brackets by
\[ \begin{align} \lbrace F, g, H\rbrace &\coloneqq -\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{1}{\rho}\frac{\delta g}{\delta\mathbf{v}}\times\frac{\delta H}{\delta\mathbf{v}}\right)d^3r = -\int_A\frac{1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\delta g}{\delta\mathbf{v}}\times\frac{\delta H}{\delta\mathbf{v}}\right)d^3r, \end{align} \]
here $g$ is a functional. For Nambu brackets applies
\[ \begin{align} \lbrace F, g, H\rbrace = \lbrace H, F, g\rbrace = \lbrace g, H, F\rbrace. \end{align} \]
Furthermore applies
\[ \begin{align} \lbrace F, g, H\rbrace = -\lbrace F, H, g\rbrace = -\lbrace g, F, H\rbrace = -\lbrace H, g, F\rbrace.\tag{10.54}\label{eq:poisson_bracket_atm_gen_deriv_5} \end{align} \]
If you put this together with the equations (10.50) - (10.51) in Eq. (10.66), you get
\[ \begin{align} \lbrace H, F\rbrace &= \lbrace H, F\rbrace_\rho + \lbrace H, F\rbrace_{q_1} + \lbrace H, h_a, F\rbrace = -\lbrace F, H\rbrace_\rho - \lbrace F, H\rbrace_{q_1} - \lbrace F, h_a, H\rbrace = -\lbrace F, H\rbrace. \end{align} \]
Now define a reduced helicity $Z$ (for the original definition of helicity see Eq. (15.104)).
\[ \begin{align} Z \coloneqq \frac{1}{2}\mathbf{v}\cdot\nabla\times\mathbf{v} \end{align} \]
and the absolute helicity $Z_a$ by
\[ \begin{align} Z_a \coloneqq \frac{1}{2}\mathbf{v}_a\cdot\nabla\times\mathbf{v}_a = \frac{1}{2}\left(\Omegabi\times\mathbf{r} + \mathbf{v}\right)\cdot\nabla\times\left(\Omegabi\times\mathbf{r} + \mathbf{v}\right), \end{align} \]
this is the helicity measured in the inertial system. The corresponding functionals are given by
\[ \begin{align} Z &\coloneqq \int_A\frac{1}{2}\mathbf{v}\cdot\nabla\times\mathbf{v}d^3r,\\ Z_a &\coloneqq \int_A\frac{1}{2}\mathbf{v}_a\cdot\nabla\times\mathbf{v}_ad^3r \end{align} \]
defined. Now let $\mathbf{h}: A \to \mathbb{R}^3$ be a test function with $\mathbf{h}\left(\partial A\right) = \mathbf{0}$ and $\epsilon > 0$. You define
\[ \begin{align} Z'\left(\epsilon\right) &\coloneqq \int_A\frac{1}{2}\left(\mathbf{v} + \epsilon\mathbf{h}\right)\cdot\nabla\times\left(\mathbf{v} + \epsilon\mathbf{h}\right)d^3r = \int_A\frac{1}{2}\mathbf{v}\cdot\nabla\times\mathbf{v}d^3r + \int_A\frac{\epsilon}{2}\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r + \int_A\frac{\epsilon}{2}\mathbf{v}\cdot\nabla\times\mathbf{h}d^3r + \mathcal{O}\left(\epsilon^2\right). \end{align} \]
The equation now applies to first order in $\epsilon$
\[ \begin{align} \frac{Z'\left(\epsilon\right) - Z'\left(0\right)}{\epsilon} &= \frac{Z'\left(\epsilon\right) - Z}{\epsilon} = \int_A\frac{1}{2}\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r + \int_A\frac{1}{2}\mathbf{v}\cdot\nabla\times\mathbf{h}d^3r = \frac{1}{2}\int_A\mathbf{h}\cdot\nabla\times\mathbf{v} + \mathbf{v}\cdot\nabla\times\mathbf{h}d^3r\nonumber\\ &= \frac{1}{2}\int_A\mathbf{v}\cdot\nabla\times\mathbf{h} - \mathbf{h}\cdot\nabla\times\mathbf{v} + 2\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r \stackrel{\href{ch-40-vector-analysis.html#eq:diff_op_rule_9}{\text{Glg. (B.55)}}}{=} \frac{1}{2}\int_A\nabla\cdot\left(\mathbf{h}\times\mathbf{v}\right) + 2\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r\nonumber\\ &= \int_A\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r. \end{align} \]
In the last step, Gauss's theorem and the fact $\mathbf{h}\left(\partial A\right) = \mathbf{0}$ were used. Therefore applies
\[ \begin{align} \frac{\delta Z}{\delta\mathbf{v}} = \nabla\times\mathbf{v} = \zetabi. \end{align} \]
The same applies too
\[ \begin{align} \frac{\delta Z_a}{\delta\mathbf{v}_a} = \nabla\times\mathbf{v}_a = \etabi. \end{align} \]
Due to
\[ \begin{align} \frac{\delta Z}{\delta\mathbf{v}} = \frac{\delta H}{\delta\mathbf{v}_a} \end{align} \]
also applies
\[ \begin{align} \frac{\delta Z_a}{\delta\mathbf{v}} = \etabi. \end{align} \]
Thus, one can use Eq. (10.46) in the form
\[ \begin{align} \frac{dF}{dt} = \lbrace F, H\rbrace &= \lbrace F, Z_a, H + \lbrace F, H\rbrace_\rho + \lbrace F, H\rbrace_{q_1}\rbrace\tag{10.66}\label{eq:poisson_bracket_atm_gen_parts} \end{align} \]
write.
The Poisson bracket formalism leads to a special form of pressure gradient acceleration. In addition to $\rho$ and $q_1$, this usually contains a third variable, and occasionally even a fourth or fifth. One of these variables is chosen as the so-called semi-prognostic variable $q_2$. Due to the thermal equation of state of ideal gases, this is no longer an independent prognostic variable, as the thermodynamic state would otherwise be overdetermined, but it is distinguished from other diagnostic variables because it plays a role in the evolution of forms of energy.
For energy forms $F$ the fact $\frac{\delta F}{\delta \mathbf{v}} \parallel \mathbf{v}$ applies, so in this case applies
\[ \begin{align} \lbrace F, H_a, H\rbrace = 0 \Rightarrow \lbrace F, H\rbrace = \lbrace F, H\rbrace_\rho + \lbrace F, H\rbrace_{q_1}. \end{align} \]
Follow from this
\[ \begin{align} \lbrace K, H\rbrace &= \int_A-\frac{\delta K}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta K}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta K}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A-\frac{\delta K}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A\frac{\delta\left(H - K\right)}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A\frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r,\\ \lbrace I', H\rbrace &= \int_A-\frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} - \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r. \end{align} \]
If you add both, you get as expected
\[ \begin{align} \lbrace K, H\rbrace + \lbrace I', H\rbrace = \lbrace H, H\rbrace = 0. \end{align} \]
Use the potential temperature instead of the temperature and define
\[ \begin{align} q_1 \coloneqq \newtilde{\theta} = \rho\theta, \end{align} \]
follows with Eq. (9.65)
\[ \begin{align} H &= H\left(\mathbf{v}, \rho, q_1\right) = H\left(\mathbf{v}, \rho, \newtilde{\theta}\right) = \int_A\frac{1}{2}\rho\mathbf{v}^2 + \rho\phi + c_d^{(v)}\frac{p\left(\rho, \newtilde{\theta}\right)}{R_d}d^3r\nonumber\\ &= \int_A\frac{1}{2}\rho\mathbf{v}^2 + \rho\phi + c_d^{(v)}\frac{p_0}{R_d}\left(\frac{R_d\newtilde{\theta}}{p_0}\right)^\kappa d^3r = \int_A\frac{1}{2}\rho\mathbf{v}^2 + \rho\phi + c_d^{(v)}\frac{p_0}{R_d}\left(\frac{R_d q_1}{p_0}\right)^\kappa d^3r. \end{align} \]
The functional derivatives of this are
\[ \begin{align} \frac{\delta H}{\delta\rho} &= \frac{1}{2}\mathbf{v}^2 + \phi,\\ \frac{\delta H}{\delta q_1} &= c_d^{(v)}\frac{p_0}{R_d}\frac{R_d}{p_0}\kappa\left(\frac{R_d q_1}{p_0}\right)^{\kappa - 1} = c_d^{(p)}\left(\frac{R_d q_1}{p_0}\right)^{\kappa - 1} = c_d^{(p)}\left(\frac{R_d q_1}{p_0}\right)^{\frac{R_d}{c_d^{(v)}}} \stackrel{\href{ch-08-first-law-in-the-atmosphere.html#eq:exner_pressure_diag}{\text{Glg. (9.71)}}}{=} c_d^{(p)}\Pi,\\ \frac{\delta H}{\delta\mathbf{v}} &= \rho\mathbf{v}. \end{align} \]
Now the time developments of the quantities $K$ and $I'$ should be examined. This includes preparation
\[ \begin{align} \frac{\delta K}{\delta\rho} = \frac{1}{2}\mathbf{v}^2, & {} & \frac{\delta K}{\delta q_1} = 0, & {} & \frac{\delta K}{\delta\mathbf{v}} = \rho\mathbf{v},\\ \frac{\delta I'}{\delta\rho} = \phi, & {} & \frac{\delta I'}{\delta q_1} = c_d^{(p)}\Pi, \frac{\delta I'}{\delta\mathbf{v}} = 0. \end{align} \]
and thus receives
\[ \begin{align} \lbrace K, H\rbrace &= \int_A\frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta K}{\delta\mathbf{v}} - \frac{\delta K}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta K}{\delta\mathbf{v}}\right) - \frac{\delta K}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A\left(\frac{1}{2}\mathbf{v}^2 + \phi\right)\nabla\cdot\left(\rho\mathbf{v}\right) - \frac{1}{2}\mathbf{v}^2\nabla\cdot\left(\rho\mathbf{v}\right) + c_d^{(p)}\Pi\nabla\cdot\left(q_1\mathbf{v}\right)d^3r\nonumber\\ &= \int_A\phi\nabla\cdot\left(\rho\mathbf{v}\right) + c_d^{(p)}\Pi\nabla\cdot\left(q_1\mathbf{v}\right)d^3r\nonumber\\ &= \int_A-\rho\mathbf{v}\cdot\nabla\phi - c_d^{(p)}q_1\mathbf{v}\cdot\nabla\Pi d^3r\tag{10.78}\label{eq:poisson_entropy_kinetic}, \end{align} \] \[ \begin{align} \lbrace I', H\rbrace &= \int_A\frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta I'}{\delta\mathbf{v}} - \frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta I'}{\delta\mathbf{v}}\right) - \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A-\phi\nabla\cdot\left(\rho\mathbf{v}\right) - c_d^{(p)}\Pi\nabla\cdot\left(q_1\mathbf{v}\right)d^3r.\tag{10.79}\label{eq:poisson_entropy_potential_internal} \end{align} \]
If you use the entropy instead of the potential temperature and define
\[ \begin{align} q_1 \coloneqq \newtilde{s} = \rho s, \end{align} \]
follows
\[ \begin{align} H &= H\left(\mathbf{v}, \rho, q_1\right) = H\left(\mathbf{v}, \rho, \newtilde{s}\right) = \int_A\frac{1}{2}\rho\mathbf{v}^2 + \rho\phi + c_d^{(v)}\frac{p\left(\rho, \newtilde{s}\right)}{R_d}d^3r\nonumber\\ &= \int_A\frac{1}{2}\rho\mathbf{v}^2 + \rho\phi + c_d^{(v)}\frac{p\left(\rho, q_1\right)}{R_d}d^3r.\tag{10.81}\label{eq:hamilton-functional_entropy} \end{align} \]
The functional derivatives of this are
\[ \begin{align} \frac{\delta H}{\delta\rho} \stackrel{\href{ch-04-statistical-physics.html#eq:chemisches_potential_prop_0}{\text{Glg. (5.91)}}}{=} \frac{1}{2}\mathbf{v}^2 + \phi + G, & {} & \frac{\delta H}{\delta q_1} \stackrel{\href{ch-04-statistical-physics.html#eq:chemisches_potential_prop_0}{\text{Glg. (5.91)}}}{=} T, & {} & \frac{\delta H}{\delta\mathbf{v}} = \rho\mathbf{v}. \end{align} \]
$G$ denotes the specific Gibbs potential. Now the time developments of the quantities $K$ and $I'$ are examined again. This includes preparation
\[ \begin{align} \frac{\delta K}{\delta\rho} = \frac{1}{2}\mathbf{v}^2, & {} & \frac{\delta K}{\delta q_1} = 0, & {} & \frac{\delta K}{\delta\mathbf{v}} = \rho\mathbf{v},\\ \frac{\delta I'}{\delta\rho} = \phi + G, & {} & \frac{\delta I'}{\delta q_1} = T, & {} & \frac{\delta I'}{\delta\mathbf{v}} = 0. \end{align} \]
and thus receives
\[ \begin{align} \lbrace K, H\rbrace &= \int_A\frac{\delta H}{\delta\rho}\nabla\cdot\left(\rho\mathbf{v}\right) - \frac{\delta K}{\delta\rho}\nabla\cdot\left(\rho\mathbf{v}\right) + T\nabla\cdot\left(q_1\mathbf{v}\right)d^3r\nonumber\\ &= \int_A\left(\phi + G\right)\nabla\cdot\left(\rho\mathbf{v}\right) + T\nabla\cdot\left(q_1\mathbf{v}\right)d^3r\nonumber\\ &= \int_A-\rho\mathbf{v}\cdot\nabla\left(\phi + G\right) - q_1\mathbf{v}\cdot\nabla Td^3r,\\ \lbrace I', H\rbrace &= \int_A\frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta I'}{\delta\mathbf{v}} - \frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta I'}{\delta\mathbf{v}}\right) - \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A-\frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} - \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A-\left(\phi + G\right)\nabla\cdot\left(\rho\mathbf{v}\right) - T\nabla\cdot\left(q_1\mathbf{v}\right)d^3r. \end{align} \]
This gives the expression for the pressure gradient acceleration
\[ \begin{align} -\frac{1}{\rho}\nabla p = -\nabla G - s\nabla T.\tag{10.87}\label{eq:entropy_form_of_pressure_gradient} \end{align} \]
This is also carried out
\[ \begin{align} -\nabla G - s\nabla T &= -c_d^{(p)}\nabla T + T\nabla s = -c_d^{(p)}\nabla T + Tc_d^{(p)}\nabla\ln\left(\theta\right)\nonumber\\ &= -c_d^{(p)}\nabla T + \frac{T}{\theta}c_d^{(p)}\nabla\theta = -c_d^{(p)}\theta\nabla\frac{T}{\theta} = -c_d^{(p)}\theta\nabla\Pi \end{align} \]
confirmed.
This section is only formulated using the entropy formula started in Section 10.1.3. In an irreversibly dry atmosphere, two irreversible brackets are defined:
\[ \begin{align} \left(F, \mathbf{f}_R\right) & \coloneqq \left\langle\frac{\delta F}{\delta\mathbf{v}}\Big| \mathbf{f}_R\right\rangle = \int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\mathbf{f}_Rd^3r\\ \left(F, Q\right) & \coloneqq \left\langle\frac{\delta F}{\delta\newtilde{s}}\Big|\frac{\rho\epsilon}{T}\right\rangle = \int_A\frac{\delta F}{\delta\newtilde{s}}\frac{\rho\epsilon}{T}d^3r \end{align} \]
According to Eq. (8.96) applies here
\[ \begin{align} \left(H, \mathbf{f}_R\right) + \left(H, Q\right) = 0. \end{align} \]
The dynamics are then determined by
\[ \begin{align} \frac{dF}{dt} = \lbrace F, H\rbrace + \left(F, \mathbf{f}_R\right) + \left(F, Q\right). \end{align} \]
According to the findings of section 9.5.1, the replacement can be carried out in a humid atmosphere without condensates
\[ \begin{align} \theta \to \theta_v \end{align} \]
take over the reversible part of the dynamics according to Sect. 10.1.2 if one assumes that dry air and water vapor are perfect ideal gases.
Condensates move at a speed $\mathbf{v}_i \not= \mathbf{v}$, which does not correspond to the wind speed because they have a superimposed fall speed. This must be taken into account in the global integral of the kinetic energy. This would generate some additional terms. However, this is not practical relative to the benefits. Instead, the existence of the condensates is linked to the dynamics with the following interventions:
When solving the radiation transfer equation, it is assumed that the radiation field is in equilibrium with the temperature and density field at all times. The time derivatives of the radiation field are therefore neglected (quasi-stationary process). If one wanted to embed the radiation field in the Hamiltonian formalism, this would lead to equations with time derivatives for the radiation field that would have to be solved on such small time scales that this is not practical for the atmosphere.