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. For the Hamilton function of the atmosphere, one then writes
\[ \begin{align} H = H\left(\rho, q_1, \mathbf{v}\right) = \int_A\frac{1}{2}\rho\mathbf{v}^2 + \rho\phi + \newtilde{I}\left(\rho, q_1\right)d^3r. \end{align} \]
$\newtilde{I}\left(\rho, q_1\right)$ 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.
As the scalar product of two scalar functions $f$, $g$, one defines
\[ \begin{align} \left\langle f \big| g\right\rangle \coloneqq \int_Af^\star\left(\mathbf{r}, t\right)g\left(\mathbf{r}, t\right)d^3r. \end{align} \]
This corresponds to the unitary product defined in quantum mechanics, Eq. (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
\[ \begin{align} \frac{dF}{dt} = \left\lbrace F, H\right\rbrace + \frac{\partial F}{\partial t}. \end{align} \]
If $F$ does not depend explicitly on time, which is usually the case, it follows
To derive an explicit expression for $\left\lbrace F, H\right\rbrace$, one begins by inserting $K = H$ into Eq. (2.73):
\[ \begin{align} \left\lbrace F, H\right\rbrace \coloneqq \sum_{i = 1}^f\left(\frac{\partial F}{\partial q_i}\frac{\partial H}{\partial p_i} - \frac{\partial F}{\partial p_i}\frac{\partial H}{\partial q_i}\right) \end{align} \]
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 the generalized coordinates are functions, the Poisson bracket must first be generalized to this case. Let $f = f\left(\mathbf{u}\right)$ be a scalar or vector valued function of the atmospheric state functions. One can now define a functional belonging to $f$
\[ \begin{align} F\left[\mathbf{u}\right] &\coloneqq \int_Af\left(\mathbf{u}\right)d^3r \end{align} \]
Differentiating this with respect to time, one obtains
\[ \begin{align} \frac{dF\left[\mathbf{u}\right]}{dt} &= \int_A\frac{\partial f\left(\mathbf{u}\right)}{\partial t}d^3r = \int_A\frac{df\left(\mathbf{u}\right)}{d\mathbf{u}}\cdot\frac{\partial\mathbf{u}}{\partial t}d^3r, \end{align} \]
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 $x \in \left\{\rho, q_1, \mathbf{v}\right\}$. Thus one has
\[ \begin{align} \frac{dF\left[\mathbf{u}\right]}{dt} &= \left\lbrace F, H\right\rbrace = \int_A\frac{\delta F}{\delta \rho}\frac{\partial\rho}{\partial t}d^3r + \int_A\frac{\delta F}{\delta q_1}\frac{\partial q_1}{\partial t}d^3r + \int_A\frac{\delta F}{\delta \mathbf{v}}\cdot\frac{\partial\mathbf{v}}{\partial t}d^3r.\tag{10.12}\label{eq:poisson_bracket_atm_gen_deriv_0} \end{align} \]
When choosing $q_1$, one now restricts oneself to quantities for which
\[ \begin{align} \frac{\partial q_1}{\partial t} = -\nabla\cdot\left(q_1\mathbf{v}\right)\tag{10.13}\label{eq:deriv_q_1_entropy} \end{align} \]
holds. Defining $q_1' \coloneqq \frac{q_1}{\rho}$, it follows from Eq. (10.13)
\[ \begin{align} \frac{\partial\left(\rho q_1'\right)}{\partial t} &= -\nabla\cdot\left(\rho q_1'\mathbf{v}\right)\nonumber\\ \Leftrightarrow q_1'\frac{\partial\rho}{\partial t} + \rho\frac{\partial q_1'}{\partial t} &= -q_1'\nabla\cdot\left(\rho\mathbf{v}\right) - \rho\mathbf{v}\cdot\nabla q_1'\nonumber\\ \stackrel{\text{continuity eq.}}{\Leftrightarrow} \rho\frac{\partial q_1'}{\partial t} &= -\rho\mathbf{v}\cdot\nabla q_1' \Leftrightarrow \md{q_1'} = 0. \end{align} \]
$q_1'$ 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 $x$. Owing to Eq. (2.71), one has
\[ \begin{align} \frac{\delta H}{\delta p_\rho} &= \frac{\partial\rho}{\partial t} = -\nabla\cdot\left(\rho\mathbf{v}\right),\tag{10.16}\label{eq:poisson_bracket_atm_gen_deriv_6}\\ \frac{\delta H}{\delta p_{q_1}} &= \frac{\partial q_1}{\partial t} = -\nabla\cdot\left(q_1\mathbf{v}\right) = -\nabla\cdot\left(\frac{q_1}{\rho}\rho\mathbf{v}\right),\tag{10.17}\label{eq:poisson_bracket_atm_gen_deriv_7}\\ \frac{\delta H}{\delta p_\mathbf{v}} &= \frac{\partial\mathbf{v}}{\partial t} = -\frac{1}{\rho}\nabla p - \frac{\etabi}{\rho}\times\rho\mathbf{v} - \nabla\phi - \nabla k,\tag{10.18}\label{eq:poisson_bracket_atm_gen_deriv_8} \end{align} \]
here $\phi$ is the geopotential and $k$ the specific kinetic energy. For the globally integrated kinetic energy $K$ of the atmosphere, one has
\[ \begin{align} K \coloneqq \int_A\frac{1}{2}\rho\mathbf{v}^2d^3r. \end{align} \]
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). Independently of the choice of $q_1$, one has
\[ \begin{align} \frac{\delta I'}{\delta\mathbf{v}} = 0 \Rightarrow \frac{\delta H}{\delta\mathbf{v}} = \frac{\delta K}{\delta\mathbf{v}} = \rho\mathbf{v}.\tag{10.21}\label{eq:poisson_bracket_atm_gen_deriv_9} \end{align} \]
For $K$, one furthermore has
\[ \begin{align} \frac{\delta K}{\delta q_1} &= 0,\tag{10.22}\label{eq:poisson_bracket_atm_gen_deriv_15}\\ \frac{\delta K}{\delta\rho} &= \frac{1}{2}\mathbf{v}^2 = k. \end{align} \]
Substituting this into Eqs. (10.16) - (10.18), one obtains
\[ \begin{align} \frac{\delta H}{\delta p_\rho} &= \frac{\partial\rho}{\partial t} = -\nabla\cdot\frac{\delta H}{\delta\mathbf{v}},\tag{10.24}\label{eq:poisson_bracket_atm_gen_deriv_10}\\ \frac{\delta H}{\delta p_{q_1}} &= \frac{\partial q_1}{\partial t} = -\nabla\cdot\left(q_1\mathbf{v}\right) = -\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right),\tag{10.25}\label{eq:poisson_bracket_atm_gen_deriv_11}\\ \frac{\delta H}{\delta p_\mathbf{v}} &= \frac{\partial\mathbf{v}}{\partial t} = -\frac{1}{\rho}\nabla p - \frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}} - \nabla\phi - \nabla\frac{\delta K}{\delta\rho},\tag{10.26}\label{eq:poisson_bracket_atm_gen_deriv_12} \end{align} \]
Now the pressure gradient must be expressed in terms of $H$. According to Eq. (5.91), one has
\[ \begin{align} p &= -\left(\frac{\partial E}{\partial V}\right)_{S, N}, \end{align} \]
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 $I$. Thus one has
\[ \begin{align} p &= -\left(\frac{\partial I/m}{\partial V/m}\right)_{s} = -\left(\frac{\partial i}{\partial V/m}\right)_{s}, \end{align} \]
where $i = I/m$ is the specific internal energy. $\frac{V}{m} = \alpha = \frac{1}{\rho}$ is the specific volume. Thus one has
\[ \begin{align} p &= -\left(\frac{\partial i}{\partial\alpha}\right)_{s} = -\frac{\partial\rho}{\partial\alpha}\left(\frac{\partial i}{\partial\rho}\right)_{s} = \frac{1}{\alpha^2}\left(\frac{\partial i}{\partial\rho}\right)_{s} = \rho^2\left(\frac{\partial i}{\partial\rho}\right)_{s}. \end{align} \]
One now has
\[ \begin{align} \left(\frac{\partial i}{\partial\rho}\right)_{s} &= \left(\frac{\partial i}{\partial\rho}\right)_{\newtilde{s}} + \left(\frac{\partial i}{\partial\newtilde{s}}\right)_{\rho}\frac{\partial\newtilde{s}}{\partial\rho}. \end{align} \]
For the internal energy density $\newtilde{I}$, one has
\[ \begin{align} \newtilde{I} = \rho i \Rightarrow i = \frac{\newtilde{I}}{\rho}, \end{align} \]
From this it follows
\[ \begin{align} p &= \rho^2\left(\frac{\partial\newtilde{I}/\rho}{\partial\rho}\right)_{\newtilde{s}} + \rho^2\left(\frac{\partial\newtilde{I}/\rho}{\partial\newtilde{s}}\right)_{\rho}\frac{\partial\newtilde{s}}{\partial\rho} = \rho\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}} - \newtilde{I} + \newtilde{s}\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}. \end{align} \]
Thus, for the pressure gradient, one obtains
\[ \begin{align} \nabla p &= \left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}}\nabla\rho + \rho\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}} - \nabla\newtilde{I} + \left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}\nabla\newtilde{s} + \newtilde{s}\nabla\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}\nonumber\\ &= \left(\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}}\nabla\rho + \left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}\nabla\newtilde{s} - \nabla\newtilde{I}\right) + \rho\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}}\newtilde{s}\nabla\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}\nonumber\\ &= \rho\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}} + \newtilde{s}\nabla\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}. \end{align} \]
Thus, for the pressure gradient acceleration, one has
\[ \begin{align} -\frac{1}{\rho}\nabla p &= -\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{\newtilde{s}} - s\nabla\left(\frac{\partial\newtilde{I}}{\partial\newtilde{s}}\right)_{\rho}.\tag{10.34}\label{eq:poisson_bracket_atm_gen_deriv_14} \end{align} \]
Owing to Eq. (10.15), $\frac{q_1}{\rho} = f\left(s\right)$, which implies
\[ \begin{align} \left(\frac{\partial i}{\partial x}\right)_s = \left(\frac{\partial i}{\partial x}\right)_{q_1/\rho} \end{align} \]
Thus, Eq. (10.34) can be generalized to
\[ \begin{align} -\frac{1}{\rho}\nabla p &= -\nabla\left(\frac{\partial\newtilde{I}}{\partial\rho}\right)_{q_1} - \frac{q_1}{\rho}\nabla\left(\frac{\partial\newtilde{I}}{\partial q_1}\right)_{\rho}. \end{align} \]
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), it follows
\[ \begin{align} -\frac{1}{\rho}\nabla p &= -\nabla\left(\frac{\delta I}{\delta\rho}\right)_{q_1} - \frac{q_1}{\rho}\nabla\left(\frac{\delta H}{\delta q_1}\right)_{\rho}. \end{align} \]
Substituting this into Eq. (10.26), one obtains
\[ \begin{align} \frac{\delta H}{\delta p_\mathbf{v}} &= \frac{\partial\mathbf{v}}{\partial t} = -\nabla\left(\frac{\delta I}{\delta\rho}\right)_{q_1} - \frac{q_1}{\rho}\nabla\left(\frac{\delta H}{\delta q_1}\right)_{\rho} - \frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}} - \nabla\phi - \nabla\frac{\delta K}{\delta\rho}\nonumber\\ &= -\nabla\frac{\delta H}{\delta\rho} - \frac{q_1}{\rho}\nabla\frac{\delta H}{\delta q_1} - \frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}}.\tag{10.39}\label{eq:poisson_bracket_atm_gen_deriv_16} \end{align} \]
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. Substituting Eqs. (10.24) - (10.25), (10.39) into Eq. (10.12), one obtains
\[ \begin{align} \left\lbrace F, H\right\rbrace &= -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}}d^3r - \int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &-\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}} + \nabla\frac{\delta H}{\delta\rho} + \frac{q_1}{\rho}\nabla\frac{\delta H}{\delta q_1}\right)d^3r.\tag{10.40}\label{eq:poisson_bracket_atm_gen_deriv_1} \end{align} \]
Rearrangement yields
\[ \begin{align} \left\lbrace F, H\right\rbrace &= -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta\rho}d^3r\nonumber\\ &- \int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right) + \frac{\delta F}{\delta\mathbf{v}}\cdot\frac{q_1}{\rho}\nabla\frac{\delta H}{\delta q_1}d^3r\nonumber\\ &-\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}}\right)d^3r.\tag{10.41}\label{eq:poisson_bracket_atm_gen_deriv_2} \end{align} \]
The following hold
\[ \begin{align} \frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta\rho} &= \nabla\cdot\left(\frac{\delta H}{\delta\rho}\frac{\delta F}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta\rho}\nabla\cdot\left(\frac{\delta F}{\delta\mathbf{v}}\right),\\ \frac{\delta F}{\delta\mathbf{v}}\cdot\frac{q_1}{\rho}\nabla\frac{\delta H}{\delta q_1} &= \nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta q_1}\frac{\delta F}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\right). \end{align} \]
Substituting this into Eq. (10.41), one obtains
\[ \begin{align} \left\lbrace F, H\right\rbrace &= -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \nabla\cdot\left(\frac{\delta H}{\delta\rho}\frac{\delta F}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta\rho}\nabla\cdot\left(\frac{\delta F}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &- \int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right) + \nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta q_1}\frac{\delta F}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &-\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}}\right)d^3r. \end{align} \]
One now starts from
\[ \begin{align} \int_A\nabla\cdot\left(\frac{\delta H}{\delta\rho}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r = \int_A\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta q_1}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r = 0 \end{align} \]
which is satisfied by Gauss's theorem if $\frac{\delta F}{\delta \mathbf{v}}\cdot\mathbf{n} = 0$ on $\partial A$. Under these conditions, in a reversible dry atmosphere, the time evolution equation holds for arbitrary functionals $F$:
\[ \begin{align} \frac{dF}{dt} = \lbrace F, H\rbrace &= -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} - \frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta F}{\delta\mathbf{v}}d^3r\nonumber\\ &- \int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &- \int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\left(\frac{\etabi}{\rho}\times\frac{\delta H}{\delta\mathbf{v}}\right)d^3r.\tag{10.46}\label{eq:poisson_bracket_atm_gen_explicit} \end{align} \]
From this, as expected, one obtains
\[ \begin{align} \lbrace H, H\rbrace &= 0. \end{align} \]
One now defines
\[ \begin{align} \lbrace F, H\rbrace_\rho &\coloneqq -\int_A\frac{\delta F}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} - \frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta F}{\delta\mathbf{v}}d^3r = -\int_A-\frac{\delta H}{\delta\mathbf{v}}\cdot\nabla\frac{\delta F}{\delta\rho} + \frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta\rho}d^3r\nonumber\\ &= -\int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta\rho} - \frac{\delta H}{\delta\mathbf{v}}\cdot\nabla\frac{\delta F}{\delta\rho}d^3r,\\ \lbrace F, H\rbrace_{q_1} &\coloneqq -\int_A\frac{\delta F}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right) - \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\right)d^3r = -\int_A-\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\cdot\nabla\frac{\delta F}{\delta q_1} + \frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta q_1}d^3r\nonumber\\ &= -\int_A\frac{q_1}{\rho}\frac{\delta F}{\delta\mathbf{v}}\cdot\nabla\frac{\delta H}{\delta q_1} - \frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\cdot\nabla\frac{\delta F}{\delta q_1}d^3r \end{align} \]
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:
\[ \begin{align} \lbrace F, H\rbrace_\rho &= -\lbrace H, F\rbrace_\rho\tag{10.50}\label{eq:poisson_bracket_atm_gen_deriv_3}\\ \lbrace F, H\rbrace_{q_1} &= -\lbrace H, F\rbrace_{q_1}\tag{10.51}\label{eq:poisson_bracket_atm_gen_deriv_4} \end{align} \]
One now defines 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, one has
\[ \begin{align} \lbrace F, g, H\rbrace = \lbrace H, F, g\rbrace = \lbrace g, H, F\rbrace. \end{align} \]
Furthermore, one has
\[ \begin{align} \lbrace F, g, H\rbrace = -\lbrace F, H, g\rbrace = -\lbrace g, F, H\rbrace = -\lbrace H, g, F\rbrace.\tag{10.54}\label{eq:poisson_bracket_atm_gen_deriv_5} \end{align} \]
Substituting this together with Eqs. (10.50) - (10.51) into Eq. (10.66), one obtains
\[ \begin{align} \lbrace H, F\rbrace &= \lbrace H, F\rbrace_\rho + \lbrace H, F\rbrace_{q_1} + \lbrace H, h_a, F\rbrace = -\lbrace F, H\rbrace_\rho - \lbrace F, H\rbrace_{q_1} - \lbrace F, h_a, H\rbrace = -\lbrace F, H\rbrace. \end{align} \]
Now define a reduced helicity $Z$ (for the original definition of helicity, see Eq. (15.104)) by
\[ \begin{align} Z \coloneqq \frac{1}{2}\mathbf{v}\cdot\nabla\times\mathbf{v} \end{align} \]
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
\[ \begin{align} Z &\coloneqq \int_A\frac{1}{2}\mathbf{v}\cdot\nabla\times\mathbf{v}d^3r,\\ Z_a &\coloneqq \int_A\frac{1}{2}\mathbf{v}_a\cdot\nabla\times\mathbf{v}_ad^3r \end{align} \]
Now let $\mathbf{h}: A \to \mathbb{R}^3$ be a test function with $\mathbf{h}\left(\partial A\right) = \mathbf{0}$ and $\epsilon > 0$. One defines
\[ \begin{align} Z'\left(\epsilon\right) &\coloneqq \int_A\frac{1}{2}\left(\mathbf{v} + \epsilon\mathbf{h}\right)\cdot\nabla\times\left(\mathbf{v} + \epsilon\mathbf{h}\right)d^3r = \int_A\frac{1}{2}\mathbf{v}\cdot\nabla\times\mathbf{v}d^3r + \int_A\frac{\epsilon}{2}\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r + \int_A\frac{\epsilon}{2}\mathbf{v}\cdot\nabla\times\mathbf{h}d^3r + \mathcal{O}\left(\epsilon^2\right). \end{align} \]
To first order in $\epsilon$, one now has the equation
\[ \begin{align} \frac{Z'\left(\epsilon\right) - Z'\left(0\right)}{\epsilon} &= \frac{Z'\left(\epsilon\right) - Z}{\epsilon} = \int_A\frac{1}{2}\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r + \int_A\frac{1}{2}\mathbf{v}\cdot\nabla\times\mathbf{h}d^3r = \frac{1}{2}\int_A\mathbf{h}\cdot\nabla\times\mathbf{v} + \mathbf{v}\cdot\nabla\times\mathbf{h}d^3r\nonumber\\ &= \frac{1}{2}\int_A\mathbf{v}\cdot\nabla\times\mathbf{h} - \mathbf{h}\cdot\nabla\times\mathbf{v} + 2\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r \stackrel{\href{ch-40-vector-analysis.html#eq:diff_op_rule_9}{\text{Eq. (B.55)}}}{=} \frac{1}{2}\int_A\nabla\cdot\left(\mathbf{h}\times\mathbf{v}\right) + 2\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r\nonumber\\ &= \int_A\mathbf{h}\cdot\nabla\times\mathbf{v}d^3r. \end{align} \]
In the last step, Gauss's theorem and the fact that $\mathbf{h}\left(\partial A\right) = \mathbf{0}$ were used. Thus one has
\[ \begin{align} \frac{\delta Z}{\delta\mathbf{v}} = \nabla\times\mathbf{v} = \zetabi. \end{align} \]
Analogously, one also has
\[ \begin{align} \frac{\delta Z_a}{\delta\mathbf{v}_a} = \nabla\times\mathbf{v}_a = \etabi. \end{align} \]
Owing to
\[ \begin{align} \frac{\delta Z}{\delta\mathbf{v}} = \frac{\delta H}{\delta\mathbf{v}_a} \end{align} \]
one also has
\[ \begin{align} \frac{\delta Z_a}{\delta\mathbf{v}} = \etabi. \end{align} \]
Thus, one can write Eq. (10.46) in the form
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 $\frac{\delta F}{\delta \mathbf{v}} \parallel \mathbf{v}$ holds, so in this case one has
\[ \begin{align} \lbrace F, H_a, H\rbrace = 0 \Rightarrow \lbrace F, H\rbrace = \lbrace F, H\rbrace_\rho + \lbrace F, H\rbrace_{q_1}. \end{align} \]
From this follow
\[ \begin{align} \lbrace K, H\rbrace &= \int_A-\frac{\delta K}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta K}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta K}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A-\frac{\delta K}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A\frac{\delta\left(H - K\right)}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A\frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r,\\ \lbrace I', H\rbrace &= \int_A-\frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} - \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r. \end{align} \]
Adding both, one obtains, as expected,
\[ \begin{align} \lbrace K, H\rbrace + \lbrace I', H\rbrace = \lbrace H, H\rbrace = 0. \end{align} \]
If one uses the potential temperature instead of the temperature and defines
\[ \begin{align} q_1 \coloneqq \newtilde{\theta} = \rho\theta, \end{align} \]
it 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
\[ \begin{align} \frac{\delta H}{\delta\rho} &= \frac{1}{2}\mathbf{v}^2 + \phi,\\ \frac{\delta H}{\delta q_1} &= c_d^{(v)}\frac{p_0}{R_d}\frac{R_d}{p_0}\kappa\left(\frac{R_d q_1}{p_0}\right)^{\kappa - 1} = c_d^{(p)}\left(\frac{R_d q_1}{p_0}\right)^{\kappa - 1} = c_d^{(p)}\left(\frac{R_d q_1}{p_0}\right)^{\frac{R_d}{c_d^{(v)}}} \stackrel{\href{ch-08-first-law-in-the-atmosphere.html#eq:exner_pressure_diag}{\text{Eq. (9.71)}}}{=} c_d^{(p)}\Pi,\\ \frac{\delta H}{\delta\mathbf{v}} &= \rho\mathbf{v}. \end{align} \]
Now the time evolutions of the quantities $K$ and $I'$ are to be examined. As preparation, one computes
\[ \begin{align} \frac{\delta K}{\delta\rho} = \frac{1}{2}\mathbf{v}^2, & {} & \frac{\delta K}{\delta q_1} = 0, & {} & \frac{\delta K}{\delta\mathbf{v}} = \rho\mathbf{v},\\ \frac{\delta I'}{\delta\rho} = \phi, & {} & \frac{\delta I'}{\delta q_1} = c_d^{(p)}\Pi, \frac{\delta I'}{\delta\mathbf{v}} = 0. \end{align} \]
and thus obtains
\[ \begin{align} \lbrace K, H\rbrace &= \int_A\frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta K}{\delta\mathbf{v}} - \frac{\delta K}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta K}{\delta\mathbf{v}}\right) - \frac{\delta K}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A\left(\frac{1}{2}\mathbf{v}^2 + \phi\right)\nabla\cdot\left(\rho\mathbf{v}\right) - \frac{1}{2}\mathbf{v}^2\nabla\cdot\left(\rho\mathbf{v}\right) + c_d^{(p)}\Pi\nabla\cdot\left(q_1\mathbf{v}\right)d^3r\nonumber\\ &= \int_A\phi\nabla\cdot\left(\rho\mathbf{v}\right) + c_d^{(p)}\Pi\nabla\cdot\left(q_1\mathbf{v}\right)d^3r\nonumber\\ &= \int_A-\rho\mathbf{v}\cdot\nabla\phi - c_d^{(p)}q_1\mathbf{v}\cdot\nabla\Pi d^3r\tag{10.78}\label{eq:poisson_entropy_kinetic}, \end{align} \] \[ \begin{align} \lbrace I', H\rbrace &= \int_A\frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta I'}{\delta\mathbf{v}} - \frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta I'}{\delta\mathbf{v}}\right) - \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A-\phi\nabla\cdot\left(\rho\mathbf{v}\right) - c_d^{(p)}\Pi\nabla\cdot\left(q_1\mathbf{v}\right)d^3r.\tag{10.79}\label{eq:poisson_entropy_potential_internal} \end{align} \]
If one uses the entropy instead of the potential temperature and defines
\[ \begin{align} q_1 \coloneqq \newtilde{s} = \rho s, \end{align} \]
it 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
\[ \begin{align} \frac{\delta H}{\delta\rho} \stackrel{\href{ch-04-statistical-physics.html#eq:chemisches_potential_prop_0}{\text{Eq. (5.91)}}}{=} \frac{1}{2}\mathbf{v}^2 + \phi + G, & {} & \frac{\delta H}{\delta q_1} \stackrel{\href{ch-04-statistical-physics.html#eq:chemisches_potential_prop_0}{\text{Eq. (5.91)}}}{=} T, & {} & \frac{\delta H}{\delta\mathbf{v}} = \rho\mathbf{v}. \end{align} \]
$G$ here denotes the specific Gibbs potential. Now the time evolutions of the quantities $K$ and $I'$ are examined again. As preparation, one computes
\[ \begin{align} \frac{\delta K}{\delta\rho} = \frac{1}{2}\mathbf{v}^2, & {} & \frac{\delta K}{\delta q_1} = 0, & {} & \frac{\delta K}{\delta\mathbf{v}} = \rho\mathbf{v},\\ \frac{\delta I'}{\delta\rho} = \phi + G, & {} & \frac{\delta I'}{\delta q_1} = T, & {} & \frac{\delta I'}{\delta\mathbf{v}} = 0. \end{align} \]
and thus obtains
\[ \begin{align} \lbrace K, H\rbrace &= \int_A\frac{\delta H}{\delta\rho}\nabla\cdot\left(\rho\mathbf{v}\right) - \frac{\delta K}{\delta\rho}\nabla\cdot\left(\rho\mathbf{v}\right) + T\nabla\cdot\left(q_1\mathbf{v}\right)d^3r\nonumber\\ &= \int_A\left(\phi + G\right)\nabla\cdot\left(\rho\mathbf{v}\right) + T\nabla\cdot\left(q_1\mathbf{v}\right)d^3r\nonumber\\ &= \int_A-\rho\mathbf{v}\cdot\nabla\left(\phi + G\right) - q_1\mathbf{v}\cdot\nabla Td^3r,\\ \lbrace I', H\rbrace &= \int_A\frac{\delta H}{\delta\rho}\nabla\cdot\frac{\delta I'}{\delta\mathbf{v}} - \frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} + \frac{\delta H}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta I'}{\delta\mathbf{v}}\right) - \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A-\frac{\delta I'}{\delta\rho}\nabla\cdot\frac{\delta H}{\delta\mathbf{v}} - \frac{\delta I'}{\delta q_1}\nabla\cdot\left(\frac{q_1}{\rho}\frac{\delta H}{\delta\mathbf{v}}\right)d^3r\nonumber\\ &= \int_A-\left(\phi + G\right)\nabla\cdot\left(\rho\mathbf{v}\right) - T\nabla\cdot\left(q_1\mathbf{v}\right)d^3r. \end{align} \]
This gives the expression for the pressure gradient acceleration
This is also confirmed by
\[ \begin{align} -\nabla G - s\nabla T &= -c_d^{(p)}\nabla T + T\nabla s = -c_d^{(p)}\nabla T + Tc_d^{(p)}\nabla\ln\left(\theta\right)\nonumber\\ &= -c_d^{(p)}\nabla T + \frac{T}{\theta}c_d^{(p)}\nabla\theta = -c_d^{(p)}\theta\nabla\frac{T}{\theta} = -c_d^{(p)}\theta\nabla\Pi \end{align} \]
This section is worked out only with the entropy formalism begun in Sect. 10.1.3. In an irreversible dry atmosphere, one defines two irreversible brackets:
\[ \begin{align} \left(F, \mathbf{f}_R\right) & \coloneqq \left\langle\frac{\delta F}{\delta\mathbf{v}}\Big| \mathbf{f}_R\right\rangle = \int_A\frac{\delta F}{\delta\mathbf{v}}\cdot\mathbf{f}_Rd^3r\\ \left(F, Q\right) & \coloneqq \left\langle\frac{\delta F}{\delta\newtilde{s}}\Big|\frac{\rho\epsilon}{T}\right\rangle = \int_A\frac{\delta F}{\delta\newtilde{s}}\frac{\rho\epsilon}{T}d^3r \end{align} \]
According to Eq. (8.96), one has 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 Sect. 9.5.1, in a moist atmosphere without condensates one can, with the substitution
\[ \begin{align} \theta \to \theta_v \end{align} \]
carry over the reversible part of the dynamics from Sect. 10.1.2, assuming that dry air and water vapor are perfect ideal gases.
Condensates move at a velocity $\mathbf{v}_i \not= \mathbf{v}$ that does not correspond to the wind velocity, since they have a superimposed fall velocity. 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:
The pressure gradient acceleration is multiplied by a prefactor $\frac{\rho_h}{\rho}$, since the condensates have to be accelerated by the pressure gradient but do not generate any pressure themselves. The same applies to frictional acceleration. The Coriolis acceleration and gravity, on the other hand, are not modified in this way, since these forces also act on condensates (the fact that the Coriolis acceleration acting on condensates differs from the Coriolis acceleration acting on the gas phase due to their falling speed is ignored).
Since the first law is formulated for the quantity $\rho_h\theta_v$, the thermal power densities must be multiplied by a prefactor $\frac{\rho_hc_h^{(v)}}{\rho c} \leq 1$.
When solving the radiative 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 Hamilton 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.