The momentum $\mathbf{p}$ of a particle $\Delta V$ with $N_h$ gas molecules and $N_c$ condensate nuclei is through
\[ \begin{align} \mathbf{p} = \sum_{i = 1}^{N_h + N_c}m_i\mathbf{v}_i\stackrel{\href{ch-05-state-of-the-atmosphere.html#eq:def_windvector}{\text{Glg. (6.2)}}}{=}\mathbf{v}\sum_{j = 1}^{N_h}m_j + \sum_{j = N_h + 1}^{N_h + N_c}m_j\left(\mathbf{v} + \Delta\mathbf{v}_j\right), \end{align} \]
Here $\Delta\mathbf{v}_j$ is the sinking speed of the corresponding condensate class, which is assumed to be independent of density. One can note Newton's second axiom in inertial systems:
\[ \begin{align} \md{\mathbf{p}} &= \rho \Delta V\md{}\mathbf{v} = \mathbf{F}_{\text{int}} + \mathbf{F}_{\text{ext}} \end{align} \]
The force is the sum of internal forces and external forces. Assuming that the internal forces consist of pairwise interactions of the particles that satisfy Newton's Third Axiom,
\[ \begin{align} \mathbf{F}_{\text{int}} &= \sum_{i, j = 1}^{N}\mathbf{F}_{i, j} = \sum_{i = 1}^{N}\sum_{j = i + 1}^{N}\mathbf{F}_{i, j} + \mathbf{F}_{j, i} \stackrel{\text{Newton III}}{=}\mathbf{0}. \end{align} \]
Here $\mathbf{F}_{i, j}$ is the force on the $i-$th particle due to the $j-$th particle. Two types of forces come into question for $\mathbf{F}_{\mathrm{ext}}$:
The first point usually only includes the gravity field. For the weight $\mathbf{F}_g$ applies
\[ \begin{align} \mathbf{F}_g = \sum_{i = 1}^{N}m_i\mathbf{g} = \rho\Delta V \mathbf{g}. \end{align} \]
The force acting due to the pressure on the open ball $\Omega \coloneqq B_r\left(\mathbf{r}_0\right)$ with radius $r$ with $\Delta V = \frac{4}{3}\pi r^3$ and center $\mathbf{r}_0$ applies
\[ \begin{align} \mathbf{F}_p &= -\int_{\partial\Delta V}pd\mathbf{A}. \end{align} \]
Here we assume a spherical particle, then it follows
\[ \begin{align} \mathbf{F}_p &= -\int_{\partial\Delta V}pr^2\mathbf{n}d\omega\approx-\int_{\partial\Delta V}\left(p\left(\mathbf{r}_0\right) + \mathbf{r}\cdot\nabla p\right)r^2\mathbf{n}d\omega \end{align} \]
with $d\omega$ as the solid angle element. To carry out the integration, you set up $\nabla p$ or similar. A. on the z-axis so that:
\[ \begin{align} \nabla p = \left|\nabla p\right|\mathbf{e}_z. \end{align} \]
This results in $p_0 \coloneqq p\left(\mathbf{r}_0\right)$
\[ \begin{align} \mathbf{F}_p &= -\int_{\partial\Delta V}\left(p_0 + \mathbf{r}\cdot\nabla p\right)r\mathbf{r}d\omega=-r\int_{\partial\Delta V}\left(\mathbf{r}\cdot\nabla p\right)\mathbf{r}d\omega = -r^2\left|\nabla p\right|\int_{\partial\Delta V}\cos\left\lbrace\nabla p, \mathbf{r}\right\rbrace\mathbf{r}d\omega\nonumber\\ &= -r^3\left|\nabla p\right|\int_{\vartheta = 0}^\pi\int_{\phi = 0}^{2\pi}\cos\left(\vartheta\right)\sin\left(\vartheta\right)\left(\begin{array}{c} \sin\left(\vartheta\right)\cos\left(\phi\right)\\ \sin\left(\vartheta\right)\sin\left(\phi\right)\\ \cos\left(\vartheta\right) \end{array}\right)d\phi d\vartheta\nonumber\\ &= -r^3\nabla p2\pi\int_0^\pi\cos^2\left(\vartheta\right)\sin\left(\vartheta\right)d\vartheta = r^3\nabla p2\pi\left[\frac{1}{3}\cos^3\left(\vartheta\right)\right]_0^\pi = -\frac{4}{3}\pi r^3\nabla p = -\Delta V\nabla p. \end{align} \]
This force is called pressure gradient force. You receive
\[ \begin{align} \rho \Delta V\md{}\mathbf{v} = \rho\Delta V \mathbf{g} - \Delta V\nabla p, \end{align} \]
If you convert this to the acceleration of the particle $\md{\mathbf{v}}$, you get
\[ \begin{align} \md{\mathbf{v}} = -\frac{1}{\rho}\nabla p + \mathbf{g}.\tag{8.10}\label{eq:newton_II_fluid} \end{align} \]
This equation is called Euler equation.
In an alternative derivation, one considers a macroscopic open set $\Omega\subseteq\mathbb{R}^3$, the momentum $\mathbf{p}$ in this set is given by
\[ \begin{align} \mathbf{p} = \int_\Omega\rho\mathbf{v}d^3r. \end{align} \]
At this point you carry out the momentum flux density tensor $\Pi$
\[ \begin{align} \Pi_{i, j} \coloneqq \rho U_iU_j \end{align} \]
for $1 \leq i, j \leq 3$ a. For $1 \leq i \leq 3$
\[ \begin{align} \Pi\cdot\mathbf{e}_i = \sum_{j = 1}^3\rho v_jv_i\mathbf{e}_j = v_i\rho\mathbf{v} \end{align} \]
the flow of the momentum density in the $\mathbf{e}_i-$ direction. For a general unit vector $\mathbf{e}$, $\Pi\cdot\mathbf{e}$ is the flux of momentum density in the $\mathbf{e}-$direction. Thus, with the findings of section 6.2.2
\[ \begin{align} \frac{\partial\mathbf{p}}{\partial t} &= \int_\Omega\frac{\partial\left(\rho\mathbf{v}\right)}{\partial t}d^3r = \int_\Omega-\nabla p + \rho\mathbf{g}d^3r - \int_{\partial\Omega}\Pi\cdot d\mathbf{n}\nonumber\\ &\Rightarrow \int_\Omega\frac{\partial\left(\rho\mathbf{v}\right)}{\partial t}d^3r = \int_\Omega-\nabla p + \rho\mathbf{g} - \nabla\cdot\Pi d^3r\nonumber\\ &\Rightarrow \int_\Omega\frac{\partial\left(\rho\mathbf{v}\right)}{\partial t} + \nabla p + \nabla\cdot\Pi - \rho\mathbf{g}d^3r = 0. \end{align} \]
So it already applies
\[ \begin{align} \frac{\partial\left(\rho\mathbf{v}\right)}{\partial t} + \nabla p + \nabla\cdot\Pi - \rho\mathbf{g} = 0.\tag{8.15}\label{eq:momentum_flux_form} \end{align} \]
For the divergence of $\Pi$ follows
\[ \begin{align} \left(\nabla\cdot\Pi\right)_i &= \sum_{j = 1}^3\frac{\partial}{\partial x_j}\left(\rho v_iv_j\right) = \sum_{j = 1}^3v_iv_j\frac{\partial\rho}{\partial x_j} + \rho v_j\frac{\partial v_i}{\partial x_j} + \rho v_i\frac{\partial v_j}{\partial x_j}\nonumber\\ &= v_i\mathbf{v}\cdot\nabla\rho + \rho\left(\mathbf{v}\cdot\nabla\right)v_i +\rho v_i\nabla\cdot\mathbf{v}, \end{align} \]
so applies
\[ \begin{align} \nabla\cdot\Pi &= \mathbf{v}\left(\mathbf{v}\cdot\nabla\rho + \rho\nabla\cdot\mathbf{v}\right) + \left(\rho\mathbf{v}\cdot\nabla\right)\mathbf{v}. \end{align} \]
Thus follows
\[ \begin{align} \rho\frac{\partial\mathbf{v}}{\partial t} + \mathbf{v}\frac{\partial\rho}{\partial t} + \nabla p + \mathbf{v}\left(\mathbf{v}\cdot\nabla\rho + \rho\nabla\cdot\mathbf{v}\right) + \left(\rho\mathbf{v}\cdot\nabla\right)\mathbf{v} - \rho\mathbf{g} = 0. \end{align} \]
With the continuity equation you finally get
\[ \begin{align} \frac{\partial\mathbf{v}}{\partial t} + \left(\mathbf{v}\cdot\nabla\right)\mathbf{v} = -\frac{1}{\rho}\nabla p + \mathbf{g}. \end{align} \]
Newton's axioms only apply in inertial systems. In relation to the Earth, one considers the center of the Earth to be unaccelerated, even if it orbits the Sun, and only considers the rotation of the Earth. Let $(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)$ denote the basis of the resting coordinates (see section D.1). If you measure accelerations in this KS, you can link them to the acting forces using Newton's second axiom. However, if you choose a base rotating at the angular velocity of the Earth, further acceleration terms arise that do not arise from physical forces. The base of the global coordinates is in resting coordinates
\[ \begin{align} \mathbf{e}_x\left(t\right) = \left( \begin{array}{c} \cos\left(\omega t\right) \\ \sin\left(\omega t\right) \\ 0 \end{array}\right), & {} & \mathbf{e}_y\left(t\right) = \left( \begin{array}{c} - \sin\left(\omega t\right) \\ \cos\left(\omega t\right)\\ 0 \end{array}\right), & {} & \mathbf{e}_z\left(t\right) = \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) \end{align} \]
with the angular velocity of the Earth's rotation $\omegabi = \left(0, 0, \omega\right)^T$. Let a particle with the position vector $\mathbf{r} = \mathbf{r}(t) = x(t) \mathbf{e}_x(t) + y(t)\mathbf{e}_y(t) + z(t)\mathbf{e}_z(t)$ be given. Then applies to the speed
\[ \begin{align} \mathbf{v} = \newdot{\mathbf{r}} = \newdot{x}\mathbf{e}_x + \newdot{y}\mathbf{e}_y + \newdot{z}\mathbf{e}_z + x\newdot{\mathbf{e}}_x + y\newdot{\mathbf{e}}_y + z\newdot{\mathbf{e}}_z. \end{align} \]
The following applies to the time derivatives of the basis vectors
\[ \begin{align} \newdot{\mathbf{e}}_x &= \omega\left( \begin{array}{c} - \sin\left(\omega t\right)\\ \cos\left(\omega t\right)\\ 0 \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \\ \omega \end{array}\right)\times \left( \begin{array}{c} \cos\left(\omega t\right)\\ \sin\left(\omega t\right) \\ 0 \end{array}\right) = \omegabi\times\mathbf{e}_x,\\ \newdot{\mathbf{e}}_y &= \omega\left( \begin{array}{c} - \cos\left(\omega t\right) \\ - \sin\left(\omega t\right) \\ 0 \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \\ \omega \end{array}\right)\times\left( \begin{array}{c} - \sin\left(\omega t\right)\\\cos\left(\omega t\right) \\ 0 \end{array}\right) = \omegabi\times\mathbf{e}_y,\\ \newdot{\mathbf{e}}_z &= \mathbf{0} = \omegabi\times\mathbf{e}_z. \end{align} \]
The velocity vector is therefore written as
\[ \begin{align} \mathbf{v} = \mathbf{v}' + x\omegabi\times\mathbf{e}_x + y\omegabi\times\mathbf{e}_y + z\omegabi\times\mathbf{e}_z = \mathbf{v}' + \omegabi\times\mathbf{r}, \tag{8.25}\label{eq:geschw_rot_system} \end{align} \]
For the speed measured in the canceled system, $\mathbf{v}' = \newdot{x}\mathbf{e}_x + \newdot{y}\mathbf{e}_y + \newdot{z}\mathbf{e}_z$ was written and the linearity of the vector product Eq.en (A.147) - (A.148) exploited. Newton's Second Axiom is a statement about the acceleration $\md{\mathbf{v}}$, so we derive the above equation again in time:
\[ \begin{align} \md{\mathbf{v}} &= \left(\md{\mathbf{v}'}\right)' + \omegabi\times\mathbf{v'} + \omegabi\times\left(\mathbf{v'} + \omegabi\times\mathbf{r}\right) = \left(\md{\mathbf{v'}}\right)' + 2\omegabi\times \mathbf{v}' + \omegabi\times\left(\omegabi\times\mathbf{r}\right), \end{align} \]
where $\left(\md{\mathbf{v}'}\right)'$ is the acceleration measured in the rotating system. One is interested in the acceleration in global coordinates, i.e. in the vector $\left(\md{\mathbf{v'}}\right)'$.
\[ \begin{align} \left(\md{\mathbf{v}'}\right)' = \md{\mathbf{v}} - 2\omegabi\times\mathbf{v}' - \omegabi\times\left(\omegabi\times\mathbf{r}\right)\tag{8.27}\label{eq:transformkraft} \end{align} \]
The location-dependent term $-\omegabi\times\left(\omegabi\times\mathbf{r}\right)$ is the centrifugal acceleration, the speed-dependent term $-2\omegabi\times\mathbf{v'}$ is the Coriolis acceleration. For the IS acceleration $\md{\mathbf{v}}$, the accelerations that result from the forces according to Newton's second axiom must be used. In particular, note that the Coriolis force does no work because it is perpendicular to the velocity vector.
stress refers to all the forces that a fluid exerts on itself. O.B.d. A. imagine a cube whose sides are parallel to the coordinate surfaces of a Cartesian coordinate system. The stress is determined by the stress tensor
\[ \begin{align} T = \left(\begin{array}{ccc} T_{x, x} & T_{x, y} & T_{x, z} \\ T_{y, x} & T_{y, y} & T_{y, z} \\ T_{z, x} & T_{z, y} & T_{z, z} \\ \end{array}\right) \end{align} \]
described. The elements of this tensor have the dimension of a pressure (force per area). The i-th line of T describes forces acting on the faces of the cube (anti)parallel to the i-axis. The entry in the jth column denotes the force acting on this surface in the j direction if the normal of the surface under consideration is parallel to the i axis, otherwise the signs are swapped. This causes compressive forces to always be negative.
First, the case of a fluid at rest is considered. In this case the stress must be isotropic. It corresponds to the pressure $p$, so in this case the stress tensor applies
\[ \begin{align} T_{i, j} = -p\delta_{i, j}. \end{align} \]
The sign comes from the convention that compressive forces are always negative. Now note the decomposition
\[ \begin{align} T = -p + \tau. \end{align} \]
Imagine a fluid particle $\left[-\frac{\Delta}{2}, \frac{\Delta}{2}\right]^3$ with $\Delta > 0$. The z-component $D_z$ of the torque acting on this particle $\mathbf{D}$ is proportional to
\[ \begin{align} D_z &\propto T_{1, 2} + \frac{\Delta}{2}\frac{\partial T_{1, 2}}{\partial x} + T_{1, 2} - \frac{\Delta}{2}\frac{\partial T_{1, 2}}{\partial x} - \left(T_{2, 1} + \frac{\Delta}{2}\frac{\partial T_{2, 1}}{\partial y} + T_{2, 1} - \frac{\Delta}{2}\frac{\partial T_{2, 1}}{\partial y}\right)\nonumber\\ &= 2T_{1, 2} - 2T_{2, 1}, \end{align} \]
where the values of $T$ are evaluated at the center of the coordinate system. Assuming vanishing torque:
\[ \begin{align} T_{1, 2} = T_{2, 1}. \end{align} \]
This assumption is comparable to the concept of quasi-static change of state in statistical physics. By changing the orientation of the coordinate system, this also applies to all other off-diagonal indices, so $T$ is symmetrical. Therefore the tensor $\tau$ must also be symmetric.
The size $\frac{\partial v_i}{\partial x_j}$ is called velocity gradient tensor. You can take notes for this one
\[ \begin{align} \frac{\partial v_i}{\partial x_j} = S_{i, j} + \frac{1}{2}R_{i, j} \end{align} \]
with
\[ \begin{align} S_{i, j} \coloneqq \frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right), & {} & R_{i, j} \coloneqq \frac{\partial v_i}{\partial x_j} - \frac{\partial v_j}{\partial x_i}.\tag{8.34}\label{eq:def_rotation_tensor} \end{align} \]
$S_{i, j}$ is called stress tensor, which is symmetrical, and $R_{i, j}$ is called rotation tensor, which is antisymmetric. The elements of the rotation tensor correspond to the rotation of fluid particles, but not to deformation or acceleration. Only the elements of the stress tensor are relevant for the stress. Therefore you can make notes
\[ \begin{align} \tau_{i, j} = \sum_{n, m}K_{i, j, m, n}S_{m, n}\tag{8.35}\label{eq:strain_to_stress} \end{align} \]
with a complex fourth-order tensor $K$, whose elements describe how the fluid reacts to deformations. The totals run from one to three. This formulation allows all elements of $\tau$ to be linearly linked in pairs with all elements of $S$.
If one assumes a medium whose material properties are independent of direction, which is the case for ideal gases, $K$ must be independent of the orientation of the coordinate system. This implies that $K$ is isotropic. An isotropic tensor of the fourth order has the form
\[ \begin{align} K_{i, j, m, n} = \lambda\delta_{i, j}\delta_{m, n} + \mu\delta_{i, m}\delta_{j, n} + \gamma\delta_{i, n}\delta_{j, m}, \end{align} \]
where $\lambda, \mu, \gamma \in \mathbb{C}$ are scalars. Since both $\tau$ and $S$ are symmetric, they do not change when $i$ and $j$ are swapped. The same must also apply to $K$. This implies
\[ \begin{align} \gamma = \mu. \end{align} \]
This leads to
\[ \begin{align} K_{i, j, m, n} = \lambda\delta_{i, j}\delta_{m, n} + 2\mu\delta_{i, m}\delta_{j, n}. \end{align} \]
$\mu$ is called dynamic viscosity and $\lambda$ is called Lamé constant. Putting this into Eq. (8.35), you get
\[ \begin{align} \tau_{i, j} = \sum_{n, m}\left(\lambda\delta_{i, j}\delta_{m, n} + 2\mu\delta_{i, m}\delta_{j, n}\right)S_{m, n} = \lambda\delta_{i, j}\sum_{m}S_{m, m} + 2\mu S_{i, j}. \end{align} \]
From Eq. (8.34) follows
\[ \begin{align} \sum_{m}S_{m, m} = \nabla\cdot\mathbf{v}. \end{align} \]
Therefore applies
\[ \begin{align} T_{i, j} = -p\delta_{i, j} + \lambda\delta_{i, j}\nabla\cdot\mathbf{v} + 2\mu S_{i, j}\tag{8.41}\label{eq:stress_tensor_deriv_0} \end{align} \]
For $i = j$ we get from this
\[ \begin{align} T_{i, i} = -p + \lambda\nabla\cdot\mathbf{v} + 2\mu S_{i, i}. \end{align} \]
If you sum this over $i = 1, 3$, it follows
\[ \begin{align} \sum_{i}T_{i, i} &= -3p + \left(2\mu + 3\lambda\right)\nabla\cdot\mathbf{v}\nonumber\\ \Rightarrow p &= -\frac{1}{3}T_{i, i} + \sum_{i}T_{i, i} + \left(\frac{2}{3}\mu + \lambda\right)\nabla\cdot\mathbf{v}. \end{align} \]
From this, in contrast to the thermodynamic pressure, one can derive a average pressure
\[ \begin{align} \newoverline{p} \coloneqq -\frac{1}{3}T_{i, i} \end{align} \]
derive. This applies to this one
\[ \begin{align} p - \newoverline{p} = \sum_{i}T_{i, i} + \left(\frac{2}{3}\mu + \lambda\right)\nabla\cdot\mathbf{v}. \end{align} \]
Now define the bulk viscosity $\mu_v$ by
\[ \begin{align} \mu_v \coloneqq \frac{2}{3}\mu + \lambda \Rightarrow p - \newoverline{p} = \sum_{i}T_{i, i} + \mu_v\nabla\cdot\mathbf{v}. \end{align} \]
The Stokes Assumption
\[ \begin{align} \mu_v = 0 \end{align} \]
is often justified, but is not done further. For the Lamé constant $\lambda$ is obtained in bulk viscosity terms
\[ \begin{align} \lambda = \mu_v - \frac{2}{3}\mu.\tag{8.48}\label{eq:lame_viscosity_relation} \end{align} \]
Putting this into Eq. (8.41), you get
\[ \begin{align} T_{i, j} = -p\delta_{i, j} + \left(\mu_v - \frac{2}{3}\mu\right)\delta_{i, j}\nabla\cdot\mathbf{v} + 2\mu S_{i, j}. \end{align} \]
Another spelling is:
\[ \begin{align} T_{i, j} = -p\delta_{i, j} + \tau_{i, j} = -p\delta_{i, j} + 2\mu\left(S_{i, j} - \frac{1}{3}\nabla\cdot\mathbf{v}\delta_{i, j}\right) + \mu_v\nabla\cdot\mathbf{v}\delta_{i, j}.\tag{8.50}\label{eq:stress-tensor} \end{align} \]
The assumption that $T$ depends linearly on $S$ is an approximation that is correct for most fluids. Such fluids are called Newtonian fluids. In meteorology it is not necessary to go beyond this.
In Cartesian coordinates with $\mathbf{v} = \left(u, v, w\right)^T$, Eq. (8.50)
\[ \begin{align} T = \left(\begin{array}{ccc} -p + 2\mu\frac{\partial u}{\partial x} + \left(\mu_v - \frac{2}{3}\mu\right)\nabla\cdot\mathbf{v} & \mu\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right) & \mu\left(\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x}\right)\\ \mu\left(\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\right) & -p + 2\mu\frac{\partial v}{\partial y} + \left(\mu_v - \frac{2}{3}\mu\right)\nabla\cdot\mathbf{v} & \mu\left(\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y}\right)\\ \mu\left(\frac{\partial w}{\partial x} + \frac{\partial u}{\partial z}\right) & \mu\left(\frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}\right) & -p + 2\mu\frac{\partial w}{\partial z} + \left(\mu_v - \frac{2}{3}\mu\right)\nabla\cdot\mathbf{v} \end{array}\right).\nonumber\\ & \end{align} \]
From this we can calculate the frictional acceleration $\mathbf{f}_R$:
\[ \begin{align} \mathbf{f}_R &= \frac{1}{\rho}\nabla\cdot\tau = \frac{1}{\rho}\left(\begin{array}{c} 2\mu\frac{\partial^2u}{\partial x^2} + \left(\mu_v - \frac{2}{3}\mu\right)\frac{\partial\nabla\cdot\mathbf{v}}{\partial x} + \mu\left(\frac{\partial^2u}{\partial y^2} + \frac{\partial^2v}{\partial y\partial x}\right) + \mu\left(\frac{\partial^2u}{\partial z^2} + \frac{\partial^2w}{\partial z\partial x}\right)\\ \mu\left(\frac{\partial^2v}{\partial x^2} + \frac{\partial^2u}{\partial x\partial y}\right) + 2\mu\frac{\partial^2v}{\partial y^2} + \left(\mu_v - \frac{2}{3}\mu\right)\frac{\partial\nabla\cdot\mathbf{v}}{\partial y} + \mu\left(\frac{\partial^2v}{\partial z^2} + \frac{\partial^2w}{\partial z\partial y}\right)\\ \mu\left(\frac{\partial^2w}{\partial x^2} + \frac{\partial^2u}{\partial x\partial z}\right) + \mu\left(\frac{\partial^2w}{\partial y^2} + \frac{\partial^2v}{\partial y\partial z}\right) + 2\mu\frac{\partial^2w}{\partial z^2} + \left(\mu_v - \frac{2}{3}\mu\right)\frac{\partial\nabla\cdot\mathbf{v}}{\partial z} \end{array} \right)\nonumber\\ &= \frac{\mu}{\rho}\Delta\mathbf{v} + \frac{1}{\rho}\left(\begin{array}{c} \mu\frac{\partial^2u}{\partial x^2} + \left(\mu_v - \frac{2}{3}\mu\right)\frac{\partial\nabla\cdot\mathbf{v}}{\partial x} + \mu\frac{\partial^2v}{\partial y\partial x} + \mu\frac{\partial^2w}{\partial z\partial x}\\ \mu\frac{\partial^2u}{\partial x\partial y} + \mu\frac{\partial^2v}{\partial y^2} + \left(\mu_v - \frac{2}{3}\mu\right)\frac{\partial\nabla\cdot\mathbf{v}}{\partial y} + \mu\frac{\partial^2w}{\partial z\partial y}\\ \mu\frac{\partial^2u}{\partial x\partial z} + \mu\frac{\partial^2v}{\partial y\partial z} + \mu\frac{\partial^2w}{\partial z^2} + \left(\mu_v - \frac{2}{3}\mu\right)\frac{\partial\nabla\cdot\mathbf{v}}{\partial z} \end{array} \right)\nonumber\\ &= \nu\Delta\mathbf{v} + \frac{\mu_v}{\rho}\nabla\left(\nabla\cdot\mathbf{v}\right) + \frac{\mu}{\rho}\left(\begin{array}{c} \frac{\partial^2u}{\partial x^2} - \frac{2}{3}\frac{\partial\nabla\cdot\mathbf{v}}{\partial x} + \frac{\partial^2v}{\partial y\partial x} + \frac{\partial^2w}{\partial z\partial x}\\ \frac{\partial^2u}{\partial x\partial y} + \frac{\partial^2v}{\partial y^2} - \frac{2}{3}\frac{\partial\nabla\cdot\mathbf{v}}{\partial y} + \frac{\partial^2w}{\partial z\partial y}\\ \frac{\partial^2u}{\partial x\partial z} + \frac{\partial^2v}{\partial y\partial z} + \frac{\partial^2w}{\partial z^2} - \frac{2}{3}\frac{\partial\nabla\cdot\mathbf{v}}{\partial z} \end{array} \right)\nonumber\\ &= \nu\Delta\mathbf{v} + \frac{\mu_v}{\rho}\nabla\left(\nabla\cdot\mathbf{v}\right) + \frac{\nu}{3}\left(\begin{array}{c} \frac{\partial^2u}{\partial x^2} + \frac{\partial^2v}{\partial y\partial x} + \frac{\partial^2w}{\partial z\partial x}\\ \frac{\partial^2u}{\partial x\partial y} + \frac{\partial^2v}{\partial y^2} + \frac{\partial^2w}{\partial z\partial y}\\ \frac{\partial^2u}{\partial x\partial z} + \frac{\partial^2v}{\partial y\partial z} + \frac{\partial^2w}{\partial z^2} \end{array} \right) = \nu\Delta\mathbf{v} + \left(\frac{\mu_v}{\rho} + \frac{\nu}{3}\right)\nabla\left(\nabla\cdot\mathbf{v}\right)\nonumber \end{align} \]
\[ \begin{align} \Leftrightarrow\mathbf{f}_R &= \nu\Delta\mathbf{v} + \left(\frac{\mu_v}{\rho} + \frac{\nu}{3}\right)\nabla\left(\nabla\cdot\mathbf{v}\right)\tag{8.52}\label{eq:friction_acceleration} \end{align} \]
It was assumed that the viscosities were homogeneous. For the friction, other values than those given in Eq. (8.52) obtained expression is used. Therefore, from now on, the derivations will mostly use a general expression
\[ \begin{align} \mathbf{f}_R = \left(\begin{array}{c} f_{R,x}\\ f_{R,y}\\ f_{R,z} \end{array}\right) \equiv \left(\begin{array}{c} \mathbf{f}_R^{(H)}\\ f_{R,z} \end{array}\right) \end{align} \]
carried for frictional acceleration. This makes sense despite the relative smallness of friction in the atmosphere due to its importance for the energy cascade (see section 8.2.4).
A scale analysis of the frictional acceleration for the synoptic scale gives $10^{-5}10^{-11} = 10^{-16}$ m/s$^2$, the typical acceleration is $10^{-4}$ m/s$^2$, so the friction on the synoptic scale is twelve orders of magnitude smaller than the acceleration and is therefore completely negligible. It nevertheless has an important thermodynamic significance as it is responsible for the dissipation of kinetic energy. The dissipated kinetic energy is converted into thermal energy. Fluids without friction are called ideal fluids.
In the case of incompressibility the equation holds because $S_{m, m} = \nabla\cdot\mathbf{v} = 0$
\[ \begin{align} T_{i, j} = -p\delta_{i, j} + 2\mu S_{i, j}. \end{align} \]
From this it follows
\[ \begin{align} \frac{1}{\rho}\nabla\cdot T & \stackrel{\href{#eq:friction_acceleration}{\text{Glg. (8.52)}}}{=} \nu\Delta\mathbf{v}.\tag{8.55}\label{eq:friction_acceleration_inc} \end{align} \]
The friction term $\mathbf{f}_R$ dissipates specific kinetic energy $k = \frac{1}{2}\mathbf{v}^2$ into internal energy $i = c^{(V)}T$. The momentum equation in an IS is
\[ \begin{align} \md{\mathbf{v}} = \frac{\partial\mathbf{v}}{\partial t} + \left(\mathbf{v}\cdot\nabla\right)\mathbf{v} = -\frac{1}{\rho}\nabla p + \mathbf{g} + \mathbf{f}_R \end{align} \]
ring. In this case, the material change in the specific kinetic energy $k$ applies
\[ \begin{align} \md{k} = \md{}\left(\frac{1}{2}\mathbf{v}^2\right) = \mathbf{v}\cdot\md{\mathbf{v}} = -\frac{1}{\rho}\nabla p\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{g} + \mathbf{v}\cdot\mathbf{f}_R = -\frac{1}{\rho}\sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} + \sum_{i = 1}^3v_ig_i + \frac{1}{\rho}\sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i}. \end{align} \]
Multiplying this by $\rho$ gives
\[ \begin{align} \rho\md{k} = -\sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} + \sum_{i = 1}^3\rho v_ig_i + \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i}.\tag{8.58}\label{eq:dissipation_deriv_0} \end{align} \]
Let $\Omega = \Omega\left(t\right) \subseteq \mathbb{R}^3$ be a connected control volume. Then the total energy in this area in the absence of heat flows (except dissipation) applies:
\[ \begin{align} \frac{d}{dt}\int_{\Omega\left(t\right)}\rho\left(k + i\right)d^3r &= \int_{\Omega\left(t\right)}\rho\mathbf{v}\cdot\mathbf{g}d^3r + \int_{\partial\Omega\left(t\right)}\mathbf{f}\cdot\mathbf{v}dA.\tag{8.59}\label{eq:dissipation_deriv_1} \end{align} \]
Here $\mathbf{v}$ is the force acting on the surface of the control volume. For these, the definitions at the beginning of Section 8.2 apply
\[ \begin{align} \int_{\partial\Omega\left(t\right)}\mathbf{f}\cdot\mathbf{v}dA &= \int_{\partial\Omega\left(t\right)}\sum_{i, j = 1}^3n_iT_{i, j}v_jdA = \int_{\Omega\left(t\right)}\sum_{i, j = 1}^3\frac{\partial\left(T_{i, j}v_j\right)}{\partial x_i}d^3r. \end{align} \]
For the integrand one obtains
\[ \begin{align} \sum_{i, j = 1}^3\frac{\partial\left(T_{i, j}v_j\right)}{\partial x_i} &= \sum_{i, j = 1}^3v_j\frac{\partial T_{i, j}}{\partial x_i} + T_{i, j}\frac{\partial v_j}{\partial x_i} = \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i} + \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} - \sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} - \sum_{i = 1}^3p\frac{\partial v_i}{\partial x_i}. \end{align} \]
Putting this into Eq. (8.59), you get
\[ \begin{align} \frac{d}{dt}\int_{\Omega\left(t\right)}\rho\left(k + i\right)d^3r &= \int_{\Omega\left(t\right)}\rho\sum_{i = 1}^3g_iv_i + \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i} + \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} - \sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} - \sum_{i = 1}^3p\frac{\partial v_i}{\partial x_i}d^3r. \end{align} \]
If you let $\Omega$ shrink to the size of a fluid particle, you get
\[ \begin{align} \md{\left[\rho\left(k + i\right)\right]} &= \sum_{i = 1}^3\rho v_ig_i + \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i} + \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} - \sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} - \sum_{i = 1}^3p\frac{\partial v_i}{\partial x_i}. \end{align} \]
If you subtract Eq. (8.58), one gets
\[ \begin{align} \md{\left(\rho i\right)} &= \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} - \sum_{i = 1}^3p\frac{\partial v_i}{\partial x_i} = -p\nabla\cdot\mathbf{v} + \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} \equiv -p\nabla\cdot\mathbf{v} + \rho\epsilon. \end{align} \]
The specific dissipation rate, or short Dissipation $\epsilon$, is defined by
\[ \begin{align} \epsilon \coloneqq \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} \stackrel{\href{#eq:def_rotation_tensor}{\text{Glg.en (8.34) - (8.34)}}}{=} \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}\left(S_{i, j} - \frac{1}{2}R_{i, j}\right) \stackrel{R_{i, j}\text{ antisymmetrisch}}{=} \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}S_{i, j}. \end{align} \]
It will be as a result of this section
\[ \begin{align} \epsilon = \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}S_{i, j} = \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i}\tag{8.66}\label{eq:dissipation} \end{align} \]
noted.
Friction is usually thought of as a braking force. This is not always the case in fluids. To see this, we use the one-dimensional flow field
\[ \begin{align} u = 2 + \sin\left(x\right) > 0 \end{align} \]
out, ignoring units for the moment. Such a field is realistic, it is a background wind with a superimposed wave. If the medium is incompressible, the frictional acceleration is $a_R$ up to a constant
\[ \begin{align} a_R = \frac{\partial^2u}{\partial x^2} = -\sin\left(x\right). \end{align} \]
Thus, the work done by friction can
\[ \begin{align} ua_R = -u\sin\left(x\right) = -\left(2 + \sin\left(x\right)\right)\sin\left(x\right) = -2\sin\left(x\right) - \sin^2\left(x\right)\tag{8.69}\label{eq:fric_td_deriv_1} \end{align} \]
have both signs. It is therefore possible in fluids for friction to produce kinetic energy. Nevertheless, the mean value of Eq. (8.75) negative, on average the friction dissipates kinetic energy in this case too. However, this is not yet a generally valid derivation of the dissipative effect of friction.
However, if the equations are integrated globally, the negative effect of friction on the integral of kinetic energy can also be seen formally. For this purpose, a one-dimensional incompressible medium is initially assumed, which is limited to the area $\left[0, L\right]$, the boundary conditions are $u\left(x = 0, L\right) 0$. Thus one obtains by means of partial integration
\[ \begin{align} \int_0^Lua_Rdx = \int_0^Lu\frac{\partial^2u}{\partial x^2}dx = \left[u\frac{\partial u}{\partial x}\right]_0^L - \int_0^L\left(\frac{\partial u}{\partial x}\right)^2dx = -\int_0^L\left(\frac{\partial u}{\partial x}\right)^2dx \leq 0. \end{align} \]
Accordingly applies
\[ \begin{align} \int_0^Lqdx = -\int_0^Lua_Rdx \geq 0, \end{align} \]
The friction destroys globally integrated kinetic energy by means of a heat power density $q$ in favor of the internal energy. Locally, however, friction can produce kinetic energy, which is associated with $q < 0$. The boundary conditions are therefore crucial for the thermodynamic role of friction.
Both trains of thought can also be generalized in three dimensions. To do this, first start with a three-dimensional wind field of the shape
\[ \begin{align} \mathbf{v}\left(\mathbf{r}\right) = \mathbf{u}\left(2 + \sin\left(\mathbf{k}\cdot\mathbf{r}\right)\right)\tag{8.72}\label{eq:diss_sample_wind_field} \end{align} \]
out of. From this it follows
\[ \begin{align} \mathbf{v}\cdot\Delta\mathbf{v} = -k^2u^2\sin\left(\mathbf{k}\cdot\mathbf{r}\right)\left(2 + \sin\left(\mathbf{k}\cdot\mathbf{r}\right)\right). \end{align} \]
This expression takes both signs, so even in the three-dimensional case the frictional acceleration can locally produce kinetic energy. The work that the frictional acceleration does can be calculated locally
\[ \begin{align} \mathbf{v}\cdot\Delta\mathbf{v} = \sum_{i, j = 1}^3v_i\frac{\partial^2v_i}{\partial x_j^2} \end{align} \]
note down. Since this expression is isotropic, further consideration can be limited to the terms with x-derivative:
\[ \begin{align} \sum_{i = 1}^3v_i\frac{\partial^2v_i}{\partial x^2}\tag{8.75}\label{eq:fric_td_deriv_0} \end{align} \]
Now look at the quantity
\[ \begin{align} \Omega \coloneqq \left[0, L_x\right] \times \left[0, L_y\right] \times \left[0, L_z\right] \end{align} \]
with the boundary conditions
\[ \begin{align} \mathbf{v}\cdot\mathbf{n} = 0, & {} & \nabla\mathbf{v}\cdot\mathbf{n} = \mathbf{0} \end{align} \]
on $\partial\Omega$, here $\mathbf{n}$ is a normal vector on $\partial\Omega$. If one now integrates Eq. (8.75) on $\Omega$, one obtains
\[ \begin{align} \int_\Omega\sum_{i = 1}^3v_i\frac{\partial^2v_i}{\partial x^2}d^3r &= \int_{z = 0}^{L_z}\int_{y = 0}^{L_y}\int_{x = 0}^{L_x}\sum_{i = 1}^3v_i\frac{\partial^2v_i}{\partial x^2}dxdydz. \end{align} \]
If you just do the integration over x, you get
\[ \begin{align} \int_0^{L_x}\sum_{i = 1}^3v_i\frac{\partial^2v_i}{\partial x^2}dx = \int_0^{L_x}v_x\frac{\partial^2v_x}{\partial x^2} + v_y\frac{\partial^2v_y}{\partial x^2} + v_z\frac{\partial^2v_z}{\partial x^2}dx. \end{align} \]
If you integrate again partially, it follows
\[ \begin{align} \int_0^{L_x}v_x\frac{\partial^2v_x}{\partial x^2} + v_y\frac{\partial^2v_y}{\partial x^2} + v_z\frac{\partial^2v_z}{\partial x^2}dx &= \left[v_x\frac{\partial v_x}{\partial x} + v_y\frac{\partial v_y}{\partial x} + v_z\frac{\partial v_z}{\partial x}\right]_0^L - \int_0^{L_x}\left(\frac{\partial v_x}{\partial x}\right)^2 + \left(\frac{\partial v_y}{\partial x}\right)^2 + \left(\frac{\partial v_z}{\partial x}\right)^2dx\nonumber\\ &= -\int_0^{L_x}\left(\frac{\partial v_x}{\partial x}\right)^2 + \left(\frac{\partial v_y}{\partial x}\right)^2 + \left(\frac{\partial v_z}{\partial x}\right)^2dx \leq 0. \end{align} \]
This implies
\[ \begin{align} \int_\Omega\mathbf{v}\cdot\Delta\mathbf{v}d^3r \leq 0. \end{align} \]
Accordingly, in this case too, friction destroys kinetic energy.
Both observations can also be generalized to the three-dimensional compressible case. To do this, one starts again with a wind field of the form Eq. (8.72), but this time with the additional condition $\mathbf{u}\cdot\mathbf{k} = 0$. This implies
\[ \begin{align} \nabla\cdot\mathbf{v} = \mathbf{k}\cdot\mathbf{u}\cos\left(\mathbf{k}\cdot\mathbf{r}\right) = 0. \end{align} \]
Thus, the term with $\nabla\cdot \mathbf{v}$ falls into Eq. (8.52) with this special wind field, which, in analogy to the three-dimensional incompressible case, means that the work done by the friction force $\mathbf{v}\cdot\mathbf{f}_R$ can take on both signs.
If you multiply Eq. (8.52) with $\rho\mathbf{v}$, one obtains
\[ \begin{align} \rho\mathbf{v}\cdot\mathbf{f}_R = \mu\mathbf{v}\cdot\Delta\mathbf{v} + \left(\mu_v + \frac{\mu}{3}\right)\mathbf{v}\cdot\nabla\left(\nabla\cdot\mathbf{v}\right). \end{align} \]
Assuming that the viscosities are homogeneous, the following applies analogously to the incompressible case
\[ \begin{align} \int_\Omega\mu\mathbf{v}\cdot\Delta\mathbf{v}d^3r \leq 0.\tag{8.84}\label{eq:friction_dissipation_deriv_0} \end{align} \]
Calculate for the second term
\[ \begin{align} \int_0^{L_x}u\frac{\partial\nabla\cdot\mathbf{v}}{\partial x}dx &= \left[u\nabla\cdot\mathbf{v}\right]_0^{L_x} - \int_0^{L_x}\frac{\partial u}{\partial x}\nabla\cdot\mathbf{v}dx. \end{align} \]
With the kinematic boundary condition $u\left(x = 0, L_x\right) = 0$ the first term is omitted. Therefore applies
\[ \begin{align} \int_\Omega\mathbf{v}\cdot\nabla\left(\nabla\cdot\mathbf{v}\right)d^3r = -\int_\Omega\left(\nabla\cdot\mathbf{v}\right)^2d^3r \leq 0. \end{align} \]
This also applies in the compressible case
\[ \begin{align} \int_\Omega\rho\mathbf{v}\cdot\mathbf{f}_Rd^3r \leq 0.\tag{8.87}\label{eq:friction_work_sign} \end{align} \]
According to Eq. (8.66)
\[ \begin{align} \epsilon &= \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}S_{i, j} \stackrel{\href{#eq:stress-tensor}{\text{Glg. (8.50)}}}{=} \frac{1}{\rho}\sum_{i, j = 1}^3\left[2\mu\left(S_{i, j} - \frac{1}{3}\nabla\cdot\mathbf{v}\delta_{i, j}\right) + \mu_v\nabla\cdot\mathbf{v}\delta_{i, j}\right]S_{i, j}\nonumber\\ &= \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 - \frac{2}{3}\frac{\mu}{\rho}\sum_{i, j = 1}^3\nabla\cdot\mathbf{v}\delta_{i, j}S_{i, j} + \frac{\mu_v}{\rho}\sum_{i, j = 1}^3\nabla\cdot\mathbf{v}\delta_{i, j}S_{i, j}\nonumber\\ &\stackrel{\href{#eq:def_rotation_tensor}{\text{Glg. (8.34)}}}{=} \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 - \frac{2}{3}\frac{\mu}{\rho}\sum_{i, j = 1}^3\nabla\cdot\mathbf{v}\delta_{i, j}\frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right) + \frac{\mu_v}{\rho}\sum_{i, j = 1}^3\nabla\cdot\mathbf{v}\delta_{i, j}\frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right)\nonumber \end{align} \] \[ \begin{align} &= \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 - \frac{2}{3}\frac{\mu}{\rho}\nabla\cdot\mathbf{v}\sum_{i, j = 1}^3\delta_{i, j}\frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right) + \frac{\mu_v}{\rho}\nabla\cdot\mathbf{v}\sum_{i, j = 1}^3\delta_{i, j}\frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right)\nonumber\\ &= \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 - \frac{2}{3}\frac{\mu}{\rho}\left(\nabla\cdot\mathbf{v}\right)^2 + \frac{\mu_v}{\rho}\left(\nabla\cdot\mathbf{v}\right)^2 = \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 + \left(\frac{\mu_v}{\rho} - \frac{2}{3}\frac{\mu}{\rho}\right)\left(\nabla\cdot\mathbf{v}\right)^2. \end{align} \]
According to Eq. (8.48) applies
\[ \begin{align} \mu_v - \frac{2}{3}\mu = \lambda, \end{align} \]
here $\lambda \geq 0$ is the Lamé constant. Therefore applies
\[ \begin{align} \epsilon \geq 0.\tag{8.90}\label{eq:dissipation_sign} \end{align} \]
According to Eq., the friction destroys (8.87) kinetic energy on a global average, but can also produce kinetic energy locally, as shown in Sect. 8.2.4. The dissipation, on the other hand, is always positive. So local applies
\[ \begin{align} \rho\mathbf{v}\cdot\mathbf{f}_R + \rho\epsilon \stackrel{\text{i. A.}}{\not=} 0. \end{align} \]
However, it is expected that on a global average the dissipation produces exactly as much internal energy as the friction destroys kinetic energy. To show this, one calculates
\[ \begin{align} \int_\Omega\rho\epsilon + \rho\mathbf{v}\cdot\mathbf{f}_Rd^3r &\stackrel{\href{#eq:dissipation}{\text{Glg. (8.66)}}}{=} \int_\Omega\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} + \rho\mathbf{v}\cdot\mathbf{f}_Rd^3r = \int_\Omega\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} + \rho\sum_{i, j = 1}^3\frac{1}{\rho}v_j\frac{\partial\tau_{i, j}}{\partial x_i}d^3r\nonumber\\ &= \int_\Omega\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} + \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i}d^3r = \int_\Omega\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} + v_j\frac{\partial\tau_{i, j}}{\partial x_i}d^3r\nonumber\\ &= \int_\Omega\sum_{i, j = 1}^3\frac{\partial\left(\tau_{i, j}v_j\right)}{\partial x_i}d^3r. \end{align} \]
As an example, you first consider only the terms with partial derivatives with respect to $x$ and assume $\Omega = \left[0, L_x\right] \times \left[0, L_z\right] \times \left[0, L_y\right]$:
\[ \begin{align} \int_{0}^{L_x}\sum_{j = 1}^3\frac{\partial\left(\tau_{x, j}v_j\right)}{\partial x}dx &= \left[\sum_{j = 1}^3\tau_{x, j}v_j\right]_0^{L_x} = \left[\tau_{x, x}u + \tau_{x, y}v + \tau_{x, z}w\right]_0^{L_x} \stackrel{u\left(x = 0, L_x\right) = 0}{=} \left[\tau_{x, y}v + \tau_{x, z}w\right]_0^{L_x}\nonumber\\ &= \left[v\mu\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right) + w\mu\left(\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x}\right)\right]_0^{L_x} \nonumber\\ &\stackrel{u\left(x = 0, L_x\right) = 0}{=} \left[v\mu\frac{\partial v}{\partial x} + w\mu\frac{\partial w}{\partial x}\right]_0^{L_x} = \mu\left[v\frac{\partial v}{\partial x} + w\frac{\partial w}{\partial x}\right]_0^{L_x} \end{align} \]
With the boundary condition
\[ \begin{align} \nabla\mathbf{v}\cdot\mathbf{n} = 0 \end{align} \]
are partial derivatives of tangential wind components orthogonal to the surface equal to zero. Thus follows
\[ \begin{align} \int_{0}^{L_x}\sum_{j = 1}^3\frac{\partial\left(\tau_{x, j}v_j\right)}{\partial x}dx &= 0. \end{align} \]
This also applies to the other spatial directions through cyclic swapping. So it is indeed
\[ \begin{align} \int_\Omega\rho\epsilon + \rho\mathbf{v}\cdot\mathbf{f}_Rd^3r = 0.\tag{8.96}\label{eq:friction_dissipation_balance} \end{align} \]
If you put Eq. (B.131) and Eq. (8.52) in Eq. (8.10), this results
\[ \begin{align} \md{\mathbf{v}} = -\frac{1}{\rho}\nabla p + \mathbf{v}\times\mathbf{f} + \mathbf{g} + \mathbf{f}_R.\tag{8.97}\label{eq:momentum} \end{align} \]
$\mathbf{f} \coloneqq 2\omegabi$ is the Coriolis vector. With a renaming, the acceleration $\left(\md{\mathbf{v}'}\right)'$ was also denoted by $\md{\mathbf{v}}$, see Eq. (B.131). $\mathbf{v}$ is the speed measured in the rotating system. The centrifugal acceleration term was absorbed in $\mathbf{g}$.
The acceleration $\md{\mathbf{v}}$ can be divided into a local time derivative $\frac{\partial\mathbf{v}}{\partial t}$ and an advective part:
\[ \begin{align} \md{\mathbf{v}} = \frac{\partial\mathbf{v}}{\partial t} + \left(\mathbf{v}\cdot\nabla\right)\mathbf{v} \end{align} \]
For the advective part, Eq. (B.57)
\[ \begin{align} \left(\mathbf{v}\cdot\nabla\right)\mathbf{v} = \frac{1}{2}\nabla\left(\mathbf{v}\cdot\mathbf{v}\right) - \mathbf{v}\times\left(\nabla\times\mathbf{v}\right). \end{align} \]
Define the specific kinetic energy $k$ by
\[ \begin{align} k \coloneqq\frac{1}{2}\mathbf{v}^2, \end{align} \]
follows
\[ \begin{align} \frac{\partial\mathbf{v}}{\partial t} = -\frac{1}{\rho}\nabla p + \mathbf{v}\times\left(\mathbf{f} + \nabla\times\mathbf{v}\right) - \nabla k + \mathbf{g} + \mathbf{f}_R.\tag{8.101}\label{eq:momentum_mod} \end{align} \]
Now equation (8.97) is written down component by component with respect to the location-dependent orthonormal basis of the spherical coordinates. For the vector $\mathbf{f}$ applies
\[ \begin{align} \mathbf{f} = |\mathbf{f}|\left(\begin{array}{c} 0\\ \cos\left(\varphi\right)\\ \sin\left(\varphi\right) \end{array}\right)\eqqcolon\left(\begin{array}{c} 0\\ f'\\ f \end{array}\right). \end{align} \]
$f \coloneqq\mathbf{k}\cdot\mathbf{f}$ bezeichnet man als den Coriolis-Parameter. Damit folgt
\[ \begin{align} \mathbf{v}\times\mathbf{f} = \left(\begin{array}{c} fv - f'w\\ -fu\\ f'u \end{array}\right). \end{align} \]
If you insert this into the equation of motion, it follows
\[ \begin{align} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z} - \frac{uv\tan\left(\varphi\right)}{a + z} + \frac{uw}{a + z} &= -\frac{1}{\rho}\frac{\partial p}{\partial x} + g_x - f'w + fv + f_{R,x} \tag{8.104}\label{eq:x_momentum},\\ \frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} + w\frac{\partial v}{\partial z} + \frac{u^2\tan\left(\varphi\right)}{a + z} + \frac{vw}{a + z} &= -\frac{1}{\rho}\frac{\partial p}{\partial y} + g_y - fu + f_{R,y} \tag{8.105}\label{eq:y_momentum},\\ \frac{\partial w}{\partial t} + u\frac{\partial w}{\partial x} + v\frac{\partial w}{\partial y} + w\frac{\partial w}{\partial z} - \frac{u^2 + v^2}{a + z} &= -\frac{1}{\rho}\frac{\partial p}{\partial z} + g_z + f'u + f_{R,z} \tag{8.106}\label{eq:z_momentum}. \end{align} \]
If you want to apply a system of differential equations to a set $\Omega$, you need boundary conditions. Let $\Omega$ be limited by condensed matter. Let $\mathbf{n}$ be a normal vector on $\partial\Omega$, then the boundary condition for the wind field is $\mathbf{v}$
\[ \begin{align} \mathbf{v}\cdot\mathbf{n} = 0, \end{align} \]
since particles cannot overcome the fixed boundary $\partial\Omega$. This is called kinematic boundary condition. This boundary condition is often also required at the upper edge of the atmosphere, which is not a physical reality but a necessary constraint. Another commonly used boundary condition is
\[ \begin{align} \mathbf{v}\left(\partial\Omega\right) = \mathbf{0}, \end{align} \]
what is called adhesion condition. This can be understood as a parameterization of the kinematic boundary condition for small-scale effects on rough surfaces. The term soil friction describes the influence of the validity of the kinematic boundary condition on the earth's surface on the dynamics of a geofluid. If a phase interface is in motion, such as the sea surface, one requires that the normal component of the velocity in each medium converges to a common limit as it is approached. For the parallel components this must be the case. A. However, this does not apply.
If the pressure field is continuous, then $A$ applies at a phase boundary
\[ \begin{align} \lim_{\mathbf{r} \uparrow A} p = \lim_{\mathbf{r} \downarrow A} p,\tag{8.109}\label{eq:boundary_cond_dynamic} \end{align} \]
where the two limit values represent the different sides of the phase interface. This applies except for any surface tension that may be present, which is precisely a discontinuity in the pressure field at an interface. Eq. (8.109) is called dynamic boundary condition.