The momentum $\mathbf{p}$ of a particle volume $\Delta V$ with $N_h$ gas molecules and $N_c$ condensate nuclei is given by
\[ \begin{align} \mathbf{p} = \sum_{i = 1}^{N_h + N_c}m_i\mathbf{v}_i\stackrel{\href{ch-05-state-of-the-atmosphere.html#eq:def_windvector}{\text{Eq. (6.2)}}}{=}\mathbf{v}\sum_{j = 1}^{N_h}m_j + \sum_{j = N_h + 1}^{N_h + N_c}m_j\left(\mathbf{v} + \Delta\mathbf{v}_j\right), \end{align} \]
Here $\Delta\mathbf{v}_j$ is the settling velocity of the corresponding condensate class, assumed to be independent of density. In inertial systems, Newton's second axiom can be written as
\[ \begin{align} \md{\mathbf{p}} &= \rho \Delta V\md{}\mathbf{v} = \mathbf{F}_{\text{int}} + \mathbf{F}_{\text{ext}} \end{align} \]
The force is the sum of internal and external forces. Assuming that the internal forces consist of pairwise interactions between 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. For $\mathbf{F}_{\mathrm{ext}}$, two types of forces are relevant:
Force fields of distant charge and mass distributions
Interactions with neighboring particles
The first point usually includes only the gravitational field. For the weight force $\mathbf{F}_g$, one has
\[ \begin{align} \mathbf{F}_g = \sum_{i = 1}^{N}m_i\mathbf{g} = \rho\Delta V \mathbf{g}. \end{align} \]
For the force due to pressure acting on the open sphere $\Omega \coloneqq B_r\left(\mathbf{r}_0\right)$ with radius $r$, center $\mathbf{r}_0$, and $\Delta V = \frac{4}{3}\pi r^3$, one obtains
\[ \begin{align} \mathbf{F}_p &= -\int_{\partial\Delta V}pd\mathbf{A}. \end{align} \]
Assuming a spherical particle, it follows that
\[ \begin{align} \mathbf{F}_p &= -\int_{\partial\Delta V}pr^2\mathbf{n}d\omega\approx-\int_{\partial\Delta V}\left(p\left(\mathbf{r}_0\right) + \mathbf{r}\cdot\nabla p\right)r^2\mathbf{n}d\omega \end{align} \]
with $d\omega$ denoting the solid-angle element. To carry out the integration, one aligns $\nabla p$ with the $z$-axis without loss of generality, so that
\[ \begin{align} \nabla p = \left|\nabla p\right|\mathbf{e}_z. \end{align} \]
This yields, with $p_0 \coloneqq p\left(\mathbf{r}_0\right)$,
\[ \begin{align} \mathbf{F}_p &= -\int_{\partial\Delta V}\left(p_0 + \mathbf{r}\cdot\nabla p\right)r\mathbf{r}d\omega=-r\int_{\partial\Delta V}\left(\mathbf{r}\cdot\nabla p\right)\mathbf{r}d\omega = -r^2\left|\nabla p\right|\int_{\partial\Delta V}\cos\left\lbrace\nabla p, \mathbf{r}\right\rbrace\mathbf{r}d\omega\nonumber\\ &= -r^3\left|\nabla p\right|\int_{\vartheta = 0}^\pi\int_{\phi = 0}^{2\pi}\cos\left(\vartheta\right)\sin\left(\vartheta\right)\left(\begin{array}{c} \sin\left(\vartheta\right)\cos\left(\phi\right)\\ \sin\left(\vartheta\right)\sin\left(\phi\right)\\ \cos\left(\vartheta\right) \end{array}\right)d\phi d\vartheta\nonumber\\ &= -r^3\nabla p2\pi\int_0^\pi\cos^2\left(\vartheta\right)\sin\left(\vartheta\right)d\vartheta = r^3\nabla p2\pi\left[\frac{1}{3}\cos^3\left(\vartheta\right)\right]_0^\pi = -\frac{4}{3}\pi r^3\nabla p = -\Delta V\nabla p. \end{align} \]
This force is called the pressure gradient force. One obtains
\[ \begin{align} \rho \Delta V\md{}\mathbf{v} = \rho\Delta V \mathbf{g} - \Delta V\nabla p, \end{align} \]
Rearranging for the particle acceleration $\md{\mathbf{v}}$ gives
\[ \begin{align} \md{\mathbf{v}} = -\frac{1}{\rho}\nabla p + \mathbf{g}.\tag{8.10}\label{eq:newton_II_fluid} \end{align} \]
This equation is called the 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, one introduces the momentum flux density tensor $\Pi$
\[ \begin{align} \Pi_{i, j} \coloneqq \rho U_iU_j \end{align} \]
for $1 \leq i, j \leq 3$. For $1 \leq i \leq 3$,
\[ \begin{align} \Pi\cdot\mathbf{e}_i = \sum_{j = 1}^3\rho v_jv_i\mathbf{e}_j = v_i\rho\mathbf{v} \end{align} \]
is the flux 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 Sect. 6.2.2, one has
\[ \begin{align} \frac{\partial\mathbf{p}}{\partial t} &= \int_\Omega\frac{\partial\left(\rho\mathbf{v}\right)}{\partial t}d^3r = \int_\Omega-\nabla p + \rho\mathbf{g}d^3r - \int_{\partial\Omega}\Pi\cdot d\mathbf{n}\nonumber\\ &\Rightarrow \int_\Omega\frac{\partial\left(\rho\mathbf{v}\right)}{\partial t}d^3r = \int_\Omega-\nabla p + \rho\mathbf{g} - \nabla\cdot\Pi d^3r\nonumber\\ &\Rightarrow \int_\Omega\frac{\partial\left(\rho\mathbf{v}\right)}{\partial t} + \nabla p + \nabla\cdot\Pi - \rho\mathbf{g}d^3r = 0. \end{align} \]
Thus, one already has
\[ \begin{align} \frac{\partial\left(\rho\mathbf{v}\right)}{\partial t} + \nabla p + \nabla\cdot\Pi - \rho\mathbf{g} = 0.\tag{8.15}\label{eq:momentum_flux_form} \end{align} \]
For the divergence of $\Pi$, it 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 one has
\[ \begin{align} \nabla\cdot\Pi &= \mathbf{v}\left(\mathbf{v}\cdot\nabla\rho + \rho\nabla\cdot\mathbf{v}\right) + \left(\rho\mathbf{v}\cdot\nabla\right)\mathbf{v}. \end{align} \]
Thus it 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, one finally obtains
\[ \begin{align} \frac{\partial\mathbf{v}}{\partial t} + \left(\mathbf{v}\cdot\nabla\right)\mathbf{v} = -\frac{1}{\rho}\nabla p + \mathbf{g}. \end{align} \]
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 Sect. D.1). If one measures accelerations in this coordinate system, one can relate them to the acting forces using Newton's second axiom. However, if one chooses a basis rotating at the Earth's angular velocity, additional acceleration terms arise that do not result from physical forces. For the basis of the global coordinates, in the resting coordinates, one has
\[ \begin{align} \mathbf{e}_x\left(t\right) = \left( \begin{array}{c} \cos\left(\omega t\right) \\ \sin\left(\omega t\right) \\ 0 \end{array}\right), & {} & \mathbf{e}_y\left(t\right) = \left( \begin{array}{c} - \sin\left(\omega t\right) \\ \cos\left(\omega t\right)\\ 0 \end{array}\right), & {} & \mathbf{e}_z\left(t\right) = \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right) \end{align} \]
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, for the velocity, one has
\[ \begin{align} \mathbf{v} = \newdot{\mathbf{r}} = \newdot{x}\mathbf{e}_x + \newdot{y}\mathbf{e}_y + \newdot{z}\mathbf{e}_z + x\newdot{\mathbf{e}}_x + y\newdot{\mathbf{e}}_y + z\newdot{\mathbf{e}}_z. \end{align} \]
For the time derivatives of the basis vectors, one has
\[ \begin{align} \newdot{\mathbf{e}}_x &= \omega\left( \begin{array}{c} - \sin\left(\omega t\right)\\ \cos\left(\omega t\right)\\ 0 \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \\ \omega \end{array}\right)\times \left( \begin{array}{c} \cos\left(\omega t\right)\\ \sin\left(\omega t\right) \\ 0 \end{array}\right) = \omegabi\times\mathbf{e}_x,\\ \newdot{\mathbf{e}}_y &= \omega\left( \begin{array}{c} - \cos\left(\omega t\right) \\ - \sin\left(\omega t\right) \\ 0 \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \\ \omega \end{array}\right)\times\left( \begin{array}{c} - \sin\left(\omega t\right)\\\cos\left(\omega t\right) \\ 0 \end{array}\right) = \omegabi\times\mathbf{e}_y,\\ \newdot{\mathbf{e}}_z &= \mathbf{0} = \omegabi\times\mathbf{e}_z. \end{align} \]
The velocity vector is therefore written as
\[ \begin{align} \mathbf{v} = \mathbf{v}' + x\omegabi\times\mathbf{e}_x + y\omegabi\times\mathbf{e}_y + z\omegabi\times\mathbf{e}_z = \mathbf{v}' + \omegabi\times\mathbf{r}, \tag{8.25}\label{eq:geschw_rot_system} \end{align} \]
Here, for the velocity measured in the primed system, $\mathbf{v}' = \newdot{x}\mathbf{e}_x + \newdot{y}\mathbf{e}_y + \newdot{z}\mathbf{e}_z$ was written as an abbreviation, and the linearity of the vector product, Eqs. (A.149) - (A.150), was exploited. Newton's second axiom is a statement about the acceleration $\md{\mathbf{v}}$, so one differentiates the above equation once more with respect to 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 $\left(\md{\mathbf{v'}}\right)'$.
The position-dependent term $-\omegabi\times\left(\omegabi\times\mathbf{r}\right)$ is the centrifugal acceleration; the velocity-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 forces that a fluid exerts on itself. Without loss of generality, 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 row of $T$ describes forces acting on the faces of the cube (anti)parallel to the $i$-axis. The entry in the $j$-th column denotes the force acting on this face in the $j$-direction if the normal of the face under consideration is parallel to the $i$-axis; otherwise the signs are reversed. This means that compressive forces are always 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, for the stress tensor, one has
\[ \begin{align} T_{i, j} = -p\delta_{i, j}. \end{align} \]
The sign comes from the convention that compressive forces are always negative. One now writes 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 $\mathbf{D}$ acting on this particle 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 symmetric. Therefore the tensor $\tau$ must also be symmetric.
The quantity $\frac{\partial v_i}{\partial x_j}$ is called the velocity gradient tensor. For it, one can write
\[ \begin{align} \frac{\partial v_i}{\partial x_j} = S_{i, j} + \frac{1}{2}R_{i, j} \end{align} \]
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 the strain tensor, which is symmetric, and $R_{i, j}$ the 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 strain tensor are therefore relevant for the stress. Therefore, one can write
\[ \begin{align} \tau_{i, j} = \sum_{n, m}K_{i, j, m, n}S_{m, n}\tag{8.35}\label{eq:strain_to_stress} \end{align} \]
with a complex fourth-order tensor $K$ whose elements describe how the fluid reacts to deformations. The sums 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 the dynamic viscosity and $\lambda$ the Lamé constant. Substituting this into Eq. (8.35), one obtains
\[ \begin{align} \tau_{i, j} = \sum_{n, m}\left(\lambda\delta_{i, j}\delta_{m, n} + 2\mu\delta_{i, m}\delta_{j, n}\right)S_{m, n} = \lambda\delta_{i, j}\sum_{m}S_{m, m} + 2\mu S_{i, j}. \end{align} \]
From Eq. (8.34), it follows
\[ \begin{align} \sum_{m}S_{m, m} = \nabla\cdot\mathbf{v}. \end{align} \]
Thus one has
\[ \begin{align} T_{i, j} = -p\delta_{i, j} + \lambda\delta_{i, j}\nabla\cdot\mathbf{v} + 2\mu S_{i, j}\tag{8.41}\label{eq:stress_tensor_deriv_0} \end{align} \]
For $i = j$, one obtains from this
\[ \begin{align} T_{i, i} = -p + \lambda\nabla\cdot\mathbf{v} + 2\mu S_{i, i}. \end{align} \]
Summing 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 define an average pressure
\[ \begin{align} \newoverline{p} \coloneqq -\frac{1}{3}T_{i, i} \end{align} \]
For it, one has
\[ \begin{align} p - \newoverline{p} = \sum_{i}T_{i, i} + \left(\frac{2}{3}\mu + \lambda\right)\nabla\cdot\mathbf{v}. \end{align} \]
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 made in the following. For the Lamé constant $\lambda$, in terms of the bulk viscosity, one obtains
\[ \begin{align} \lambda = \mu_v - \frac{2}{3}\mu.\tag{8.48}\label{eq:lame_viscosity_relation} \end{align} \]
Substituting this into Eq. (8.41), one obtains
\[ \begin{align} T_{i, j} = -p\delta_{i, j} + \left(\mu_v - \frac{2}{3}\mu\right)\delta_{i, j}\nabla\cdot\mathbf{v} + 2\mu S_{i, j}. \end{align} \]
Another way of writing this is:
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) becomes
\[ \begin{align} T = \left(\begin{array}{ccc} -p + 2\mu\frac{\partial u}{\partial x} + \left(\mu_v - \frac{2}{3}\mu\right)\nabla\cdot\mathbf{v} & \mu\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right) & \mu\left(\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x}\right)\\ \mu\left(\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\right) & -p + 2\mu\frac{\partial v}{\partial y} + \left(\mu_v - \frac{2}{3}\mu\right)\nabla\cdot\mathbf{v} & \mu\left(\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y}\right)\\ \mu\left(\frac{\partial w}{\partial x} + \frac{\partial u}{\partial z}\right) & \mu\left(\frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}\right) & -p + 2\mu\frac{\partial w}{\partial z} + \left(\mu_v - \frac{2}{3}\mu\right)\nabla\cdot\mathbf{v} \end{array}\right).\nonumber\\ & \end{align} \]
From this, the frictional acceleration $\mathbf{f}_R$ can be calculated:
\[ \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} \]
It was assumed that the viscosities are homogeneous. For friction, expressions other than the one obtained from Eq. (8.52) are often used. Therefore, from now on, the derivations mostly carry along a general expression
\[ \begin{align} \mathbf{f}_R = \left(\begin{array}{c} f_{R,x}\\ f_{R,y}\\ f_{R,z} \end{array}\right) \equiv \left(\begin{array}{c} \mathbf{f}_R^{(H)}\\ f_{R,z} \end{array}\right) \end{align} \]
for the frictional acceleration. This is sensible despite the relative smallness of friction in the atmosphere, owing to its importance for the energy cascade (see Sect. 8.2.3).
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, because $S_{m, m} = \nabla\cdot\mathbf{v} = 0$, the equation
\[ \begin{align} T_{i, j} = -p\delta_{i, j} + 2\mu S_{i, j}. \end{align} \]
holds. From this it follows
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 reads
\[ \begin{align} \md{\mathbf{v}} = \frac{\partial\mathbf{v}}{\partial t} + \left(\mathbf{v}\cdot\nabla\right)\mathbf{v} = -\frac{1}{\rho}\nabla p + \mathbf{g} + \mathbf{f}_R \end{align} \]
In this case, for the material change of the specific kinetic energy $k$, one has
\[ \begin{align} \md{k} = \md{}\left(\frac{1}{2}\mathbf{v}^2\right) = \mathbf{v}\cdot\md{\mathbf{v}} = -\frac{1}{\rho}\nabla p\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{g} + \mathbf{v}\cdot\mathbf{f}_R = -\frac{1}{\rho}\sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} + \sum_{i = 1}^3v_ig_i + \frac{1}{\rho}\sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i}. \end{align} \]
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, for the total energy in this region in the absence of heat fluxes (except dissipation), one has:
\[ \begin{align} \frac{d}{dt}\int_{\Omega\left(t\right)}\rho\left(k + i\right)d^3r &= \int_{\Omega\left(t\right)}\rho\mathbf{v}\cdot\mathbf{g}d^3r + \int_{\partial\Omega\left(t\right)}\mathbf{f}\cdot\mathbf{v}dA.\tag{8.59}\label{eq:dissipation_deriv_1} \end{align} \]
Here $\mathbf{v}$ is the force acting on the surface of the control volume. For it, with the definitions at the beginning of Sect. 8.2, one has
\[ \begin{align} \int_{\partial\Omega\left(t\right)}\mathbf{f}\cdot\mathbf{v}dA &= \int_{\partial\Omega\left(t\right)}\sum_{i, j = 1}^3n_iT_{i, j}v_jdA = \int_{\Omega\left(t\right)}\sum_{i, j = 1}^3\frac{\partial\left(T_{i, j}v_j\right)}{\partial x_i}d^3r. \end{align} \]
For the integrand one obtains
\[ \begin{align} \sum_{i, j = 1}^3\frac{\partial\left(T_{i, j}v_j\right)}{\partial x_i} &= \sum_{i, j = 1}^3v_j\frac{\partial T_{i, j}}{\partial x_i} + T_{i, j}\frac{\partial v_j}{\partial x_i} = \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i} + \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} - \sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} - \sum_{i = 1}^3p\frac{\partial v_i}{\partial x_i}. \end{align} \]
Substituting this into Eq. (8.59), one obtains
\[ \begin{align} \frac{d}{dt}\int_{\Omega\left(t\right)}\rho\left(k + i\right)d^3r &= \int_{\Omega\left(t\right)}\rho\sum_{i = 1}^3g_iv_i + \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i} + \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} - \sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} - \sum_{i = 1}^3p\frac{\partial v_i}{\partial x_i}d^3r. \end{align} \]
Letting $\Omega$ shrink to the size of a fluid particle, one obtains
\[ \begin{align} \md{\left[\rho\left(k + i\right)\right]} &= \sum_{i = 1}^3\rho v_ig_i + \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i} + \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} - \sum_{i = 1}^3v_i\frac{\partial p}{\partial x_i} - \sum_{i = 1}^3p\frac{\partial v_i}{\partial x_i}. \end{align} \]
Subtracting Eq. (8.58) from this, one obtains
\[ \begin{align} \md{\left(\rho i\right)} &= \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} - \sum_{i = 1}^3p\frac{\partial v_i}{\partial x_i} = -p\nabla\cdot\mathbf{v} + \sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} \equiv -p\nabla\cdot\mathbf{v} + \rho\epsilon. \end{align} \]
The specific dissipation rate, or dissipation $\epsilon$ for short, is defined by
\[ \begin{align} \epsilon \coloneqq \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} \stackrel{\href{#eq:def_rotation_tensor}{\text{Eqs. (8.34) - (8.34)}}}{=} \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}\left(S_{i, j} - \frac{1}{2}R_{i, j}\right) \stackrel{R_{i, j}\text{ antisymmetric}}{=} \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}S_{i, j}. \end{align} \]
As the result of this section, one has
Friction is usually thought of as a braking force. This is not always the case in fluids. To see this, one starts from the one-dimensional flow field
\[ \begin{align} u = 2 + \sin\left(x\right) > 0 \end{align} \]
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 $a_R$, up to a constant, reads
\[ \begin{align} a_R = \frac{\partial^2u}{\partial x^2} = -\sin\left(x\right). \end{align} \]
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) is negative; on average, friction therefore 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, a one-dimensional incompressible medium is again assumed, confined to the interval $\left[0, L\right]$; 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, one has
\[ \begin{align} \int_0^Lqdx = -\int_0^Lua_Rdx \geq 0, \end{align} \]
Thus, when integrated globally, friction destroys 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 this end, one first starts from a three-dimensional wind field of the form
\[ \begin{align} \mathbf{v}\left(\mathbf{r}\right) = \mathbf{u}\left(2 + \sin\left(\mathbf{k}\cdot\mathbf{r}\right)\right)\tag{8.72}\label{eq:diss_sample_wind_field} \end{align} \]
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. For the work done by the frictional acceleration, one can write locally
\[ \begin{align} \mathbf{v}\cdot\Delta\mathbf{v} = \sum_{i, j = 1}^3v_i\frac{\partial^2v_i}{\partial x_j^2} \end{align} \]
Since this expression is isotropic, one can restrict further consideration to the terms with the $x$-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 set
\[ \begin{align} \Omega \coloneqq \left[0, L_x\right] \times \left[0, L_y\right] \times \left[0, L_z\right] \end{align} \]
with the boundary conditions
\[ \begin{align} \mathbf{v}\cdot\mathbf{n} = 0, & {} & \nabla\mathbf{v}\cdot\mathbf{n} = \mathbf{0} \end{align} \]
on $\partial\Omega$, where $\mathbf{n}$ is a normal vector on $\partial\Omega$. Integrating Eq. (8.75) over $\Omega$, one obtains
\[ \begin{align} \int_\Omega\sum_{i = 1}^3v_i\frac{\partial^2v_i}{\partial x^2}d^3r &= \int_{z = 0}^{L_z}\int_{y = 0}^{L_y}\int_{x = 0}^{L_x}\sum_{i = 1}^3v_i\frac{\partial^2v_i}{\partial x^2}dxdydz. \end{align} \]
Carrying out only the integration over $x$, one obtains
\[ \begin{align} \int_0^{L_x}\sum_{i = 1}^3v_i\frac{\partial^2v_i}{\partial x^2}dx = \int_0^{L_x}v_x\frac{\partial^2v_x}{\partial x^2} + v_y\frac{\partial^2v_y}{\partial x^2} + v_z\frac{\partial^2v_z}{\partial x^2}dx. \end{align} \]
Integrating by parts again, 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 this end, one again starts from 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, for this special wind field, the term with $\nabla\cdot\mathbf{v}$ in Eq. (8.52) drops out, 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.
Multiplying Eq. (8.52) by $\rho\mathbf{v}$, 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, in analogy to the incompressible case, one has
\[ \begin{align} \int_\Omega\mu\mathbf{v}\cdot\Delta\mathbf{v}d^3r \leq 0.\tag{8.84}\label{eq:friction_dissipation_deriv_0} \end{align} \]
For the second term, one calculates
\[ \begin{align} \int_0^{L_x}u\frac{\partial\nabla\cdot\mathbf{v}}{\partial x}dx &= \left[u\nabla\cdot\mathbf{v}\right]_0^{L_x} - \int_0^{L_x}\frac{\partial u}{\partial x}\nabla\cdot\mathbf{v}dx. \end{align} \]
With the kinematic boundary condition $u\left(x = 0, L_x\right) = 0$ the first term drops out. Thus one has
\[ \begin{align} \int_\Omega\mathbf{v}\cdot\nabla\left(\nabla\cdot\mathbf{v}\right)d^3r = -\int_\Omega\left(\nabla\cdot\mathbf{v}\right)^2d^3r \leq 0. \end{align} \]
Thus, also in the compressible case, one has
According to Eq. (8.66), one has
\[ \begin{align} \epsilon &= \frac{1}{\rho}\sum_{i, j = 1}^3\tau_{i, j}S_{i, j} \stackrel{\href{#eq:stress-tensor}{\text{Eq. (8.50)}}}{=} \frac{1}{\rho}\sum_{i, j = 1}^3\left[2\mu\left(S_{i, j} - \frac{1}{3}\nabla\cdot\mathbf{v}\delta_{i, j}\right) + \mu_v\nabla\cdot\mathbf{v}\delta_{i, j}\right]S_{i, j}\nonumber\\ &= \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 - \frac{2}{3}\frac{\mu}{\rho}\sum_{i, j = 1}^3\nabla\cdot\mathbf{v}\delta_{i, j}S_{i, j} + \frac{\mu_v}{\rho}\sum_{i, j = 1}^3\nabla\cdot\mathbf{v}\delta_{i, j}S_{i, j}\nonumber\\ &\stackrel{\href{#eq:def_rotation_tensor}{\text{Eq. (8.34)}}}{=} \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 - \frac{2}{3}\frac{\mu}{\rho}\sum_{i, j = 1}^3\nabla\cdot\mathbf{v}\delta_{i, j}\frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right) + \frac{\mu_v}{\rho}\sum_{i, j = 1}^3\nabla\cdot\mathbf{v}\delta_{i, j}\frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right)\nonumber \end{align} \] \[ \begin{align} &= \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 - \frac{2}{3}\frac{\mu}{\rho}\nabla\cdot\mathbf{v}\sum_{i, j = 1}^3\delta_{i, j}\frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right) + \frac{\mu_v}{\rho}\nabla\cdot\mathbf{v}\sum_{i, j = 1}^3\delta_{i, j}\frac{1}{2}\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right)\nonumber\\ &= \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 - \frac{2}{3}\frac{\mu}{\rho}\left(\nabla\cdot\mathbf{v}\right)^2 + \frac{\mu_v}{\rho}\left(\nabla\cdot\mathbf{v}\right)^2 = \frac{2\mu}{\rho}\sum_{i, j = 1}^3S_{i, j}^2 + \left(\frac{\mu_v}{\rho} - \frac{2}{3}\frac{\mu}{\rho}\right)\left(\nabla\cdot\mathbf{v}\right)^2. \end{align} \]
According to Eq. (8.48), one has
\[ \begin{align} \mu_v - \frac{2}{3}\mu = \lambda, \end{align} \]
here $\lambda \geq 0$ is the Lamé constant. Thus one has
According to Eq. (8.87), friction destroys kinetic energy on a global average, but, as shown in Sect. 8.2.3, can also produce kinetic energy locally. The dissipation, on the other hand, is always positive. Locally, one thus has
\[ \begin{align} \rho\mathbf{v}\cdot\mathbf{f}_R + \rho\epsilon \stackrel{\text{in general}}{\not=} 0. \end{align} \]
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
\[ \begin{align} \int_\Omega\rho\epsilon + \rho\mathbf{v}\cdot\mathbf{f}_Rd^3r &\stackrel{\href{#eq:dissipation}{\text{Eq. (8.66)}}}{=} \int_\Omega\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} + \rho\mathbf{v}\cdot\mathbf{f}_Rd^3r = \int_\Omega\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} + \rho\sum_{i, j = 1}^3\frac{1}{\rho}v_j\frac{\partial\tau_{i, j}}{\partial x_i}d^3r\nonumber\\ &= \int_\Omega\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} + \sum_{i, j = 1}^3v_j\frac{\partial\tau_{i, j}}{\partial x_i}d^3r = \int_\Omega\sum_{i, j = 1}^3\tau_{i, j}\frac{\partial v_j}{\partial x_i} + v_j\frac{\partial\tau_{i, j}}{\partial x_i}d^3r\nonumber\\ &= \int_\Omega\sum_{i, j = 1}^3\frac{\partial\left(\tau_{i, j}v_j\right)}{\partial x_i}d^3r. \end{align} \]
As an example, one first considers only the terms with partial derivatives with respect to $x$ and assumes $\Omega = \left[0, L_x\right] \times \left[0, L_z\right] \times \left[0, L_y\right]$:
\[ \begin{align} \int_{0}^{L_x}\sum_{j = 1}^3\frac{\partial\left(\tau_{x, j}v_j\right)}{\partial x}dx &= \left[\sum_{j = 1}^3\tau_{x, j}v_j\right]_0^{L_x} = \left[\tau_{x, x}u + \tau_{x, y}v + \tau_{x, z}w\right]_0^{L_x} \stackrel{u\left(x = 0, L_x\right) = 0}{=} \left[\tau_{x, y}v + \tau_{x, z}w\right]_0^{L_x}\nonumber\\ &= \left[v\mu\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right) + w\mu\left(\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x}\right)\right]_0^{L_x} \nonumber\\ &\stackrel{u\left(x = 0, L_x\right) = 0}{=} \left[v\mu\frac{\partial v}{\partial x} + w\mu\frac{\partial w}{\partial x}\right]_0^{L_x} = \mu\left[v\frac{\partial v}{\partial x} + w\frac{\partial w}{\partial x}\right]_0^{L_x} \end{align} \]
With the boundary condition
\[ \begin{align} \nabla\mathbf{v}\cdot\mathbf{n} = 0 \end{align} \]
are partial derivatives of tangential wind components are equal to zero. Thus it follows
\[ \begin{align} \int_{0}^{L_x}\sum_{j = 1}^3\frac{\partial\left(\tau_{x, j}v_j\right)}{\partial x}dx &= 0. \end{align} \]
By cyclic permutation, this also follows for the other spatial directions. It is therefore indeed the case that
Inserting Eqs. (B.131) and (8.52) into Eq. (8.10) yields
$\mathbf{f} \coloneqq 2\omegabi$ 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 velocity 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) gives
\[ \begin{align} \left(\mathbf{v}\cdot\nabla\right)\mathbf{v} = \frac{1}{2}\nabla\left(\mathbf{v}\cdot\mathbf{v}\right) - \mathbf{v}\times\left(\nabla\times\mathbf{v}\right). \end{align} \]
If one defines the specific kinetic energy $k$ by
\[ \begin{align} k \coloneqq\frac{1}{2}\mathbf{v}^2, \end{align} \]
it follows that
Now Eq. (8.97) is written componentwise with respect to the location-dependent orthonormal basis of spherical coordinates. For the vector $\mathbf{f}$, one has
\[ \begin{align} \mathbf{f} = |\mathbf{f}|\left(\begin{array}{c} 0\\ \cos\left(\varphi\right)\\ \sin\left(\varphi\right) \end{array}\right)\eqqcolon\left(\begin{array}{c} 0\\ f'\\ f \end{array}\right). \end{align} \]
$f \coloneqq\mathbf{k}\cdot\mathbf{f}$ is called the Coriolis parameter. It follows that
\[ \begin{align} \mathbf{v}\times\mathbf{f} = \left(\begin{array}{c} fv - f'w\\ -fu\\ f'u \end{array}\right). \end{align} \]
Substituting this into the equation of motion yields
\[ \begin{align} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z} - \frac{uv\tan\left(\varphi\right)}{a + z} + \frac{uw}{a + z} &= -\frac{1}{\rho}\frac{\partial p}{\partial x} + g_x - f'w + fv + f_{R,x} \tag{8.104}\label{eq:x_momentum},\\ \frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} + w\frac{\partial v}{\partial z} + \frac{u^2\tan\left(\varphi\right)}{a + z} + \frac{vw}{a + z} &= -\frac{1}{\rho}\frac{\partial p}{\partial y} + g_y - fu + f_{R,y} \tag{8.105}\label{eq:y_momentum},\\ \frac{\partial w}{\partial t} + u\frac{\partial w}{\partial x} + v\frac{\partial w}{\partial y} + w\frac{\partial w}{\partial z} - \frac{u^2 + v^2}{a + z} &= -\frac{1}{\rho}\frac{\partial p}{\partial z} + g_z + f'u + f_{R,z} \tag{8.106}\label{eq:z_momentum}. \end{align} \]
If one wants to apply a system of differential equations to a set $\Omega$, boundary conditions are needed. Let $\Omega$ be bounded by condensed matter. Let $\mathbf{n}$ be a normal vector on $\partial\Omega$, then the following holds as the boundary condition for the wind field $\mathbf{v}$
\[ \begin{align} \mathbf{v}\cdot\mathbf{n} = 0, \end{align} \]
since particles cannot overcome the fixed boundary $\partial\Omega$. This is called the 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 the adhesion condition. This can be understood as a parameterization of the kinematic boundary condition for small-scale effects on rough surfaces. The term surface friction describes the influence of the validity of the kinematic boundary condition at 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, however, this need not hold in general.
If the pressure field is continuous, then at a phase boundary $A$ one has
\[ \begin{align} \lim_{\mathbf{r} \uparrow A} p = \lim_{\mathbf{r} \downarrow A} p,\tag{8.109}\label{eq:boundary_cond_dynamic} \end{align} \]
where the two limit values represent the different sides of the phase interface. This holds 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 the dynamic boundary condition.