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# <p class="text-center">2024/03/04</p>
## <p class="text-center">Review</p>
### Momentum equation
1. vectorial form
$$
\begin{aligned}
\frac{d\vec{V}}{dt} = -\frac{\nabla p}{\rho}+\vec{g^*}+\nu \nabla^2 \vec{v} - 2\vec{\Omega}\times\vec{v}+\Omega^2\vec{R}
\end{aligned}
$$

2. Component equation
$$
\begin{aligned}
\frac{du}{dt} &= -\frac{1}{\rho}\frac{\partial p}{\partial x}+\nu\nabla^2 u+2\Omega v \sin\phi-2\Omega w \cos\phi+\frac{uv}{a}\tan\phi-\frac{uw}{a}\\
\frac{dv}{dt} &= -\frac{1}{\rho}\frac{\partial p}{\partial y}+\nu\nabla^2 v-2\Omega u \sin \phi-\frac{u^2}{a}\sin\phi\\
\frac{dw}{dt} &= -\frac{1}{\rho}\frac{\partial p}{\partial z}-g +\nu\nabla^2 w+2\Omega \cos \phi + \frac{u^2}{a}
\end{aligned}
$$
Generally, by scale analysis, the horizontal velocity is much larger than vertical velocity. With this sense, the Coriolis force can be simplified as:
$$
\begin{aligned}
(\frac{du}{dt})_{Co} &= 2\Omega v sin\phi \\
(\frac{dv}{dt})_{Co} &= -2\Omega u sin\phi
\end{aligned}
$$

### Angular velocity vector

This figure shows the components of the angular velocity likes below:
$$
\begin{aligned}
\vec{\Omega} = \Omega<0, cos\phi, sin\phi>
\end{aligned}
$$

With the vectorial form of Coriolis force, $-2\vec{\Omega}\times\vec{v}$, which components can be written as:
$$
\begin{aligned}
-2\vec{\Omega}\times\vec{v} = -2\Omega
|\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
0 & cos\phi & sin\phi\\
u & v & w
\end{vmatrix}| = -2\Omega<wcos\phi-vsin\phi, usin\phi, -ucos\phi>
\end{aligned}
$$

If only consider horizontal component of velocity and Coriolis force, the components can be simplified as:
$$
\begin{aligned}
\frac{du}{dt} &= fv \\
\frac{dv}{dt} &= -fu\\
\end{aligned}
\Rightarrow \frac{d\vec{V_h}}{dt} = -f\hat{k}\times\vec{V_h}
$$

## <p class="text-center">Derivation of Vectorial form of Momentum equation</p>

Suppose following equation is known that:
$$
\frac{D_a \vec{A}}{Dt} = \frac{D\vec{A}}{Dt}+\vec{\Omega}\times\vec{A}
$$

where $\vec{A}$ is an arbitrary property of atmosphere.
Consider position vector $\vec{r}$:
$$
\frac{D_a \vec{r}}{Dt} = \frac{D \vec{r}}{Dt}+\vec{\Omega}\times\vec{r}
$$

$$
\Rightarrow \vec{v_a} = \vec{v}+\vec{\Omega}\times\vec{r}
$$

Then consider the absolute velocity vector:
$$
\begin{aligned}
\frac{D_a \vec{v_a}}{Dt} &= \frac{D \vec{v_a}}{Dt}+\vec{\Omega}\times\vec{v_a} \\
\Rightarrow \frac{D_a \vec{v_a}}{Dt} &= \frac{D (\vec{v}+\vec{\Omega}\times\vec{r})}{Dt}+\vec{\Omega}\times(\vec{v}+\vec{\Omega}\times\vec{r}) \\
\Rightarrow \frac{D_a \vec{v_a}}{Dt} &= \frac{D \vec{v}}{Dt} + 2\cdot\vec{\Omega}\times\vec{v} + \Omega^2\vec{r}
\end{aligned}
$$

In this vectorial equation, the term $\frac{D_a \vec{v_a}}{Dt}$ is absolute acceleration in inertial frame, which follows the second law of Newton motion, including PGF, GF, VF.
And the $\frac{D \vec{v}}{Dt}$ is acceleration in rotating frame, which is we need to obtain.
Therefore, there is:
$$
\frac{D\vec{v}}{Dt} = -\frac{\nabla p}{\rho}+\vec{g}+\nu \nabla^2 \vec{v}-2\vec{\Omega}\times \vec{v}
$$

## <p class="text-center">Component Equation in spherical coordinate</p>

(Sor for the lazy QAQ)

## <p class="text-center">Hydrostatic and Hyposmetric Equation</p>

### Hydrostatic equation

Vertical momentum equation, ignoring VF, Coriolis force and curvature terms:
$$
\frac{Dw}{Dt} = -\frac{1}{\rho}\frac{\partial p}{\partial z}+\vec{g}
$$

When the vertical acceleration is 0, then the condition is hydrostatic atmosphere. And the vertical momentum equation can be rewritten as:
$$
\begin{aligned}
0 &= -\frac{1}{\rho}\frac{\partial p}{\partial z}-g
\end{aligned}
$$

$$
\Rightarrow \frac{\partial p}{\partial z} = -\rho g
$$

The equation is hydrostatic equation.

### Hyposmetric Equation

By hydrostatic equation:
$$
dp = -\rho g dz
$$

By ideal gas law:
$$
\begin{aligned}
dp &= -\frac{p}{RT}g dz \\
\Rightarrow -\frac{RT}{p} dp &= d\Phi
\end{aligned}
$$
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