Alutsyah Luthfian | Create Your Badge
While seeing the weatherman said "It gonna be rain in Jakarta tomorrow", a critical person should think what mechanism driving the air to put down some of it moisture. He would run for YouTube, chasing the lapse rate videos showing the grow of cloud over time. Then he could also run for satellite images seeing the case of it. Then he would got a simple conclusion that the water vapor rises high to the sky and it cools down, and after that it rains. Then he asked again to himself, how a cloud could rise? Why not
all cloud rising and pour rain? Then he may google it up and found articles talking about it. Some articles said that convective energy available in a specific area governs that. It may say that convective energy, is related to static instability.
Here, i just trying to talk about static instability. In meteorology, especially the one talking about weather phenomenon happening at the spatial scale of 2 - 2000 km (mesoscale), there are 5 types of instability: Static instability, Inertial instability, Shear instability, Symmetric instability, and Centrifugal instability. Each type of instability has their own use. The static instability is used to determine whether a place is potential for thunderstorm buildup or no. Even in tornadogenesis, static instability plays a great role in determining how powerful the tornado will be.
Formulation of Buoyancy
Let we consider a piece of air parcel trying to rise in the sky. The vector working at this air parcel is buoyancy B defined as these.
\[\overrightarrow B = \overrightarrow g \cdot \frac{{\left( {T - \overline T } \right)}}{{\overline T }}\]
For sure, because of buoyancy, the parcel of air is rising. Let's see that Newton's Law stated a force has acceleration, so then we can infer that the buoyancy force
accelerates the air parcel
upward against gravitational force, $\overrightarrow B = \frac{{d\overrightarrow w }}{{dt}}$, $\overrightarrow w$ is upward velocity. We also know that the upward velocity $\overrightarrow w = \frac{{d\overrightarrow {\Delta z} }}{{dt}}$ for ${\overrightarrow {\Delta z} }$ is displacement along the troposphere's depth. Then, we could describe the buoyancy force in more intricating form like these.
\[\frac{{{d^2}\overrightarrow {\Delta z} }}{{d{t^2}}} = g \cdot \frac{{\left( {T - \overline T } \right)}}{{\overline T }}\]
Now its the time to talk about the two kinds of T in equation above. The
$T$ stands for
air parcel temperature and the
$\overline T $ stands for
environmental temperature.
Breaking Apart the $T$ and $\overline T $
In meteorology, the $T$ and $\overline T $ are actually changing with height. We usually says that temperature decreases with height, but let us change that view: in case of air inversion caused when hot air is advected to a cooler region, air aloft is
warmer than that of surface, so we better describe temperature as a continuous function of z; $T\left( z \right)$ and $\overline T \left( z \right)$. Then, if we fly from height z
0 to height a in troposphere,
- \[T\left( a \right) - T\left( z_0 \right) = \int\limits_{{z_0}}^a {T'\left( z \right)dz} \]
- \[\overline T \left( a \right) - \overline T \left( z_0 \right) = \int\limits_{{z_0}}^a {\overline T '\left( z \right)dz} \]
Analyzing the integral like a sir: $\int {UdV} = UV - \int {VdU} $. We define $U = T'\left( z_0 \right)$ or that $ \Rightarrow$ $U = \overline T '\left( z_0 \right)$, and $V = - (a - z)$ so that $dV = dz$, then equation (1.) and (2.) looks like these.
- \[T\left( a \right) - T\left( {{z_0}} \right) = T'\left( z_0 \right)\left( {a - {z_0}} \right) + \int\limits_{{z_0}}^a {\left( {a - z} \right)T''\left( z \right)dz} \]
- \[\overline T \left( a \right) - \overline T \left( {{z_0}} \right) = \overline T '\left( z_0 \right)\left( {a - z_0} \right) + \int\limits_{{z_0}}^a {\left( {a - {z}} \right)\overline T ''\left( z \right)dz} \]
You can analyze the integration above further, and you may found that it is never end. A note when doing integration above is
treating $V = (a - z)$ just like $V = - ({z^ * })$ , as we know that whenever we add or substract the domain z, it still located at the z-line but with another quantity. If you forgot to treat $V = (a - z)$ as $V = - ({z^ * })$, the integration result will be different with below.
- Temperature of air parcel at height a
\[T\left( a \right) = T\left( {{z_0}} \right) + T'\left( {{z_0}} \right)\left( {a - {z_0}} \right) + ... + \frac{{{T^n}\left( {{z_0}} \right)}}{{n!}}{\left( {a - {z_0}} \right)^n}\]
- Temperature of environment at height a
\[\overline T \left( a \right) = \overline T \left( {{z_0}} \right) + \overline T '\left( {{z_0}} \right)\left( {a - {z_0}} \right) + ... + \frac{{{{\overline T }^n}\left( {{z_0}} \right)}}{{n!}}{\left( {a - {z_0}} \right)^n}\]
To simplify, we can ignore the small figures of $\frac{{{T^n}\left( {{z_0}} \right)}}{{n!}}{\left( {a - {z_0}} \right)^n}$ and $\frac{{{{\overline T }^n}\left( {{z_0}} \right)}}{{n!}}{\left( {a - {z_0}} \right)^n}$ so that the equations above are limited to the second figure. Defining
${\Gamma _p} = T'\left( z_0 \right) \to \frac{{\partial T}}{{\partial z}}$, the change of air parcel temperature with the change of height and
$\gamma = \overline {T} '\left( {z_0} \right) \to \frac{{\partial \overline T }}{{\partial z}}$, the change of environmental air temperature as the height changes, so $T$ and $\overline T $ equals to:
- \[T = {T_0} + {\Gamma _p}\left( {\Delta z} \right)\]
- \[\overline T = {\overline T _0} + \gamma \left( {\Delta z} \right)\]
$T$ is parcel temperature at height a, ${T_0}$ is parcel temperature at height ${z_0}$, ${\Delta z} = a - {z_0}$, $\overline T$ is environmental temperature at height a, ${\overline T _0}$ is environmental temperature at height ${z_0}$.
The first state of air parcel in atmosphere is always equilibrum, means that air parcel temperature is the same with environmental temperature (${T_0} = {\overline T _0}$), so that equation (7.) and (8.) finally become:
- \[T = {T_0} + {\Gamma _p}\left( {\Delta z} \right)\]
- \[\overline T = {T _0} + \gamma \left( {\Delta z} \right)\]
Stability of Environmental Lapse Rate
Reviewing the equation for $\overrightarrow B$, let us analyze it in depth.
- \[\overrightarrow B = \overrightarrow g \cdot \frac{{\left( {T - \overline T } \right)}}{{\overline T }}\]
Then we replace the $T$ and $\overline T$ with ones from equation (9.) and (10.).
- \[\frac{{{d^2}\overrightarrow {\Delta z} }}{{d{t^2}}} = \overrightarrow g \frac{{\left( {{T_0} - {\Gamma _p} \cdot \Delta z - {T_0} + \gamma \cdot \Delta z} \right)}}{{{T_0} - \gamma \cdot \Delta z}}\]
- \[\frac{{{d^2}\overrightarrow {\Delta z} }}{{d{t^2}}} = - \overrightarrow g \Delta z\frac{{\left( {{\Gamma _p} - \gamma } \right)}}{{{T_0} - \gamma \cdot \Delta z}}\]
If the ${\Delta z}$ is small, then the ${\gamma \cdot \Delta z}$ is very small compared to ${T_0}$ so it is possible for us to ignore ${\gamma \cdot \Delta z}$. After that, the equation (13.) becomes this.
- \[\frac{{{d^2}\overrightarrow {\Delta z} }}{{d{t^2}}} + \overrightarrow g \Delta x\frac{{\left( {{\Gamma _p} - \gamma } \right)}}{{{T_0}}} = 0\]
The equation (14.) above is a form of
second order ordinary differential equation. Some people likes to use the term
linear instead of
ordinary but this is just a matter of convenience. For brevity, you can click
here if you wanna know much about theorems and solving the second order ordinary differential equation. The main point is the equation (14.) has the solution as presented below.
- \[\Delta z\left( t \right) = {C_1} \cdot {e^{i{{\left[ {{\textstyle{g \over {{T_0}}}}\left( {{\Gamma _p} - \gamma } \right)} \right]}^{\frac{1}{2}}}t}} + {C_2} \cdot {e^{ - i{{\left[ {{\textstyle{g \over {{T_0}}}}\left( {{\Gamma _p} - \gamma } \right)} \right]}^{\frac{1}{2}}}t}}\]
$C_1$ and $C_2$ is a constant depending on the magnitude and direction of initial parcel displacement. We don't use vector here because we emphasize the quantity of the vector as a function of time.
If ${\Gamma _p}>{\gamma }$, the equation (15.) preserves its form. To know the characteristics of equation (15.) we should convert the exponentials into a complex function, based from ${e^{i\left( \theta \right)}} = \cos \left( \theta \right) + i \cdot \sin \left( \theta \right)$, as invented by Euler via Maclaurin series expansion. Then, we just consider the real part of ${e^{i\left( \theta \right)}}$, that is $\cos \left( \theta \right)$ (because we can't see imaginary waves!
).
After changing the equation (15.) into complex form and taking just the real part, then we obtain this very (16.) equation.
- \[\Delta z\left( t \right) = \left( {{C_1} + {C_2}} \right) \cdot \cos \left\{ {{{\left[ {{\textstyle{g \over {{T_0}}}}\left( {{\Gamma _p} - \gamma } \right)} \right]}^{{\textstyle{1 \over 2}}}}t} \right\}\]
Seeing the equation (16.) above, reminds us about common wave formula, the $\left( {{C_1} + {C_2}} \right)$ is the amplitude, and ${{{\left[ {{\textstyle{g \over {{T_0}}}}\left( {{\Gamma _p} - \gamma } \right)} \right]}^{{\textstyle{1 \over 2}}}}}$ is angular frequency of parcel motion, namely
Brunt–Väisälä frequency. This means, when parcel lapse rate is bigger than environmental lapse rate, the air parcel would fail to rise. Instead of rising, the air parcel will trigger a gravity wave that propagates away from the location of displacement. The distortion seen above car's body in a very hot summer is a good approximation. Another approximation is the lines of altocumulus cloud form at afternoon after a stable day.
If ${\Gamma _p}<{\gamma }$, then $i{\left[ {{\textstyle{g \over {{T_0}}}}\left( {{\Gamma _p} - \gamma } \right)} \right]^{{\textstyle{1 \over 2}}}}$ becomes real. When the t is very big, the C
1 can be ignored, and the equation would be like this.
- \[\Delta z\left( t \right) = {C_2}{e^{{{\left[ {{\textstyle{g \over {{T_0}}}}\left( {\gamma - {\Gamma _p}} \right)} \right]}^{{\textstyle{1 \over 2}}}}t}}\]
From equation (17.) above, it is obvious that when parcel lapse rate is smaller than environmental lapse rate, the air parcel would rise as the time going. The rising air parcel will release its heat and condenses into cloud and rain as it cools.
Convective Available Potential Energy (CAPE)
From the buoyancy equation above, an air parcel
can only rise automatically if the temperature of air parcel is
warmer than the environmental temperature. The level where the air parcel becomes warmer than the environmental temperature is called
Level of Free Convection or LFC. Other point important in determining CAPE is equilibrum level, is when the air parcel temperature cools down to be the same with environmental temperature again.
This is the skew-T log-P plot showing CAPE, level of free convection, and equilibrum level.
Deriving CAPE:
- \[\frac{{d\overrightarrow w }}{{dt}} = \overrightarrow B \]
- Multiplying (18.) with $\overrightarrow w \equiv \frac{{d\overrightarrow z }}{{dt}}$
\[\overrightarrow w \frac{{d\overrightarrow w }}{{dt}} = \overrightarrow B \frac{{d\overrightarrow z }}{{dt}}\]
- \[\frac{d}{{dt}}\left( {\frac{{{{\overrightarrow w }^2}}}{2}} \right) = \overrightarrow B \frac{{d\overrightarrow z }}{{dt}}\]
- Canceling the $dt$, and integrating (20.) from LFC from EL.
\[\int\limits_{LFC}^{EL} {d{{\overrightarrow w }^2}} = 2\int\limits_{LFC}^{EL} {\overrightarrow B \cdot d\overrightarrow z } \]
- Vector dot product is scalar. So then,
\[w_{EL}^2 - w_{LFC}^2 = 2\left[ {CAPE} \right]\]
- Since the vertical velocity $w$ at level of free convection equals 0, so vertical velocity at equilibrum level is considered maximum. Then,
\[w_{EL}^2 = 2\left[ {CAPE} \right]\]
- \[\frac{1}{2} \cdot w_{EL}^2 = CAPE\]
Putting All Above Together
What drive cloud to pour rain? It is
instability. Instability is driven by difference between air parcel temperature and environmental temperature. The more moisture an air parcel has, its lapse rate would be smaller (remember that water vapor is a good heat storage). Also, sky above a suburb or city is
normally has bigger lapse rate compared to a field or forest. Steep lapse rate also experienced in frontal boundary. In mountains, the environmental lapse rate experienced there is a mix between horizontal and vertical environmental lapse rate, and that two are always big there. Big lapse rate of environment added with small lapse rate of air parcel would make the air parcel rise. The rising air parcel cools into water dropplet or ice (if the buoyancy is great), graupel or snow (if buoyancy is just positive).
A tremendous buoyancy will turn the air parcels into a tremendous castle of cumulonimbus cloud - that's why forecasters uses CAPE as a parameter for determining thunderstorm. Buoyancy can even tilt a horizontally aligned rotational motion of air into an EF-5 wedge tornado.
The buoyancy of air parcel in sea is relatively smaller than that of land (except in equator where the air parcel is forcedly lifted), because the environmental lapse rate is small, and so does the parcel lapse rate. In desert, the air parcel is mostly dry, so its lapse rate is big, along with the environmental lapse rate.
What about hurricane forming in the sea? It is the case of shear instability, that later turned on the static instability and moreover, the inertial instability.
Not all cloud pouring rain, because:
- Clouds have different rate of inner instability depending on their origin. The most unstable one will perform convection and possible to pour rain.
- The environmental lapse rate should be adequate to fire convection if no cloud is sufficiently unstable.
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