# Modeling the Atmosphere, pt. 1

There’s a physics joke where a dairy farmer calls up the local university asking for some help improving his cow business. He ends up on the phone with a physicist, who assures the man that a solution can be found. Months later the farmer gets a call back. It’s the physicist, who says excitedly, “I can can help you, but my model only works for the case of spherical cows in a vacuum.”

It’s a horrible joke. Non-physicists don’t find it funny, and physicists just nod and think, “yes, that’s exactly how we work.” But as an anecdote it proves useful, particularly for physicists about to present what will look like an incredibly reductionist model. Let’s take a look at one of physics’ most simplistic but useful representations of an atmosphere.

The exponential atmosphere model, also called the isothermal or barotropic atmosphere model, rests on three assumptions: the atmosphere is composed of an ideal gas, temperature is constant regardless of height, and acceleration due to gravity remains the same at all altitudes. These assumptions allow for a model in which pressure is a function of height:

$P(z)$

which means that atmospheric pressure $P$ depends only on the altitude $z$.

Attribution: Greg L at the English language Wikipedia

By definition, an ideal gas is composed of a very large number (at least one mole, i.e. $6.022\times 10^{23}$) of identical, spherical particles. These particles follow Newton’s laws of motion and don’t interact with each other or the system’s boundaries except for perfectly elastic collisions (i.e. collisions in which no kinetic energy is lost). Particle size measures very small (at least ten times as small) compared to the average distance between particles. The particles distribute uniformly and randomly throughout an ideal gas, and in equilibrium conditions some density (mass per unit volume) $\rho$ characterizes it. These assumptions taken together lead to the ideal gas law, stated mathematically as

$PV = NkT$

where $P$ represents pressure in Pascals $\text{Pa} = \frac{\text{N}}{\text{m}^2}$, $V$ represents volume in $\text{m}^3$, $N$ equals the number of particles (not to be confused with the $N$ for units of force in Newtons), $k$ is the Boltzmann constant $1.38 \times 10^{-23} \frac{\text{N}\text{ m}}{\text{ K}}$, and $T$ represents the temperature in Kelvin $K$.

Working from the ideal gas model, we can begin to construct a decently useful atmospheric model. To start, consider a thin slice of atmosphere $S$ parallel to the ground. We want this thin slice $S$ of atmosphere to be general, so instead of specifying a particular altitude for $S$ to occupy we say that it resides at some arbitrary height $z$ from the ground. The slice $S$ measures very thin, only some small amount that we will call $\Delta z$ from bottom to top.

What can we say about our slice $S$ of atmosphere? Although $S$ is paper-thin, it extends very far in the $x$ and $y$ directions. We can visualize $S$ as a large plane with some surface area $A$ (Technically, $S$ gradually curves, extending all the way around the Earth to form a thin spherical shell around the planet, but this model ignores that detail). Our slice $S$ of atmosphere may be thin in $z$, but its great extent in $x$ and $y$ means that, in total, $S$ contains many particles of air. Each particle contains some mass, and the sum of all these masses totals a surprisingly substantial aggregate mass $M$ for our slice $S$ of atmosphere.

We now know that $S$ is a large object of mass $M$ a height $z$ above the ground. Working from these facts, a reasonable question to ask is whether $S$ is falling downward. After all, most large, massive objects standing in midair do. But the answer is that it can’t be! Remember, our slice $S$ of atmosphere is general; it occupies an arbitrary altitude $z$ and represents any slice of atmosphere we might consider. Whatever holds true about our slice $S$ of atmosphere must hold true for any and all slices of atmosphere, regardless of altitude, otherwise $S$ wouldn’t be general. So, if $S$ is falling downward, then so must all slices of atmosphere. Under these circumstances, the entire atmosphere collapses into the ground. None of the atmospheres we observe display this behavior, so our arbitrary slice $S$ of atmosphere must not be falling.

We can ask the same question but in the opposite direction: is our slice $S$ of atmosphere rising? Again, it cannot be. If any given slice of atmosphere rose away from the ground, then the entire atmosphere would fly off into space. We do not observe this phenomenon, so it seems that an arbitrary slice of atmosphere does not rise. Therefore, because an arbitrary slice $S$ of atmosphere neither falls nor rises, we must conclude that $S$ is stationary – it’s suspended in midair.

How is it possible that an object $S$ of mass $M$ stands stationary in midair? The object must feel a downward force from to gravity in accordance with Newton’s second law of motion:

$\vec{F} = m\vec{a}$

In this system, the acceleration vector $\vec{a}$ would be the acceleration due to gravity $\vec{g} = g (\hat{-z}) = -g (\hat{z}) \approx - 9.81 \frac{\text{m}}{\text{s}^2} \hat{z}$, and the mass $m$ would be the mass $M$ of our slice of atmosphere $S$. Therefore, the force due to gravity on our slice $S$ of atmosphere is given by:

$\vec{F} = M \vec{g} = -M g \hat{z}$

(Remember, this atmospheric model assumes a constant acceleration due to gravity, i.e. $\vec{g}$ remains the same at all heights).

Clearly, our slice $S$ of atmosphere experiences a downward force. So why doesn’t it fall? The answer lies with pressure.

Pressure is a force per unit area (units $\frac{\text{N}}{\text{m}^2}$); an ideal gas exerts pressure at its boundaries. In our model, the slice $S$ forms a two-dimensional boundary of surface area $A$ for all of the air beneath it. Therefore, it experiences a force upwards due to pressure from the air underneath it:

$\vec{F} = P(z) A \hat{z}$

Air resides above our slice $S$ of atmosphere as well, exerting a force due to pressure downwards. Since $S$ has a thickness of $\Delta z$, the pressure from above differs slightly from the pressure from below:

$\vec{F} = P(z + \Delta z) A (\hat{-z}) = -P(z + \Delta z) A \hat{z}$

An object stands motionless only when all forces acting on that object sum to zero. All three forces must sum to zero in order to ensure that our slice $S$ of air remains in place. Since the vectors for all of our forces are in the same direction $\hat{z}$, we can work with their scalars:

$P(z) A - P(z + \Delta z) A - Mg = 0$

Rearranging gives

$P(z + \Delta z) - P(z) = \frac{-Mg}{A}$

If we divide both sides by $\Delta z$, some interesting things start to happen

$\frac{P(z+\Delta z) - P(z)}{\Delta z} = \frac{-Mg}{A\Delta z}$

The left side, $\frac{P(z+\Delta z) - P(z)}{\Delta z}$, may be replaced by $\frac{dP(z)}{dz}$ using the definition of the derivative (see equation 6). The change on the right side is less opaque. In the denominator, the product of the surface area $A$ of our slice $S$ of atmosphere with its height $\Delta z$ yields the volume $V = A\Delta z$ of $S$:

Now our equation has the form

$\frac{dP(z)}{dz} = \frac{-Mg}{V}$

There remains one final simplification. On the right side of the equation, $\frac{-Mg}{V}$, the fraction $\frac{M}{V}$ represents a particular quantity. Let’s examine the units in order to decide the nature of this quantity. Mass $M$ has units $\text{kg}$, and volume $V$ has units $\text{m}^3$. What quantity represents mass per unit volume $\frac{\text{kg}}{\text{m}^3}$? Density! Using density $\rho = \frac{M}{V}$, the equation becomes

$\frac{dP(z)}{dz} = -\rho g$

A small, single equation that describes the entire atmosphere! At least, to a reasonable approximation. Remember, this model assumes an ideal gas, constant temperature, and uniform acceleration due to gravity – none of which are true for any atmosphere! That reminds me of a physics joke…

At this point, we can describe the change in pressure $\frac{dP(z)}{dz}$ with respect to altitude $z$ in terms of density $\rho$ and acceleration due to gravity $g$. Although this relation is short and sweet, its form is not particularly useful. Future posts will delve into what exactly this equation means and explore the applications of its modified (more user-friendly) form, the barometric equation.