# Modeling the Atmosphere, pt. 3 Pressure (y-axis) as a function of height (z-axis)

Parts one and two of modeling the atmosphere derived a way to describe atmospheric pressure as a function of altitude, assuming a constant temperature and acceleration due to gravity throughout the ideal gas. In this installment of Modeling the Atmosphere, the same function will be derived, this time using the Boltzmann distribution.

The Boltzmann distribution states that the fraction of particles $f(E_i)$ in an energy state $E_i$ is given by the equation $f(E_i) = \frac{g_i e^{\frac{-E_i}{kT}}}{Z}$

where $g_i$ represents the degeneracy of the energy level $E_i$ and the partition function $Z = \displaystyle \sum_i g_i e^{\frac{-E_i}{kT}}$

This distribution works for discrete energy levels, so a continuum approximation must be made for classical systems. By using a density of states $g(E)$, which supposes that there are $g(E) dE$ states in the energy interval $E$ to $E+dE$, the probability distribution for the energy $P(E) dE = \frac{g(E) e^{\frac{-E}{kT}} dE}{\int_0^\infty g(E) e^{\frac{-E}{kT}} dE}$

The first goal in when using the Boltzmann distribution to model an atmosphere is to find an expression for the energy $E$.

The equipartition theorem states that each degree of freedom for a particle receives $\frac{1}{2} k T$ of energy. So, in an isothermal atmosphere of ideal gas, each particle receives the same amount of kinetic energy $E_K$. The potential energy of a particle depends on its altitude and is given by the equation $E_P = mgz$, where $m$ is the mass of a particle, $g$ is the acceleration due to gravity (once again assumed to be constant), and $z$ represents altitude. Thus, the total energy $E$ of an ideal gas particle in an isothermal atmosphere is given by $E = E_K + E_P = E_K + mgz$

Assuming that the density of states $g(E)$ is a constant $g_0$, then the continuous distribution $P(E) dE = P(z) dz = \frac{g_0 e^{\frac{-E_K - mgz}{kT}} dz}{\int_0^\infty g_0 e^{\frac{-E_K - mgz}{kT}} dz} = \frac{g_0 e^{\frac{-E_K}{kT}} e^{\frac{-mg}{kT}z} dz}{\int_0^\infty g_0 e^{\frac{-E_K-mgz}{kT}} dz}$

Calculating the integral: $\int_0^\infty g_0 e^{\frac{-E_K-mgz}{kT}} dz = g_0 e^{\frac{-E_K}{kT}} \int_0^\infty e^{\frac{-mg}{kT}z} dz = g_0 e^{\frac{-E_K}{kT}} \left[\frac{-kT}{mg} e^{\frac{-mg}{kT}z}\displaystyle\right]_0^\infty$

Evaluate at zero: $e^{\frac{-mg}{kT}0} = 1$

And at infinity: $\displaystyle\lim_{z\to\infty} e^{\frac{-mg}{kT}z} = 0$

Thus: $g_0 e^{\frac{-E_K}{kT}} \frac{-kT}{mg} \left[e^{\frac{-mg}{kT}z} \displaystyle\right]_0^\infty = g_0 e^{\frac{-E_K}{kT}} \frac{-kT}{mg} [0-1] = g_0 e^{\frac{-E_K}{kT}} \frac{kT}{mg}$

Therefore $P(z)dz = \frac{g_0 e^{\frac{-E_K}{kT}} e^{\frac{-mg}{kT}z} dz}{\int_0^\infty g_0 e^{\frac{-E_K-mgz}{kT}}dz} = \frac{g_0 e^{\frac{-E_K}{kT}} e^{\frac{-mg}{kT}z} dz}{g_0 e^{\frac{-E_K}{kT}} \frac{kT}{mg}} = \frac{mg}{kT} e^{\frac{-mg}{kT}z} dz$

For a large system, this equation describes the fraction of particles $f(z)$ at a height between $z$ and $z+dz$: $f(z)dz = \frac{mg}{kT} dz \cdot e^{\frac{-mg}{kT}z}$

Notice that the coefficient $\frac{mg}{kT} dz$ has units $\frac{\text{kg }\frac{\text{m}}{\text{s}^2}}{\frac{\text{J}}{\text{K}}\text{ K}} \text{ m}$, which cancels out to be unitless.

How can this expression for the factional number of particles at a height $z$ be transformed to describe pressure as a function of altitude? The ideal gas law allows pressure $P$ to be expressed in terms of the number of particles $N$: $P = \frac{NkT}{V}$

The number of particles $N(z)dz$ between the height $z$ and $z+dz$ is given by the fraction of particles $f(z)$ between $z$ and $z+dz$ times the total number of particles $N_T$ in the system: $N(z)dz = N_T f(z)dz$

So the pressure as a function of altitude is given by: $P(z) = \frac{kT}{V} N_T f(z)dz = \frac{kT}{V} N_T \frac{mg}{kT} dz e^{\frac{-mg}{kT}z} = \frac{m g N_T dz}{V} e^{\frac{-mg}{kT}z}$

Notice that the coefficient $\frac{mgN_Tdz}{V}$ now has units of pressure: $\frac{\text{kg }\frac{\text{m}}{\text{s}^2}1\text{ m}}{\text{m}^3} = \frac{\text{N}}{\text{m}^2}$

Using the boundary condition at $z=0$, this coefficient is equal to the initial pressure $P_0$: $P(z) = P_0 e^{\frac{-mg}{kT}z}$

This equivalence $P_0 = \frac{m N_T g dz}{V}$

makes sense conceptually. To see why, note that $m N_T = M$, or the mass of all of the particles in the atmosphere, and $V = A dz$, where $A$ is the surface area of the ground. Then $\frac{m N_T g dz}{V} = \frac{M g dz}{A dz} = \frac{M g}{A}$

which describes the force $F = Mg$ of the entire atmosphere across the surface.

Thus, the exact same expression for pressure as a function of altitude (previously derived using a different method) has been found, this time starting from the Boltzmann distribution!