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## Why is scale height expressed in terms of 1/e?

Yesterday (Monday 27 June) I wrote a blogpost entitled “What is the scale height of water vapour in the Earth’s atmosphere?“. In that blogpost I said that the scale height of the gas (nitrogen, oxygen) in the Earth’s atmosphere was 7.64km, and that this means that every 7.64km it reduces by a factor of $1/e$, which is a factor of $0.368$.

This means that, at 7.64km, the atmosphere has a thickness (pressure) of  $0.368 (= 36.8\%)$ of its value at sea level. If we go to twice this height, 15.28km, the pressure of the atmosphere is $0.135\% \; (= 0.368 \times 0.368)$ of its value at sea level. At $3 \times 7.64 = 22.92 \text{ km}$ it is $0.368^{3} = 0.05 = 5\%$ of its value at sea level, etc.

The question was asked, why is it a factor of $1/e$ that we quote for the scale height, and not $1/2$, or $1/\text{something else}$? The answer is the way that the formula for the scale height $H$ is derived. It comes about from integrating an infinitesimal change in pressure $dP$ divided by the pressure $P$, that is the integral of $(dP/P)$, and this is what leads to the exponential.

## Deriving the equation for the scale height of the atmosphere

We are going to determine the pressure of the atmosphere as a function of altitude. Pressure is defined as the force per unit area, and for the atmosphere the pressure is due to the weight (not mass) of the overlying atmosphere. This is why pressure goes down with altitude. Let us assume that at a particular height $z$ the atmosphere has a density (mass per unit volume) of $\rho$ and a pressure $P$.

If we have a small slab of volume $dV = Adz$, the mass of this slab will be $dm=\rho dV = A \rho dz$. The weight of any object is given by $dW=gdm$, so the weight of this slab is $Ag\rho dz$. But, pressure is force (weight) per unit area, so

$dP = \frac{dW}{A} = \frac{Ag \rho dz}{A} = g \rho dz \text{ (1)}$

Consider a slab of thickness $dz$ and area $A$, so the volume $dV=Adx$.

Re-arranging Equation (1) we get

$\frac{dP}{dz} = - g \rho \text{ (2)}$

where $g$ is the acceleration due to gravity at that particular altitude. In theory $g$ is a function of $z$, but because the change in $g$ is so small for the changes in $z$ that we will consider, we are going to assume it is constant. The minus sign in the above expression is because $P$ decreases as we increase $z$.

For an ideal gas, we can write

$pdV = NkT$

where $N$ is the number of molecules, $k$ is Boltzmann’s constant, and $T$ is the temperature in Kelvin. If the mass of each molecules is $M$, then the total mass of the slab of gas is $NM$, and so we can say that the density is

$\rho = \frac{NM}{dV} = \frac{NM}{1} \cdot \frac{P}{NkT} = \frac{MP}{kT} \text{ (3)}$

If we combine equations (2) and (3) we get

$\frac{dP}{dz} = -g \frac{MP}{kT} \text{ (4)}$

Re-arranging Equation (4) we get

$\frac{dP}{P} = -\frac{Mg}{kT}dz \text{ (5)}$

Equation (5) is a differential equation, so to solve it we integrate

$\int{ \frac{dP}{P} }= - \frac{Mg}{kT} \int{ dz }$

which becomes

$\ln{P} = -\frac{Mg}{kT} z +C \text{ (6)}$

where $C$ is a constant which we determine by the boundary conditions. $\ln{P}$ is the natural logarithm of the pressure $P$, that is the logarithm to the base $e$. The boundary conditions are that at $z=0, \; P(z)=P_{0}$, the pressure at sea level, so we can write

$\ln{ P_{0}} = 0 + C \rightarrow C = \ln{ P_{0} }$

Putting this back in Equation (6) we have

$\ln{P} = -\frac{Mgz}{kT} + \ln{ P_{0} } \rightarrow \ln{P} - \ln{ P_{0} } = -\frac{Mgz}{kT} \rightarrow \ln{ \left( \frac{ P }{ P_{0} } \right) } = -\frac{Mgz}{kT} \text{ (7)}$

We get rid of the logarithm in Equation (7) by taking the exponent, so it becomes

$\frac{ P }{ P_{0} } = e^{ -\frac{Mgz}{kT} } \text{ (8)}$

Finally, we define the scale height $H$ as $H = \frac{ kT }{Mg}$ so we have
$\boxed{ \frac{P}{P_{0}} = e^{-\frac{z}{H}} \text{ or } P=P_{0}e^{-\frac{z}{H}} \text{ (9)} }$

As we can see, the pressure varies with altitude in the sense that the ratio of pressure at any altitude $P$ to its value at sea level $P_{0}$ is given by an exponent; the negative sign in the exponent tells us that pressure will decrease with increasing altitude.

## The variation of pressure with altitude

If we plot Equation (9) we get the following (with a value of $H=7.64 \text{ km}$)

The variation of pressure with altitude assuming a scale height of 7.64km

This shows the exponential drop off of atmospheric pressure with altitude, as given in Equation (9) above. We can, however, plot the pressure (y-axis) on a logarithmic scale. We take Equation (9) and write

$\ln{ \frac{P}{P_{0} } } = - \frac{z}{H} \rightarrow \ln{P} - \ln{P_{0}} = -\frac{z}{H}$

which we can re-arrange to give

$\boxed{ \ln{P} = -\frac{1}{H}z + \ln{P_{0}} \text{ (10)} }$

This is the equation of a straight line (c.f. $y=mx+c$), so the intercept of our straight line is $\ln{P_{0}}$ and our gradient is $-(1/H)$. It is because the integration of our expression $dP/P$ (Equation (5) above) produces an exponential that the scale height $H$ is expressed as the altitude one needs to ascend for the pressure to drop by a factor of $1/e$ and not, e.g. 1/2.

If we plot the pressure as a function of altitude with the pressure (on the y-axis) plotted on a logarithmic scale, we get a straight line. The equation of this line is $\ln{P} = - \frac{1}{H}z + \ln{P_{0}}$ The gradient of the line is $-1/H$, where $H$ is the scale height of the atmosphere. So, on this linear-log plot, if we increase the altitude by $H$, the natural log of the pressure will drop by 1.