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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.

## What is the scale height of water vapour in the Earth’s atmosphere?

Someone recently asked me what was the scale height of water vapour in the Earth’s atmosphere, so I decided to see if I could find out. The scale height of water vapour is particularly important for infrared, sub-millimetre, millimetre and microwave astronomy, as it is the water vapour in the Earth’s atmosphere which prevents large fractions of these parts of the electromagnetic spectrum from reaching the ground. This is why we can, for example, only study the Cosmic Microwave Background from space or from a few particularly dry places on Earth such as Antarctica and the Atacama desert in Chile.

## What does the term ‘scale height’ mean?

First of all, let me explain what the term “scale height” means. It is the altitude by which one needs to go up for the quantity of something (water, nitrogen, oxygen, carbon dioxide) to go down by a factor of $1/e$, where $e$ is the base of natural logarithms, and $e=2.71828.....$. The scale height, usually written as $H$, is dependent upon the temperature of the gas, the mass of the molecules, and the gravity of the planet. We can write that

$H = \frac{ kT }{ Mg } \text{ (1)}$

where $k$ is Boltzmann’s constant, $T$ is the temperature (in Kelvin), $M$ is the  mass of the molecule and $g$ is the value of the acceleration due to gravity. If we were to plot the atmospheric pressure as a function of altitude we find that it follows an exponential, this is because of the differential equation which produces Equation (1) above (I will go into the mathematics of how Equation (1) is derived in a separate blog).

In the case of air, which is some 80% nitrogen molecules and 20% oxygen molecules, the scale height has been well determined and is $7.64 \text{ kilometres}$ (or, to put it another way, it drops by a half every $5.6 \text{ km}$). So, if one were at an altitude of $5.6 \text{ km}$, half of the atmosphere would be below you. Go up another $5.6 \text{ km}$ and it drops by a half again, so at $11.2 \text{ km}$ 75% of the atmosphere is below you.

## What is the scale height of water vapour in the Earth’s atmosphere?

Determining the scale height of water vapour in the Earth’s atmosphere is, I have discovered, essentially impossible. Or, to put it better, it is a meaningless figure. This is because it varies too much. It depends on temperature, so even in a given place it can vary quite a bit. So, instead, we talk of precipitable water vapour (PWV) at a particular place (both location and altitude). PWV is the equivalent height of a column of water if we were to take all the water vapour in the atmosphere above a particular location and it were to precipitate as rain.

The Mount Abu Infrared Observatory in India, for example, is at an altitude of 1,680 metres, and quotes a PWV of 1-2mm in winter. The PWV would be higher in summer, as water sinks in the atmosphere when it is cold. For Kitt Peak in Arizona, which is at an altitude of 2,090 metres, the PWV varies between about 15mm and 25mm. This is why very little infrared astronomy is done at Kitt Peak. For Mauna Kea in Hawaii, which is at an altitude of just over 4,000 metres, it varies between 0.5mm and 2mm. This is why there are a number of infrared, sub-mm and millimetre wave telescopes there.

At the South Pole, which is at an altitude of 2,835 metres, the PWV is measured to be between 0.25 and 0.4 in the middle of the Austral winter (June/July/August). Why is this so much lower than Mauna Kea, even though it it is at a lower altitude? It is because it is so much colder.

The average Precipitable Water Vapour at the South Pole averaged over a 50-year period from 1961 to 2010. Even in the Austral summer it is low, but in the Austral winter (June/July/August) it drops to as low as 0.25 to 0.35mm, one of the lowest values found anywhere on Earth.

High in the Atacama desert, on the Llano de Chajnantor (the Chajnantor plateau), which is at an altitude of 5,000 metres and where ALMA and other millimetre and microwave telescopes are being located, the PWV is typically about 1mm, and drops to as low as 0.25mm some 25% of the time (see e.g. this website). This is why Antarctica and the Atacama desert (in particular the Chajnantor plateau) have become places to study the Cosmic Microwave Background from the Earth’s surface; we need exceptionally dry air for the microwaves to reach the ground.

## Summary

To summarise, it is meaningless to talk about a scale height for water vapour in the Earth’s atmosphere, as the vertical distribution of water vapour not only varies from location to location, but varies at a given location. So, instead, we talk about Precipitable Water Vapour (PWV); the lower this number the drier the air is above our location. To be able to do infrared, sub-millimetre, millimetre and microwave astronomy we need the PWV to be as low as possible, the best sites (Antarctica and the Atacama) get as low as 0.25mm and are usually below 1mm. The exceptionally dry air above Antarctica and the Atacama desert enable us to study the Cosmic Microwave Background from the ground, something we usually have to do from space.