The principle of hydrostatic equilibrium is that the pressure at any point in a fluid at rest (whence, “hydrostatic”) is just due to the weight of the overlying fluid.

As pressure is just force per unit area, the pressure at the bottom of a fluid is just the weight of a column of the fluid, one unit of area in cross-section.

This principle is simple to apply to incompressible fluids, such as most
liquids (e.g., water). [Note that water and other common liquids are not
strictly incompressible; but very high pressures are required to change
their densities appreciably.]
If the fluid is incompressible, so that the density is independent of the
pressure,
the weight of a column of liquid is just proportional to the height of the
liquid above the level where the pressure is measured. In fact, the mass
of a unit-area column of height **h** and density **ρ** is just
**ρh**; and the weight of the column is its mass times
the acceleration of gravity, **g**. But the weight of the unit-area
column is the force it exerts per unit area at its base — i.e., the
pressure. So

For example, the density of water is 1000 kilograms per cubic meter (in SI
units), so the weight of a cubic meter of water is 1000 kg times
**g**, the acceleration of gravity (9.8 m/sec^{2}), or 9800 newtons.
This force is exerted over 1 m^{2}, so the pressure produced by
a 1-meter depth of water is 9800 pascals (the Pa is the SI unit
of pressure, equal to 1 newton per square meter).

The unit of pressure used in atmospheric work on Earth is the hectopascal; 1 hPa = 100 Pa. So the pressure 1 m below the surface of water (ignoring the pressure exerted by the atmosphere on top of it) is 98 hPa. Standard atmospheric pressure is 1013.25 hPa, so it takes 1013.25/98 = 10.33 meters of water to produce a pressure of 1 atmosphere. (That's about 34 feet, for those who like obsolete units.)

The pressure in the ocean increases by about 1 atmosphere for every 10 meters of depth. The average depth of the ocean is about 4 km, so the pressure on the sea floor is about 400 atmospheres.

The density of air, under standard conditions, is about 800
times less than the density of water — almost 1.3 kg per cubic meter.
So the height of a column of air needed to exert the standard atmospheric
pressure of 1013.25 hPa would be about 8 thousand meters (8 km), **if** it
were all the same density — i.e., **homogeneous**.
That height is “the height of the homogeneous atmosphere.”

You can see from the relation above that this height, **H**, is just
**P/(gρ)**.
Even though the atmosphere isn't really homogeneous, this 8-km height is a
useful characteristic length that keeps turning up in calculations of
atmospheric refraction.
(A better name for this concept is
**Radau**'s
term, “reduced height”.)

Hydrostatic equilibrium is a little more complicated to apply to air, because air is very compressible. The same principle still applies, but we now have to deal with a density that varies with pressure and temperature.

is the differential equation that expresses hydrostatic
equilibrium. [Remember that **g** is the local acceleration of gravity,
which is needed to convert the element of mass ( **ρ dh** )
into the
force (i.e., its weight) it adds to the unit area beneath it. The minus
sign is there because **g** acts in the negative direction along the height
scale. We are implicitly assuming that the range of **h** is so small
compared to the radius of the Earth that **g** can be assumed constant.]
Its elementary solution is

and if **ρ** = const., we can take it outside the integral:

or

In other words, the pressure is just proportional to the
height, **h**, of the column of fluid, as we already knew.
(The minus sign comes from the fact that we measure **h** positive upward,
but the atmosphere's weight is directed downward.)

For air, things are not so easy. The density, **ρ**, depends on
both **P** and **T**, the absolute temperature.
The **equation of state** is the function that tells us the density.
For air, it's a very good approximation to use the equation of state for
an ideal gas,

where **μ** is the (dimensionless) molecular weight — about 29 for
dry air — and
**R** is the “gas constant” that takes care of the units.

Even though this relation is very
simple, it still complicates the integrand of the hydrostatic equation.
First of all, it gets **P** involved in the integrand, which is no longer a
simple function of **h**.
Secondly, it introduces a new independent variable, **T**.

We'd like to express everything inside the integral as a function of a
single variable.
To do this, we need some *additional* relationship among **P**,
**ρ**, and **T**,
which would allow us to get rid of the second independent variable.

Unfortunately, there is no additional physical relationship available, in
general. The actual dependence of **P**, or **ρ**, or **T**,
on height, is quite
variable in the real world. This dependence is what's meant by
the phrase “the structure of the atmosphere.” [In the
astronomical literature of the 18th and 19th Centuries, it was often
called the “constitution” of the atmosphere, which is
confusing to a modern reader; “constitution” today means
“composition” rather than “structure.”] The
structure of the real atmosphere varies considerably from place to place
and from time to time.

It's often convenient (though somewhat unrealistic) to assume that the structure of the atmosphere is polytropic; this is explained on the polytropes page.

But, even though we can't integrate the hydrostatic equation until some
additional information (such as the run of **T** with height, or the
dependence of **P** on **ρ**) is available,
we can still evaluate the height of the *homogeneous* atmosphere.
That's just the height the atmosphere would have if it had the same
density everywhere (i.e., the density at the surface of the Earth),
and the same pressure at the bottom as the real atmosphere.

Given the temperature and pressure at the surface, and the composition of
the gas there (which is what determines the mean molecular weight,
**μ**),
we can find the density of air at the surface. Then the height of the
homogeneous atmosphere is just

because **P** is the weight of a column of gas of height **H** and
density **ρ** in the
gravitational field of the Earth, with acceleration **g**.

If you'd rather think in terms of the mass of one molecule, **m**,
then the gas law is **ρ = μP/(NkT)**, where **N** is
Avogadro's number, **k** is Boltzmann's constant, and **μ** is
the (dimensionless) molecular weight.
(The molecular mass **m** is **μ** times the atomic mass unit,
**u**.)
Then the height of the homogeneous atmosphere is

but we can use the ideal gas law, **ρ = μP/(R T)** ,
to get rid of **ρ**. So, write **dP** as

Now, integrate this. On the left side, we get
∫ **dP/P** ,
which is just ln** P** ;
on the right, **T** is constant,
so we get some constants times ∫**dh**,
which is just **h**. Of course there is a constant of integration; we
see that it has to be the value of ln** P** at **h** = 0.
So:

Combine the two logarithms to get ln (**P/P _{0}**) on the left.
Then get rid of the logarithm by exponentiating both sides:

Now, remember the
homogeneous atmosphere? Its height **H** was
**RT/μg**. Notice that this is the reciprocal of the coefficient of
**h** in the argument of the exponential. So we have

In this case, the reduced height **H** is usually called the
“scale height” of the atmosphere.

Finally, because **T** is constant, the density decreases
exponentially with height exactly as the pressure does.
This is important for the
bending
of rays near the horizon, because the bending is proportional to the
density gradient.

There's a fine webpage on Hydrostatics at the University of Denver, if you'd like to know more about this subject.

Copyright © 2003 – 2006, 2010, 2014, 2020 Andrew T. Young

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