Hydrostatic balance

32. Hydrostatic balance#

In order to calculate the optical depth we need n, the number concentration of molecules at a particular height in the atmosphere. This means that we need to know the density as a function of height.

At large scales it is a very good approximation to assume hydrostatic balance, which is the statement that the vertical pressure differential across a layer provides exactly the amount of force necessary to balance gravity, so the atmosphere is not rising or sinking:

../../_images/hydrostat.png

Fig. 32.1 Hydrostatic balance for a 1 \(m^2\) cross section of a layer#

In symbols, the balance shown by Fig. 32.1 implies that:

(32.1)#\[ dp = - \rho g dz \]

Question: is this the same pressure p as the local pressure given by the equation of state: \(p=\rho R_d T\)?

32.1. Scale Heights#

In Chapter 1 (p. 8) Roland introduces the idea of a characteristic length scale for the variation of pressure in the atmosphere, and on page 14 he defines the equation of state for dry air. To get weighting functions for our sounding retrievals we need to now the optical depth \(\tau\) vs. height, which means that we need to know the gas density \(\rho\) as a function of height. Below I show how to get \(\rho(z)\) for typical atmospheric conditions.

\[ p = p_0 \exp \left ( \frac{-z}{H_p} \right ) \]

and on page

In words – \(H_p\) gives the height at which gravity wins out over the kinetic energy of the air molecules – warmer atmospheres take longer to decrease their pressure by a factor of \(\exp(-H_p/H_p) = \exp(-1)\) (the e-folding height) because the air molecules have more energy to travel higher.

  1. Do pressure first: Rewrite (32.1) using the ideal gas law:

    (32.2)#\[ dp = - \rho g dz = - \frac{p}{R_d T} g dz = \frac{p}{H_p} \]

    where \(H_p=R_d T/g\) is the pressure scale height.

    \[ d\ln p = - \frac{dz }{H_p} \]
    \[ \int_{p_0}^{p}\!\,d \ln p = - \int_{0 }^{z}\!\frac{1}{H_p} dz^\prime \]

    Since temperature (and gravity) are changing with height, we need to know profiles of g(z) and T(z) to go any further. If we don’t have those profiles, then a useful approximation is to define a vertically averaged \(H_p\), since the average involves a large mass of air (about 10 metric tons/cubic meter), it will vary more slowly than the temperature at any one level on any given day, and we can still get a useful profile of pressure vs. height without having a specific sounding:

    (32.3)#\[ \frac{ 1}{\overline{H_p}} = \overline{ \left ( \frac{1 }{H_p} \right )} = \frac{\int_{0 }^{z}\!\frac{1}{H_p} dz^\prime }{z-0} \]

    Now we can replace the z integral by the average:

    \[ \int_{p_0}^{p}\!\,d \ln p = - \int_{0 }^{z}\!\frac{1}{H_p} dz^\prime = -\frac{z}{\overline{H_p}} \]

    and integrating the left-hand side:

    \[ \ln p/p_0 = - \frac{z }{\overline{H_p}} \]

    Taking exp of both sides gives the pressure with height:

    \[ p = p_0 \exp \left ( - \frac{z }{\overline H_p} \right ) \]

    Values for \(\overline{H_p}\) are relatively constant for a particular climate regime, like midlatitude winter.

  2. Now repeat this for density \(\rho\). We need to use the chain rule for the equation of state:

    \[ \frac{dp }{dz} = \frac{d }{dz} (\rho R_d T) = R_d \left ( \frac{d\rho }{dz} T + \rho \frac{ dT}{dz} \right ) = - \rho g \]
    \[ \frac{d\rho }{dz} = -\frac{\rho }{T} \left ( \frac{g }{R_d} + \frac{ dT}{dz} \right ) = - \rho \left ( \frac{1 }{H} + \frac{1 }{T} \frac{dT }{dz} \right ) = - \frac{\rho}{H_\rho} \]

    Bottom line – with this definition of the density scale height we’ve got an equation that looks like (32.2):

    (32.4)#\[ \frac{d\rho }{\rho} = - \left ( \frac{1 }{H} + \frac{1 }{T} \frac{dT }{dz} \right ) dz = - \frac{dz }{H_\rho} \]

    Pull the same trick for the vertical average:

    (32.5)#\[ \frac{ 1}{\overline{H_\rho}} = \overline{ \left ( \frac{1 }{H_\rho} \right )} = \frac{\int_{0 }^{z}\!\frac{1}{H_\rho} dz^\prime }{z-0} \]

    We use this vertical average in exactly the same way as before to get the density profile:

    \[\begin{split} \begin{aligned} \ln \rho/\rho_0 =& - \frac{z }{\overline{H_\rho }} \\ \rho =& \rho_0 \exp \left ( - \frac{z }{\overline H_\rho} \right )\end{aligned} \end{split}\]

    In the hydrostatic.ipynb notebook I show that the midlatitude summer sounding gives \(H_p\approx7.8\) km and \(H_\rho \approx 9.5\) km. We can use the equations:

    \[ p = p_o \exp ( -z/H_p) \]

    and

    \[ \rho = \rho_0 \exp ( -z/H_\rho) \]

    To calculate the optical depth for various atmospheric conditions.

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