Meteo 465

Radiation in the middle atmosphere

Radiation plays two important roles:

heating and cooling by infrared and heating by UV/visible.

Generally, heating is by solar UV absorption by O₂ in the mesosphere and thermosphere and by O₃ in the stratosphere.

Generally, cooling is by infrared cooling by CO₂ (15 mm), O₃ (9.6 mm), and H₂O (80 mm).

photochemical dissociation – stratospheric ozone, ionospheric ions

Less than1% of the solar radiation is in the UV, yet it is responsible for most of the photochemical processes (ionization, dissociation, etc.)

The goal of this section is to look at these two different types of processes. I will assume that you all are familiar with the basic structure of shortwave and longwave radiation from you radiative transfer courses.

Solar absorption and atmospheric heating.

The figure on atmospheric absorbers shows the important absorbers and the wavelengths at heights at which they absorb.

The most important are the wavelength regions listed the table below.

wavelength (nm)	atmospheric absorber
121.6	solar Lyman a line. Absorption by mesospheric O2, but not O3
100 - 175	O2 Schumann Runge continuum. Absorption by thermospheric O2
175 - 200	O2 Schumann Runge bands. Absorption by mesospheric and upper stratospheric O2. O3 effects important only in the stratosphere
200 - 242	O2 Herzberg continuum. Absorption by stratospheric O2. O3 Hartley band. Absorption by stratospheric O3.
242 - 310	O3 Hartley band. Absorption by stratospheric O3, leading to O(^1D).
310 - 400	O3 Huggins bands. Absorption by stratospheric and tropospheric O3, leading to O(^3P).
400 - 850	O3 Chappuis bands. Absorption by tropospheric O3 causes photodissociation even at the surface.

We can see from the figure on the solar spectrum where the solar energy is.

To look at how this absorption works, we start with the Beer Lambert Law:

dI(l) = -k_a(l) I(l) ds

where k_a = s_a(l) n, where s_a(l) is the absorption cross section (cm²) and n is the molecule number density (molecules cm^-3).

We integrate this absorption over a path through the atmosphere:

I(l) = I_o(l) exp{-ò s_a(l) n(s) ds}

where ds is the differential path through the absorbing layer.

The optical thickness is t_a = ò s_a(l) n(s) ds .

The transmission through the path s is T(l,s) = exp{-t_a(l,s)}

How do we get the path, s, in terms that we can understand?

We start with the local solar zenith angle, χ. By spherical geometry,

cos(χ) = cos(f) cos(d) cos(HA) + sin(f)sin(d),

where f = latitude,

d = solar declination angle (±23.5^o for solstice, 0^o equinox)

HA = hour angle, for which 0^o is local noon

If we assume a parallel path atmosphere, ds = dz sec(χ). This approximation is good until we have c > 75^o. At that solar zenith angle, we need to take Earth curvature into account.

Recall that n(z) = n(z=0) exp(-z/H)

Thus,

I(z) = I(¥) exp{-sec(χ) ò_z^¥ s_a n_o exp(-z/H) dz}

= I(¥) exp{-sec(χ) s_a n_o H exp(-z/H)}

where I(¥) is the solar intensity above Earth’s atmosphere.

The rate of energy disposition by absorption is proportional to the rates of other processes, such as photodissociation. If there is any fluorescence, then the rate of energy deposition will be somewhat less.

r = - dI/ds = -dI/dz cos(χ) = s_a n_o I(¥) exp{-(z/H + t_oexp(-z/H)}

where t_o = s_a n_o H sec(χ).

The altitude with the maximum absorption, which is found by differentiating the expression r and setting it equal to zero and then solving for z, is:

z_m = H ln(t_o) = H ln(s_a n_o H sec(χ))

For an overhead sun, the altitude of maximum absorption is:

z_o = H ln(s_a n_o H)

so that the altitude of maximum energy absorption at any other solar zenith angle is:

z_m = z_o + H ln(sec(c))

The rate of energy deposition at the maximum is given by:

r_m = s_a n_o I(¥) cos(χ) exp(-1 – z_o/H)

The variation of the rate of energy deposition is given by the equation:

r/r_o = exp{1 – Z – sec(χ) exp(-Z)}, where Z = (z-z_o)/H

where r_o is the rate of energy absorption for c = 0.

r_o = s_a n_o I(¥) exp(-1 – z_o/H)

The figure on maximum deposition rate vs altitude shows the effect of solar zenith angle.

z_m is

independent of the intensity;

but strongly dependent on:

the nature and distribution of the absorbing gas

the solar zenith angle

the wavelength of the radiation

 Since the solar spectrum and absorption occur over a range of wavelengths, several strongly absorbing layers will occur at different altitudes.

Oxygen and ozone are examples, since they absorb strongly at different wavelengths:

t_a(l,z,c) = sec(c) {ò_z^¥ s_a(O₂, l) [O₂(z)] dz + ò_z^¥ s_a(O₃, l) [O₃(z)] dz}

We can see this in the figure on the solar absorption of O₂ and O₃.

Solar heating rates.

How does this energy deposition translate into hearing rates? What we need to know is the heat capacity of the air at different altitudes.

If we use for I the units W m^-2, which is really a flux, then the heating rate can be related directly to the rate of energy deposition. Let’s go back to the original form of the flux:

I = I(¥) ò exp(-òs_a(l) n(s) ds)dl

Now we are integration over a range of wavelengths. Only O₂ and O₃ really contribute to the heating:

I = ò_l I(¥) exp{-ò s_O2(l) [O₂] ds -ò s_O3(l) [O₃] ds}dl

The rate of energy deposition is:

r = -dI/ds = ò_l I {s_O2(l) [O₂] + s_O3(l) [O₃] }dl

r has units of J s^-1 m^-3.

If all of this energy goes into heating, then the rate of energy deposition equals the rate of heating, by the First Law of Thermodynamics:

dU/dt = Q, which when using our notation, rc_p dT/dt = r, or

dT/dt = (r c_p)^-1 ò_l I {s_O2(l) [O₂] + s_O3(l) [O₃] }dl

where r is the mass density, and c_p is the specific heat, or specific enthalpy = 1005 J kg^-1 K^-1

We can see this heating rate in the figure on shortwave and longwave effects and the meridional heating rates. Why does the heating rate peak where it does?

Infrared Heating and Cooling

Infrared heating plays a relatively small role in stratospheric heating. The main infrared heating is by absorption of the 2.7 mm and 4.3 mm bands of CO₂. The 9.6 mm band of O₃ provides some heating near the tropopause.

We can see this in the right-hand side of the figure on shortwave and longwave effects.

Now, on to infrared cooling.

Infrared cooling is primarily due to the 15 mm band of CO₂. The second most important infrared cooling is in the 9.6 mm band of O₃, which is most important near the stratopause.

We can use the “cooling-to-space approximation” to estimate the cooling rate. In this approximation, we ignore the downward flux of infrared and look only at the upward flux. This is not a bad approximation, as can be seen in the figure on upward and downward fluxes.

The temperature change with time is given by the equation:

dT/dt = - C B_n(T)

where B_n(T) is the Planck’s Law. C is a constant that depends on the infrared lines, their shape and overlap, and their oscillator strengths. For this case, in which the absorption is strong, the expression becomes:

dT/dt = - {5.4x10⁵ (χ_CO2 / 3.3x10^-4)^1/2} 3.53x10^-4 exp(-960/T),

where the second half of the right-hand-side is an approximation of the Planck function for the CO₂ 15 mm band.

What is the time constant for cooling? It varies from ~15 days in the upper troposphere to 3-5 days in the upper stratosphere. Throughout the stratosphere, the radiative lifetimes are much shorter than the typical transport at midlatitudes.

This can be seen in the figure on radiative, photochemical, and diffusive transport.