Diurnal variations

Introduction

By day, the ground (and so the air near it) warms as the surface absorbs heat from sunlight. When the Sun goes down, the ground cools by re-radiating that heat to “cold space”, thus cooling the adjacent air. This diurnal cycle of radiative heating and cooling of the surface drives changes in the atmospheric boundary layer that cause corresponding diurnal changes in atmospheric refraction.

Here's an example of the change in the temperature profile in an average April day, measured by W. D. Flower at Ismailia, Egypt (near the Suez Canal).

Flower's mean temperature profile for 5 am in April On the left you have the mean temperature profile for 5 a.m., just before sunrise. The lowest 60 meters of the nocturnal atmosphere are thermally inverted, and very stable: the Standard Atmosphere profile (shown dashed) slopes the opposite way from the measurements, and is bent, not straight.

Flower's data show even more extreme differences if only clear nights are averaged. The Standard Atmosphere is obviously a poor model to use for calculating astronomical refraction.

But at 1 p.m. the atmosphere is convective. The air near the ground has more than ten times the adiabatic lapse rate, and only the air above 15 meters has about the same slope as the Standard Atmosphere.

These two images are frames from an animation of the Ismailia measurements.

Air itself absorbs very little heat from sunlight. So very little of its temperature change is due to sunlight directly; the variations are mostly due to temperature changes of the Earth's surface, which absorbs most of the solar energy. Because heat conduction in air is very weak, these surface temperature changes move up through the boundary layer mainly by turbulent mixing.

Unfortunately, boundary-layer turbulence is a complicated problem that depends on wind shear, surface roughness and slope, and stability. Stably-stratified layers can isolate the air above them from fluctuations at the surface. So we cannot predict the diurnal variations in a simple way. However, some general principles allow us to get a partial understanding of this complex situation.

Let's look at the various pieces separately.

Radiative heating and cooling

Every surface at thermodynamic temperature T radiates heat at a rate that's proportional to T⁴. Thermodynamics tells us that the emission must balance the absorption — otherwise we could make a perpetual-motion machine — and that there is a maximum emission from a perfectly absorbing surface that would appear perfectly black: an idealized “black body”.

Black-body radiation extends over the whole electromagnetic spectrum, but is concentrated in a couple of octaves around a wavelength that is inversely proportional to the temperature T. At room temperature (a little below 300 K), the middle of this interval is in the “thermal infrared”, near 14 microns wavelength. On the other hand, the Sun's effective temperature is a little less than 6000 K, and its thermal radiation is centered near 0.7 microns, at the red end of the visible spectrum. Notice that the ratio of temperatures is about 20, and so is the ratio of middle wavelengths.

How black are real surfaces? Most surfaces are pretty black. The blackness of a real surface is measured by its emissivity, usually denoted by the letter ε, which is 1 for the perfect black body. The emissivity depends on wavelength; so the effective emissivity of a surface depends on its temperature, because most of its thermal radiation is at wavelengths that change with the temperature.

Thermal emissivities for real surfaces near 300 K are usually close to 1, so a rough calculation can ignore this issue. In particular, organic materials like plant leaves and plastics and paint are nearly perfect absorbers and emitters at ordinary temperatures, because the hydrogen atoms in their molecules absorb at vibrational wavelengths in the thermal infrared. (That means that white paint, water and ice, and snow can be treated as black surfaces for our thermal work here.) Some polished metal surfaces, on the other hand, are so nearly perfect reflectors that they can be treated as nearly “white” for thermal radiation. That's why the mirrors of infrared telescopes are usually gold-coated.

So the incoming (visible) sunlight that heats the Earth's surface is mostly absorbed, unless the surface is clean (fresh) snow. And the thermal radiation from the surface is mostly transmitted by the atmosphere.

Some numbers

Let's see how much radiation comes in, and how much goes out. In a steady state, these must be equal.

The radiated power per unit area from a black surface at temperature T is about

5.67 × 10⁻⁸ W · m⁻² · T⁴

for T in kelvins; so at 300 K, about 460 watts per square meter is emitted by the surface, day and night, if the temperature is that warm (about 27°C or 80°F).

That overestimates the net radiative loss of the surface, because the infrared opacity of the air makes the sky appear to be roughly 250 K, so that at least half of the outgoing radiant heat is returned by the atmosphere. Maybe 200 W per square meter is a more realistic net loss.

But the daytime solar radiation is much bigger. If the Sun were at the zenith, some 1360 watts per square meter would reach the top of the atmosphere, and about a kilowatt per square meter would reach the ground.

However, in most places the Sun is never at the zenith; and even where it sometimes is (in the tropics), it's only there momentarily on one or two days a year. So we need to consider the Sun's zenith distance.

Effect of solar Z.D.

There are two different effects that need to be included. One is the geometric effect of spreading a square meter cross section of sunlight over more than a square meter when its light does not fall squarely on the Earth's surface. That's Lambert's cosine law at work: the power per unit area falling on the surface is proportional to the cosine of the Sun's zenith distance. (A beam of sunlight one square meter in cross-section is spread over more than one square meter when it falls on the surface obliquely.)

The second effect is attenuation of the beam of sunlight by atmospheric absorption and scattering. If the attenuation were the same throughout the spectrum, the transmission of the atmosphere would just decline exponentially with increasing airmass. But in reality, the absorption is concentrated in molecular bands, and the scattering varies with wavelength; the radiation in the absorption bands is removed high in the atmosphere, leaving mostly the radiation in the more transparent spectral regions to be absorbed near the ground. So the attenuation decreases more slowly with increasing airmass. That's known as the “Forbes effect”, after J. D. Forbes, the man who pointed this out in his Bakerian Lecture (1842).

The Forbes Effect is important for precise calculations, especially at large airmasses. But here, we are being rather sloppy and not trying for high accuracy; and we are also not concerned with large zenith distances. So I'll just assume a transmission of about 0.7 per airmass, and even approximate the airmass by sec Z. That makes the transmission at some zenith distance Z equal to

exp(− 0.36/cos Z) .

Example: solar heating at 30° altitude

The transmission is about 0.5 when the Sun is 30° above the horizon, where Z = 60°, and the airmass is about 2.

The geometric effect is proportional to cos Z, which is 0.866 when Z is 60°.

The two effects (geometric and extinction) together make the power per unit area about 1360 × 0.5 × 0.866 = 590 watts per square meter — considerably larger than the cooling rate of the surface. So the Sun must be lower in the sky than 30° above the horizon to compensate the cooling of the surface exactly.

Why should we care where the Sun is when its heating just balances the radiative loss of heat at the surface? Because that's when the surface layer switches from stable to unstable in the morning, and from unstable to stable in the evening.

Transition

Because the solar heating rate of the surface when the Sun is well above the horizon is much larger than the surface cooling rate at night, and the total daytime heating must balance the nighttime cooling, the transition between heating and cooling must occur when the Sun is some distance above the horizon. The calculation above shows that this must be less than 30°.

A lower limit to the Sun's altitude at transition can be found by assuming that heat is rapidly distributed over the whole surface of the Earth. Assume the planet is a sphere of radius R. It receives a beam of sunlight equal to its cross-sectional area, which is πR², but radiates the absorbed heat from the whole surface area, which is 4 πR². Then the balance between incoming and outgoing radiation would occur when the Sun is low enough in the sky to spread one square meter of beam cross-section over 4 square meters of a horizontal surface. That requires the sine of its altitude to be 1/4 = 0.25, which would make the altitude about 14.5°.

But we know that the heating is not distributed evenly over the surface. So a better estimate of the Sun's altitude at crossover would be found by treating the heat exchange as restricted to a particular latitude zone.

The area of a narrow zone of latitude is proportional to the width of its projection on a plane perpendicular to the incoming solar radiation; at the equator, that's just the diameter of the Earth. But the cooling occurs over the circumference of the zone, which is π times larger. So at low latitudes, the crossover occurs when the sine of the Sun's altitude is 1/π, making the altitude about 18.6°.

At some latitude φ, the area of the zone is increased by the secant of the latitude. So the solar heating per unit area is decreased by this factor, but the cooling rate is increased by it. That explains (roughly) the decrease in average daily temperature from the equator to the poles. But it doesn't change the day/night ratio, which remains the same (apart from neglected effects like the increase in atmospheric extinction with solar zenith distance, and the meridional transport of heat by winds). So the crossover solar altitude remains about the same, except near the poles.

As the diurnal motion is about 15° per hour of time, that would put the transitions a little over an hour after sunrise, and before sunset. Consequently, the surface layer has an inverted lapse rate more often than not — contrary to the unrealistic “Standard Atmosphere” models.

Observations of crossover

The transitions are readily apparent to an attentive observer. For example, the surface wind is largely due to variations in surface temperatures, so the wind tends to die down about an hour before sunset. I noticed this as a boy: the hour before sunset was the best time to fly hand-launched gliders outdoors. Later, the reduced temperature contrast between air and surface was noticed by people studying the effects of turbulence on laser beams: the hour before sunset was when the turbulent scintillations were smallest.

Quantitative temperature profiles show the effect as well. The literature on crossovers was reviewed in 2020 by

W. M. Angevine, J. M. Edwards, M. Lothon, M. A. Lemone, S. R. Osborne
“Transition periods in the diurnally-varying atmospheric boundary layer over land”
Boundary-Layer Meteorology , 177, No. 2–3, pp. 205–223 (2020)

which is available at DOI: 10.1007/s10546-020-00515-y

They point out that the morning crossover occurs sooner than expected, because part of the heat that destroys the nocturnal inversion comes from entrainment, particularly when strong wind shear produces mixing.

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