Guest essay by Clyde Spencer
The total reflectivity of Earth is important in energy budget calculations and in Global Circulation Models used to predict climate. If the value used is too low, it will contribute to models predicting more warming than is correct. I’ll demonstrate below why measured albedo is in fact too low as an estimate of total reflectivity.
One frequently sees reference to the nominal 30% albedo of Earth with respect to the energy budget and alleged anthropogenic global warming. Although, the CRC Handbook of Chemistry and Physics lists a value of 36.7% in recent editions. Albedo is a measure of the apparent reflectivity, of visible light, of celestial bodies such as the moon, Mars, and asteroids. This works well to estimate the type of surface cover and particle size for objects with only bare rocks and regolith. To properly measure albedo over an observed hemisphere, the sun, measured body, and the observer (Earthlings!) should be in approximate alignment. Earth’s albedo is estimated by measuring the albedo of the moon (12%) during a Full Moon, and then measuring the Earthshine near the time of a New Moon. This measurement mostly captures diffuse-reflectance of vegetation, soil, regolith, and especially clouds. However, it includes some minor specular reflections from smooth objects (such as rocks, but especially water waves) whose surfaces are oriented such that incoming rays reflect towards the moon.
However, the illuminated portion of the Earth will always be slightly different, varying with time and location of the measurements, and this probably contributes to the observed variation in measured albedo over time. Because the moon is not in the plane of the ecliptic, any measurements of Earthshine over time are going to be from different viewing angles, and of different regions of Earth’s surface, along with changes resulting from weather and seasons. That is, the albedo, as estimated from Earthshine measurements, can be expected to vary considerably for geometric reasons. These measurements provide an apparent reflectivity integrated over an entire hemisphere of the Earth.
Now, I say “apparent” reflectivity because not all light arriving at Earth is reflected back towards the sun. Albedo is, at best, a lower-bound on the amount of light reflected from Earth. It would only be appropriate to use if Earth were completely covered by clouds, like Venus (albedo = 65%), or completely covered by regolith, like Mars (albedo = 15%), with no water, clouds, or vegetation.
Additionally, albedo is an instantaneous measurement. To the extent that vegetation and clouds change with the seasons, a further, more complicated correction has to be made for those effects to derive an annual average.
For purposes of illustration (or perhaps I should say illumination) consider what the situation would be if Earth were a Hollywood Waterworld (ala Kevin Kostner). Assume that there is only a sediment-free ocean, no clouds, and gentle winds, minimizing whitecaps. Water strongly reflects specularly (as any fisherman or sailer knows only too well), in contrast to the strong diffuse-reflectance of most other things on Earth.
Specular-reflectance is quantitatively calculated with Fresnel’s Reflectance Equation ,using the index of refraction and the extinction coefficient (sometimes called absorption coefficient), which constitute the Complex Refractive Index (CRI). The index of refraction that most readers are probably familiar with is defined simply as the ratio of the speed of light in a vacuum to the speed of light in the material being measured. The extinction coefficient for water varies with wavelength – being larger for red and infrared than the other end of the spectrum – but we can consider it to be negligible for the following illustration. Thus, the CRI for water has an imaginary component that can be considered to be zero. This somewhat simplifies calculations. The CRI varies with temperature, salinity, and the wavelength of the incident light. I’ve used an approximation (n = 1.34) for typical seawater at 550nm wavelength, which is the peak emission of sun light.
Remember that the angle of reflection equals the angle of incidence, where incidence is measured from the surface normal. Assume that we are looking at our Waterworld from a geostationary orbit and that this Waterworld is a perfect sphere instead of the oblate spheroid (Geoid) that the real Earth is. Light is coming from the sun in rays that are parallel at this distance. A ray hitting normal to the surface of the water has an angle of incidence of 0°. All other rays intercepting the surface of the illuminated hemisphere have larger angles of incidence, out to a maximum of 90° at the limbs, or terminator. Ninety degrees is grazing the surface; there is 100% reflectance, and the character of the light is spectroscopically the same as the source (the sun).
The graph below shows what the reflectance is for all angles of incidence for seawater. Note that above about 40° the reflectance starts to rise noticeably, by 60° the reflectance is rising sharply, and above 79° it is above the nominal albedo (30%) commonly cited in Global Warming energy budget discussions.
Figure 1. Total Reflectivity of seawater
From the position of the geostationary satellite, looking down, one would see a spot of light reflecting back from the bundle of rays approximately normal to the surface of the water. That would be about 2% reflectance. Everything else would look black, save for some random white caps reflecting some flashes of light. However, that doesn’t mean that all the light is being absorbed! The orbital position would just be wrong for observing the light reflected out into space, away from the sun. The rays hitting at greater distances from the normal ray will experience a larger angle of incidence because of the curvature of the Waterworld. Thus, they will experience a greater percentage of reflectance. The importance of this is that as the reflectance increases with increasing angle of incidence, the absorption, and thus warming, decreases proportionately.
Fresnel’s equation is not integrated easily in its native form; at least I’m not up to it! A 6th-degree polynomial gives a reasonably good fit, which would allow me to somewhat mechanistically integrate it and obtain the area under the curve. However, some readers have expressed objections to high-order polynomials for fitting curves. I think it has something to do with a phobia about wiggling elephant trunks. Therefore, I took the easy way out and just weighed the total area of the graph on paper, and then cut out the area under the curve and weighed it. The ratio of the two weights gives the proportion of the two areas.
The area under the reflectance curve for seawater (Fig. 1, above) is about 9%. Because the sun moves with a constant angular velocity, the value is the time-averaged reflectance of a point on the surface of Waterworld for a 6-hour period. That means that for any particular spot on the surface of the ocean, starting at the terminator at sunrise, on the Equator, it would have an initial instantaneous reflectance of 100%. As the planet rotates under the sun, the percentage of light reflected from the point will decrease until it reaches the minimum of ≈2% at local solar noon. The process will then be reversed until the point again reaches maximum reflectance at the position of the sunset terminator. The total light reflected from that point, over a nominal 12-hour sunlit period, will be about 18% of the incident light. This is more than twice the value listed for the diffuse-reflectance of water and is in the range of values given for vegetation.
Furthermore, it should be noted that a point on the terminator at sunrise, at 60° latitude (N&S), would experience a decline in reflectivity until local noon, just as at the equator, but it will never get below six percent! The minimum reflectivity increases with increasing latitude, leading to larger average reflectances.
However, that doesn’t adequately describe the situation because the surface areas with the greatest reflectivity have larger areas than the lowest reflectivity areas. Because the Earth is rotating, it is important where the water is with respect to land and clouds. While an average reflectivity value may be useful for an initial first-order approximation, the variation with time/location has to be taken into account for accurate modeling.
I also did a discreet summation of the frustums of a hemisphere (Af = 2πR DX). Multiplying the normalized (to a unit area for the hemisphere) frustum areas by the average reflectivity for the angle of incidence, for each of the frustums, gives the area-weighted reflectivity for each frustum. Summing them gives an area-weighted average reflectivity of about 18%. This is the instantaneous area-averaged reflectance over a hemisphere. This is almost an order of magnitude larger than the sunlight reflected from a small spot on the surface of the ocean directly below the local noon sun during an equinox. It is far greater than the apparent albedo (≪2%) of our hypothetical Waterworld!
Even on the real Earth, these specular-reflection effects are significant because about 71% of the surface is covered by water, which NASA claims has a reflectance of about six percent. That means the actual total reflectivity of Earth must be higher than the estimate of diffuse reflectivity obtained from Earthshine albedo measurements (See “Terrestrial Albedo”).
Now, things get more complex as we add land and clouds because they behave differently than water. Bare land and vegetation generate a combination of specular and diffuse-reflection that is best described with a Bidirectional Reflectance Distribution Function (BRDF). It is a mathematical description of how light is distributed with the two effects of diffuse and specular-reflection, with varying angles of incidence. There is always a strong forward lobe of reflectance (anisotropy) for oblique illumination, even for snow. That is because snowflakes (and plant leaves) are smooth, planar, and tend to be sub-parallel to the surface of the Earth. Thus, measurements of albedo from directly above will result in values that are lower than the true total reflectance that would be obtained by integrating the BRDF over an entire hemisphere.
There has been work done with modeling CERES satellite measurements; however, judging from the following illustrations, they don’t have it right. The right-hand illustration of Fig. 2 shows a hemisphere with a large amount of land. The oceans are shown as darker than vegetated land (8% –18% albedo). Indeed, a value of 6% for open ocean seems to be totally inappropriate. Therefore, this illustration appears to be primarily the diffuse-reflectance water-albedo, and not total reflectivity. Because the value is so low, it apparently does not take into account how the specular-reflectance changes with time and position. That is, it is an average of the small incidence-angle (nadir) specular-reflectance (≈2%) of direct sunlight, as obtained from satellites; whitecaps; bottom reflectance from shallow water; diffuse-reflectance of suspended particles near the water surface; and scattered skylight coming from all angles.
Figure 2. Terra/CERES views the world in outgoing longwave radiation (left) and reflected solar radiation (right). Image Credit: NASA
Vegetation behaves differently from inorganic reflectors. First, leaves tend to be smooth, with a waxy coating that favors specular reflection. Next, while plants look green because red and blue light are absorbed (It is also highly reflective in the near-infrared.), not all incident absorbed light contributes to warming. The plant chlorophyll converts incident light to carbohydrates and it does not result in warming. The estimates of efficiency with which this conversion takes place vary with the plant and the growing conditions. However, it is generally thought to be in the low single-digit percentage range. Thus, the effective reflectance of visible light by vegetation should be adjusted upward in warming calculations to account for the lack of warming. Phytoplankton and algae in the oceans similarly capture sunlight and convert it to biomass instead of warming the water to the extent that a first-order estimate from reflectivity alone would suggest.
Common dry sand can have a diffuse-reflectance as high as 45%! Regolith, sand, and soils typically have a dominantly diffuse-reflectance. Soil exposed by agriculture can vary from very bright yellow or whitish, calcium-rich desert soils, to dark organic-rich soils; when the soils are wet they get darker and have a stronger specular component. And, of course, little if any soil is exposed during the growing season. Any ‘average’ value assigned to bare soils will probably be wrong more often than right. Estimates for soil reflectivity should be derived from soil maps to better obtain area-weighted values for different regions of the world, and take into account the growing season.
The other extreme is diffuse-reflectance from clouds, which approaches true Lambertian Reflectance. Lambertian Reflectance is a condition where light is reflected equally in all directions and the clouds appear uniformly white from all directions, except under the clouds and in shadow areas between them. Clouds can vary widely in their albedo, but a commonly accepted average value is around 50%.
What clouds contribute to the energy balance is the difference between their reflectance and the reflectance of the surficial materials they are covering. Clouds also re-direct light, making calculations more challenging. However, clouds have their greatest impact on reflectivity and albedo when they are within about 50 degrees of the surface-normal pointing towards the sun.
The Polar Regions are notoriously cloudy. That is one reason the Vikings invented the use of the sunstone to help them navigate where a compass and stars were unusable. Much of the year, it doesn’t really matter whether there is open water or ice because clouds interfere with sunlight reaching the surface during the Arctic Summer, and there is no sunlight during the Arctic Winter! When sunlight does reach the surface, the 100% reflectivity at the Earth’s limbs helps explain, in part, why the poles are so cold. (This also occurs around the perimeter of the oceans at the planet’s terminators, not just at the poles.) Snow on top of ice actually scatters light in all directions, including downward. Thus, there is actually more absorption at a glancing angle with the presence of snow than there would be with calm, open water! There are tradeoffs that complicate the situation and I don’t think they are being taken into account by most climate modelers. The simplistic explanation of decreasing Arctic ice being responsible for reinforcing warming is probably a stretch by those unfamiliar with Fresnel’s equations.
Clouds are the best evidence that the naive “Science is settled” claim is false. Clouds are highly variable in their position, extent, and albedo, often moving rapidly under the influence of winds. There is reason to believe that a decrease in cloudiness has had more impact on the retreat of glaciers than has the supposed average increase in air temperature (≪1°C/century at mid-latitudes). Melting of ice is not directly proportional to the increase in temperature if the temperature never gets above the melting point of ice. At best, one would expect that glacier retreat would follow the lapse rate if air temperature were the controlling factor. What one finds is that the retreat of glaciers is more pronounced on mountain slopes with south-facing aspects. That suggests that there has been a decline in clouds that formerly protected the snow and ice from direct sunlight.
More frustrating yet, the phase-change energy exchanges that take place in clouds cannot be modeled at a scale necessary to capture details, for inclusion into Global Circulation Models. There simply aren’t any computers powerful enough to handle the calculations, and simplifying assumptions must be made. There is an old aphorism that “The Devil is in the details.” That applies quite aptly to the problem of energy exchange in clouds. Because clouds change so rapidly, and have an important impact on both albedo and total reflectivity, it is probably inappropriate to talk about the Earth’s albedo as though it were some constant. Clouds are changing all the time!
In summary, albedo is commonly defined as the “whiteness of a surface.” Water is commonly characterized as having low reflectivity because albedo is used. Water can be illuminated obliquely and have a reflectivity higher than snow, but it will appear black to an observer not in the plane of reflection, and observing at the angle of reflection. The albedo commonly used by climatologists is on the low-end of a range of measured values. Even the high-value albedos, are too low to be used as an estimate of total terrestrial reflectivity because albedo excludes almost all specular reflections. Albedo is an under-estimate of the reflectivity of Earth surficial materials; it is only appropriate for clouds, and to a lesser extent, for bare sands. At the very least, specular-reflection has to be taken into account, as well as diffuse-reflectance, for the oceans. The most accurate representation would be through measured BRDF for all angles of solar incidence for the major surficial materials found on Earth.
I incorrectly stated that the average reflectance for a parcel of water on the equator of the hypothetical Waterworld was about 18%. I initially made two mistakes. I estimated the area under the curve of the Fresnel equation incorrectly, determining it to be about 9% for a 6-hour period. I then incorrectly doubled that for the full period of daylight. That was unjustified because the symmetry of the curve still only provides an average reflectance for the full period of daylight that is the same as the morning-to-noon average.
I have subsequently calculated the area under the curve, and I find it to be about 12.2% for the range of 0 degrees angle-of-incidence to 90 degrees. I also calculated the instantaneous reflectance for the ideal Waterworld hemisphere, and I find it to be about 17.6%. The reason that the numbers are not the same is because a spot near the Equator has a lower reflectance at noon (0°) than a spot at higher latitudes, which only sees a minimum reflectance appropriate for an angle of incidence equal to the latitude.
While the numbers have changed some, I stand by the argument of the implications for the NASA estimates of open ocean reflectance, and unaccounted for loss-of-light from the surface of Earth.
Incidentally, I want to emphasize that for any calm water on the terminator, the reflectance will be 100%.