Reflectance and Emission from Planetary Surfaces

Ge151: Spring 2011

Reflectance

Reflected visible and near infrared light from planetary surfaces is arguably the most important remote sensing window available to us. Not only are most planetary images taken in the visible (where the sun provides a convenient abundant illumination source) but reflected light in the near-infrared provides important spectral information capable of distinguishing many minerals. This tutorial will cover a few basic things to make sure everybody is starting from the same level.

When dealing with reflection the first thing to consider is the spectrum of the source (the sun for planetary observations). The solar luminosity is roughly 4x10²⁶ Watts, emitted isotropically over 4pi ster-radians. The solar constant for each planet is the solar flux at that planets position (for Earth its around 1400 Watts). The solar flux dies off as an inverse square law, Jupiter at ~5 AU receives about 1/25^th (~4 %) of the Earth's solar flux. The shape of the solar spectrum follows that of a 6000 K blackbody. It peaks in the visible region (not surprisingly since our eyes have evolved to take advantage of that) at about half a micron. To convert the Planck function to specific flux (flux per hertz) you have only to multiply it by the solid angle subtended by the sun. The solar constant is then just the integral of the specific flux over all frequencies.

Most surface images of planetary bodies are taken with a broadband filter which covers most of the visible range. Several things affect the brightness received by each pixel in the CCD array or each element in a vidicon (light sensitive vacuum tube) image (believe it or not for some bodies this is still the best we have). The flux falling on the surface is the primary factor and is the same for all pixels unless the image covers such a large fraction of the planet that the solar incidence angle varies considerably from one part of the image to another. The albedo of the surface making up that particular pixel is a measure of what fraction of the incident light is reflected, there are many types of albedo and we'll discuss some of them later. Albedo varies from pixel to pixel, e.g. sand is darker than dust (for reasons we'll also discuss later). Local slopes can also affect the incidence angle and so affect the brightness of that pixels, e.g. the sunlit side of a hill is brighter than the shadowed side. Before we go any further, let's define our observational geometry to prevent any confusion.

The above figure is taken from Hapke, 'Theory of Reflectance and Emittance Spectroscopy' (Cambridge University Press, 1993). The incidence angle is denoted by i and is the angle the incident solar radiation makes with the local vertical. The emission angle is denoted by e and is the angle made by the emitted radiation toward the detector with the local vertical. Nadir pointing spacecraft observe only what's directly below them, so the emission angle is usually very small. The phase angle is denoted by g and is the angle between the incident and emitted rays. For nadir observations (where e is zero) the phase angle is equivalent to the incidence angle. When g is zero the sun is directly behind the observer this can lead to a surge in brightness known as the opposition effect, which we'll talk about later.

For a spherical planet the incidence angle on the sun is given by the following formula, where L is the latitude, H is the hour angle and D is the solar declination.

cos(i)=sin(L)·sin(D) + cos(L)·cos(D)·cos(H)

Taking into account the position of the sun, planet and spacecraft it is possible to derive an I/F value for each pixel. This is the fraction of light which hits the surface contained within the pixel that was reflected. It depends both on the albedo of the surface at that point and on the local slopes which may have affected the incidence angle. That's fine for bodies like the Moon, Mercury and the Galilean satellites but not for bodies like the Earth or Mars which have atmospheres. The atmosphere can scatter light out of the incoming solar radiation which reaches the surface providing diffuse illumination rather than collimated and it can also scatter radiation out of the outgoing beam of radiation. In short the presence of an atmosphere is bad news when trying to interpret I/F values. We'll restrict ourselves to how the surface affects the I/F values but keep in mind that any atmosphere is playing a major role.

Albedos

Before describing the different kinds of albedo it is necessary to describe the concept of a Lambert Surface. A Lambert surface is one which appears equally bright from whatever angle you view it (any value of e). The Moon is a very good approximation of a Lambert surface, it appears equally bright at its edges and at its center making it appear almost like a two dimensional disk rather than a sphere. The flux from 1 m² of a Lambert surface is proportional to cos(e) due to the geometrical effect of foreshortening, however the same angular size includes more surface area at grazing angles and so these effects cancel out leaving the brightness independent of e. There are many kinds of albedo, all of which were defined for use with the Moon and then generalised to cover other bodies. Some of the more common ones you may come across are:

Normal Albedo: Defined as the ratio of the brightness of a surface element observed at g=0 to the brightness of a Lambert surface at the same position but illuminated and observed perpendicularly i.e. i=e=g=0.

Geometrical (Physical) Albedo: It is the weighted average of the normal albedo over the illuminated area of the body. Defined as the ratio of the brightness of a body at g=0 to to the brightness of a perfect Lambert disk (not sphere) of the same size and distance as the body illuminated and observed perpendicularly i.e. i=e=g=0.

Bond (Spherical) Albedo: Defined as the total fraction of incident irradiance scattered by a body in all directions. This quantity will be important when considering the total amount of energy absorbed by a surface for things like thermal balance etc...

Slopes

We've already mentioned that local slope can serve to enhance or reduce the apparent brightness of a patch of ground. If we assume an albedo for each pixel then we can figure out what the solar incidence angle is for each pixel based on the I/F value. We can subtract the incidence angle for a flat piece of ground at that time,latitude and season, so we are left with the slope of each pixel but only in one direction! This procedure is known as photoclinometry, it works best when you can remove the albedo easily e.g. if the surface is covered with some uniform albedo material like frost or dust. This can only tell you what the slopes toward and away from the sun are, if you want to reconstruct the entire topographic surface then you need at least two observations illuminated from different directions. Again the atmosphere wreaks havoc with this sort of technique and make it difficult to get quantitative results however if you could constrain the answer with other methods such as a laser altimeter in the case of Mars then you can generate very high resolution topography maps for small areas.

Reflection Phenomena

Photometric and Phase Functions: Surfaces behave differently when viewed under different geometries. The phase function gives the brightness of the surface viewed under some arbitrary phase angle divided by the brightness expected if the surface were viewed at zero phase. The photometric function is the ratio of surface brightness at fixed e but varying i and g to its value at g=0. For a Lambert surface the photometric function is given by cos(e).

Grainsize Effects: Scattering from a particulate medium in the visible and near infrared tends to be dominated by multiple scattered light i.e. photons that have been scattered within the surface more than once. Photons pass through grains getting absorbed along the way, they get scattered by imperfections within the grains and by grain surfaces. The more surface area there is the more scattering there is and the bigger the grains (more volume) the more absorption there is. The surface area to volume ratio of the grains therefore has an effect on the amount of light scattered by the surface. The surface area to volume ratio scales linearly with the grain size so for the same material surfaces with larger grain sizes will in general be darker. The most familiar example of this is probably sand on the beach, dry sand is bright but when sand get wet it clumps into larger grains and turns darker. In reflection spectra larger grains produce broader, deeper absorption bands.

Opposition Effect: When observed at zero phase surfaces experience a sharp increase in brightness known as the opposition effect (also known as: opposition surge, heiligenschein, hot spot and bright shadow). The pictures below show the opposition surge in the case of the moon, look at the edges of the astronauts shadow (which is at zero phase angle), can you see the extra brightness compared to the surrounding regolith. The plot shows the same effect but in a more quantitative way.

The physical explanation offered for the opposition effect is that of shadow hiding. If the illumination source is directly behind you then you cannot see any of shadows cast by surface grains (since they're all hinding behind the grains themselves), whereas if you were looking toward the illumination source then you can see all the shadows cast by the surface making it appear darker. Coherent backscatter is another mechanism offered as an explanation and it is likely that both mechanisms play some role although from recent work on Clementine data it looks like shadow hiding is the dominant mechanism at least in the case of the Moon.

Reflection from particulate mediums is an area of research in its own right, which has been largely pioneered by Bruce Hapke in recent years. This is only meant as a taste of whats really out there. Anyone wishing to go deeper into the (gruelling) mathematical modeling of reflection can come to me for references and/or help.

Questions

(1) What is the flux at Venus for a filter with a bandwidth of 0.05 microns centered on 0.7 microns?

(2) At equinox on Mars the MOC camera aboard MGS is making observations at 60 N at 2pm local time. The pixel size is 1.4 meters across, how much detectable power is falling on the surface within the area of each pixel ? The narrow angle band is centered on 0.7 microns and has a bandwidth of 0.4 microns.

Emission

In 1800, Sir William Herschel, the royal astronomer to the King of Great Britain, conducted an experiment to study the heating effects of sunlight. He used a prism to separate light into the colors of the spectrum and used a thermometer to measure the temperature in each color. As he moved from the violet to the red region, the temperature increased; however, when he placed the thermometer in the region just beyond the color red, the temperature continued to increase, even when no light was visible to the naked eye. William Herschel had just discovered the portion of the electromagnetic spectrum known as the infrared.

The term infrared spans a huge range of wavelengths. In this tutorial we're going to be talking about the thermal infrared. By thermal infrared we really mean heat since the 10 micron region corresponds to the peak of a 300 K blackbody, which is a nice cozy summer day for most of us.

We'll start off with the planck equation in both the wavelength and frequency forms. The units of B are J s^-1 m^-2 ster^-1 hz^-1 or m^-1. (We use S.I. units exclusively in these tutorials). This quantity is known by several names: Radiance, Emittance, Brightness (or more correctly specific brightness), surface brightness or specific intensity. This is a measure of power flowing from 1 square meter in some direction with a spread of propagation vectors of 1 ster-radian. Since no real surface emits as a blackbody, B is usually modified by a factor known as the emissivity which varies between 0 and 1. Just as there are many kinds of albedo there are many kinds of emissivity. However for our needs this introduces unnecessary complications.

The great thing about these equations is that they depend only on temperature. A quick note on nomenclature though. Terms like radiance, flux, irradiance and brightness are often bandied about in annoyingly imprecise ways as if they all mean the same thing [some of them do and some don't]. In an effort to set the record straight we'll try and provide exact definitions.

• Radiance: (also known as: specific brightness, emittance,surface brightness or specific intensity) is given by the Planck equations above. When a spacecraft measures the luminosity (just energy per second) of a surface it is necessary to convert this to radiance, i.e. correct the quantity so that it comes from 1 m², divided by the bandwidth so that it is now per hertz, and divide by the collecting area of the instrument (in ster-radians) so that it is now per ster-radian. From the radiance we can derive things like surface temperature or emissivity in that band.

• Intensity: (also known as brightness) This is simply the radiance integrated over all wavelengths or frequencies. It still contains a per solid angle term so the amount of energy you can collect depends on the amount of the sky your detector covers as seen by the source. The units are J s^-1 m^-2 ster^-1. Note the intensity does not vary with distance from the source. The amount of energy one receives changes with distance because the solid angle subtended by the detector as seen from the source changes. However the intensity remains constant.

• Irradiance: (also known as specific flux) This is the radiance integrated over solid angle. It is usually used to describe collimated or near-collimated beams such as sunlight on a planets surface. The units are J s^-1 m^-2 hz^-1 or m^-1. This is still a function of wavelength / frequency.

• Flux: This is the radiance integrated over all solid angles and over all wavelengths/frequencies. It is a measure of the total energy coming from 1 m² of a surface. Its not so easy to measure this quantity you could measure the radiance but then you'd have to make some assumption about how the surface emits over all other emission angles and over all other wavelengths. The units are J s^-1 m^-2.

• Luminosity: This is the final quantity, it is the radiance integrated over all solid angles, all wavelengths and over the surface area of the whole body. The temperature may vary from surface area to surface area but for things like the sun you could assume its pretty constant. The units are J s^-1.

Example: A spacecraft orbiting 400 km above the planet's surface has a far infrared instrument with a circular collecting area 50cm across. It observes a patch of the surface 3 km by 3 km at 50 microns through a filter 1 micron wide. The instrument records a luminosity of 2.25*10^-6 Watts. What is the radiance and hence the temperature of that patch of ground ?

• The area observed is 3000x3000 meters, or 9x10⁶ m².

• The area of the collector is pi*0.25² m². It is at a distance of 400 Km from the surface so the solid angle it subtends is pi*0.25² / 400000², which is 1.23x10^-12 ster-radians.

• The bandwidth is 1 micron or 10^-6 meters.

We can divide the luminosity recorded by the spacecraft by these three quantities to derive the radiance. This gives us a radiance of 204000 J s^-1 m^-2 ster^-1 m^-1. We can rearrange the Planck formula as shown below and use it to get the temperature.

So the answer is 273 K, some major assumptions that went into this are : (1) The surface is assumed to emit isotropically, (2) The emissivity is assumed to be 1 (more on this later), (3) The atmosphere is assumed to play no role in the thermal infrared. The last assumption is the most serious. However accounting for the atmosphere would be a course in itself so we'll ignore it for now.

We saw in the example that you can't recover the emissivity and the temperature from a single measurement. One way to break the connection is to take a spectrum which resolves the black body peak. The wavelength of peak emission is directly connected to the temperature through the Wien's Displacement Law, quoted below. The derivation basically follows the steps of taking the derivative of the Plank formula with respect to wavelength and setting it equal to zero to find the maximum. The resulting equation cannot be solved analytically using standard special functions of mathematical physics, but can be solved in terms of Lambert's W-Function.

If the surface emits isotropically then we can derive an expression for the flux emitted solely in terms of the temperature. Remember the flux is just the radiance integrated over solid angle and wavelength. For a patch of flat ground emitting into a hemisphere you may be tempted into thinking that the solid angle is simply half a sphere or 2 pi ster-radians. This would be true for a point.  However because the emission is coming from a finite area (1 m² in this case) the effective solid angle is just pi due to geometrical effects. After multiplying by the solid angle we have to integrate over all wavelengths. The result is given by the Stefan-Boltzmann Law, given below.

Here sigma is the Stefan-Boltzmann constant, 5.67x10^-8 J s^-1 k^-4. Knowing the total flux emitted from a surface will be important when we consider the thermal state of the subsurface and heat balance of surfaces later in the course. As with Planck's Law, the emissivity factor must be taken into account to find the flux from any real surface.

The emissivity may vary with wavelength in the same way as albedo does (see tutorial one). Below is shown a thermal spectrum of quartz along with its ideal blackbody spectra.

If we were to divide the real spectra by the hypothetical blackbody then we would recover the emissivity as a function of wavelength (spectral emissivity ). This can (and is) done for planetary surfaces. Different rock types show different emission spectra. Below are some examples. As mentioned before removing the effect of the atmosphere (especially in the case of Mars which has a lot of suspended dust) is a formidable challenge and is a permanent thorn in the side of workers in this field.

Questions:

(3a) Suppose you fly a mission to Mercury, and you go into polar orbit so you can sample both the day and night sides which have temperatures of ~700 and ~100 Kelvin respectively. For the orbit and instrument described in the example what is the range of powers that you would expect to record? Assume an albedo of 0.16 and an emissivity of 0.84 and a Lambert surface.

(3b) At what wavelengths does the intensity of reflected and emitted light (day and night sides) peak?

(3c) For observations on the day-side what wavelengths would you have to observe at to ensure that what you are looking at is dominated by emitted rather than reflected radiation (by at least a factor of 10)? You will need to use also the methods described in tutorial one.

(4) The stability of liquid water on and within the surface of Mars is key to many aspects of martian science.  In this exercise we ask you to consider some aspects of the frozen to liquid transition of water within the martian surface.

Heat flow is a key aspect of this problem and many others and, in uniform media, can easily be approximated by the formula:

On Mars the heat flow has been guessed by theoretical work to be approximately 30 mW per square meter.

Rearrange the above equation so that you can estimate the depth to the melting temperature of water in terms of a surface temperature.

Assume a conductivity typical of rock (~3 W/m/K) and and global average Martian surface temperature (~200 K).  How far down would you expect to find liquid water (assuming there is any water to find).

What about under the polar cap? Typical ice conductivities are 2.25 W/m/K and typical polar surface temperatures are about 160K. The northern polar cap rises about 3Km above the surrounding terrain.  Do you think its likely that there is liquid water underneath it?  Why ?

What would be the difference if 50% of the polar cap were made up of dirt (rock conductivity). If there were a significant concentration of ionic salts at the base would you expect a significantly different conclusion?

In all these cases we neglect the effect of increased pressure at depth. What effect does increased pressure have on the melting point and how big is that effect (no need for the exact relationship, order of magnitude will do)

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