[This web page is taken, lightly edited, from a single lecture given as part of a module consisting of a series of lectures, one each by a number of colleagues, based around suitable topics in applied mathematics, topped and tailed by lectures about the modelling process. Each lecture was followed by a supervised period in which the students (around 30 in a typical year, working in groups of around four) expanded on the topic just delivered. For their assessment, the students had to write essays and do projects based around chosen topics.]

           

The Sun is distinctly darker -- almost dirty! -- near the edge. Children who paint the Sun as a uniform yellow circle are misled! Neptune too is a somewhat deeper blue near the edge, but that could be an artifact of the banding, and a uniform blue circle would not be a bad picture. The Moon, however, is definitely not darker at the edge. If anything, it's lighter, especially on the right-hand side [where the edge as viewed from Earth is sunlit -- the Moon here is just past full]. So, it's perfectly reasonable to draw a full Moon as a uniform silver circle [though dirty grey would be closer].

This picture of a half-Moon shows the effect much more strongly! Sadly, it's not real -- actually, it's a simulation as developed for the film Apollo 13. But the lighting is meant to be realistic.

Something else that you may have noticed, but that is not apparent from photographs, is that the full Moon is very much brighter than a crescent Moon. That's not because it's closer, nor merely because of the larger lit area -- it really is a whole lot brighter.

OK, how can we account for these phenomena? The easiest one first: the diagram shows how photons emerging from the Sun after traversing a similar distance inside the Sun do so from deeper layers when emerging normally than when emerging at a shallow angle. Deep inside the Sun, it is so `foggy' that you can expect to see only a centimetre or so, not enough for this effect to be noticeable. But near the surface, it's only slightly misty, so when you look at the centre of the Sun's disk, you see into regions where the Sun is quite a lot hotter than actually at the surface, whereas when you look at the edge, all the photons you see come from very near the surface. The effect is accentuated by the fact that the energy radiated by matter at temperature T varies as T⁴.

Project 1: This does not have much to do with reflexion [though one model for the way in which energy gets from the centre of the Sun to the outside is via `Brownian motion' of the photons over many thousands of years], but nevertheless makes a reasonable investigation. Find out about `stellar atmospheres', and the Sun in particular. You can look in the Library under classmarks QB800 to QB809. You can assume that matter inside the Sun radiates photons isotropically [uniformly in all directions], and that photons are absorbed at a rate proportional to the `opacity' [fogginess] at each point inside the Sun. In real life, the opacity is an extremely messy physical function [it varies strongly with temperature, pressure, chemical composition and wavelength -- which is why some materials are opaque and some transparent and of varying colours]; but a simple model in which you assume it to be constant will show you most of the features.

Next, let's look at Neptune. Neptune, like the Moon [and most other bodies in the Solar System], shines entirely by reflected light. How does reflexion work? In general, we have an incident ray coming in direction i, hitting a surface with normal n, and a reflected ray going in direction r. We can assume angles of incidence and reflexion i and r as shown; note, we do not assume that i = r, nor even that i, n and r are in the same plane. We will be interested in the case where i is the direction from the Sun, representing therefore sunlight, and r is the direction towards Earth, so this is the light we see. The incident light is partially absorbed [especially red light, in the case of Neptune, which is why its colour is blue], and partially reflected in all directions [otherwise we would be able to see Neptune only from favoured angles!]. The amount reflected in the direction r depends in a complicated way on the directions i, n and r, depending on the nature of the surface.

There are two extreme cases: (i) Perfect reflexion, as in a mirror. In this case, there is no reflected light except when r happens to be the `mirror image' of i, following the usual laws, so that i, n and r are coplanar and i = r; Neptune would look like a bright spot, with most of its surface invisible. [Anyone interested: for a less extreme version of this, look up Phong illumination in any book on computer graphics; this describes a law in which the intensity of illumination drops off according to a power law in the cos of the angle between r and the mirror-image direction.] (ii) Diffuse reflexion, which we're used to seeing from matt [or `unshiny'] surfaces. In this case, of `Lambertian reflexion', the apparent intensity of the light is independent of r. This is a consequence of two cancelling factors: from a Lambertian surface, the light emitted is proportional to cos r [Lambert's law]; but the viewing angle means that a beam of cross-section A comes from an area A sec r of the surface. In this model, if you keep the Sun and Neptune fixed, but you move around [but always at the same distance from Neptune], then any particular part of Neptune's surface will always have the same brightness, depending only on the relationship between i and n. Noting that a beam of sunlight of cross-section A has to illuminate an area A sec i of the surface, in the simplest cases we expect the apparent brightness of Neptune to vary across the surface proportionally to cos i. This would account for a full Neptune looking bright near the centre [i = 0] and dark near the edge [i = 90°]; but actually Neptune is only a little less light near the edge, and the full Moon certainly isn't dark near the edge. So what else is going on?

Possibly surprisingly, the answer turns out to be somewhat similar, though for different reasons, in both cases. The figure shows a beam of sunlight striking the atoms in a gas. If the atoms are tightly packed, as in a solid or liquid, at the surface [the sloping dotted line], then the angle of the surface matters -- you get a few atoms strongly lit if the surface is orthogonal to the beam, or many atoms weakly lit if it is at a very shallow angle. But if they are widely spaced, then it doesn't matter very much -- atoms near the surface are fully lit, those further back are partially hidden, and those a long way in are completely hidden. The numbers and proportions are virtually independent of the angle, until the angle [either of incidence or reflexion] becomes so shallow that the surface atoms start to shield each other.

Thus, the view from Earth of a full [and gaseous] Neptune is of an almost uniform disc [apart from actual surface changes on Neptune!], except very near the edge. Viewed at an angle, Neptune would be much fainter, partly because part of each atom would be in shadow from the Sun, but also because each lit portion may be shielded by other atoms. [Physically, this is oversimplified -- atoms and simple molecules are too small to reflect light, so we're really talking about `scattering', but the effect is much the same.]

But the Moon's surface is solid, so the above doesn't apply? This was a real puzzle, and all attempts to match the Moon's brightness patterns to known rocks failed, until it was realised that the Moon's surface [the `regolith'] is very dusty. The dust particles in the top few millimetres form a highly porous but fixed structure, and many of the same features apply. The two differences are that the dust particles are big enough to act as reflectors, and that they are touching and fixed rather than moving. Full details are beyond the scope of this study, but the result is that the brightness of a given surface is proportional to k / ( 1 + cos r / cos i) where i and r are as above, and k depends on the phase angle between incident and reflected rays, getting smaller as the angle increases, the exact dependence being a function of the porosity. It is found that a good fit can be obtained if the porosity is 38%, that is, if the regolith is 38% dust and 62% empty space, and if the dust is Lambertian. However, even this doesn't make the full Moon bright enough; for that, we need extra back-scatter [like `cat's eyes', common with crystalline structures].

Project 2: For attempts in class! We can model this structure very crudely as follows. Imagine an infinite chessboard. Now replace the black squares by white cubes [or, more generally, blocks of arbitrary height]. Now the tops of the cubes form a Lambertian surface in the usual way; but, depending where the Sun is, the vertical surfaces and the lower squares [see diagram] may be wholly or partially in shadow or be lit [at a different angle of incidence for the verticals]. The observer, again depending on angle, now sees half of the surface [the top] in sunlight. The other half comprises partly the vertical surfaces, and partly the lower horizontal surfaces [unless the angle is too great for this] So the observer sees a certain fraction [at least half] of the horizontal surface in sunlight, a certain fraction of vertical surface [perhaps none] in sunlight [at different angles of incidence and reflexion], and a certain fraction of shadow. Calculate these fractions for various angles, and hence estimate the brightness. What would the Moon [full or otherwise] look like on this model?

Note: For a given configuration of the Sun--Moon--Observer, the relationship between incident and reflected rays, i and r is fixed, as the Moon is small compared with interplanetary distances. In particular, for the full Moon, i and r are parallel, and for a quarter Moon they are perpendicular. Different points on the Moon's surface are characterised by different directions of the normal, n.

Project 3: Do project 2 `properly', using spherical and random `dust'. Good opportunity for some computer modelling, or for some Net surfing for relevant information.

Rainbows

Project 4: How bright should the drop appear from each viewing angle? [Note that at every surface, some of the light is reflected and some refracted: you will have to guess or look up the fractions as a function of i to get an accurate result.] If you have a uniform cloud of raindrops, what should you see? [If you can get a rainbow with red and blue in the right order, and if you can explain why the inside of the bow is much brighter than the outside, you will be doing quite well! Scope for more surfing of the Web.]