The horizon is in the window and almost unlit

The sky inside a cone treated Snell’s window as what it is geometrically: a picture surface, in which the whole upper hemisphere of directions is pressed into a cone of half-angle 48.61°, with an area scale that runs to zero at the rim and no other picture surface’s kind of distortion. In the section on what is actually visible inside it, that essay said in passing that the outer part of the disc is dim. It did not measure that, because the essay’s question was where each direction goes. The question left over is how much of each direction’s light gets there.

That is not a detail. Geometry alone says a person standing on the bank, with their feet at the water’s edge, is visible to a fish at the rim of the window. The fish’s view of that person depends on how much of their light the surface lets through. A picture surface whose scale runs to zero at its rim, and whose light also runs to zero at the same rim, has a very different horizon from one that merely squeezes it.

What the surface passes #

At a water surface, a ray of light divides. Fresnel’s equations give the share reflected for each of the two polarisations, and for unpolarised light the share is their average. A ray from straight overhead loses 2.04 per cent, which is $((n-1)/(n+1))^2$. As the ray gets further from the zenith the share reflected rises, slowly at first. Near grazing it climbs to all of it.

Read at the fish’s end, each angle under water corresponds to one direction in the sky, so the transmitted share is a function of the angle off the zenith under water.

98.0 per cent straight up. 97.5 at 30° under water, which is sky 41.8° from the zenith. 94.4 at 40°, 86.1 at 45°, then half at 48.19°, where the sky’s angle is 83.5°. It reaches nothing at 48.61°. Near its end the curve falls with a vertical tangent, because in the rule near grazing the transmitted share depends on the cosine of the sky’s angle, and that cosine falls as the square root of the remaining distance to the critical angle.

The two curves’ sum is a reciprocity that is worth checking, and the check is direct: a ray from below meeting the surface at a given angle reflects exactly the share a ray from above fails to pass along the same path. The two are computed separately here, and they agree to nine digits at every angle tested. So a surface can be described as a single split, in which light coming to the eye along a given line is partly from above and partly from below, in shares that sum to one. It is not that the sky stops at the rim and the bottom’s reflection begins. The two trade places smoothly over a very small angle.

Brighter than the sky, for most of the way #

Going through the surface does not only lose light. It also concentrates it. The whole of a hemisphere of sky is compressed into a much smaller solid angle, so the same flux passes through a smaller cone of directions, and its radiance rises by the square of the index. For water that is 1.78. So a uniform sky seen from below is $1.78\,T$ times as bright as the same sky seen in air, with $T$ the share transmitted.

At the zenith that is 1.74. A diver looking straight up at an overcast sky is looking at something 74 per cent brighter than what is overhead for a person on the bank. This is not extra light. A small patch of the sky’s light has been squeezed into a smaller range of directions by the bending of its rays, with the total conserved, so what the eye records per unit of direction goes up. It is the same accounting that conformal is not undistorted applies to scale, here applied to brightness. It holds near that value across most of the disc: 1.73 at 30° and 1.68 at 40°. Only over the last few degrees does it begin to fall.

The crossing, where the sky through the surface is as bright as the sky in air, is at 48.01° under water, which is sky at 82.2° from the zenith. So only the bottom eight degrees of sky appear dimmer than they would in air, and all of those eight degrees lie within the outermost 0.59° of the window’s radius. By any standard, all the action happens at the very edge.

That gives the window its appearance. Most of its area is bright and rather featureless, with its scale changing steadily, and then there is a rim. The rim holds a remarkable number of things at a remarkably small size and low contrast. A dive photographer who has shot upwards through a calm surface will recognise the look: a clear bright disc whose rim appears to hold nothing at all, though trees and people on the bank are all geometrically there.

The horizon’s share of the picture #

Now put numbers on the claim that the horizon is in the window and has little to show. Take bands of sky above the horizon and ask both questions at once: how much of the disc’s radius does each band get, and how much of its light arrives?

Twenty to thirty degrees above the horizon takes 4.31° of the radius at 92 per cent. Ten to twenty degrees takes 2.80° at 80 per cent. Five to ten takes 0.73° at 57 per cent. The bottom five degrees, in which a bank and its trees would stand for anybody not lying right at the water’s edge, takes 0.25° at 32 per cent.

A fish looking at a person on the bank therefore gets a person with their feet at the rim and their head a quarter of a degree in. It would need the resolution of a good human eye, looking up through perfectly calm water, to resolve anything of their shape at all, about 15 arc minutes at under a third of its light against whatever else is at the rim. The advice commonly given to anglers, to stay low and back from the bank, can be read from these bands. It is geometry and transmission together, and what a ray does at a surface is what makes the bank visible there at all. The bands above the five-degree line are the ones the angler should be worried about, since a head at any distance that stands ten degrees above the horizon is out of the worst of both.

It also explains why a real horizon in the window is very seldom visible as a line. In clear calm water the geometric rim is perfectly sharp. The picture at the rim has no contrast left to show it with.

What a level patch receives #

The picture is the fish’s view of the rim, in the same sense that the ceiling on apparent distance is a driver’s view of a mirror’s limit. There is a second way of counting the same light, and it answers a different question: how much of the light that reaches the bottom comes through the rim?

A level surface weights each direction by its cosine. So the low sky, which is both dim through the surface and arriving at a slant, counts for very little: beyond 80° from the zenith, 1.59 per cent of what reaches a level patch; beyond 85°, 0.24 per cent. A reader working out how much light a pond’s bottom gets, or a solar panel under water, can safely treat the low sky as absent, and can use a cosine-weighted transmission of a single number. For water that number is 0.934: of all the light from a uniform sky, 93.4 per cent comes through the surface.

So two ways of asking give the same answer. Whether the question is about seeing the rim or being lit by it, the sky’s last few degrees hardly exist below the surface, however exactly geometry puts them there. This is an instance of something that keeps recurring in optics. A map tells where things go and is silent on what arrives. The penumbra is the lamp’s image has the same separation for a shadow, and where a surface spends its pixels has it for a picture: a scale says how much of the surface a direction takes, and nothing else.

The edge belongs to the bottom #

So far, the picture has been of a disc against a dark surround. The surround is not dark, of course. Outside the critical angle the surface is a perfect mirror reflecting the bottom, and the disc’s rim is really a boundary between two things: the sky coming through and the bottom coming back.

Their brightnesses have to be compared. Suppose the pool is shallow and clear, with a matte bottom of albedo $a$. It is lit by the sky through the surface, and that sends light back up. Some of that returns from the surface. Let $\bar T$ be the cosine-weighted transmission of the surface from above, which is 0.934, and let $\bar r$ be its cosine-weighted reflectance from below, which is 0.475, large because a good part of the upward light is beyond the critical angle. Summing the repeated bounces gives the bottom’s brightness relative to the sky:

$$L_b = \frac{a\,\bar T}{1 - a\,\bar r}.$$

Sand, with an albedo of 0.3, gives 0.326. White tile at 0.8 gives 1.20. Set $a = 1$ and the formula gives $0.934/(1 - 0.475) = 1.777$, and that value is the square of the index, to three places. It has to be. A perfectly white, perfectly closed box of water under a uniform sky is in equilibrium, and the brightness inside it must be $n^2$ times what is outside, or light would be pumped through a transparent surface, one way, for nothing. This is the check that matters most here, since two integrals that have nothing obvious to do with each other have to come to one exact number. They do.

Now look at what an eye receives at each angle: $L_{\text{sky}}\,n^2 T + L_b R$, with $T + R = 1$.

A dark bottom has a brightness of 0.1 and leaves a very sharp fall at the rim, from 1.7 to 0.1 over less than a degree. Sand is similar, at a ratio of about 5 to 1. White tile has a brightness of 1.20, so there is a dip, and what it dips from is not much higher than what lies outside it. And with a perfectly white bottom, the formula becomes $n^2(T + R) = n^2$, with no dependence on the angle at all: a perfectly white pool has no Snell’s window. Every direction looks equally bright, and the rim that dominates every description of the effect is simply not there.

That reorders the phenomenon. The geometric rim remains, in the sense that directions inside it come from above and those outside it do not. But what a person sees as the window’s edge is a contrast, set by the bottom’s reflectance. In a swimming pool with a pale floor, the edge is far weaker than in dark open water. Anyone who has looked up in both has seen it, and the reason is this formula. So the window is less a property of the surface alone than of the surface and what lies under it, which in hindsight is exactly what a picture of two things divided by a boundary should be.

What polarisation does #

Fresnel’s two polarisations separate at the surface, and the transmitted light is polarised to an extent that also depends on the angle.

At Brewster’s angle, where the sky arrives at 53.1° and the p-polarised reflection vanishes, the light in the water is only 4.1 per cent polarised, at 36.9° under water. It is the reflected light that Brewster’s angle polarises entirely, and very little of that is the light in question. The transmitted light keeps rising past that point, reaching 26.8 per cent a hundredth of a degree inside the rim and tending to $(n^2-1)/(n^2+1) = 0.28$, as the two polarisations’ transmissions fall to zero in the fixed ratio $n^2$.

A reader who knows that many aquatic animals can sense polarisation might find that suggestive, and it may be. Some species do appear to use the underwater polarisation pattern. Most of that pattern, however, comes from scattering by the water itself, and from the sky’s own polarisation, neither of which is modelled here. The surface’s own contribution is what this figure shows. It is a small, smooth, radial pattern, and its strongest part is the one with the least light in it.

Only glass-smooth water has a sharp rim #

A calm surface. Any ripple changes the local normal, and a fraction of a degree of slope moves a ray a good way across the rim, so the sharp rim of this essay becomes a shimmering one on any real water. Every statement here about the last half a degree is a statement about glass-smooth water. On a real pond with small waves the disc’s edge flickers over several degrees, and the transmitted brightness is an average over the slopes.

A uniform unpolarised sky. The sky is brighter near the sun and is itself polarised. A low sun’s light arrives near grazing and mostly reflects away, and a high sun’s arrives near the zenith and mostly passes, which is why a bright low sun through the surface makes so little of an impression compared with the same sun at noon.

Clear water, and a shallow pool. Absorption and scattering reduce the contrast with distance and add a veil of light that is independent of all this geometry. A diver at a few metres in clear water sees roughly what is described; at twenty, the rim has gone into the blue. The closed form for the bottom assumes the light bounces with no loss between bottom and surface.

The index is 1.333 at all wavelengths. Water’s index varies slightly across the visible, so the critical angle varies by a little under half a degree between red and blue. That makes the rim coloured on very calm water, a slight effect that this essay has not measured. A straight stick in water has more on what an unaided index assumption can hide, and a ball of water shows what happens once the surface stops being flat.

Still open: what a ripple does to the window’s area #

The rim here is the sharpest thing in the essay, and it is the first thing a real surface removes. A small slope on the surface tilts the window. A wave of slope $s$ moves the local critical cone by about $s(n-1)/n$ in the underwater direction, so a surface with a distribution of slopes blurs the disc’s rim by a certain amount, and the disc does not stay a disc: it breaks into a moving patchwork with bright focused lines and gaps.

The measurement that settles what a ripple does generates a sea of small waves with a stated slope distribution, traces rays from a point at a stated depth to the surface and on into the sky, and asks three things: how the mean brightness at each angle differs from the calm surface’s curve; how large a slope it takes for the disc’s rim to spread over more than the bottom five degrees of the sky occupy; and, since a wave is also a lens, how much of the bright patchwork is real focusing rather than noise, and at what depth it is strongest.

A map says where a direction goes and nothing about how much of it arrives. Snell’s window holds the whole horizon and shows almost none of it, and its famous edge is not in the surface but in the pool.