The ray that bends without a surface

The shimmer of water on a hot road is not a reflection of the sky in anything. There is no surface there. The rays that arrive at the eye from below the horizon have come from the sky, turned round in mid-air, and come back up, and every part of that journey took place in perfectly ordinary air with nothing in it.

Fermat’s principle when the index has no steps in it #

Fermat’s principle says a ray takes a path along which the optical path length

$$L = \int n(\mathbf{r})\, ds$$

is stationary. For two uniform media with a boundary between them, making that stationary gives Snell’s law by differentiating with respect to the crossing point. For an index that varies continuously there is no crossing point to differentiate with respect to, and the variational problem has to be done properly. The Euler–Lagrange equation gives the ray equation,

$$\frac{d}{ds}\left(n \frac{d\mathbf{r}}{ds}\right) = \nabla n,$$

which says that a ray accelerates toward higher index — sideways, at a rate set by the gradient of $n$ across it.

The case with a boundary is worth having in view first. Draw many candidate paths from a point in one medium to a point in another, compute the optical path length of each, and the one that is stationary is the one obeying Snell’s law — the law is not an extra rule, it is what stationarity produces at a step in the index. The generalisation about to be made is to let $n$ change everywhere instead of once, at which point the family of candidate paths becomes a family of curves rather than a family of corners.

The invariant, and what it forbids #

Suppose the index depends on height alone, $n = n(z)$ — which is the case for an atmosphere, a road, or any medium stratified by gravity. Then $\nabla n$ has no horizontal component, so the horizontal part of $n\,d\mathbf{r}/ds$ is constant along the ray:

$$n(z)\cos\theta(z) = C,$$

with $\theta$ the angle of the ray from the horizontal. That is the invariant, and everything follows from it.

It plays the part angular momentum plays in an orbit. A conserved quantity, fixed by the initial conditions, that bars the ray from regions it would otherwise reach: since $\cos\theta \le 1$, the ray can only be at heights where $n(z) \ge C$. Where $n(z)$ falls to $C$ the ray is horizontal, and it can go no lower. It turns.

Snell’s law is the same statement for a medium made of slabs. Writing $n_1\sin i_1 = n_2\sin i_2$ with angles from the normal is $n_1\cos\theta_1 = n_2\cos\theta_2$ with angles from the layers, and stacking slabs of decreasing index gives a ray that bends further and further from the vertical until, at the slab where $n$ has fallen to $C$, it goes horizontal — which is the critical angle, arrived at as a limit rather than as a special case.

A mirage is total internal reflection spread over a hundred metres. At a boundary the refracted angle reaches ninety degrees at one particular incidence and nothing gets across beyond it; in a graded medium there is no boundary, and instead a continuous decline of the index in which the ray’s angle from the horizontal falls to zero. Nothing is reflected in either case. The ray simply has nowhere left to refract to, and turns.

Built out of wavelets the bend has an obvious cause. A front travelling into a slower medium has its far side held back, so the front swings round — and in a graded medium every part of the front is held back by a slightly different amount, continuously, which is why the ray curves rather than kinking. Refraction is not a rule about angles. It is what a front does when one edge of it is slower than the other.

The numbers a hot road actually has #

The air just above sun-heated tarmac may be 50 °C while the air at head height is 25 °C. Hotter air is less dense, and the index of air minus one is proportional to density, so the index is lower near the road and rises with height over a few centimetres. Modelling that as

$$n(z) = n_\infty - \Delta\, e^{-z/h}, \qquad \Delta \approx 3\times 10^{-5}, \quad h \approx 5\ \text{cm},$$

the ray from an eye at height $H$ launched at angle $\alpha$ below the horizontal has $C = n(H)\cos\alpha \approx n_\infty(1 - \alpha^2/2)$, and it turns where $\Delta e^{-z/h} = n_\infty \alpha^2 / 2$. Setting $z = 0$ gives the largest angle that can be turned:

$$\alpha_{\max} = \sqrt{\frac{2\Delta}{n_\infty}} = 0.444°.$$

Under half a degree. Every ray steeper than that hits the road; every ray shallower turns and comes back, and the horizontal distance it takes to do so is $H/\alpha \approx 200$ m for a standing observer.

The index deficit is worth deriving rather than asserting, because it is the one number in the calculation that sounds as though it were fitted. Air at ordinary conditions has $n - 1$ of about $2.8\times10^{-4}$, and that excess over vacuum is proportional to the density of the air. At constant pressure — and the pressure a few centimetres above a road is the same as the pressure at head height to a part in $10^5$ — density is inversely proportional to temperature, so

$$\frac{\mathrm{d}(n-1)}{\mathrm{d}T} = -\frac{n-1}{T} \approx -1\times10^{-6}\ \text{per kelvin}.$$

Twenty-five kelvin of difference between the road and head height therefore buys about $2.4\times10^{-5}$, which is the value used above to within the accuracy of the temperature guess. Nothing was tuned: the whole of the mirage is one part in forty thousand of the index of air, arrived at from the ideal gas law and a tabulated constant.

That is the whole of the mirage: sky, arriving from below, at angles no greater than a fiftieth of the Moon’s width. It looks like water because the sky is what a puddle would show, and it shimmers because the layer is turbulent and $\Delta$ fluctuates — the same fluctuating column that makes a star twinkle and sets the limit on what a telescope on the ground can resolve.

Why a mirage has no colours, and where the colour hides #

A phenomenon built entirely out of refraction ought to separate colours, and a mirage does not. It is worth seeing why, because the exception is one of the prettiest sights in the subject.

The index of air is dispersive, but barely: $n-1$ is about one per cent larger for violet than for red. The critical angle goes as the square root of $\Delta$, so it differs between the two ends of the spectrum by half a per cent of 0.444° — a couple of arcseconds, against an effect that is already at the limit of what an eye separates. Nothing in the shimmer is coloured because the whole angular range of the phenomenon is smaller than the blur of the eye that watches it.

Now run the same reasoning on the atmosphere as a whole, where the total refraction at the horizon is not half a degree of spread but half a degree of deflection. One per cent of that is about twenty arcseconds, which is a tenth of the Sun’s diameter and comfortably resolvable. So the setting Sun is not one image but a stack of them: the blue image sits highest, the red lowest, each displaced by refraction according to its own index. The stack is invisible while the whole disc is up, because the images overlap almost exactly. At the instant the last sliver disappears, the red image has already set and the blue is scattered out of the beam by the long path through the atmosphere — leaving green, alone, for a second or so. That is the green flash, and it is this essay’s arithmetic with dispersion put back in.

The condition for seeing it is the condition for the refraction to be large and steady, which is a clean horizon and a stable air mass. A layer with the strong near-surface gradient of the road in it adds shimmer and destroys it — the two effects on this page are competitors rather than partners.

The invariant is a symmetry #

The conserved quantity used throughout has a provenance worth naming, because it explains why one appeared at all rather than being lucky.

Fermat’s principle is a variational problem, and the quantity to be made stationary is an integral of $n$ along the path. In a medium stratified by height, that integrand does not depend on $x$: the medium is the same however far along it one goes. A variational problem whose integrand is independent of a coordinate has a conserved momentum conjugate to that coordinate, and here it is exactly $n\cos\theta$. The invariant is the horizontal translation symmetry of the road, written down.

Which is why the same object appears in every stratified problem and never in a general one. An index varying in both $x$ and $z$ has no such symmetry and no such conserved quantity, and rays in it must simply be integrated. A spherically stratified atmosphere has a rotational symmetry instead, and the invariant becomes $n r\cos\theta$ — the extra factor of $r$ being precisely what turns the flat-earth expression into the one an astronomer uses for refraction near the horizon, and precisely what an orbit’s angular momentum has that a straight-line momentum does not.

It also says what would break the mirage. Not a weaker gradient — that only shrinks the critical angle — but a gradient that varies along the road, which destroys the symmetry and with it the clean turning point. That is what the shimmer is: the invariant failing to be invariant, at the scale of a turbulent eddy.

Beside the atmosphere, and beside the planet #

The road’s gradient is enormous by atmospheric standards, and the comparison worth making is not with zero but with two other numbers.

The Earth’s curvature is the number to compare against because a ray that bends at exactly $1/R_\oplus$ per metre follows the ground round. The free atmosphere reaches about a sixth of that, so a horizontal ray falls away from the surface more slowly than the surface falls away from it — and the visible horizon is further off than geometry says, by about 8%. Surveyors carry the correction as an “effective Earth radius” of about 1.2 times the real one, and radio engineers carry a larger one — four thirds — because water vapour contributes to the radio index and not to the optical one, so radio rays bend more than light does through the same air.

The correction is not a small technicality in two places where it is routinely met. The sun is visible when it is geometrically below the horizon, by about half a degree at sunrise — which is very nearly its own angular diameter, so the whole disc that appears at the moment of sunrise is an image of a sun that has not risen. And the disc is visibly flattened at that moment, because the ray from its lower limb passes through more of the gradient than the ray from its upper limb and is bent more; the vertical diameter shrinks by about a fifth while the horizontal one does not change at all.

Occasionally the gradient near the ground goes the other way and gets large: cold air trapped beneath warm, over ice or a cold sea. Then $n$ falls with height, the ray curves downward, and if the gradient beats $1/R_\oplus$ the ray follows the curvature of the planet and objects below the horizon become visible. That is a superior mirage, and the extreme form — a ray trapped in a layer, bouncing between two turning points — is the same confinement that an optical fibre achieves with a step in the index instead of a gradient.

The same mathematics turns up where no medium exists at all. Light passing a mass is deflected, and one way to compute the deflection is to give the vacuum an effective refractive index that rises toward the mass — at which point the ray equation used above applies unchanged and the bending is a gradient-index problem. The factor of two separating the Newtonian answer from the correct one is, in that language, a statement about what the effective index has to be.

Grading across the beam instead of along it #

Stratify the index across a beam rather than along its path and something cleaner happens. Take a rod whose index falls parabolically from the axis,

$$n^2(r) = n_0^2\left(1 - A r^2\right),$$

and the paraxial ray equation becomes $d^2r/dz^2 = -A r$ — the harmonic oscillator, exactly. Every ray is therefore a sinusoid of the same spatial period $2\pi/\sqrt{A}$, whatever height it entered at.

A rod cut at a quarter of the period images; cut at half a period it returns the beam parallel again, inverted. This is how the lens in a photocopier bar, a borescope relay and a fibre coupler are built, and the specification is a length — a “quarter-pitch rod” — where a conventional lens is specified by curvatures.

What a graded rod is competing with is a lens, and the division of labour is exactly opposite. A conventional lens does all its bending at two surfaces and nothing in between; the graded rod does all of it in between and nothing at the surfaces. The two produce the same image and fail differently — a lens’s surfaces introduce aberrations that depend on where a ray strikes them, while a graded rod’s departures come from its profile not being exactly parabolic far from the axis.

The eye’s lens is graded too, from about 1.406 at its core to 1.386 at its edge, and the gradient contributes a substantial part of its power. A lens of uniform index with the same shape would be noticeably weaker, and would suffer more of the aberration a single curvature cannot avoid.

A ray and an orbit, written the same way #

The invariant $n\cos\theta = C$ is worth pushing a little further, because it turns the ray problem into one already solved elsewhere.

Write the ray as $z(x)$ in a stratified medium. The invariant gives $\cos\theta = C/n(z)$ directly, and since $\tan\theta = dz/dx$,

$$\frac{dz}{dx} = \pm\sqrt{\frac{n^2(z)}{C^2} - 1}.$$

Compare that with a particle of energy $E$ in a potential $V(z)$, whose speed is $\sqrt{2(E - V)/m}$: the same structure, with $n^2(z)/2$ playing the part of $-V(z)$ and $C^2/2$ the part of $-E$. Rays bend toward high index the way particles accelerate toward low potential, and the height at which a ray turns is the height at which a particle would run out of kinetic energy.

A particle in a potential well turns where its energy meets the potential, and the mirage is that picture laid on its side: the index profile is the well, the ray invariant is the energy, and the turning point is where the ray goes horizontal. What the analogy delivers is that everything known about turning points transfers — including that a wave, unlike a particle, does not stop exactly there but leaks a little way past, which is why a mirage’s boundary is soft rather than sharp.

The correspondence is not a decoration. It was the route by which optics and mechanics were seen to be the same subject: Hamilton wrote his mechanics in 1834 by taking the formalism he had built for rays and applying it to particles, and the identification of Fermat’s principle with the principle of least action is the same identification that makes a wave attached to a particle an idea somebody could have.

The ocean has a channel in it #

Air is the familiar graded medium and it is not the most consequential one. Sound in seawater travels through a gradient that runs both ways, and the result is a waveguide a thousand kilometres long.

The speed of sound in the sea rises with pressure and rises with temperature. Going down from the surface, temperature falls and pressure rises, so the two work against each other: the temperature term dominates near the top and the pressure term dominates lower down, and somewhere around a kilometre deep the speed passes through a minimum.

Everything on this page then applies. A ray that strays upward from that depth enters faster water and is refracted back down; a ray that strays downward enters faster water and is refracted back up. The minimum is a trough in the invariant’s landscape, and rays launched near it oscillate about it instead of leaving — turning points above and below, and no boundary anywhere.

The consequence is that sound put into that layer spreads in two dimensions rather than three, so its intensity falls as $1/r$ rather than $1/r^2$, and small explosions have been detected across an entire ocean basin. Ewing and Worzel identified the channel during the Second World War as a way for a downed airman to be located from a single small charge, and it is what carries the calls of fin and blue whales over distances nothing else in the sea can manage.

Where the model stops #

The ray picture assumes the index varies slowly on the scale of a wavelength. Everything above is geometrical optics. Where $n$ changes appreciably within a wavelength the rays stop being the right description and the wave equation has to be solved — which is exactly the situation at an ordinary boundary, and is why a step index has to be handled by matching fields across it rather than by tracing curves.

Near a turning point the ray approximation fails on its own terms. The invariant makes $\cos\theta \to 1$ and the ray horizontal, and the wave solution there is not a smooth turn but an Airy function, with an evanescent tail reaching below the turning height. The tail is microscopic for light in air. It is not microscopic for radio waves in the ionosphere, where the same equations govern and the turning point is kilometres thick.

The profile is a model. The exponential used here is a convenient smooth fit to a real thermal boundary layer that is turbulent, unsteady and not horizontally uniform. The critical angle it gives is right to a factor of order one, and the shimmer is what its unsteadiness looks like.

What the pictures cannot show #

The mirage figure exaggerates heights against distances by more than a hundred to one. At true scale every ray in it would be indistinguishable from the horizontal axis, and the figure would be a horizontal line — which is the honest picture and conveys nothing. The caption states the factor for that reason.

None of the figures shows what an observer sees. A ray diagram shows paths; the image is what the brain constructs by projecting arriving rays backwards in straight lines, and it appears below the road because the arriving ray is travelling upward. Nothing in the tracing knows about that step.

And the profile figure plots a gradient, which is a derivative, on a logarithmic axis — so the sign is lost. The road’s gradient is positive and the atmosphere’s negative, and the whole difference between an inferior and a superior mirage is that sign. The caption carries it because the axis cannot.

Where the ladder goes next #

The rung below asked what makes a ray take the path it takes; this one asked what happens when the medium refuses to be uniform. What appeared was a conserved quantity, and with it the vocabulary of orbits: turning points, forbidden regions, trapped rays.

The obvious next rung is trapping — a ray confined between two turning points, which is a waveguide, and the discovery that only certain paths survive many round trips because the others interfere with themselves. That is where geometrical optics hands back to the counting argument that produces modes, and where the ray picture earns its keep by predicting how many of them there are.

The habit worth taking: when a law is stated about a boundary, ask what it becomes when the boundary is smeared out. Often the answer is a differential equation with a conservation law attached, and the conservation law is more useful than the original rule.