where n is the relative index of refraction (of water, in this case). This relative index is defined as
Since the speed of light in air is almost that “in vacuo,” we will refer to n for simplicity as the refractive index; its generic value for water is n ≈ 4/3, but it does depend slightly on wavelength (this is the phenomenon of dispersion, and without it we would only have bright “white bows”!)
How does the deviation angle D vary with the angle of incidence i? Note that for rays that enter the sphere along the axis of symmetry and are reflected from the back surface, D(0) = π. The classic “rainbow problem” is to find whether there are maxima or minima in this deviation angle as a function of the angle i. On differentiating equation (A11.1) with respect to i, and using equation (A11.2), it is found that D′(i) = 0 when
for n = 4/3, on converting to degrees of arc. This means that D is stationary at i = ic. Physically, this corresponds to a concentration of deviated rays in a small angular region about
It can be shown that D″(ic) > 0, so that the deflection D(ic) is a minimum (in fact it is a global minimum in (0, π/2) because D″(ic) > 0). This minimum angle of deviation is often referred to as the rainbow angle. Its supplement 180° − D(ic) ≈ 42° is the semi-angle of the rainbow “cone” formed with apex at the observer’s eye, the axis being the line joining the eye to the shadow of her head (the antisolar point).
A comment about the color dispersion in a rainbow is in order. While it is not standardized to the satisfaction of everyone, the visible part of the electromagnetic spectrum extends from the “red” end (700–647 nm, or 0.700μ–0.647μ, etc.) to the “violet” end (424–400 nm), a nanometer (nm) being 10−9 m. For red light of wavelength λ ≈ 656 nm, the refractive index n ≈ 1.3318, whereas for violet light of wavelength λ ≈ 405 nm, the refractive index n ≈ 1.3435, a slight but very significant difference! All that has to be done is the calculation of ic and D(ic) for these two extremes of the visible spectrum, and the difference computed. Then voilà! We have the angular width of the primary rainbow. In fact since D(ic) ≈ 137.8° and 139.4° for the red and violet ends, respectively, the angular width is ΔD ≈ 1.6°, or about three full moon angular widths.
There is also a secondary rainbow that is frequently visible along with the primary bow as a result of an extra reflection inside the raindrops. As a result it is the fainter of the two, slightly wider, and about 9 degrees “higher” in the sky than the primary bow.
What about the common halo mentioned above? Have you ever noticed a circular ring around the sun (or, for that matter, the moon) when the sky is clear except for wispy thin cirrus clouds in the vicinity of the sun? I read on a now-defunct website that such beautiful displays, known as ice crystal halos, can be seen on average twice a week in Europe and parts of the United States, and certainly my own experience is not terribly different, though I would estimate that I notice them about three times a month on average. The “radius” of a ring around the sun or moon is naturally expressed in terms of degrees of arc, subtended at the observer’s eye by the apparent radius, just as for the rainbow. The most frequently visible one is the 22° circular halo, followed by parhelia (or sundogs, colored “splotches” of light on one side or other of the sun, and commonly both). In my own rather limited experience, the sundogs are as common as the halo if not more so, at least as I walk to work in southeastern Virginia. These are found at the same altitude as the sun, close to (but just beyond) the 22° halo. A convenient and literal “rule of thumb” is that the outstretched hand at arm’s length subtends about the same angle, so that if one’s thumb covers the sun, one’s little finger extends to about the 22° halo (or the sundog). There are many other types of ice crystal displays; for more information consult the encyclopedic and highly regarded website by atmospheric optics expert Les Cowley [46].
Halos are formed when sunlight is refracted, reflected, or both from ice crystals in the upper atmosphere and enters the eye of a careful observer: be careful, because even the common types are easily missed (but never look directly at the sun, of course). As already noted, they are often produced when a thin uniform layer of cirrus or cirrostratus cloud covers large portions of the sky, especially in the vicinity of the sun. Surprisingly, perhaps, they may occur at any time of the year, even during high summer, because above an altitude of about 10 km it is always cold enough for ice crystals to form. In particularly cold climes, of course, such crystals can form at ground level (though we are not thinking here of snow crystals). Such halos can arise from diamond dust, which is essentially ground-level cloud composed of tiny ice crystals. It can form wherever the temperature is well below freezing. Of course, some types of halo (such as a circumhorizontal arc) are very latitude-dependent, and may therefore be seen only rarely in higher latitudes. Very many of the crystals producing halos are hexagonal prisms; some are thin flat plates while others are long columns, and sometimes the latter have bullet-like or pencil-like ends. A significant feature of all these crystals is that while any given type may have a range of sizes, the angles between the faces are the same. Although they do not possess perfect hexagonal symmetry, of course, they are sufficiently close to this that simple geometry based on such idealized forms suffices to describe the many different arcs and halos that are seen. The halos that result from cirrus cloud crystals depend on two major factors: their shape and their orientation as they fall. Their shape is determined to a great extent by their history, that is, the temperature of the regions through which they drift as they are drawn down by gravity and buffeted around by winds and convection currents.
In order to explain the reasons for the various angles, for example, 22°, it is necessary to examine the crystal geometry in more detail. It is clear from Figure A11.2 that hexagonal ice crystals can be thought of as presenting “prism angles” of 60°, 90°, or 120° to rays entering them in the planes indicated, depending on the orientation of the crystal. There are both similarities and differences with the formation of rainbows in raindrops, and noting that a sphere is (infinitely) more symmetrical than a regular hexagonal plate, a result from optics that is extremely important for such halos is that the deviation angle for light refracted through a prism is a minimum for symmetric ray paths.
Figure A11.2. Ray paths through hexagonal ice crystal prisms for both the 22° and the rarer 46° halos.
Recall that a rainbow arises when the angle through which a ray is deviated on passing through the raindrop is an extremum. The above result tells us that in the case of prisms, the corresponding extremum—also a minimum, in fact—occurs when the ray path through the crystal is symmetric. A second result from elementary optics defines the magnitude of this minimum deviation angle (Dm)—the location of the halo relative to the line joining the sun and the observer—in terms of the refractive index (of ice here) and the apex angle of the prism.
We can use these results to explain the occurrence of the 22° and the less common 46° halos. As shown in Figure A11.2 (b,c), there are three prism angles in a hexagonal ice crystal prism: 60° (light entering side 1 and exiting side 3); 90° (light entering a top or bottom face and exiting through a side) and 120° (light entering side 1 and being totally internally reflected by side 2). The first two of these create color by dispersion of sunlight; the last contributes nothing directly to a halo, at least of interest to us here. The refractive index of ice for yellow light is n ≈ 1.31 For an apex angle of 60°, Dm ≈ 22°, and for an apex angle of 90°, Dm ≈ 46°. As in the case of the rainbow, all possible deviations are present in reality, but it is the “clustering” of deviated rays near the minimum that provides the observed intensity in the halos (but unlike the case of the rainbow, no reflection contributes to their formation in these two cases).
Appendix 12
THE EARTH AS VACUUM CLEANER?
How effective is the Earth at clearing a path through “space matter” in its vicinity? This is of interest, of course, with regard to possible close encounters with asteroids, as discussed in
Chapter 24. In order to get a handle on this problem, one major task is to determine how effective the Earth is at capturing “errant” asteroids. In order to do so, we will use some elementary physical principles and properties of conic sections.
Because of the gravitational attraction of the Earth (or any other sufficiently massive body), it can in principle “pull in” objects that are not traveling directly toward a head-on collision. This results in a capture cross section (CCS) that is at least as large as its geometric cross section (GCS). It is clear that this will depend (in particular) on the speed of the object relative to the Earth; something “whizzing by” our planet at high speed is less likely to be captured than a much slower object on the same path.
We can reduce this to a simple “two-body” problem by ignoring, for now, the rest of universe (as I so often do). The approach here is based on the article by Tatum (1997) [47]. Essentially, we consider the trajectory of the asteroid to be such that the gravitational attraction of the Earth is the dominant mechanism in this encounter. This is reasonable since the force of attraction of the Earth is ten times that of the sun at a distance of about 52,000 miles, or 13 Earth radii from its center, and about 1700 times that of the moon at this same distance. Under this assumption we can regard the trajectory of the asteroid as a hyperbolic one about a stationary Earth. Of course, both objects are in elliptical orbits around the sun, but in this “geocentric model” a hyperbolic orbit around the Earth is perfectly adequate for a “back of the envelope” calculation such as this.
The notation is as follows (see Figure A12.1). The impact parameter B (no pun intended) is the closest distance of approach to Earth’s center that the asteroid would have it were on a straight line path, that is, one unaffected by Earth’s gravitational attraction (or more accurately, by their mutual gravitational attraction). Its initial speed (at “infinity”) is v0 and the equation of its hyperbolic path is
a being the x-intercept and b the semi-transverse axis of the conjugate hyperbola.
Exercise: Using Figure A12.1, show that the impact parameter
The closest distance of approach on the hyperbolic orbit is called the perigee distance p, and if R is the radius of the Earth, the asteroid will collide with the Earth if p < R (the case p = R corresponds to grazing incidence). The GCS of the Earth is πR2 and the potential CCS is defined to be πB2. How do these two areas compare as a function of the initial speed v0?
First we will need to calculate the potential energy Φ() of the asteroid under the action of the Earth’s gravitational field, where is the position vector of the asteroid relative to the Earth. The gravitational force , described by the famous inverse-square law, is directed along the radius vector joining the two bodies, and for a spherically symmetric force Φ is given by
Figure A12.1. The asteroid’s hyperbolic orbit drawn for “grazing incidence” with the Earth. The impact parameter B is the perpendicular distance of the asymptote from the Earth”s center of attraction at C. The radius of the Earth is R and the speed of the asteroid “at infinity” is v0.
where ma is the mass of the asteroid, Me is the mass of the Earth, and G is the gravitational constant. At the perigee distance we suppose the asteroid speed relative to the Earth to be vp, and we apply the principle of the conservation of energy to the asteroid, namely, that the
that is,
or, simplifying,
Another fundamental principle is the conservation of angular momentum,
that is,
Elimination of vp from these two equations results in
In this equation ve = (2GMe/R)1/2 is the escape velocity from the surface of the Earth. To see this we equate the “escape” kinetic energy with the potential energy of a particle of unit mass at infinity. Thus
from which the result follows. Incidentally, ve ≈ 11.2 km/s. Equation (A12.5) gives that the ratio of capture-to-geometric cross sections has a lower bound of one, as would be expected. It differs significantly from this value only when v0 < ve; thus for v0 = 0.7ve, B2/R2 ≈ 3.
As pointed out by Tatum [47], there is a class of asteroids that approach our planet quite closely, and with small relative speeds, so the CCS/GCS ratio for these may be quite large. One such object is 1991 VG, which at that time was, at eight meters, the smallest astronomical object ever discovered (in orbit). The semi-major axis of its orbit is 1.04 astronomical units (≈ 1.50 × 108 km; the mean Earth-sun distance is approximately 1 A.U.). The eccentricity of its orbit is 0.067 (Earth’s is 0.0167), which brings it inside the Earth’s orbit for part of the time. With this in mind, let the Earth move again (Oh, the power of mathematics!), in a circular orbit for simplicity, and imagine such an asteroid in a nearby circular orbit. Obviously we expect that these orbits will not be concentric in practice, but they may be close enough for a significant portion of the orbit to make this simplification useful. If the distance apart of the two orbits is less than the radius of the CCS, then the object can be captured. The closer the orbit is to that of the Earth, the smaller the relative speed and the larger the CCS becomes.
Suppose that the speed of the Earth in its orbit is V with orbit radius a, and the corresponding speed and orbit radius for the asteroid are V + δV and a + δa, respectively. Now Kepler’s third law of planetary motion states that the square of the orbital period P of a planet is directly proportional to the cube of the semi-major axis of its orbit (here a), or in terms of a constant of proportionality K, P2 = Ka3. For a circular orbit, P = 2πa/Ve so that on eliminating P, Kepler’s law reduces to
Using differentials it follows that the relative difference in speed between the two objects is
From the earlier discussion we can identify δV, the relative difference in speeds, as v0 and the difference in orbital radii, δa, as the impact parameter b. Using equation (A12.7) in (A12.5) we obtain (in the current notation)
The positive real root of this biquadratic equation is
In performing these calculations I found it simplest to write the term
since a/V = T/2π, where T is the period of the Earth’s orbit in seconds. We have already calculated the quantity 2GM/R from equation (A12.6). The result is δa ≈ 8.45 × 105 km (δa ≈ 5.63 × 10−3 A.U.). In terms of the Earth’s equatorial radius R ≈ 6.38 × 103 km, this is about 132 Earth radii, or approximately 2.2 times the Earth-moon semi-major axis.
From equation (A12.7) we find that the corresponding difference in orbital speed |δV| ≈ 0.085 km/s.
Exercise: Show from equations (A12.2) and (A12.5) that, for a grazing collision,
This shows that the perigee speed is quite sensitive to the impact parameter, varying as its square. Using this result we calculate the impact speed at a grazing collision to be ≈11.1 km/s, just a tiny bit less than the escape speed from the Earth. Since we have shown that the Earth sweeps out a toroidal volume of radius approximately 0.056 A.U. in its path around the sun, it can be thought of as a giant vacuum cleaner with CCS (132)2 ≈ 1.74 × 104 times that of the Earth’s GCS. Therefore any object moving in a circular orbit with radius between 0.9944 A.U. and 1.0056 A.U. will collide with (be captured by) the Earth. That’s some vacuum cleaner!
ANNOTATED REFERENCES AND NOTES
NOTES
[1] Batty, M. and Longley, P. (1994). Fractal Cities. Academic Press, San Diego, CA.
[2] Polya, G. (1954). Mathematics and Plausible Reasoning. I. Induction and Analogy in Mathematics. Princeton University Press, Princeton, NJ.
[3] Hern, W.M. (2008). “Urban malignancy: similarity in the fractal dimensions of urban morphology and malignant neoplasms.” International Journal of Anthropology 23, 1–19.
[4] See Adam (2006) for a slightly different version of the Princess Dido story.
[5] Barshinger, R. (1992). “How not to land at Lake Tahoe!” American Mathematical Monthly 99, 453–455.
[6] Taxi rides, squirrels, and light bulbs: I am grateful to my friend, colleague, and co-author Larry Weinstein, an NYC boy, for his insights into city life; I have drawn on hi
s expertise for these questions.
[7] Weinstein, L. and Adam, J.A. (2008). Guesstimation: Solving the World’s Problems on the Back of a Cocktail Napkin. Princeton University Press, Princeton, NJ.
[8] Eastaway, R. and Wyndham, J. (1999). Why do Buses Come in Threes? The Hidden Mathematics of Everyday Life. Wiley, London.
[9] Skyscrapers and harmonic motion: see http://www.cpo.com/pdf/tpst_pfc%20ch19%20connections.pdf.
[10] Burghes, D., Galbraith, P., Price, N. and Sherlock, A. (1996). Mathematical Modelling. Prentice-Hall, Hempstead, Hertfordshire.
[11] Barnes, G. (1990). “Food, eating, and mathematical scaling.” Physics Teacher 28, 614–615.
[12] Burton, R.F. (1998). Biology by Numbers. Cambridge University Press, Cambridge.
[13] Edwards, D. and Hamson, M. (1989). Guide to Mathematical Modelling. CRC Press, Boca Raton, FL.
[14] Lowson, M.V. (2004). “Idealised models for public transport systems.” International Journal of Transport Management 2, 135–147.
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