X and the City: Modeling Aspects of Urban Life

by Adam, John A.

  m being the electron mass, a being its acceleration, and A being a constant of proportionality. This is recognizable as an inhomogeneous second-order differential equation with constant coefficients. It is in fact the equation of forced simple harmonic motion.

  Exercise: Show that the solution s(t) to equation (20.2), satisfying the simplest initial conditions s(0) = 0 and s′(0) = 0, is given by

  provided that A ≠ mω2. In the event that these quantities are equal—a case known as resonance, which will not be pursued here—the solution can be found directly from the original differential equation or by applying L’Hôpital’s rule to the solution (20.3).

  Exercise: Show that the solution for s(t) when A = mω2 is given by

  Returning to equation (20.3), we note that, because of the “tight binding” of electrons in most aerosols, the term so the second term usually can be neglected. Then the electron acceleration can be approximated by the expression

  The amplitude of this acceleration is proportional to ω2E0. Recall that the electric field (and hence acceleration of the electron) is perpendicular to the direction of the incident wave. Reversing this, we anticipate that the electric field of the scattered wave will be proportional to the perpendicular component of the acceleration at large distances from the particle, that is, to a sin θz, where θz is the angle between the direction of the scattered light and the z-axis (see Figure 20.4), which is the direction of the incident electric field. We also expect that the total electric field is proportional to the number of electrons present in the particle, and therefore to its volume V. The intensity Is of this scattered light is proportional to the square of the electric field, and allowing for the usual inverse square fall-off with distance r, consistent with energy conservation, we arrive at the proportionality relation

  where . The scattered wave makes an angle θx = π/2 − θz with the forward direction (x-axis). We have therefore obtained a fundamental result for Rayleigh scattering, namely that the intensity of scattered light is inversely proportional to the fourth power of the wavelength. More accurately, the probability that a photon of sunlight will be scattered from its original direction by an “air molecule” is ∝ λ−4. In any case, a higher proportion of blue light (λ ≈ 0.4μ) than red light (λ ≈ 0.7μ) is scattered, in fact about (7/4)4 ≈ 9 times more.

  Figure 20.4. Light is scattered from a particle at the origin in a direction that makes an angle θz with respect to the z-axis. The observation distance is r.

  There are some other things to note from this formula. The scattered radiation is proportional to V2; this results of course in a reduction of intensity in the forward direction. Note that the dependence on V2 corresponds to one on the sixth power of radius R. This dependence on V can be contrasted with intensity reduction arising from absorption within the particle. The latter will be proportional to the number of absorbing molecules present, and therefore to V. Therefore, in relative terms the importance of scattering (as opposed to absorption) for intensity reduction in the forward direction will be greater for larger particles (still satisfying the requirement that R/λ 1). Also it is clear that the intensity of scattered light depends on the angle θz, but remember, this is for the special case of polarized light. For scattering in the x-y plane, there is no angular dependence because of symmetry about the z-axis, and so for this case.

  Sunlight is unpolarized [35] because the electric field in light from the sun vibrates in all possible directions. We can use the existing result (20.6) and generalize it to this case as follows. We define the scattering plane to be the plane containing both the forward and scattered directions. An unpolarized incident wave can be written as the arithmetic mean of two independent linearly polarized components, one parallel to and the other perpendicular to the scattering plane. Essentially, to do this we average the scattered intensity by replacing the term sin2θz by (sin2θz + sin2θy)/2, where θy is the angle the direction of observation makes with the y-axis. By including the corresponding angle with the x-axis, we note that the squares of the direction cosines sum to one, that is,

  This is equivalent to

  Figure 20.5. Angular variation of the polarization intensity components for the scattered light. The central curve, I, represents the component as given by equation (20.6); curve II is for the component as given by equation (20.7); and curve III is for the total intensity polarization, , as given by equation (20.8). The forward direction is θx = 0°.

  Therefore in total

  for unpolarized sunlight. This total scattering intensity is just the sum of the two polarization intensities. This is illustrated in Figure 20.5. The perpendicular component of polarization () is the central figure, the middle figure is the parallel component (), independent of angle because this is scattering in the x-y plane, perpendicular to the electric field vector), and the outer figure () is the sum of these two. Therefore the scattered light is most polarized (and least bright) perpendicular to the incoming light, and least polarized (and most bright) in the forward and backward directions.

  The angular dependence of the polarization for unpolarized incident light is conveniently expressed in terms of the quotient

  The graph of P is shown in Figure 20.6 (in Cartesian form with θx in radians), illustrating afresh that (for ideal molecular scatterers) the light is 100% polarized at 90° to the incoming radiation and completely unpolarized in the forward (0°) and backward (180°) directions. (In practice this will not be the case because molecules are not perfect scatterers [36].)

  Figure 20.6. Degree of polarization for unpolarized incident light as given by equation (20.9).

  We have stated that the results above for Rayleigh scattering apply when the particle size is much less than the wavelength of incident light, that is, R/λ 1. But what does the phrase “much less” really mean? There is no universal definition, and as usual, it depends on the context. Several books and articles on the topic of scattering use the inequality R < λ/20, or even R < λ/10, and we shall be content with this. Since the wavelength of visible light lies in the range 400–800 nm (0.4μ–0.8μ), this means that R < 0.04μ should suffice.

  For particles larger than this the processes of reflection, refraction, and diffraction cannot be neglected. It turns out that most of the light is scattered near the forward direction, but for particles whose size is comparable with the wavelengths of light, the situation is quite complex (but very interesting!) both mathematically and physically. The exact solution to this problem for spherical particles is often referred to as the Mie solution, from a 1908 paper by Gustav Mie; Debye also solved the problem independently in 1909, but the solution had already been published in 1863 by Clebsch, and again independently by Lorenz in 1890 and 1898. The rediscovery of existing results is a rather common theme in the history of science and mathematics.

  The total amount of light scattered by a spherical particle is quite sensitively dependent on R/λ as already noted. This dependence can be characterized by a quantity Cs, the scattering area coefficient. It is defined to be the amount of light that would be intercepted by a particle (not necessarily a sphere) of cross-sectional area CsπR2. If, for example, the particles are very small, that is, R/λ 1, the dominating mechanism is Rayleigh scattering, with its dependence on R6 (or V2, see equation (20.8)), then Cs will increase as (R/λ)4 initially. For scattering by sufficiently large particles, Cs approaches two. This is the source of a fascinating paradox, the extinction paradox: why is Cs not equal to one? The debate involves some very sophisticated mathematical physics that will not be discussed here except to say that it involves interactions between the external and internal radiation fields of the particle. However, it should be pointed out that the paradox has not been entirely resolved if recent literature on the topic is anything to go by!

  Figure 20.7 shows a graph of Cs vs. 2πR/λ (the ratio of circumference to wavelength) for a spherical particle with the same density as water. As noted above, Cs rises toward its first (and largest) maximum, it does so initially
as the power function (2πR/λ)4. This rapid increase in scattering “efficiency” reaches a maximum when R/λ ≈ 6.3/2π ≈ 1. Recall that the Rayleigh scattering formula was derived assuming R/λ is small. For large enough particles the electric field within it is no longer approximately uniform, and this detracts from the cumulative effect for smaller ones, leading to a dramatic decline in Cs. The subsequent oscillatory behavior is a consequence of the particle size periodically meeting the requirements for destructive and constructive interference of the electric field within it.

  Figure 20.7. Dependence of the coefficient Cs on the ratio 2πR/λ for a spherical particle with refractive index equal to that of water, that is, ≈ 4/3. (The incident light is assumed to be monochromatic for simplicity.) Redrawn from Williamson (1973).

  Consider the graph on the left side of the first maximum, say 0 < 2πR/λ < 4π ≈ 12.6, and suppose that R is fixed. Then for longer wavelengths, corresponding to orange or red light, the scattering (as measured by Cs) is less than for shorter wavelengths, such as in blue or violet light, for that size particle (because Cs increases as R/λ increases). Now we let R be a somewhat larger (but still fixed) value corresponding to the decreasing part of the graph after the first maximum. This time the scattering efficiency is reversed: Cs decreases as λ decreases. It should be apparent that, given a suitable size distribution of particles—such as those from forest fires or volcanic eruptions—the sun or moon may appear strangely colored—blue moons and even green ones have been reported [36].

  While the amount of light scattered by a particle is of theoretical interest, the amount scattered by a given mass concentration of particles is of more direct interest in gauging the effects of air pollution. For simplicity, we consider the particles to be of uniform size and density ρ. Then the mass of each particle is m = 4πρR/3. If M is the particle mass concentration (the mass of pollutant per unit volume, for example, c mg/m3), then a cylinder of cross-sectional area 1 m2 will contain M/m = 3M/4πρR3 particles for every meter of its length. Multiplying this number by the effective particle cross-sectional area CsπR2 gives a relative measure of the amount f of the incident beam that is scattered, that is,

  for a given wavelength of light. Figure 20.8 shows a graph of this ratio denoted by ; valid for any prescribed wavelength, it shows how scattering of light for a given mass concentration is most efficient when the wavelength of light is close to the particle size, that is, λ ≈ R.

  Figure 20.8. Dependence of the relative scattering coefficient on the ratio 2πR/λ for a spherical particle with refractive index ≈4/3. Redrawn from Williamson (1973).

  X = xB: RAINBOWS AND “ROADBOWS” IN THE CITY

  We are familiar with rainbows and, perhaps to a lesser extent, with atmospheric halos formed during the day when (very nearly) parallel light from the sun is “scattered” into the observer’s eye by raindrops and ice crystals, respectively. The drops or crystals producing the phenomena are located on the surface of an apparent cone; in the case of the rainbow the axis extends from the observer’s eye toward the so-called anti-solar point (i.e., extending to the shadow of the observer’s head and beyond) as shown in Figure 20.9. In the case of, say, the common 22° ice crystal halo (Figure 20.10), the axis of the cone extends in a direction from the observer’s eye to the sun.

  The reader is encouraged to consult Appendix 11 if he or she needs to brush up on the basic mathematics behind the formation of these beautiful images.

  But have you ever noticed a rainbow-like reflection from a road sign when you walk or drive by it during the day? Tiny, highly reflective spheres are used in road signs, sometimes mixed in paint, or sometimes sprayed on the sign. And sometimes, after a new sign has been erected, quantities of such “microspheres” can be found on the road [37] near the sign. It is possible to get samples of these tiny spheres directly from the manufacturers, and reproduce some of the reflective phenomena associated with them. In particular, for glass spheres with refractive index n ≈ 1.51 scattered uniformly over a dark matte plane surface, a small bright penlight provides the opportunity to see a beautiful near-circular bow with angular radius about 22°! (Don’t confuse this with the 22° ice crystal halo; again, see Appendix 11.) This bow appears to be suspended above the plane as a result of the stereoscopic effects we now proceed to examine.

  Figure 20.9. Formation of a (primary) rainbow.

  Figure 20.10. Formation of the common 22° ice crystal halo.

  Figure 20.11. Geometry for the “microsphere” bow.

  In Figure 20.11, representing the x-z plane, the (point) source of light is at the point S(0, 0, h) and the observer’s right (R) and left (L) eyes, aligned along the x-axis, are located at (±D, 0, H), respectively, where H > h. The intersection of the bow with this plane is at the point B(xB, 0, zB). The ray path SxRR corresponds to a deviation of δ = 180° − (θ1 − θ2) from the original direction. It is an interesting exercise to show that for a sphere of refractive index n this deviation is

  For n ≈ 1.51. δ ≈ 2.76 radians, or approximately 158°.

  Exercise: Use equations (A11.3) and (A.11.4) in Appendix 11 to establish the result in equation (20.11). (The angle δ here is D(ic) in equation (A11.4).)

  We shall work with the angle ψ = θ1 − θ2 ≈ 22° in what follows. From Figure 20.8 we note that

  Similarly, for the left eye,

  After some rearrangement it follows that

  The resulting quadratic equation for the location of xR is

  All the coefficients in this equation are known, or in principle measurable, so an explicit analytic expression for xR can be obtained. However, care is required in choosing the correct root, since xR must be unique; it is therefore probably better to solve (20.12) directly instead. By changing the sign of D in both these equations we can obtain those for xL also.

  Using similar triangles, points (x, z) on the line RxR satisfy the equation

  and, changing the sign of D for xL on the line LxL,

  Since these two lines intersect at the perceived bow, the point B, we can equate the two expressions and solve for x = xB. Thus we find

  and substituting this value into either of the equations (20.16) or (20.17), we obtain

  Note that, from equations (20.12) and (20.13), both xL and xR are functions of h, H, and D.

  Regarding the shape of the bow, for an observer directly over the light source it may appear to be exactly circular because of the bilateral symmetry of the human face. In the above scenario the eyes are separated by a distance 2D along the x-axis, but have no separation in the y-direction. Obviously, if D → 0 the shape will approach that of a circle, but of course the stereoscopic effect will also disappear (and we would lose an eye!). If D were to increase, the bow would become more ellipse-like. In practice, for most observers the departure from a circular shape will be small unless the eyes are not above the light source, in which case the bow has an ellipse-like shape, and appears tilted relative to the x-y plane. (See Crawford 1988.)

  * But note that, if the smoke is exhaled, it appears to be whiter because moisture from the air in the lungs has coated the smoke particles. This increases their effective size, and alters the wavelength dependence of the scattering. Light is now scattered more uniformly than before, and hence the smoke appears whiter.

  Chapter 21

  NIGHTTIME IN THE CITY—I

  What is the difference between night and day in the city? That may appear to be a silly question, but humor me here. During the day, the sun provides the light we need, outdoors at least. Artificial lighting may be necessary inside a building, depending on the number and location (or even existence) of windows. In the country at night, the only sources of light are the moon, planets, and stars, when they are visible, barring the occasional encounter with a UFO. By contrast, most of the heavenly bodies are not seen in the city (with the possible exception of the moon and a bright planet or star). The sources of exterior light—in particular, street lamps—are artificial and, compared with the sun, relativ
ely close by! This means that the light “rays” cannot be considered parallel, unlike those from the sun, and so the light emanating from them is divergent. This fact can give rise to some interesting modifications of certain daytime optical phenomena, as we shall see below. But let’s see what the night brings by concentrating first on some of the less exotic but more familiar “night light” phenomena.