Atmospheric Radiation. January 3, 2018

Transcription

1 Atmospheric Radiation January 3,

2 Contents 1 Electromagnetic waves 3 2 Light as quanta Atmospheric application: Oxygen dissociation in the stratosphere Flux and Intensity Flux Intensity Atmospheric application: Solar intensity Relationship between flux and intensity Blackbody radiation Atmospheric example: Earth s energy balance Local Thermodynamic Equilibrium Kirchoffs Law Why is the sky blue? 18 7 Why are clouds white? 19 8 The mathematics of scattering and absorption and emission Absorptance and transmittance Optical depth Transmittance in a plane parallel atmosphere Size parameter Single-scattering albedo Phase function Energy transitions of molecular absorption The simplest case: The Hydrogen atom Polyatomic molecules and rotational transitions Polyatomic molecules and vibrational transitions Vibrational-rotational spectra Molecular absorption profiles Line strengths Line profiles Line absorption Weak and strong lines Molecular absorption, the greenhouse effect, and climate change 39 2

3 1 Electromagnetic waves Figure 1.1: Electromagnetic wave The fundamental equations that describe electromagnetic radiation are Maxwell s equations. The solutions to the equations are sinusoidal of form ~E = ~E 0 exp i~k ~x iwt (1.1) ~H = ~H 0 exp i~k ~x iwt where E is the electric field and H is the magnetic field, the real component of~k is the wavenumber 2p/l,~x is the direction of wave propagation and w is the angular frequency of the radiation 2pn. Some fundamental properties of the waves are that they do not diverge, that H and E are normal to each other, and that radiation of one frequency or wavelength does not interact with radiation of another frequency. The last point is particularly useful because it means we can analyse the separate contributions of a range in frequency or wavelength to the total energy input from radiation as it accumulates over time. The symbols e and µ stand for the electric permittivity and magnetic permeability, respectively. They are merely properties of matter. Curiously, unlike say sound or water waves, E-M waves do not require a medium and can travel in a vacuum. The permittivity and permeability of a vacuum (1.2) 3

4 have the symbols e 0 and µ 0. The instantaneous flux density of the electric field in the direction of the wave is defined by the Poynting vector In a vacuum this can be simplified to ~S = ~E ~H (1.3) S = E2 0 µ 0 c cos2 (wt) Although averaged over time it is equal to the average Poynting Flux S av = 1 2 Re( ~E 0 ~H 0 ) or S av = 1 2 ce 0E 2 0 (1.4) S av has units of W m 2 (or J s 1 m 2 ) and is a function of wavelength. This is units of flux (quantity per length squared per second). In atmospheric sciences, we more commonly use the symbol F. The point here is that the flux of energy F is proportional to the square of the magnitude of an electromagnetic wave. It is the flux density of electromagnetic waves that is of primary interest to climate studies. Figure 1.2: Electromagnetic wave quantities Okay, but let s look at interactions of radiation with something that is not a vacuum. Remember from Eq. 1.1 ~E = ~E 0 exp i~k ~x iwt 4

5 Assume that the wavenumber vector~k is a complex number with a real and imaginary component, i.e. ~k = ~ k 0 + i~k (1.5) Therefore, the consequence of k being complex is the following ~E = E 0 exp k ~ ~x exp i ~k0 ~x wt (1.6) Note that if ~k = 0 then the amplitude of the wave is constant. Figure 1.3: Atmospheric radiation and transmission in the atmosphere as a function of wavelength Thus the phase of the wave is given by f = ~ k 0 ~x wt (1.7) At constant phase df = 0 = ~ k 0 d~x wdt (1.8) implying that the phase speed of radiation in the direction the waves propagate is given by v = d~x dt = w ~ k 0 (1.9) or alternatively v = nl = w 2p l (1.10) 5

6 The point here is that frequency and wavelength are linked through the phase speed of the wave. In a vacuum the phase speed is c = m/s, the speed of light, where c = nl = w 2p l = w k (1.11) It is common practice in the atmospheric sciences to divide n by the speed of light to obtain the wavenumber ñ = n c which is most commonly expressed in units of cm 1. The only reason for dividing by c is that we end up with numbers around 1000 that our poor little brains can handle more easily than So, for example, we can express visible light with a wavelength of l = 0.3 µm ( cm or m) as having a frequency of n = c/l = Hz or ñ = 1/l = cm 1. 2 Light as quanta A paradox of modern physics is that particles sometimes appear to be waves, and sometimes appear to be particles, depending on how you look at them. Sometimes we ll treat radiation as waves and sometimes as particles. As far as we know, neither is any more wrong or right. But sometimes one of the two is more useful. Figure 2.1: Electron two slit experiment, showing that photons exhibit a duality in their behaviour as both waves and particles. From Eq. 1.11, the frequency of light is related to its wavelength l through the speed of light c n = c l (2.1) 6

7 The speed of light in a vacuum is about ms 1, regardless of how fast you are moving relative to another frame of reference (this is the principle behind Einstein s special theory of relativity, published in 1905, which led to the famous E = mc 2 ). Another definition often used by atmospheric scientists is the wavenumber, usually described in terms of cm 1. The wavenumber is related to the frequency through a factor of c. ñ = 1 l (2.2) Strangely though, energy comes in discrete, although very small, packets or particles, whose energy is related to the frequency of a wave! E = hn (2.3) where h is Planck s constant, J s. While Planck was the first to postulate, as much as a mathematical trick as anything, the quantization of energy, Einstein was the first to make a compelling theoretical argument, also in 1905, in his explanation of what is known as the photoelectric effect. It won Einstein the Nobel prize, and is immortalized in this song. 2.1 Atmospheric application: Oxygen dissociation in the stratosphere In the formation of ozone, only radiation with wavelengths smaller than µm is capable of dissociating molecular oxygen into atomic oxygen, according to the reaction O 2 + photon! O + O Based on this information, how much energy is apparently required to break the molecular bond of a single molecule of O 2? E = hn = hc/l = / = J A very small number, but in the atmosphere these short wavelength photons are the only ones with sufficiently high energy to break molecular bonds and thereby play a role in chemical reactions. The number of photons raining down on us is huge. Assuming the flux density of sunlight reaching the surface on a cloudy day is 100 W m 2 with a mean wavelength of 0.5 µm the flux of photons is N = F hn = Fl hc = s 1 m 2 Huge! Since this huge number will wash out any sense of photons being discretized events, in atmospheric sciences we usually treat radiation as a wave, except in the case of chemical photolysis, or in certain computational approaches to radiative transfer. 3 Flux and Intensity 3.1 Flux Radiative flux has units of W m 2, and is perhaps more appropriately called the flux density, although atmospheric scientists never call it this. Radiative flux may be coming from one direction 7

8 (the sun), or on a cloudy day at the surface from all directions. The only restriction is that the flux refers to radiation incident on a surface normal to a hemisphere enclosing the surface. Electromagnetic waves typically represent a superposition of many many wavelengths l or frequencies n (remember the two are related by c), each with its own amplitude F 0. By superposition we mean that we can describe the electromagnetic wave as a time series F (t) (think of rapidly pulsed light striking you with varying intensity) A Fourier transform can convert this time signature to frequency space: F (n)= Z 0 F (t)e 2pint dt (3.1) This function contains all the original information present in the original time series but displays it in the frequency domain. The original time series can be restored exactly by taking the inverse Fourier transform F (t)= 1 Z F (n)e 2pint dn (3.2) 2p 0 We now define a spectral flux F n or F l with units W m 2 s or W m 2 µm 1. F(l,l + Dl) F l = lim Dl!0 Dl and F(n,n + Dn) F n = lim Dn!0 Dn The broadband flux is the integrated flux over a range of wavelengths (3.3) (3.4) F (l 1,l 2 )= Z l2 l 1 F l dl (3.5) So, as a function of time, a sensor receives an oscillating intensity associated with the sinusoidal nature of the electromagnetic waves, e.g. F (t). Applying a Fourier transform to this signal yields the intensity of the waves as a function of frequency F (w). In real life we sense this as an object looking red or green, even though our eyes are only receiving F (t). Our brain is converting a temporal signal to information in frequency space. Note that the broadband flux could also be calculated as F (l 1,l 2 )=F (n 1,n 2 )= Z n2 n 1 F n dn (3.6) provided that n i = c/l i. That is F l dl = F n dn = df even if F l 6= F n. The flux of energy F must be the same whether we express it within a band of n or l. However, the spectral flux F l and F n is dependent on the co-ordinate system considered. More on this in an assignment. 3.2 Intensity Intensity I (or commonly radiance) has units of W m 2 per steradian, where a steradian is an definition of a solid angle dw. An infinitesimal increment of solid angle is dw = sinqdqdf = A/r 2 (3.7) 8

9 Figure 3.1: Spectrum as a function of wavelength where in the atmosphere, the convention is that q is the zenith angle (0 if directly overhead), and f is the azimuthal angle (0 in the direction of an incoming beam). A hemisphere covers 2p steradians and a full sphere 4p steradians. A is the cross-sectional area normal to radius vector r. 3.3 Atmospheric application: Solar intensity The intensity of radiation coming from any direction is a function of angle W =(q,f) and is defined as I (W)= df (3.8) dw For example, the sun - earth distance is D s = km and the sun radius is R s = km. Compute the angular diameter subtended by the sun, its solid angle, and the intensity of solar radiation if S 0 = 1370 W m 2. The solution requires recognizing that dw = A/D 2 s = pr 2 s/d 2 s = sr Thus, I = df/dw = 1370/ = Wm 2 sr 1 9

10 δθ θ δφ Figure 3.2: Illustration of a solid angle 3.4 Relationship between flux and intensity From any given level in the atmosphere, we often talk about the upwelling and downwelling flux. This is because when studying radiative transfer in the atmosphere we are primarily concerned with the amount of energy entering or leaving a system through its upper and lower boundaries. The upwelling flux is simply the sum of the upward component of all intensities integrated over the upward hemisphere F " = Similarly for the downwelling flux F # = 4 Blackbody radiation Z 2p Z p/2 0 0 Z 2p Z p 0 p/2 The basic principles of thermal emission are as follows: I " (q,f)cosq sinqdqdf (3.9) I # (q,f)cosq sinqdqdf (3.10) An object of temperature T radiates energy at all wavelengths (or equivalently frequencies). The amount of energy radiated at a specific frequency follows a relation known as the Planck function. 10

11 δ ω R I r r F r F R R E r E R Figure 3.3: Relationships between intensity I, flux density F and flux E of isotropic radiation emitted from a spherical source with radius r, indicated by the blue shading, and incident upon a much larger sphere of radius R, concentric with the source. Thin arrows denote intensity and thick arrows denote flux density. Fluxes E R = E r and intensities I R = I r. Flux density F decreases with the square of the distance from the source. The wavelength of peak radiation is inversely proportional to the object temperature. Equivalently, the peak frequency is proportional to the temperature. The total amount of energy radiated, summed over all wavelengths, is proportional to the temperature to the fourth power F µ T 4 With certain caveats (described later) an object absorbs as effectively as it emits By definition, a blackbody absorbs all radiation incident upon it. An example is the sun. The Planck function, which describes the intensity of radiation emitted from a blackbody is (W m 2 sr 1 frequency 1 ) 2hn 3 B n (T )= c 2 e hn/kt (4.1) 1 which, in terms of wavelength is B l (T )= 2hc 2 l 5 e hc/klt 1 (4.2) 11

12 Figure 4.1: A blackbody defined by a cavity where emission and absorption are in equilibrium so as to maintain a constant temperature What is notable about this function is that it has a peak. We can derive the location of the peak (Wien s Law) by taking the first derivative of B l with respect to l and setting to zero to get l max = 2902/T (4.3) Note that this is the peak for B l not for B n. If we were to find the maximum of B n with respect to n and convert back to l the formula would be l max = 5107/T (4.4) Quite different! But why is there a peak? Where does this function come from? A first statement here is that the derivation of the Planck function is not straightforward, and was approached initially through combinations of classical, statistical and quantum mechanics. But it is an extremely important result, perhaps the first important result of the newly developed quantum theory. Its ingredients have much to say about other topics in atmospheric physics. Here we will hand-wave some of the more central concepts. The way this problem was first approached was to think of the radiating body as a bunch of individual oscillators. The total internal energy of the oscillators would be distributed among translational, vibrational, rotational, and electronic energy u tot = u trans + u rot + u vib + u elec (4.5) 12

13 B (MW m 2 µ m 1 sr 1 ) K 6000 K 5000 K (µ m) Figure 4.2: The blackbody spectrum. Note that colder temperatures correspond to lower energies of emission and longer wavelengths. Each of these types of energy could be further subdivided into distinct independent modes, each having total energy kt, where k is the Boltzmann constant ( J/K). Now all this jiggling associated with the temperature corresponds to displacements of electric charge. As discussed previously this generates an induced radiation field the dipole radiation whose magnitude varies as frequency n (or w) squared. Intuitively, the more jiggles that happen, the more energy that is lost and the cooler the object must get. Absent external inputs, a stove doesn t stay hot. In a so-called black-body, we don t let dipole radiation just radiate away into space. Rather it is trapped inside an enclosed black (non-reflecting) box. Thus, any radiation that is emitted and lost also acts as a source for creating more jiggles. Accordingly, we can imagine an idealized equilibrium state in which the amount of energy that enters the system (say through a small aperture in the box) is the same as the amount of energy that escapes it. What is this equilibrium state? Well first, we can think of this as being like any linear system in a condition of steady-state: the rate of change of a substance is proportional to coefficient a times the substance (the source) minus coefficient b times the substance (the sink), e.g., for temperature dt = at bt dt In equilibrium, a = b. The final step is to determine the absorption of incoming radiation, because the absorbed incoming radiation must be equivalent to the loss rate for there to be an equilibrium. 13

14 In our case, the energy sink is proportional to the amount of energy of the system (kt) and to the power of absorption and emission. Things get more complicated here but classical theory suggests that a = b µ w 2. Ultimately, the central feature is that the radiant intensity is a product of the thermal energy kt and its frequency squared w 2 µ n 2. This classical approach gives us the so-called Rayleigh-Jean s law B nrj = 8pn2 c 3 kt (4.6) which looks like the long tail of the Planck function, and where, of course w = 2pn. The problem here is that the Rayleigh-Jeans law just goes to infinity for high frequencies and short wavelengths. We don t get fried by X-rays, thankfully. Here s the safety catch: remember, energy is quantized, and the amount of energy in a oscillator mode kt can never be less than hn. When kt hn, the Rayleigh-Jeans assumption that total oscillator energy is kt is no longer correct. Now, a quantum description must take over. It accounts for the rapid dropoff in energy at short wavelengths, high frequencies in B. If energies are sufficiently high that hn kt, quantum mechanics must take over from the classical mechanics perspective because it is impossible that anything is smaller than hn. From this perspective, radiant energy decreases with increasing frequency. Combining the Rayleigh-Jeans and Quantum mechanical approaches, the desired expression for Black-body radiation (Eq. 4.1) is B n (T )= 2hn 3 c 2 (exp(hn/kt) 1) Which implies a peak in the spectrum as prescribed by Wien s law (Eq. 4.3). To get the total intensity averaged over all frequencies we integrate B n to get (W m 2 sr 1 ) where B(T )= Z 0 B n (T )dn = bt 4 b = 2p 4 k 4 / 15c 2 h 3 Since blackbody radiation is isotropic the blackbody flux is (W m 2 ) F BB (T )=pbt 4 = st 4 (4.7) where s = Jm 2 sec 1 deg 4 is the Stefan-Boltzman constant. Ultimately there are three methods for obtaining F µ T 4 : 1. Stefan showed that F = st 4 based on experimental measurements 2. Boltzmann showed that F µ T 4 based on thermodynamic arguments 3. Planck showed that F = st 4 by assuming that energy is quantized, obtaining the Planck function, and integrating this to obtain s, where s = pb is a mix of fundamental constants. 14

15 F F Figure 4.3: A planetary blackbody where emitted thermal radiation from the sphere is in balance with absorbed solar radiation. 4.1 Atmospheric example: Earth s energy balance Exercise 4.6 in Wallace and Hobbs Calculate the equivalanet blackbody temperature of the Earth, assuming a planetary albedo of Assume that the Earth is in radiative equilibrium. Solving for T E we get F E = st 4 E = (1 A)F S 4 = (1 0.30) = 239.4Wm 2 T E = 5 Local Thermodynamic Equilibrium 1/4 FE /4 = s = 255K (4.8) A key atmospheric concept is Local Thermodynamic Equilibrium. If the atmosphere was not in LTE we couldn t define temperature. Here we describe what this means. Local thermodynamic equilibrium means that over time scales of interest fluctuations in the Boltzmann distribution between energetic modes are so fast that effectively all modes are in equilibrium with each other. This sounds very abstract but it is critically important because it means that we can describe a volume of air as having a temperature and pressure. With respect to the atmosphere, the energy associated with a molecule has two major components: the kinetic energy associated with translational motions (molecular movement in space), and the kinetic energy associated with molecular scale energy transitions u tot = u trans + u mol 15

16 The molecular scale energy can in turn be broken up into rotational, vibrational, vibrationalrotational and electronic transition components (plus a few others that we won t concern ourselves with here) u tot = u trans + u rot + u vib + u elec +... Each of these components, as we discussed, has several, modes or degrees of freedom. Translational energy has three degrees of freedom: one each in the x, y, and z directions. For internal energy, CO 2 for example has two rotational degrees of freedom, three distinct vibrational degrees of freedom, and many, many vibrational-rotational degrees of freedom. Translational energy is what we sense as temperature. A classical treatment works fine here. Each degree of translational freedom has kinetic energy for a total in the x, y, and z direction (3 DOFs) u transx = 1 2 kt u x,trans = 3 2 kt This energy is what we interpret as temperature in daily life (more on this later). It is the kinetic energy of the molecules that causes the pressure on our skin that we interpret as heat. Molecular energy is more complicated. If we look at the exponential expression that is part of the Planck radiation equation 2hn 3 B n = c 2 e hn/kt 1 you will notice that this radiation depends on temperature also, but in a way that depends not at all on the translational motions of the molecules but rather only the energy transitions (or jiggles) within the molecules themselves. So really, we have two completely different temperatures Thermal Temperature 3 2 kt therm Planck Temperature B n (T Planck ) Local Thermodynamic Equilibrium, the precondition for Kirchoff s Law, and what applies to the bottom km of our atmosphere, requires that How does this happen? Imagine the following sequence of events T therm T Planck (5.1) 1. A molecule maintains a Boltzmann distribution that is a function of its radiating temperature T Planck 2. The molecule is bombarded by electromagnetic radiation that, if it is at frequencies corresponding to molecular modes, disturbs the Boltzmann distribution, raising molecular energies over all to a higher state. 16

17 3. Two things can happen here. If left to themselves, the molecule will reestablish a Boltzmann distribution by releasing photons (energy). This is what causes the Northern Lights. However, if the pressure of the gas is high enough, the molecules collide before this release of energy can happen. Instead, through collisions, the molecular energy gets passed between molecules and gets turned into kinetic energy. 4. Through continual absorption of radiative energy, and redistribution as kinetic energy, an equilibrium is maintained between u trans and u mol such that T therm T Planck For atmospheric pressure above 0.05 mb (i.e % of the atmosphere) this condition applies. Maintenance of LTE is usually glossed over in introductory texts because it applies to most of the atmosphere. However, understanding why it occurs is integral to understanding why our planet is livable, i.e. how it is that radiative energy is converted to the thermal energy that keeps our planet warm. The concept of temperature really is conditional on an assumption that a system is in LTE. 5.1 Kirchoffs Law Under conditions of Local Thermodynamic Equilibrium, we often hear that Kirchoff s Law applies, which states that the emissivity of a layer is equivalent to the absorptivity e l = a l (5.2) The premise here is that if an object has a temperature T and an absorptivity a l < 1 then it is not a blackbody. Something may have an absorptivity less than unity at a particular wavelength if it is transparent or shiny. A blackbody absorbs all radiation incident upon it at a particular wavelength, hence its emissivity is unity. To see why Kirchoff s law holds consider that at LTE there must be equilibrium between total absorption and emission. Otherwise temperature would be changing. Thus absorption equals emission and Z Z a l B l dl = e l B l dl implying that 0 0 e l = a l Note that an object can still absorb more radiation than it emits at a particular wavelength - if, for example, the radiation it absorbs is coming from another object with a higher temperature. For example, consider object a and object b with T b T a. Object a radiates energy F a = st 4 a Object b absorbs a fraction of energy from a at a particular wavelength according to its absorptivity: F bl (absorbed)=a l st 4 a 17

18 It reemits the energy with emissivity but at temperature T b. Thus e l = a l F bl (emitted)=e l st 4 b Notice that F bl (emitted) F bl (absorbed), even though the emissity and absorptivity are the same, because T b T a. 6 Why is the sky blue? The atmosphere is made up of oxygen and nitrogen mostly. Oxygen absorbs some solar radiation, but mostly these two molecules only scatter light to make the sky blue. How do they do this? To understand how, we need to understand what is known as dipole radiation. Consider an electric field incident on an atom represented by an electron surrounding a nucleus, i.e. an electromagnetic dipole. From Eq. 1.1, the incident electric field at a fixed point can be described by ~E = ~E 0 e iwt (6.1) The incident electric field makes the electric field of the atom oscillate. There s a fair bit of handwaving here, but it makes sense that the amplitude of displacement of electric charge within the molecule should be proportional to the incident electric field, i.e. ~x(t)=~x 0 e iwt µ ~ E = ~E 0 e iwt (6.2) But, now the molecule is vibrating and moving its electric charge back and forth, it generates its own electric field, distinct from the incident electric field. This is new electromagnetic field represents what we term the scattered radiation. Remember that force required to accelerate the charge q e of an electron, which has mass m e, is F = q e E = m e a An electric field is created by the acceleration of the electric charge created by the incident radiation. Therefore the scattered field, or dipole radiation, is determined by: ~E dipole (t) µ a(t) But we know that acceleration of the electron is simply the second derivative of the displacement ~x(t), so from Eq. 6.2 a(t)=ẍ(t)= w 2 ~x 0 e iwt Therefore the scattered electric field - the dipole radiation - is given by ~E dipole µ w 2 ~E 0 (6.3) Both the intensity and flux of an electromagnetic wave are proportional to the square of the electric field. Therefore, because I µ E 2, we can write ~I dipole µ w 4 ~I 0 (6.4) 18

19 In other words, the intensity of scattered dipole radiation is proportional to the intensity of the incoming radiation, and the fourth power of the frequency of the radiation. Now, noting that Nitrogen and Oxygen molecules respond as dipoles to incoming solar radiation, we can explain why the sky is blue. If ~I dipole µ w 4 then, because w/2p = n = c/l, then ~I dipole µ 1/l 4 Blue has a wavelength of 0.4 µm, and red has wavelength of 0.7 µm. Therefore I blue /I red = /0.4 4 = 9 All wavelengths of light are scattered, but blue light is scattered 9 times more effectively than red light. Of course, this then begs the question, if short wavelengths are what are favored, than why is the sky not violet? To answer this question, we must consider human physiology and the diminished sensitivity of the eye to the color violet. 7 Why are clouds white? The concentration of water molecules in clear and cloudy air is about the same. Why then, can we see clouds, and why do they look white? To address why we can see clouds, consider the following. The amount of dipole scattering we see in clear air is proportional to the amount of light scattered and the concentration of molecules N ~I total =~I dipole1 +~I dipole2 +~I dipole3 +...= N~I dipole (7.1) This is why the sky gets darker and darker the higher we go in the atmosphere, until it becomes completely black. N is getting exponentially smaller with height. When molecules are in the condensed phase, as they are in water droplets, the molecules are so close together (less than 1 nm) that any electrical vibrations are no longer independent of each other. Molecules that are close to each other instead prefer to vibrate sympathetically, or in phase. This amplifies the response of the dipoles such that ~I total = N~I dipole N~I dipole = N 2 ~I dipole (7.2) The number of water molecules in even a tenuous cirrus cloud is probably about N = 10 30, so obviously N 2 o N, and the cloud is visible where clear is not. But this still doesn t tell us why clouds are white. We are still looking at dipole radiation in a cloud, even if the dipoles are all very close together and vibrating in phase. 19

20 Figure 7.1: Interference pattern created by a plane wave with wavelength l passing through two slits separated by distance 2r. The phase difference between the two wavefronts Df is a function of angle from the forward direction q. Radiation scattered by a single dipoles only always vibrates in phase in the forward scattering direction. Off the forward scattering direction, there is positive and negative interference, and which it is is a function of particle size and the incident wavelength. The phase difference Df is a function of the angle between the scattered waves q Df = 2pr l (1 cosq) (7.3) The relationship between the intensity of the signal for two interfering light sources I 1 and I 2 and Df is I 1+2 = I 1 + I p I 1 I 2 cosdf (7.4) Let s consider two examples and the special case that I 1 = I 2. If q ' 0,2p,4p..., or if 2pr l, or 2pr (1 cosq)/l ' 0,2p,4p..., then Df ' 0,2p,4p... and I 1+2 ' 4I. This is the case of constructive interference. If, however, q ' p,3p,5p..., or 2pr (1 cosq)/l ' p,3p,5p..., then Df ' p,3p,5p..., and I 1+2 ' 0. This is the case of destructive interference.. 20

21 Now let s consider a given separation between sheets of dipoles r, so 2pr/l is fixed. Then there is no cancellation of forward scattered light (1 cos q = 0) and maximum cancellation at back angles (1 cosq = 2), as shown in Fig Construc)ve Interference Destruc)ve Interference Figure 7.2: Interference by sheets of dipoles. 21

22 Figure 7.3: Interference pattern created by a droplet. A similar argument can be applied to stacked sheets of dipoles. If two sheets of dipoles are separated by a distance 2r such that 2pr/l 1, then then sheets vibrate close to in phase, acting effectively as a single sheet of dipoles. But if the sheets are much farther apart and 2pr/l 1, then there are interference effects and cancellation of dipole radiation from one sheet by another sheet. 22

23 Figure 7.4: Interference pattern created by a cloud of droplets, the Corona. Can you explain the color separation? Since the wavelength of visible light is about 0.5 µm, and the distance between water molecules is about 1000th of that, so, very roughly, we can get about = 1 trillion water molecules to vibrate in phase. If a particle gets very much bigger than 0.05 µm across, then dipole radiation by the molecules is no longer coherent or all in phase with constructive interference. As a particle grows, some combination of constructive and destructive interference occurs as any two sheets of dipoles span a larger range of separations. Supposing that q = p, then from Eq. 7.3, Df = 4pr/l. Suppose too that there is destructive interference so that Df = p = 4pr/l. As the particle grows, destructive interference affects the shortest (blue) wavelengths first before it affects the longer wavelengths of light. Normally, blue light is scattered preferentially. But now, the scattered color becomes less and less blue, and more and more red. Individual dipoles still scatter blue preferentially, but this is offset by interference effects. Of course, as the particle grows further, blue shifts back to constructive interference and the red undergoes destructive interference, so this explanation is not enough for why clouds are white. If the particle is large enough compared to the wavelength, then all colors are scattered roughly equally for particles about 2 µm across (for sunlight). This is difficult to explain, but the best approach is to think that for large particles, there are an similarly large number of possible separations of all dipoles with all other dipoles. Ultimately, there is a roughly even combination of constructive and destructive interference for all colors and for all combinations of dipoles. This interference scattering signature dominates the blue scattering signature for single dipoles. Cloud droplets are typically at least 10 µm across so clouds scatter all wavelengths roughly equally, i.e., white. The words we use for this phenomenon are Rayleigh scattering I µ I0 /l 4, which applies to 23

24 particles that vibrate entirely in phase. In the atmosphere this works for gases, and very very small aerosols we can t see. Mie scattering goes roughly as I µ I 0 /l 2, and applies to most aerosol particles that are still small enough to preferentially scatter blue but less effectively so. Geometric scattering applies to particles where a wave treatment is no longer necessary, and which scatter light of all wavelengths equally (i.e. I µ I 0 ). Clouds particles fall into this regime. A nice demonstration of various scattering regimes is to make a solution of sodium thiosulfate, into which you titrate sulfuric acid. A precipitate will form that should initially scatter blue and transmit red, but scatter white and transmit little as the particles grow. Figure 7.5: Milk is blue if you dilute it with enough water, meaning it is not white due to interference phenomena. 24

25 But the reason why clouds are white in fact actually has nothing to do with the fact that all colors are scattered nearly equally by cloud droplets. Rather it is due to multiple scattering. To prove this to yourself, milk is white even though really dilute milk scatters blue preferentially. 8 The mathematics of scattering and absorption and emission 8.1 Absorptance and transmittance The transmittance of an layer depends on its optical depth t, which in turn depends on how much of the substance the radiation has to pass through, and how dark the substance is to the radiating wavelength. A stout (cola) transmits less light from a direct beam of radiation than a lager (gingerale), even if in the same sized glass. The direct intensity I of the radiation is attenuated exponentially, so that the brightness of an object seen through the medium drops of in proportion according to the most simple of first-order differential equations di l dt l = I l This has the standard exponential decay solution Thus the transmittance is and the absorptance is I l (t l )=I l (0)exp( t l ) T l = I (t l ) I (0) = exp( t l ) (8.1) A l = 1 exp( t l ) (8.2) Note that as t l!, the transmittance goes to 0 and the absorptance goes to 1. Note that Beer s Law expresses only extinction from a direct beam of radiation. Milk, which scatters rather than absorbs light, might have the same optical depth as a cola, and attenuate the direct radiation equally. Beer s Law says nothing about whether a direct beam of radiation is attenuated due to scattering (as in the case of milk) or absorption (as in the case of a cola). However, the milk looks much brighter because it is scattering radiation in all directions, and this radiation becomes multiply scattered. Multiple scattering mean that the radiation is just re-directed and homogenized (made more isotropic). This is a more advanced topic. 8.2 Optical depth How do we express t in terms of physically meaningful parameters? There are a number of different ways that are used, all related. At the most fundamental level, every molecule has an effective absorption cross-section s a or scattering cross-section s s that has units of area. What this represents is the effective cross-sectional area of the molecule that absorbs all radiation incident upon it. For example a glass of cola is black, and probably has an absorption cross-section close to it s actual cross-sectional area normal to the plane of incident radiation. The same glass filled with ginger-ale would have a much smaller absorption cross-section, even though the glass itself has 25

26 the same physical cross-sectional area. Ginger-ale is lighter in color. The absorption and scattering cross-sections are related to the geometric cross-section s through a scattering or absorption efficiency K l, which is dimensionless. Thus s sl = K sl s (8.3) s al = K al s (8.4) so, for example K s = s s (8.5) s The volume absorption (or scattering) coefficient b a is the product of s a and the concentration of molecules or particles N in units of #/m 3 b al = Ns al = NK al s (8.6) b sl = Ns sl = NK sl s (8.7) Thus b a has units of 1/m. Since the optical depth t is dimensionless it follows that dt al = b al ds = Ns al ds = NK al sds (8.8) dt sl = b sl ds = Ns sl ds = NK sl sds (8.9) where ds is the pathlength of the radiation. The more soda or beer you peer through, the harder it is to see to the other side. It is also common to express b a in terms of a mass absorption coefficient k a in units of m 2 /kg such that dt al = k a rwds (8.10) where r is the density of air and w is the mixing ratio of the gas in the air. This is very close to a final, very commonly used formulation that defines the t a in terms of the mass path u with units of gm 2 such that du = rwds (8.11) and dt a = k a du (8.12) Of course, it is important to recognize that in the atmosphere u = u(z) and therefore, if the mass path is defined with respect to a vertical column. t a (z)=k a Z z du(z) dz (8.13) dz where refers to the top of the atmosphere. Note that dz = cosqds where ds is the slant path. 26

27 Figure 8.1: Illustration of the slantpath through a vertical layer. 8.3 Transmittance in a plane parallel atmosphere Thus according to Eq. 8.1 T (z)=exp( k a u) or, coming back to the expression for exponential decay of intensity, we must modify di l dt l = I l to I l (t l )=I l (0)exp( t l ) µ di l dt l = I l I l (t l )=I l (0)exp( t l /µ) (8.14) where, µ = cosq, where q = 0 and µ = 1 for the up direction, q = 90 and µ = 0 for the side direction. and What this shows is that for a particular atmospheric optical depth, radiation is 27

28 attenuated most when it is coming it at side angles, because it must pass through much more atmosphere before coming out the other side than if it is passing straight through. There are a few other important single-scattering parameters that are necessary to describe the full radiation field: 8.4 Size parameter The size parameter x is a dimensionless number that express the size of the droplets relative to the wavelength of incident radiation x = r k = 2pr/l (8.15) where a is the particle radius, and l is the wavelength of the light. Remember that we have shown that the size of a particle relative to incident radiation that is quite important in determining how light is scattered. r (µ m) Solar radiation Geometric optics Terrestrial radiation x = 1 Mie scattering Rayleigh scattering Negligible scattering Weather radar (µ m) Raindrops Drizzle Cloud droplets Smoke, dust, haze Air molecules Figure 8.2: In this plot, the size parameter increases as one moves from the Rayleigh Scattering regime to the geometric optics regime. 8.5 Single-scattering albedo The single-scattering albedo w 0 ranges from 0 to 1 and represents the fraction of energy incident on a particle that is scattered rather than absorbed. i.e. w 0 = K s K s + K a = s s s s + s a = 28 t s t s + t a (8.16)

29 5 m i = 0 Scattering efficiency K m i = 0.01 m i = 0.1 m i = Size parameter x Figure 8.3: The extinction efficiency as a function of size parameter for a range of values of the complex index of refraction. More absorption brings the extinction efficiency closer to unity and a single scattering albedo of 0.5. In the visible, clouds and Rayleigh scattering gases like N 2 and O 2 have a single scattering albedo unity (i.e. no absorption). In their IR absorption bands, gases have a single-scattering albedo of zero (i.e no scattering). 8.6 Phase function If light is scattered (w 0 > 1), the probability it is scattered in any given direction is expressed by the phase function p(µ), where µ = cos q, where q is relative to the direction of the incoming beam. µ = 1 corresponds to forward scattering (the same direction as the incoming beam), and µ = 1 corresponds to back-scattering (towards the light source). With the exception of such things as mirrors, light is scattered over a distribution of angles. Cloud particles are strongly forward scattering of visible light, but they have more isotropic Rayleigh scattering of radar wavelengths. Gases have isotropic Rayleigh scattering in the visible also. In general p(µ) is normalized such that Z 1 1 p(µ)dµ = 1 (8.17)

30 Incident Beam (a) (b) (c) Forward Figure 8.4: A graphical representation of the phase function with forward scattering at zero degrees to the right for a) Rayleigh Scattering b) Mie Scattering and c) the Geometric Optics regime. Note that the larger the size parameter, the more strongly forward scattering is the particle. 9 Energy transitions of molecular absorption We have discussed how the internal energy of a molecule can be separated into translational, rotational and vibrational modes. Energy can be added or subtracted from each of these modes by the absorption/emission of a photon of electromagnetic radiation. The energy can also change the charge distribution within the molecule, or even break apart the molecule if the photon is sufficiently high energy (as in the case of O 2 and UV radiation). 9.1 The simplest case: The Hydrogen atom For example the absorption of a photon by a hydrogen atom will raise the energy level of the electron from its ground state to an excited state. The frequency of the photon that is associated with this energy transition is DE = hn (9.1) Now, at the molecular level energy is quantized. Without going into details, what this means is that only certain discrete frequencies can be absorbed in order to excite a transition in energy level. For a hydrogen atom these are 1 1 n = R H j 2 k 2 where j and k are integers associated with the energy level of the electron (ground state = 1). 30

31 Figure 8.5: The phase function of water droplets with dominant optical features (M. King et al.). 9.2 Polyatomic molecules and rotational transitions But we don t care about monatomic molecules, since the only monatomic molecules in the atmosphere are Nobel gases like argon and krypton that require really really high energies to be excited (as seen in neon signs). In addition to the three translational degrees of freedom (whose energy depends only on temperature not radiation), there is rotational and vibrational energy. For example monatomic molecules have zero degrees of rotational freedom. Symmetric linear triatomic molecules like CO 2 have one degree of rotational freedom (two have the same energy), and asymmetric molecules such as H 2 O have three independent modes. In quantum mechanics the angular momentum associated with molecular rotation is L = h p J (J + 1) (9.2) 2p where J is an integer rotational quantum number. In classical mechanics and quantum mechanics, the energy associated with rotation is E = L 2 /2I (9.3) where I is the moment of inertia. Therefore, the discrete energy associated with a particular molecular mode of rotation is E J (J)= 1 h 2 J (J + 1) (9.4) 2I 2p 31

32 Figure 8.6: The phase function of ice crystals with dominant optical features (M. King et al.). The frequencies associated with molecular transitions are where DE = E J 0 E J = hn (9.5) n = B n J 0 J B n J (J + 1) (9.6) where B n = h/ 8p 2 ci. It turns out that in the rules of quantum mechanics energy transtions that might be associated with absorption of EM radiation are only associated with DJ = J J 0 = ±1. The frequencies associated with pure rotational transitions tend to be low energy in the 10 to 100 cm 1 range. An important note here is that the molecule must have a dipole moment. In other words, one end of the molecule must be more positive than the other. If this is not the case, there is no potential for an incoming electromagnetic wave to exert torque on the molecule. As examples, nitrogen, oxygen and carbon dioxide are linearly symmetric, so they have no dipole moment. Consequently they have no purely rotational molecular transitions. The greenhouse gas N 2 O on the other hand, with structure N-N-O, does have a dipole moment. It is linear but asymmetric. It has two modes of rotation, one along an axis lying vertically along this page, and a second lying along an axis pointing into the page. These two modes are called degenerate because they differ not in the amount of energy required but only in what direction you are looking at the molecule. A third mode, with axis lying horizontal to this page, goes straight thround the length of the molecule, so it has zero energy. 32

33 Minimum angle of deviation 46 Refracting angle halo Cloud of hexagonal ice columns Light from sun halo Refracting angle 60 Minimum angle of deviation 22 Observer Figure 8.7: Professor Peter V. Hobbs with his ice crystals halos Water, however, which has structure H / O \ H has a different rotational mode along all three axes. In the atmosphere, purely rotational water modes matter most at very long wavelengths greater than 20 µm, where water has a very complex rotational structure. This is because several modes of rotation can be excited simultaneously. 9.3 Polyatomic molecules and vibrational transitions Vibrational transitions are possible provided there is a molecular bond. A molecular bond is like a spring. The classical analogy is Hooke s law where F = k r 0 r (9.7) where r 0 is the equilibrium separation, and the spring resonates like a simple harmonic oscillator with resonant frequency n 0 = 1 p 0 k/m (9.8) 2p where m 0 is the reduced mass and the energy at each frequency is E = hn 0. In the quantum world, allowable vibrational transitions have energy where the quantum mechanical rule is DE n = v 0 hn 0 v hn 0 Dv = v 0 v = ±1 33

34 v is ±1. Thus, the frequency of photons that excite a transition is given by or DE n = hn = Dv hn 0 n = Dv n 0 With a simple molecule like O-O, there is only one vibrational mode. With triatomic molecules there are normally three: the v 1 mode, or symmetric stretch mode, the v 2 or bending mode and the v 3 or asymmetric stretch mode. Vibrational transitions are higher energy than rotational transitions and are in the 1000 cm 1 range. A nice illustration of vibrational motions is here. 9.4 Vibrational-rotational spectra Figure 9.1: Vibrational - rotational transitions for Dv =+1 and DJ =+1 It gets more complicated. CO 2, as stated previously, has no purely rotational modes. But clearly it can have vibrational modes due to vibrational stretching of the molecular bonds. For example, the greenhouse effect associated with CO 2 is due to the n 2 bending mode. If the stretching is asymmetric, than it can induce a dipole moment in the molecule, which then makes the molecule susceptible to rotational transitions. There are huge numbers of options for such vibrational-rotation transitions, and they account for much of the complexity in the absorption bands seen by various gases in the atmosphere. A simple illustration of these transitions is shown in Fig The important 15 µm (670 cm 1 ) absorption band for CO 2 and the 6.7 µm band for water vapor (used for water vapor satellite imagery) are both vibrational-rotational bands The CO 2 band is shown in Fig

35 Figure 9.2: Vibrational - rotational transitions for CO 2 10 Molecular absorption profiles 10.1 Line strengths The strength of any particular absorption line (i.e. the amount of energy associated with any particular molecular transition) can be derived from quantum theory or measured in a lab. Databases exist for these quantities. The amount of energy absorbed in the atmosphere is going to depend on the number of molecular transitions that occur, which in turn will depend on the concentration of molecules. The symbol used for the line strength is S. We ll show later how this affects atmospheric absorption Line profiles But the line is not infinitely narrow, rather it has a characteristic shape or profile that means that a range of frequencies are absorbed for any individual molecular transition. Imagine Hooke s law again (Eq. 9.7). F (x)= kx = mẍ Solving the ode, the oscillator has a single natural frequency that it can be forced at w 0 = p k/m 0 (10.1) However there is always a frictional damping force which is usually proportional to the velocity v. In the atmosphere this frictional damping force is associated with collisions between molecules. 35

36 This induces pressure broadening of the absorption line, smearing the range of frequencies at which the molecule will undergo a molecular transition. Note that w 0 = 2pn 0 or w 0 = 2pcn 0. Doppler Lorentz k ν (ν ν 0 )/α Figure 10.1: Lorentz and Doppler profile expressed in terms of the absorption coefficient k n where the absorption optical depth is t = k n u where u is the column amount of the absorbing gas. Expressed in terms of a frequency n, the line shape takes on a characteristic shape known as the Lorentz profile defined by a L /p f (n n 0 )= (n n 0 ) 2 + al 2 (10.2) where a L (which replaces g) is the half width of the spectrum at half-maximum and is proportional to temperature and pressure p 1/2 T0 a L = a 0 (10.3) p 0 T a 0 comes from lab studies. Typical values of a 0 lie in the range 0.01 to 0.1 cm 1. Note that the factor of p in the numerator is there so that Z f (n n 0 )dn = 1 (10.4) Essentially, this normalization means that adding up the energetic contributions to single line due to all frequencies gives the total. 36

37 There is also something called Doppler broadening associated with the fast motions of the molecules. It spreads out frequencies the same way that the siren from an ambulance changes pitch depending on whether it is coming towards you are moving a way from you. But this only begins to matter where atmospheric pressures are very low. For most of the atmosphere below about 20 km, pressure broadening dominates. Finally, there is inescapable natural broadening due to the Heisenberg uncertainty principle. If you are familiar with this, the upshot is that quantum mechanics dictates that you can t narrow down your estimate of the amount of energy absorbed without broadening your uncertainty in just which frequency this energy is getting absorbed at. In the atmosphere this effect is generally small compared to either Doppler or pressure broadening. 11 Line absorption As we describe earlier, individual lines actually cover a range of frequencies, due in particular to pressure broadening. There are a number of ways to deal with this particularly thorny problem. We will deal with just one here, because it illustrates a particularly interesting property of the atmospheric greenhouse effect. First we need to recognize that for one single absorption line (one individual molecular transition) the mass absorption coefficient is spread over a wide range of wavenumbers. From Eq Sa L /p k a (n)=sf(n)= (n n 0 ) 2 (11.1) + al 2 where S is the strength of the line, and has the same units as k a (meters squared per kilogram). Note that in the hypothetical instance that there were no broadening of any kind then S and k a would be equivalent because the integral of f (n) is 1. Now if we want to find the transmittance associated with an individual line, it wouldn t be sensible just to limit our selves to the center of the profile. Rather we would want to calculate a band - averaged transmittance T representative of a range of frequencies Dn that covers at least one or two line widths a L on either side of the centerline n 0. From Eq. 8.1 T = 1 Z exp( t n )dn (11.2) Dn n where The band averaged absorptance is t n = k a u = Sf(n)u (11.3) A =1 T (11.4) 11.1 Weak and strong lines As u increases, the transmittance goes down and the absorptance goes up. If u is sufficiently large, the value of t at the centerline n 0 becomes large enough that the transmittance is 0 and the absorptance is 1. That particular wavelength becomes saturated. As u continues to increase the wings 37

38 Transmittance Transmittance Transmittance Transmittance ZENITH ATMOSPHERIC TRANSMITTANCE UV VIS Near IR Thermal IR 1 CO 1 0 N O CH O O CO H 2 O 1 0 Total Wavelength [µm] Figure 11.1: Absorption profiles for various atmospheric gases. Transmittance Transmittance Transmittance Transmittance of the Lorentz profile become saturated also, starting from the center and moving progressively further outward. So it seems that after a certain point you don t get as much bang for your buck as you increase u. We define weak lines as those that are nowhere saturated and strong lines as those that are saturated at the very least at the center line. How do we quantify this effect? From Eqs and 11.3, the band averaged transmittance of a line, as described above is T = 1 Dn Z n exp( Sf(n)u)dn If we substitute the Lorentz profile for f (n) (Eq. 10.2) we end up with a rather hairy integral! T = 1 Z a L /p exp S Dn n (n n 0 ) 2 u dn + al 2 38

39 It has a solution called the Landenberg-Reiche function (named after the folks who solved it) that involves Bessel functions of the first kind of order 0 and 1. You are probably just as happy not to deal with this. The important thing to recognize is that in the limit of small values of u (weak lines), the above integral has the solution T = 1 Su Dn (11.5) and for large values of u (strong lines) T = 1 2 p Sa L u Dn (11.6) Thus for weak lines absorptance increases linearly with u and for strong lines it increases as the square root of u. Changes to greenhouse gas concentrations are most potent if the lines are initially weak. 12 Molecular absorption, the greenhouse effect, and climate change The greenhouse effect is a measure of how much atmospheric gases trap terrestrial radiative flux, reducing the amount of energy that escapes to outer space, and increasing the equilibrium temperature of the planet. The primary greenhouse gases are water vapor (75 % of the total), CO 2 (32 %), ozone (10 %) and methane CH 4 (8 %). Note that there is overlap between bands, which is why the total is greater than 100%. Now lets compare CO 2 and CH 4, which are increasing rapidly in the atmosphere due to human activities. The major source of CO 2 is fossil-fuel combustion, which powers the world economy. Its current concentration is about 380 ppm. Methane is produced by microbes of the archaea family in the guts of livestock (particularly cattle), in landfills, rice paddies, and whenever some area is flooded. Its current concentration is 1.85 ppm. A great many of the absorption lines in the atmosphere are already saturated at their centers. This is particularly true of CO 2 in its 15 µm vibrational-rotational absorption band, where the broadband emissivity is (it is unity around the centerline). Thus CO 2 is a strong line. CO 2 is well-mixed in the atmosphere, so the implication of this result is that by doubling CO 2 concentrations in the atmosphere - something we are well on our way to doing - we only increase the greenhouse effect due to CO 2 by a factor of p 2 or about 40%. An additional complication to this calculation is that there are also water vapor rotational bands at 15 µm, so the increase is attenuated even further. Other greenhouse gases in the atmosphere are not yet saturated. Perhaps most important of these is methane CH 4. At 7.6 µm methane has a vibrational-rotational band that is not quite yet saturated but is located at a region of the terrestrial blackbody spectrum where there is still significant energy. The broadband emissivity of this wavelength band is much lower than CO 2. Since the lines are not quite yet saturated the greenhouse climate forcing by methane follows 39

40 Figure 12.1: The Keeling Curve showing the increase in CO2 concentrations in the atmosphere the weak line rule, and it is linearly proportional to its concentration. A doubling of methane concentrations will approximately double the methane greenhouse forcing. We need to note however that the above arguments have an oversimplification. The wings of individual lines overlap. This further reduces the impact of doubling u. The extent of this effect is beyond the scope of this course. The climate sensitivity expresses the temperature change per doubling of CO 2 concentrations. Figure 12.3 summarizes some estimates of climate sensitivity based on numerical model and paleoclimate estimates. (From Skinner 2012): Symbols indicate estimates of past global average radiative forcing anomalies and the corresponding global average temperature change, based on published data. Radiative forcing estimates include land albedo and greenhouse gas effects (and sometimes the impacts of atmospheric dust). Anomalies are relative to pre-industrial values, except where indicated. Uncertainties are 1sv or conservative estimates where published uncertainties are unavailable. Yellow shading indicates a summary probability distribution for numerical modelderived climate sensitivity values (18); dashed gray lines indicate 1 C increments. The solid black line is a linear regression on all the data, forced through the origin and with 95% confidence limits (dotted black lines). The ball-park paleoclimate sensitivity estimates shown here range widely between 0.6 C and 6.5 C, but taken together imply a climate sensitivity of ~3 C, which is in very close agreement with the best estimate derived from numerical models. However, this agreement may mask evidence for nonlinear feedbacks and abrupt climatic transitions that are not captured in the climate sensitivity as commonly defined. LGM, Last Glacial Maximum; PETM, Paleocene- Eocene Thermal Maximum. 40

41 Figure 12.2: Trends in methane in the atmosphere 41

42 Figure 12.3: 42