Atmospheric Physics. August 10, 2017

Transcription

1 Atmospheric Physics August 10,

2 Contents I Equilibrium Thermodynamics 6 1 Molecular Thermodynamics Atmospheric application: water vapor and dry air Intensive and extensive variables 13 3 Heat, Work and the First Law of Thermodynamics Atmospheric Application: the general circulation Internal energy in an ideal gas 19 5 Total energy or the Enthalpy Atmospheric Application: the dry and moist static energy Equilibrium solutions Case 1: Equilibrium solution for a materially closed system at constant entropy Atmospheric application: the geopotential Case 2: Equilibrium solution for a materially closed system at constant pressure Atmospheric application: the dry adiabatic lapse rate Case 3: Equilibrium in a materially closed system at constant temperature Atmospheric application: the hydrostatic equation Atmospheric application: atmospheric thickness Quasi-equilibrium transitions Case 1: An ideal gas Atmospheric application: potential temperature Case 2: Phase changes Atmospheric application: saturation vapor pressure of water Equilibrium distributions Log-log plot with a positive slope: Power Laws at constant temperature Log-linear plot with a negative slope. The Boltzmann distribution when jumps are large Atmospheric example: the hypsometric equation Log-Log plot with a negative slope. Power Laws at constant pressure Atmospheric example: aerosol size distributions Conservation of matter and energy Atmospheric application: Adiabatic liquid water content of a cloud Atmospheric application: Saturated adiabatic lapse rate and the equivalent potential temperature Atmospheric application: Stability

3 II Non-equilibrium Thermodynamics Second Law of Thermodynamics and Entropy Examples of changes in entropy Atmospheric Application: Entropy production in the atmosphere Heat engines Atmospheric Application: A hurricane Atmospheric Application: The general circulation Available energy The Gibbs free energy A hand waving derivation of available energy A formal derivation of the available energy Components of the available energy The available energy at equilibrium Non-equilibrium flows The Universal Staircase Hand-waving Formal solution Flows between discrete surfaces Continuum flows expressed in terms of a velocity Continuum flows to particles expressed in terms of a diffusivity Non-equilibrium flows and the diffusion (or heat) equation III Cloud Physics Evaporation Atmospheric example: potential difference of the relative humidity Atmospheric example: evaporating puddles Pure droplet formation Solution droplet formation Kohler equation Activity of solution droplets Cloud droplet growth Rain Production Collection Terminal velocity

4 19 Ice nucleation Homogeneous nucleation Heterogeneous nucleation Ice crystal diffusional growth Observed cloud microstructures Vertical structure Horizontal structure Continental versus maritime clouds Observed rain and snow distributions Atmospheric mixing Mixing of conserved variables Adiabatic Mixing without Condensation Atmospheric application: mixing across a front Atmospheric turbulence Characteristics Length Scales in Turbulent Flows IV Atmospheric Radiation Electromagnetic waves Light as quanta Atmospheric application: Oxygen dissociation in the stratosphere Flux and Intensity Flux Intensity Atmospheric application: Solar intensity Relationship between flux and intensity Dipole radiation Blackbody radiation Atmospheric example: Earth s energy balance Local Thermodynamic Equilibrium Kirchoffs Law Why is the sky blue? Why are clouds white? 134 4

5 33 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 Radiative transfer equation Energy transitions of molecular absorption 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 152 5

6 Part I Equilibrium Thermodynamics 1 Molecular Thermodynamics Perhaps the most basic equation in atmospheric thermodynamics is the ideal gas law p = rrt where p is pressure, r is the air density, T is temperature, and R is the gas constant for dry air. While the derivation of this equation takes some effort, it is very much worth it because it gives us a deep understanding of what pressure and temperature really are. Any work the atmosphere can do to move around air is derived from the internal energy of its component gases. How much of this energy does a unit volume of gas have? Consider the following diagram: Figure 1.1: A piston Say you are pushing on a piston filled with a gas. The pressure is simply p = F/A (1.1) and the amount of work you would have to do to compress the gas is dw = F ( dx) (1.2) but so from (1.2) and (1.3) F = pa (1.3) dw = padx = pdv (1.4) 6

7 where dv = Adx since you are pushing an area through a differential length to get a differential volume. Alternatively, we could switch the sign to get not the amount of work done on the gas but the amount of work done by the gas dw = pdv (1.5) Where does the pressure come from? The pressure arises from the motion of the molecules of gas inside the piston banging against the piston wall. Each time they bang against the piston wall, the piston acts like a perfect reflector. If the piston does not move due to this constant banging, then there must be a force balancing the momentum p associated with the banging of the molecules (note bold font to distinguish from pressure momentum is a vector), that is imparted to the piston wall with every collision. This force is F = dp/dt (1.6) We are assuming here that every molecule that hits the piston leaves with the same energy, i.e. the same speed v and mass m, and therefore the same momentum. The piston is a perfect reflector. But momentum is a vector! Therefore, the momentum imparted to the piston by a single molecule is the change in momentum associated with coming in and bouncing out: Dp x = mv x ( mv x )=2mv x (1.7) What is Dt though? Suppose we have N molecules in volume V, or n = N/V as the concentration 1. Only molecules within distance v x Dt hit the piston 2. The volume occupied by these molecules is DV = v x ADt 3. The number of molecules that hit the piston is DN = ndv = nv x ADt, where n = DN/DV is the molecular number concentration 4. The number of molecules that hit per unit time is DN/Dt = ndv /Dt = nv x A Therefore, for all molecules, we combine number 4 and (1.7) to get So F x = DN Dp x Dt F x = nv x A 2mv x = DN Dt Dp x Now, taking (1.1), this leads to the expression for pressure exerted by all molecules in the x- direction p = F x /A = 2nmv 2 x 7

8 But if we only want the one half of the x-direction that is in the direction of the piston then we need to divide by two to get the pressure on the piston p = F x /A = nmv 2 x (1.8) There is a complication here though, which is that molecules have an equilbrium distribution of speeds (un-normalized here) that obeys a Boltzmann distribution m 1/2 f (v x )= exp mv 2 2pkT x /2kT For our purposes, we are interested in the average speed p = nm v 2 x We could also multiply and divide by two to write p = 2n mv 2 x/2 (1.9) The term in brackets is obviously merely the average kinetic translational energy of each gas molecule in the x-direction u trans,x = mv 2 x/2 (1.10) (the dot refers to per molecule ). Since it will come up later, we will also note that, at equilibrium, the kinetic energy equals the potential energy of the molecule. We now have p = 2n u trans,x (1.11) We also note the gaseous motion is obviously three-dimensional, and there is kinetic energy in each of these directions: v 2 x v 2 y v 2 z m 2 = m 2 = m 2 and although it is moderately difficult to prove, it is easy to believe that if we have three degrees of freedom (i.e. f = 3), that, like kinetic and potential energy, all degrees should share the energy equally, meaning each gets a third of the total v 2 x = v 2 /3 so so v 2 u trans = f u trans,x = 3 u trans,x = m 2 p = 2 3 n u trans (1.12) This equation should start to be looking vaguely familiar. We should really be calling u trans the temperature of the gas molecule. However, hindsight is 20/20, and due to some historical missteps we have defined a temperature T that is related to u trans by a constant factor u trans = f 2 kt = 3 kt (1.13) 2 8

9 where, k = J/K/molecule is Boltzmann s constant and 3kT/2 = mv 2 /2. The average kinetic (or potential) energy per degree of freedom f is kt /2. Therefore, p = nkt (1.14) It is revealing here to consider that pressure p has units of J/m 3, and therefore is an expression of energy density the number density of molecules n times kt. What does kt represent? It is not the internal energy density per molecule, which for translational energy alone is u trans = 3 2kT > kt, and there is vibrational and rotational energy too, which makes the internal energy even bigger. Perhaps the best way to think about kt is that it is the average total energy (kinetic plus potential) carried along the direction normal to a surface. Note k.e. = 1 2 kt and p.e = 1 2kT. Thus kt is the total average energy per molecule that is available to do pdv work. p = nkt is then the density of this available energy. We ll call this average energy µ 0 = kt (1.15) It will be useful later. There are several ways to express the ideal gas law, which relates pressure temperature and density in the atmosphere (among other places). The most common formulations are pv = NR T where N is the number of kilomoles. In the atmospheric sciences, we typically make some simplifications. We replace R by R = R /M, where M is the molecular weight of the gas in kg/kmol. In this case Alternatively, we can replace r by the specific volume a = 1/r: Main points p = rrt (1.16) pa = RT (1.17) Pressure is a force over an area where the force is due to the change in momentum associated with molecules bouncing off their surroundings Pressure has units of energy density Temperature has units of energy. The amount of energy kt/2 is the average amount of translational (or potential) energy per degree of freedom per molecule. There are three degrees of translational freedom so internal translational energy is 3kT/2. Pressure has in effect two degrees of freedom due the molecular bounce The amount of energy kt is the average energy per molecule available to do pdv work. Pressure is the density of energy available to do pdv work on the surroundings 9

10 Question What if the volume of gas is an air parcel in a surrounding atmosphere? Do the equations above need to change in any way if the box is removed? 1.1 Atmospheric application: water vapor and dry air The pressure exerted by a mixture of chemically inert gases is equal to the sum of partial pressures of the gases. The partial pressure is proportional to the molecular density. As we showed in Eqs and 1.16, in total p = nkt = rrt But what if there is a mixture of gases, as in the atmosphere? Dalton s Law states that we can treat the gases as if they are independent from one another, making the good assumption that all the molecules have the same temperature (or translational kinetic energy). In other words, we can just add up the molecular densities n i, or energy densities p i, so that p = ktsn i = T Sr i R i = S i p i (1.18) So it is convenient to separate gases into their major constituents, which in the case of the atmosphere, is mostly dry air and moist air. For dry air (which is mostly nitrogen and oxygen) For water vapor we use the symbol e for pressure and p d = r d R d T (1.19) e = r v R v T (1.20) The mean molar weight for dry air is M d =28.97 kg/kmol so R d = R /M d = 287Jdeg 1 kg 1, and for moist air M v = 18.0 kg/kmol so R v = R /M v = 461Jdeg 1 kg 1. Thus the total pressure is p = p d + e =(r d R d + r v R v )T Note again, that we are saying that the water vapor molecules have the same temperature (kinetic energy) as the dry air molecules. We ll introduce an important parameter here e = R d R v = M v M d = (1.21) We define the mixing ratio (i.e. the ratio water vapor mass to dry air) in the atmosphere as Substitute the ideal gas law (1.19) and (1.20), we get w = m v m d (1.22) w = r v e/r v T = r d (p e)/r d T 10

11 e = e p e w ' e e p (1.23) An important adjustment that can be made to the ideal gas equation is to adjust for the added buoyancy that is associated with water vapor being lighter than dry air so that within a given volume, the density of air is given by r = m d + m v V = r d + r v = p e R d T + e R v T or, with some algebra r = p apple 1 R d T apple e p (1 e) = 1 p R d T (1 e) w e Rearranging p = h 1 rr d T (1 e) e w i = rr d T v where T v, defined this way, is the virtual temperature. It works out that to a good approximation T v ' T e w = T ( w) (1.24) e Dry air: N 2,O 2 Moist air: N 2,O 2,H 2 0 ρ dry > ρ moist If T dry = T moist then T v,dry < T v,moist Figure 1.2: Moist versus dry air 11

12 This is a very useful quantity to know since the density of moist air is given by: r = p p = R d T v R d T ( w) At constant pressure, air that is either moist or hot will be less dense than air that isn t, and be more inclined to rise (Fig. 1.2). Question Dalton s law states that partial densities and pressures are additive. Can you provide a compelling reason why temperatures are not additive as well? Why is there not a partial temperature? 12

13 2 Intensive and extensive variables T,p m,v u,h U,H T,p m/2,v/2 u,h U/2,H/2 Figure 2.1: Intensive (lower case) versus extensive (upper case) variables When considering physical systems, it is often very useful to stay aware of whether the property being considered is intensive or extensive. An intensive variable is one that does not depend on the volume of the system, and an extensive variable is one that does. Intensive variables are sometimes called the bulk properties of the system. For example, consider the following expression of the ideal gas law p = rrt (2.1) Say we had a box with each of the properties p, r and T, and we subdivided the box, it would not change the values of these properties, so each of these properties is intensive. However, r = m/v (2.2) Subdividing the box, into small boxes with smaller V would lead to a corresponding reduction in the mass in each subdivided box m. Thus, m and V are extensive variables. The distinction is important, because often we want to know how much we have of a particular thing. The nature of the thing is defined by one or two of the intensive variables p, r, and T (note that from 2.1, we only need two to define the third). The amount we have is defined by the mass m. In atmospheric sciences, when we consider the ideal gas equation p = rrt where r is an intensive variable and mass m = rv is the extensive variable, where m normally refers to either the mass of dry air or the mass of water vapor. The convention used in Wallace and Hobbs, one that is commonly employed is to use the lower case for intensive quantities and the upper case for extensive quantities. Thus, for example, the internal energy per unit mass is given the symbol u and the internal energy is given the symbol U. In atmospheric sciences, we very often 13

14 thing about intensive quantities, because we reference them to a parcel of air in which the mass of the air (but not the volume) does not change with time. Thus u = U m In the notes here, I will express intensive variables that are referenced with respect to volume in bold lower case, i.e. the internal energy per unit volume is u. Where the intensive variable is per molecule it is expressed, e.g., as u. Main points Intensive variables are independent of how much matter there is. Extensive variables depend on how much matter there is. Question This distinction between extensive variabiles and intensive variables is particularly important whenever we talk about flows, of air, radiation, or whatever. Why? 14

15 3 Heat, Work and the First Law of Thermodynamics We can express Eq for the ideal gas equation in terms of per unit mass rather than per molecule: p = 2 3 ru trans (3.1) where u trans = 3 RT (3.2) 2 is the translational energy per kilogram due to molecules whizzing around in three dimensions. There are other types of internal energy we can consider (and will later), but suffice to say we can talk about a total internal energy per kilogram u. Note that, like T, u is an intensive variable, so the extensive property is U = mu where m is the mass of air. We call U, the total internal energy. Our question, is how do we change the total internal energy? Some outside agency adds energy to the system (e. g. condensation, absorption of sunlight or terrestrial radiation), through flows of some foreign material that does not define U. We call this energy heat The system changes its energy through flows of whatever material does define U (e. g. through expansion pdv or lifting mgdz). We call this work. Whatever happens, there must be conservation of energy. We express this as the First Law of Thermodynamics DU = DQ DW (3.3) where, DQ is the heat added to the system, and DW is the work done by the system. Both heat and work can be either positive or negative, so what is the difference between the two? What is most important to recognize is that heat and work are just names for energy. Our goal is foremost to keep an accurate account of the amount of energy that flows into and out of a system. That said, there are different types of energy. The best way to distinguish heat from work is that heat is associated with with flows of a different type of matter than is associated with internal energy U. So if U = mu, where m is the mass of the material we are interested in (say the air in a parcel of air), there is also an external energy associated with a different type of material that we call heat (say photons associated with warming sunlight). For example, radiative heating is associated with photons. Radiative heating can heat a parcel of air composed of molecules. Heating does not directly change the extensive variable of the air parcel m. In the atmosphere, we think about gases. For a gas, an easy way of visualizing work is to examine a piston Imagine there being a chemical explosion in the piston that releases external energy DQ to the gas. The external material flow here is due to a redistribution of electrons in a chemical reaction. This flow of external energy or heating adds to the internal energy of the gas. An increase in the 15

16 dx p A p1 p p2 dw = pdv V1 dv V2 Figure 3.1: A piston at work internal energy of the gas is an increase in the amount of energy it has that is available to do work. As a consequence, the differential rate of doing work by the gas on the piston is but pv = mrt so and noting that x 1 dx/dt = d lnx/dt dw dt dw dt dw dt = pa dx dt = pdv dt = p dv dt = mrt V dv dt = mrt d lnv dt The amount of work that is done by the gas in the piston is the amount of area under the curve. or, at constant temperature DW = DW = mrt A similar derivative can be applied to the heating DQ = Z V2 V 1 Z t Z t pdv d lnv dt 0 dt 0 dq dt dt0

17 Thus, the change in internal energy is DU = Z t 0 dq dt dt0 mrt Z t 0 d lnv dt 0 dt 0 (3.4) Heating adds to internal energy so that it can do work through expansion, which allows the internal energy to relax. An important point to be aware of is that the total amount of internal energy we have DU is a consequence of a time integral of heating and doing work. The extensive property of how much we have (e.g. U or N) depends on past flows. The present is linked to the changes in the past. Main points Total energy is conserved, meaning it can only be shifted from one form to another When energy flows into a system, two things can happen. One is that the internal energy can go up. The other is that the system can do work on something else. Work, heating, and changes to internal energy happen over time. Current states evolve from past states. Question When a system does work on something else, where does that energy go? Is one system s work another systems heat? 3.1 Atmospheric Application: the general circulation Since in the atmosphere we treat most things per unit mass, we can rewrite the 1st law in terms of Jkg 1. du dt = dq dw (3.5) dt dt since a = V /m dw = pda du dt = dq dt Thus, u is an intensive variable, as it independent of the amount of air that is considered. Substituting p = RT /a, the change in internal energy is p da dt (3.6) du dt = dq dt RT a da dt = dq dt RT d lna dt Du = Dq Dw = Dq RT 17 Z t 0 d lna dt 0 dt (3.7)

18 Figure 3.2: Balance between incoming shortwave radiation and outgoing longwave radiation at the top of the atmosphere Within the context of Figure 3.2 and Eq. 3.4, we can think about why the atmosphere is able to do work. Due to an imbalance between incoming shortwave radiation and outgoing longwave radiation, there is net heating at the equator and there is a net cooling at the poles. Presumably these two things balance in total, so that the internal energy of the atmosphere as a whole does not change, so Du = 0 and Dq = RT Z t 0 d lna dt 0 dt 0 (3.8) This equation pretty much sums up climate! Radiative heating Dq permits work to be done through the expansive processes of convection and winds R t d lnv 0 dt 0 dt 0. Cooling at the poles leads to work done on the poles through a shrinking. Meanwhile, the globally-averaged atmospheric temperature T stays constant. Energy that is available to do work takes the form of meridional motions that we call the wind. Energy will not be created or destroyed, but it will be transported from the equators to the poles because a high energy density is maintained by the imbalance at the equator and a low energy density is maintained at the poles. The one thing that is missing in this discussion is that air does not leave the poles once it gets there, so there must be a return circulation. 18

19 4 Internal energy in an ideal gas What if the internal energy is not constant when a gas is heated? In terms of mass, Eq can be expressed as u trans = 3 2 RT where the number three represents the number of degrees of freedom associated with the kinetic energy in the x, y, and z directions. By extension, the total internal energy per kilogram is u = f RT (4.1) 2 where f is the number of degrees of freedom for a molecule. The total internal energy is not just the energy derived from the translational motions of the individual gas molecules, but also from internal motions by the molecules. Now this is central to atmospheric sciences because it allows us to express how the temperature of gas is linked to how much we heat it (through, for example, radiative absorption), i.e. the specific heat. Remember, heating doesn t change extensive quantities associated with how much of a particular thing we have (e.g mass or number of molecules). What it does change is the intensive property of the amount of internal energy per unit stuff, or equivalently, the temperature. But, by how much? From the First Law of thermodynamics (Eq. 3.6), du dt = dq dt Rearranging dq dt = du dt + dw = du d lna + RT dt dt dt If a substance that is defined by the intensive property u is heated at constant specific volume a (e.g,. we add heat to a box filled with air, then d lna/dt = 0, and the internal energy change is equal to du dq = dt dt a We would expect the temperature to go up in the box at some rate dt/dt. Dividing both sides by dt/dt, we get du dq = = c v dt dt a where c v is what we call the specific heat at constant volume (or, equivalently, constant a). Note that it is much more intuitive to think of it in inverse dt = 1 dq c v Framed this way, c v tells us how much the temperature will rise for a given amount of heating. Now, we can substitute for u, our expression from Eq. 4.1, leading to q du c v = = = f T dt 2 R (4.2) a 19 dw dt

20 Figure 4.1: Heating at constant volume leads to a temperature and internal energy rise Just by inspection, we can see that, if we add energy to a volume through heating, and the volume doesn t change, then the temperature will rise. But it will not rise by as much as might be expected: there are other degrees of freedom in the substance that get a share of the energy. f is at least three due to the translational modes of the molecules. If f is more then temperature must rise even less. If you were a French engineer in the 19th century, this would be rather important to know because increasing temperature is what is going to drive the piston, and you would want to know what the increase in temperature is for a given release of energy from burning coal. Since the engineers needed to figure out how much temperature (or internal energy rises), this led them to the concept of the specific heat at constant volume c v. From an atmospheric science standpoint, radiative heating drives temperature perturbations, although it is not quite the same because the perturbations are not normally considered at constant volume (unless we are considering the planet as a whole), but rather at constant pressure (see the section on enthalpy). Still there is a fundamental problem, as will be shown below, which is that there is basic disagreement between what classical physics would predict the value of f to be and the value of c v that is actually measured experimentally. In fact, this was a key point that led 19th century physicists to be concerned that the wonderful successes of the kinetic theory of gases weren t all that. In commenting on this problem in 1869 Maxwell said in a lecture I have now put before you what I consider to be the greatest difficulty yet encountered by the molecular theory, a conundrum that paved the way for the quantum theory of matter. 1. Monotomic Molecule (example Ar) 3 translational degrees of freedom f = 3,U tot = 3 2 R T. Value of f for Ar implied by observations at atmospheric temperatures f = Diatomic molecule (examples: N 2 and O 2 ) 20

21 3 translational degrees of freedom 2 rotational degrees of freedom (whole body) 2 vibrational degrees of freedom (P.E. plus K.E. (kt/2 each) associated with change in bond length) 7 degrees of freedom f = 7,U tot = 7 2 R T. Value of f for N 2 and O 2 implied by observations at atmospheric temperatures f = 5. There is a big discrepancy here! The classical theory works great for Argon but terrible for nitrogen and oxygen. Air is mostly a diatomic gas (N 2,O 2 ), so for per kilogram of gas u = 5 2 RT = c vt (4.3) where we would expect u = 7RT /2 based on the kinetic theory of gases. A curious thing happens though which is that if the temperature of the gas goes up a lot, say to several thousand degrees, then f does indeed approach a value of 7 as initially predicted. How can f be a function of temperature? Nitrogen is nitrogen is nitrogen, no matter the temperature. Right? Off hand, it makes no sense. One of the first successes of the quantum theory was to resolve this problem. The value kt/2 (like RT /2) is a fine representation of the kinetic energy per independent mode provided that kt is not much less than the smallest possible energy, the quantum energy hn, where n is the oscillation frequency for that mode! Here, h is the Planck constant. Vibrational modes are higher energy than rotational or translational modes. What happens is that if the natural frequency of a vibrational mode n is sufficiently high that kt hn then these modes simply will not respond to an addition of energy to the gas. Effectively these vibrational modes are frozen because the bonds are too stiff. If the temperature gets sufficiently high, then these modes can be activated again. Otherwise, these degrees of freedom are unavailable at normal atmospheric temperatures. Our atmosphere is composed primarily of nitrogen and oxygen, both of which have stiff bonds with high natural vibration frequencies. These vibrational modes are simply too high potential energy to interact with the translational modes we sense as temperature. They are way up in the hills while the translational modes live in the valley. In any case, from Eq. 4.3, we can now rewrite our expression for the first law to be dq T dt = c v a + p a Heating leads to some combination of a temperature rise at constant volume and an expansion at constant temperature Main points Heating doesn t all go into expansion work. Some goes into raising the temperature T (4.4) Energy is shared among all available degrees of freedom. The more degrees of freedom the less temperature rises. Air has 5 degrees of freedom, three translational and two rotational. 21

22 Question Eq. 4.4 is not actually used very often in the atmospheric sciences to examine temperature changes in an air parcel. What is a limitation for applying the equation to the atmosphere that wouldn t have been a concern of most 19th century engineers and physicists? 22

23 5 Total energy or the Enthalpy So now we can think of there being two types of energy associated with the volume of a gas. There is the internal energy per unit mass u that is due to the rotational and translational motions of the molecules. Since each degree of freedom has a potential energy RT /2, then we could generalize even further to say that the internal energy is the sum of all internal potentials of the system u = f  i=1 g int i (5.1) where g int i is each internal potential. There is also energy per unit mass associated with the pressure of the molecules on their surroundings a p. The total energy is normally called the enthalpy h = u + a p (5.2) This pressure volume energy is sometimes referred to as an external potential. Thus h = f  i=1 g int i + g ext (5.3) One way to think about this is that the external potential is the potential for flows that we can see, and the internal potential is for those we can t. For an ideal gas, since pa = RT h = c v T + RT or h =(c v + R)T = f + 2 RT (5.4) 2 where f is the number of degrees of freedom for the molecules. For air f = 5, so dh dt = 7 2 RdT dt (5.5) To see why we have introduced the enthalpy, let s go back to the first law again (Eq. 3.7) dq = du+ dw = du+ RT d lna (5.6) Now, previously we evaluate ( q/ T) a to obtain the specific heat at constant volume c v (Eq. 4.2). What about the specific heat at constant pressure? From Eq. 5.6, taking the derivative with respect to temperature: q = T p u + RT T p but, since a = RT /p then (d lna) p =(d lnt ) p, so q = T p lna T u lnt + RT T p T p 23 p

24 Recognizing that u = c v T and RT d lnt = RdT, we get q = c v + R (5.7) T p This is the same expression dervied in Eq. 5.4! Thus dh dt = q p (5.8) This is why we introduced the term enthalpy. It is because heating at constant pressure increases the enthalpy of a substance. This result is important because we can define the specific heat at constant pressure to be Thus c p = dh dt = q T p (5.9) h =(c v + R)T = c p T (5.10) Therefore, it is easy to see that c p = 7 2 R. For dry air, R = 287J/deg/kg, c v = 717J/deg/kg, so c p = 1004J/deg/kg. For water vapor R = 461J/deg/kg, c v = 1463J/deg/kg, so c p = 1952J/deg/kg. For air, the internal energy has five degrees of freedom and pressure two (due to the bounce). Thus c p : c v : R ' 7 : 5 : 2. This very important, because what it means that if we add energy to air, five parts of seven go into increasing the internal energy density, and two parts in seven go into increasing the pressure. The reason we have introduce enthalpy as well as heating, even if they change at the same rate, is because enthalpy is a property of the system (or a state variable) where as heat is not. You cannot point to the heat of a substance but you can point to its enthalpy. Enthalpy is one of the most important variables in atmospheric sciences, because we cannot control for the volume of an air parcel, but we can control for its pressure. Combining Eq with Eq. 3.6, and noting that d (pa) = pda + adp gives an alternative expression of the First Law dq T dt = c p p a p T If there is heating the temperature rises and the pressure drops. Main points The enthalpy is the internal energy plus the energy the system exerts on its surroundings The energy exerted on the surroundings is a pressure volume energy (5.11) The specific heat is related to the number of degrees of freedom associated with the internal energy plus the number of degrees of freedom associated with the pressure The total number of degrees of freedom is = 7. Therefore c p : c v : R = 7:5:2 for air. 24

25 Questions Why are we generally more interested in the enthalpy of an air parcel than its internal energy? 5.1 Atmospheric Application: the dry and moist static energy In the atmosphere, we need to consider that there are other forms of internal energy that a parcel of air can have, other than just those associated with translational, vibrational and translational motions of a gas. A second form of internal energy is gravitational energy. In our atmosphere, air can move up and down. Thus, add to the total internal energy the gravitational potential per unit mass u = Âg int i = c v T + gz (5.12) i Adding pa = RT to this we get a revised enthalpy called the dry static energy so heating at constant pressure yields q h d = gz + c v T + RT = gz + c p T (5.13) p = dh d dt dt = c p dt + gdz dt (5.14) A second form of internal energy is that associated with the thermodynamic phase of a substance. The latent heat per unit mass associated with a change in phase from liquid to vapor is L v = J/kg at 0 degrees C. Thus the energy associated with some fraction of dry air being in the vapor phase is L v w where w = m v /m d (Eq. 1.22). Thus, adding the latent heat to the total internal energy, we get u = c v T + gz + L v w (5.15) which leads us to the moist static energy (MSE in Wallace and Hobbs) h m = c p T + gz + L v w (5.16) so q = dh m dt = c p p dt dt + gdz dt + L dw v (5.17) dt Of course, we could keep going and specify any number of things depending on the question we were interested. As shown in Fig. 3.2, heating can be related to the radiative flux divergence through 1 F net q = (5.18) r z where F net is the net flux (up - down) of radiation in the positive vertical direction z. The energy that is available to create changes in T, z and w comes from the absorption and emission of radiation. The conversion of radiative flux divergence to a change in h m is the starting point for radiative and climate processes. Questions Why is there a negative sign in Eq. 5.18? 25 p p

26 6 Equilibrium solutions Very often we want to consider situations that are adiabatic, which strictly means that the internal energy and enthalpy do not change because there is no heating. But this is a bit boring, and in fact it is never possible (as will be shown). All objects radiate for example. What we look at far more frequently is equilibrium conditions where energetic flows are in balance. Most of the textbook equations in atmospheric sciences are derived from this principle. The starting point is three expressions of the first law (Eq. 3.8), all basically equivalent. The first expression is or, if the work is done by an ideal gas. du dt = Dq Dt dw dt (6.1) And from Eq du dt = Dq Dt p da dt dh dt = Dq Dt + a dp dt So again, as in Eq. 5.8, if energy is added to a parcel of air at constant pressure, it s enthalpy will go up by the same amount. 6.1 Case 1: Equilibrium solution for a materially closed system at constant entropy As will be shown, constant entropy s is a short hand for no net heating of the system, i.e. is is adiabatic, but work is being done on or by the system. In this case, from Eq. 6.1, we obtain u = s w s (6.2) (6.3) (6.4) or, from Eq. 5.1, Main points Âi g int i = s w s (6.5) For a materially closed system, the adiabatic or equilibrium condition is that work done on a system raises its internal or potential energy. Work done by the system lowers it. Alternatively, a rise in the potential energy can be balanced by an ability to do work on something else. Potential energy is an ability to do work. 26

27 6.1.1 Atmospheric application: the geopotential What is the potential energy of the atmosphere? There are many possible forms of potential energy but here we look at the gravitational potential. Remember from Eq. 1.2 that work is required to move a system a distance through a force field. Here, the force field is F g = mg, and the distance might be dz. Thus, the work done on the system to increase its height, per unit mass is gdz. At constant temperature, from Eq. 5.12, du = gdz. Thus, we can term a potential energy that is the amount of energy required to lift a mass m through a gravitation acceleration g through a distance dz. The intensive variable is the geopotential, which has units of energy per mass F(z)= Z z 0 g(z)dz Basically, the geopotential is the potential energy per unit mass that comes from lifting through a variable gravitational field at constant temperature (see Table 3.1 in Wallace and Hobbs). For most purposes, we can assume gravity is a constant, so that F(z)=g or equivalently, we can take the derivative to get Z z 0 dz df(t )=gdz (6.6) 6.2 Case 2: Equilibrium solution for a materially closed system at constant pressure Now we can consider a second system in which net heating of the system at constant pressure is zero. Energy may be entering the system through heating, and leaving the system through dissipation or cooling, but there is no net flux convergence or divergence. In other words, from Eq. 6.3: or, from Eq. 5.3, d  i dh dt = g int i dt q = 0 (6.7) p + g ext i = q p Transformations can still occur however, but they must obey the adiabatic condition above. Adiabatic here doesn t mean no heating, but rather no net heating. Main points (6.8) For a materially closed system at constant pressure, the adiabatic or equilibrium condition is that energetic flows in must be balanced by energetic flows out, such that total energy per unit mass h (in whatever form) does not change. The different components of total energy can still vary, but there must be a balance between so that gains in one are offset by loss of another 27

28 Figure 6.1: An abstract illustration of where heating of a system is balanced by an equivalent rate of dissipation or cooling, so that all primary system state variables stay fixed Atmospheric application: the dry adiabatic lapse rate For example, if the air is dry, and we consider the dry static energy, a change in h d at equilibrium is given by (Eq. 5.14) dh d dt = c p dt dt + gdz dt = 0 A parcel that rises or falls without exchanging energy with its surroundings at a given pressure will have dh d /dt = 0 and will change its temperature at rate T c p = g or p z p T = g (6.9) z p c p This is the dry adiabatic lapse rate of about G d = g/c p = 9.8 K/km. Of course we could do something similar for the moist static energy, but the derivation is more complicated (and outlined in Wallace and Hobbs problem 3.50). Note that c p is not a function of pressure, and g is only a very very weak function of pressure (the atmosphere largely disappears by the time we care), so to a very close approximation: dt dz = g c p 28

29 Figure 6.2: Heating at constant pressure leads to an enthalpy or total energy rise 6.3 Case 3: Equilibrium in a materially closed system at constant temperature The final condition is constant temperature. A system doesn t have to be adiabatic to be in equilibrium. Alternatively, the system might be heated, but it sufficiently fast that the temperature, or energy per degree of freedom, stays constant. In this way heating balances working. From Eq. 6.1 q T = w T = p a Substituting the ideal gas law and the relationship between work and potential Âi g i lna = RT T T or Âi g i ln p = RT T T or T (6.10) (6.11) (6.12) Âi g i = a T p T (6.13) 29

30 Figure 6.3: An abstract illustration of where heating of a system leads to working by a system while the internal energy or enthalpy stay fixed so we have a range of relationship now q w Âi g i lna = = = RT T Main points T T T ln p = RT T a p = p = a T (6.14) For a materially closed system at constant temperature, the adiabatic or equilibrium condition is that energetic flows in must be balanced by expansion work, such that total energy per unit mass h or u (in whatever form) does not change. If temperature does not change, then net heating leads to an increase in volume or a drop of pressure Atmospheric application: the hydrostatic equation From the first law (Eq. 3.7) q u w = + T T T Since the total internal energy is given by u = c p T + gz (Eq. 5.13), at constant temperature du = gdz. And, from Eq. 6.14, we get If there is no heating, then q T = g z T a gdz T = adp T 30 p T T

31 Figure 6.4: Heating at constant temperature leads to an equivalent amount of working since a = 1/r, we obtain the equation p = rg z T This equation might best be interpreted as Dp = rgdz which suggests that pressure of a layer of fluid of constant density increases with its depth. But it the expression dp/dz also implies a pressure gradient which would drive motions of the air. In this case it suggest that the pressure increases with height which would require that the atmosphere falls down because it is being pushed from above. But, the atmosphere doesn t fall down. There must be a balancing pressure pushing upwards as shown in Fig No vertical motions requires that the total vertical pressure gradient at any location is zero. By changing sign we obtain the hydrostatic equation p = rg (6.15) z T Since density is always positive, the air pressure must decrease with height Atmospheric application: atmospheric thickness We can now substitute the ideal gas law. Assuming that the atmospheric layer is isothermal (i.e. constant temperature), then the density is (Eq. 1.24) r = p(z) RT = p(z) R d T v 31

32 Column with unit cross-sectional area δz z δp gρδz Pressure = p + δp Pressure = p Ground Figure 6.5: Fig 3.1 from Wallace and Hobbs. A downward gravitational pressure must be balanced by an upward pressure in order to maintain the air mass in hydrostatic equilibrium Thus, substituting we get p = p(z)g (6.16) z T R d T v Suppose we want to get the amount of potential energy per unit mass found between two layers, we are looking for a geopotential difference F 2 F 1. This is useful, because air flows from high to low potential energy, so we can use this difference in two locations to figure out where the air will go. How do we find out this quantity? Re-arranging dp p = gdz R d T v Since at constant temperature df = gdz (Eq. 6.6) and dp/p = d ln p R d T v d ln p = gdz (6.17) Integrating between two levels Z z2 Z F2 Z p2 Z p1 gdz = df = R d T v d ln p = R d T v d ln p z 1 F 1 p 1 p 2 Giving us DF = F 2 F 1 = R d T v ln p 1 p 2 (6.18) 32

33 This is really important because differences in DF gives us the amount of energy per unity mass that is available to do thermodynamic work through atmospheric flows. It is one possible form of the potential g we described earlier. Note that if we integrate Eq and change sign to account for the hydrostatic equation we can also obtain Eq DF = gdz = R d T v ln p 1 p 2 (6.19) Hurricanes are driven by a net heating of an atmospheric column, through a flux of energy from the ocean and a radiative cooling to space. The heating permits work to be done by way of an expansion of the column and a pressure drop. An expression that is related to Eq is to talk about the thickness of an atmospheric layer between two pressure surfaces. To get this we just divide Eq by gravity to get DZ = Z 2 Z 1 = R dt v g ln p 1 p 2 (6.20) It is easy to see from this equation a very intuitive result. A warm air column with higher T v will be thicker than a cold air column, provided that the column is defined as lying between two pressure surfaces p 1 and p 2. Warm air is less dense. 33

34 7 Quasi-equilibrium transitions Figure 7.1: An abstract illustration of transitions from one equilibrium state to another We have defined a two important equilibrium conditions. The first is one is heating at constant pressure and dq = dh (7.1) dt p dt Assuming equilibrium, we were able to derive, for example the dry adiabatic lapse rate ( T/ z) p = g/c p. The second sense is that there is heating at constant temperature is balanced by working through a change in pressure at constant temperature. This leads to the following key result q = T w = T Âi g i = RT T lna T ln p = RT T = p a = a T From which we were able to show that heating increases the atmospheric thickness. Now we ask what will happen to the relationship between heating and enthalpy if the pressure is changed a little bit for the constant pressure condition (Eq. 7.2) or what will happen to the p (7.2) T 34

35 relationship between heating and working if the temperature is changed a little bit for the constant temperature condition (Eq. 7.2). Effectively, we are asking what is the sensitivity of these equilibrium conditions to small changes. We are still looking for an equilibrium condition. However, we now want to know how the equilibrium solution is different at different temperatures. Effectively what we are seeking is an expression for quasi-equilibrium transitions. Conditions are changing slowly enough that something very close to equilibrium is always maintained. 7.1 Case 1: An ideal gas Suppose for example, there is no heating at constant pressure, then, from Eq. 7.1 and Eq. 5.11, c p dt p adp T = 0 But if pressure is constant then we only get the trivial solution that temperature is constant too. Not very interesting. But, let s integrate to see how a very small change in pressure would relate to a very small change in temperature at equilibrium: c p (T T 0 )=a (p p 0 ) Now, we keep pressure p 0 constant as we promised, and T 0 too, but now we take the derivative of a small change with respect to temperature We end up with the expression c p d (T T 0 ) dt = a d (p p 0) dt c p = a dp T dt p (7.3) that could have obtained easily before, but how we got there matters as will become more clear below. Since a = RT /p d ln p T = c p (7.4) d lnt p R or more generally from Eq. 5.9 d ln p T d lnt p = q T p = dh d(pa) d (pa) dt Both the numerator and the denominator on the RHS are related to the number of degrees of freedom in the system. Eq. 7.5 is a bit of a master equation that we can take to almost any situation. It s really useful. (7.5) 35

36 7.1.1 Atmospheric application: potential temperature Integrating Eq. 7.4 from initial state i to final state f gives us T f = T i p f p i R/cp We can use the above equation to derive an expression for the potential temperature q, which represents the temperature a parcel would have if it were brought adiabatically from from (p,t ) to standard pressure p 0, which is typically set to 1000 mb. In this case R/cp p0 q = T p The potential temperature is an important concept as it represents a line of thermodynamic equilibrium for dry air in which there is no net exchange of energy with the environment through radiative absorption or emission, or through turbulent mixing, such that dh d = 0. In quasi-equilibrium transitions an air parcel may rise or sink, or move around horizontally, simultaneously changing both its temperature and pressure. But if there is no energy exchange, q is constant. Effectively, all parcels lying along a constant theta surface are in local thermodynamic equilibrium (Fig. 7.3). For example, suppose an air parcel rises from the surface at 1000 mb to 700 mb without mixing with its surroundings. If it s surface temperature is 25 C, what is its final temperature? T f = T i p f p i R/cp = 269K Actually this equation can be related to the dry adiabat on a Skew T log p plot, if we consider from above that lnt = R/c p ln p + const On the Skew T log p plot ln p is the ordinate. Since T not lnt is the abscissa, adiabats are not straight but slightly curved from the lower right to upper left of the diagram Dry adiabats on a Skew-T are the same as lines of constant q. It is interesting to look at potential climatologically too since air motions tend to follow surfaces of constant potential temperature. This is often used in air quality modeling to see where air is going or coming from (look up HYSPLIT for example). At least, the potential temperature is very useful for being a tracer of air motions in systems that are not changing rapidly. What do we mean by rapidly? Well, consider the following important timescales in the atmosphere. Free troposphere with no condensational processes 1 to 2 weeks is the timescale for radiative cooling to have an influence Synoptic disturbances: timescale 1 to 2 weeks the same as the above, which is not a coincidence since radiative cooling and heating drive atmospheric motions Moist convection: timescales minutes to hours The implication is that an assumption of thermodynamic equilibrium, and motion along constant q surfaces is a useful guide over shorter time scales of perhaps a few days, provide there is no moist convection. Where would this work? Not the tropics! 36

37 7.2 Case 2: Phase changes Figure 7.2: A Skew T diagram for SLC But what about phase changes where changes are more abrupt? What we have discussed thus far is the internal energy associated with sensible heat, i.e. the energy that can be associated with m < v 2 /2 > temperature T. A second form of internal energy is that associated with the thermodynamic phase of a substance. It plays an extremely important role in the atmosphere because water is found in the liquid, ice and vapor phase, and rapidly changes between all three, releasing or absorbing vast amounts of energy. From Eq. 7.2, the equilibrium solution is that the change in volume for a given heating is equal to inverse of the current pressure a = 1 q T p When we look at phase changes, we are concerned with the amount of heat that must be applied to a liquid to cause it go from a dense liquid phase to a gas that occupies much more space per molecules. Because we are looking at a jump, since this is a phase change and the difference between a gas and a liquid is very large, we are interested in D differences. Thus, the relationship is no longer a differential, but rather Da = 1 Dq T p 37

38 Figure 7.3: Zonal mean temperature and potential temperatures in the troposphere (From Houghton, 3rd Ed.). The amount of heat that must be applied to enable a phase change is the latent heat of vaporization Dq = L v At equilibrium, the pressure in the gas and liquid phase are equal. Taking the pressure in the vapor phase, p v a v = R v T, so Da T = L va v R v T How does this equilibrium expression for constant temperature change if the temperature changes a small amount? Taking the derivative with respect to temperature and keeping the pressure fixed dda T dt p = L va v R v T 2 Here, we make a simplification. For specific volume, the difference between vapor and liquid is Da = a v a l = 1 r v 1 r l But the density of vapor is typically tiny compared to that of liquid, by about a factor of 1000, so Da ' 1 r v = a v Thus, the expression for the quasi-equilibrium transition to another temperature becomes or da v T dt p = L va v R v T 2 d lna v T d lnt p = L v R v T 38 (7.6)

39 Normally, this is expressed a little differently. Recognizing that d lna v T = d ln p T, we get d ln p T d lnt p = L v R v T (7.7) which again is similar to Eq. 7.5, except d ln p T = Dh (7.8) d lnt p pa v Main points We are often not only interested in the equilibrium solution at constant pressure or temperature, but how the equilibrium solution changes as pressure or temperature change. The quasi-equilibrium solution for an ideal gas is related to the ratio of the gas constant to the specific heat at constant pressure The quasi-equilibrium solution for a phase change is related to the ratio of the the gas constant times the temperature to the latent heat Either expression is dimensionless Question What are the similarities between Eq. 7.5 and Eq. 7.7? Can you get to Eq. 7.7 directly from Eq. 7.5? Atmospheric application: saturation vapor pressure of water The famous Clausius Clapeyron equation gives the sensitivity of the saturation (equilibrium) vapor pressure over a plane surface of pure water e S (T ). From Eq. 7.7, we have d lne S d lnt = L v R v T One would be tempted to derive a solution for this equation as Lv 1 1 e S (T )=e S0 (T )exp T 0 T R v (7.9) (7.10) At freezing the latent heat is J kg 1 and R v = 461 J/K/kg. This is fine, but only if the difference between T and T 0 is small. The expression d ln p/d lnt is extraordinarily powerful, but it must not be solved without care, because it is a sensitivity and other things that we have assumed to be constant might be changing with temperature also. The latent heat L v does vary somewhat by a few percent over the normal range of atmospheric temperatures, and this does bad things to the accuracy of our nice exponential solution for e(t ). A more accurate solution takes into account a solution for L v (T ). 39

40 e s (T )=611.2exp( 17.67T T ) where, T is in C. The reason for the slightly different form of the equation is that L v itself is a weak function of temperature e s (mb) T (C) Figure 7.4: Clausius-Clapeyron Equation for water vapor over liquid water Often you will hear the expression that at warm temperatures the air can hold more water vapor. Certainly from the C-C equation this seems to be correct, but semantically it is misleading. Nowhere in our discussion of the derivation have we needed to mention air at all. Rather, the C-C equation would hold even in the absence of dry air. That said, in our atmosphere at least, having air present is what enables the temperature to be as high as it is, since it is molecular collisions that enable radiative absorption to show up as a measurable temperature. 40

41 8 Equilibrium distributions Eq. 7.5 seems to imply that it is possible to experience a wide range on pressures and temperatures at equilibrium. This ends up being really important because it leads to distributions, something that is explored throughout the sciences. It is very likely that you will spend much of your career looking at distributions of one sort or another. It may help immensely if you are able to understand their origins. There are two types of distributions that are commonly discussed, power laws and exponential distributions. These show up everywhere. Certainly throughout all of atmospheric sciences. But even your social networks will follow one of these two distributions. Note for example that where we asssumed zero net heating at constant pressure, we obtained d ln p T = c p d lnt p R This could be plotted on a log-log plot with T on the x-axis and p on the y-axis, and for all combinations of pressure and temperature there would be a straight line and the slope would be c p /R or 7/2 for dry air. Integrating would yield p = T c p/r + const. which, of course, is the potential temperature expression. Alternatively, where we looked at zero at heating at constant temperature, and there was a big jump, we obtained the expression d ln p T = L v d lnt p R v T This would not be suitable for a log-log plot because T would be part of the slope. But, we do get d ln p T d (R v T ) p = L v (R v T ) 2 which suggests a log-linear plot. Integrating we get Lv p = p 0 exp exp R v T 0 Lv R v T which is an exponential distribution. Plotting logp versus T would yield a straight line with a slope of L V /R v. This is great, but we can take this even further. What is d ln p T? Here, it helps to think of the ideal gas equation in the molecular form. We have already stated that p = rrt where RT = g ext is a potential per unit mass. From Eq. 1.14, we could write the analogous expression in number form: p = nkt where µ ext = kt. Thus d ln p T d lnn µ ext d lnt p d ln µ ext n where n is the number density. The suggestion is that we now think of size distributions with a number concentration n on the y-axis and a potential energy available to do work µ ext on the x-axis. There are three basic types of size distributions to consider. 41

42 8.1 Log-log plot with a positive slope: Power Laws at constant temperature This type of distribution describes quasi-equilibrium transitions where the jump is small, as we showed when deriving the potential temperature distribution from which can be written otherwise as d ln p T = c p d lnt p R d lnn µ ext d ln µ ext n = dµtot dµ ext where µ tot = µ int + µ ext. The slope is dµ tot /dµ ext, which for air is (c v + R)/R = 7/2. The general solution is n = n 0 (µ/µ 0 ) dµtot /dµ ext + const. (8.1) 8.2 Log-linear plot with a negative slope. The Boltzmann distribution when jumps are large For quasi-equilibrium transitions where the jump is big, we obtained the equation for phase changes But, we do get d ln p T = L v d (R v T ) p (R v T ) 2 which can be generalized as d lnn µ ext dµ ext n = Dµtot µ ext (µ ext ) 2 the solution for the size distribution is n µ ext = n 0 exp Dµ tot µ ext µ ext The slope on a log-linear plot of n versus Dµ tot is 1/µ ext. Expressed in terms of mass Dg tot g ext r g ext = r 0 exp What we are expressing here is how the number concentration changes when one potential changes and the other is fixed. One example that has already been derived is the saturation vapor pressure (Eq. 7.10). g ext Atmospheric example: the hypsometric equation How does the density of air molecules at one level compare to the air density at another level? If g ext = RT v is fixed and Dg tot T = gdz is the potential that is changing, then gdz r T = r 0 exp (8.4) RT v 42 (8.2) (8.3)

43 This is the hypsometric equation. Assuming the temperature of the atmosphere is constant, the density of the air drops of exponentially with height. Since we are assuming that temperature is constant, it follows too that Very often, one sees the expression p T = p 0 exp p T = p 0 exp gdz RT v Dz H where H = RT v /g is the Scale Height. The scale height is about 8 km at mid-latitudes Question Is the atmospheric temperature constant? Is it approximately so, at least in the troposphere? 8.3 Log-Log plot with a negative slope. Power Laws at constant pressure For internal distributions within a system that is at equilibrium (i.e. not one where there are quasiequilibrium transitions), we also get a power law. Since p = nµ ext, we can take the derivative with respect to time to obtain (8.5) (8.6) dp dt = d (nµext ) dt n µ = µ ext ext + n (8.7) µ ext n If the pressure is constant and nothing is changing then dp/dt = 0 and The solution is d lnn µ ext d ln µ ext n = 1 (8.8) n = n 0 (µ/µ 0 ) 1 + const. This is a power law. So, on a log-log plot of n versus µ ext the slope is -1 or n µ 1/µ ext. There can be a large number of low potential things or a low number of high potential things. 43

44 8.3.1 Atmospheric example: aerosol size distributions Figure 8.1: Junge aerosol size distribution (Junge, 1955) Normally, aerosol size distributions have what is known as a Jungian size distribution with a slope of -3 on a log-log plot, provided that the plot is n(r) where r is the aerosol radius. Thus d lnn d lnr = 3 44

45 or n µ 1/r 3. Note, however, that the volume is given byv µ r 3. Thus d lnn d ln µ = 1 d lnn 3 d lnr = 1 What does this tell us? Assuming that aerosols are formed by condensation, then we might assume (correctly) that the aerosol volume is related to an accumulated potential through the sum of all condensational work done. Question The equations for the power law and the Boltzmann distribution should be the same in the limit of small jumps. Can you show this? Main points Power laws and exponential distributions are observed everywhere in nature. They can be seen as a consequence of basic thermodynamic principles The distributions can be expressed as a function of many different variabiles, but most fundamentally as a number concentration versus some form of potential energy µ ext. Exponential distributions tend to show up when there are big jumps Power laws tend to show up when there are small changes. The slope is positive for quasiequilibrium transitions and negative with a slope of -1 at equilibrium. 9 Conservation of matter and energy From Eq. 6.7, we stated that the enthalpy (or total energy) of an adiabatic parcel does not change with time. We also alluded to the fact that the system is materially closed, requiring that the amount of matter in the system is also constant. This could be written generally as Âm i = const. i Âg int i + g ext i = const. i 9.1 Atmospheric application: Adiabatic liquid water content of a cloud The basic physics we need to consider for understanding the formation of cloud in a parcel that rises and falls adiabatically is that heat and energy in the parcel are conserved Q = c + w = const. (Conservation of mass) h m = c p T + gz + L v w = const. (Conservation of energy) 45

46 where w is the water vapor mixing ratio and c is the liquid water mixing ratio. Q is the total water content, the quantity that is conserved if the parcel does not gain or lose mass. In a pseudoadiabatic process, as the air expands, conservation of mass requires that the saturated mixing ratio decreases by an amount dw s as vapor is converted to condensed liquid. If the air remains saturated Q = c + w s (9.1) dq = 0 = dc + dw s (9.2) The liquid water content has units of g/m 3, so dc = dw s (9.3) LWC = rc (9.4) where r is the air density. Typical values range from 0.05 g/m 3 in thin stratus to in excess of 1g/m 3 in cumulus towers. In a saturated adiabatic process dh m = 0 = c p dt + gdz + L v dw s (9.5) Therefore L v dw s = c p dt + gdz dw s dz = c p L v apple dt dz + g c p or, from Eq. 9.3 and 9.4 dw s dz = dlwc dz c p L v (G d G s ) ' const. (9.6) = rc p L v (G d G s ) ' const. (9.7) So we have just shown that LWC increases approximately linearly with height. Lets try some typical values for stratocumulus G s ' G d = r ' 1 c p = 1004 L v = which gives us dlwc ' g/m 3 /m dz which is about right. A 200 m thick stratus cloud typically has a cloud top value of LWC of 0.3 g/m 3. 46

47 9.2 Atmospheric application: Saturated adiabatic lapse rate and the equivalent potential temperature But what is the moist adiabatic lapse rate G s? Again, in a moist adiabatic process where mass is also conserved dh m = c p dt + gdz + L v dw s = 0 We can use this to derive the saturated adiabatic lapse rate, G s dt dz = g + L v dw s c p c p dz but w s is a function of temperature and pressure since w s ' ee s (T )/p, where e s (T ) is given by the Clausius Clapeyron equation (Eq. 7.9) Thus ws dw s = p dw s dz = ws p de s dt = L ve s R v T 2 (9.8) T T ws dp+ T dp dz + ws T But the hydrostatic equation tells us that dp/dz = rg. Also G s = dt/dz dw s dz = ws ws rg G s (9.9) p T T p Equating this with Eq. 9.6, and solving for G s, we get G s = g c p 1 rl ws p 1 + L c p ws T T dt p And recognizing w s ' ee s (T )/p, and the C-C equation for e s (T ) we get G s = g c p p p dt dz 1 + Lw s R d T (9.10) 1 + el2 w s c p R d T 2 It is often useful to refer to a thermodynamic chart for a first guess. The point of learning them is to see how the essential ingredients of cloud physics can be combined: conservation of energy and mass, the hydrostatic equation, and the Clausius-Clapeyron equation for the saturation vapor pressure over a liquid surface. Another expression, that is useful, one that is derived in Wallace and Hobbs, is the Equivalent Potential Temperature, which is conserved along moist adiabats. As before when we said that it is equivalent to say 47

48 Here and dh d dt = 0, dq dt = 0 R/cp apple p0 ws L v q e = T exp p c p T apple ws L v q e = q exp c p T dh m = 0, dq e = 0 dt dt On the Skew-T plot shown in Fig. 7.2, moist adiabats are the purple lines. Note that they do not cool with height as rapidly as the orange dry adiabats. Note too that in the region around 600 mb where the dewpoint and the temperature are almost the same, the temperature nearly parallels the nearest moist adiabat line. This suggests a well-mixed cloud layer, perhaps altocumulus. 9.3 Atmospheric application: Stability We are going to try to address the problem of how a parcel of air rises within its environment when there is no exchange of energy or matter. The basic conceptual idea is that a parcel starts somewhere in the atmosphere with some updraft velocity, perhaps due to orographic lifting or localized convergence of some sort. As the parcel rises its further acceleration or deceleration is due to its temperature relative to its environment. If the parcel is warmer than its environment, a buoyancy force will accelerate the parcel upwards towards regions of similar low environmental density aloft. However, as the parcel rises, it cools. In an idealized model in which the parcel does not mix with its surroundings, it cools with height at the dry adiabatic lapse rate. Since the environment is cooling at slower rate with height, eventually the parcel is colder and less buoyant than its surroundings, and it begins to sink. What we will show is that an unstable dry atmosphere is characterized by dt/dz = G > G d or dq/dz < 0. A stable dry atmosphere is characterized by G < G d or dq/dz > 0 The physics is as follows: Within both the parcel and the environmental air, from the hydrostatic equation (Eq.??) dp dz = rg But the density r is different between the parcel and the environmental air. If r 0 and T 0, and r and T are the density and temperature of the parcel and the environment respectively. The upward positive force experienced by the parcel is r 0 r g per unit volume, or (r 0 r) g r 48

49 per unit mass. In other words, a parcel denser than its environment will accelerate downwards. Substituting the ideal gas equation, and assuming that the environmental temperature is close to the parcel temperature, we get a b = (T 0 T ) g T for the upward buoyancy force per mass (or acceleration). Figure 9.1: The period of the oscillations resulting from an initial perturbation is 2p/N Actually, we should really be taking into account the fact that humidity also contributes to the buoyancy of the parcel, and use the virtual temperature T v instead of the temperature T (Eq. 1.24) T v ' T ( w) where w is the mixing ratio. So the buoyancy acceleration is then a b = (T v 0 T v ) g This leads us to Eq in Wallace and Hobbs, since acceleration is the second time derivative of position d 2 z dt 2 = (T v 0 (z) T v (z)) g (9.11) T v (z) We can improve on this expression by noting that as the parcel goes up or down, its going to cool or warm at the dry adiabatic lapse rate G d. Thus T v T 0 v = T v0 G d z 0 where z 0 = z z 0 is the displacement from the equilibrium position. Similarly, the environmental temperature at the same level is given by T v = T v0 Gz 0 49

50 This leads to (Tv 0 T v ) = (T v0 G d z 0 ) (T v0 Gz 0 ) T v T v0 Gz 0 = (G d G)z 0 T v0 Gz 0 Provided that the vertical temperature perturbation is small compared to the actual temperature we get (Tv 0 T v ) = (G d G)z 0 T v T v0 Gz 0 ' (G d G)z 0 T v0 Substituting now into Eq. 9.11, we get d 2 z 0 dt 2 = (G d G)z 0 g (9.12) T v0 So notice here that if the displacement z 0 is positive and the parcel is above its equilibrium point, and G < G d (which is the normal case), then the RHS is negative because the air parcel is colder than its surroundings. The parcel has inertia, of course, but eventually the negative acceleration or buoyancy pushes the parcel back towards its equilibrium point. Again because it has inertia, it will overshoot the equilibrium, and the displacement will be negative, making the parcel positively buoyant, which accelerates the parcel upwards. And so on. If this sounds like an oscillation, this is true. We can rewrite Eq in the form of the wave equation d 2 z 0 dt 2 + N2 z 0 = 0 (9.13) where N 2 = g T v (G d G) (9.14) is the square of the Brunt-Vaisala frequency. This is a hugely important number in atmospheric sciences, as it is like the second hand in the atmospheric clock. Rule of thumb is that N = 0.01radianss 1. r 9.8 N = 255 (9.8 6) 10 3 = So if a full cycle is 2p radians then an oscillation period is t = 2p N ' 600s or about 10 minutes. Clouds are the easiest place to see this. They rise and decay roughly over this time scale, as they go through a full buoyancy oscillation. The solution to Eq is, of course, sinusoidal. If the amplititude of the wave is z 0 0, and the parcel is at peak positive displacement at t = 0, then in complex notation z 0 (t)=z 0 0e int = z 0 0 (cosnt isinnt) Looking just at the real component z 0 (t)=z 0 0 cosnt 50

51 The advantage of using the complex solution, is that something becomes very apparent for the special case that the atmosphere is absolutely unstable, in which case G > G d. From Eq r r g N = (G d G)= ( 1) g r g (G G d )=i (G G d ) T v T v T v where i = p 1 is the imaginary number. Since i 2 = 1, this then implies that apple r appler g g z 0 (t)=z 0 0e int = z 0 0 exp i 2 (G G d )t = z 0 0 exp (G G d )t T v T v Since the expression in the brackets is positive, any initial displacement upwards results in the parcel continuing to rise exponentially. There is no restoring force to bring the parcel back towards equilibrium. In terms of potential temperature Earlier, in an assignment, we derived the expression relating the dry static energy to the potential temperature dh d = c p T d lnq dt dt This can also be expressed in terms of a gradient with respect to height Since h d = c p T + gz, it follows that dh d dz = c pt d lnq dz dt c p dz + g = c pt d lnq dz Dividing by c p T and substituting expressions for G and G d d lnq dz = 1 T (dt dz + g c p )= 1 T (G d G) So dq dz = q T (G d G) This is very useful since we know that if G < G d then the atmosphere is stable and meaning that dq/dz > 0. So we can look at a sounding on a Skew-T plot and compare the temperature profile to the lines of constant potential temperature. If the temperature sounding is going to higher values of potential temperature q as it goes up with height, then the atmosphere is stable. Of course, if dq/dz < 0 then the atmosphere is unstable, and a storm is sure to happen because air must rise to restore stability, taking high q air up with it, and restoring the stable situation of dq/dz. Because the atmosphere is moist, the more general expression for moist static stability is dq e dz = q e T (G s G) > 0 51

52 It s actually very rare to see situations where dq/dz< 0 which are called absolutely unstable, since they cannot last long at all. However, we can see situations quite frequently when dq e /dz < 0. These are easily identified on an atmospheric sounding on a Skew-T plot by comparing the temperature sounding to the moist-adiabatic. We call such situations convectively unstable, because they lead to cloud formation. 52

53 Part II Non-equilibrium Thermodynamics 10 Second Law of Thermodynamics and Entropy Reversibility and the Second Law Figure 10.1: Transfer of heat from the system to its environment is spontaneous if entropy production is positive, requiring that the system has a higher temperature. We start with a definition. Imagine a closed system. The system is not isolated, but rather exists in within a universe, or perhaps to keep things on a more manageable scale, a bigger system. Lets say through a transfer of heat to the system, we change its state from (defined by its state variables) from S i to S f (S just being some arbitrary symbol defining the state). If by taking an equivalent amount of heat away from the system the system returns exactly to its original state S i, then the system is said to be reversible. Any process that cannot satisfy these requirements is said to be irreversible. An example of a reversible process might be an purely cyclical process like the buoyancy oscillation described by the wave equation (Eq. 9.13). However, the second law of thermodynamics states that all natural processes are irreversible. Ultimately, any wave is damped. It loses energy through friction or radiative cooling, and its amplitude must be restored through some other external forcing like orographic forcing or radiative heating. A corollary of the Second Law of Thermodynamics (which we shall not prove) is 53

54 Another: In a reversible transformation, heat can only be converted to work by moving heat from a warmer to a colder body In the absence of external work done on a body, heat can only move from warm to cold. For example, absorbing heat Q 2 from a cold reservoir and releasing Q 1 to a hot reservoir gives us a refrigerator, which requires an amount of work Q 1 Q 2 to have been done by some outside agency. We can term the change in entropy DS of the system during a transformation. or per unit mass DS = DQ rev T (10.1) Ds = Dq rev (10.2) T In a complete reversible process, the system is returned to its original state, so the entropy does not change and there is no change in the entropy of the surrounding universe (i.e. Ds = 0). Since all natural processes are irreversible, the total entropy of the universe always increases. SDS > 0 The zero in this case is simply a recognition that the net change of energy in the universe as a whole is always zero. The reason for the inequality is predicated on conservation of energy and the condition of an amount of energy DQ moving from a high potential, high temperature system to a low potential, low temperature environment. SDS = DQ T sys + DQ T env 0 The inequality holds provided T sys > T env. The inequality is zero under assumed reversible conditions of local thermodynamic equilibrium for which T sys = T env. Note that in the above example that it is possible for the entropy of the system to go down even while the entropy of the universe and the environment to go up. In general, the change in entropy (per unit mass) is ds = dq/t Under this assumption, the first law can be written or Tds= dh adp= c p dt adp (10.3) ds = c p d lnt Rd ln p (10.4) but we can show that from the equation for the potential temperature that d lnq = d lnt 54 R c p d ln p

55 from which we can see that ds = c p d lnq (10.5) Note, that in the presence of condensation it is better to use q e ds = c p d lnq e (10.6) If we are assuming adiabatic motions (which is sometimes a fair assumption in synoptic meteorology, particularly at high altitudes where phase changes don t affect entropy as much) then dq/t = ds = 0 and dq = 0. That is why when we say air moves along an isentropic surface (isoentropic = constant entropy), it is equivalent to saying it is moving along lines of constant q. For dry air in the absence or balance of diabatic heating (from the sun for example) and cooling (due to radiation of heat to outer space), this is a decent approximation Examples of changes in entropy 1. Increase the sensible heat of an object at constant pressure Tds= c p dt adp s lnt = c p p p or, another way this could be expressed, from Eq. 6.7 is s = 1 h = 1 q p T p T p (10.7) (10.8) Heating at constant pressure increases the entropy of the system. 2. Decrease in the pressure at constant temperature Tds= c p dt adp s ln p = R T p (10.9) This is very important because it means we can interpret the second law intuitively as a requirement that, absent some external force, the pressure must decrease with time. Remember that maintaining constant temperature requires that heating equals working, i.e. from Eq q T = w dt We showed previously in Eq that work is done when, at constant pressure, the volume goes up or the pressure goes down w = RT d ln p = RT d lna dt dt T 55 T

56 So you can see now that entropy goes up when heating enables work to be done so that the pressure goes down. s = 1 q = 1 w ln p = R (10.10) T T T T T T 3. Move mass at constant temperature and pressure. Entropy increases in a materially closed system as things get hotter at constant pressure, or if there is expansion at constant temperature. The only thing that remains to be changed is the amount of mass at constant temperature and pressure. If the system is materially open, so that matter can flow in and out of the constant T and p surface, this is just the same as saying S T,p = ( Q/) T,p T = ( H/) T,p T = c p m T,P = c pt ( m/) T,p T (10.11) So, entropy production, provided it is along a surface of constant pressure and temperature, is related to an increase in enthalpy due to an increase in mass! When you (a surface of constant temperature and pressure) get fatter, your entropy goes up not down! There is so much confusion here, since many (including some quite famous physicists) think of us as ordered structures, whose growth must correspond to a decrease in entropy. Quite the opposite! Looking at us, by ourselves, it is an increase in entropy. An example of how this works for phase changes is to write: S = ( Q/) T,p T = ( H/) T,p T T,p = L T m The entropy rises when there is latent heat release due to condensation Atmospheric Application: Entropy production in the atmosphere T,P (10.12) The three possibilities for entropy production described above can show up in many different ways in the atmosphere. Obvious examples are as follows. Along a constant pressure surface, the atmosphere responds to radiative heating by increasing molecular motions to give a rise in temperature. Along a constant temperature surface, hot air expands so that the pressure decreases. Locally, the entropy can increase due to convergence of winds to increase the air mass along a surface of constant temperature and pressure (or isentrope). Another possibility is that entropy is produced through cloud production, since latent heat is released through condensation. 11 Heat engines Nineteenth century engineers like Carnot and Watt were very interested in designing engines that efficiently burned coal in order to do work. These heat engines were the basis of the European 56

57 and later the American Industrial evolution. Carnot imagined an idealized form of the heat engine called the Carnot Cycle. In the atmospheric sciences, this cycle is particularly useful to help us imagine how the atmosphere behaves as a heat engine. Figure 11.1: Carnot cycle: heating does positive expansion work at constant temperature; cooling does negative contraction work; in between there is adiabatic expansion and compression. Fig shows an illustration of the Carnot cycle. There are four steps in the cycle. 1. From point A to B, there is some adiabatic process that raises the temperature by way of compression and an increase in pressure, without any increase in entropy. From Eq or, more generally c p DT = adp Dh = adp 2. Then, from point B to C, there is diabatic heating at a constant high temperature. Energy flows into the engine. From Eq we know this leads to positive work being done by way of expansion and a drop in pressure. Dq T = Dw = adp T The remaining steps deal with the return phases of the cycle. The same expressions apply, but there is a negative sign. 57

58 3. In step C to D, the heating stops, but there is continued expansion through a decrease in pressure so the temperature adjusts by falling to a lower value. 4. Finally, in step D to A, there is diabatic cooling at a constant low temperature. Energy flows out of the engine. negative work is done by the heat engine. Main points There are some key messages to remember here: In a complete cycle, the temperature does not change. The internal energy might change in any given leg of the cycle, but averaged over the course of a complete cycle, there is no change in internal energy Work done by the system in step B to C is offset by work done on the system in step D to A. Since the work is driven by heating at constant temperature, and there is no change in internal energy the total amount of work done by the system is Dw = ÂDw = ÂDq = Dq in Dq out Note that conservation applies: Dq in = Dq out + Dw. Only a portion of the energy that flows into the system is available to do work. This implies that we can express an engine efficiency h = Dw = Dq in Dq out (11.1) Dq in Dq in The Second Law prescribes that the overall entropy of the universe must increase for any process: T ds(universe) > du(system)+dw(system) Therefore, if the state of the system doesn t change (Du = 0), the second law says you must always put more heat Dq in = T Ds into a system than the amount of work Dw you get out of it - the efficiency of any process must be less than 100%. h < 1 (11.2) A Carnot Cycle ends up exactly where it started, as if nothing had happened. Entropy is created in the universe because there is a flow of heat from high to low temperature. The total increase in entropy is Ds = Dq out T 2 But, within the engine, there is no change in entropy at all. Thus, the Carnot Cycle is an idealization, albeit a very useful one. Work is only the type of energy that we are interested in it is your choice! There is only ever energy in and energy out, and these flows are always in balance. There is no objective definition of what counts as work. It may seem silly, but the exhaust of a car does work by forcing away surrounding air. From this perspective, any energy that is wasted in propelling the vehicle forward would be termed waste heat. 58 Dq in T 1

59 Adiabat Adiabat T 1 B C Isotherm Temperature, T T 2 A D Isotherm X Entropy, S Y Figure 11.2: The Carnot Cycle in S-T space, from Wallace and Hobbs. Figure 11.3: Illustration of the Carnot cycle Figures 11.2 and 11.3 show the Carnot Cycle graphically, arguably from a cleaner perspective. The goal of these figures is to set up two orthogonal co-ordinates, something we should always try to do when setting up a problem. In Figure 11.2, the cycle axes are set up in a space of entropy and temperature. The legs of the Carnot Cycle carve out a rectangle whose area is the work done. To see how, consider that the area of the rectangle is Area = (T 1 T 2 )(Ds) = T 1 Ds T 2 Ds = Dq in Dq out = Dw From this perspective, because a rectangle is carved out, the following axiom is clear A Carnot Cycle represents the maximum amount of work that can be done in a cycle that returns the system to its initial state. Figure 11.3 gives the Carnot Cycle a more 3D perspective in which the legs of the Carnot Cycle are not explicitly resolved, but instead are represented by circulations along isentropes. The circulations are sustained by energetic flows in and out of the system. Some fraction of the incoming flow goes to work Atmospheric Application: A hurricane In the atmosphere, we can easily think of the sun heating the Earth. This lowers the air s density, and then the heated airmass does work by expanding upwards or outwards. Meanwhile it radiatively cools to space, and then it contracts. Circulations are along isentropic surfaces. Crossisentropic flows are due to diabatic heating, as derived in an assignment question. A good example of this is a hurricane as shown in Fig The ocean heats the air at a high constant temperature, following which there is ascent along isentropes within the hurricane 59

60 Figure 11.4: The Carnot Cycle in a Hurricane. Note that point D above corresponds to point A in Figs and eye-wall. Ascent in the eye-wall really is nearly isentropic. There is cooling as the air expands. Then the hurricane cools efficiently to space through thermal radiation. Finally the cooled air mass sinks slowly and compresses quasi-isentropically. It is through this cycle of heating and cooling that the hurricane is able to perform thermodynamic work in the form of horizontal and vertical motions Atmospheric Application: The general circulation You eat food to do work, and the rest of the energy gets lost as heat. The Earth gets heated by the sun, and it uses this Dq in energy to do work in the form of atmospheric motions, with the remainder of the energy Dq out getting lost to space through thermal radiation at the top of the atmosphere. The efficiency in this case is only about 1 to 2% if work is defined in terms of a capacity to create atmospheric motions. 60

61 12 Available energy Let s imagine that there are two potential surface (like a high and low pressure) and that the total enthalpy in the system is H 1+2 = H 1 + H 2 = m 1 c p T + m 2 c p T and the total internal energy is U 1+2 = U 1 +U 2 = m 1 c v T + m 2 c v T By itself, this doesn t tell us anything about whether mass flows from 1 to 2, or vice versa, or there s no flow at all. It does give us an indication of the total kinetic (or potential) energy in the system. But that s it. Because there is no reference to pressure differences, the total enthalpy doesn t allow for flows. We could express these ideas in the context of the entropy, which tells us that through the almighty Second Law that stuff flows from low to high pressure, or high to low density at constant temperature. A problem with using entropy as a variable is that it is not a particularly intuitive concept. The mechanics of using entropy for evaluating system evolution is well developed, but it sometimes feels a bit like magic. Even with the link to mass we outlined, it is still not particularly obvious how to point at the entropy of an object. Nor are entropy changes easily described for such important irreversible processes in atmospheric sciences as the scattering of light (although they can be). What is more intuitive, and equally valid thermodynamically, is to express irreversibility, or the direction of time, in terms of free or available energy. Why introduce a new variable? Think of it this way. In a materially closed system. Internal Energy U is the extensive state variable that changes due to heating at constant volume at rate dq/dt Enthalpy H is the extensive state variable that changes due to heating at constant pressure at rate dq/dt Available energy G is the extensive state variable that represents a potential and it increases due to a heating at constant pressure, it decreases due to a density drop at constant temperature, and it increases if there is material convergence at constant temperature and pressure The Gibbs free energy Josiah Willard Gibbs, unpolluted by an engineer s desire to build better steam engines, figured out how to represent potential energy changes thermodynamically in the 1870s. These are now expressed in terms of a concept called the Gibbs free energy G. Gibbs defined this energy in terms of G H TS (12.1) Gibbs was one of the greatest American physicists of all time, on par with Boltzmann and Maxwell at a time when American physics barely existed. Einstein called him the greatest mind in American history. What is remarkable about his work is that it is largely his formulations for statistical mechanics, thermodynamics and vector calculus that have lasted to this day. 61

62 12.2 A hand waving derivation of available energy Frustratingly, the expression for G is always introduced a bit ad hoc, and rarely clearly explained in standard texts. Let s try to hand-wave how much available energy there might be by using the normal enthalpy of a gas as an example. We have already showed that dq dt T = RT d ln p dt Heating per mass leads to a pressure change by way of work. We can call RT d ln p the amount energy per unit mass that is available to do work. We know that matter flows from high to low pressure. But there are other things that can change the total available energy that we should be able to hand-wave based on intuition. Imagine a ski hill. The total available potential energy might easily be defined as the total number of skiers multiplied by the vertical distance of the ski slope. A change in the potential energy could be due to an increase in a number of skiers at a constant elevation (i.e. mass), or an increase in the mountain size (i.e. pressure difference). A third component is that we would have to take into account that many of the skiers would have to be raised to the top of the ski lift before their potential energy became available. A way to interpret Eq is that H is the total amount of energy in the system, in both levels, associated with all degrees of freedom plus those associated with pressure (H =(1 + f /2)mRT ). TSis the portion of this enthalpy that is unavailable for work: total skiers minus those at the bottom of the slope or at the lift. To keep things simple at this stage we ll do everything in terms of unit mass, i.e. h = u+ pa = c v T + RT = c p T. Suppose again that we have two air masses with total energies h 1 and h 2, and with internal energies u 1 and u 2. We want to know how much energy is available to drive flows between airmass 2 and airmass 1. As a starting point, suppose that u 1 = u 2 because temperatures of the two systems are equivalent. Then the available (or Gibbs free) energy Dg = h 2 h 1 = adp. Well, that is easy, and makes sense. There is energy available to drive flows if there is a pressure difference at constant temperature. But, now suppose that the two air masses not only have different pressures, but different temperatures as well. How much energy is available then? Suppose that the airmass 1 is colder than airmass 2. Then, the amount of available energy is less than adp because a fraction of this energy must go into first raising the temperature of airmass 1 until it is in equilibrium with airmass 2. In other words Or, in differential form Dg = adp (h 2 h 1 ) = adp Dh = adp T c p DT p dg = dh p + adp h = c p dt p + adp T (12.2) Ah, very nice. Notice that Eq is just the negative of Eq. 10.3! In other words dg dt = T ds dt The law of the universe is that the entropy increases and the available energy decreases. (12.3) 62

63 12.3 A formal derivation of the available energy In Gibbs original treatment Z H = TdS Taking as a reference point that S 2 = 0 in the upper level, and S = S 1 > 0 in the unavailable lower level, then DS = S 1 S 2 = S 1 and Z TdS= T DS = TS 1 = H 1 H 1 is the portion of the total enthalpy that is associated with high entropy matter with S = S 1. Thus, G = H TS= H 1+2 (S = 0) H 1 (S = S 1 ) G is the amount of enthalpy (which could always be revised to moist static energy or whatever) that remains as being available to do work through flows from the upper level to the lower level. Now lets say we have two systems, each with their own lower level S 1 and their own temperature and pressure. And suppose that we are interested in the amount of energy that is available for flows between the two systems. We get this by taking the total derivative of G to obtain DG, assuming that DG is small. DG = DH D(TS) G = (DU + pdv p +V Dp V ) T DS T SDT S + Dm m T,p = (DQ +V Dp V ) DQ T SDT S + g(t, p)dm = SDT S +V Dp V + g(t, p)dm where the D refers to the (small) difference between the two levels, and DU = DQ that we are allowing for difference in mass too through Dm so that pdv p. Note DG(T, p)=( G/ m) T,p Dm = g(t, p)dm Gibbs expressed g(t, p) as the chemical potential µ (T, p)=( G/ N) T,p. Indeed this is what is used most commonly in other sciences. However, in the atmospheric sciences, we tend to reference potential with respect to mass rather than molecular number, hence g(t, p) not µ (T, p). Notice that the nice thing about this expression is that it makes clear that, in the absence of any contrasts in mass temperature or pressure, DG is a minimum of 0. Remember too that g(t, p) can include many different sources of energy, such as gz or L v q v. Also, as we mentioned before, we can make the transformation DS(T, p)=c p Dm, or starting from zero: S = c p m(t, p) In which case, we can write the more general expressions for the the amount of energy that is available to do enable energetic transformations DG = G 2 (T 2, p 2,m 2 ) G 1 (T 1, p 1,m 1 )= DH m,p +V (T,m)Dp m,t + g(t, p)dm T,p (12.4) 63

64 The total differential for DG can be expressed as Substituting Eq into Eq. 12.4, we obtain DG = mdg m + g(t, p)dm g(t,p) (12.5) mdg m = DH m,p +V (T,m)Dp m,t (12.6) which is known as the Gibbs-Duhem equation. Dividing by mass, the available energy per unit mass is Dg = Dh p + adp T (12.7) which is the same as the expression we got through hand-waving in Eq What this formalizes is how the potential per unit mass of a system can be changed by a change in pressure or temperature without changing the potential energy DG of the system. In the atmospheric sciences, we might write dg = dh p + 1 r dp T (12.8) where, as always, we could replace h with h d or h m by adding other potentials. Note that the derivative in Eq could be, for example, with respect to space dx or time dt Components of the available energy The energy barrier We ve got three expressions on the right of Eq The first DH (m, p) represents the potential energy barrier or activation energy that must first be overcome before flows can happen. Like lifting skiers up the lift before they can ski, we must break bonds by raising the temperature before the bonds become chemically accessible. If you think of pressure and temperature being two orthogonal axes, we have to get the temperature difference DT back to the origin (aka T ) before we have diffusional equilibrium and the pressure difference Dp becomes available to enable material flows. In a Carnot Cycle, it might represent steps A to B or steps C to D. Perhaps in the atmosphere, you could interpret this as the energy that needs to be added through convection before air can diffuse from high to low pressure from the equator towards the poles in the Hadley cell. In a car, there is an enthalpy barrier that must be overcome through external ignition before chemical energy becomes available to create a pressure change. (The fact that one has to happen first is a reason to object to differential expressions like dp/dt which imply the changes happen at the same time). The amount of energy per unit matter required to over come the enthalpy barrier is: Dg(p)=c p DT p = Dh p (12.9) 64

65 The amount of energy available to do work through material flows The second term V Dp(m,T ) for DG states that, like a bigger mountain, we get greater contrasts and available potential energy if there are large pressure differences. Suppose that we are in a materially closed system without any (or very small) temperature contrasts, then the available energy per unit matter is G V Dg(T,m)= = Dp T,m = adp = 1 Dp = RT Dln p = RT Dlnr (12.10) m T m T r We have seen this expression previously. If we are at constant temperature (i.e. no heating), then the gibbs energy per unit material goes down when work is done by way of expansion and a drop of pressure. Effectively dg is the potential energy available per unit material that is available to do pda work. Since global entropy Ds = RDln p always increases with time, this means that, absent an external inflow of energy, the total atmosphere spontaneously loses potential energy as the pressure drops and the specific volume expands. Stuff flows downhill from high to low pressure and spreads out. A low pressure system has mass flowing to it from the high pressure system (pressure has units energy per volume). One expression we ll come back to a lot when discussing evaporation is the molecular form of Eq Dµ(T, N)=kTDlnp (12.11) The amount of material available to dissipate energy through material flows The third term yields g(t, p)dm, the increase in potential associated with an increase in mass. On the face of it, this is pretty obvious. The more we have the more potential we have The available energy at equilibrium At equilibrium, where DG is not changing, Eq must be equal to zero. Thus we get a very important balance equation for equilibrium conditions, mdg m = gdm g(t,p) (12.12) Increasing the potential Dg must be balanced by a decrease Dm in the amount of mass at that potential. Expressed in terms of a number and potential, this would be ndµ n = µdn µ(t,p) (12.13) which is the same as the expression we previously derived in Eq. 8.7 that yielded the power law expression d lnn/d ln µ. The important point here is that we may not be able to resolve internal gradients along isentropes defined by surfaces of a constant relationship between T and p. However, we can assume a certain statistical distribution for the likelihood that something has a certain amount of energy. 65

66 13 Non-equilibrium flows 13.1 The Universal Staircase If both pressure and temperature (or enthalpy) are fixed then from Eq g = 0 (13.1) T,p Unlike the extensive variable G, which depends on mass or number, the intensive variables g cannot change at constant T and p. So constant g is the same as saying we re along an isentrope, where processes are in the idealized state of being reversible. Nothing exciting happens along isentropes. Temperature and pressure are in a fixed relation. In fact, being reversible, time isn t even resolved. But this is super useful because if we define this surface, then all we have to (and can) look at is irreversible, time dependent flows of matter in and out of this surface. So Eq may seem rather boring, but it really lets us hone in on problems in a way that was more difficult before. Another phrase you will see for describing such surfaces is that they are at Local Thermodynamic Equilibrium. I find that a helpful way to think of the universe is as a staircase in a space of g(t, p), where the size of each step m is constantly changing due to flows in to the step from higher values of g and out of the step to lower values of g or chemical potential µ. Spontaneous processes occur, by the second law, when total potential energy drops (or entropy increases), so that for the system as a whole dg dt = d  i m ig i (T, p) dt < 0 (13.2) Let s suppose that the system as a whole is materially closed. The only way the second law can be satisfied is for matter to get distributed to lower specific potential surfaces g i (T, p)=( G/ m) T,p. There can always be an increase in the total potential G along any given specific potential g(t, p). As things flow downhill, there could be net convergence of matter m at any given level. So, from the general expression for G, along a constant potential surface, the potential energy increases with mass. G = g(t, p) T,p m T,p (13.3) Remember, at constant T and p, g (or µ) is constant. What we have now is the result that the extensive variable G is increased due to net convergence of material into a surface at constant temperature and pressure. This is super-important because it will call for us to apply the divergence theorem later Hand-waving We established previously that, at any given level of temperature (where temperature could be expanded more broadly to include things like gravitational potential) and pressure g, the potential energy G is determined by g(t, p)m. 66

67 The amount of mass at any given potential level m is determined by flows. Flows through space are driven by gradients in the density of potential energy. It is the divergence of flows around a point of constant potential (pressure and temperature) that determines whether the amount of material there is increasing or decreasing. We are now getting to the question of why and how things happen. What matters? Figure 13.1: Illustration of flows through potential energy levels leading to convergence along As in Fig. 13.1, think of a cascading waterfall with a series of three pools, each with its own g(t, p), and g points upwards. Intuitively we might say the following The balance of flows of water into and out of the pool is determined by m = ~j in ~j out g(t,p) where ~j is a current (units mass/time) defined so that it is positive pointing downhill. More generally, in a continuum of space, we could say that the change is determined by the divergence of flows m = g(t, p) ~j = g(t, p) d ~j g(t,p) dg We see expressions like this all the time. For example g could be a height co-ordinate in a gravitational potential field gz. Notice that if flows ~j are decreasing as we move down the 67

68 spatial co-ordinate defined by then more must be coming into intermediate levels than flowing out. Hence the negative sign, since this leads to dm/dt being positive. The Second Law demands that the average matter-weighted potential of all pools hgi = Â i m i g i /m = ÂG i /m must go down with time. We showed that Dh + RT Dln p = Dg T,p so at constant temperature or internal enthalpy (i.e., there is no energy barrier that must first be overcome to enable flows), the potential difference is related to a pressure difference between pools through Dg = RT Dln p = Dp/r. Since the pressure difference is what enables hgi to drop, we d expect the current to be proportional to the pressure difference between the pools ~j µ Dp where the negative is because the current is in the opposite direction of the pressure gradient. Note that we have discretized the potential difference. The potential difference could also be treated as a continuum in space, in which case ~j µ p Putting the above two points together, and noting that at constant T, dp= RT dr m µ ~j µ ( p) 2 p µ 2 r g(t,p) So there is a LaPlacian involved here. A gradient is required to drive flows. But unless the gradient itself changes with distance, then there is no convergence or divergence of flows at any given level. This is what we call the steady-state condition, for which 2 r T = 0 We might expect some cross-sectional interface between the two pools to determine the total magnitude of flows There would be a speed of flow that should matter too 13.3 Formal solution The Gibbs-Duhem relationship gives us the relationship between pressure and temperature gradients and potential gradients SDT S,p +V Dp T,V = mdg If there is no energy barrier to initiating flows, we can constrain further for constant temperature mdg = V Dp or Dg = 1 Dp (13.4) r 68

69 or as, a continuum expression g = 1 p (13.5) r So higher pools have higher pressure. How is this linked to down gradient flows? This is where we must add some additional physics. What we would like is an expression relating the flow of mass into level g(t, p) due to a pressure gradient. To get an idea of what this should look like, previously we showed that in a steady-state system at constant temperature where total potential isn t changing: m g m = g m T,p Any increase in potential dg must be offset by a decrease in mass at that potential leads to the equilibrium statement that d lnm dt which, since Dg µdp, dm can be rewritten as = d lng dt (13.6) dm/dt. This dm T,p = m p dp T,m (13.7) Increasing pressure is offset by a decrease in mass at that pressure. Also, since m = rv, and d ln p T = d lnr T dm T,p = rv p dp T,m = Vdr T,m So, the rate of increase in mass along a given potential level should be of form m = rv T,p p p T,m (13.8) or alternatively m = V T,p r T,m (13.9) These two equations are saying that as things fall down through pressure or density surfaces in the system as a whole (meaning at constant temperature and mass), the mass in the lower surface increases. We now examine three ways of expressing Equations 13.8 and 13.9, all essentially equivalent. 69

70 13.4 Flows between discrete surfaces Often we treat things as if our steps are discretized (like a high and low pressure zone), such that we have fixed intervals in p that are crossed with characteristic time t m = rv Dp T,p p t = V Dr (13.10) t or r = Dr T,m T,p t (13.11) If Dr = r r 0 is positive where r 0 is the bottom reference level, then the mass or density will increase in the top level where the density is r. For a volume encompassing everything, local mass or density can change due to flows down potential gradients. The above two equations will prove essential to understanding why whatever evaporates in sub-saturated environments, where there is a jump in vapor pressure or density. Note that we can t say anything in the above about what determines the time scale t. Figuring out this time scale requires we know more about how the atmosphere works Continuum flows expressed in terms of a velocity What if we approach the problem as a continuum in space? More accurately we might consider a flux due to a speed versus forced by a pressure (or density) gradient so that m = rv p = rv dp T,p p T,m p d~x d~x dt = rv ~u p = V~u r (13.12) p So if the local density or pressure gradient is positive in the direction of ~x, then flows must have a negative velocity ~u for there to be a local convergence of matter. In terms of density r = ~u r (13.13) T,p which is the advection equation. If the local gradient Notice that if we bring the RHS to the LHS, then we get the expression for the total derivative used in dynamics. It expresses how fluid flows at speed ~u down a density gradient lead to a local convergence of fluid Continuum flows to particles expressed in terms of a diffusivity Normally we think of flows as being through a cross-sectional area density A/V in the total volume V, in which case we replace the velocity with a diffusivity D (units area/time), so that AD V~u. The diffusivity is like a velocity, except that it is a scalar rather than a vector quantity, and is always towards to system of interest, so that there is a reversal of sign. m = AD r (13.14) T,p 70

71 In discrete form we get m = D ADr T,p Dx (13.15) where Dx is some length scale separating high and low density r. This is the equation we will use to explore growth of a droplet Non-equilibrium flows and the diffusion (or heat) equation To be more complete, we need to recognize that we may have gradient that itself is changing thus the current ~j = AD r may be spatially variable along some differential distance d~x. Imagine that flows are across a cross-section A in and out of an elemental unit of volume DV = ADx that contains a differential of mass at constant temperature and pressure Dm = rdv (e.g. a sphere with cross-section 4pr 2 has DV = 4pr 2 Dr). A is normal to the flow and Dx is in the direction of the flows. Further, consider that we can reference the the flows in terms of the cross-section so that we can define a flux ~F by ~F = ~j/a We ll eventually treat radiation this way. From Gauss s divergence theorem Z m = ~F dv Thus, since and then Thus, T,p V ~F = ~ j A = AD r A ( r)= 2 r = D r Z m = ~F dv = DV D 2 r T,p V r = T,p (m/dv ) T,p = D 2 r (13.16) This is the diffusion equation used throughout physics. There can only be a local convergence of matter when the LaPlacian of density 2 r is positive, meaning that the density gradient is increasing away from the object. If the gradient itself isn t changing with distance, then flows into an isentrope will be present, but they will be equivalent to flows out. This is the Steady State condition described by: r = D 2 r = 0 (13.17) T,p The time scale t that is required to reach steady-state can be estimated using dimensional analysis: Dr t D Dx 2 71

72 or, since AD V~u t Dx2 D t ADx2 Vu Dx u These expressions are presented because very often it makes things much easier if we can assume steady state. What is required is that t be short compared to the timescales of interest. Main points These are the key equations, all of which say the same thing in different ways. In a closed system, the density drops as mass collects at lower levels: m r = V T,p T,m (13.18) Equilibrating density differences has a characteristic time t r = Dr T,m t Equilibrating density gradients is done with speed ~u r = ~u r Particle mass with area A at a lower potential grows due to a density difference at rate m = AD r T,p T,p T,p Where there are flows in and out of a system, it is the gradient of gradients, or LaPlacian that determines whether things grow or shrink: r = D 2 r T,p The above expressions all came from the same place the Gibbs-Duhem equation. All could (and perhaps should) be modified to include an energy barrier in enthalpy or temperature, but assuming this barrier has been crossed, the above equations are seen everywhere. The key thing to recognize is that they all express the same idea given by the balance equation m g m = g m Mass flows are from high to low potential energy per unit mass. T,p (13.19) 72

73 Part III Cloud Physics 14 Evaporation How does evaporation work? Well, let s take what we know. The Second Law (Eq. 13.2) requires that for the system as a whole dg = d  m i ig i (T, p) < 0 dt dt The Gibbs-Duhem relationship (Eq. 12.6) gives us the relationship between pressure and temperature gradients and potential gradients DH m,p +V Dp H,V = mdg H,p If there is no energy barrier to initiating flows, we can constrain further for constant temperature (or enthalpy) mdg = V Dp or (Eq. 13.4) or as, a continuum expression (Eq. 13.5) Dg = 1 r Dp g = 1 r p Finally, pressure (or density) gradients drive flows (Eq ): m = D ADr T,p Dx 14.1 Atmospheric example: potential difference of the relative humidity Show that over a plane surface of pure water at constant temperature, where RH is the relative humidity, that the difference in the chemical potential µ between the vapor and liquid phase is where Dµ = µ v (T ) µ l (T )=kt ln(rh/100) DG = mdg NDµ where N refers to the amount of matter (typically molecules) and Dµ is the amount of potential energy per molecule. The trick in the above is recognizing that, just above the liquid surface, vapor is in saturation (or equilibrium), such that its pressure is the same as the liquid water beneath it and 73

74 Dp. Consequently, moving a vapor molecule from just above the liquid surface to the air above it is equivalent to moving it from the liquid phase to the vapor phase. Essentially this transfer is an example of the isothermal expansion (or compression) of an ideal gas as it goes from saturation to RH, where the gas we are interested in is water vapor. We are intested in the change of potential energy per molecule as it undergoes this transition. From the above Dg = 1 r Dp or, in terms of chemical potential, using the symbol e for vapor pressure and noting that e = nkt Dµ = 1 De De = kt ' ktd lne n e The change in chemical potential required to move a vapor molecule from right at the liquid surface to some distance farther away phase is thus Dµ = µ v µ sat v = kt ln e RH = kt ln esat 100 The value µ v sat refers to the chemical potential of water vapor at saturation where it is in equilibrium with the liquid phase and µ v sat = µ l Atmospheric example: evaporating puddles On the basis of the above, why does a puddle evaporate in dry conditions? Note that a system always work towards equilibrium. In this case a system always works towards a state where p v = p l. The chemical potential of vapor molecules just above a flat surface is what the chemical potential would be at saturation. To understand evaporation, draw a box around the vapor and liquid system, and do an accounting of the total gibbs free energy in the box for the transfer of molecules, with their respective chemical potentials µ, from the liquid phase to the vapor phase. The system acts to minimize the potential difference Dµ, such that in a spontaneous process d Dµ < 0. If lne/e sat is non-zero, then the imbalance between µ v and µ l must be reduced according to the second law. The way the system can do this is through material transfer from high potential at saturation to low potential at RH<100%. Remember we derived that r = Dr T,m T,p t so a gradient in vapor density from the puddle to the dry air leads to a less dense system with more vapor in dry air. Alternatively, consider a volume that contains dry air and liquid.the liquid potential is G l = N l µ sat l In the rest of the volume G v = N v µ v 74

75 The the total potential is G = N v µ v + N l µ sat l At constant temperature dg = dn v dt dt µ v + dn l dt µsat l But molecules are neither created or destroyed so, in a close system, what leaves the surface adds to the rest of the volume (and vice versa) So dg dt = dn v dt dn l dt µ v µ sat l = dn v dt = dn v dt Dµ = dn v kt lne/esat dt Since dg/dt < 0 through the second law, if e < e sat then Dµ = kt lne/e s < 0 and dn v /dt must be positive (i.e. evaporation). Likewise, if e > e sat, then Dµ = kt lne/e s > 0 and dn v /dt must be negative (i.e. condensation). Main points The relative humidity can be related to a difference in potential energy between the air and a liquid surface at saturation Water vapor flows from high to low potential or high to low relative humidity Question It is often stated that the air can hold more water on a warm day. Does this wording make sense? 75

76 15 Pure droplet formation The basic question is what makes the existence of a droplet thermodynamically preferable to the existence only of water vapor. We have already derived an expression for the saturation vapor pressure over water. But what is the vapor pressure over a solution droplet? Atmospheric liquid clouds nucleate on tiny aerosol particles with dry diameters less that 1 µm diameter. These aerosol particles are the seeds of clouds. Without them, clouds could not exist in anything like the form we observe. Below we discuss some of the physics that is relevant to cloud formation Step 1 Gas molecules must collide and stick, to form a nucleus of a few molecules that is thermodynamically stable. It turns out that the stability of such a nucleus is favored by low temperatures (low particle energies). Thirteen molecules seems to be the magic number required to get things going. The physics of nucleation formation is beyond the level of this course. Step 2 Once, a nucleus is formed, a certain supersaturation is required to maintain the droplet in equilibrium with its environment, and a higher supersaturation to enable it to grow. We have shown that that Dµ = µ v µ l = 0ife = e sat (T ). This is strictly true only over a flat surface of water. Over a curved surface, it turns out that e > e sat (T ) for equilibrium. Qualitatively, the reason is that the total energy of the liquid surface includes not just the energy of water molecules themselves, but in addition the potential energy associated with the surface tension due to stretching the surface of the water into a curve. Consider that if the droplet were to burst, the surface tension would provide available energy to do dynamic work on the droplet. Therefore, at the same temperature, a curved droplet is at a higher potential with respect to water vapor than is a flat surface of water. This means that water vapor that ordinarily would be at higher potential than a flat surface water, and drive condensation, is now at a lower potential and is associated with evaporation. Starting from the Gibbs-Duhem relationship (Eq. 12.6), expressed in terms of Dµ DH m,p +V Dp H,V = NDµ H,p We showed previously that, if there is no enthalpy barrier and DH = 0, then this can be expressed as Dg = Dp/r, or, per molecule (Eq.??): Dµ = 1 n Dp = µ v µ v,l sat e (T ) = ktdlne = kt ln e sat (T ) where e sat (T ) is determined by the Clausius-Clapeyron Equation. If e > e sat then the current is from vapor to the droplet. If e = e sat, then the system is in equilibrium since Dµ = 0. We must now consider that there is an energy barrier to droplet formation due to surface tension forces. Suppose that enthalpy of surface tension is given by H = sa 76

77 where A is the surface area of the droplet and s is the surface tension (units Newtons per meter or Joules per meter squared) which is about s = Nm 1 = Jm 2 Then the magnitude of the energy barrier produced, per molecule of droplet growth is dh/dt dn l /dt = dh dn l Defining the molecular enthalpy dh/dn l as Dḣ and n l = dn l /dv d as the molecular number density in a droplet of volume V d and radius r, Dḣ = dh dn = s da = 8prdr s n l dv d 4pr 2 = 2s dr n l n l r (15.1) Thus, from the Gibbs-Duhem expression (which we divide by N) we have a new expression for Dµ that includes the energy barrier from the surface tension for a droplet of radius r so Dµ tot = µ v µ sat l,v (T,r) Dḣ + 1 n Dp = Dµ tot Substituting our expressions for Dḣ (Eq. 15.1) and Dp, we now have 2s e + kt ln n l r e sat l,v (T ) = Dµ tot (15.2) Evaporation requires Dµ tot < 0 and µ v < µ l,v sat, allowing molecules to fall down, away from the liquid surface. Similarly, surface tension represents an energy barrier that always forces molecules away from the liquid surface. Thus, for Dµ tot to be positive, which would allow for condensation onto a curved droplet, the vapor pressure e must be greater than it would be over a flat surface. In fact what is required is e ln e sat l,v (T ) > 2s n l ktr For a curved droplet to be in equilibrium with its environment, Dµ tot = 0 and we get the Kelvin Eq., which expresses the saturation vapor pressure over a curved surface with radius r 2s e sat l,v (T,r)=esat l,v (T )exp 1 (15.3) n l kt r or, perhaps more familiarly in terms of per mass e sat l,v (T,r)=esat l,v (T )exp 2s r l R v T 1 r (15.4) A couple of things to notice about this equation. First as r approaches infinity, the curvature effect on the saturation vapor pressure becomes insignificant. In other words, even if the droplet 77

78 is a sphere, from a thermodynamic standpoint, it can basically be considered to be a flat surface. For atmospheric applications, the Kelvin effect can be neglected, and we can consider droplets to be essentially flat if they are larger than about 1 µm. Second, at the other extreme, for extremely small droplets, huge supersaturations are required for the droplet to be in equilibrium with its environment. For recently nucleated particles, say 0.1 µm across, the supersaturation at equilibrium is in excess of 100% (i.e. 200% relative humidity). We never get relative humidities this high in the atmosphere, and since any pure droplet forming from vapor must start from an embryo of a few molecules, clearly pure water droplets cannot form in the atmosphere T = 5 C 10 4 Supersaturation (%) mer R (µm) Figure 15.1: Dependence of the supersaturation of water vapor at equilibrium over a pure water droplet. The size of a 13 molecule nucleus is shown for reference. Main points We are now considering the energy barrier component of the Gibbs-Duhem equation The energy barrier is the energy required to form the droplet surface against surface tension forces that work to shrink the droplet The equilibrium vapor pressure for a pure curved droplet is higher due to a flat droplet due to the extra energy require to maintain a curved surface 78

79 Question What are possible factors that might reduce the equilibrium vapor pressure for a constant droplet size? Are any of these plausible considerations in the atmosphere? 79

80 16 Solution droplet formation 16.1 Kohler equation The Kelvin Equation (Eq. 15.3) has the form e sat l,v (T,r)=esat l,v (T )exp 2s n l rkt which was obtained by adding an energy barrier due to surface tension so that the total potential is Dµ tot = kt ln e e sat (T ) Thus we can interpret the Kelvin Eq. as being the equilibrium solution where Dµ tot = 0. We found that maintain equilibrium requires extraordinarily high values of supersaturation S = e/e sat 1 in order to compensate for the surface tension potential when the droplets are just nucleated and small. Somehow, we need to increase Dµ tot so that the vapor is at a higher or equal potential than the potential level of the droplet molecules µ v,l (T,r), so that being lazy, the droplet molecules are happiest to stay in the liquid phase. An effective way to do this is to add solute to the liquid. What this does effectively is to reduce the density of water molecules at the interface between the liquid and vapor phase, since the water molecules are separated by solute ions. Consequently the density of water vapor needs to be less in order for there to be equilibrium. Let s say we define a solution activity such that 2s n l r µ l (T,a w )=µ sat l (T,a w = 1)+kT lna w (16.1) where a w is bounded by 0 and 1. If a w < 1 then the chemical potential of liquid is depressed compared to what it would be at equilibrium over a flat surface of pure water. If a w = 1 then the chemical potential of the liquid is what we normally assume for pure water, and µ l (T,a w = 1)=µ sat l (T ) (16.2) Thus, since, right at the surface of the water, vapor is at equilibrium with the underlying liquid surface, and µ v sat (T,a w ) µ l (T,a w )=µ l sat (T )+kt lna w (16.3) we can alter our expression for Dµ from before to include the energy barrier representing the energy per molecule required to take the step upwards from the solution potential to the pure liquid potential, i.e. Dḣ = µ sat l (T ) µ l (T,a w ) µ sat v (T ) µ l (T,a w )= kt lna w (16.4) The energy barrier that must be overcome to bring water vapor into equilibrium with a flat surface of pure water is Dµ = µ v sat µ v = kt lnrh. Thus, the water vapor is in equilibrium with the solution when the two energy barriers are equal and kt lnrh = kt lna w (16.5) 80

81 where, a w corresponds to the relative humidity over a flat solution that brings the potential µ v into equilibrium with µ l (T,a w ). From the Gibbs-Duhem relation Dḣ + 1 n Dp = Dµ tot the combined potential difference per molecule between the vapor and the liquid phase, for a curved solution droplet, is now Dµ tot = µ v µ l (T,r,a w )=kt ln e e sat (T ) 2s n l r kt lna w (16.6) The second and third terms represent the sum of the energy barriers to diffusional flows Dḣ. If the activity of the solution can be lowered to a w < 1 then the last term is positive and this increases the energy barrier between the liquid and vapor phase Dµ. This makes existence of the liquid phase more stable and likely, despite the curvature effect. There is a bigger wall for the liquid molecules to jump over to get into the vapor phase, or a smaller wall required to initiate condensation. Our revised equilibrium expression then for the vapor pressure over a solution droplet is obtained by solving for Dµ tot = 0 This is known as the Kohler equation Activity of solution droplets apple 2s e sat (r,a w,t )=e sat (T )exp n l r + kt lna w 2s e sat (r,a w,t )=e sat (T )a w exp n l rkt 2s e sat (r,a w,t )=e sat (T )a w exp r l R v Tr For an ideal solution (usually true if very dilute) /kt (16.7) (16.8) a w = n 0 n + n 0 (16.9) where n 0 is the molecules of water and n molecules of solute. This is known as Raoult s Law. For dilute solutions a w = 1 n/n 0 (16.10) But we must remember that salts dissociate, i.e. NaCl dissociates to Na + and Cl. We represent this using what is really a bit of a kluge, but popular nonetheless - the so called Van t Hoff factor i. n! in (16.11) where, for NaCl, and NH 4 HSO 4 (ammonium bisulfate) i = 2, and for (NH 4 ) 2 SO 4 (ammonium sulfate), i = 3. i is essentially the number of solute ions that the molecule dissolves into. The 81

82 problem with this approach is that the number i stops meaning anything physical when one starts talking about organic non-ionic species, which nonetheless are soluble. What do we see from this? 1. The higher the concentration of the solute, the lower the equilibrium relative humidity 2. The more dissociation, the higher the solution effect. Effectively what is happening is that, while curvature makes it easier for molecules to escape a liquid droplet, dissolving a solute makes it harder for them to escape. Effectively there are fewer liquid molecules per unit surface area for them to escape. The Kohler curves are shown below. These show the relative humidity relative to saturation over a flat surface of pure liquid water that would be in equilibrium with a droplet of radius r containing dissolved solute. Effectively, it is e sat (r,a w,t )/e sat (T ) (NH 4 ) 2 SO 4 (r s =0.02 µm) (NH 4 ) 2 SO 4 (r s =0.03 µm) (NH 4 ) 2 SO 4 (r s =0.04 µm) (NH 4 ) 2 SO 4 (r s =0.06 µm) (NH 4 ) 2 SO 4 (r s =0.08 µm) (NH 4 ) 2 SO 4 (r s =0.1 µm) (NH 4 ) 2 SO 4 (r s =0.15 µm) (NH 4 ) 2 SO 4 (r s =0.3 µm) S r (µm) Figure 16.1: Kohler curves for ammonium sulfate. The location of peak saturation ratio S is at r, the critical radius. Below this size the solution effect dominates the Kelvin effect, and solution droplets are in equilibrium with their environment (i.e Dµ = 0). If the particle grows spontaneously, it is then too large for the ambient equilibrium saturation for the droplet, so it shrinks to its original size. Likewise, if it shrinks it grow again to its equilibrium size. These are haze particles, which are in stable equilibrium. Above this size the haze particles are said to be activated. The Kelvin effect dominates and Dµ < 0 if the particles grow spontaneously. A particle that grows spontaneously will be have an equilibrium saturation lower than the ambient saturation, so it keeps growing spontaneously by vapor diffusion. The total 82

83 Gibbs free energy of the system will continue to decrease. Once the droplets get big enough, the Kelvin effect stops mattering, and the droplet acts effectively like a plane surface of water. Therefore, the function of a solute in droplet formation is to lower the amount of chemical potential energy of the water vapor (i.e. supersaturation) required to get the droplet to the size r where it can start to grow spontaneously of its own accord. The supersaturations required are typically less than 1%, which is quite manageable in clouds. Main points While surface tension increases the energy barrier required to maintain a curved surface, presence of a solute in the water decreases the energy barrier. For a salt, the magnitude of the effect is proportional to the number of solute ions Solution droplets can have an equilibrium relative humidity less than 100%. The equilibrium is stable. A droplet is in unstable equilibrium where the curvature effect and the solute effect are in balance. Any further growth is unstable and leads to spontaneous condensation. 83

84 17 Cloud droplet growth Once droplets are activated, how do they grow? There are two primary components to this problem. Growth by vapor diffusion, and growth by collision-coalescence. We showed that one of many legitimate forms for the non-equilibrium solution for flows from high to low potential is a current ~j that adds to the amount of mass m in a lower potential surface at constant T and p. From Eq ~j = m T,p = A RT D p = AD r where, the sign convention here is for the direction to be from the high potential to the low potential, so that a negative density gradient in this direction corresponds with positive flow. Well, of course this could apply to the diffusional growth of a droplet, where the droplet is a low potential surface, and its super-saturated environment is a higher potential surface. Evaluated at the droplet surface, where the droplet has radius r, the mass increase m in condensed molecules, at the expense of vapor m v, and to surface area A = 4pr 2 follows mv T,p = md T,p = 4pr 2 D r x x=r where, the diffusivity D for the water vapor in air is about m 2 s 1. Also, here we ve switched the direction x so that it points radially away from the droplet. But, what is r/ x x=r? If we assume the ambient vapor field is in steady-state over time scales relevant to instantaneous flows, which is extremely close to being true, then from Eq dr/dt = 2 r = 0 If we assign x as the radial dimension, in spherical coordinates, we get which has the general solution 2 r x r 2x x = 0 r (x)=c 1 C 2 /x Applying the boundary conditions x!, r! r ; x! r, r! r r, we find the vapor field can be described by r (x)=r r x (r r r ) Thus, taking our flux solution, dm dt dm dt = 4pr 2 D r x x=r = 4prD (r v r vr ) (17.1) So rate of mass accumulation of droplets is proportional to their size. Note that the above equation is for a single droplet only, and that we would need to multiply the above equation by 84

85 the droplet number concentration to get total mass density for a cloud. What stops droplets from growing indefinitely? Naturally, as droplets grow, they will deplete the available water from their surroundings, thereby reducing the vapor concentration gradient between the ambient air and the surface of the droplet. How fast do droplets grow in size? Starting from m = 4 3 pr lr 3 Using the ideal gas law for water vapor dm dt = 4pr l r 2 dr dt dr dt = D rr l (r v r vr ) dr dt = D rr l R v T (e e r ) dr dt = Dr v rr l e (e e r ) which, assuming e ' e sat (T ), such that the perturbation is not large, we can show that e e r e ' e e s e s S 1 = s where S is the saturation ratio and s is the supersaturation. So r dr dt = G ls (17.2) where G l = Dr v = Desat (T ) r l r l R v T We have made a number of assumptions in these derivations that we should be aware of. (17.3) 1. All water vapor molecules that hit the droplet stick. The reality may be that only 3% do. This matters mostly during initial growth of the droplet. 2. The temperature at the surface of the droplet is the same as the ambient air. In reality it is warmer due to the latent heat associated with condensation. It requires energy to transport this extra heat away from the droplet. This is an energy barrier that retards the rate of droplet growth 3. The droplet is stationary with respect to its environment. A droplet that falls as it grows is ventilated which changes the distribution of water vapor around the droplet and carries away latent heat. Apparently, however, this effect is very small, and can usually be neglected. If we integrate the growth rate equation we get r (t)= q r G lst (17.4) 85

86 Figure 17.1: The assumed supersaturation is 0.05%, p = 900 mb and T = 273K. where r 0 = r, the activation radius if we are starting from a haze particle. A table of the amount of time it takes for a particle to reach a given size is given below. Note that droplets grow very rapidly at first, and as the droplets grow their initial size matters less and less. The same equations can be used to estimate the evaporation rate of droplets below cloud base It is easy to see from this table why the edges of clouds are so sharp. Only falling drops larger than about 0.1 mm will give any blurring to the cloud, and droplets must be significantly bigger in order to have any chance of hitting the ground, particularly if cloud base if very high (as it often is here in Utah in Summer). It is important to recognize that when a water drop evaporates, it does not completely disappear, but rather becomes a haze particle whose size is determined by the the Kohler equation according to the mass of solute it contains, and the ambient relative humidity (aka saturation ratio). Main points Through the diffusion equation, the growth of droplet mass is proportional to the available water vapor and the droplet radius The rate of droplet growth is proportional to the inverse of the droplet radius so droplets grow very quickly at first but much slower as they get larger. Droplets forget their initial conditions as they get larger The above principles apply equally to evaporation 86

87 18 Rain Production 18.1 Collection Figure 17.2: Droplet evaporation How do we get rain? So far we ve discussed droplet growth by vapor diffusion, but this is not the process that by itself is primarily responsible for precipitation in warm clouds. The primary production mechanism for growth of precipitation sized droplets is collision-coalescence. In this process droplets collide due to differential settling velocities, and coalesce. If we say the big collector drop is r 1 and the smaller drops it is collecting have radius r 2 and r 1 r 2 then the rate of collection depends on 1. The collision cross-sectional area is 2. The relative collection velocity is p (r 1 + r 2 ) 2 ' pr 2 1 v T 1 v T 2 ' v T 1 3. The coalescence efficiency (i.e. the probability that two colliding droplets stick), which is approximately unity if r 1 > 10r 2 4. The collision efficiency E (r 1,r 2 )(i.e. the probability that two droplets that are within the sweep area of the collector drop actually collide. 5. The LWC of the collected droplets Therefore the growth rate of the collector drop is dm 1 dt = pr 2 1E (r 1,r 2 )v T 1 LWC 2 (18.1) 87

88 Notice that we could have gotten Eq. 18.1) from the non-equilibrium thermodynamic solution we derived earlier (Eq ) m = V~u r T,p which, since the volume swept out by a falling droplet is V = ADx, and r = Dr/Dx, it could be re-written as m = Av T Dr or, T,p m = Av T LWC T,p and noting that the effective cross-section here is A = Epr1 2 m1 = pr 1Ev 2 T LWC T,p Now, noting that and we can show that dr 1 dt dm dt = dm dr dr dt dm dr = 4prr2 = E (r 1,r 2 )v T 1 LWC 2 4r l (18.2) Note that r 1 > 20 µm appears to be a critical initial size for initiation of the collision-coalescence process. If droplet are unable to grow to this size in sufficient concentrations by vapor diffusion, then typically rain does not form (Fig. 18.1) Terminal velocity For very particles smaller than about 40 µm radius, we can use a solution that can be derived from first principles, where the drag force is proportional to radius 6pµv T r = 4 3 prr3 g v T = 2rr2 g 9µ = k 1r 2 (18.3) where k 1 ' cm 1 s 1. However this assumes Stokes drag. However, for rain drops with radius 0.6mm <r < 2mm the drag force is in fact proportional to r 2. The solution then for v T is v T ' k 2 r 1/2 (18.4) 88

89 Figure 18.1: Growth rate by collision compared to diffusion. where k 2 ' cm 1 s 1. In the intermediate range 40 µm <r < 0.6mm v T ' k 3 r (18.5) where k 3 ' s 1 Terminal velocities as a function of droplet size are given in Table 18.2 Collision Efficiency The computation of the collision efficiency from first principles is a very difficult problem owing to the complexities of the flow field around a sphere falling through a viscous medium. Rogers and Yau state For any size of collector drop, the collision efficiency is small for small values of r 2 /r 1. The collected droplets are then small, have little inertia, and are easily deflected by the flow around the collector drop. The inertia of the droplets increases with r 2 /r 1 accounting for an increase in collision efficiency up to a radius ratio of about 0.6. Two counteracting effects come into play as r 2 /r 1 increases beyond this value. Because the difference in the size of the drops is getting smaller, the relative velocity between the drops is reduced, prolonging the time of interaction. The flow fields interact strongly, and the time can be sufficient for the droplet to be deflected around the drop without collision. On the other hand there is a possibility for a trailing droplet to be attracted into the wake of a drop falling close by at nearly the same speed. This effect can lead to wake capture and to collision efficiencies that exced unity for values of r 2 /r 1 ' 1 (Fig. 18.3). There is also a coalescence efficiency. Big drops can bounce of each other. Really, though it is 89

90 Figure 18.2: Table of terminal fall speed the collision efficiency that is of most interest for collision-coalescence, because the concentrations of these bouncy droplets are so low. Question: Rain production in cumulus Imagine a cumulus cloud with and LWC of 0.4 g m 3 1 km thick with a monodisperse droplet mode at 10 µm radius and a second mode with concentrations of 1/litre at 100 µm radius at cloud top. The updraft velocity w in the cloud is 0.05 m/s. 1. Derive an expression for the size of the drizzle drops as a function of distance as they descend through the cloud dr dt = w lv T E (r 1,r 2 ) 4r l This distance the droplet falls from cloud top is h = v T t, so Thus, dr dt /dh dt = dr dh = LWCv T E (r 1,r 2 ) (w v T )4r l Z H 0 w l dh = 4r l Z rh r 0 w v T v T E dr 1 Okay. This is a bit nasty, but it can be solved for r 1 (H) if we know v T (r) and we assume w l 6= w l (h). apple Z rh Z w rh w l H = 4r l v T (r 1 )E dr r 0 E dr 1 r 0 90

91 Figure 18.3: Collision efficiency (Rogers and Yau) If we assume that in general w v T (which maybe isn t so good), integrating we get the radius in meters at cloud base as 2. What is the droplet size at cloud base? Just substitue h = 1000m to get r = 180µm 3. What is the precipitation flux in mm/hr r = r = r 0 + w l 4r l h = H The drizzle drops are in the intermediate size range for falls speed v T = k 3 r, so Flux = LWC precip v T precip Flux = 4/3prr 3 1 N k 3r 91

92 Flux = 4/3p (1000) = kg/m 2 /s with appropriate conversions (divide by r l = 1000 kg/m3 (for m/s) and multiply by 3600 s in an hour (for m/hr) and 1000 mm per m), I get 0.1 mm/hr. 4. In the absence of updrafts generating new cloud, what is the time scale for depletion of all the liquid water in the cloud? LWC = LWC 0 exp( jt) lossrate = j = The cloud water depth in units of length is Thus precip rate cloud water depth cloud depth LWC = = 0.4mm r l 1000 lossrate = j = precip rate cloudwaterdepth = ' 0.25/hr This leads to a time scale of about 4 hours. It indicates that the cloud exhausts itself of water every few hours. Another interesting conclusion is that j is the same as the loss rate of CCN from the entire boundary layer. How is this the case? 92

93 19 Ice nucleation Liquid water can exist in clouds at temperatures below -30 C even. How then do we get ice in clouds? We discuss in class two basic formation mechanisms for ice Homogeneous nucleation Homogeneous nucleation is the freezing of a haze or water droplet at temperatures generally below 35 C. The physical principle is similar to that for homogeneous nucleation of water droplets, except that it occurs within in a water droplet rather than a supersaturated vapor field. If a droplet is sufficiently cold, by chance a few of the water droplet molecules will be moving slow enough to stick, and form an ice crystal embryo from which the entire droplet can turn to ice. Homogeneous nucleation appears to be the most important process responsible for producing large ice crystal concentrations in excess of 1 cm 3 in cirrus clouds, and is dominant particularly in the upper reaches of cloud with strong updrafts such as cumulonimbus. Recent studies show that a haze or water droplet freezes only when the saturation ratio with respect to ice is about 1.6 (compared to about for water clouds). Once a haze particle or water droplet freezes, it can grow directly by vapor diffusion. In general nucleation is the conversion of a liquid sphere to solid ice form. In this process a fraction of droplet will be frozen, with concentration N f, and a fraction will be unfrozen, with concentration dn u dn u = dn f dt dt The rate at which the conversion from liquid to solid takes place is given by the nucleation rate J (units number per volume per second) and the droplet volume and the number of unfrozen droplets remaining. Thus dn u dt dn u = = N u V d J (T ) N u V d J (T )dt N u = N 0 exp( V d J (T )t) (19.1) The nucleation rate J (T ) itself is a thorny problem. However, studies by Koop in 2000 have shown a particularly interesting result. We know from Eq that, at equilibrium µ v = µ l sat + kt lna v w Similarly µ i = µ l sat + kt lna i w where a i w = a v w/s i, where S i = e s (T )/e si (T ). Note that the saturation vapor pressure over water is higher than that over ice, so S i can be greater than 1. Koop showed empirically that J = J a v w a i w (19.2) Well this is super useful, because at equilibrium, the value of a w should be the same as the ambient relative humidity (Eq. 16.5). Also, we know the saturation vapor pressure over ice and water. Thus, bingo, from the formula Koop provided (Eq. 19.2) and Eq. 19.1, we can calculate whether or not a haze aerosol should freeze to form an ice crystal at a given time t. 93

94 19.2 Heterogeneous nucleation Heterogeneous nucleation on the other hand involves an ice nucleus. An ice nucleus is usually composed of a material that has a similar crystalline structure to ice. By way of mimicking ice when such an aerosol particle comes in contact with a supercooled (<0 C) droplet, the droplet gets tricked into freezing. This process can occur at any temperature below freezing, but some ice nuclei require colder temperatures than others too initiate freezing. Perhaps the most important ice nucleus in the atmosphere is kaolinite (blackboard chalk) which is a common component of windblown dust and acts as an ice nucleus at an average temperature of -9 C. However concentrations of ice nuclei in the atmosphere are typically only about 1/litre, compared to at least 100/litre in clouds warmer than -35 C. It remains a puzzle exactly how to resolve this discrepancy. However, it appears that large concentrations of ice crystals are associated with clouds with initially high concentrations of liquid water droplets larger than about 20 µm. One possible mechanism for ice crystal multiplication that produces high crystal concentrations is the Riming-Splintering mechanism or Hallett-Mossop process. In this process an ice nucleus causes the outer shell of a droplet to freeze first, which then shatters to form more ice nuclei that can themselves initiate ice crystal production. However, this process appears to be most important between -3 and -8 C, and large concentrations of ice crystals have been observed at both colder and warmer temperatures. Although Hallett-Mossop is included in many mesoscale models, this is a big area of debate currently and far from resolved. 94

95 20 Ice crystal diffusional growth Generally air saturated with respect to water is supersaturated with respect to ice. This sets up a vapor pressure gradient such that in mixed-phased clouds, ice grows at the expense of water. As before if the ice crystal is spherical, from Eq dm dt = 4prD (r v r vr ) where r vr represents the vapor pressure at the surface of the ice crystal. But as we know, ice crystals can be some truly crazy shapes. We can address this problem by drawing an analogy between the flux of vapor in and out of a crystal with charge leakage from a capacitor of the same size and shape. The capacitance C of a spherical capacitor is C = 4pre 0 (20.1) where e 0 = C/N/m 2 is the permittivity of free space. Forgetting about e 0, substituting Eq into Eq yields dm = CD (r v r vc ) (20.2) dt where r vc is the water vapor density at the surface. Following the same arguments as described for water droplets dm = CG i s i (20.3) dt where G i = Dr v ( ) and s i = e( ) e si e si Wallace and Hobbs outlines in Table 6.1 the types of growth that dominate. in some regimes it is the basal face that dominates and we get columnar shapes. In others, it is the prism face, and we get more plate like structures. In general, the greater the supersaturation with respect to ice, the more complicated the shape. Looking at diffusional growth more closely, consider Wulff s theorem, which is that for slow diffusional growth of ice crystals h b = s b ' 0.92 (20.4) h p s p where s is the surface tension of the ice crystal basal (b) or prism (p) face, and h is the height of the face. To understand this better consider the Kelvin equation we came up with, which is that e sat µ exp(ks) where k is a constant and ks is related to the energy barrier associated with surface tension that must be overcome before there is condensation. So the greater the surface tension the greater the saturation vapor pressure. This means that if we have a given vapor pressure e, then the speed of condensation will go as dm µ e e sat dt 95

96 Therefore if the surface tension of the basal facet is is high compared to the prism face, then saturation vapor pressure is higher relative to the basal face, and the vapor pressure gradient e e sat is lower. This means that if s p is higher than s b, it follows that vapor will preferentially condense on the basal face: simply it is harder to condense on the prism face due to the higher surface tension there. It s a little counter-intuitive (just draw it to convince yourself), but the faster the basal face grows, the large h p becomes, explaining Wulff s theorem. If s p > s b then h p > h b. Wulff s theorem explains the shapes of ice crystals well when supersaturations are very small and temperatures between -10 C and -22 C. The problem is that Wulff s theorem doesn t take us particularly far in explaining the wide variety of crystal aspect ratios that are observed at other temperatures and supersaturations. Growth is not necessarily very slow, extremely thin liquid layers can form on ice during the growth process, and ice crystals do not grow through simple vapor deposition but rather in steps and spiral formations. The true physics that controls ice crystal growth largely remains a mystery. 96

97 21 Observed cloud microstructures Here we discuss some the observed properties of warm (liquid water) clouds such as stratus and stratocumulus Vertical structure There are four basic parameters that are usually examined in measurements of clouds. Figure 21.1: Number (left) and LWC (right) size distributions in clean stratocumulus and stratocumulus polluted by ships. 1. Size distributions (either, number, surface area, or volume). These look very similar to aerosol size distributions except the modes in the distributions are found at much larger 97

98 sizes. The mode in the number distribution is typically between 10 and 25 µm diameter. The volume distribution is typically bimodal if the cloud is precipitating, with the second mode at somewhere between 100 and 1000 µm diameter. The precipiation mode has the largest sizes near cloud base, and the cloud mode (the small mode) has the largest sizes typically near cloud top. 2. The number concentration of droplets N (z)is usually nearly constant with increasing height. 3. The liquid water content LWC (z)usually increases linearly with height, starting near zero at cloud base. 98

99 99

100 100 Figure 21.2: Vertical structure in stratus and stratocumulus off the coast of California

101 4. The volume mean particle size r v also increase with height, but starts at about 4 µm near cloud base, and increases less than linearly with height to somewhere between 5 and 15 µm diameter. Figure 21.3: Liquid water content profile in cumulus The vertical structure tends to reflect entrainment and precipitation which causes the LWC to drop off from its adiabatic value nearer to cloud top. 101

102 21.2 Horizontal structure Figure 21.4: Horizontal transect in cumulus The horizontal structure of clouds can be relatively homogenous, as in the case of stratus or fogs, or highly variable in stratocumulus or cumulus clouds. The degree of inhomogeneity appears to be linked to how turbulent the cloud is, and how it is precipitating. Observations show that the number concentration and liquid water content are much more highly variable horizontally than the droplet size Continental versus maritime clouds It is generally observed that continental clouds have smaller and more numerous droplets than maritime clouds. The degree to which this is the case appears to be closely linked to the concentrations of aerosols in the environment. 22 Observed rain and snow distributions Much like aerosol size distributions it turns out that rain and snow distributions can be closely parameterized by a simple exponential fit. The fit for rain drops is called the Marshall-Palmer distribution after the two scientists who first recognized this distribution based on a summer s observations in Ottawa Canada published in This distribution of droplets is characterized by the following equation dn(d) dd = N 0e LD 102

103 where D is the droplet diameter and N the concentration and L (units cm 1 ) is the slope of the rain distribution of a semilogarithmic plot. Marshall and Palmer showed that the slope can be related to the rainfall rate by L(R)=41R 0.21 where R is the rainfall rate and has units of mm hr 1. Strangely N 0 appears to be nearly independent of R and has an approximate value of N 0 = 0.08cm 4 (see Fig in Rogers and Yau) The equivalent equations for snow turn out to be L(R)=25.5R 0.48 N 0 = R 0.87 Figure 22.1: Snowflake shapes photographed by the muli-angle snowflake camera at Alta. Top row graupel, middle row rimed snow and bottom row aggregates. 103

104 1 Aggregates Rimed Graupel dn/dd max λ agg =0.69 mm 1 λ rim =1.19 mm 1 λ grp =1.99 mm D max (mm) Figure 22.2: Size distributions of snowflakes obtained with a Multi-Angle Snowflake Camera at Alta base. Concentrations are normalized. An example of snowflake shapes is shown in size distributions of snow measured at Alta is shown in Figure Associated size distributions are shown in Fig Note how riming produces more compact particles. 104

105 23 Atmospheric mixing This section introduces some elementary concepts associated with mixing and turbulence in the environment Mixing of conserved variables Studies of mixing of different airmasses often approach the problem by considering the conserved variables associated with each. Thus far we have discussed several of these 1. Conserved under unsaturated adiabatic processes q, h d, w 2. Conserved under saturated adiabatic processes q e, h m, Q Note that w s and c are not conserved in general, and variables that are conserved under dry processes are not conserved under moist processes. Now suppose that we have two airmasses that meet and mix, without interaction with the surrounding environment. Such mixing is a fundamental aspect of for example heat transport from the equator to the poles, or temperature and humidity gradients along fronts, or entrainment of dry air into clouds. How do we determine the state variables associated with the mixed parcels? By way of illustration if we take mass fraction f from one air parcel A and mass fraction (1 f ) from a second parcel B and there is no saturation during the mixing process then the dry static energy and mixing ratio associated with the new parcel is h d = fh da +(1 w = fw A +(1 f )h db f )w B Since h d = c p T + gz, if one knows the height of the new parcel, one now knows the temperature and humidity of the new parcel Adiabatic Mixing without Condensation We derive here the basic equations dealing with adiabatic mixing without condensation. Assumptions The two parcels with masses m 1 and m 2 mix adiabatically The two parcels are at the same level (d (gz)=0) (e.g. mixing across a front). 105

106 Since the two parcel are at the same level we are mixing only enthalpy and water vapor. Therefore, m 1 h 1 + m 2 h 2 =(m 1 + m 2 )h The change in enthalpy is then DH = 0 = m 1 (h h 1 )+m 2 (h h 2 ) so the new temperature is m 1 Dh 1 + m 2 Dh 2 = 0 m 1 c p (T T 1 )+m 2 c p (T T 2 )=0 T = m 1T 1 + m 2 T 2 m 1 + m 2 Similarly w = m 1w 1 + m 2 w 2 m 1 + m 2 e = m 1e 1 + m 2 e 2 m 1 + m 2 q = m 1q 1 + m 2 q 2 m 1 + m 2 106

107 23.3 Atmospheric application: mixing across a front Figure 23.1: Two subsaturated parcels combining to produce a saturated parcel. Note that both e and T mix linearly, but that clearly RH does not. Two parcels of air mix thoroughly across a cold front at 1000 mb. Parcel one has a temperature of 23.8 C and w = 16g/kg, and parcel two has a temperature of T = 12.4C and a mixing ratio of 5 g/kg. 1. If both parcels of air mix equally, what is the final temperature and mixing ratio of the combined air masses? Since the parcels mix equally, the new temperature and mixing ratio is simply the average temperature and mixing ratio, which are 18.1 and 10.5 g/kg. 2. What is the initial RH of each parcel and final RH? It works out that e s (23.8 )=29.5mb e s (12.4 )=14.4mb e s (18.1 )=20.7mb and that the saturated mixing ratio is w s = e s p 107

108 So the relative humidities are w s (23.8 )= w s (12.4 )=0.009 w s (18.1 )= RH (23.8 )=87% RH (12.4 )=56% RH (18.1 )=82% So unlike T and w, RH does not mix linearly! This is an important consideration since it leads to the observation that two subsaturated parcels of air can combine to produce a cloud Figure 23.2: Sea-smoke A most striking example of this sort of this is when a cold dry air mass moves over a warm moist air or surface to produce sea-smoke. Another example is when warm moist exhaust from an airplane mixes with cold dry air to produce a contrail. 108

109 24 Atmospheric turbulence 24.1 Characteristics Examples of Turbulence Cumulus clouds Jet streams Dust storms Characteristics irregularity diffusivity - no spreading no turbulence vorticity dissipative due to viscous losses (waves are non-dissipative although they can be dispersive) characteristic of fluid flows Associated with large values of the Reynolds number Re where Re = ul/n, where L is the characteristic length scale of an object moving with speed u through a viscous medium with kinematic viscosity n. Sources shear in a mean flow buoyancy The transition from laminar to turbulent flow is one of the more poorly understood aspects of physics. On his death bed, the famous fluid dynamicist Horace Lamb said, ""I am an old man now, and when I die and go to heaven there are two matters on which I hope for enlightenment. One is quantum electrodynamics, and the other is the turbulent motion of fluids. And about the former I am rather optimistic" 24.2 Length Scales in Turbulent Flows Turbulent flows have a wide range of length scales covering many orders of magnitude in size ranging from the dimensions of the flow itself at the high end, to the length scale associated with the diffusive action of molecular viscosity at the low end. What are the relevant scales for laminar, low-shear flow? The Navier-Stokes equations for steady flow are u du dx = 1 r dp dx + n 2 u x 2 (24.1) 109

110 The first term is inertia, the second the pressure gradient force, and there is a non-linear third term, the viscous term with n = 0.15 cm 2 s 1, which is needed to create turbulence. For viscosity to be sufficiently large to damp turbulence the viscous and inertia terms have to be of the same magnitude. Doing scale analysis we would require U 2 L nu L 2 Dividing the first term by the second, we obtain the Reynolds number UL n Re (24.2) If the Reynolds number is of order unity than turbulence is damped and flow is laminar. But Reynolds numbers typical of the atmosphere are typically on the order of a few thousand, so the atmosphere tends to be turbulent. In clouds the source of energy is mostly latent heat release, and this creates turbulent motions while sending air upwards. With respect to the turbulence, it is common to talk about an energy dissipation rate e with units of energy per mass per second (m 2 /s 3 ). It represents the rate at which energy is transferred from large-scale motion to small scales. Eddies start big and then become progressively smaller. But how small before the flows become laminar because viscosity takes over? Combining the rate at which the smallest scales receive energy with the viscosity, using dimensional relationships we can derive length, time, and velocity scales for the smallest scales as follows h = n 3 /e 1/4 (24.3) t =(n/e) 1/2 (24.4) v =(ne) 1/4 (24.5) These are called the Komolgorov microscales of length, time, and velocity. The corresponding Reynolds number formed with these numbers is equal to one hv/n = 1 Small-scale motion is highly viscous, and that the viscous dissipation of turbulence as heat adjusts itself to the energy supply e by adjusting length scales h. Can we estimate what the supply rate of energy e is to small scales? Well the energy comes from large scale turbulence. This energy is proportional to u 2. It seems appropriate to assume that the time scale T associated with dissipation of the energy is then u/l which is roughly one overturning of the eddy. Therefore the rate of energy dissipation is of order u 2 /T = u 2 u/l. e u 3 /l (24.6) 110

111 Figure 24.1: Analysis of turbulence data Let s say you measure a time series of wind speed. By doing a fourier transform on this time series you can obtain a power spectrum of the data. It works out that for there to be a cascade of energy from large scales to small scales requires that the slope of this energy spectrum be - 5/3. At one end of the spectrum are the eddies of speed u with length scale l. At the other end is the Kolmogorov microscale with speed and length scales v and h. The rate at which energy is transferred between the two is e, and this is also the rate at which large scale turbulence gets converted to heat. The power spectrum continues leftward to larger length scales, but these tends to be associated with gravity waves first, and synoptic scale waves second. Main points Dimensional analysis is powerful! Isotropic turbulence spans a range of scales that is determined by primarily by the viscosity of air and less so by the turbulent energy dissipation rate 111

112 Question: Power dissipation by a cumulus cloud Figure 24.2: A cumulus cloud Estimate the energy dissipation rate in a cumulus cloud, both per unit mass and for the entire cloud. Base your estimates on velocity and length scales typical of cumulus clouds. Compute the total dissipation rate in kilowatts. Also estimate the Kolmogorov microscale h. Use r = 1.25kg/m 3 and n = m 2 /s. We can estimate the dissipation rate using e u 3 /l In a cumulus cloud a typical turret (arguably the length scale of the largest eddies) is about 250 m. The updraft velocity is about 1 m/s. This gives values of e m 2 /s 3 for the energy dissipation rate per unit mass. The energy dissipation rate of the entire cloud, assuming fairly suitable dimensions of 1 km 3 and r = 1.25 kg/m 3 is m e = = Js 1 = 5000KW (How does this compare to energy production associated with latent heat release during ascent?). The Kolmogorov microscale is h = n 3 /e 1/4 = ! 1/ = = 1mm So energy gets converted to heat in cloud at spatial scales on the order of 1 mm. Pretty small! Typical values of e range from 10 4 to 10 2 m 2 /s 3 in convective clouds to 10 5 to 10 4 m 2 /s 3 in stratiform clouds. 112

113 References Tennekes, H. and Lumley, J. L., A First Course in Turbulence, MIT Press, 1972, pp 300. Seinfeld, J. H., and Pandis, S. N. Atmospheric Chemistry and Physics, Wiley Interscience, pp

114 Part IV Atmospheric Radiation 25 Electromagnetic waves An electric generator rotates a magnet to create electricity. An magnetic resonance imager found in a hospital creates a magnetic field by passing a large electric current in a circle. A changing electric field induces a magnetic field and a changing magnetic field induces an electric field. The magnetic field is in each case normal to the electric field. Figure 25.1: 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 (25.1) ~H = ~H 0 exp i~k ~x iwt (25.2) 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 radition 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 114

115 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 (25.3) 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 (25.4) 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. Okay, but let s look at interactions of radiation with something that is not a vacuum. Remember from Eq ~E = ~E 0 exp i~k ~x iwt Assume that the wavenumber vector~k is a complex number with a real and imaginary component, i.e. ~k = ~ k 0 + i~k (25.5) Therefore, the consequence of k being complex is the following ~E = E 0 exp k ~ ~x exp i ~k0 ~x wt (25.6) Note that if ~k = 0 then the amplitude of the wave is constant. Thus the phase of the wave is given by At constant phase f = ~ k 0 ~x wt (25.7) df = 0 = ~ k 0 d~x wdt (25.8) implying that the phase speed of radiation in the direction the waves propagate is given by or alternatively v = d~x dt = w ~ k 0 (25.9) v = nl = w 2p l (25.10) 115

116 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 (25.11) For something that is not a vacuum, we can introduce a complex refractive index N for describing a medium. ~k = ~ k 0 + i~k = w c N = w c n0 + in (25.12) where N is the (complex) refractive index of the material with real and imaginary components n 0 and n 00, respectively. It can be shown that the real component of the refractive index is just a measure of how much the phase speed of light is slowed by a medium n 0 = c v (25.13) We showed previously that Therefore ~F µ ~E 2 ~F = ~F 0 exp 2~k ~x (25.14) Thus, the flux decays exponentially with distance through an absorbing medium. Now, since from Eq we can define an absorption coefficient such that, from Eqs and 33.6 ~k = w c Im(N)=w c n = 2pn l (25.15) b a = 4pn /l (25.16) ~F = ~F 0 exp( b a x) (25.17) This is known as Beer s Law. The flux of radiation is attenuated in an absorbing medium as the distance travelled through it. Main points Light can be expressed mathematically using the language of waves The phase speed of light in a medium can be related to the speed of light in a vacuum through a complex refractive index The complex component of the refractive index can be related to attenuation of the radiative flux Question What is the characteristic distance that light propagates before it is attenuated in an absorbing medium? 116

117 26 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 26.1: Electron two slit experiment, showing that photons exhibit a duality in their behaviour as both waves and particles. c From Eq , the frequency of light is related to its wavelength l through the speed of light n = c l (26.1) 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 (26.2) Strangely though, energy comes in discrete, although very small, packets or particles, whose energy is related to the frequency of a wave! E = hn (26.3) 117

118 where h is Planck s constant J s. Einstein figured this one out too, 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 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. Main points Light can also be treated in a quantized form where the energy per unit light is proportional to the frequency of the radiation High energy photons are better able to dissociate chemical bonds 118

119 27 Flux and Intensity 27.1 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 (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 (27.1) 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 (27.2) 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 (27.3) (27.4) F (l 1,l 2 )= Z l2 l 1 F l dl (27.5) 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 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 (27.6) 119

120 Figure 27.1: 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 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 (27.7) 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 1 120

121 δθ θ δφ Figure 27.2: Illustration of a solid angle 27.4 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 Z 2p Z p/2 0 0 I " (q,f)cosq sinqdqdf (27.8) F # = Z 2p Z p 0 p/2 I # (q,f)cosq sinqdqdf (27.9) Main points Radiation is integrated over frequency to obtain intensities and over the cosine weighted solid angle to obtain fluxes 121

122 δ ω R I r r F r F R R E r E R Figure 27.3: 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. Question Why can you see the headlight of a distant car clearly even if it is too far away to make you brighter? 122

123 28 Dipole radiation The atmosphere is made up of oxygen and nitrogen mostly, but also greenhouse gases. Greenhouse gases absorb radiation. Oxygen absorbs some solar radiation, but mostly nitrogen and oxygen 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. The incident electric field can be described by ~E = ~E 0 e iwt (28.1) An electric field exerts a force. Remember that force required to accelerate the charge q of a dipole, which has mass m, is ~F = q~e = m~a = m d2 ~x dt 2 (28.2) By doing thermodynamic work by displacing a charge through a force field, the incident electric field makes the electric field of the atom oscillate. The full expression for the displacement of the charge is the equation for a damped simple harmonic oscillator m d2 ~x dt 2 = K~x b d~x dt + ~x wd3 dt 3 + qe 0 exp( iwt) (28.3) = restoring + damping + recoil + f orcing You might be unfamiliar with the recoil or jerk term. It is a recoil associated with the radiative emission. There must be radiative emission to balance the radiative forcing and maintain equilibrium. How? The molecule is vibrating and moving its electric charge back and forth, so it generates its own electric field, distinct from the incident electric field. This new electromagnetic field represents what we term the scattered radiation. 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 (28.4) and if we try Eq as a solution in Eq. 28.3, then we obtain for the amplitude of the displacement ~x 0 = q E 0 m w0 2 w 2 (28.5) igw where w 0 = p K/m and where the damping term due to absorption is g = g a + g s w 2 /w 2 0 and the damping term due to scattering is g a = b m g s = w 0 w m 123 (28.6) (28.7)

124 The instantaneous rate at which the incoming electric field does work on the dipole oscillator is the force times the rate of displacement P = q~e d~x dt (28.8) The time averaged power hpi is distributed between absorption and scattering so that hpi = hp a i + hp s i (28.9) where hp a i = q2 2m E2 0 w 2 g a w 2 0 w g 2 w 2 (28.10) and hp s i = q2 2m E2 0 w 2 g s w 2 w0 2 w g 2 w 2 w0 2 (28.11) Energy is conserved since incident radiation is either absorbed or scattered. g s and g a determine how it is partitioned. Note that the relationship of power dissipated to the incident flux ~F is through the absorption and scattering cross-sections s Question hp a i = s a ~F (28.12) hp s i = s s ~F (28.13) Eq is complex. What is the implication for the relationship of~x(t) to ~E (t)? 124

125 29 Blackbody radiation Figure 29.1: A blackbody defined by a cavity where emission and absorption are in equilibrium so as to maintain a constant temperature The basic principles of thermal emission are as follows: 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. 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 ) similar to a gamma function of form f (x) µ x n exp( x) B n (T )= 2hn 3 c 2 e hn/kt 1 (29.1) 125

126 which, in terms of wavelength is B l (T )= 2hc 2 l 5 e hc/klt 1 (29.2) What is notable about this function is that it has a peak. We can derive the location of the peak B (MW m 2 µ m 1 sr 1 ) K 6000 K 5000 K (µ m) Figure 29.2: The blackbody spectrum. Note that colder temperatures correspond to lower energies of emission and longer wavelengths. (Wien s Law) by taking the first dervitation of B with respect to l and setting to zero to get l max = 2897/T (29.3) 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 (29.4) 126

127 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 dipole radiation that is emitted (our E 0 in dipole radiation) 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. 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, which from Eq suggest that a = b µ w 2. Things get more complicated here, but ultimately, we do not lose the central feature that the radiant intensity is a product of the thermal energy and its frequency squared. This classical approach gives us the so-called Rayleigh-Jean s law B nrj = 8pn2 c 3 kt (29.5) 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. A final consideration is that, when there are jumps, equilibrium systems follow a distribution of states in proportion to their potential energy called the Boltzmann distribution. This is only the exponential distribution we showed previously from Eq. 8.2 Dµ tot µ ext n µ ext = n 0 exp (29.6) or, n = n 0 exp( µ ext P.E./kT) where we showed the most familiar example of this is that in a gaseous medium (e.g. the atmosphere) density decreases with height. Making an isothermal approximation r = r 0 exp( 127 mgh/kt)

128 Because quantum mechanical energy is quantized (i.e. u n = nhn, where n = 0,1,2...). But we get something similar anyway which is that the energy distribution follows u qm = hn exp(hn/kt) 1 (29.7) Notice that this is much like e hn/kt and actually would be if energies weren t quantized. The main point here is that if energies are sufficiently high that hn kt, quantum mechanics must take over from the classical mechanics perspective. And in this perspective, radiant energy decreases with increasing frequency. Combining the Rayleigh-Jeans and Quantum mechanical approaches, we end up with our desired expression for Black-body radiation (Eq. 29.1) 2hn 3 B n (T )= c 2 (exp(hn/kt) 1) Which implies a peak in the spectrum as prescribed by Wien s law. So, as you can see, the derivation is involved, even in our hand-waved approach. However, it touches on a number of concepts that we will get back to repeatedly, so are worth keeping in mind. 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 (29.8) where s = Jm 2 sec 1 deg 4 is the Stefan-Boltzman constant. Note that, we can show that F = st 4 using three methods 1. Stefan showed that F = st 4 based on experimental measurements 2. Boltzmann showed that F µ T 4 based on thermodynamic arguments (for an assignment) 3. Planck showed that F = st 4, based on quantum mechanical arguments, where s = pb is a mix of fundamental constants 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 T E = A)F S 4 = (1 0.30) = 239.4Wm 2 1/4 FE /4 = s = 255K (29.9) 128

129 F F Figure 29.3: A planetary blackbody where emitted thermal radiation from the sphere is in balance with absorbed solar radiation. Main points The Planck function is similar to a gamma function implying it has a peak The integral of the function is proportional to temerature to the fourth power Question The Planck function seems to be most elegantly expressed in terms of frequency instead of wavelength. Does this mean that we should always express light in terms of frequency rather than wavelength? When can a case be made for using wavelength instead? 129

130 30 Local Thermodynamic Equilibrium 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 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. Every DOF is associated with an energy transition. At an equilibrium we find that, averaged over a large number of molecules, the probability that we will find a molecule with a particular energy level is determined by what is known as the Boltzmann distribution. Ignoring that some modes are degenerate (different physically but with the same energy), consider a hypothetical molecule with two energy levels 0 and 1 (0 being the ground). Then the relative population of state 1 to state 0 is given by n 1 n 0 = exp( hn/kt) i.e. there s exponentially fewer in a particular energy state, the higher the energy level. In fact we see this exponential expression as part of the Planck radiation equation B n = 2hn 3 c 2 e hn/kt 1 130

131 In fact, maintenance of this distribution is what creates the Planck blackbody radiation. 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 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 (30.1) 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. 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 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 (30.2) 131

132 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 (30.3) implying that 0 0 e l = a l (30.4) 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: It reemits the energy with emissivity but at temperature T b. Thus F bl (absorbed)=a l st 4 a 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. Question Is Eq necessarily true? Could there be another possible solution implied by Eq. 30.3? 132

133 31 Why is the sky blue? From Eq , we showed that the scattered power of dipole radiation is given by hp s i = e2 2m E2 0 w 2 g s w 2 w0 2 w 2 + g 2 w 2 w0 2 In the high frequency regime where w w 0, then absorption is negligible and the scattered electric field - the dipole radiation - is given by hp s i ' e2 w 4 g 2m E2 s 2 0 w0 6 (31.1) 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 (31.2) 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. 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 (31.3) 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. Question If short wavelengths are what are favored, than why is the sky not violet? 133

134 32 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 (32.1) 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 (32.2) 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. 134

135 Figure 32.1: 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) (32.3) so, for a given separation between sheets of dipoles r, there is no cancellation of forward scattered light (1 cosq = 0) and maximum cancellation at back angles (1 cosq = 2), as shown in Fig A similar argument can be applied to stacked sheets of dipoles. If two sheets of dipoles are separated by a distance r 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. 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 135

136 vibrate in phase. If a particle gets very much bigger than 0.5 µm across, then water molecules in one part of the particle no longer know about water molecules in the other part - they do not vibrate in synch. As a particle grows, this 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, which affect blue wavelengths. first. If the particle is large enough compared to the wavelength, then all colours are scattered roughly equally for particles about 2 µm across (for sunlight). 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 I 0 /l 4, which applies to 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. But the reason why clouds are white in fact 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. Question What is an example of an atmospheric phenomenon where interference is clearly visible? 33 The mathematics of scattering and absorption and emission 33.1 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 = This has the standard exponential decay solution Thus the transmittance is I l I l (t l )=I l (0)exp( t l ) T l = I (t l ) I (0) = exp( t l ) (33.1) 136

137 and the absorptance is A l = 1 exp( t l ) (33.2) 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 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 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 (33.3) s al = K al s (33.4) so, for example K s = s s (33.5) 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 (33.6) b sl = Ns sl = NK sl s (33.7) 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 (33.8) dt sl = b sl ds = Ns sl ds = NK sl sds (33.9) 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. 137

138 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 (33.10) 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 (33.11) and dt a = k a du (33.12) 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 (33.13) dz where refers to the top of the atmosphere. Note that dz = cosqds where ds is the slant path. Figure 33.1: Illustration of the slantpath through a vertical layer. 138

139 33.3 Transmittance in a plane parallel atmosphere Thus according to Eq 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) (33.14) 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 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: 33.4 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 (33.15) 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 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 = t s t s + t a (33.16) 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). 139

140 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 33.2: In this plot, the size parameter increases as one moves from the Rayleigh Scattering regime to the geometric optics regime 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 Question Z 1 1 p(µ)dµ = 1 (33.17) 2 1 What is another example of a dimensionless parameter like the size parameter we have discussed? What is the utility of dimensionless numbers like these? Question Why is haze brightest when looking generally towards the sun but clouds are brightest from the side when looking generally away from the sun? 140

141 5 m i = 0 Scattering efficiency K m i = 0.01 m i = 0.1 m i = Size parameter x Figure 33.3: 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 Radiative transfer equation The full statement of radiative transfer must acknowledge that while, according to Beer s Law, radiation is extinguished by particles within a layer, the particles also are a source for radiation. These sources are scattering of incoming radiation from layers above an below into the observer angle µ scattering of the incoming flux from the sun S into the observer angle µ isotropic emission of blackbody radiation by the layer Thus the entire equation for attenuation of radiation is di(t, µ) µ = I (t, µ) dt Some things to note w 0 2 Z 1 1 I t, µ 0 p µ, µ 0 dµ 0 w 0 4p Sp(µ, µ 0)e t/µ 0 (1 w 0 )B[T (t)] (34.1) Our convention here is that, by convention, t is positive down from the top of the atmosphere while µ = cosq is positive up. For this reason, the first term is a sink term, and the other terms on the RHS are source terms, even though they have negative signs in front. 141

142 Incident Beam (a) (b) (c) Forward Figure 33.4: 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. The equation is highly non-linear, and includes both an integral and a differential. It is mathematically impossible to solve without employing some approximations. You get Beer s Law out of the first term on the RHS The second term represents multiple scattering. Radiation comes into the layer from all directions µ 0. If it is not absorbed (fraction w 0 ) it will be scattered. It is scattered into the viewer angle µ with probability p(µ, µ 0 ). Radiation from all directions µ 0 can be scattered into direction µ, so we must integrate over all directions µ 0. A difficulty with the phase function is that it is extremely focused into the very forward scattering direction, so elaborate mathematical tricks are required to approach the phase function computationally. For climate modeling however, the only directions of interest are up and down, so what is done is to parameterize the phase function in terms of the asymmetry parameter defined as the mean cosine of the scattering angle g = hµi such that perfect forward scattering has g = 1, even scattering (dipole being an example) has g = 0 and perfect backscattering has g = 1. In terms of the phase function g = 1 2 Z 1 1 p(µ)µdµ (34.2) Note the cosine weighting. Values of g for water droplets are about 0.88, for ice crystals measurements point to values around 0.75, aerosols are around 0.7 and gases have g =

143 The second term represents scattering of the solar beam. It is obviously proportional to the flux of solar radiation at the top of the atmosphere S. It is also proportional to the amount the solar beam has already been attenuated by the atmosphere above exp( t/µ 0 ). µ 0 here is the zenith angle of the sun. It is positive downward by convention, whereas µ and µ 0 are positive upward by convention. Only radiation that isn t absorbed gets scattered, hence the w 0. 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 = cosq, 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 = The final term is the isotropic blackbody radiation. Since t here refers to the optical depth from both scattering and absorption, only fraction (1 w 0 ) is available to be absorbed and reemitted with intensity B[T (t)]. All negative terms on the RHS are source terms. That is, they add to the intensity along the direction of propagation of the beam. Question What first order simplifications would you try to make in order to approximately solve Eq. 34.1? 143

144 35 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) 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 (35.1) 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) 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) (35.2) 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 (35.3) 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) (35.4) 2I 2p 144

145 The frequencies associated with molecular transitions are where DE = E J 0 E J = hn (35.5) n = B n J 0 J B n J (J + 1) (35.6) 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. 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 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 (35.7) 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 (35.8) 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 DE n = N 0 hn 0 N hn 0 145

146 where the quantum mechanical rule is DN = N 0 N = ±1 N is ±1. Thus, the frequency of photons that excite a transition is given by or DE n = hn = DN hn 0 n = DN 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 Vibrational-rotational spectra 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. The important 15 µm 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. Main points Absorption of energy by a molecule corresponds to an energy transition at a frequency determined by the Planck constant Only specific transitions are allowed Molecular energy transitions can be electronic, rotational, vibrational, or some combination of rotational and vibrational Linear symmetric molecules have no pure rotational modes 146

147 36 Molecular absorption profiles 36.1 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. 35.7). 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 (36.1) 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. 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. We showed previously that the time averaged power hpi associated with dipole absorption or radiation (Eq ) is given by hp a i = q2 2m E2 0 w 2 g w 2 0 w g 2 w 2 where g = g a + g s but in this case g s g a. Since the flux of incident radiation is proportional to E 2 0 and the absorption cross-section is determined by hp ai = s a ~F (Eq ). The absorption cross-section is given by the expression s a = gs am gw 2 w 2 0 w g 2 w 2 (36.2) where s am is the maximum possible absorption cross-section and the rest is related to a line profile for the absorption as a function of forcing frequency w. If g w 0 then! gsam g s a = 4 (w 0 w) 2 (36.3) + g 2 /4 147

148 Doppler Lorentz k ν (ν ν 0 )/α Figure 36.1: 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. The first bracket expresses the line strength, and the second bracket the line shape. 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 (36.4) + al 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 (36.5) 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 (36.6) Essentially, this normalization means that adding up the energetic contributions to single line due to all frequencies gives the total. 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 148

149 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. Main points There is no absorption at a particular frequency only within a band The width of the band for a particular absorption mode is related to the damping In the lower atmosphere, damping is primarily due to the loss of energy that comes from molecules colliding 149

150 37 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 (37.1) + 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 where The band averaged absorptance is 37.1 Weak and strong lines T = 1 Z exp( t n )dn (37.2) Dn n t n = k a u = Sf(n)u (37.3) A =1 T (37.4) 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 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 37.3, 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. 36.4) 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 2 150

151 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 (37.5) and for large values of u (strong lines) T = 1 2 p Sa L u Dn (37.6) 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. Main points Weak lines are linear with the amount of absorbing gas in the atmosphere Strong lines go as the square root of the amount of absorbing gas in the atmosphere Question As a silly question, could we solve global warming by sucking all the air out of the atmosphere, thereby reducing the width of absorption modes? 151

152 38 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 1.7 ppm. Figure 38.1: The Keeling Curve showing the increase in CO2 concentrations in the atmosphere 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 152

153 Figure 38.2: Trends in methane in the atmosphere 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 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 38.3 summarizes some estimates of climate sensitivity based on numerical model and paleoclimate estimates. 153

154 Figure 38.3: (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 model-derived 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,