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The Atmosphere

These notes contain everything necessary to run a flipped course on"The Atmosphere"at an introductory undergraduate level. There are notes for the students to read before each course meeting and problems for them to work on in small groups during course meetings. Topics include (1) atmospheric composition, structure, and thermodynamics; (2) solar and terrestrial radiation in the atmospheric energy balance; (3) atmospheric dynamics and circulation. I include 10 problem sets, four practice midterms, and two practice finals. Problems are drawn from the atmospheres of modern and past Earth, solar system planets, and extrasolar planets. You are free to use these notes in your classes, and to expand them as you please. If you catch any typos or errors, please send them to me. Enjoy!


Acknowledgements
Daniel Koll and Jonah Bloch-Johnson were my teaching assistants for the first two years and helped develop many of the questions in the problem sets. Daniel was particularly good at suggesting questions. Jonah has been my teaching assistant for four years now, and has been extremely helpful in developing the flipped course. He has excellent physical insight and contributed to the descriptions in the lesson introductions in a number of places. Jade Checlair and David Plotkin have also both served as teaching assistants and helped me develop the course a great deal.
I would also like to acknowledge the undergraduate students at the University of Chicago who have taken the course. The students have been extremely bright and hard-working. It's been a pleasure to work with them and seeing them learn has made my work on this course worth it. Their questions have helped me improve my understanding of the course material and the descriptions in the lessons tremendously.
I used figures from a variety of sources in these notes: some from Wallace and Hobbs, some from papers, some from the internet, and some I made myself. To focus student attention on the most important references, I didn't reference the source for most figures. But I'd like to acknowledge the work people put into making them and say thanks! Finally, I am grateful to Wallace and Hobbs for developing such an excellent textbook, which these notes are partially based on.

A sample syllabus
Overview: This course introduces the physics and phenomenology of the Earth's atmosphere, with an emphasis on the fundamental science that underlies atmospheric behavior and climate. Topics include (1) atmospheric composition, structure, and thermodynamics; (2) solar and terrestrial radiation in the atmospheric energy balance; (3) atmospheric dynamics and circulation. This is a hard course. Understanding the concepts of the atmosphere requires mathematics. If you struggle with mathematics and abstraction, expect to devote extra time to this course. That said, no one who has ever shown up to every class, worked hard, and not cheated, has ever gotten less than a C. So passing this course is within all of your capabilities.
Prerequisites: MATH 13200 (math through basic calculus) Format: This will be a "flipped" course. Before each class meeting you should read the introduction of the appropriate class worksheet. Note that the most important equations will be set out of the text in numbered equations. If you have any questions on the material, please ask them at the start of class. In class you will work in small groups on the problems in the worksheet, and we will be available for questions. At the end of each class we will discuss the problems together and give the answers. The worksheets will not be graded.
In this course we are opening the door to knowledge for you. We cannot force you through it. If you do not understand something, it is your responsibility to ask about it. If you didn't understand the reading, you need to raise your hand at the start of class. If your group is moving too quickly through a problem that you don't understand, you need to stop them. If you have questions about course material that aren't resolved in class, you need to come to office hours and ask them.
Problem Sets: There will be one problem set per week. These are designed to take the average student 1.5-3 hours per week. There is a fair amount of variation in the time it takes students, so you will have to calibrate for yourself. Also, some of the problem sets will take longer than others. Make sure to start all of them early, and go to office hours if you have questions.
You may discuss problems with each other, but you must each write up your own solutions separately. It is not appropriate to write up solutions in the presence of other students, to discuss your solutions with others as you write them up, or to share your written-up solutions with other students. Also, if you are asked to explain something, don't just Google it and copy the first thing that pops up. Problem set solutions will be due at the start of class on the specified due date, and will not be accepted late. This is necessary so that all problem sets can be graded and returned quickly.
Tests: There will be two midterms and a final. The notes provide example tests from previous years. The best way to use these is to print them out without looking at them, go to the library with only a pen, and take them for the appropriate amount of time. You can then grade them yourselves using the solutions, and use the results to determine how to focus your study.
Grading: The problem sets will represent 50% of the final grade (5% each), the midterms will be worth 10% of the final grade, and the final will be worth 30% of the final grade. The grade scheme will be as follows:

Orientation and Estimation
We're going to start with some exercises where you will look at observations to get oriented to the important features of Earth's atmosphere. You may not know all of the terms involved, but you can learn by doing the exercises. Here's a diagram of tropical atmospheric circulation that might be helpful: An important skill that I hope you will learn a bit about in this class is the ability to estimate the answer to a problem quickly. Scientists use this when they encounter a problem to see if a potential solution is plausible before making a complex calculation. We also often make complicated calculations on computers nowadays, and estimation is a great way to check if the computer is telling you something crazy or maybe true. In this course I may either ask you to estimate the "order of magnitude" or the answer to "one significant digit." For example, a typical man is about 1.75 m tall. The order of magnitude of his height is 1 m and his height is 2 m to one significant digit.
For an example, let's estimate the order of magnitude of the total mass of beard grown by a man per year. There are many ways to do this, but let's try the following. The total mass of beard, m, should be the number of beard hairs, N, multiplied by the cross-sectional area of a beard hair, σ , multiplied by the length each hair grows in a year, l, multiplied by the density of beard hair ρ: m " Nσ lρ.
To get the answer now we just need to estimate each term in this equation. The spacing between hairs is about 1 mm, so there are 100 hairs cm´2. Consider the head to be a sphere of radius 10 cm. The area of the beard is a quarter of the total surface area, or 1 4ˆ4 πr 2 =πˆ10 2 cm 2 . So N=(πˆ10 2 cm 2 qˆ10 2 hairs cm´2 = πˆ10 4 hairs. The smallest thing the human eye can see is about 0.1 mm (you can check this with a ruler). We can just make out the thickness of a beard hair, so it's diameter must be about 0.1 mm=10´4 m. This means that σ " πr 2 " πp0.5ˆ10´4mq 2 hair´1 « 10´8 m 2 hair´1. Notice that we set π 4 « 1. This is appropriate since we are only trying to estimate the order of magnitude. Hair, including beard hair, grows about 0.5 in per month, or about 1 cm per month. Many of you probably know this. If you didn't, you probably know immediately that a beard grows faster than 1 mm per month and slower than 10 cm per month. When you come to something you really don't know how to estimate, one trick is to find a lower and upper limit, then take the geometric mean. In this case, it would give us 1 cm, which turns out to be the right answer. There are order of 10 months per year, so l " 10 cm year´1 " 0.1 m year´1. Hair sinks in water (if it breaks the surface tension), but just barely. So the density of hair must be similar to that of water, or ρ " 10 3 kg m´3. We can put this all together to find: m "(π1 0 4 hairs)ˆ(10´8 m 2 hair´1)ˆ(10´1 m year´1)ˆ(10 3 kg m´3)=πˆ10´2 kg year´1 « 30 g. For comparison, the mass of an adult hummingbird is about 3 g. So a typical man shaves off 10 hummingbirds worth of beard every year! See the picture below for an example of what happens if you let all that mass accumulate for multiple years. Notice that I kept track of the factor of π in this exercise, even though this isn't strictly necessary to estimate the order of magnitude. Two critical things to take away from this exercise are: (1) Always check your units and make sure they come out right, and (2) use scientific notation.

Exercises
(1) Watch this video (https://youtu.be/gpKzp3Br6-k) a couple times. It contains pictures of Earth from above taken by an orbiting satellite stitched together.
A. Find the extratropical cyclones (eddies), which occur in the mid-latitudes. Notice that they are roughly symmetric between the northern and southern hemispheres. These are the storms that cause most weather in Chicago. Can you tell which direction they spin (clockwise or counter-clockwise) in each hemisphere? You can slow the video down to make it easier to see what's happening.
B. The Inter-Tropical Convergence Zone (ITCZ) is the region where the northern and southern Hadley cells converge, near the equator. This causes convection, clouds, and precipitation. Can you find the ITCZ? Is it north or south of the equator?
C. Tropical cyclones are called hurricanes, typhoons, or cyclones, depending on the ocean basin they occur in. Can you find some tropical cyclones? How do they compare with extratropical cyclones?
D. This video goes through about one year. Find at least two ways to determine the season in Chicago as time progresses.
(2) Pressure is force per area. Weight is a force equal to the gravitational acceleration multiplied by the mass. The surface pressure of the atmosphere on Earth is about 10 5 Pa, where 1 Pa=1 N m´2. Earth's radius is 6,400 km. Estimate the total mass of Earth's atmosphere without using a calculator. Keep only one significant digit. The total mass of Earth is 6ˆ10 24 kg and the mass of Earth's oceans is 10 21 kg. How does the mass of Earth's atmosphere compare to the total mass and the mass of the oceans?
(3) As Caesar lay dying in the Theater of Pompey on the Ides of March 44 BC, he breathed one last breath. As you inhale now, how many molecules from Caesar's last breath do you breath in? Again, make an "order-of-magnitude" estimate and do not use a calculator. Remember that Avogadro's number is 6ˆ10 23 molecules mol´1. Air has a molecular mass of about 30 g mol´1. Assume that Caesar's last breath has now been mixed throughout the entire atmosphere. You will have to estimate the volume of a breath, and you can assume that the density of air near the surface is 1 kg m´3.

More Orientation and Estimation
Today we're going to continue our orientation to Earth's atmosphere and make some calculations to get a rough understanding of modern fossil fuel usage. Reread the introduction from last class to make sure you understand estimation. Remember to use scientific notation and pay careful attention to units for the estimation problems.
To build up vocabulary, take a look at the figure below. It shows an idealized typical "vertical temperature profile" of Earth's atmosphere. The lowest region, where we live, is called the troposphere. The surface is strongly heated by sunlight and transfers this heat to the lower atmosphere. This causes the air near the surface to become less dense and rise in a process called convection. Convection happens throughout the troposphere so that air in it is well-mixed. Above the troposphere is the stratosphere. The stratosphere is "stratified," with hotter air on top of colder air as a result of heating by absorption of sunlight by ozone. This makes vertical motion and mixing of air much harder in the stratosphere than in the troposphere. The tropopause separates the troposphere from the stratosphere. The troposphere and stratosphere contain 99.9% of the atmosphere, and will be the focus of this course. The other regions of the atmosphere only become important in certain specialized areas. 12
A. For Jan/Feb, focus on the northern hemisphere, poleward of 30˝N. Why are there two large cold regions?
B. Why is there such a persistent blue spot at (30˝N,90˝E)?
C. For Jul/Aug, focus on the northern hemisphere, poleward of 30˝N. Why are the regions that were previously cold now warm?
D. If you watch the two movies carefully you can detect large, eastwards moving swirls (weather systems) poleward of 30˝N. When are the swirls most active in each hemisphere?
(2) What is the total mass of CO 2 put into the atmosphere by a 1-GW coal power plant per year? You may assume that the plant has an efficiency of 1/3 and that burning coal releases 10 7 J kg´1. Remember that coal is basically just carbon. Total world power consumption is 10 5 TW-h per year. How many 1-GW power stations would be needed to supply this power, assuming that they run continuously (1 TW=10 3 GW)? The total mass of CO 2 in the atmosphere is 3ˆ10 15 kg. Assuming the entire world's power supply is provided by coal, how many years would it take to double the amount of CO 2 in the atmsophere? Assume that no other process adds or removes CO 2 from the atmosphere and assume that humanity's power usage stays constant.
(3) What is the mass of CO 2 emitted by a typical car in the US per year? Assume it is driven 10 4 mi and gets 20 mpg. You may approximate gasoline as made entirely of carbon. If on average there is a car for every two people in the US, what is the total mass of CO 2 emitted by cars in the US per year? 13

Pressure and Cyclones
A cyclone is a relatively small part of the atmosphere (maybe on a scale of about 1000 km) that rotates around a region of low pressure (see the diagram below). This rotation is due to the Coriolis force, which we will learn about later in the course. The direction of rotation is counterclockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere. In either case, this direction is referred to as "cyclonic." Anticyclones, on the other hand, have high pressure at their center and rotate in the opposite direction. Cyclones have air converging near the surface, which generally leads to air rising, which causes precipitation and the formation of clouds. Extratropical cyclones, also called baroclinic eddies, are the storms we are familiar with in Chicago. They are particularly common in the winter. The reason is that the North-South gradient in surface temperature is ultimately what drives them, and this gradient is stronger in the winter. We usually get a new one coming through every 3-5 days. Tropical cyclones are driven by a different mechanism, which we will learn more about later. Tropical cyclones are called hurricanes in the Atlantic ocean and eastern Pacific, typhoons in the western Pacific, and cyclones in the Indian and South Pacific. Cyclones generally get blown around by the prevailing winds. There are also larger scale lows and highs in the atmosphere that persist in roughly the same place for entire seasons. These can deviate winds and lead to seasonal shifts in weather. 14

Compositon and the Ideal Gas Law
The ideal gas law is where P is the pressure in Pa, V is the volume in m 3 , n is the number of moles, R˚=8.3 J mol´1 K´1 is the ideal gas constant, and T is the temperature in K. In atmospheric science we usually rewrite this equation as P " ρRT, where ρ " n˚M V is the gas density in kg m´3, M is the molecular mass of the gas mixture in kg mol´1 (you probably learned molecular masses in g mol´1, so don't forget to convert to kg mol´1), and is called the gas constant. The gas constant, R, is measured in J kg´1 K´1, and is specific to the particular gas mixture under consideration (it depends on the molecular mass). If we are considering a mixture of gases, then the apparent molecular weight is the molar-fraction-weighted average of the molecular weights M " where f i is the molar fraction of the ith component and M i is the molecular weight of the ith component. 17

Exercises
(1) Stars increase their luminosity with time, such that the Sun was about 25% less luminous 4 billion years ago than it is today. Nevertheless, we have strong evidence that Earth's climate was fairly similar to how it is now, rather than completely frozen over. Carl Sagan, an alumnus of the University of Chicago, was one of the first people to point out that this is an interesting mystery, and to try to solve it [2]. Most people now think that the solution is that some greenhouse gases (CO 2 , CH 4 , and even H 2 ) had much higher concentrations on early Earth than they do today. F. Calculate the number of moles of CO 2 in the entire column of air per unit area of this atmosphere.
G. Calculate the mass of CO 2 in the entire column of air per unit area of this atmosphere. You should have found that the mass of CO 2 in the entire column of air per unit area is greater than the partial pressure of CO 2 at the surface divided by the surface gravity. The reason is that the partial pressure of CO 2 at the surface results from the weight of both N 2 and CO 2 throughout the atmosphere, and N 2 has a lower molecular mass.
H. What is the mass fraction of CO 2 in the atmosphere?
I. Suppose we want to increase the amount of CO 2 so that it represents 50% of the atmosphere by molar fraction. What would the new partial pressure of CO 2 and total atmospheric pressure have to be?

Virtual Temperature
One issue that complicates a lot of things on Earth is that the atmosphere has a variable amount of water vapor, depending on the temperature and atmospheric motion. This means that the gas constant changes with time and as you move around. This makes it complicated to calculate the density, which we often need to figure out if air is going to convect, or mix vertically because denser air is above less dense air (atmospheric convection is slightly more complicated than this, but we'll get to that later). Here we'll try to deal with this. First, note that the ideal gas law applies to the dry component of air: where P d is the partial pressure of dry air, m d is the mass of dry air, and R d is the dry air gas constant. The ideal gas law also applies to the moist component of air: e " m v V R v T , where e is the partial pressure of water vapor, m v is the mass of water vapor, and R v is the gas constant of water vapor. By Daulton's law, we also know that the total pressure is the sum of the partial pressures: P " P d`e . Finally, we can write the total density as: ρ " m d`mv V . If we solve these equations for density we get: where we have defined the virtual temperature, T v , as: Using the virtual temperature allows us to use the ideal gas law with a constant gas constant (R d ) rather than a variable gas constant (that depends on the amount of water vapor) in order to calculate the air density. The cost is that we have to define a new temperature, the virtual temperature, which depends on the amount of water vapor. Physically, the virtual temperature is the temperature dry air would need in order to have the same density as the moist air under consideration. On modern Earth M v ă M d , so adding moisture to air makes it less dense. This means that on modern Earth the virtual temperature is always greater than or equal to the temperature.

Hydrostatic Balance and the Scale Height
First let's derive the hydrostatic equation. This works for an atmosphere in hydrostatic balance (not accelerating vertically). It describes a balance between a net effect of pressure pushing up and the weight (force due to gravity) of air pulling down. Consider the diagram below: 68 Atmospheric Thermodynamics vertical force that acts on the slab of air between z and z ϩ ␦z due to the pressure of the surrounding air. Let the change in pressure in going from height z to height z ϩ ␦z be ␦p, as indicated in Fig. 3.1. Because we know that pressure decreases with height, ␦p must be a negative quantity, and the upward pressure on the lower face of the shaded block must be slightly greater than the downward pressure on the upper face of the block. Therefore, the net vertical force on the block due to the vertical gradient of pressure is upward and given by the positive quantity Ϫ␦p, as indicated in Fig. 3 That is, the pressure at height z is equal to of the air in the vertical column of sectional area lying above that level. If t the Earth's atmosphere were distributed over the globe, retaining the Earth's in its present form, the pressure at sea l be 1.013 ϫ 10 5 Pa, or 1013 hPa, which is as 1 atmosphere (or 1 atm).

Geopotential
The geopotential ⌽ at any point in t atmosphere is defined as the work tha done against the Earth's gravitational fie a mass of 1 kg from sea level to that poin words, ⌽ is the gravitational potentia mass. The units of geopotential are J kg Ϫ1 The force (in newtons) acting on 1 kg above sea level is numerically equal to (in joules) in raising 1 kg from z to z ϩ therefore or, using (3.18), Let's do a force balance on the darkened slice of the column of air. The bottom of this slice has a pressure force pushing up of PpzqˆA, where A is the horizontal area of the column. The slice has a pressure force pushing down on it of´Ppz`δ zqˆA. The gravitational force of the air pushes down and is´ρˆAˆδ zˆg (mass times gravitational acceleration). So assuming that there is no net vertical acceleration, the vertical force balance is: We can cancel A and rearrange this equation as Ppz`δ zq´Ppzq δ z "´ρg, or in the infinitesimal limit dP dz "´ρg.
This is the hydrostatic equation and it describes a force balance between the vertical gradient of pressure and the weight of the air. Since the pressure goes to zero as the height goes to infinity, we can integrate vertically to find that Ppzq " A related concept is the geopotential, Φ, which is defined as the work that must be done against Earth's gravitational field to raise a mass of 1 kg from sea level to the height under consideration. dΦ " gdz, and we can integrate to find that since g is roughly constant throughout the atmosphere. We will hear more about the geopotential later in the course. Finally, let's derive an equation for how the pressure decays with height. We can combine the hydrostatic equation ( dP dz "´ρg) and the ideal gas law (P " ρRT ) to eliminate ρ: dP dz "´g RT P. If we assume that T is constant, we can define H " RT g (9) and integrate this equation to find

The Lapse Rate and Potential Temperature
Energy comes in different flavors. The first law of thermodynamics says that energy is conserved. If you lose some in one flavor, you have to gain it in another flavor. One of the flavors is internal energy. Even if a parcel of air seems motionless on a large scale, it's molecules still have kinetic and potential energy, which we call internal energy. Temperature characterizes the kinetic internal energy of molecules. Other flavors of energy are heat added to the system and work done by the system. We can write energy conservation in differential form for a parcel of air as follows where dq is the heat added to the air parcel, du is the change in internal energy of the air parcel, and dw is the work done by the air parcel. This equation says that if we add heat to an air parcel, it goes into some combination of increasing the temperature of the parcel or allowing the parcel to do work on its surrounding environment.
To think about work done by an air parcel, let's consider the air in the piston shown in the figure below. The work done by the air in the chamber is the external force it is pushing against multiplied by the distance it pushes (dW " Fdx). If we assume that the force pushing in on the chamber is from the pressure of the air outside, then we find that F " pA, where A is the area of the stopper. The change in volume of the parcel is just its area multiplied by the distance moved perpendicular to the stopper (dV " Adx), so we can write dW " pdV . We won't prove it here, but this also works for an air parcel that isn't in a chamber. If w is the work per unit mass and α " 1 ρ is the specific volume of the air, or volume per unit mass, then we can also write this as dw " pdα and we can rewrite the 1st law of thermodynamics as dq " du`pdα.
The specific heat is the amount of heat we have to add to a substance to increase its temperature by 1 K. The specific heat at constant volume is c v " p dq dT q v , where the subscript v means we take the derivative at constant volume. At constant volume dq " du (all the heat goes into increasing the temperature, and none goes into doing work) so that c v " p du dT q v . For an ideal gas u depends only on T (we didn't show this, but trust me), so we can drop the constant volume assumption and get c v " du dT , or du " c v dT . This means we can rewrite the 1st law as dq " c v dT`pdα.
The specific heat at constant pressure is defined as c p " p dq dT q p . It is often more useful than the specific heat at constant volume in atmospheric sciences because gases in the atmosphere tend to be at constant pressure rather than volume. Using the product rule we can rewrite the 1st law as dq " c v dT`dppαq´αd p. Using the ideal gas law, dppαq " dp p ρ q " dpRT q " RdT , so that dq " pc v`R qdT´αd p. Or at constant pressure (d p " 0), dq " pc v`R qdT , which means that c p " p dq dT q p " c v`R . This means we can also write the 1st law as dq " c p dT´αd p.
Now let's think about the temperature lapse rate, or decrease in temperature with height in the atmosphere. Let's consider an air parcel that changes height with no heat added or subtracted (dq " 0), which means the motion is an "adiabatic process." We now have the tools to infer the lapse rate in this situation, which is called the "dry adiabatic lapse rate" and is written Γ d " p dT dz q q"0 . We use the word "dry" because the condensation of moisture releases a lot of heat. The atmosphere tends to roughly follow this dry adiabatic lapse rate at least sometimes in regions that are dry like deserts, and cold regions at the poles or high in the atmosphere. With dq " 0, we have 0 " c p dT´αd p. Using the hydrostatic equation (d p "´ρgdz "´g α dz) we find that 0 " c p dT`gdz so that we can write the dry adiabatic lapse rate as Note that deriving the dry adiabatic lapse rate required the 1st law of thermodynamics (conservation of energy), the ideal gas law, and the hydrostatic equation. In Earth's atmosphere, c p « 10 3 J K´1 kg´1 and g « 10 m s´2, so that Γ d « 10´2 K m´1 " 10 K km´1. Finally, let's define a new quantity called potential temperature. Potential temperature is nice because an air parcel has constant potential temperature if no heat is added or subtracted from it. Let's return to the 1st law with no heat exchange (0 " c p dT´αd p). The ideal gas law tells us that α " RT p so that 0 " c p dT T´R d p p . Integrating we find that const. " c p logpT q´R logppq which we can exponentiate to rewrite as T " pconst.qp R c p . This equation tells us that for an adiabatic process the temperature and pressure have to be related in this way for some constant that we can find. At surface pressure (p " p 0 ), we will define the temperature as T " θ , which will be the potential temperature. So the constant equals θ p´R c p 0 . Notice that since p 0 is a constant, θ needs to be a constant as well. We can rearrange to find which defines the potential temperature. The bottom line is that any air parcel that moves adiabatically will have a constant potential temperature (θ ). If it goes to lower pressure and decompresses, it's temperature will decrease, but it's potential temperature will stay constant. If it goes to higher pressure and compresses, it's temperature will increase, but it's potential temperature will stay constant. 25

Exercises
(1) Explain the following: A. Air released from a tire is cooler than its surroundings.
B. The potential temperature in a dry convecting atmosphere is constant.
C. In Greek mythology, Icarus falls from the sky because he flew too high and close to the Sun so the wings his father, Daedalus, made for him get too hot and melt. Assuming that Icarus was flying in the troposphere, does this make sense?
D. Students are taught that when air is heated it expands. Then they are taught (but not in the same lesson) that when air expands it cools. Some students find these two explanations to be contradictory. Please resolve the apparent contradiction.
E. An airplane flies in the troposphere of a dry atmosphere. The air outside the airplane is very cold. When this air is brought into the pressurized cabin of the airplane, it has the same pleasant temperature for the passengers that they experienced on the ground before taking off.
F. How does the potential temperature of air at the surface, in the atmosphere at the plane's altitude, and inside the pressurized cabin compare?

Stability
The lapse rate and potential temperature are important concepts, so we're going to continue learning about them today. The problem will continue to focus on these general issues, and these notes will focus on how you can use the lapse rate and potential temperature to think about atmospheric stability, or whether vertical convection will occur. If an air parcel moves adiabatically vertically, it will follow the dry adiabatic lapse rate and have a constant potential temperature. If the actual atmosphere has a lapse rate that is greater than the dry adiabatic lapse rate, then a parcel lifted from the surface will have a higher potential temperature than the surrounding atmosphere at height (and therefore a higher temperature). This means it will be bouyant (less dense that the surroundings). This is an unstable situation and convection (strong vertical mixing) will occur until the lapse rate is equal to the dry adiabatic lapse rate. This means that an atmosphere cannot sustain a lapse rate higher than the dry adiabatic lapse rate, or such an atmosphere unstable (see first figure below). During the day in the troposphere the Sun is continually heating the surface, increasing the lapse rate and destabalizing the atmosphere so that convection occurs. On the other hand, if an atmosphere has a lapse rate less than the dry adiabatic lapse rate, as it does in the stratosphere, this is a stable situation and no convection needs to occur. The tropical troposphere can also have a lapse rate less than the dry adiabatic lapse rate because of water vapor condensation, which we will learn about soon. As the second figure below shows, we can see this as an increase in the potential temperature with height in the zonal mean (averaged over longitude) potential temperature profile. This means it is stable to dry convection, and the potential temperature increases with height. As we will learn, the tropical troposphere can still be unstable to moist convection in this configuration. The figure shows that the extratropics are also generally stable to dry convection, which results from atmospheric heat transport from lower latitudes and altitudes to higher latitudes and altitudes.

Exercises
(1) The Galileo probe descended into Jupiter's atmosphere in July, 1995 [3]. The figure below shows temperature data (blue line) retrieved from the probe's descent. The data start when the probe entered Jupiter's atmosphere. The data end at the bottom right of the plot because the probe failed. The pressure was about 23 bar and the temperature was about 425 K when the probe failed. The notional altitude is taken to be zero at a temperature of 166 K in this plot. The pressure at this altitude is equal to Earth's surface pressure.
A. Notice that Jupiter has a stratosphere and a troposphere. At roughly what temperature and notional altitude is the tropopause located? F. The heat capacity at constant pressure (C p ) of an ideal diatomic gas in this temperature range is 3.5R, where R is the (compound specific) gas constant in J K´1 kg´1. Make a prediction for the temperature lapse rate on Jupiter. Now estimate the temperature lapse rate from the plot and compare to your prediction.

Water Vapor in the Atmosphere
Today we're going to think more about water vapor in the atmosphere. Water vapor is important because it affects the air density and the ability of humans to lose excess heat through sweating. It is highly variable in the atmosphere and we need to define variables to keep track of it. The partial pressure of water vapor in the atmosphere, or vapor pressure, is often denoted e. The mass of water vapor (m v ) divided by the mass of dry air (m d ) is called the mixing ratio of water vapor (w), The specific humidity (q) is the ratio of water vapor mass to total air mass, including both dry air and water vapor, If the air is fairly dry, then q « w. The vapor pressure is the molar fraction of water vapor multiplied by the total pressure (p), or e " n v n d`nv p. Using the masses (little m) and molecular masses (big M), we can rewrite this as e " where ε " M v M d . Liquid water molecules are in constant motion. This kinetic energy is measured by temperature. Sometimes the molecules burst across the liquid surface into the surrounding air, which is a process we call evaporation (left panel of figure below). On the other hand, water molecules in gaseous form also have kinetic energy, which sometimes causes them to bash into the liquid water and join it. If we put liquid water in an enclosed container, these two processes will eventually balance each other (right panel of figure below). In other words in any given amount of time, as many liquid water molecules will be crossing the liquid threshold and becoming gaseous as gaseous water molecules will be crashing into the liquid and becoming liquid. The water vapor partial pressure when this equilibrium is established is called the saturation vapor pressure (e s ).
The general term for this type of phase transition boundary is a Clausius-Clapeyron relationship, and this term is used for the water vapor pressure in Earth's atmosphere. The Clausius Clapeyron relationship tells us that the saturation vapor pressure is a strong function of temperature (figure below). For temperatures below 0˝C, e s is very close to zero, and the air is almost completely dry. For a temperature of 100˝C, e s is equal to the standard atmospheric surface pressure. In fact, this is what defines the boiling point of water. At normal temperatures on Earth's surface, the saturation vapor pressure increases about 7% per degree Celcius of warming. Another way to measure water is in terms of the relative humidity, or the percentage of water vapor mixing ratio relative to the saturation water vapor mixing ratio.
Water vapor and dry atmosphere satisfy the ideal gas law idependently, so The dew point (T d ) is the temperature to which air must be cooled at constant pressure in order for it to be saturated in water vapor. So for air with a water vapor mixing ratio of w, The wet bulb temperature (T w ) is the temperature measured by a thermometer with a wet cloth covering its bulb. If there is enough circulation for the ambient air to pass through the cloth, water will evaporate into the ambient air until it is saturated, lowering the air temperature. So the wet bulb temperature is the temperature of a parcel of air cooled to saturation by the evaporation of water into it. In other words, the wet bulb temperature is the lowest temperature that can be achieved by evaporative cooling of a surface into ambient air (notice the analogy with human cooling via sweat). The dark lines in the figure below show contours of the wet bulb temperature at surface pressure for different ambient temperatures and relative humidities. As you can see, the drier the air, the lower the wet bulb temperature. The lower left of this plot is cut off because the fit being used is unstable in that region, not for any physical reason.
To understand the relationship between the dew point and the wet bulb temperature, let's imagine air with temperature T and a mixing ratio (w 0 ) that is less than saturation approaching a wet cloth. The dew point is defined by w 0 " w s pT d q. If the air passes slowly through the cloth, it reaches saturation via evaporation of water into it, and has a new mixing ratio (w f ). Note that w 0 ă w f . The new temperature of the air is the wet bulb temperature, which is defined by w f " w s pT w q. Since w 0 ă w f , T d ă T w . Both the dew point and the wet bulb temperatures must be less than the initial temperature, so T d ă T w ă T . In the special case where the air was initially saturated, T d " T w " T .

Exercises
(1) Explain why hot weather causes more human discomfort when the air is humid than when it is dry.
(2) How much liquid water must evaporate to lower the temperature of a person by 5˝C? Give your answer as a percentage of the mass of the person and assume that the latent heat of evaporation of water is 2.5ˆ10 6 J kg´1 and the specific heat of the human body is 4.2ˆ10 3 J kg´1 K´1.
(3) Solve for vapor pressure in terms of relative humidity, total pressure, and saturation vapor pressure. Use this to write the virtual temperature (T v " T 1´e p p1´εq ) in terms of relative humidity and saturation vapor pressure, instead of in terms of vapor pressure. Do this exactly, then make an approximation assuming air with a low specific humidity.

Moist Convection
Let's consider an air parcel with water vapor in it that rises and experiences colder temperatures as it does. The vapor pressure is such a strong function of temperature that this will eventually lead to condensation if the parcel rises enough. Condensation releases heat, so the adiabatic assumption we used to derive the dry adiabatic lapse rate will be violated. In such a situation the actual lapse rate will actually be less than the dry adiabatic lapse rate because the heat released by condensation will raise the temperature of the air aloft where condensation occurs. We can see this in the plot below, which shows typical atmospheric temperature profiles in different regions of Earth and seasons. In all cases the observed lapse rate is less than the dry adiabatic lapse rate (10 K km´1). We call the lapse rate in an atmosphere where the only non-adiabatic process we allow is the condensation of water vapor the moist adiabatic lapse rate. The moist adiabatic lapse rate is smaller if the water vapor mixing ratio is higher because more condensation and heating occurs when the air is lifted. Now let's derive a moist analog to the potential temperature. We will call it the equivalent potential temperature, and it will be roughly constant in an atmosphere experiencing moist convection. Based on Lesson 06, we know that the 1st law of thermodynamics can be written as If the heat is released by condensation of water vapor, then dq "´L v dw s , where L v is the latent heat, so that dθ θ "´L v c p T dw s . Now notice that dp w s T q " w s T p dw s w s´d T T q « dw s T , using dT T ! dw s w s because a small fractional change in temperature causes a large fractional change in mixing ratio along the saturation curve. Therefore, assuming that L v and c p are constant, we can write dθ θ "´d´L v w s c p T¯, which we can integrate to find log θ`const. "´L v w s c p T . We choose the constant to be log θ e , where θ e is the equivalent potential temperature, using the fact that the θ e " θ if w s " 0. Now we can solve for the equivalent potential temperature as The equivalent potential temperature is the potential temperature that an air parcel would have if all of the water vapor it contains were condensed and the heat from the condensation used to heat the parcel. It is also the temperature that an air parcel would have if it were slowly raised in the atmosphere to a low enough temperature that essentially all of the water vapor in it had condensed, such that it retained all the heat that the condensation released, then slowly and adiabatically lowered to the surface.
(1) The plot below shows results from numerical calculations using a one dimensional (vertical) model of the atmosphere [4]. The model can be run with convection artificially repressed, even if a real atmosphere would be convectively unstable. This is called radiative equilibrium. The model can also be run including a simple parameterization of moist convection. This is called radiative-convective equilibrium.
A. Is the temperature lapse rate larger in the lowest few km with or without convection? B. Would it be possible for an atmosphere to exist in the radiative equilibrium state?
C. For the atmosphere with convection, estimate the lapse rate from the plot. Is the lapse rate larger or smaller than the dry adiabatic lapse rate? Why?
(2) The figure below shows average profiles of temperature variables in the tropics.
A. Explain why the moist potential temperature is roughly constant within the troposphere.
B. Explain why the dry potential temperature increases with height.
C. Explain why the moist potential temperature is always at least as large as the dry potential temperature.
D. Explain why the moist potential temperature is equal to the dry potential temperature above a pressure of about 200 mb.
E. Is the atmosphere stable or unstable to dry convection?  The atmosphere is nearly always stable to dry processes. A parcel displaced upwards (downwards) in an adiabatic process moves along a dry adiabat [the dotted line] and cools down (warms up) at a rate that is faster than that of the environment, ∂T E /∂z. Since the parcel always has the same pressure as the environment it is not only colder (warmer), but also denser (lighter). The parcel therefore experiences a force pulling it back toward its reference height.

The Carnot Cycle and Hurricanes
A heat engine is a mechanism that does work by taking heat from a hot source (Q 1 ) and dumping it in a cold sink (Q 2 ). Heat engines work through cycles, and all the quantities in this paragraph are averaged over a cycle. By the conservation of energy, if the internal energy of the heat engine stays constant, the work it does has to be equal to the heat it takes from the source minus the heat it dumps into the sink (W " Q 1´Q2 ). The efficiency of a heat engine is the work it can do divided by the heat it takes from the source (η " W Q 1 " Q 1´Q2 Q 1 ). Carnot imagined an idealized heat engine with the components shown in the figure below. The upper panel is a cylinder with a conducting base and insulating sides and top. It has a piston on top that can move without friction and can either be pushed down or pulled up. There is a gas inside the cylinder that can expand or contract. The bottom three panels are a hot reservoir at temperature T 1 , an insulating stand that doesn't allow heat transfer, and a cold reservoir at temperature T 2 . The reserviors are so large that their temperature doesn't change if heat is taken out or put into them.

Atmospheric Thermodynamics
Carnot's cycle consists of taking the working substance in the cylinder through the following four operations that together constitute a reversible, cyclic transformation: i. The substance starts with temperature T 2 at a condition represented by A on the p-V diagram in Fig. 3.20. The cylinder is placed on the stand S and the working substance is compressed by increasing the downward force applied to the piston. Because heat can neither enter nor leave the working substance in the cylinder when it is on the stand, the working substance undergoes an adiabatic compression to the state represented by B in Fig. 3.20 in which its temperature has risen to T 1 .
It follows from (3.36) th done by the working su cycle is equal to the area ABCD in Fig. 3.20. Also stance is returned to its done is equal to Q 1 Ϫ Q engine is given by (3.78) engine has done work by tity of heat from a warm One way of stating the s ics is "only by transferri colder body can heat b cyclic process." In Exer engine can be more effic working between the sam that all reversible engine temperature limits have th ity of these two statem Carnot's theorems, depen ond law of thermodynam Exercise 3. 15 Show t ratio of the heat Q 1 abso voir at temperature T 1 to the cold reservoir at to T 1 ͞T 2 .

Solution:
To prove thi let the substance in the C ideal gas and we take i ABCD shown in Fig. 3.20

Fig. 3.19
The components of Carnot's ideal heat engine. Red-shaded areas indicate insulating material, and white areas represent thermally conducting material. The working substance is indicated by the blue dots inside the cylinder.
The Carnot cycle is shown in pressure-volume space in the diagram below. Imagine that we start at point B with the gas compressed (low volume). We place the cylinder on the hot reservoir, and it expands isothermally (at a constant temperature T 1 ) to point C, drawing heat Q 1 . We then place it on the insulating platform and allow it to expand adiabatically (no heat exchange) to point D, when it has temperature T 2 . Then we move it to the cold reservoir and it contracts isothermally at temperature T 2 , dumping heat Q 2 to the cold reservoir. Finally, we move it back to the insulating platform and it contracts adiabatically until it has temperature T 1 . We can then repeat the cycle if we want to. We already saw that dW " pdV , so the work done by or to the cylinder on each component is W " ş pdV . This means that the total work done by the cylinder over the cycle is represented by the blue area in the diagram. It turns out that the Carnot cycle represents the most efficient heat engine possible. That means that the efficiency of the Carnot cycle is the theoretical maximum efficiency of a heat engine. Now let's calculate the efficiency of the Carnot cycle. The ideal gas law can be written pV " . During this process oes work by expanding to the piston. to the nonconducting ubstance undergoes an g book web site in ature falls to T 2 . ance does work to the piston. aced on the cold ing the force applied g substance is along DA back to its ransformation the up a quantity of heat where ␥ is the ratio of the specific heat at constant pressure to the specific heat at constant volume. For mRT , where m is the mass of gas in the cylinder and we have used ρ " m V . Two components of the cycle occur at constant temperature (isothermal), and the mass of gas in the cylinder is constant. This means that the quantity pV is constant for the isothermal components of the cycle, so that p B V B " p C V C and p A V A " p D V D . On the adiabatic components of the cycle the potential temperature, θ , is a constant. So θ " T´p 0 p¯R c p "´p ρR¯´p 0 p¯R c p is constant, where we have used the ideal gas law. Since the density is inversely proportional to the volume (V " const. ρ ) for a constant amount of gas, V p 1´R c p " V p c v c p is constant, where we have used c p " c v`R . We can rewrite this to find that pV γ is constant, with γ " We now have four equations that we can use to eliminate the pressures and find that V C V B " V D V A . Returning to the 1st law of thermodynamics, dq " c v dT`dW , we note that for the isothermal components of the cycle dq " dW . This means Q 1 " This is the maximum possible efficiency of a heat engine, and it is not 100% efficient. The smaller the temperature of the cold reservior (T 2 ) and the larger the temperature of the hot reservior (T 1 ), the more efficient the Carnot cycle is.

Exercises
The Carnot cycle can be used as a model to understand hurricane strength [5]. The diagram below shows the radial motion of air in a hurricane. Imagine that the hurricane is axially symmetric like a doughnut. The diagram is a slice through the doughnut from the center to the outside. The horizontal axis is radius and the vertical axis is height. The clouds at small radius in the diagram represent the hurricane eye wall. transfer would increase linearly with wind speed, but the increasing roughness of the sea surface leads to a somewhat greater dependence on wind.
The actual rate of heat transfer is a subject of much controversy and research. The dependence of the transfer rate on wind is the principal feedback mechanism that allows hurricanes to develop. In its essence, the hurricane may be thought of as a wind-induced surface heat exchange instability, in which increasing surface winds lead to increased heat transfer from the sea, which leads to intensification of the storm winds, and so on. The energy cycle of the mature hurricane has been idealized by the author (Emanuel 1986) as a Carnot engine that converts heat energy extracted from the ocean to mechanical energy. In the steady state, this mechanical-energy generation balances frictional dissipation, most of which occurs at the air-sea interface. The idealized Carnot cycle is illustrated in Figure 1. Carnot's theorem may be easily derived from Bernoulli's equation and the first law of thermodynamics. The former states that along streamlines or absolute vortex lines in a steady system, www.annualreviews.org/aronline Annual Reviews A. Assume this hurricane is in the northern hemisphere. Based on your study of cyclones in the first week, will the surface winds of the storm be into or out of the page?

Electromagnetic Radiation
Energy can be transferred between objects by a variety of mechanisms. The diagram below shows four types of energy transfer that occur when a pot is heated on a stove. We've learned about convection and latent heat in the context of the atmosphere. If you don't know what conduction is, look it up online now. We're going to start talking about electromagnetic radiation now.
Electromagnetic radiation is propagating oscillations in electric and magnetic fields. As we will learn over the next few weeks, everything emits electromagnetic radiation all the time. The amount and type of electromagnetic radiation that an object emits depend on its temperature and radiative properties. Electromagnetic radiation has a frequency (ν) and a wavelength (λ ). In a vacuum, it travels at a constant speed of c " λ ν " 3.0ˆ10 8 m s´1.
There are many types of electromagnetic radiation (see figure below). We can use the wavelength of electromagnetic radiation, or light, to classify it. Visible light has a wavelength of 400-750 nm. Ultraviolet radiation has a shorter wavelength and infrared radiation has a longer wavelength. The Sun's radiation is mainly in the visible and near infrared (almost visible) part of the spectrum, and is often referred to as shortwave radiation in Earth science. The radiation Earth emits is mainly in the infrared part of the spectrum, and is often referred to as longwave radiation in Earth science.
Quantum mechanics tells us that we can also think of electromagnetic radiation as a particle, a photon, in addition to as a wave. The energy of a photon is proportional to its frequency. Thinking in terms of photons with discrete energy will be necessary for us to understand the greenhouse effect.

Exercises
(1) This question concerns the pot in the first figure of this handout.
A. What is the difference between conduction and convection?
B. The pot is being heated from below and convection in the water is moving this heat upward.
If the pot were heated from above, would it move energy downward through conduction or convection? Why?
C. Is the troposphere or the stratosphere more like a pot being heated from above?
D. The right hand of the person holding the pot experiences heat by conduction through the metal of the handle. Suppose the person puts their left hand over the pot. Describe mechanistically how heat can be moved from the hot pot to the left hand.
E. In the picture heat is leaving the burners via radiation. If the person stuck their left hand over a burner, would they sense heat? Explain qualitatively how radiation can move energy between two objects that don't touch each other.
F. Given that we can see a red color coming from the burners, in what part of the spectrum are they emitting radiation?
G. If you touch the ground on a typical day, does it feel hotter or colder than the burner?
H. The part of the spectrum that something emits in is related to the temperature it emits at. We can see the ground on a typical day in various colors. Combine this fact with your answer from the previous part to argue whether we can see direct emission of radiation from the ground. If we can't see direct emission from the ground, how do we see it?
I. Is the radiation emitted by the ground considered longwave or shortwave? Is the radiation you see from the ground considered longwave or shortwave?

Blackbody Emission
A blackbody is a perfect emitter and absorber of electromagnetic radiation. A blackbody emits electromagnetic radiation at different wavelengths according to the Planck function, as shown in the diagram below. The most important things to remember about the Planck function is that hotter bodies emit more radiation and have a peak emission at a shorter wavelength.
The decrease in the wavelength of peak emission (λ m ) as a function of temperature is captured in Wein's displacement law, which states that You can use the figure above to check that this works (just remember to convert between nm and µm). The radiative flux emitted by an object is the integral over wavelength of the Planck curves shown in the figure above. The increase in radiative flux (F, in W m´2) emitted as a function of temperature is captured in the Stefan-Boltzmann law, where σ " 5.67ˆ10´8 W m´2 K´4. Notice that the radiative flux increases with the fourth power of temperature, so hotter objects emit much more radiation. To use both of these laws, always remember to convert the temperature into degrees Kelvin. Most objects are not blackbodies. Instead they emit and absorb radiation less efficiently than a blackbody. At a given wavelength, a real object emits a fraction ε λ of the radiation a blackbody would emit. ε λ is called the emissivity. Similarly, at a given wavelength a real object absorbs a fraction a λ of the radiation a blackbody would absorb. a λ is called the absorptivity. Something called Kirchhoff's law tells us that the absorptivity equals the emissivity at every wavelength

The Greenhouse Effect
This is a link to a cool video (https://www.youtube.com/watch?v=c-AHtUsO6Wc) that demonstrates a lot about radiation. It is filmed with infrared cameras that are super-cooled so they don't measure themselves. The cameras measure infrared as black and white and the colors are added falsely. The dancers are in a cold barn during winter to maximize the temperature contrast between their bodies and the surroundings. Watch the video and try to notice as many interesting things as you can. Here are some observations I made: You can see hot and cold areas of their bodies. I think the hot corresponds to areas where blood is near the skin surface. Muscle and fat deposits show up as cold. You can also see the arms get progressively colder toward the hand, reflecting cooling as the blood flows out from the heart. There are a couple times when the dancers lay on the floor and when they move the floor stays hot for a while. The walls of the barn are lined with aluminum foil, which reflects infrared. At some point they dance near the walls and you can see their heat reflected off the crinkly surface. I think the "trippy tracers" are due to an extended exposure rather than a physical effect. I think the vertical reflections are a post-processing effect, rather than due to an infrared mirror. Let me know if you see anything else cool worth mentioning.

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Exercises Today we're going to build a simple model for Earth's greenhouse effect. This is a long, but important worksheet. The figure below is a simple schematic of a one-layer model of the greenhouse effect. S is the solar flux, A is the surface albedo, G is the ground thermal flux, and H is the thermal flux emitted by the one atmospheric layer. The ground directly absorbs a solar flux of (1-A)S. All fluxes are in units of W m´2. (1) A. Find a formula for G in terms of T s , the surface temperature. Assume that the ground has an emissivity of 1.
B. Assume the atmospheric layer has an emissivity of ε and a temperature of T a . Find a formula for H.
C. S is the solar flux in W m´2 at the distance of Earth's orbit. So S Watts pass through 1 m 2 of area oriented perpendicular to the Sun's rays at Earth's distance from the Sun. Write the total amount of solar energy absorbed by Earth's surface (S abs , in W) in terms of S, the surface albedo (A), and Earth's radius (R). Hint: Think about the area of sunlight that the Earth blocks (Earth's shadow), not the surface area of Earth.
D. Write the total amount of energy emitted by Earth's surface (E em ) in terms of T s , σ , and R.
E. Earth's surface can also absorb energy from the atmospheric layer (flux H). This is the Greenhouse effect. Write the total amount of atmospheric radiation energy absorbed by Earth's surface (E abs ) in terms of ε, T a , σ , and R.
F. For Earth to be in equilibrium we must have S abs`Eabs " E em . Plug in your formulas for each of these terms to determine a formula for the surface energy balance. R should cancel, but a geometric factor of 1 4 should remain on one of the terms.

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G. Calculate the total energy emitted by the atmospheric layer in terms of ε, T a , σ , and R.
H. Calculate the total energy absorbed by the atmospheric layer in terms of ε, T s , σ , and R. You must use Kirchhoff's law, which states that the emissivity of a material equals its absorbtivity. Also, notice that we are assuming that solar energy passes directly through the atmospheric layer without being absorbed.
I. For Earth to be in equilibrium the energy emitted and absorbed by the atmospheric layer must be equal. Use this to find a formula for the atmospheric energy balance. R should cancel.
J. Combine your formulas for surface and atmsopheric energy balance to eliminate T a and solve for T s .
K. If the CO 2 is increased, which term in your solution for T s does this effect, and how? Use your model to predict what will happen to Earth's surface temperature if the CO 2 is increased.

Absorption and Scattering
As light passes through the atmosphere, it can be absorbed, scattered in different directions, or transmitted unaffected. Often some combination of the three occurs. Atmospheric gases, clouds, and aerosols (solids and liquids floating in the atmosphere) can all lead to absorption and scattering of light. Absorption by a particular species (a gas, cloud, or aerosol) at a particular wavelength is characterized by the mass absorption coefficient (κ a λ ), which is measured in m 2 kg´1. You can think of κ a λ as the area that 1 kg of the particles of the material would take up if they were laid out in a single layer on a flat surface. This isn't exactly right because individual particles can weirdly cast a shadow that is not always equal to their geometric cross section, but it's approximately correct. What's important to remember is that the higher κ a λ is, the more that species absorbs at that wavelength. Similarly, scattering by a particular species at a particular wavelength is characterized by the mass scattering coefficient (κ s λ ). For either absorption or scattering, we can define something called optical thickness for each species (labled i) where M i is the mass path length, or total amount of mass of the species in the way of the light. We have also indexed κ λ i to make it clear that absorption and scattering coefficients are different for different species. M i is measured in kg m´2 and we can it between two heights as ρ i here is the density of the species under consideration, not the total air density. If we wanted to consider the mass path length of the entire atmosphere, then we would write M " ş 8 0 ρ i dz. Now suppose we are considering the special case of a well-mixed gas (such as CO 2 ) with a mass fraction of r, so that at any level ρ i " rρ, where ρ is the total density of air. The total path of this gas over the entire atmosphere is M " where we have used the hydrostatic relation (d p "´ρgdz) and P s is the surface pressure. If there are multiple species that can absorb and/or scatter, we just have to add up the individual optical thicknesss to get the total optical thickness We use the optical thickness in something called Beer's law, where T λ is the fraction of light transmitted. In words, the fraction of light transmitted through the atmosphere at a particular wavelength is an exponentially decaying function of optical thickness at that wavelength. Absorption by gases depends strongly on wavelength. Quantum mechanics tells us that gases can only absorb and emit electromagnetic radiation at certain discrete energies. We can think of electromagnetic radiation as particles (photons) in addition to waves. The energy of photons is propotional to their frequency. This means that only particular frequencies, and wavelengths, of light can interact with a particular gas. We can see this in the diagram below. If we take light from a blackbody and shine it through a prism, the colors are separated out. If we heat up a gas to the same temperature as the blackbody and shine the light it emits through a prism, only certain colors will be visible. The wavelength of the colors we see in this case corresponds to the energies that the gas can emit at. This is how neon lights work (different gas mixtures are used to make different colors). If instead we shine the original blackbody through the same gas at a colder temperature and then through the prism, we see absorption features at the same wavelengths where we previously saw emission in the resulting spectrum. Absorption by clouds and aerosols tends to not have as fine structure in wavelength as absorption by gases, although there are often broad differences between absorption of longwave and shortwave radiation.
Scattering by particles much smaller than the wavelength of electromagnetic radiation is called Rayleigh scattering. Atmospheric gases Rayleigh scatter sunlight. Rayleigh scattering is proportional to the inverse of the wavelength of light to the fourth power This means that light at shorter wavelengths is scattered much more than light at longer wavelengths. Given that blue light has a wavelength of about 400 nm and red light has a wavelength of about 700 nm, this means that the atmosphere should scatter blue light about`7 4˘4 " 9.4 times more than red light. As the diagram below shows, this is why the sky appears blue. Sunlight that is not pointed directly at us can be scattered toward us by the atmosphere. This scattering is much more efficient for blue light than red light, so we see the atmosphere as blue. Larger particles, such as those that make up clouds and aerosols, scatter light through a process called Mie scattering. The dependence of Mie scattering on wavelength and direction is much more complicated than Rayleigh scattering, and is not something we have time to get into. Just remember that it exists and that scattering by clouds and aerosols can have a large effect on the climate of a planet.

Exercises
(1) The figure on the next page shows absorption coefficients for CO 2 and H 2 O as a function of wavenumber (one divided by wavelength). It also shows a blackbody curve at 260 K, which gives you a sense of which wavenumbers are important for terrestrial radiation.
A. An atmosphere is generally considered optically thick at a given wavenumber if τ Á 1 (optically thick means the atmosphere interacts strongly with radiation). Approximately what fraction of radiation makes it through an atmosphere if τ " 1?
B. On a typical day water vapor represents on the order of 1% of the mass of Earth's atmosphere. Use this to estimate the order of magnitude of M, in kg m´2 for water vapor. What value of κ is necessary for water vapor to make Earth's atmosphere optically thick at a given wavenumber?
C. The mass fraction of CO 2 in the atmosphere is about 6ˆ10´4. What value of κ is necessary for CO 2 to make Earth's atmosphere optically thick at a given wavenumber?
D. Based on these calculations and the figure, do you think CO 2 or H 2 O provides more greenhouse warming for Earth?
E. Use the figure to explain why CO 2 is an important greenhouse gas on Earth.
(2) Explain why the Sun appears red during sunrise and sunset.
(3) Large, explosive volcanic eruptions loft sulfate aerosols into the stratosphere. Particularly vibrant sunsets are often observed around the world after such events. Explain this.
for applications across engineering and atmospheric sciences. The database is freely available at http://www.cfa.harvard .edu/hitran. A simple, flexible Python-language interface to HITRAN is included in the online software supplement to reference 5, available at http://geosci.uchicago.edu/~rtp1/ PrinciplesPlanetaryClimate. Measurements of absorption cross sections allow one to relate the absorption-emission properties of a layer of atmosphere to its composition. Figure 2, for example, shows absorption cross sections for CO 2 and water vapor as a function of wavenumber, which is proportional to frequency; the thermal IR ranges from 200 cm −1 to 2000 cm −1 . The spectra there are computed from the HITRAN database with lines broadened by collision with air at a pressure of 5 × 10 4 Pa, which corresponds to about the middle of Earth's atmosphere by mass. The CO 2 molecule has four main groups of absorption features in the thermal IR, of which the most important for Earthlike conditions is the one with the wavenumber near 667 cm −1 . (The corresponding wavelength is 15 µm.) The feature arises from vibrational bending modes of the linear triatomic molecule, which are given a fine structure by mixing with rotational transitions; the inset to figure 2 shows the corresponding lines. Water vapor is a polar molecule, and its richer set of vibrational and rotational modes allows it to absorb effectively over a much broader range of frequencies than CO 2 .
Gases exhibit continuum absorption, as does the con-densed matter making up clouds of all kinds. In some cases continua result from the overlap of nearby lines, but in other cases continua appear where no lines are in the vicinity. Loosely speaking, those continua arise because, over the finite duration of a collision, a pair of colliding molecules acts somewhat like a single, more complex molecule with transitions of its own. Equivalently, they result from the overlap of the tails of remote collisionally broadened lines. The statistical mechanics governing the far regions of long line tails is not at all well understood; 6 nonetheless, the continua have been quite well characterized, at least for those cases relevant to radiative transfer in Earth's atmosphere. For present-day Earth, the only important continuum is the water vapor continuum in the window around 1000 cm −1 . Carbon dioxide continua are unimportant for conditions that have prevailed on Earth during the past several billion years, but they are important for plugging the gaps in the line spectra for the dense CO 2 atmospheres of Venus and early Mars. Diatomic homoatomic molecules like N 2 , which are transparent to IR in Earthlike conditions, have collisional continua that become important in cold, dense atmospheres. For example, the continuum makes N 2 one of the most important greenhouse gases on Saturn's largest moon, Titan. The wavenumber, defined as the reciprocal of the wavelength, is proportional to frequency. If a layer of atmosphere contains M kilograms of absorber for each square meter at the base of the layer, then light is attenuated by a factor exp(−κM) when crossing the layer, where κ is the absorption coefficient. The horizontal dashed lines on the CO 2 plot give the value of absorption coefficient above which the atmosphere becomes very strongly absorbing for CO 2 concentrations of 300 ppm and 1200 ppm; the green rectangle shows the portion of the spectrum in which the atmosphere is optically thick for the lower concentration, and the orange rectangle indicates how the optically thick region expands as the concentration increases. The inset shows fine structure due to rotational levels.

Radiation and Vertical Structure
We can measure electromagnetic radiation emitted by a planet as a function of wavelength to produce a spectrum. We might do this for Earth using a satellite instrument, or another planet in the solar system using an instrument on a fly-by mission or that we have put into orbit around that planet. We can even make crude measurements of the electromagnetic spectra of extrasolar planets orbiting stars light years away using fancy space telescopes. Last time we saw how the electromagnetic spectrum of a planet can reveal crucial information about the gases present in the planet's atmosphere. Today we are going to learn how we can use the electromagnetic spectrum of a planet to learn about its vertical temperature structure.
The mass absorption/emission coefficients (κ a λ ) of the gases in a planet's atmosphere depend strongly on wavelength. This means that the mass necessary for an atmosphere to be optically thick (τ « 1) depends strongly on wavelength. Emission to space is dominated by the region of the atmosphere where τ « 1. At higher altitudes there isn't enough of the emitting gas for much emission to space. At lower altitudes almost all of the emitted radiation is absorbed by the atmosphere before it reaches space. The emitted radiation we measure from space at a given wavelength ends up looking like blackbody emission with a temperature corresponding to the atmospheric temperature where τ « 1. We call this equivalent blackbody temperature the brightness temperature, and it is very important to remember that it depends on wavelength. At wavelengths where the atmosphere absorbes well (large κ a λ ), τ « 1 will occur high up in the atmosphere and the brightness temperature will correspond to the atmospheric temperature at these high altitudes. At wavelengths where the atmosphere absorbs poorly (small κ a λ ), τ « 1 will occur low in the atmosphere and the brightness temperature will correspond to the atmospheric temperature at these low altitudes. There may even be regions of the spectrum where the atmosphere absorbs so poorly (κ a λ « 0) that we can see emission from the planet's surface. In this way we can use a measurement of the electromagnetic spectrum of a planet to infer things such as its vertical temperature lapse rate and whether it has a stratosphere.

Exercises
(1) The figure below shows measured radiative emission as a function of wavenumber for Earth, Mars, and Venus as well as reference blackbody curves. G. In 1960, as part of his PhD thesis at the University of Chicago, Carl Sagan estimated the surface temperature of Venus [6]. He did this using observations of the planet in the microwave part of the electromagnetic spectrum. Explain why microwave emission might be different for Venus and Mars, despite the fact that infrared emission is so similar.
separately on a grid of millions of frequencies and the results are summed to obtain net fluxes. Climate models, however, require greater computational efficiency; one needs to compute the frequency-averaged radiation flux at each of several thousand model time steps for each of several thousand grid boxes covering a planet's atmosphere. Modelers use various approximations to represent the aggregate effects of spectral lines averaged over bands about 50 cm −1 wide. Such approximations are validated against line-by-line codes that have, in turn, been validated against laboratory and atmospheric observations. When averaged over a broad band, radiative flux decays algebraically rather than exponentially with distance traversed, because the progressive depletion of flux at strongly absorbed frequencies leaves behind flux at frequencies that are more weakly absorbed. 5

Confirmation by observed spectra
The Sun radiates approximately like a blackbody having a temperature of 6000 K, even though the temperature of the solar interior is many millions of degrees. That's because the visible-wavelength and IR photons that predominate in solar radiation can escape from only the cooler outer layers of the Sun. Similarly, the 2.7-K cosmic microwave background radiation gives the temperature of the radiating layer of the very early universe, redshifted down from its original, much higher temperature. The radiating layer of a planet is the IR equivalent of the Sun's photosphere. When a planet is viewed from above, the emission seen at a given frequency originates in the deepest layer that is optically thin enough for significant numbers of photons to escape. The effective emission temperature for that frequency is a suitably weighted average temperature of that layer. If the atmospheric temperature varies with height, variations of the absorption coefficients of atmospheric constituents with frequency show up in planetary emission spectra as variations of emission temperature; the more transparent the atmosphere is, the deeper one can probe.
For atmospheres heated partly from below-either as a consequence of solar absorption at the ground as in the case of Earth, Mars, and Venus, or due to internal absorption and escaping interior heat as with Jupiter and Saturn-the lower layers of the atmosphere are stirred by convection and other fluid motions, and the constant lifting and adiabatic cooling establish a region whose temperature decline with height approximates that of an adiabat. That region is the troposphere. At higher altitudes, heat transfer is dominated by radiative transfer instead of fluid motions; the corresponding region is the stratosphere. Stratospheric temperature is constant or gently decaying with height for pure IR radiative equilibrium, but in situ absorption of solar radiation can make the stratospheric temperature increase with height. Ozone facilitates such absorption on Earth, and organic hazes have a similar effect on Titan. Typical temperature profiles for Earth, daytime Mars, and Venus are shown in the right-hand column of figure 3.
The  The panel to the left shows a summer-afternoon emission spectrum for Mars observed by the TES instrument on the Mars Global Surveyor. Its accompanying temperature profile was obtained from radiooccultation measurements corresponding to similar conditions. (c) The panels here show a Venusian equatorial night thermal spectrum as measured by the Venera 15 orbiter 14 together with a typical temperature profile for the planet. The upper portion (dashed curve) of the temperature sounding is based on radiooccultation observations from the Magellan mission; the lower portion (solid curve) was observed by a Pioneer Venus descender probe. For all three planets, squiggly arrows on the temperature profiles indicate the range of altitudes from which IR escapes to space.

Global Energy Balance
The figure below is a diagram of Earth's global-mean energy budget both at the surface and at the top-of-atmosphere (TOA). This budget was put together through a painstaking compilation of available surface, aircraft, balloon, and satellite measurements. Study the diagram a bit to get a sense of what the big players in Earth's energy budget are.
Today we're going to think about annual-mean maps of TOA radiation measurements from the CERES satellite instrument. We can look at these maps to understand a lot about Earth. The maps show longwave emission, net shortwave (downward minus upward), and net radiation (net shortwave minus longwave). Longwave radiation is directly emitted by terrestrial sources. Shortwave radiation is emitted by the Sun and impinges on Earth. Some shortwave radiation is reflected back to space, and the plot shows the net shortwave. The net radiation tells us about the overall impact of both shortwave and longwave radiation. Where the net radiation is positive, radiation has a warming effect on the planet in the annual-mean. Where the net radiation is negative, radiation has a cooling effect on the planet in the annual mean.

Partial Derivatives and the Streamfunction
Today we will start thinking about atmospheric dynamics. In order to understand atmospheric dynamics we will need to review some concepts from calculus. The partial derivative is the derivative of a function of multiple variables with respect to just one of them. For example, if we have a function of two variables f px, yq, we can write the partial derivative with respect to x as B f px,yq Bx and the partial derivative with respect to y as B f px,yq By . When we take B f px,yq Bx , we treat y as if it were just another constant. Similarly, when we take B f px,yq By , we treat x as if it were just another constant. One useful place partial derivatives come up is with something called the streamfunction. Under certain conditions a streamfunction can be found for a two-dimensional flow. The streamfunction (ψ) is defined such that where u is the velocity in the x direction and v is the velocity in the y direction. The velocity vector points along contours of a streamfunction (lines in px, yq space where the streamfunction is constant), so the streamfunction is a scalar field that can be used to represent a vector field. A analogy to help you think about the streamfunction is as follows. Imagine a wanderer wandering in hilly terrain. If we choose the way she walks carefully, then the topography would be the streamfunction of her flow. In particular, imagine that the wanderer always wanders at constant altitude, or at a constant value of the streamfunction in this analogy. She always wanders with her right hand facing the slope, and she moves with a speed proportional to the steepness of the slope. If she moves like this, then the topography is her streamfunction. You have probably seen a topographic map before. A topographic map is an example of a contour plot, and we will be working with contour plots of the streamfunction below.

The Divergence
The divergence is an operator on a vector field that tells us about the source of the field. We will consider two-dimensional flow, so we can think of a velocity vector v with two components pu, vq.
The divergence can be written as If ∇¨ v ą 0 at a point then the field is diverging and we can think of the flow as spreading out from that point. If ∇¨ v ă 0 at a point then the field is converging and we can think of the flow as flowing toward that point. To consider a concrete example, let's think about the bathtub pictured below. Consider the horizontal, two-dimensional flow of water in the bathtub. If the drain is plugged and we turn the spigot on, the horizontal flow will have to diverge from where the water from the spigot enters the tub because this is a source of water. The divergence of the horizontal flow will be positive where the water enters the tub. Now imagine we fill the tub and turn the spigot off. If we unplug the drain, the drain will be a sink of water, and the horizontal flow will have to converge toward it. The divergence of the horizontal flow will be negative where the water enters the drain.
A. From looking at the plot, is ∇¨ v positive, negative, or zero at px, yq " p0, 0q? B. Calculate ∇¨ v at px, yq " p0, 0q.  (6) In the last class we learned that for certain flows we can define a streamfunction (ψ) such that u "´B ψ By and v " Bψ Bx . Calculate the divergence ∇¨ v in terms of the streamfunction, ψ. Your answer should indicate that flows for which we can define a streamfunction have a special property. What is it?

The Curl and Vorticity
The curl is an operator on a vector field that tells us about the infinitesimal rotation at different points in the field. Qualitatively, the curl of a velocity field at a particular point tells us how much "spinniness" or "whirling" there is at that point. If we imagine putting a tiny water wheel or turnstyle into the velocity field at a given point, the curl of the velocity field would tell us which direction and how fast the water wheel would turn. The curl of a velocity field is also called the vorticity. The vorticity is incredibly important for understanding atmospheric circulation because cyclones are clumps of spinniness in the flow. We will consider two-dimensional flow, so we can think of a velocity vector v with two components pu, vq. For two-dimensional flow, the curl of the velocity vector can be written A trick we can use to assess the sign of the vorticity is the "right-hand rule." To use the righthand rule, start with a vector plot of velocity. Put your right hand down where you want to assess the vorticity. Fold your fingers toward the palm of your hand in the direction that the velocity is swirling. If your thumb is pointed away from the page, then the vorticity is positive. If your thumb is pointed toward the page, then the vorticity is negative. If the velocity isn't swirling, then the vorticity is zero.
C. Calculate ∇ˆ v at px, yq " p1, 1q. Does it make sense from the plot why it is the same or different as ∇ˆ v at px, yq " p0, 0q? A. From looking at the plot, is ∇ˆ v positive, negative, or zero at px, yq " p0, 0q? B. Calculate ∇ˆ v at px, yq " p0, 0q.
C. Calculate ∇ˆ v at px, yq " p1, 1q. Does it make sense from the plot why it is the same or different as ∇ˆ v at px, yq " p0, 0q?

The Coriolis Force
The Coriolis force is an apparent force felt in a rotating reference frame. Watch this video (https://youtu.be/dt XJp77-mk) to understand the Coriolis force better. The Coriolis force is equal to F c "´fkˆ v (39) on a planet, wherek represents the vertical direction, is the Coriolis parameter, φ is the latitude, and Ω " 2π rad day´1 " 7.3ˆ10´5 s´1 on Earth. We can write the Coriolis force in the x direction (F x c ) and the y direction (F y c ) as Notice that a positive velocity in the x direction, u, causes a Coriolis force in the negative y direction (as long as we are in the Northern Hemisphere, so f is positive). Also, a positive velocity in the y direction, v, causes a Coriolis force in the positive x direction. Notice that the Coriolis force is always perpendicular and to the right of the flow in the Northern Hemisphere (see the diagram below). In the Southern Hemisphere the Coriolis force is always perpendicular and the the left of the flow, since f is negative.
We can also use the right-hand rule and the formula for the Coriolis force ( F c "´fkˆ v) to determine the direction it acts in. To use the right-hand rule for a cross product ( Aˆ B " C), point the fingers of your right hand toward vector A and curl them toward vector B. Your thumb will now point in the direction of vector C. In the Northern Hemisphere, f is positive, so the righthand rule tells us that the Coriolis force pushes an object to the right of its direction of travel (see the diagram above, and remember the negative sign in the formula for the Coriolis force). In the Southern Hemisphere, f is negative and the Coriolis force bends an object to the left of its direction of travel.
The Rossby number (Ro) is used to characterize the importance of the Coriolis force in atmospheric flows. Ro where U is the typical speed of air or the object under consideration and L is the typical length scale of the flow. The Coriolis force is dominant if Ro ! 1 and the Coriolis force can be neglected if Ro " 1. If Ro « 1, then the Coriolis force is important, but not dominant. The Rossby number is extremely important. It is named after Carl-Gustaf Rossby, one of the most important atmospheric scientists of all time. He was so amazing that he appeared on the cover of Time magazine! Rossby was the chair of the Department of Meteorology at the University of Chicago in the 1940's. The Department of Meteorology merged with the Department of Geology in the 1960's to form the Department of the Geophysical Sciences, and Hinds building was built to house it.
in the y direction. Similarly, if we want v, the velocity in the y direction, we need to look at derivatives in the x direction. That's how a streamfunction works, and it ultimately works for our particular streamfunction, which is proportional to pressure, because the Coriolis force points to the right of the velocity.

Exercises
(1) The figure below is the weather map from Problem Set 07. The solid contours show the geopotential height (z) of the 500 hPa pressure surface on a particular day at a particular time. The numbers are contour levels in decameters (570 decameters = 5700 meters) and the contour spacing is 60 m. Bands of latitude and longitude are shown as dotted lines.
the other protrudes northwa Pronounced troughs (along wh equatorward) are evident over the central Pacific, and the Plains, and several weaker trou at other locations. The typical cessive troughs (counting the w longitude or 4000 km, which co retically predicted wavelength o Time-lapse animations of w one shown in Fig. 8.2 reveal move eastward at a rate of ϳ1 sponds to the wintertime clim wind speed around the 700strength of the westerlies gen height within the extratropical t above this so-called steering l waves from west to east, while level are overtaken by the wa (or troughs) typically pass a fi intervals of roughly 4 days, bu day or two apart if the steerin  Fig. 8.1 Plotting convections used in synoptic charts.

Geostrophic Balance 2
Geostrophic balance is really important for understanding atmospheric flow in the midlatitudes. It's so important that we're going to spend another class getting familiar with it! Last time we found that we can define a geostrophic streamfunction as ψ " Φ f " gz f such that u "´B ψ By " and v " Bψ Bx , where z is the geopotential height. Alternatively, we can write the pressure gradient force as F p "´´1 ρ BP Bx ,´1 ρ BP By¯"´1 ρ ∇P. Performing a force balance between the pressure gradient force and the Coriolis force leads to the definition of an alternative streamfunction, This streamfunction assumes that the density is approximately constant horizontally at a given level, and again neglects variations in f , the Coriolis parameter, in the y direction.
(1) Watch this movie (https://youtu.be/n4rJX4IIRaU). Focus on the extratropical cyclone above Chicago. Estimate the pressure differential between the storm center and far away from the center. How big is this differential compared to standard sea level pressure? Hints: You may assume an air density of 1 kg m´3. You will need to estimate the distances involved and read the video caption to estimate the amount of time that elapses over the course of the video.
(2) Consider an intense tropical cyclone located at 15˝latitude. It's maximum azimuthal wind speed is 60 m s´1, which occurs just outside the eyewall (10 km from the storm center).
A. Calculate the Coriolis force (really an acceleration) experienced by the air as it rotates around the cyclone's center. In which direction (toward or away from the center) does the Coriolis force point?
B. Calculate the centrifugal acceleration ( v 2 r ) experienced by the air as it rotates around the cyclone's center. In which direction (toward or away from the center) does the centrifugal acceleration point?
C. Is one of these accelerations larger than the other? What is it balanced by? D. How is this different from the situation for the extratropical cyclone in the last question? 92

Potential Vorticity
Today we're going to learn about the "potential vorticity." It is a conserved quantity for fluid in a rotating reference frame. Potential vorticity conservation is the fluid equivalent of angular momentum conservation for solids. To remember how the conservation of angular momentum works, watch this video (https://youtu.be/FmnkQ2ytlO8). Conservation of angular momentum allows the skater in the video to increase her angular velocity (spininess) by pulling her legs and arms into her body.
For fluids we can define a quantity called the potential vorticity, where f is the Coriolis parameter and represents planetary vorticity, is the relative vorticity (the vorticity of the fluid itself, also ζ is pronounced "zeta"), and H is the thickness of the fluid. Technically this definition applies for a constant density fluid, but it gives us a good qualitative picture of what happens in the atmosphere, and can be generalized to the atmospheric case with a bit more math. The point is that the potential vorticity, Q, is conserved (constant) during flow. This is a very powerful statement because it allows us to predict aspects of the flow easily. Atmosphere geeks get so excited about potential vorticity, or PV, that they even write songs about it (https://youtu.be/g2-RpvGGQU4). To use potential vorticity in the atmosphere, we just have to think about the thickness, H, as the thickness between two layers of constant potential temperature. The figure below shows how potential vorticity conservation works. Assume we have a parcel of air with some vorticity and thickness associated with it. For now let's assume that all of it's vorticity is relative and none of it is planetary (say the parcel is at the equator), so you can see all of the spinning directly. This air parcel is represented on the left of the figure below. Now let's assume that the air is stretched vertically. To conserve mass it will have to squeeze in toward it's center. Just like how the skater spun up as she pulled her arms and legs toward her center, the air spins up as it stretches vertically and pulls in toward its center. This is represented by the right diagram in the figure below. The higher vorticity on the right is represented by more lines curling around the air parcel. The fact that the two parcels have the same potential vorticity, even though they have different vorticities, is the statement of potential vorticity conservation that we can use to understand atmospheric flow. We just have to remember that Q " f`ζ H is conserved.

Exercises
(1) A cylindrical column of air with a radius of 100 km starts at the equator at rest. It moves northward to Chicago maintaining a constant height.
A. Is the relative vorticity, ζ " Bv D. Let's assume that the column has a tangential velocity of U in the counter-clockwise direction at it's radius, R. The tangential velocity is zero at the center of the column. Show that we can approximate Bv Bx « U R and Bu By «´U R . E. Use this to estimate the mean tangential velocity at the perimeter of the air column when it reaches Chicago.

Tornadoes
Today we're going to learn how tornadoes form using our knowledge of potential vorticity. The relative vorticity is so high in tornadoes that we can neglect the planetary vorticity in this discussion. There are three steps to the formation of tornadoes. First, a large amount of vorticity needs to be generated. This is done by a wind shear (winds that are higher up in the atmosphere than at the surface). It is natural for a wind shear to develop near the surface because surface friction brings the wind to near zero right at the surface. The special condition we need is a large wind shear, so really fast horizontal winds relatively close to the surface. The diagram below shows how a wind shear leads to vorticity. Notice that it is horizontal vorticity.
Next this horizontal vorticity is tilted vertically. This is accomplished by convection. See the diagram below. To understand how this works, watch this video (https://youtu.be/sb82tVOq2dY) of high divers. At the start of most of these jumps the high divers cause rotation of their body around the axis from their feet to their head. They then allow this axis to tilt in a range of directions. The body keeps spinning around the axis of the trunk, but the trunk can be pointed in different directions. A similar thing is happening when horizontal vorticity is tilted upward by convection to become vertical vorticity.
Finally, the air column is stretched vertically, which increases its vorticity. We can understand this through the conservation of potential vorticity (see the diagram below, which is from last class). The convection that caused the tilting also causes the stretching.
The diagram below puts the three steps together. In panel A wind shear generates horizontal vorticity. In panel B vertical convection transforms the horizontal vorticity into vertical vorticity. In panel C the convection stretches the vertical vorticity, which causes the vertical vorticity to increase drastically, which we observe as a tornado.

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Finally, you may have heard of the "Fujita scale," which is used to measure tornado intensity. It is named after Ted Fujita, who was a member of the Department of Meteorology and later the Department of the Geophysical Sciences at the University of Chicago from the 1950's until his death just before the turn of the century. Fujita was a young professor in Japan during the second World War in the mid-20th century. After the United States dropped an atomic bomb on the city of Hiroshima, Japan, as part of that war, the Japanese government sent Fujita in to figure out what had happened. By carefully analyzing burn marks on objects such as gravestones around the city, he was able to reconstruct the three-dimensional location and power of the explosion fairly accurately. He later used similar methods to study tornadoes in North America by analyzing the wreckage they create. This led to the development of his famous scale, as well as his nickname, "Mr. Tornado." You can see Mr. Tornado in action (and some vintage footage of Hinds building as well as 20th century technology) in this video (https://youtu.be/YGrhwlYSWUU).

Problem Set 3
(1) This problem is based on a method that has been used to reconstruct the atmospheric pressure on ancient Earth [7]. Vesicles in crystallized lava flows form because air bubbles were trapped in the lava as it crystallized. The figure shows a schematic diagram of this. of Manua Loa along many cracks and fissures near the crater rim. Many recent flows have been well mapped, and there are small lobes available for sampling. These exhibit vesicularity profiles that reflect simple emplacement history, the lava having been apparently removed from the main channels of the lava flows.
Sampling sites are also available at low elevation in Hilo and elsewhere. Roadcuts, stream channels, and natural cliffs and fissures provide many sampling sites near the terminus of the flows or in breakout lobes. With the large number of potential sampling sites available at the base and summit, the only constraint on the number of sampl funds for analysis. However, sampling sit much more limited along the flanks of Maun Roadcuts and natural fissures are generally a so fully exposed cross sections are unava Nevertheless, we found a site on the SE flank a roadcut exposed a flow with vesicularity r ing simple emplacement history.
To avoid potential complexities in size bution and to obtain maximum pressure resol it is best to collect samples from near the to bottom of the flow to capture the initially e modal bubble size before bubble rise and c A. The diagram shows that vesicles near the top of a lava flow will tend to have a larger volume than vesicles near the bottom. Explain physically why this is.
B. Suppose that the atmospheric pressure is P a and the lava density is ρ l . Derive a formula for the pressure felt by the air bubbles when the vesicles are formed as a function of depth (z) in the lava. (1) Explain why a liquid boils when its saturation vapor pressure equals ambient atmospheric pressure. Imagine we bring water to a boil in a pot in Chicago and in Denver. We then put and egg into the boiling water. In which city will the egg be fully cooked (hard boiled) first? Why?
(2) The figure below shows the saturation vapor pressure of water as a function of temperature. When you exhale, the air coming from your lungs is at body temperature and close to being saturated (point "A"). Now you go outside on a warm day (when ambient air is at point "B"), and on a cold day (when ambient air is at point "C"). Do you expect to see your breath on either day? Why? Hint: consider what the mixing of two parcels of air would look like on this plot.
(3) A modern, pressurized jet airplane is flying in the upper troposphere in the tropics. Air is brought into the cabin from the atmosphere outside the airplane. For the passengers to experience a similar cabin air temperature to the surface air temperature, does the air brought into the cabin from outside need to be cooled, heated, or neither? If you think the air needs to be cooled or heated, estimate by how many degrees to one significant digit.

Problem Set 5
(1) The figures on the next page show the spectrums of a Sun-like star (a G-star) and a smaller M-star. M-stars represent about 75% of stars in the galaxy, and possible terrestrial planets in the habitable zone of nearby M-stars (Proxima Centauri and TRAPPIST-1) have been found recently. The spectrums are only approximate black bodies because of non-uniform absorption and emission by gases in the upper atmospheres of the stars. For this problem, however, you can assume that the spectrums are blackbodies.
A. Estimate the photospheric (where the light is emitted to space) temperatures of the Sun-like star and the M-star.
B. Estimate the energy flux emitted by each star.
C. Estimate the power emitted by each star (the luminosity). You may assume that the Sun-like star has a radius of 700,000 km and the M-star has a radius of 430,000 km.
D. The Sun has a habitable planet (Earth) orbiting it at a distance of 1 astronomical unit. For a planet orbiting the M-star to receive the same amount of stellar radiation (in W m´2) as Earth, at what distance (in astronomical units) would it need to be orbiting the M-star? Compare this to the distances of planets in the solar system to the Sun. Hint: A star emits a roughly constant amount of energy per unit time (it's luminisity). As you go to larger distances from the star, this energy is spread out over the surface of an imaginary sphere of larger radius. 108

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(1) The flux of sunlight that strikes both the Earth and the Moon when the Sun is directly overhead is about 1360 W m´2. The moon has a very low albedo (about .1), very little thermal inertia, and very little ability to move energy horizontally, so that every point on the surface immediately comes into equilibrium with the sunlight that is striking it. The Earth's surface has a large thermal inertia, and the oceans and atmosphere have the ability to transport heat horizontally. As a result, it can smooth out differences in incoming sunlight between different locations. One example of this is the fact that the temperature on Earth only drops a little over night even though no sunlight is being absorbed, so that the equilibrium nighttime temperature is near absolute zero. (1) The figure below is the type of map used in weather forecasting. It is a contour plot of the geopotential height of the 500 hPa pressure surface on a particular day at a particular time. If we imagine a surface through the atmosphere that has a constant pressure of 500 hPa, sometimes this surface will go up and sometimes it will go down as we move horizontally. The plot shows the height of this surface above sea level. To orient you, the plot is a polar projection, with the north pole at the center. Continents are shown in blue and oceans in white. The reason this plot is useful is that the 500 hPa geopotential height is approximately a streamfunction of the atmospheric flow. So we can directly infer atmospheric velocities by looking at it. The numbers are contour levels in decameters (570 decameters = 5700 meters) and the contour spacing is 60 m. "H" represents a local maximum of the height field and "L" represents a local minimum. the other protrudes northwa Pronounced troughs (along wh equatorward) are evident over the central Pacific, and the Plains, and several weaker trou at other locations. The typical cessive troughs (counting the w longitude or 4000 km, which co retically predicted wavelength o Time-lapse animations of w one shown in Fig. 8.2 reveal move eastward at a rate of ϳ1 sponds to the wintertime clim wind speed around the 700strength of the westerlies gen height within the extratropical t above this so-called steering l waves from west to east, while level are overtaken by the wa (or troughs) typically pass a fi intervals of roughly 4 days, bu day or two apart if the steerin   transfer would increase linearly with wind speed, but the increasing roughness of the sea surface leads to a somewhat greater dependence on wind. The actual rate of heat transfer is a subject of much controversy and research. The dependence of the transfer rate on wind is the principal feedback mechanism that allows hurricanes to develop. In its essence, the hurricane may be thought of as a wind-induced surface heat exchange instability, in which increasing surface winds lead to increased heat transfer from the sea, which leads to intensification of the storm winds, and so on. The energy cycle of the mature hurricane has been idealized by the author (Emanuel 1986) as a Carnot engine that converts heat energy extracted from the ocean to mechanical energy. In the steady state, this mechanical-energy generation balances frictional dissipation, most of which occurs at the air-sea interface. The idealized Carnot cycle is illustrated in Figure 1. Carnot's theorem may be easily derived from Bernoulli's www.annualreviews.org/aronline Annual Reviews structure seen in Fig. 5 provides evidence for a regime of gravity waves similar to those observed in the Voyager radio occultation data 31,32 . Turbulence due to shear instability (Kelvin-Helmholtz instability) is expected wherever the vertical shear of the wind speed is large. The wind shear measured by the Doppler Wind Experiment 33 is sufficiently large that the features present between 50 and 150 km are likely to be related to turbulence. The vertical resolution of the temperature measurement was sufficient to resolve the instantaneous structure of the planetary boundary layer. On the basis of the nearly constant values of the The electric field due to natural wave emissions was investigated during the descent, using the receiving dipole of the mutual impedance probe in two frequency ranges, 0-11.5 kHz and 0-100 Hz (Fig. 8b, c, passive mode). This provided a unique opportunity to investigate in situ lightning and related phenomena (for example, corona discharges) on Titan 21 that would produce electromagnetic waves 38 , excite global and local resonance phenomena in the surfaceionospheric cavity 39,40 and could drive a global electric circuit 22 . Several impulsive events have been observed during the descent, for example at 2,800 s. The narrow-band wave emission seen near 36 Hz is reminiscent of a possible resonance generated by lightning activity in the spherical waveguide formed by the surface of Titan and the inner boundary of its ionosphere, but should be interpreted with caution. A comparison of the records presented in Fig. 8a and b shows that the first spectrogram (active mode) not only displays the Figure 5 | Temperature profile of the lower atmosphere as measured by the temperature sensors, TEM 27 (expanded from Fig. 2). Temperature uncertainty is^0.25 K in the range from 60 to 110 K, and^1 K above 110 K. The temperature minimum of 70.43 K is reached at the tropopause (about 44 km; 115^1 hPa). HASI temperatures are in very good agreement (within the error bars) with data obtained by Voyager radio occultation 2 (ingress, circles; egress, crosses) assuming a pure nitrogen atmosphere. The error bars for Voyager data are reported:^15 K (egress)^10 K (ingress) near the 200-km level,^0.5 K at the tropopause. At the tropopause, HASI measured temperature values ,1 K colder than Voyager 2 , but reanalysis of these data 3 suggested a similar temperature value (70.5 K) assuming a stratospheric composition of 98.5% N 2 plus 1.5% CH 4 .
atmosphere down to the 400-km altitude level, decreasing down to 1 km at the 160-km level 24 . The profile shows in general a cut-off at the dry adiabatic lapse rate (dotted line), implying that fluctuations may lead to convective instabilities. The line at zero temperature variation is shown in white against the black curve. Six inversion layers in the upper atmosphere (at about 510, 600, 680, 800, 980 and 1,020 km) could be detected by strong peaks towards positive values. The strong lower inversion layer (4 K km 21 at ,510 km) corresponds to the feature already observed from the ground during Titan's stellar occultations 10 . The strong peaks between the 160-and 110-km levels correspond to the parachute deployment sequence.   Fig. 5 provides evidence for a regime of gravity waves similar to those observed in the Voyager radio occultation data 31,32 . Turbulence due to shear instability (Kelvin-Helmholtz instability) is expected wherever the vertical shear of the wind speed is large. The wind shear measured by the Doppler Wind Experiment 33 is sufficiently large that the features present between 50 and 150 km are likely to be related to turbulence.
The vertical resolution of the temperature measurement was sufficient to resolve the instantaneous structure of the planetary boundary layer. On the basis of the nearly constant values of the Figure 7c shows that the altitude of the maximum in the conductivity (60 km) is confirmed by the mutual impedance probe measurements. This instrument gives the impedance of the medium at 45 Hz and yields a phase shift, which is sensitive to the presence of electrons only. The quadrupolar probe also records the spectrum of the electric signal induced in the probe environment by the 45 Hz stimulus, in the bandwidth 0-9.22 kHz (Fig. 8a, active mode).
The electric field due to natural wave emissions was investigated during the descent, using the receiving dipole of the mutual impedance probe in two frequency ranges, 0-11.5 kHz and 0-100 Hz (Fig. 8b, c, passive mode). This provided a unique opportunity to investigate in situ lightning and related phenomena (for example, corona discharges) on Titan 21 that would produce electromagnetic waves 38 , excite global and local resonance phenomena in the surfaceionospheric cavity 39,40 and could drive a global electric circuit 22 . Several impulsive events have been observed during the descent, for example at 2,800 s. The narrow-band wave emission seen near 36 Hz is reminiscent of a possible resonance generated by lightning activity in the spherical waveguide formed by the surface of Titan and the inner boundary of its ionosphere, but should be interpreted with caution. A comparison of the records presented in Fig. 8a and b shows that the first spectrogram (active mode) not only displays the Figure 5 | Temperature profile of the lower atmosphere as measured by the temperature sensors, TEM 27 (expanded from Fig. 2). Temperature uncertainty is^0.25 K in the range from 60 to 110 K, and^1 K above 110 K. The temperature minimum of 70.43 K is reached at the tropopause (about 44 km; 115^1 hPa). HASI temperatures are in very good agreement (within the error bars) with data obtained by Voyager radio occultation 2 (ingress, circles; egress, crosses) assuming a pure nitrogen atmosphere. The error bars for Voyager data are reported:^15 K (egress)^10 K (ingress) near the 200-km level,^0.5 K at the tropopause. At the tropopause, HASI measured temperature values ,1 K colder than Voyager 2 , but reanalysis of these data 3 suggested a similar temperature value (70.5 K) assuming a stratospheric composition of 98.5% N 2 plus 1.5% CH 4 . Figure 3 | The temperature lapse rate. The temperature gradient dT/dz was computed from the HASI temperature (T(z)) profile, and the altitude (z) was derived from the Huygens trajectory reconstruction 26 . The spatial resolution of the HASI measurements is of the order of 20 km from the top of the atmosphere down to the 400-km altitude level, decreasing down to 1 km at the 160-km level 24 . The profile shows in general a cut-off at the dry adiabatic lapse rate (dotted line), implying that fluctuations may lead to convective instabilities. The line at zero temperature variation is shown in white against the black curve. Six inversion layers in the upper atmosphere (at about 510, 600, 680, 800, 980 and 1,020 km) could be detected by strong peaks towards positive values. The strong lower inversion layer (4 K km 21 at ,510 km) corresponds to the feature already observed from the ground during Titan's stellar occultations 10 . The strong peaks between the 160-and 110-km levels correspond to the parachute deployment sequence.  28 . Measurements (solid line) corrected for dynamic effects are shown together with values obtained by Voyager 1 radio occultation 2 (ingress, circles; egress, crosses). HASI pressure values are determined with an uncertainty of 1% along the entire descent.

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(1) Estimate the scale height in the lowest 50 km of Titan's atmosphere from the probe data.
(4) Notice that the temperature lapse rate in the probe data appears to be somewhat less than the dry adiabatic lapse rate, particularly above the lowest 5-10 km of the atmosphere. Explain how condensation of CH 4 in the atmosphere could produce a lapse rate lower than the dry adiabatic lapse rate.
(5) We could define a virtual temperature to account for variable amounts of CH 4 on Titan, just like the virtual temperature on Earth accounts for variable amounts of H 2 O. Suppose we compare two air parcels with the same pressure and temperature, but one has a higher CH 4 mixing ratio than the other. Which air parcel would have a higher value of Titan's virtual temperature? Which air parcel would be more dense?
121 Humans need to lose the heat they are continually creating to the surrounding environment, or else they would overheat. An important way that they do this is through the evaporation of sweat from the surface of the skin. The skin surface has a temperature of about 35˝C. If the ambient air has some form of temperature higher than the skin surface temperature, humans will not be able to lose heat, so they will overheat and die in a few hours. If humans continue to emit CO 2 on current projections, many regions of Earth would be in this regime and we would have to go live on Chill Zone 274b instead.
(9) Which type of temperature (temperature, potential temperature, virtual temperature, dew point, wet bulb temperature) is the relevant one to compare the skin surface to in order to determine whether humans will overheat? (10) The latent heat of evaporation of water is about 3ˆ10 6 J kg´1. A typical human loses 100 W of power. Let's assume that this power is lost exclusively through evaporation of sweat. If a year's worth of this sweat were condensed and collected, what would it's volume be in multiples of the volume of a typical human? Useful facts: There are about 3ˆ10 7 s in a year. A typical human has a volume of 100 L or 0.1 m 3 .

Midterm 2, 2017
In a paper published in the journal Science this year, Zhai et al. described a new material they had invented that might lead to a cheaper and more efficient way to cool homes [9]. The material consists of SiO 2 microspheres embedded in a thin layer of polymethylpentene (TPX) (see figure  below). The polarizable microspheres interact strongly with infrared light, making the metamaterial extremely emissive across the full atmospheric transmission window while remaining transparent to the solar spectrum. (B) Normalized absorption (blue), scattering (red), and extinction (black) cross-sections of individual microspheres as functions of size parameter (k0a). The extinction, the sum of the scattering and absorption, peaks at a size parameter of 2.5, which corresponds to a microsphere radius of 4-µm. The inset shows the electric field distributions of two microspheres with 1-and 8-µm diameters, illuminated at a 10-µm wavelength. The scale bar is 4 µm. The smaller microsphere resonates at the electric dipolar resonance while higher order electric and magnetic modes are excited in the larger microsphere. (C) Angular diagram for the scattering far-field irradiance of an 8-µm-diameter microsphere with 10-µm wavelength illumination. The incident field is polarized along the y-direction d ti l th di ti TPX is nearly transparent to solar radiation and the SiO 2 microspheres have a size and distribution chosen so that they interact strongly with terrestrial radiation. The material is then coated with a thin (200 nm) layer of silver on the bottom side. The material would be used by placing it on roofs with the silver side pointing down.
In the diagram below green represents TPX, the yellow circles represent SiO 2 microspheres, and gray represents the silver (Ag) layer.  The top panel of the diagram below shows the typical solar radiation reaching the surface (red) and the typical surface terrestrial radiation from a blackbody reaching space (blue). The bottom panel shows the emissivity of the TPX/SiO 2 microsphere mixture as a function of wavelength, both theoretically and as measured experimentally. Fig. 3. Spectroscopic response of the hybrid metamaterial. (A) Schematic of the hybrid metamaterial backed with a thin silver film. The silver film diffusively reflects most of the incident solar irradiance while the hybrid material absorbs all incident infrared irradiance and is highly infrared emissive. (B) Three-dimensional confocal microscope image of the hybrid metamaterial. The microspheres are visible due to the autofluorescence of SiO2. (C) Power density of spectral solar irradiance (AM1.5) and thermal radiation of a blackbody at room temperature. The sharply varying features of both spectra are due to the absorbance of the atmosphere. The radiative cooling process relies on strong emission between 8 and 13 µm, the atmospheric transmission window. (D) The measured emissivity/absorptivity (black curve) of the 50-µm-thick hybrid metamaterial from 300 nm to 25 µm. Integrating spheres are employed for the measurement of both solar (300 nm to 2.5 µm) and infrared (2.5 µm to 25 µm) spectra. Theoretical results for the same hybrid metamaterial structure (red curves) are plotted for comparison. Two different numerical techniques, RCWA and incoherent transfer matrix methods, are employed for the solar and infrared spectral ranges, respectively. (2) Why is the solar radiation generally at shorter wavelengths than the terrestrial radiation?
(3) Why is the surface terrestrial emission from a blackbody that reaches space nearly zero at certain wavelengths?
(4) Why has the material been chosen to have as small an emissivity as possible for wavelengths shorter than about 3 µm and an emissivity as close to one as possible for wavelengths larger than about 3 µm?

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The diagram below shows the temperature and power emitted by the material in testing as a function of time when it was put outside and exposed to the dirunal cycle in sunlight and ambient atmospheric conditions. Fig. 4. Performance of scalable hybrid metamaterial for effective radiative cooling. (A) A photo showing the 300-mm-wide hybrid metamaterial thin film that was produced in a roll-to-roll manner, at a speed of 5 m per minute. The film is 50 µm in thickness and not yet coated with silver. (B) A 72-hour continuous measurement of the ambient temperature (black) and the surface temperature (red) of an 8-in-diameter hybrid metamaterial under direct thermal testing. A feedback-controlled electric heater keeps the difference between ambient and metamaterial surface temperatures less than 0.2°C over the consecutive three days. The heating power generated by the electric heater offsets the radiative cooling power from the hybrid metamaterial. When the metamaterial has the same temperature as the ambient air, the electric heating power precisely measures the radiative cooling power of the metamaterial. The shaded regions represent nighttime hours. (C) The continuous measurement of radiative cooling power over three days shows an average cooling power > 110 W/m 2 and a noon-time cooling power of 93 W/m 2 between 11am -2pm. The average nighttime cooling power is higher than that of the daytime, and the cooling power peaks after sunrise and before sunset. The measurement error of the radiative cooling power is well within 10 W/m 2 (32).  (5) The air conditioner in a typical house draws on the order of 1000 W. If the roof of a typical house were covered with this new material, would it be able to keep the house cool without running an air conditioner? Only an order of magnitude estimate is required to answer this question.
The article doesn't specify this, but a friend at NASA told me that the shades will be set up to rotate so that they will always be pointed directly at Earth as it orbits the Sun. She also told me that the shades will transmit 20% of light that hits them, equally at all wavelengths.
Interestingly, we may have already observed an example of an alien civilization trying to increase the dankitude of its star with some shades. The star KIC 8462852, also known as Boyajian's Star, is famous for having strange dips in its lightcurve. This means that it gets much darker sometimes for unknown reasons, possibly because of an alien superstructure such as some sweet shades.
We'll be using the simple energy balance model of climate from class to understand the effects of NASA's mission on planet Earth. To remind you, the model has a surface and one atmospheric layer. The surface has a longwave emissivity of 1 and a shortwave albedo of A. The atmosphere has a longwave emissivity of ε and is transparent to shortwave radiaton. Here is the diagram of the model: In class we rewrote this model in terms of the solar flux (S, in W m´2), the surface temperature (T s , in K), the atmospheric temperature (T a , in K), and the Stefan-Boltzmann constant (σ " 5.671 0´8 W m´2).
(4) It is possible to estimate the surface temperature of an object by measuring its electromagnetic radiation emission spectrum. After NASA puts the shades in place, will our estimate of the Sun's temperature using this method change?
(5) If for some reason NASA's shade plan fails, an alternative plan for counteracting global warming with geoengineering is to put sulfate aerosols into the stratosphere, either with big guns or airplanes. These sulfate aerosols would scatter incoming sunlight, cooling the planet. Estimate the total mass of sulfate aerosols that would have to be put into the stratosphere to counteract global warming. Give your answer in multiples of the mass of full 747 airplanes (« 4ˆ10 5 kg). You may assume that global warming alters radiative balance by 4 W m´2, the solar constant is 1365 W m´2, the mass scattering coefficient for sulfate aerosols is about κ « 4 m 2 g´1, the planetary albedo is about 0.3, and the planetary radius is about 6,300 km. You may also find the following expansion useful: e´τ « 1´τ.

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In 2010 Jacob Bean, who is now a professor of astronomy at the University of Chicago, published a famous paper measuring composition of the atmosphere GJ1214b using a technique called transit spectroscopy [11]. Transit spectroscopy involves looking at the planet as the light from the star it is orbiting passes through the planet's atmosphere on its way to us. The plot below shows Jacob's transit measurements (black dots with error bars) of the planetary radius as a function of wavelength compared to what would be predicted by models incorporating various atmospheric compositions. The wiggles in the models represent absorption by atmospheric constituents, which change the apparent radius of the planet.
photochemical processes remain a viable prospect and are currently unconstrained by models. Transit observations of GJ 1214b in the infrared could shed light on the question of clouds and hazes in the planet's atmosphere: scattering from cloud and haze particles is far less efficient at these wavelengths, and radius variations in the planet's transmission spectrum could be more apparent if the atmosphere was actually composed primarily of hydrogen. Indeed, a recent study has suggested there are variations in the planet's transmission spectrum between the near-infrared J and K bands that are qualitatively consistent with the expectations for a hydrogen-dominated atmosphere (B. Croll, personal communication). Further observations are needed to clarify this issue.
The only other known transiting planet in the same mass regime as GJ 1214b, CoRoT-7b 15,16 , could also harbour an atmosphere despite the extreme level of insolation it receives from its host star 17 . However, this planet orbits a much larger star (R ?~0 :87 R 8 ; ref. 14) than GJ 1214b and, thus, it is unlikely that additional constraints on its atmosphere could be obtained with existing facilities owing to the unfavourable planet-to-star radius ratio. In contrast, our results confirm previous 18  Theoretical predictions of the transmission spectrum for GJ 1214b 6 are shown for atmospheres with a solar composition (that is, hydrogen-dominated; orange line and squares), a 100% water vapour composition (blue line and triangles), and a mixed composition of 70% water vapour and 30% molecular hydrogen by mass (green line and stars). The points for the models give the expected values for the transmission spectrum in each of the spectrophotometric channels. All of the features in the model spectra arise from variations in the water vapour opacity, with the exception of the feature at 890 nm that is due to methane absorption. The measurements and their uncertainties (black circles) were estimated by fitting the spectrophotometric data using five Markov chains with 2.5 3 10 5 steps. The uncertainties, which are valid for the relative values only, are the 1s confidence intervals of the resulting posterior distributions, and are consistent with the estimates we obtained from a residual permutation bootstrap analysis. The uncertainty in the absolute level is 0:11 R + , and is due mainly to the uncertainty in the host star mass. The data are consistent with the model for the water vapour atmosphere (x 2 5 5.6 for 10 degrees of freedom) and inconsistent with the model for the solar composition model at 4.9s confidence (x 2 5 47.3). The predictions for a solar composition atmosphere with CH 4 removed due to photodissociation (not shown) are equally discrepant with the data. The mixed water vapour and molecular hydrogen atmosphere model contains the most hydrogen possible to still be within 1s (x 2 5 11.5) of the measurements. The data are furthermore consistent with a hydrogendominated atmosphere with optically thick clouds or hazes located above a height of 200 mbar (not shown). We obtain similar results when comparing the models to measurements obtained using smaller or larger channel sizes.

LETTER RESEARCH
(4) Suppose we are looking at a wavelength where GJ1214b's atmosphere strongly absorbs the stellar light passing through it. Will the planetary radius appear relatively large or relatively small at this wavelength? Tad Komacek graduated from the Department of the Geophysical Sciences in 2013, and recently defended his PhD thesis at the University of Arizona. As part of his PhD thesis, he simulated the atmospheric circulation of idealized planets like WASP-33b [13]. The plots below show output from two of these simulations assuming either high atmospheric friction or low atmospheric friction. The plots show velocity vectors and colormaps of atmospheric temperature (yellow is hot and blue is cold). The sub-stellar point (directly under the central star) is located at (longitude,latitude)=(0,0). The anti-stellar point is located at (longitude,latitude)=(˘180,0). ) -table that the linear limit is reached at di↵erent T eq,top values depending on the strength of the drag applied. With a spatially constant ⌧ drag = 10 5 sec, the models respond linearly to forcing at T eq,top . 10 Kelvin, and are very nearly linear throughout the range of T eq,top considered. This causes the U rms -T eq,top relationship for ⌧ drag = 10 5 sec to be visually indistinguishable from a linear slope in Figure 3. However, without a spatially constant drag, the linear limit is not reached until T eq,top . 0.1 Kelvin, and the dynamics are nonlinear for T eq,top & 1K.
The main grid presented involves a parameter study mimicking that of Perez-Becker & Showman (2013) The figure below shows measurements and models of planetary radiative emission normalized by stellar emission as a function of wavelength for WASP-33b [14]. (10) The grey curve shows a modeled spectrum with an isothermal temperature profile. Either the red or the blue curve shows a modeled spectrum including a stratosphere, and the other one doesn't have a stratosphere (the temperature just decreases as you go up). The green dots are the data and the other colored dots are the models averaged over the same bins as the data. Does the model that produced the red or the blue curve include a stratosphere? 172