### Climate Physics I

Mean composition of the Earth’s atmosphere (by volume):

 N2 78.08% Basic ‘ingredients’- these homopolar molecules have no dipole moment, so no strong absorption O2 20.95% Noble gases and H2 each ~10-3->10-5% Boring CO2 >0.033% Important (see later) All increasing in atmosphere CH4 >2×10-4% N2O >5.0×10-5% O3 variable (~5×10-5%) Important H2O variable(~0-0.1%)

Standard pressure is defined as 760 mm Hg/1 atm / 1.013 ba r= 1013 mbar = 1.013×105 Pa

Atmospheric pressure is the weight of air above you (1 kg cm-3)- that’s one ton per square foot!

Pressure vs. height: as you would expect, pressure decreases with height. Still, this being physics, we want to express that more quantitatively.

Hydrostatic equation $dp=-\rho g dz$

Ideal Gas Law $pV=RT$ per mole

$\rho = \frac{M}{V} = \frac{Mp}{RT}$

$dp=\frac{-Mpg}{RT}dz$

$\frac{-dp}{p}=\frac{dz}{H}$

Where $H=\frac{RT}{Mg}$ is the scale height- the height over which the pressure drops by 1/e

$p=p_0e^{-\int_0^z \frac{dz}{H}}$

If the atmosphere were isothermal (T=const) as we find approximately in the lower stratosphere, then we have $p=p_0e^{\frac{-z}{H}}$

Troposphere- vertical temperature profile: the troposphere is the lowest ‘layer’ of the atmosphere (0->10km).

Take a ‘parcel’ of air rising adiabatically (same pressure as surroundings, but T may change).

For one mole of ideal gas $pV=RT$ (in case you hadn’t guessed, we’ll be using this a lot).

$pdV+Vdp=RdT=(C_p-C_V)dT$

Now to use our other good friend, the First Law of Thermodynamics.

$dQ=C_VdT+pdV=0$ (adiabatic)

$latex pdV=-C_VdT -> $Vdp=C_pdT$ Now back to the hydrostatic equation $dp=-\rho g dz$ $\frac{C_p}{V}dT=-\rho g dz$ $\frac{dT}{dz}=\frac{-\rho V g}{C_p}=\frac{-Mg}{C_p}=\frac{-g}{c_p}=-\Gamma$ Γ is called the adiabatic lapse rate. Tropospheric lapse rate: Γ=9.7 K km-1 for dry air. Cp for water vapour is almost twice that of dry air- that and latent heat effects mean that Γ~6-7 K km-1 in reality. As the atmosphere is mostly heated from below, often $\frac{dT}{dz}<-\Gamma$, inducing convection (more on this later). Vertical temperature profile of stratosphere: above 10km, large-scale convection ceases and the temperature stops falling. This optically thin, radiative layer of atmosphere is known as stratosphere. Stratosphere- optically thin, radiative, stratified (hence the name), T const Troposphere- optically thick, convective, $\frac{dT}{dz}=\frac{-g}{c_p}$ Radiative transfer will be discussed further later. Potential temperature is the temperature that a parcel of air at the point of observation would achieve if it were moved adiabatically to a standard pressure p0 (usually 1 bar). Using the First Law of Thermodynamics and the Ideal Gas Law as earlier- $C_pdT=Vdp=\frac{RT}{p}dp$ $\frac{C_p}{R}\frac{dT}{T}=\frac{dp}{p}$ Integrate from p,T (point of observation) to p0, θ (standard pressure) $\frac{C_p}{R}\log\frac{\theta}{T}=\log\frac{p_0}{p}$ $(\frac{\theta}{T})^{\frac{C_p}{R}}=\frac{p_0}{p}$ $\theta=T(\frac{p_0}{p})^{\frac{R}{C_p}}$ Differentiate wrt to z $\frac{d\theta}{dz}=\frac{dT}{dz}(\frac{p_0}{p})^{\frac{R}{C_p}}-T(\frac{R}{C_p})(\frac{p_0^{\frac{R}{C_p}}}{p^{\frac{R}{C_p}+1}})\frac{dp}{dz}$ $\frac{d\theta}{dz}=\frac{dT}{dz}\frac{\theta}{T}-\frac{R}{C_p}\frac{1}{p}\theta\frac{dp}{dz}$ $\frac{d\theta}{dz}=\theta(\frac{1}{T}\frac{dT}{dz}-\frac{R}{C_p}\frac{1}{p}\frac{dp}{dz})$ To clean this up a bit more, recall that $\frac{dp}{dz}=-\rho g=\frac{-Mpg}{RT}$ So $\frac{d\theta}{dz}=\frac{\theta}{T}(\frac{dT}{dz}+\frac{g}{C_p})$ $\frac{d\theta}{dz}=\frac{\theta}{T}(\frac{dT}{dz}-\Gamma)$ Stability: a parcel of air is stable if, when it undergoes a small vertical displacement from equilibrium, it experiences a buoyancy force restoring it to its original position. The atmosphere is stable if $\frac{d \theta}{dz}>0$, $|\frac{dT}{dz}|<\Gamma$ and unstable if $\frac{d \theta}{dz}<0$, $|\frac{dT}{dz}|>\Gamma$ Entropy: now we are in a position to find the entropy profile of the atmosphere.$latex dQ=C_VdT+pdV=C_pdT-Vdp for one mole of ideal gas

$dS=\frac{dQ}{T}=C_p\frac{dT}{T}-\frac{V}{T}dp$

$\frac{V}{T}=\frac{R}{p}$

$dS=C_p\frac{dT}{T}-R\frac{dp}{p}$

Integrating

$S=C_p\ln T-R\ln p+const$

$\frac{\theta^{C_p}}{p_0^R}=\frac{T^{C_p}}{p^R}$

$S=C_p\ln \theta-R\ln p_0+const$

Now absorb Rln p0 into the constant

$latex S=c_p\ln\theta+const’ $\Delta S=C_p(\Delta \ln\theta)$ so ln θ can be used as an entropy scale. Atmospheric humidity: the concentration of water varies with latitude and height from 0-10%. There are several different ways to quantify humidity. Mass mixing ratio, xm, is the ratio of the mass of water vapour to the mass of dry air in a volume V with total pressure p. Partial pressure and density of water pH2O, ρH2O Partial pressure and density of air pa, ρa $p=p_{H_2O}+p_a$ $\rho=\frac{Mp}{RT}$ $\frac{M_{H_2O}}{M_a}=0.622$ $x_m=\frac{\rho_{H_2O}}{\rho_a}=\frac{0.622p_{H_2O}}{p_a}$ Volume mixing ratio, xV, is the ratio of the volume of water vapour to the volume of dry air if the constituents were separated at total pressure p. $x_V=\frac{p_{H_2O}}{p_a}=\frac{n_{H_2O}}{n_a}$ Relative humidity is the ratio of x to its value at saturation. Dew/frost point is the temperature at which saturation occurs. Molecular mass of a mixture: we’ll be considering one mole of damp air from now on. Hydrostatic equation for a mixture of gases $dp=dp_1+dp_2$ $dp=-(\rho_1+\rho_2)gdz$ $dp=-(\frac{M_1p_1}{RT}+\frac{M_2p_2}{RT})gdz$ $dp=-\frac{M'p}{RT}gdz$ Where $M'=\frac{M_1p_1+M_2p_2}{p}$ is the effective molecular mass (pressure weighted mean) Stability of moist air $\frac{dT}{dz}=-\frac{g}{c_p}=-\Gamma$ without phase changes First Law $dQ=C_VdT+pdV+\frac{L}{M_{H_2O}}dm=0$ adiabatic (the last term on the LHS accounts for condensation- L is of course latent heat). From the Ideal Gas Law $pdV+Vdp=(C_p-C_V)dT$ $C_pdT-Vdp+\frac{L}{M_{H_2O}}dm=0$ $C_p\frac{dT}{dz}-V\frac{dp}{dz}+\frac{L}{M_{H_2O}}\frac{dm}{dz}=0$ We’ll now go ahead and find an expression for $\frac{dm}{dz}$ that we can substitute back in here. $m=M_{H_2O}n_{H_2O}$ $m=\frac{M_{H_2O}p_{H_2O}V}{RT}$$latex m=\frac{M_{H_2O}p_{H_2O}}{p}

$dm=M_{H_2O}(\frac{dp_{H_2O}}{p}-\frac{p_{H_2O}}{p^2}dp)$

$dm=M_{H_2O}(\frac{1}{p}\frac{dp_{H_2O}}{dT}-\frac{p_{H_2O}}{p^2}dp)$

$\frac{dm}{dz}=M_{H_2O}(\frac{1}{p}\frac{dp_{H_2O}}{dT}\frac{dT}{dz}-\frac{p_{H_2O}}{p^2}\frac{dp}{dz})$

Clausius-Clapeyron equation $\frac{dp_{H_2O}}{dT}=\frac{L}{T(V_v-V_l)}$

Vv>>Vl

$\frac{dp_{H_2O}}{dT}=\frac{L}{TV_v}$

$\frac{dp_{H_2O}}{dT}=\frac{L p_{H_2O}}{RT^2}$

$\frac{dm}{dz}=M_{H_2O}(\frac{1}{p}\frac{Lp_{H_2O}}{RT^2}\frac{dT}{dz}-\frac{p_{H_2O}}{p^2}\frac{dp}{dz})$

Whew- so we’re finally ready to substitute.

$C_p\frac{dT}{dz}-V\frac{dp}{dz}+\frac{L}{p}\frac{Lp_{H_2O}}{RT^2}\frac{dT}{dz}-\frac{Lp_{H_2O}}{p^2}\frac{dp}{dz}=0$

$(C_p+\frac{L}{p}\frac{Lp_{H_2O}}{RT^2})\frac{dT}{dz}=(V+\frac{Lp_{H_2O}}{p^2})\frac{dp}{dz}$

$\frac{dT}{dz}=\frac{(V+\frac{Lp_{H_2O}}{p^2})(-\rho g)}{C_p+\frac{L}{p}\frac{Lp_{H_2O}}{RT^2}}$

$\frac{dT}{dz}=\frac{\frac{-\rho Vg}{C_p}-\frac{\rho Vgp_{H_2O}L}{C_pRTp}}{1+\frac{L^2p_{H_2O}}{C_ppRT^2}}$

$\frac{\rho Vg}{C_p}=\frac{g}{c_p}=-\Gamma_{dry}$

$\frac{dT}{dz}=\frac{-\Gamma_{dry}(1+\frac{Lp_{H_2O}}{RTp})}{1+\frac{L^2p_{H_2O}}{C_ppRT^2}}$

$\frac{dT}{dz}=-\Gamma_{sat}$

Γsat is always less than Γdry

If $|\frac{dT}{dz}|<\Gamma_{sat}$, moist air is absolutely stable.

If $\Gamma_{sat}<|\frac{dT}{dz}|<\Gamma_{dry}$, moist air is conditionally stable (so it depends on the relative humidity).

If forced above the condensation level, moist air becomes unstable and continues to rise, forming clouds.

Saturated vapour pressure over a convex surface of radius r.

$SVP(r)=SVP(\infty)e^{\frac{2\sigma}{r}\frac{M}{(\rho_l-\rho_v)RT}}$

Where $SVP(\infty)$ is the saturated vapour pressure over a plane surface, and σ is surface tension.

Saturation ratio: $S=\frac{SVP(r)}{SVP(\infty)}=e^{\frac{a}{r}}$

Supersaturation: $S'=(S-1) \times 100\%$

Formation of cloud droplets: a droplet can only grow by condensation once it has reached a critical radius r*; smaller droplets just evaporate unless the surrounding air is highly supersaturated. It is very unlikely, however, that a drop of critical size would just form at random from a collection of water molecules- this would require the simultaneous collision of ~17 molecules at S’~750%. Even so, clouds form on a regular basis despite the fact that we don’t live in quite such a waterlogged atmosphere, so there must be some other mechanism at work.

In fact, soluble particles in the atmosphere act as condensation nuclei, making the formation of large droplets possible. They reduce the SVP, thus attracting vapour.

Effect of solute on vapour pressure (Raoult’s Law)

$S(r)\sim \frac{n_0}{n_0+n}=(1+\frac{b}{r^3})^{-1}$

Where n0 is the number of water molecules and n is the number of solute molecules.

Critical radius: saturation ratio over a droplet is the product of two terms.

$S(r)=e^{\frac{a}{r}}(1+\frac{b}{r^3})^{-1}$

The first term accounts for curvature and tends to increase S, the second term is the solute term and tends to decrease S.

We’ll now expand this as a power series-

$S(r)=(1+\frac{a}{r}+\ldots)(1-\frac{b}{r^3}+\ldots)$

$S(r)\sim 1+\frac{a}{r}-\frac{b}{r^3}$ conveniently neglecting higher order terms.

S has a maximum S* at critical radius r*.

$\frac{ds}{dr} \simeq \frac{-a}{r^2}+\frac{3b}{r^4}$

$\frac{ds}{dr}=0$ at r*

$r*=\sqrt{\frac{3b}{a}}\simeq 0.1\mu m$

Above this radius, the droplet will tend to grow in order to reduce the vapour pressure above its surface (smaller droplets will tend to shrink).

Sources of condensation nuclei

1. The most common natural source is sea spray.
2. Anthropogenic (man-made) aerosols such as sulphates are also important.
3. Electrical charges can work in the same way, but this effect is rarely important in the atmosphere.

Rate of droplet growth is given by

$\frac{dm}{dt}=-4\pi r^2D \frac{d\rho}{dr}$

$\frac{dm}{dt}\int_R^{\infty}\frac{dr}{r^2}=-4\pi D \int_{\rho_R}^{\rho_{\infty}}d\rho$

$\frac{dm}{dt}[\frac{-1}{r}]_R^{\infty}=-4\pi D (\rho_{infty}-\rho_R)$

$\frac{dm}{dt}=-4\pi RD(\rho_{infty}-\rho_R)$

$1-S=\frac{\rho_{infty}-\rho_R}{\rho_{infty}}$

$\frac{dm}{dt}=\rho_l\frac{dv}{dt}=4\pi R^2\frac{dR}{dt}\rho_l$

Where ρl is the liquid density.

Now substitute for $\frac{dm}{dt}$ with the expression we just found.

$4\pi R^2\frac{dR}{dt}\rho_l=-4\pi D (\rho_{infty}-\rho_R)$

But $\rho_{\infty}=\rho_v$, the vapour density, so

$\frac{dR}{dt}=\frac{D}{R}{\rho_v}{\rho_l}(S-1)$

$\frac{dR}{dt}=D{\rho_v}{\rho_l}(\frac{a}{r^2}-\frac{b}{r^4})$

Heat loss/gain of a droplet

Heat gain due to latent heat L

$\frac{dQ}{dt}=L\frac{dm}{dt}=-4\pi RLD (\rho_{\infty}-\rho_r)$

Heat lost due to thermal conductivity of air, λ

$\frac{dQ}{dt}=4\pi R \lambda (T_{\infty}-T_R)$

In equilibrium $T_R-T_{\infty}=\frac {LD (\rho_{\infty}-\rho_r)}{\lambda}$

Coalescence: the growth rate of cloud droplets is very slow- it takes over half a day to make a 40μm drop! But once the droplet begins to fall, larger particles will collide with smaller ones and grow by coalescence. This can be a very efficient process, especially if updrafts are also present in the clouds.

Growth of ice crystals: the SVP over ice is less than that over water, so at a given temperature an ice crystal will grow more rapidly at the expense of any water droplets present. Under the right conditions, crystals can grow to greater than 100μm in only a few minutes. At mid-northern latitudes, most rain actually originates as ice particles which melt before reaching the ground.

Ozone layer: ozone, O3 is continually formed and destroyed in the upper atmosphere.

$O_2+\gamma \rightarrow O+O$

$O+O_2+M \rightarrow O_3+M$

γ is a UV photon; M is an arbitrary molecule that carries off excess energy.

$O_3+\gamma \rightarrow O_2+O$

$O_3+O \rightarrow O_2+O_2$

Oxides of N, H, Cl and Br (mostly produced by human activities) catalyse the breakdown of ozone.

Ocean- hydrological cycle: 71% of the Earth is covered by ocean (6% ice), and it is the main source of atmospheric water vapour. Precipitation from the atmosphere provides surface freshwater fluxes such as streams and rivers that of course flow back into the ocean and affect ocean currents.

Ocean- global energy balance: equatorial regions receive much more energy from the Sun than polar regions, although energy lost by radiation is much the same everywhere. The ocean plays an important part in transporting the excess heat around, although the atmosphere/condensation have equally important roles. The top 3.2m of the ocean has the same heat capacity as the entire atmosphere, with total ocean heat content being ~1000 times that of the atmosphere.

Ocean- amelioration of global warming takes place in two ways

1. Ocean takes up ~35% of carbon released from burning fossil fuels.
2. Large heat capacity and slow deep mixing delay the effects of warming trends.

Salinity: ocean salt comes from the weathering of rocks over geological timescales. Near the surface, salinity varies from 3.3-3.8%, 85% of which is sodium chloride. In the deeper ocean, salinity is much less variable.

Note: unlike freshwater, saltwater gets denser as it gets colder. At high T $\rho \sim \rho(T)$, at low T $\rho \sim \rho(r)$ (dominant contributions).

Ocean equation of state: unfortunately, there is no ‘ideal ocean law’. It is usually conventional to linearise the equation of state.

$\frac{\rho}{\rho_0}\simeq 1-\alpha(T-\bar{T})+\beta(S-\bar{S})+$higher order terms

$\alpha=\frac{-1}{\rho}(\frac{d\rho}{dT})$

$\beta=\frac{1}{\rho}(\frac{d\rho}{dS})$

Ocean measurements

1. The basic method (and yes, I do mean basic). Fill a bucket with seawater and quickly take measurements before its temperature changes.
2. Slightly more advanced method. Lower a cluster of sampling bottles (there, doesn’t that sound better than ‘bucket’) into the water and fill them at different depths.
3. Acoustic thermometry. The ocean is practically opaque to other types of radiation, but the speed of sound in water can be used to work out T, S and p (temperature dependence dominates).
4. Surface temperature and altimetry can be measured by satellites.

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Vertical structure of the ocean

1. 0-10m. Mixed layer: rapid overturning by wind. As you can guess from the name, the layer is well mixed, so the temperature is much the same throughout.
2. 10-100m. Thermocline layer: slow overturning leads to steep gradients- temperature decreases with depth.
3. 100m-4km. Deep layer: small gradients, almost no vertical mixing. Temperature pretty much constant throughout.

Temperature gradient and depth of the thermocline layer depend on diffusion, surface winds, surface heat budget etc.

Estimate of \$latex \frac{\partial T}{\partial t} in the mixed layer

$\frac{\partial T}{\partial t}=-\omega \frac{\partial T}{\partial z}+\kappa \frac{\partial^2 T}{\partial z^2}$

Where the first term on the RHS corresponds to advection, and the second to diffusion.

In the steady state $\omega \frac{\partial T}{\partial z}=\kappa \frac{\partial^2 T}{\partial z^2}$

Dimensional analysis

$\frac{\omega \Delta T}{D}=\frac{\kappa \Delta T}{D^2}$