### Climate Physics II

Deep ocean circulation is predominantly adiabatic and salt conserving, allowing the use of soluble tracers. Results show that communication between surface and depths occurs only in isolated cold regions (North Atlantic, Antarctica).

Ocean structure and circulation- heat flux: solar energy is absorbed in the few top tens of metres; long wave radiation is then re-radiated. Latent heat fluxes also play an important part, and as you would expect, there are strong regional, seasonal and diurnal variations.

Ocean structure and circulation- freshwater flux: salinity is altered by freezing (‘rejects’ brine), river runoff and net evaporation minus precipitation.

Ocean structure and circulation- momentum flux: surface winds generate waves and cause turbulent mixing.

Geostrophic flow and thermohaline circulation: temperature and salinity gradients produce pressure gradients. These gradients are balanced by the Coriolis force (acts perpendicular to velocity). This balances creates a geostrophic flow, which gives rise to the thermohaline circulation. This is a kind of global ‘conveyor belt’ that moves a lot of heat around the planet. The key features are massive sinking in the north Atlantic and Antarctic, and upswelling almost everywhere else.

Sources of energy

Energetic particles: 0.001 Wm-2

Geothermal: 0.06 Wm-2

Anthropogenic: 0.02 Wm-2

As you might expect, the Sun is the main source of energy, and the nuclear reactions involved therein are covered in great detail in the Astrophysics section.

$B(\lambda, T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kT}}-1}$

Optically thick slab

Optically thin slab

Atmospheric absorption

1. Ozone absorbs UV, visible and IR (9.6μm).
2. O2 absorbs UV (ozone production).
3. H2O absorbs IR and far IR.
4. CO2 absorbs near IR and IR.
5. CH4, CO, N2O and NO also absorb at various wavelengths.
6. There is also continuum absorption by water dimers, aerosols, cloud particles etc.

Sinks of energy: where does the Sun’s energy go?

1. 50% is absorbed by the surface of the Earth, which re-emits long wave radiation- hence the atmosphere is mostly heated from below.
2. ~20% is absorbed by polar molecules in the atmosphere.
3. ~30% is reflected by clouds, dust and the Earth’s surface.

Radiance is energy per unit time, area, wavenumber and solid angle; if we put it all together, we can see that it is measured in units of W m-1 sr-1

Radiative transfer equation: take monochromatic radiance passing through an element of absorbing medium.

θ = temperature

ρ = density

κ = absorption coefficient

$dR = -R\kappa \rho dz + B \kappa \rho dz$

where the first term on the RHS corresponds to absorption (assuming no scattering) and the second term corresponds to emission.

$d\tau =-\kappa \rho dz$

$\frac{dR}{d\tau}=R-B$ Schwartzchild’s equation

$\frac{dR}{d\tau}e^{-\tau}-R e^{-\tau}=-B e^{-\tau}$

$\frac{d}{d\tau}(R e^{-\tau})=-B e^{-\tau}$

$[R e^{-\tau}]_0^{\tau_g}= -\int_0^{\tau_g}B e^{-\tau}d\tau$

At the top of the atmosphere, τ=0 and R=Ratm (we see the radiance of the whole atmosphere)

At the Earth’s surface, τ= τg and R=0 (since there’s no atmosphere below us, we don’t see any radiance).

$R_{atm}=-\int_0^{\tau_g}B e^{-\tau}d\tau$

Transmission $T= e^{-\tau}$

$dT=- e^{-\tau}d\tau$

-> $R_{atm}=\int_{e^{-\tau_g}}^1 BdT$

Or, in terms of z where z=0 when τ= τg and z=∞ when τ=0

$R_{atm}=\int_0^{\infty} B \frac{dT}{dz} dz$

Now, to get the total radiance, we need to add the black body radiance emitted by the surface and measured at z=∞

$R_{\infty}=\int_0^{\infty} B \frac{dT}{dz} dz+B_0T_0$

Weighting function $\frac{dT}{dz}$ peaks high or low in the atmosphere depending on whether the opacity is large or small at wavelength λ.

These weighting functions and their peaks are useful for working out atmospheric temperature profiles from measured radiances.

Instruments for remote sensing (obviously we us a satellite to house them).

Basic idea

Measured $R_n=B(\theta(z_n))$ where θ is the temperature at height z where the weighting function peaks.

$\theta(z_n)=B^{-1}(R_n)$

Take the weighted mean over all n, using $\frac{dT}{dz}$ to get a complete temperature profile.

Infrared radiometer: used for remote sounding.

1. Radiometer views Earth through a plane mirror. The mirror can be rotated to view two calibration targets- cold space (zero radiance) and a calibration blackbody at temperature TC>Tearth. The calibration targets are used to produce a linear calibration curve of signal vs. radiance.
2. Chopper: mechanical shutter that interrupts the beam to ‘chop’ it into pulses of frequency 100-1000 Hz. This avoids low frequency noise and ‘drifting’ due to background, as well as generating an A.C. signal for the amplifier. Radiation can also be labelled by pulse.
3. Optics focus energy onto the detector. IR wavelengths are readily absorbed by many materials so usually reflecting optics are used. Where transmitting components are required, IR transparent materials such as Ge or diamond are used.
4. Filter: selects the wavelength range Δλ to which the instrument is sensitive. Consists of a stack of Fabry-Perot etalons comprised of ~100 thin layers of dielectric material deposited on a substrate of IR transmitting material. Can be cooled to cut down background.
5. Detector: either thermal detector or photon detector. Thermal detectors respond to the heating effect of the radiation (thermal resistance or thermoelectric effect). Photon detectors are semiconductors- the incoming photons excite transitions across the band gap. These need to be cooled to cut down on thermal transitions.
6. Amplifier etc: amplifies and outputs signal.

Output signal in frequency band ν -> ν+Δν

$S=gA\Omega \frac{\int_0^{\infty}\alpha_{\lambda}R_{\lambda}d\lambda}{\int_0^{\infty}\alpha_{\lambda} d\lambda}+ const$

g= linear gain factor (VJ-1)

A= aperture

Ω= solid angle ( is also known as the etendu)

αλ= filter transmission at wavelength λ

We usually take R=B, blackbody radiance.

$S=X\bar{R}+Y$ or $S=gP\delta t$

where X is gain, Y is offset and P is power ($P=RA\Omega$ per Δν)

Optical throughput/transmission: a factor to take into account any optical losses in the system; usually we do this by replacing with AεΩ (or τ) in any expression. A word of warning, however, in some books the term ‘optical throughput’ is used to mean what we are calling etendu.

Main sources of noise

1. Thermal/Johnson noise due to resistance. Proportional to square root of absolute temperature.
2. Shot noise due to finite electron charge. Proportional to square root of current.
3. Photon noise– proportional to square root of flux.
4. Flicker noise– proportional to 1/f.

As noise is random, overall it is proportional to $\frac{1}{\sqrt{t}}$

Signal to noise ratio

$\frac{S}{N}=\frac{P(\Delta t)^\frac{1}{2}}{NEP}$

Noise Equivalent Power (NEP) is the power that must fall on the detector to produce a signal equivalent to the sum of all sources of noise per unit bandwidth. So, despite its name, it’s not exactly a power as it has units of WHz-1/2

Noise $N=g(NEP)(\Delta t)^{\frac{1}{2}}$

$NEP=\frac{\sqrt{A\Omega}}{D*}$ where D* is the quality factor.

Noise Equivalent Radiance (NER) is the change in the target spectral radiance that produces a change in signal equal to the noise. Consider the power, P, falling on the detector in a particular frequency range ν -> ν+Δν

$P=RA\Omega \Delta \nu$

-> $(NEP)=(NER)A\Omega \Delta \nu (\Delta t)^{\frac{1}{2}}$

Noise Equivalent Temperature (NET) is the change in the temperature of the target that produces a change in signal equal to the noise. If the instrument is viewing a region emitting with black-body radiance, B, then

$\Delta B=\Delta T \frac{dB}{dT}$

-> $(NER)=(NET) \frac{dB}{dT}$

Integration time

$\Delta t =(\frac{NEP}{\epsilon A |omega \delta \nu \frac{dB}{dT}(NET)})^2$

Greenhouse effect: a popular term for the energy balance at the Earth’s surface, which affects its mean temperature. This balance is affected by the abundance of minor constituents in the atmosphere (H2O vapour, CO2, etc). Of course, these abundances are increasing because of human activities- theory predicts this may lead to an increase in mean temperature, although the exact changes are not easy to predict.

Solar constant: the flux at Earth due to radiation from the Sun.

Total flux at Sun’s surface $F=\sigma T^4$

Flux at Earth $f=\sigma T^4 \frac{R_S^2}{D^2}=1368 Wm^{-2}$ (D=1 a.u.)

Albedo: a measure of cloud and dust cover, usually given as a fraction A. Only the fraction (1-A) of the Earth’s surface receives power from the Sun. Albedo is difficult to measure exactly (A~0.3).

Effective temperature of the Earth

Total power arriving at Earth $P_1=(1-A)f\pi R_E^2$

Note that it’s not \$latex 4\pi R_E^2 because the Earth only presents a disc to solar radiation (after all, sunlight isn’t bombarding the Earth from all directions or we would never have night).

Power radiated by Earth $P_2=\sigma 4\pi R_E^2 T_{eff}^4$

In equilibrium, P1=P2

$(1-A)f=4\sigma T_{eff}^4$

$T_{eff}\simeq 250K$

But we know just from living here that the mean surface temperature is not -23°C, it’s about 10°C. This is because the atmosphere acts like a blanket.

Simple greenhouse model

Model the atmosphere as a slab at temperature Ta.

Say the atmosphere is completely transparent to short-wave radiation from the Sun (λ<4μm) but completely opaque to thermal radiation (λ>4μm).

Balance fluxes in equilibrium

At surface $\frac{1}{4}f(1-A)+\sigma T_a^4=\sigma T_s^4$

Ta is the effective temperature Teff.

$2t_a^4=T_s^4$ (from above we know that $\frac{1}{4}f(1-A)=\sigma T_a^4$ at the top of the atmosphere).

$T_S=2^{\frac{1}{4}}T_a \simeq 297K$

Despite its simplicity, this model gives a reasonable answer- but we can make better models without too much more effort.

Improved greenhouse model

Stratosphere: optically thin, completely transparent to shortwave radiation (λ<4μm). Tstrat=constant.

Troposphere: optically thick, completely opaque to longwave radiation (λ>4μm). $\frac{dT}{dz}=-\Gamma$

At tropopause $2\sigma \epsilon T_{strat}^4=\sigma\epsilon T_E^4=\frac{1}{4}f(1-A)$

At surface $T_{ground}=T_{strat}+\Gamma \Delta z$

Yet another greenhouse model

At each ‘level’ the fluxes must balance when in equilibrium

$F_S=\frac{1}{4}f(1-A)$

$F_{st}=\epsilon \sigma T_{st}^4$

Top of atmosphere $F_S=\epsilon F_{st}+(1-\epsilon)F_A$

Tropopause $F_S+\epsilon F_{st}=F_A$

Surface $F_S+F_A=F_G$

Should we wish to get more involved, we could keep on adding layers to build up a more complete profile.

Stratosphere temperature profile: up until now,  we have always assumed that the stratosphere is at constant temperature, but in fact this is not quite the case. Ozone in the upper stratosphere absorbs UV radiation from the Sun, leading to an increase in temperature with height.

Main factors affecting climate change

1. Atmospheric pollution caused by human activity; this has three main effects

a) Increased CO2 levels contribute to the greenhouse effect. A global temperature rise has been observed, but it is still unclear how much of this is due to natural variation.

b) Increased sulphate concentration may contribute to increased cloud cover.

c) CFCs, CLOx, BrOx and NOx catalyse the breakdown of ozone in the stratosphere, so there is less absorption of solar UV radiation (note that as CFCs can remain in the atmosphere for up to 40 years, they are still present despite the widespread ban on their use).

1. Ocean variations.

a) Short-term (decadal) effects include oscillations of wind driven circulation such as the El Niño phenomenon, which has definite local impact.

b) Of great importance in the long term is the stability of the thermohaline circulation, which moves a lot of heat around the planet. THC shutdown would have a significant global impact.

1. Albedo variations: cloud cover works in opposition to the greenhouse effect. Although it is very difficult to measure albedo accurately, preliminary measurements indication short-term fluctuations of ~20%.
2. Astronomical (Milankovich) cycles: variations in the Earth’s eccentricity, obliquity and precession of the equinoxes take place with a period of ~104 years. Any associated climate change will be on a similar timescale.
3. Solar constant variations

a) In the short term, this is mainly due to sunspots.

b) There also appear to be longer term variations of <0.1%, with an estimated global response of 0.03%. Any globally important changes would be on a timescale of ~108 yr.

1. Natural disasters: A B-movie style asteroid impact or large volcanic eruption would have drastic consequences in both the long and short term (the short term being the impact/eruption itself and the long term being changes in albedo and so forth).

Ice ages: the ice age periods saw some rapid changes in climate that are difficult to explain. The only mechanisms which could act fast enough are ocean circulation changes and possibly enormous volcanic eruptions. Independent studies favour the ocean circulation mechanism.

The implication is that there must be multiple ‘stable’ states for the ocean-atmosphere; the climate system can rapidly switch between these states.

Return of the simple radiative model

$C\frac{dT}{dt}=\frac{1}{4}F_S(1-\alpha)-\sigma T^4$

(α(T) is albedo).

In equilibrium $\frac{dT}{dt}=0$

-> $F_S(1-\alpha)=4\sigma T^4$

Assume that α(T) is constant (high) for an ice-covered Earth, constant (low) for an ice free Earth and a linear function of T in between.

We get three solutions, A, B and C.

In regions 1 and 3, $\sigma T^4 < \frac{1}{4}F_S(1-\alpha)$, $\frac{dT}{dt} > 0$

In regions 2 and 4, $\sigma T^4 > \frac{1}{4}F_S(1-\alpha)$, $\frac{dT}{dt} < 0$

-> points A and C are stable fixed points, point B is unstable

A: ice cover

B: partial ice cover

C: ice free

Obviously we don’t live in A or C conditions, so this model is not very accurate (and after we did all that work!).

Stommell’s 2-box model of ocean circulation

Each box is a reservoir of well-mixed water.

q, E are volume fluxes.

E is net evaporation minus precipitation.

Assume a linear equation of state.

$\rho=\rho_0(1-\alpha T+\beta S)$

$q=\frac{-k}{\rho_0}\Delta \rho$

k=friction coefficient, Δx=x1x2 for any variable x.

$q=k(\alpha \Delta T-\beta \Delta S)$

Now we want to set some simple, physically motivated boundary conditions for T and S.

We know that-

a) Air-sea heat exchange tends to restore ocean temperature to equilibrium values over short time scales.

b) Evaporation and precipitation rates do not depend on S.

So we have mixed boundary conditions- restoring for T, fixed flux for S.

Assume T maintained by air sea fluxes.

Salt budget for either box gives

$|q|S_1=(|q|+E)S_2$

(we’re using modulus signs because the flux can be either way).

$|q|(S_1-S_2)=ES_2$

ΔS is small, so S1~S2~S0

$q\Delta S \simeq ES_0$

$q\simeq k(\alpha T-\frac{\beta E S_0}{|q|})$

Let $q_0 \equiv k\alpha \Delta T$, $q_s \equiv \frac{\beta E S_0}{\alpha \Delta T}$

$q|q|-q_0|q|+q_0q_S=0$

$q=q_0-\frac{q_sq_0}{|q|}$

Solutions at $q=\frac{q_0\pm\sqrt{q_0^2\pm 4q_sq_0}}{2}$

Real solutions are possible given q0, qs >0, only ever a maximum of three- 2 positive and one negative.

The two solutions in the +ve quadrant are thermally direct.

Sinking at high latitudes

The solution in the –ve quadrant is salinity driven.

Note that if qs or E increase, then the thermally direct solutions disappear.

Critical value $q_s=\frac{q_0}{4}$

Extra calculations- how do we know we only get three solutions to that quadratic equation? Well, if you have the time and/or like to thing about these things, let’s go through it.

If q is +ve

$|q|^2-q_0|q|+q_0q_S=0$

$|q|=\frac{q_0\pm\sqrt{q_0^2-4q_sq_0}}{2}$

If q is -ve

$|q|^2+q_0|q|-q_0q_S=0$

$|q|=\frac{-q_0+\sqrt{q_0^2-4q_sq_0}}{2}$ (can’t have the other root because |q| must be positive by definition).

$q=\frac{q_0-\sqrt{q_0^2-4q_sq_0}}{2}$