Math 5490 10/13/2014 Math 5490 October 13, 2014 Topics in Applied - PDF document

Math 5490 10/13/2014 Math 5490 October 13, 2014 Topics in Applied Mathematics: Isotopes as Climate Proxies Introduction to the Mathematics of Climate Mondays and Wednesdays 2:30 – 3:45 http://www.math.umn.edu/~mcgehee/teaching/Math5490-2014-2Fall/ Streaming video is available at http://www.ima.umn.edu/videos/ Click on the link: "Live Streaming from 305 Lind Hall". Participation: https://umconnect.umn.edu/mathclimate How do we know the past climates? Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Summary Biology Matters and is complicated. phase 1 p δ 1 atmosphere ocean total δ 0   CO  H O  H CO  H  HCO phase 2 q 2 2 2 3 3 δ 2     HCO  H CO 3 3    Ca CO  CaCO 3 3 mass balance foraminifera      p q Temperature dependent fractionation occurs at every step. 0 1 2 The result: the δ 18 O in foram shells is about +30‰ compared with the equilibrium fractionation ε surrounding water (depending on temperature). ( δ 18 O)/dT ≈ ‐ 0.25 ‰/ ⁰ C           q p 1 0 2 0 (Reference: Pierrehumbert’s book) And then there’s carbon. Math 5490 10/13/2014 Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Biology Matters Carbon 14 and is yet still more complicated. Carbon 14 is created by cosmic Carbon 14 decays in the biosphere rays in the upper atmosphere. with a half ‐ life of about 6000 years. photosynthesis Nitrogen Carbon 14 Carbon 14 Nitrogen    6CO 6H O C H O 6O 2 2 6 12 6 2 electron neutron proton δ 1 = δ 13 C δ 2 = δ 13 C neutrino Fractionation is about ‐ 25‰.     0.025 2 1 Result: Plants, animals, coal, and oil are all lighter in 13 C than inorganic carbon. Carbon 14 is the basis of carbon dating. Good for about 50,000 years. At 60,000 years, it is down to 2 ‐ 10 of its original level. http://en.wikipedia.org/wiki/Carbon-14 Math 5490 10/13/2014 Math 5490 10/13/2014 Richard McGehee, University of Minnesota 1

Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Carbon 14 The rate of production of carbon 14 in the past is largely Carbon isotope ratios and unknown. It depends on cosmic events producing neutron atmospheric oxygen depletion reaching the upper atmosphere and it depends on variations in indicate that the increase in the Earth’s magnetic field. atmospheric CO 2 comes from burning fossil fuels. Carbon 14 is also fractionated by biology, so the concentration in the atmosphere depends on biological activity. Changes in Atmospheric Constituents and in Radiative Forcing , IPCC AR4, Chap. 2, p.138 http://ipcc- wg1.ucar.edu/wg1/Report/AR4WG1_Print_CH02. pdf Math 5490 10/13/2014 Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Recent History Recent History 1982 – 2002 1982 – 2002 atmospheric δ 13 C: C0 2 : ‐ 7.6‰ to ‐ 8.1 ‰ 340 ppm to 372 ppm Math 5490 10/13/2014 Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Carbon Measurements Carbon Measurements 1 gigatonne (Gt) = Atmospheric pressure at sea level: 10 9 metric tonnes = 14.7psi = (14.7psi)/((6.45cm 2 /in 2 )(2.205lb/kg)) = 1.03 kg/cm 2 10 12 kilograms (kg) = Mass of atmosphere above one square meter: 10 15 grams = 10300 kg 1 petagram (Pg) Surface area of Earth: 1 tonne = 1000 kg = 2205 lb = 1.102 tons 5.10x10 14 m2 Example: Total mass of atmosphere: current fossil carbon emissions: about 9 GtC/yr (1.03x10 4 )(5.10x10 14 ) = 5.25x10 18 kg = 5.25x10 6 Pg atomic wt of carbon: 12 Approximate composition of atmosphere: atomic wt of oxygen: 16 N 2 : 80% molecular wt 14x2 = 28 molecular wt of CO 2 : 44 (= 12 + 16 + 16) O 2 : 20% molecular wt 16x2 = 32 Current emissions of CO 2 : (44/12)x9 = 33 GtC0 2 /yr Mean molecular wt of atmosphere: How do gigatonnes convert to ppm in the atmosphere? 0.8x28 + 0.2x32 = 28.8 Math 5490 10/13/2014 Math 5490 10/13/2014 Richard McGehee, University of Minnesota 2

Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Carbon Measurements Atmospheric Carbon Total mass of atmosphere: 5.25x10 6 Pg Mean molecular wt of atmosphere: 28.8 g/mole 850 Gt Number of moles in atmosphere: (5.25x10 21 g)/(28.8g/mole) = 1.82x10 20 moles Number of moles in one millionth of the atmosphere: 1.82x10 14 800 Gt Atomic wt of C: 12 g/(mole of CO2) Mass of 1 ppm CO2: (12 g/mole)(1.82x10 14 mole/ppm) = 2.19x10 15 g = 2.19 Pg 750 Gt Actual conversion: 1 ppm by volume of atmosphere CO 2 = 2.13 Gt C 700 Gt Source: CDIAC (Carbon Dioxide Information Analysis Center) cdiac.ornl.gov/pns/convert.html Math 5490 10/13/2014 Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies δ 13 C Budget 1982 – 2002 724 GtC air: δ 1 δ 13 C: -7.6‰ air: δ 0 792 GtC increase in atmospheric CO2 δ 13 C: -8.1‰ 340 to 372 ppm coal?: δ 2 68 GtC 724 GtC to 792 GtC : 68 GtC emissions p = 724/792 = 0.914 6 GtC/yr for 20 yr : 120 GtC q = 68/792 = 0.086 About 57% stayed in the atmosphere.      p q 0 1 2             ( ) ( 8.1 0.914 ( 7.6)) / 0.086 13.4 p q δ 13 C: -7.6‰ to -8.1 ‰ 2 0 1 coal?: δ 13 C = -13.4‰ Other processes are important. Math 5490 10/13/2014 Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Paleocence ‐ Eocene Thermal Maximum (PETM) Paleocence ‐ Eocene Thermal Maximum (PETM) http://static.palaeontologyonline.com/Figure1.jpg Hansen, et al, Target atmospheric CO2: Where should humanity aim? Open Atmos. Sci. J . 2 (2008) Math 5490 10/13/2014 Math 5490 10/13/2014 Richard McGehee, University of Minnesota 3

Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Paleocence ‐ Eocene Thermal Maximum (PETM) http://maya-gaia.angelfire.com/mammals_rise_title_thm.jpg Zachos, et al , Science 292 (2001), p. 689 Math 5490 10/13/2014 Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Paleocence ‐ Eocene Thermal Maximum (PETM) Sharp decrease in δ 18 O, interpreted as large increase in temperature Sharp decrease in δ 13 C, interpreted as massive oxidation of organic carbon Zachos, et al , Science 292 (2001), p. 689 Zachos, et al , Science 292 (2001), p. 689 Math 5490 10/13/2014 Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies Paleocence ‐ Eocene Thermal Maximum (PETM) Paleocence ‐ Eocene Thermal Maximum (PETM) Sharp decrease in δ 18 O, interpreted as a rapid increase in temperature. Site 690 ‐ 1 ‐ 3 Sharp decrease in δ 13 C, interpreted as massive oxidation of ‐ 0.5 sequestered organic carbon. ‐ 2 0 ‐ 1 0.5 Site 690 d18O d13C ‐ 1 ‐ 3 1 0 d18O ‐ 0.5 1.5 ‐ 2 d13C 1 0 2 ‐ 1 2 0.5 2.5 d18O d13C 1 0 d18O 3 3 1.5 ‐ 56 ‐ 55.8 ‐ 55.6 ‐ 55.4 ‐ 55.2 ‐ 55 ‐ 54.8 ‐ 54.6 ‐ 54.4 ‐ 54.2 ‐ 54 d13C 1 Myr 2 2 2.5 δ 1 = 1.9, δ 0 = -0.8 3 3 ‐ 56 ‐ 55.8 ‐ 55.6 ‐ 55.4 ‐ 55.2 ‐ 55 ‐ 54.8 ‐ 54.6 ‐ 54.4 ‐ 54.2 ‐ 54 Myr Math 5490 10/13/2014 Math 5490 10/13/2014 Richard McGehee, University of Minnesota 4

Math 5490 10/13/2014 Isotopes as Proxies Isotopes as Proxies δ 13 C Budget Preindustrial Carbon ocean + p ocean + atmos: δ 13 C = 1.9‰ atmos: δ 1 δ 0 δ 13 C = -0.8‰ q ??: δ 2 ocean and atmosphere: δ 13 C = ?? 8000 + 700 + 600 = 39300 GtC                 (1 ) ( ) p q q q q 0 1 2 1 2 1 2 1 Let’s say 40000.     0 1 q    2 1 If organic carbon was oxidized, δ 2 = -25‰.     0.8  1.9    0.10 0 1 q      25 1.9 2 1 Sigman & Boyle, Nature 207 (2000), p.860 Math 5490 10/13/2014 Math 5490 9/3/2014 Isotopes as Proxies Isotopes as Proxies δ 13 C Budget Preindustrial Carbon p ocean + ocean + δ 13 C = 1.9‰ atmos: atmos: 40000 GtC δ 13 C = -0.8‰ δ 1 Where did 4400 Gt of δ 0 organic carbon come q ??: δ 2 from? δ 13 C = ?? Known coal reserves If organic carbon was oxidized, δ 2 = -25‰. today: 1000 Gt     0.8  1.9 Preindustrial terrestrial    0 1 0.10 q      25 1.9 carbon: 2100 Gt 2 1 4400 Gt is a lot. If x Gt of organic carbon was oxidized, x x   0.10 0.90  4000  4444 q x x 40000  It would take about 4400 Gt of organic carbon to produce the PETM δ 13 C spike. Sigman & Boyle, Nature 207 (2000), p.860 Math 5490 10/13/2014 Math 5490 9/3/2014 Isotopes as Proxies Isotopes as Proxies δ 13 C Budget PETM vs Today p Can we burn 4000 – 5000 Gt of carbon? ocean + ocean + δ 13 C = 1.9‰ atmos: Might be difficult: only 1000 Gt of known coal deposits, less of atmos: δ 13 C = -0.8‰ 40000 GtC δ 1 oil. δ 0 q But we are getting better at finding and extracting it. ??: δ 2 δ 13 C = ?? Do we have to burn that much carbon, since we are burning it faster than the deep ocean can absorb it? If methane was oxidized, δ 2 = -60‰.     0.8  1.9    0 1 0.44 q      60 1.9 2 1 If x Gt of organic carbon was oxidized, x x   0.044 0.956  1760  1800 q x x 40000  It would take about 1800 Gt of methane to produce the PETM δ 13 C spike. Math 5490 10/13/2014 Math 5490 10/13/2014 Richard McGehee, University of Minnesota 5