Math 5490 October 8, 2014

Size: px

Start display at page:

Download "Math 5490 October 8, 2014"

Gyles Waters
5 years ago
Views:

Math 5490 October 8, 204 Topics in Applied Mathematics:

Wednesdays 2:30 3:45 http://www.math.umn.

1 Math 5490 October 8, 204 Topics in Applied Mathematics: Introduction to the Mathematics of Climate Mondays and Wednesdays 2:30 3:45 Streaming video is available at Click on the link: "Live Streaming from 305 Lind Hall". Participation:

2 Isotopes as Climate Proxies How do we know the past climates?

3 What is this? δ 8 O ( ) Hansen, et al, Target atmospheric CO2: Where should humanity aim? Open Atmos. Sci. J. 2 (2008)

4 Ocean Sediment Cores

8 O as a Climate Proxy The isotope 6 O preferentially evaporates from the ocean and is sequestered in glaciers, leaving the heavier isotope 8 O more highly concentrated in the ocean.

5 8 O as a Climate Proxy The isotope 6 O preferentially evaporates from the ocean and is sequestered in glaciers, leaving the heavier isotope 8 O more highly concentrated in the ocean. Thus oceanic concentration of the isotope 8 O is higher during glacial periods. Foraminifera absorb more 8 O into their skeletons when the water temperature is lower and when more 8 O is in the water. Thus higher concentrations of 8 O in foraminifera fossils indicate lower ocean temperatures and higher glacier volume.

6 What is this? δ 8 O ( ) : per mil, per thousand 000 = 00% = 0 = % = 0.0 = 0.% = O: Oxygen 8: 8 protons 8 electrons 0 neutrons 7 O: Oxygen 7: 8 protons 8 electrons 9 neutrons 6 O: Oxygen 6: 8 protons 8 electrons 8 neutrons Most of the oxygen atoms on Earth are 6 O. About in 500 atoms is 8 O. About in 2500 is 7 O. There are other oxygen isotopes, but they are unstable.

7 What is this? δ 8 O ( ) Example Given a sample of calcium carbonate (CaCO 3 ) from a foraminifera fossil, suppose that the ratio of 8 O atoms to 6 O is r = = How would we report this finding? How would we measure it in the first place? The instruments measure the difference between two samples. Typically, one measures the difference between the sample of interest and a standard sample. A common standard is something called Vienna Standard Mean Ocean Water (VSMOW), for which the ratio of 8 O atoms to 6 O is s = Then 8 r s r O s s So we would report δ 8 O = 3.9

8 What is this? δ 8 O ( ) Conversely The 8 O content of a sample is reported as δ 8 O = 3.9 using the VSMOW standard. What is the proportion of 8 O in the sample? r 8 8 O, r s( O) s r ( ) Note r is the ratio of 8 O to 6 O. The proportion of 8 O in the sample is r r For small values of r, these are approximately equal.

9 Common Standards Isotopes Ratio Standard Source D:H VSMOW Pierrehumbert 3 C: 2 C PDB Wikipedia 2 8 O: 6 O VSMOW Pierrehumbert 8 O: 6 O VPDB Pierrehumbert Standards: VSMOW: Vienna Standard Mean Ocean Water PDB: Pee Dee Belemnite VPDB: Vienna Pee Dee Belemnite Pierrehumbert, Principles of Planetary Climate, Cambridge U Press, New York,

10 What does δ 8 O ( ) tell us? Fractionation Example: Evaporation of Water condensation vapor evaporation liquid r = ratio of 8 O: 6 O in vapor r 2 = ratio of 8 O: 6 O in liquid At equilibrium, r = f r 2 where f is the fractionation factor. (depends a lot on temperature)

11 What does δ 8 O ( ) tell us? Fractionation What about δ? r = ratio of 8 O: 6 O in vapor r 2 = ratio of 8 O: 6 O in liquid r = f r 2 r fr fs 2 2 f 2 s s s Note that the standard drops out. f is usually close to, so let f Since ε and δ are typically small, εδ is even smaller, so 2 often expressed as

12 What does δ 8 O ( ) tell us? Example: Evaporation of Water before dry air air + vapor δ after water δ = δ 0 water δ 2 δ 8 O(water) = δ 0 δ 8 O(vapor) is undefined δ 8 O(water) = δ 2 f = 0.99 = +ε δ 8 O(vapor) = δ ε = 0.0 = 0 Assume only a small amount of vapor forms. 2 0, The 8 O content of the vapor is 0 less than that of the ocean.

13 Example: Melting Glaciers glaciers before ocean ocean after common hydrogen: H = H heavy hydrogen: 2 H = D VSMOW D:H = = (very small) Assumptions current ocean: δd = 0 current glaciers: δd = % of all the water is in glaciers. Question If all the glaciers melted, what would the deuterium content of the ocean become?

14 Example: Melting Glaciers Let M be the total number of hydrogen (and deuterium) atoms in the ocean and glaciers (usually computed in moles). Let p = 0.02 be the proportion of water in the glaciers, and let q = 0.98 be the proportion of water in the oceans. Let x i denote moles of deuterium and y i denote moles of hydrogen, according to this table: Glaciers Oceans Total D x x 2 x 0 H y y 2 y 0 Isotope ratios r x y r x y r ( x x ) ( y y ) glaciers oceans total

15 Example: Melting Glaciers Solve for combined ratio ratio moles y y glaciers oceans x ry x2 r2y 2 x y pm x y qm 2 2 pm x r pm r r qm x r qm r2 r2 Solve for moles r 0 rpm rqm 2 x x2 r r2 ( r2) rp ( r) rq 2 rp rrp 2 rq 2 rrq 2 y pm qm y2 ( r2) p( r) q pr2pqrq r r 2 r 0 rp rq 2 rr 2 rprq 2

16 Example: Melting Glaciers r 0 rp rq 2 rr 2 rprq 2 Since r and r 2 are very small, a good approximation is r pr qr 0 2 s( ) ps( ) qs( ) 0 2 p q 0 2 Recall p 0.02 q ( 0.42) If all the glaciers melted, the deuterium content of the ocean would decrease by about 8.4

17 More Generally phase phase 2 total Phase Phase 2 Total Rare isotope x x 2 x 0 Common isotope y y 2 y 0 Isotope ratios r x y r x y r ( x x ) ( y y ) x y so r i i i

18 Same Computation Solve for combined ratio ratio moles y y glaciers oceans x ry x2 r2y 2 x y pm x y qm 2 2 pm x r pm r r qm x r qm r2 r2 Solve for moles r 0 rpm rqm 2 x x2 r r2 ( r2) rp ( r) rq 2 rp rrp 2 rq 2 rrq 2 y pm qm y2 ( r2) p( r) q pr2pqrq r r 2 r 0 rp rq 2 rr 2 rprq 2

19 Same Computation r 0 rprqrr rprq Since r and r 2 are very small (rare isotope assumption), a good approximation is r pr qr 0 2 s( ) ps( ) qs( ) 0 2 p q 0 2

20 Example: Melting Glaciers before glaciers δ ocean δ 2 ocean δ 0 after What about 8 O changes? Assumptions current ocean: δ 8 O = 0 current glaciers: δ 8 O = -50 2% of all the water is in glaciers. 0 pq ( 0.05) 0.00 If all the glaciers melted, the 8 O content of the ocean would decrease by about

21 Fractionation phase r phase 2 r 2 total r 0 Assume that fractionation is at equilibrium. fractionation: mass balance: r r 2 f r pr qr 0 2

22 r Fractionation f r0 pr qr2 r 2 r pr qr pfr qr ( pf qr ) r2 r f r pf q r pf q 0 0 If f is close to : f f pf q p p p q pf q pf q r r r p q r 2 0 0

23 Fractionation r ( q) r r ( p) r delta notation r i s( ) s( ) ( q) s( ) s( ) ( p) s( ) q q p p i ε and δ 0 are both small, so εδ 0 is even smaller, so it can be ignored. q p 0 2 0

24 Example: Deuterium in Water Vapor before dry air air + vapor δ after water δ 0 water δ 2 A small amount (compared to the ocean) of water vapor forms. What is the deuterium content of the vapor? Assumptions current ocean: δ 0 = δd = 0 fractionation: ε = q 2 0 p p 0 q q 0 (0.08) Under these assumptions, the deuterium content of the vapor is less than that of the ocean by 80.

25 Everything depends on temperature. Water Evaporation Fractionation Factors for 8 O Temperature ( K) Temperature ( C) Temperature ( F) δ 8 O Pierrehumbert, Principles of Planetary Climate, Cambridge U Press, New York, 200

26 Everything depends on temperature. 0 Water Evaporation Fractionation Factors 0.02 fractionation O D Formulae from Gerrit Lohmann, 2007 ⁰C

27 Example: Deuterium in Rain before vapor δ 0 vapor δ rain δ 2 after 40% of the water vapor condenses to rain. What is the deuterium content of the rain and of the remaining vapor? Assumptions vapor before: δ 0 = δd = -80 fractionation: ε = 90 0 q 2 0 p p 0.6 q0.4 0 q 0.08 (0.09) p 0.08 (0.09) Remaining vapor: δd = -6 Rain: δd = -26

28 Example: Deuterium in Rain before vapor δ 0 vapor δ rain δ 2 after Repeat 60% of the remaining water vapor condenses to rain. Assumptions vapor before: δ 0 = δd = -6 fractionation: ε = 00 0 q 2 0 p p 0.4 q0.6 0 q 0.6 (0.) p 0.6 (0.) Remaining vapor: δd = -76 Rain: δd = -76

29 Example: Deuterium in Rain before vapor δ 0 vapor δ rain δ 2 after Repeat again 80% of the remaining water vapor condenses to rain. Assumptions vapor before: δ 0 = δd = -76 fractionation: ε = 05 0 q 2 0 p p 0.2 q0.8 0 q 0.76 (0.05) p 0.76 (0.05) Remaining vapor: δd = -260 Rain: δd = -55

30 Example: Deuterium in Snow before vapor δ 0 vapor δ rain δ 2 after This time it snows. 90% of the remaining water vapor condenses to snow. Assumptions vapor before: δ 0 = δd = -260 fractionation: ε = 0 0 q 2 0 p p 0. q0.9 0 q (0.) p 0.76 (0.) Remaining vapor: δd = -359 Rain: δd = -249

$And So It Goes fractionation fractionation Pierrehumbert,$

31 And So It Goes fractionation fractionation Pierrehumbert, Principles of Planetary Climate, Cambridge U Press, New York, 200

32 Vostok and Dome C Differ Pierrehumbert, Principles of Planetary Climate, Cambridge U Press, New York, 200

33 Biology Matters and is yet still more complicated. atmosphere ocean CO HOHCO H HCO HCO H CO 3 3 Ca CO CaCO 3 3 foraminifera Temperature dependent fractionation occurs at every step. The result: the δ 8 O in foram shells is about +30 compared with the surrounding water (depending on temperature). (δ 8 O)/dT 0.25 / ⁰C (Reference: Pierre Humbert s book) And then there s carbon.

34 Biology Matters and is yet still more complicated. photosynthesis 6CO2 6H2O C6H2O6 6O2 δ = δ 3 C δ 2 = δ 3 C Fractionation is about Result: Plants, animals, coal, and oil are all lighter in 3 C than inorganic carbon.

35 Zachos, et al, Science 292 (200), p. 689

36 Paleocene-Eocene Thermal Maximum (PETM) Sharp decrease in δ 8 O, interpreted as a rapid increase in temperature. Sharp decrease in δ 3 C, interpreted as massive oxidation of sequestered organic carbon. d8o Site d8o d3c Myr d3c

Math /8/2014. Richard McGehee, University of Minnesota 1. Math 5490 October 8, Isotopes as Climate Proxies

Math /8/2014. Richard McGehee, University of Minnesota 1. Math 5490 October 8, Isotopes as Climate Proxies Math 5490 Octobe 8, 04 Topics in Applied Mathematics: Intoduction to the Mathematics of Climate Mondays and Wednesdays :30 3:45 http://www.math.umn.edu/~mcgehee/teaching/math5490-04-fall/ Steaming video