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Math 5490October 8, 2014

Isotopes as Climate Proxies

How do we know the past climates?

Math 5490 10/8/2014

Isotopes as Proxies

What is this? δ18O (‰)

Hansen, et al, Target atmospheric CO2: Where should humanity aim? Open Atmos. Sci. J. 2 (2008)

Math 5490 10/8/2014

http://eo.ucar.edu/staff/rrussell/climate/paleoclimate/sediment_proxy_records.html

Ocean Sediment Cores

Isotopes as Proxies

Math 5490 10/8/2014

18O as a Climate Proxy

Foraminifera absorb more 18O into their skeletons when the water temperature is lower and when more 18O is in the water.

Thus higher concentrations of 18O in foraminifera fossils indicate lower ocean temperatures and higher glacier volume.

The isotope 16O preferentially evaporates from the ocean and is

sequestered in glaciers, leaving the heavier isotope 18O more highly concentrated in the ocean. Thus

oceanic concentration of the isotope 18O is higher during glacial periods.

Isotopes as Proxies

Math 5490 10/8/2014


‰ : “per mil,” “per thousand”1000‰ = 100% = 110‰ = 1% = 0.01

1‰ = 0.1% = 0.001

18O: Oxygen 18: 8 protons 8 electrons 10 neutrons17O: Oxygen 17: 8 protons 8 electrons 9 neutrons16O: Oxygen 16: 8 protons 8 electrons 8 neutrons

Most of the oxygen atoms on Earth are 16O. About 1 in 500 atoms is 18O. About 1 in 2500 is 17O.

There are other oxygen isotopes, but they are unstable.

Isotopes as Proxies

Math 5490 10/8/2014


ExampleGiven a sample of calcium carbonate (CaCO3) from a foraminifera fossil,

suppose that the ratio of 18O atoms to 16O is r = 0.002013 = 2.013‰. 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 18O atoms to 16O is s = 0.0020052. Then

18 0.002013O 1 1 0.00390.0020052

r s rs s

So we would report

δ18O = 3.9 ‰

Isotopes as Proxies

Math 5490 10/8/2014

What is this? δ18O (‰)Conversely

The 18O content of a sample is reported asδ18O = 3.9 ‰

using the VSMOW standard.What is the proportion of 18O in the sample?

18 181 O, (1 O)

0.0020052(1 0.0039) 0.002013

r r ss

r

Noter is the ratio of 18O to 16O. The proportion of 18O in the sample is

Isotopes as Proxies

0.002013 0.0020081 1.002013

rr

For small values of r, these are approximately equal.

Math 5490 10/8/2014

1Pierrehumbert, Principles of Planetary Climate, Cambridge U Press, New York, 2010.2http://en.wikipedia.org/wiki/%CE%9413C

Isotopes Ratio Standard Source

D:H 0.0001558 VSMOW Pierrehumbert1

13C:12C 0.0112372 PDB Wikipedia2

18O:16O 0.0020052 VSMOW Pierrehumbert1

18O:16O 0.0020672 VPDB Pierrehumbert1

Standards:VSMOW: Vienna Standard Mean Ocean WaterPDB: Pee Dee BelemniteVPDB: Vienna Pee Dee Belemnite

Common Standards

Isotopes as Proxies

Math 5490 10/8/2014

What does δ18O (‰) tell us?

r1 = ratio of 18O:16O in vaporr2 = ratio of 18O:16O in liquid

At equilibrium,r1 = f r2

where f is the fractionation factor. (depends a lot on temperature)

FractionationExample: Evaporation of Water

liquid

vaporevaporationcondensation

Isotopes as Proxies

Math 5490 10/8/2014

r1 = ratio of 18O:16O in vaporr2 = ratio of 18O:16O in liquid

r1 = f r2

FractionationWhat about δ?

21 21 2

11 1 1 1 1

fsr fr fs s s

Note that the standard drops out.f is usually close to 1, so let

1 2 2 21 1 1

Since ε and δ are typically small, εδ is even smaller, so

1 2

Isotopes as ProxiesWhat does δ18O (‰) tell us?

1f often expressed as ‰

Math 5490 10/8/2014

δ18O(water) = δ2 f = 0.99 = 1+εδ18O(vapor) = δ1 ε = ‐0.01 = ‐10‰

Example: Evaporation of Water

waterδ = δ0

dry air

afterbefore

waterδ2

air + vaporδ1

δ18O(water) = δ0δ18O(vapor) is undefined

1 2

2 0 1 0 00, 0.01

The 18O content of the vapor is 10‰ less than that of the ocean.

Isotopes as ProxiesWhat does δ18O (‰) tell us?

Assume only a small amount of vapor forms.

Math 5490 10/8/2014

ocean

glaciers

afterbefore ocean

common hydrogen: 1H = H heavy hydrogen: 2H = DVSMOW D:H = 0.0001558 = 0.1558 ‰ (very small)

Assumptionscurrent ocean: δD = 0 current glaciers: δD = -420 ‰

2% of all the water is in glaciers.Question

If all the glaciers melted, what would the deuterium content of the ocean become?

Isotopes as ProxiesExample: Melting Glaciers

Math 5490 10/8/2014

Isotope ratios

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 xi denote moles of deuterium and yi denote moles of hydrogen, according to this table:


Glaciers Oceans TotalD x1 x2 x0

H y1 y2 y0

1 1 1 2 2 2 0 1 2 1 2( ) ( )r x y r x y r x x y y

glaciers oceans total

Math 5490 10/8/2014


1 1 1 2 2 2

1 1 2 2

glaciers oceansratio

molesx r y x r y

x y pM x y qM

11 1

1 1

22 2

2 2

1 1

1 1

pM r pMy xr r

qM r qMy xr r

1 2

1 2 2 1 1 2 1 2 1 2 1 21 20

1 2 2 1 2 1

1 2

1 2 1 20

2 1

(1 ) (1 )1 1(1 ) (1 )

1 1

1

r pM r qMx x r r p r r q r p r r p r q r r qr rr pM qMy y r p r q p r p q r q

r r

r p r q r rrr p r q

Solve for moles

Solve for combined ratio

Math 5490 10/8/2014


1 2 1 20

2 11r p r q r rr

r p r q

Since r1 and r2 are very small, a good approximation is

0 1 2

0 1 2

0 1 2

(1 ) (1 ) (1 )r pr qr

s ps qs

p q

1 20.02 0.98 0.42 0p q Recall

0 0.02 ( 0.42) 0.0084

If all the glaciers melted, the deuterium content of the ocean would decrease by about 8.4 ‰

Math 5490 10/8/2014

phase 2

phase 1

total

Isotopes as ProxiesMore Generally

Phase 1 Phase 2 TotalRare isotope x1 x2 x0

Common isotope y1 y2 y0

1 1 1 2 2 2 0 1 2 1 2( ) ( ) so 1i i i

r x y r x y r x x y yx y r

Isotope ratios

Math 5490 10/8/2014

Isotopes as ProxiesSame Computation

1 1 1 2 2 2

1 1 2 2

glaciers oceansratio

molesx r y x r y

x y pM x y qM

11 1

1 1

22 2

2 2

1 1

1 1

pM r pMy xr r

qM r qMy xr r

1 2

1 2 2 1 1 2 1 2 1 2 1 21 20

1 2 2 1 2 1

1 2

1 2 1 20

2 1

(1 ) (1 )1 1(1 ) (1 )

1 1

1

r pM r qMx x r r p r r q r p r r p r q r r qr rr pM qMy y r p r q p r p q r q

r r

r p r q r rrr p r q

Solve for moles

Solve for combined ratio

Math 5490 10/8/2014

Isotopes as Proxies

1 2 1 20

2 11r p r q r rr

r p r q

Since r1 and r2 are very small (rare isotope assumption),a good approximation is

0 1 2

0 1 2

0 1 2

(1 ) (1 ) (1 )

r pr qr

s ps qs

p q

Math 5490 10/8/2014

Same Computation

oceanδ2

glaciersδ1

afterbefore oceanδ0

What about 18O changes?


Math 5490 10/8/2014

Assumptionscurrent ocean: δ18O = 0 current glaciers: δ18O = -50‰

2% of all the water is in glaciers.

0 1 2 0.02 ( 0.05) 0.001p q

If all the glaciers melted, the 18O content of the ocean would decrease by about 1 ‰

phase 2r2

phase 1r1

totalr0

Isotopes as ProxiesFractionation

Assume that fractionation is at equilibrium.

Math 5490 10/8/2014

1

2

0 1 2

r frr pr qr

fractionation:

mass balance:


Math 5490 10/8/2014

10 1 2

2

r f r pr qrr

0 1 2 2 2 2

2 1

0 0

( )

1

r pr qr pfr qr pf q r

r r fr pf q r pf q

If f is close to 1:

2 1

0 0

111 1 1 1

1 1

ffpf q p p p q

pf q pf q

r rp qr r


Math 5490 10/8/2014

1 0 2 0(1 ) (1 )r q r r p r

delta notation

1 0 2 0

1 0 0 2 0 0

(1 )(1 ) (1 ) (1 ) (1 ) (1 ) (1 )1 1 1 1

i ir ss q s s p s

q q p p

1 0 2 0q p

ε and δ0 are both small, so εδ0 is even smaller, so it can be ignored.

Example: Deuterium in Water Vapor

waterδ0

dry air

afterbefore

waterδ2

air + vaporδ1

Isotopes as Proxies

Math 5490 10/8/2014

A small amount (compared to the ocean) of water vapor forms.What is the deuterium content of the vapor?

1 0 2 0

1 0

0 10 (0.08) 1 0.08

q p p qq

Assumptionscurrent ocean: δ0 = δD = 0 fractionation: ε = -80‰

Under these assumptions, the deuterium content of the vapor is less than that of the ocean by 80 ‰.

Everything depends on temperature.

Isotopes as Proxies

Math 5490 10/8/2014

Temperature (°K)

Temperature (°C)

Temperature (°F)

δ18O

273 0 32 -11.7‰290 17 62 -10.1‰350 77 170 -6.0‰

Pierrehumbert, Principles of Planetary Climate, Cambridge U Press, New York, 2010

Water Evaporation Fractionation Factors for 18O

Everything depends on temperature.

Isotopes as Proxies

Math 5490 10/8/2014

Formulae from Gerrit Lohmann, 2007

Water Evaporation Fractionation Factors

‐0.12

‐0.1

‐0.08

‐0.06

‐0.04

‐0.02

0

0 10 20 30 40 50 60 70 80 90 100

fractio

natio

n ‐1

⁰C

18O

D

Example: Deuterium in Rain

vaporδ0

afterbefore

rainδ2

vaporδ1

Isotopes as Proxies

Math 5490 10/8/2014

40% of the water vapor condenses to rain.What is the deuterium content of the rain and of the remaining vapor?

1 0 2 0

1 0

2 0

0.6 0.40.08 (0.09) 0.4 0.1160.08 (0.09) 0.6 0.026

q p p qqp

Assumptionsvapor before: δ0 = δD = -80‰ fractionation: ε = 90‰

Remaining vapor: δD = -116‰Rain: δD = -26‰


vaporδ0

afterbefore

rainδ2

vaporδ1

Isotopes as Proxies

Math 5490 10/8/2014

Repeat60% of the remaining water vapor condenses to rain.

1 0 2 0

1 0

2 0

0.4 0.60.116 (0.1) 0.6 0.1760.116 (0.1) 0.4 0.076

q p p qqp




vaporδ0

afterbefore

rainδ2

vaporδ1

Isotopes as Proxies

Math 5490 10/8/2014

Repeat again80% of the remaining water vapor condenses to rain.

1 0 2 0

1 0

2 0

0.2 0.80.176 (0.105) 0.8 0.2600.176 (0.105) 0.2 0.155

q p p qqp



Example: Deuterium in Snow

vaporδ0

afterbefore

rainδ2

vaporδ1

Isotopes as Proxies

Math 5490 10/8/2014

This time it snows.90% of the remaining water vapor condenses to snow.

1 0 2 0

1 0

2 0

0.1 0.90.260 (0.11) 0.9 0.3590.176 (0.11) 0.1 0.249

q p p qqp



Isotopes as Proxies

Math 5490 10/8/2014

And So It Goes

fractionationfractionation


Isotopes as Proxies

Math 5490 10/8/2014

Vostok and Dome C Differ


Isotopes as Proxies

Math 5490 10/8/2014

Biology Matters

2 2 2 3 3

3 3

3 3

CO H O H CO H HCO

HCO H CO

Ca CO CaCO

atmosphere ocean

foraminifera

Temperature dependent fractionation occurs at every step.The result: the δ18O in foram shells is about +30‰ compared with the

surrounding water (depending on temperature).(δ18O)/dT ≈ ‐0.25 ‰/ ⁰C

(Reference: Pierre Humbert’s book)

And then there’s carbon.

and is yet still more complicated.

Isotopes as Proxies

Math 5490 10/8/2014

Biology Matters

and is yet still more complicated.

Fractionation is about ‐25‰.

2 2 6 12 6 26CO 6H O C H O 6O

photosynthesis

δ1 = δ13C δ2 = δ13C

2 1 0.025

Result: Plants, animals, coal, and oil are all lighter in 13C than inorganic carbon.

Isotopes as Proxies

Math 5490 10/8/2014

Zachos, et al, Science 292 (2001), p. 689

Isotopes as Proxies

Math 5490 10/8/2014

Paleocene-Eocene Thermal Maximum (PETM)

Sharp decrease in δ18O, interpreted as a rapid increase in temperature.Sharp decrease in δ 13C, interpreted as massive oxidation of

sequestered organic carbon.

‐3

‐2

‐1

0

1

2

3

‐1

‐0.5

0

0.5

1

1.5

2

2.5

3‐56 ‐55.8 ‐55.6 ‐55.4 ‐55.2 ‐55 ‐54.8 ‐54.6 ‐54.4 ‐54.2 ‐54

d13C

d18O

Myr

Site 690

d18O

d13C