SME
Climatology
Paleoclimate proxies (δ¹⁸O, ice cores)
Use this for “Ice from the last glacial has δ¹⁸O of −40 ‰; today’s snow is −35 ‰. How much colder was it?”
Chart
δ(−7.463) = −40
This is an example. Type your own values to start a new problem.
ΔT (°C)
=
δₚₐₛₜ (‰)
−
(
δnow (‰)
)
s (‰/°C)
R
Isotope ratio of the past ice
worked out (not typed)
Formulas
ΔT
=
δₚₐₛₜ − δnow
s
−7.4627
=
−40 − (−35)
0.67
R
=
2.0052 ×
10
-3
×
(1
+
δₚₐₛₜ
1,000
)
0.001925
=
2.0052 ×
10
-3
×
(1
+
−40
1,000
)
Assumptions
•
δ¹⁸O = 1000(R ÷ Rstd − 1) in per mil (‰), where R is ¹⁸O ÷ ¹⁶O and Rstd = 2.0052 × 10⁻³ (VSMOW, mean ocean water).
•
Colder air rains out more ¹⁸O before the snow falls, so polar snow gets more negative as it cools.
•
The slope s is calibrated from today’s pattern across places (about 0.67 ‰ per °C in Greenland) and is assumed to hold in the past.
Step-by-step
WE KNOW
δnow = −35 ‰, δₚₐₛₜ = −40 ‰, s = 0.67 ‰/°C
FIND
temperature change (ΔT), isotope ratio of the past ice (R)
Step 1 · Find temperature change (ΔT)
Use ΔT = (δₚₐₛₜ − δnow) ÷ s
Each
degree
of
cooling
lowers
δ
18
O
by
s
per
mil,
so
divide
the
shift
by
s
.
ΔT
=
−40 − (−35)
0.67
(−40) − (−35) = −5
ΔT = −7.4627 °C
Step 2 · Find isotope ratio of the past ice (R)
Use R = 2.0052 × 10⁻³ × (1 + δₚₐₛₜ ÷ 1,000)
δ
18
O
is
the
per-mil
gap
from
the
standard,
so
scale
the
standard’
s
ratio
by
1
plus
δ
1000
.
R
=
2.0052 ×
10
-3
×
(1
+
−40
1,000
)
R
=
(2.0052
×
10
-3
)
×
(1
+
(−0.04))
R
=
(2.0052
×
10
-3
)
×
0.96
R
=
0.001925
Check
Put the numbers back in. Both sides should match.
−7.4627
=
−40 − (−35)
0.67
✓
0.001925
=
2.0052 ×
10
-3
×
(1
+
−40
1,000
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.