All?gre Claude J. Isotope Geology

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8.3.3 Time-dependent formation

Let us now consider a slightly more complexcase where th e formation term decreases with

time. This is geologically fairly logical as we know that the creation of radioactive energy

has decreasedwithtime.The equationbecomes:

¼ J

qt

 kM:

Thisisaverygeneralequationfoundinmanycases(Alle

greandJaupart,1985).Forexam-

ple, for continentalgrowth, wherethe equations found canbe modi¢edas ac onsequence or

alsofordegassingofthe mantle, wh ichwehavelookedat.

Integrating gives:

M ¼

k  q

qt

 e

kt



Therefore M depends onthe two constants k and q.

To getour ideas straight, letus takethe case ofdegassingof

Ar from the upper mantle,

and study the

Ar contentoftheupper mantle:

¼ l

lt

 G

0.1

0.3

0.5

0.7

4 3 2

= 0.83 10

–9

–1

R = 1.2 Ga

Remaining fractions of each age

Age (Ga)

1 0

0.90

0.1

0.3

0.5

0.7

4 3 2

= 0.5 10

–9

–1

R = 2 Ga

1 0

Figure 8.6 Spectrum calculation for

¼0.83  10

9

1

and 0.5  10

9

1

. Notice the very rapid

growth towards recent times. This is because the probability of double or triple recycling increases very

rapidly with time.

449 Non-steady states

where l

is the fraction of

Kgiving

Ar; l is the total decay constant of

K; G is the

constantofoutgassing (Figure8. 7).The equation canbe easily integrated:

Ar ¼

G  l

lt

 e

Gt



We are going to study the behavior of

Ar for the di¡erent values of G,arbitrarilysetting

¼1.

The qualitative behavior is understandable.When G is quite large, the quantity of

reaches a maximum and then declines. So let us try to apply this information to the actual

upper mantle.

The present-day potassium level i n the non-depleted mantle is K ¼250 ppm, an d

K ¼1.16 10

4

or 116 ppm. The decay constant is l ¼0.554 10

9

1

.Some4.5510

yearsago, the

K levelwas12 timeshigherandso equalto0.35 ppm.

Given that the mass of the upper mantle is 110

g, 4.55 Ga ago, there was therefore

0.35 10

gof

K.Fromthepreviouscalculationweobtainthefollowingvaluesof

Artoday:

Time (Ga)

with K extraction by landmass

G = 0

= 0.25 10

–9

–1

5×10

G = 0.75 10

–9

–1

0.05

20×10

0.02

0.01

0.04

0.03

44.5 3 2 1 0

2×10

Time (Ga)

0.05

0.1

4 3 2 1 0

1.5×10

3.2×10

1.1×10

5×10

4.5

G = 0

= 0.25 10

–9

–1

G = 0.375 10

–9

–1

G = 0.75 10

–9

–1

without K extraction

Figure 8.7 Mantle degassing model. (a) Evolution of

Ar in the mantle over time, assuming that

varies by radioactive decay alone. The scale on the right is drawn assuming

¼1. The scale on the left

is in 10

gof

Ar and corresponds to the total mass of

Ar on the hypotheses made in the main text.

(b) Evolution of

Ar in the mantle assuming

K decreases both by extraction from the continental

crust and by radioactive decay. The scale on the left is in 10

gof

Ar. On the right, it is assumed

that

¼1.

450 Isotope geology and dynamic systems analysis

Thepresent-day m ass is estimatedat2^3 10

g, which is notbad!

Thereisoneassumptioninthemodelthatisnotverysatisfactorybec auseitwasconsidered

that

K decreased in the upper mantleby radioactive decayalone. Now, weknow that K has

alsodecreasedintheuppermantlebecaus eithasbeenextractedatthesametimeascontinen-

talcrustinwhich itisenriched.Letus therefore consider that

Kdecreasesbythe law:

K ¼

ðlþÞt

where  is the constant for extraction of K from the mantle to the continental crust. (Such

extraction does notobeyan exponential lawexactly, but itis a ¢rst approximation.)

We take  ¼0.35 10

9

1

because, with this constant, total K evolves from 250 ppm

to 50 ppm in 4.5 10

years, which is about the value estimate d for the degree of depletion

of the upper mantle in K.The change in

Ar contentof the mantle c an therefore be calcu-

latedby replacing thevalu e of l ¼0.5 10

9

1

by (l þ) ¼0.85 10

9

1

Ar ¼

G  l þ ðÞ

 lþðÞt

 e

Gt



As can be seen from the curves, we obtain an acceptable value: 2.77 10

gof

Ar in the

upper mantle for G ¼0.7 5 10

9

1

.Now,whatdoes0.7510

9

1

represent? The

value1/G ¼1.3 10

is aboutthe residence time of plates subducting into the mantle. If it is

taken that the o ceanic lithosphere degasses entirely by being fed through the mid-ocean

ridges, a plausible scenario can be reconstructed for degassing the upper mantle.Which is

noguarantee atall thatthe model is aproper re£ectionofreality!

8.3.4 Effects of cyclic ﬂuctuations

Generalequations

We saw when examining cli matic variations that the Earth’s temperature varied cyclic ally

(Milankovitch cycles).We can readily imagine thatthe Earth’s tectonic activity varies cycli-

cally (Wilson’s plate tectonic cycles). How will such variations a¡e ct a dynamic system, a

reservoir which receivesand exchanges matter?

Let us consider the equation for the dynamic evolution of a reservoir as we have already

examinedit:

dM=dt ¼ J  kM;

but in wh ichwe write that the input £ ux varies periodically with the formula J ¼J

þb sin

!t,where! ¼2 p/T,T beingthe oscillationperiod and k ¼1/ R,R being th e residencetime.

G (10

9

1

)

0.25 0.375 0.75 0

Ar(g) 1.5 10

1.1 10

0.49 10

3.2 10

This problem has been addressed by Richter and Turekian (1993) and then by Lasaga and Berner (1998).

The latter presentation has inspired the one given here.

451 Non-steady states

The equation for mass evolution M (orconcentration i n an element)istherefore:

¼ J

þ b sin !t  kM :

The equation c anbewritten intheform:

kM þ

¼ J

þ b sin !t:

Th e ¢rst ter m of th is equation d epen ds on M alone, the second term being the forcing

imposed on the system from outside (we shall also c all this the source term).The question

is, of course, how does the system react and what is its response to external stimulus

(Figure 8.8).

We integrate this ¢rst-order di¡erential equation with constant coe⁄cients in the

standard manner: integrate the e quation without the second member dependent on t,

th en calculate the ¢rst integration ‘‘constant.’’ In the course of calculation, we get on the

integral:

sin !t dt:

This is the only minor mathematicaldi⁄culty.

sin x dx

is integratedby parts, integrating twice.Thisleads to :

sin x dx ¼ e

sin x  e

cos x 

sin x dx;

h ence the calculationof

sinx dx.Integrating thisgeneral equationthereforegives:

M ¼ M

kt

1  e

kt



þ !

kt

þ !

sin !t





þ !

cos ! t



Source Response

Forcing

System

Figure 8.8 Model of response of a system subjected to external action.

452 Isotope geology and dynamic systems analysis

It canbeseen thatif M

¼0andb ¼0,we¢ndthe equation inSection 8.3. 1 :

M ¼

1  e

kt



Whenevert is quitelarge (say morethan 3/k), the equation is simpli¢ed andcanbewritten:

M ¼

þ !

sin !t 

þ !

cos !t:

If !<< k,thatis,ifR << T (the residence time is very much less than the oscillation

period), the cosineterm is negligible.The equation canthenbewritten:

M ¼

sin !t

where J

/k is the equilibriumvalue aroundwhichoscillation occurs.

The system response is in phase with the source oscillation.The amplitude of £uctuation

aroundequilibriumdependsonthe ratio J

/b,thatis,onthe relativevaluebetweenthe equi-

librium value and the pulse value, but if this ratio is not too large, the amplitud e remains

large.

Conversely, if !>> k,thatis,R >> T (the residenc e time is very mu ch g reater than oscil-

lationperiod), the equationbecomes:

M ¼



cos !t:

But we have the trigonometric equality cos !t ¼^sin(!t ^ p/2). Therefore the system

oscillation is delayed by (p/2), that is, by a quarter of a period relative to the source oscilla-

tion. Moreover, the amplitude is subtracted from a term J

/k, which is very large since k is

verysmall, which greatlydamps the signal. Between the two extremes we ¢nd intermediate

behavior.

Let us examine the consequ ences of this mathemati cal result. Suppose ¢rst that the

source oscillation period is set, say at 100 ka. C hemical elements whose residence time in

the oceans is much less than100 ka, for example neodymium (R ¼1000 years), will be sub-

jected to £uctuations (if the geochemical processes governing them are involved) in

phase with the source oscillations and whi ch may be of large amplitude depending on the

correspondingJ

/b ratio (Figure 8.9).

The che mical elements with very long residence times like strontium (R ¼2^4 Ma)) wi l l

be subjected to £uctuations with a time lag of half a period c ompared w ith the source, but

thataregreatlydampedtoo.If,forexample, thedisturbancere£e ctstheglacial^inte rglacial

cycles, thestrontium£uctuationswillbe o¡setand damped.

Let us examine the e¡ect of the oscillation period.We stay with strontium with its resi-

dence time of 2^4 Ma. If the £uctu ations are of the order of 100 or 1000 ka (climate, for

example) the results will be as alreadydescribed. Ifthe £uctuations are from some geologi-

cal source of the orde r of 20^30 Ma (tectonic), the strontium £uctuations will follow the

source £uctuations, in phase and without any attenuation. Of course intermediate cases

arisebetween these two extremes.

453 Non-steady states

The relative importance of the sine and cosine terms depends on the ratio k/!.We shall

seethatthese properties extend tothe£uctuationsof isotopic compositions.

Exercise

We consider sea water and the disturbance to its chemical composition as a result of climatic

disruptions introducing different concentrations in rivers. We consider a sinusoidal distur-

bance with a period of 10 000 years.

We assume the disturbance follows the law:

sin !

For four chemical elements Pb, Nd, Os, and Sr whose residence times in sea water are

¼10

years,

¼3  10

years,

¼3  10

years, and

¼2  10

years, respectively, we

represent the ﬂuctuations by changing the values of

and

so that

¼1 and

¼1 (this

greatly ampliﬁes the Os and Sr ﬂuctuations).

Draw the response curves for the different elements and estimate the phase-shift values.

Try to identify any quantitative relationship between residence time and phase shifts. Next

try to evaluate damping, taking

¼1 and

¼1 for all of the elements.

Answer

The equation describing the response curve is written:

a þ b

sin !



Integrating gives:

C ¼

sin !



cos !

Short residence time

C / C

Time

Long residence time

Artificially amplified

response curve

Forcing

curve

Forcing

curve

Response

curve

Figure 8.9 Relationship between residence time and lag in the response curve. The response curve at

the top has been greatly ampliﬁed vertically so the effect can be seen.

, concentration;

, equilibrium

concentration.

454 Isotope geology and dynamic systems analysis

We take 10

years as the time unit and ! ¼2p/

¼0.628.

For Pb,

¼1; for Nd,

¼0.033; for Os,

¼0.033; and for Sr,

¼5  10

4

The curves are illustrated in Figure 8.10 ignoring damping. They show the offset in terms of

residence time. When the logarithm of

(residence time) is plotted against the phase shift,

we get a curve similar to a hyperbola

with two asymptotes corresponding to the two limiting

cases where

/! is very large and

/! is very small (Figure 8.11).

To evaluate damping we take

¼1 for all the elements, therefore

is 1, 3, 30, and 2000

for Pb, Nd, Os, and Sr, respectively. We evaluate the percentage variation:

0.1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2

Δt

Forcing

Response

0.1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2

Δt

0.1

0.2 0.3 0.4

Forcing

Time (10

yr)

Time (10

yr)

Time (×10

yr)

C / C

x 2000 C / C

x 1.5

Response

= 1000 yr

= 3000 yr

= 3 x 10

= 2 x 10

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2

Δt

Figure 8.10 Response curves of Pb, Nd, Os, and Sr concentrations in the ocean, if the sources vary

sinusoidally with the characteristics given in the main text. C ¼concentration – equilibrium concentration.

In fact, it is the curve tg !.t ¼!R.

455 Non-steady states



¼ D



We ﬁnd 90%, 65%, 0.2%, and 0.15% for the ﬂuctuations of Pb, Nd, Os, and Sr, respectively.

We shall see that these properties extend to ﬂuctuations in isotope composition.

Generalization

Extensionto all forcingfunctions

The great interest of this study of response to a sinusoidal source is that a response can be

obtained for anysource function byde composing it into its Fourier sine functions an d then

¢nding theresponsefor each s ine function andsumming them.

A LITTLE HISTORY

Joseph Fourier

When he was prefect of France’s Ise

re department under Napoleon, Joseph Fourier (after

whom Grenoble’s science university is named) proposed one of the most important theorems

in mathematics. He showed that any periodic function could be represented by the sum of the

sine and cosine functions with appropriate amplitudes and phases. He then showed that

when a function was not periodic ‘‘by nature,’’ it could be made periodic by truncating it and

then repeating the sampled portion. On this basis, any function can be separated into its

Fourier components and so represented as a spectrum (frequency, amplitude). Milankovitch

cycles are an example that we have already mentioned.

0.05 0.1 0.15 0.2 0.25 0.3

Ln R

Phase shift

500 years

Figure 8.11 Relationship between residence time as a logarithm and phase shift, expressed in intervals

of 10 000 years.

456 Isotope geology and dynamic systems analysis

Exercise

We take a system (borrowed from Lasaga and Berner) obeying the previous equations with

an imposed ﬂuctuation of

(

) ¼4 þsin(5

) þsin(

). The residence time constant is

taken as 1 in the arbitrary time units (

). Calculate and represent the function at the reservoir

outlet.

Answer

See Figure 8.12. Notice the phase shift, the damping of high frequencies, and the slight

alteration of low frequencies.

Th e cas e of radioact iveis otope s

We have s aid that for radioactive systems, the residence time should take account of the

mean l ife oftheradioactiveelement,which is a sortofintrinsicresidenc e time.The dynamic

equations we have seen are identical, except for this modi¢cation of residence time. In

some cases, when the residence time of the res ervoir is verylarge, it is the isotope’s lifetim e

thatdetermines the overall residence time.

The same reasoning applies d epending on the value of the l/!ratio, where ! is th e fre-

quency of the periodic disturbance and l is the decay constant.The £u ctuations i mposed

by the source shall be taken into account di¡erentially with or without a phase shift and

withor withoutamplitude damping.

5.0

4.0

3.0

2.0

1.0

5.0 10.0 15.0

Time

Quantity in reservoir

Forcing imposed by source

Response

Figure 8.12 Quantity in the reservoir as a function of time. After Lasaga and Berner (1998).

457 Non-steady states

EXAMPLE

Cosmogenic isotopes in the atm osphere

There are three types of cosmogenic isotope in the atmosphere we can look at:

Be,

H, and

For

¼5.57  10

2

1

,for

Be,

¼4.62  10

7

1

,andfor

¼1.209  10

4

1

When we observe ﬂuctuations in the abundance of these isotopes,

H reﬂects rapid

ﬂuctuations in the atmosphere whereas

C largely damps these variations. However,

Be,

with a much smaller decay constant, should damp them even more than

C. In fact, it reﬂects

rapid oscillations. Why? Because its residence time in the atmosphere of 1–3 years is very

short and it is this constant that prevails.

For similar reasons when it comes to deciphering the Earth’s early history, the information

provided by extinct forms of radioactivity is not the same as that provided by long-lived forms,

as we have seen.

8.3.5 Non-linear processes

In all the examples examined so far, thebasic di¡erential equation has been linear, even if the

forcingtermwasnon-linearovertime,asinthe previous example.Letusnowconsidera model

ofgrowthoftheEarth’score. LetthevariableNrepresentthequantityofironinthecore.Then:

¼ K 1  NðÞ

where (1 ^ N)is the mass fractionof iron in the core,the total quantityof ironbeing normal-

ized to unity. It can be considered that the attraction of this iron dispe rsed in the form of

small minerals isproportionallystronger whenthequantityofiron inthe coreishigh(gravi-

tational attraction).To characterizethe quantity N,we canthereforetake K ¼QN.

Thisgivesthe equation:

¼ QN 1  NðÞ:

This is awell-known equation, especially inpopulation dynamics. Itisthelogistic equation

(see Haberman,1997):

N 1  NðÞ

¼ Q dt:

Byseparatingoutthesimple elements, we canwrite:

1  NðÞ



dN ¼ Q dt:

Integrating gives:

ln Njjln 1  Njj¼Qt þ C;

henc e :

458 Isotope geology and dynamic systems analysis