Math 236 Project 1: Detecting Art Forgeries
Authors:
Luke Crabtree, Talyn Groos, Sam Myers
In this project we were asked to determine the authenticity of three Vermeer's paintings
Lace Maker, Laughing Girl and Woman Reading Music. And seven Delacroix's paintings: The
Death of Sardanapalus, Ovid Among the Scythians, Arabian Horses Fighting in a
Stable,Orphan Girl in a Cemetery, The Massacre at Chios, The Abduction of Rebecca, Combat
of Giaour and Hassan. We would do this by finding the analytical solution of N(t₀)=N₀ then we
determine the decay constant λ by Using the ½ life of Lead-210 then given the equation
y’(t)=-λy(t) + r(t), y(t₀)= y₀ we solve for t assuming r(t) is constant then we check or equation by
using the data from the first table then we compute the separation factors of multiple paints
over different centuries and then we can assume that the separation factor is 100 and compute
the estimated ages of the different paintings to verify their authenticity of detreman if they were
a fraud. First, we start by examining the given information. N(t) represents the number of atoms
present at time t.N’(t) represents the rate at which atoms decay in units of Decays Per Minute
(dpm). Lambda (λ) represents the positive decay constant.
(Equation 1)​ N’(t)=-λN(t)
White Lead-210 (210/82 Pb) ½ life: 22 years (when not in radioactive ore)
Polonium-210 (210/84 Po)
½ life: We know it has the same decays per minute as Lead
Radium-226 (226/88 Ra)
½ life: 1600 years
The Lead (Pb) comes from ore containing Uranium and Radium where the Uranium
decays into Radium-226 (Ra). Also, the Radium (Ra) decays into Lead (Pb) : (1600 yr ½ life). It
is important to note that in the ore, the Lead is in radioactive equilibrium with the Radium; that
1
is, the amount of radium decaying to lead, per unit time, is equal to the amount of lead
decaying, per unit time.
However, most of the uranium and its descendants are removed when extracting the lead for
the paint (see the separation factor in Table 1), at which point the Lead-210 is no longer
supported by the Radium-226 and begins decaying with ½ life of 22 years.
1. Given the initial condition N(t₀) = N₀, find the analytical solution of (1).
(Equation 1) N’(t) = -λN(t) this is a first-order differential which can be rewritten as
dN/dt = -λN
now performing separation of variables to obtain
dN/N = -λdt
now integrating both sides to obtain
∫dN/N = ∫-λdt taking this indefinite integral to obtain
ln |N| = -λt+C now we will exponentiate both sides to get N by itself
N(t) = e^-λt+C which can also be written as N = Ce^-λt
Now evaluating at the given initial condition N(t₀) = N₀, gives the equation
N₀ = Ce^-λt₀ where we can rearrange to obtain C = N₀e^λt₀. Now plugging in C,
N(t) = [N₀e^λt₀]*[e^-λt]
(Equation 1.1) ​
→ simplifies to
N(t) = N₀e^-λ(t-t₀)
2. Using the ½ life of Lead-210, determine the decay constant λ. What are the units of λ?
​
N(t₀+t₁₋₂) = N₀/2 = N₀e^-λ((t₀+t₁₋₂)-t₀) = N₀e^-λ(t₁₋₂)
​
N₀/2 = N₀e^-λ(t₁₋₂)
​
-ln(½) = λ(t₁₋₂) therefore λ=ln(2)/t₁₋₂
→ ½ = e^-λ(t₁₋₂) then taking the natural log of both sides yields,
For Lead: λ=ln(2)/22 years ≈ 0.0315 yr^-1
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3. Let y(t) denote the amount of Lead-210 (in atoms) per gram of ordinary lead (Pb) at time t.
Then
(Equation 2)​ y’(t)=-λy(t) + r(t), y(t₀)= y₀
where y(t₀) =y₀ is the amount of Lead-210 (in atoms) per gram of ordinary lead at the time of
manufacture/painting and r(t) is the rate of decay of Radium-226 to Lead-210 per gram of
ordinary lead at time t. Since the half-life of Radium-226 is 1600 years, is it reasonable to
assume that r(t) is constant over the time period (150-350 years) that we are interested in?
[Hint: What is the change in an initial quantity of Radium-226 over 350 years? How does this
compare to the precision of our measurements? Assume that r(t) = r̄ and find the solution to (2),
where r is the measured rate of Radium-226 decay (dpm/g of Pb).
​
For Radium the decay constant is λ=ln(2)/1600 years ≈0.00043 (per year).
​
Evaluating at 150 years: Using N(t)=N₀e^-λt → N(150)/N₀=e^(-0.00043*150)≈0.937.
This means the fraction remaining after 150 years is 93.7% of the starting amount.
Evaluating at 350 years ≈ 0.860. This means the difference in the fraction remaining
between 150-350 years is (93.7-86.0)% or just 7.7% (0.077). This small change in decay
rate approaches the limits of what many modern measurements tools are capable of.
Because of this, we can set r(t)=r̄ for this range of years.
Since r(t) is a constant we can write it as r.
Given y’(t)+λy=r
The integrating factor is e^∫λdt which equals e^λt
Multiplying everything by the integrating factor gives y’e^λt+λye^λt=re^λt
Recognizing the LHS is the derivative of ye^λt we can rewrite this as
(d/dt)(ye^λt)=re^λt. Now we can integrate both sides to obtain
ye^λt = ∫re^λt
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ye^λt = (r/λ)(e^λt)+C
next we solve for y
y(t)= (r/λ)+Ce^(-λt) Now that we have y(t) we can use the initial condition y(t₀)=y₀.
y₀=(r/λ)+Ce^(-λt₀)
solving for C
C=(y₀-(r/λ))*e^(λt₀)
Thus, the final solution is
(Equation 2.5)
y(t)=(r/λ)+(y₀-(r/λ))e^(-λ(t-t₀))
Equation 2.5 describes the amount of Lead-210 over time, accounting for both its
decay and the continuous production from Radium-226.
4. We can easily measure the number of decays of Polonium-210 and Radium-226 per minute
per gram of ordinary lead (dpm/g of Pb). Is this enough information to determine the age of the
painting? If not, what additional information do we need?
No, this is not enough information to extrapolate the painting's age. To properly
determine the age of the painting We need the following information;
●​ INITIAL CONDITIONS - We need to know the initial concentration of
Lead-210 when the painting was made (y₀) or at some reference level.
●​ DECAY RATE (r) - This is also needed because r represents the
measured decay rate of Radium-226 per gram of Pb after separation,
which directly affects the long-term Lead-210 production in the paint.
Measured in decays per minute per gram of Lead (dpm/g of Pb)
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5. Rewrite your solution to (2) by solving for t. Then express t in terms of y' (0) and y'(t),
measured in dpm/g of Pb. Compute the age of a few paintings in Table 1 to check if your
formula is correct.
Since the paint manufacturing date is relatively close to when the painting was made, we
can assume that t₀=0. This will allow our calculations to be in time relative to the time
when the paint and painting were created.
When solving for t y(t)=(r/λ)+(y₀-(r/λ))e^(-λt)
first we will subtract (r/λ)
y(t)-(r/λ)= (y₀-(r/λ))e^(-λt)
then we divide by (y₀-(r/λ))
y(t)-(r/λ)/(y₀-(r/λ))=e^(-λt)
we take the natural log of both sides
ln(y(t)-(r/λ)/(y₀-(r/λ)))=-λt
next we can divide by -λ to get t by itself
ln(y(t)-(r/λ)/(y₀-(r/λ)))/-λ=t
by the negative logarithmic rule we reduce
(Equation 3) ln((y₀-(r/λ))/y(t)-(r/λ))/λ=t
Equation 3 gives the time that has elapsed since the paint and painting were created as a
function of its initial amount y₀ and current amount y as well as accounting for the decay
constant λ and separation factor r.
Next, we need to express t in terms of y’(0) and y’(t). Starting with the given expression
y’(t)=-λy(t) + r(t)
we can evaluate this at t=0 to obtain the expression
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y’(0)=-λy₀+r
another form for this, if we factor out a (-λ) term we get the form
y’(0)=-λ(y₀-r)
Now we will obtain an expression for t in terms of y’(t) and y’(0).
Starting with equation 2.5 of the form
y(t)=(r/λ)+(y₀-(r/λ))e^(-λt)
by differentiating both sides with respect to t to get
y’(t)=(d/dt)[(r/λ)+(y₀-(r/λ))e^(-λt)]
where (r/λ) is a constant whose derivative is zero
y’(t)=(d/dt)[(y₀-(r/λ))e^(-λt)]
using the product rule of differentiation
y’(t)= [(d/dt)(y₀-(r/λ))]•e^(-λt)+[(d/dt)e^(-λt)]•(y₀-(r/λ))
y’(t)= 0•e^(-λt) +(y₀-(r/λ))•(-λe^(-λt)) This yields a new expression for y’(t).
y’(t)=-λ(y₀-(r/λ))e^(-λt)
From here we can recognize this is of the form
y’(t)=y’(0)•e^(-λt)
and getting the exponent by itself through division
y’(t)/y’(0)=e^(-λt)
taking the natural log of both sides we get
ln(y’(t)/y’(0))=-λt
now to get t by itself we divide by -λ
t=(-1/λ)ln(y’(t)/y’(0))
now through the negative logarithmic rule
(Equation 3.1) t=(1/λ)ln(y’(0)/y’(t))
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●​ Applying the Equation 3.1 to “Portrait by Resco”
𝑡 =
1
30±5
𝑙𝑛( 9.2±1.2 )
0.0315
Evaluating at the upper limits of Po concentration yields around 39 years of age in 1966,
indicating a production date around 1927.
Evaluating at the lower limits of Po concentration yields around 36 years of age in 1966,
indicating a production date around 1930.
Evaluating at the lower limits of Po concentration at the date of manufacture and the upper limits
at the time of measurements yields around 28 years of age in 1966, indicating a production date
around 1938.
Evaluating at the upper limits of Po concentration at the date of manufacture and the lower limits
at the time of measurements yields around 47 years of age in 1966, indicating a production date
around 1919.
These results are in agreement with the given production date of 1923.
●​ Applying the Equation 3.1 to “Landscape (England)”
𝑡 =
1
48±11
𝑙𝑛( 2.1±0.3 )
0.0315
Evaluating at the upper limits of Po concentration yields 102 years of age in 1966, indicating a
production date around 1864.
Evaluating at the lower limits of Po concentration yields 96 years of age in 1966, indicating a
production date around 1869.
Evaluating at the lower limits of Po concentration at the date of manufacture and the upper limits
at the time of measurements yields 87 years of age in 1966, indicating a production date around
1879.
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Evaluating at the upper limits of Po concentration at the date of manufacture and the lower limits
at the time of measurements yields 111 years of age in 1966, indicating a production date
around 1855.
These results are in agreement with the given production date of 1850-60.
●​ Applying the Equation 3.1 to “Eicholz (US)”
𝑡 =
1
22±8
𝑙𝑛( 0.35±0.08 )
0.0315
Evaluating at the upper limits of Po concentration yields 138 years of age in 1966, indicating a
production date around 1828.
Evaluating at the lower limits of Po concentration yields 127 years of age in 1966, indicating a
production date around 1839.
Evaluating at the lower limits of Po concentration at the date of manufacture and the upper limits
at the time of measurements yields 112 years of age in 1966, indicating a production date
around 1854.
Evaluating at the upper limits of Po concentration at the date of manufacture and the lower limits
at the time of measurements yields 151 years of age in 1966, indicating a production date
around 1815.
These results are in agreement with the given production date of 1817.
6. Let us define the separation factor (SF) by the ratio of Polonium-210 to Radium-226 at the
time of manufacture, where both are measured in dpm/g of Pb. So, the SF of the portrait by
Resco in Table 1 is
​
(30 dpm/g of Pb)/(0.36 dpm/g of Pb) = 83
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Likewise, if we assume that the SF of “Lace Maker” and “The Death of Sardanapalus” (Table 2)
is 100, then the decay rates of Polonium-210 when they were painted were 140 and 440 dpm/g
of Pb, respectively. Based on this assumption, compute dates for “Lace Maker” and “The Death
of Sardanapalus.”
Defining Separation Factor as: (Equation 4)​ SF=(Po-210)/(Ra-226) at the time of
manufacture,
●​ Using equation 3.1 t=(1/λ)ln(y’(0)/y’(t)) to compute the date of manufacture for “Lace
Maker”
1
(
𝑡 = 0.0315 𝑙𝑛
140±20
1.5±0.3
)
Where y‘(0) is the initial decay rate of Po-210 at the time of manufacture.
Where y’(t) is the current measured decay rate of Po-210 in the year 2010.
Where λ=0.0315, the decay constant for Lead/Radium.
Evaluating at the upper bounds, t=142 years before 2010 which equals the year 1867.
Evaluating at the lower bounds, t=146 years before 2010 which equals the year 1863.
Evaluating at the upper initial and lower current rates, t=155 years with inception of 1854.
Evaluating at the lower initial and upper current rates, t=133 years with inception of 1876.
Evaluating at the central bounds, t=144 years before 2010 which equals the year 1865.
●​ Now computing the date of manufacture for “The Death of Sardanapalus”
1
(
𝑡 = 0.0315 𝑙𝑛
440±40
6.6±0.8
)
Evaluating at the upper bounds, t=132 years before 2010 which equals the year 1877.
Evaluating at the lower bounds, t=134 years before 2010 which equals the year 1875.
Evaluating at the upper initial and lower current rates, t=140 years with inception of 1869.
Evaluating at the lower initial and upper current rates, t=126 years with inception of 1883.
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Evaluating at the central bounds, t=133 years before 2010 which equals the year 1876.
7. Is it reasonable to assume that the SF is 100? Assuming the SF is 100, compute the origin
years for the paintings from the 18th, 19th and 20th centuries in Table 1. How well do your
results compare to the actual age of the painting? [Note: The data in Table 1 is from 1966.]
​
Yes, I believe it is reasonable to assume that the SF is 100. We can see this first by
noticing that the median of the known separation factors given in Table 1 is near the value of
100. We can further test this by trying other separation factors and seeing how it affects the
predicted time. In the example with “The Death of Sardanapalus” we will try out the following
separation factors; 50,100,200. This results in the estimated date of manufacture with results
ranging from 1854 until 1876 for a total of 22 years difference.
​
Next, computing the origin years for the paintings in Table 1, we can obtain the following:
Painting
Date
Calculated (median)
Date
Difference (years)
Stack Process
1948
1950
2
Portrait by Resco
1923
1928
5
Flowers
1920
1922
2
Landscape by J.
Sloan
1910
1911
1
Landscape (US)
1850-60
1864
4
Landscape (England)
1850-60
1866
6
Primitive
1830
1834
4
Eicholz
1817
1834
17
Portrait (France)
1780-88
1839
51
Portrait of Claypoole
1746
1822
76
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SF = 100 works well for the 19th/20th century but fails for the 18th century, suggesting SF
changes over centuries due to ore processing changes or other factors.
8. Compute estimated dates for each painting in Table 2. Based on your calculations, make a
recommendation to the auction house Christie’s.
●​ Now computing the date of manufacture for “Laughing Girl”
1
(
𝑡 = 0.0315 𝑙𝑛
600±90
5.2±0.8
)
Evaluating at the upper bounds
Evaluating at the upper bounds, t=150 years before 2010 which equals the year 1860.
Evaluating at the lower bounds, t=151 years before 2010 which equals the year 1859.
Evaluating at the upper initial and lower current rates, t=160 years with inception of 1850.
Evaluating at the lower initial and upper current rates, t=141 years with inception of 1869.
●​ Now computing the date of manufacture for “Woman Reading Music”
1
(
88±9
𝑡 = 0.0315 𝑙𝑛 1.9±0.2
)
Evaluating at the upper bounds, t=121 years before 2010 which equals the year 1889.
Evaluating at the lower bounds, t=121 years before 2010 which equals the year 1889..
Evaluating at the upper initial and lower current rates, t=128 years with inception of 1882.
Evaluating at the lower initial and upper current rates, t=115 years with inception of 1885.
●​ Now computing the date of manufacture for “Ovid Among the Scythians”
1
(
𝑡 = 0.0315 𝑙𝑛
400±40
8.7±0.8
)
Evaluating at the upper bounds, t=121 years before 2010 which equals the year 1889.
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Evaluating at the lower bounds, t=121 years before 2010 which equals the year 1889..
Evaluating at the upper initial and lower current rates, t=128 years with inception of 1882.
Evaluating at the lower initial and upper current rates, t=116 years with inception of 1884.
●​ Now computing the date of manufacture for “Arabian Horses Fighting in a Stable”
1
(
𝑡 = 0.0315 𝑙𝑛
3900±200
61±2
)
Evaluating at the upper bounds, t=133 years before 2010 which equals the year 1877.
Evaluating at the lower bounds, t=131 years before 2010 which equals the year 1879..
Evaluating at the upper initial and lower current rates, t=134 years with inception of 1876.
Evaluating at the lower initial and upper current rates, t=129 years with inception of 1881.
●​ Now computing the date of manufacture for “Orphan Girl in a Cemetery”
1
(
𝑡 = 0.0315 𝑙𝑛
180±10
3±.2
)
Evaluating at the upper bounds, t=128 years before 2010 which equals the year 1882.
Evaluating at the lower bounds, t=127 years before 2010 which equals the year 1883..
Evaluating at the upper initial and lower current rates, t=134 years with inception of 1876.
Evaluating at the lower initial and upper current rates, t=126 years with inception of 1884.
●​ Now computing the date of manufacture for “The Massacre at Chios”
1
(
5±1
𝑡 = 0.0315 𝑙𝑛 .54±.05
)
Evaluating at the upper bounds, t=74 years before 2010 which equals the year 1936.
Evaluating at the lower bounds, t=66 years before 2010 which equals the year 1944.
Evaluating at the upper initial and lower current rates, t=79 years with inception of 1931.
Evaluating at the lower initial and upper current rates, t=61 years with inception of 1949.
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●​ Now computing the date of manufacture for “The Abduction of Rebecca”
1
(
33±3
𝑡 = 0.0315 𝑙𝑛 .5±.05
)
Evaluating at the upper bounds, t=133 years before 2010 which equals the year 1877.
Evaluating at the lower bounds, t=133 years before 2010 which equals the year 1877.
Evaluating at the upper initial and lower current rates, t=139 years with inception of 1871.
Evaluating at the lower initial and upper current rates, t=127 years with inception of 1883.
●​ Now computing the date of manufacture for “Combat of Giaour and Hassan”
1
(
𝑡 = 0.0315 𝑙𝑛
140±20
4.6±.04
)
Evaluating at the upper bounds, t=113 years before 2010 which equals the year 1897.
Evaluating at the lower bounds, t=104 years before 2010 which equals the year 1906.
Evaluating at the upper initial and lower current rates, t=113 years with inception of 1897.
Evaluating at the lower initial and upper current rates, t=103 years with inception of 1907.
____________________________________________________________________________
Most Delacroixs and especially all Vermeers have significant discrepancies between calculated
and actual ages. A possible cause is the replenishment of polonium caused by radium
presence, which would be exacerbated by sampling discrepancies in both polonium and radium
values. This is plausible because the radiometric ages of all Vermeers are similarly lower than
the actual production dates (about two centuries), and all Delacroixs are also similarly lower
than the production date (about half a century). However, there are some outliers with only
marginal error. We could consider The Abduction of Rebecca(1871), Orphan Girl in a
Cemetery(1876) and Arabian Horses Fighting in a Stable(1876) in a grey area which could
make these few paintings real.
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Resolution
Advise Christie's that these paintings’ ages don’t match historical records—likely forgeries or
require further checks. Occam’s razor suggests that most of the Vermeers and Delcroixs were
forged during specific time frames by two different prolific forgers, rather than discrepancies
being due to polonium replenishment or other factors. Suggesting further chemical analysis
(e.g., testing additional pigments) to confirm whether the discrepancies are due to repainting,
contamination, or outright forgery.
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