A surprising application of Differential Equations — The Van Meegeren art forgeries part I

Normally, differential equations are associated with Newton’s laws of motion; Maxwell’s equations are PDE’s. Most people are under the impression that differential equations are used in physics/engineering and of course, what salivates many are their applications to finance/econometrics/stock-market algorithms.

My aim, in the present blog, is to raise the awareness that any problem which can be mathematically modelled (under certain assumptions) can be solved mathematically !! The trick or validity of the  mathematical model lies on your knowledge of Math, both the width and depth, and the knowledge of other areas of the problem at hand.

Let us look at an application of differential equations to arts! (This is a classic example which I picked up  from available literature on differential equations and their applications and I would like to share with you) It was proved that the beautiful painting “Disciples at Emmaus”‘ which was bought by the Rembrandt Society of Belgium for $170,000 was a modern forgery. The story is as given below: After the liberation of Belgium in World War II, the Dutch Field Security began its hunt for Nazi collaborators. They discovered, in the records of a firm which had acted as an intermediary in the sale to Goering of the painting “Woman Taken in Adultery” by the famed 17th century Dutch painter Jan Vermeer. The banker in turn revealed that he was acting on behalf of a third rate Dutch painter H. A. Van Meegeren, and on May 29, 1945 Van Meegeren was arrested on the charge of collaborating with the enemy. On July 12, 1945 Van Meegeren startled the world by announcing from his prison cell that he had never sold “Woman Taken in Adultery” to Goering. Moreover, he stated that this painting and the very famous and beautiful “Disciples at Emmaus”, as well as four other presumed Vermeers and two de Hooghs (a 17th century Dutch painter) were his own work. Many people, however, thought that Van Meegeren was only lying to save himself from the charge of treason. To prove his point, Van Meegeren began, while in prison, to forge the Vermeer painting “Jesus Amongst the Doctors” to demonstrate to the skeptics just how good a forger of Vermeer he was. The work was completed when Van Meegeren learned that a charge of forgery had been substituted for that of collaboration. He, therefore, refused to finish and age the painting so that hopefully investigators would not uncover his secret of aging his forgeries. To settle the question an international panel of distinguished chemists, physicists and art historians was appointed to investigate the matter. The panel took x-rays of the paintings to determine whether other paintings were underneath them. In addition, they analyzed the pigments (coloring material) used in the paint, and examined the paintings for certain signs of old age. Now, Van Meegeren was well aware of these materials. To avoid detection, he scraped the paint from old paintings that were not worth much, just to get the canvas, and he tried to use pigments that Vermeer would have used. Van Meegeren also knew that old paint was extremely hard, and impossible to dissolve. Therefore, he cleverly mixed a chemical, phenoformaldehyde, into the paint, and this hardened into bakelite when the finished painting was heated in an oven. However, Van Meegeren was careless with several of his forgeries, and the panel of experts found traces of the modern pigment cobalt blue. In addition, they also detected the phenoformaldehyde, which was not discovered until the turn of the 19th century, in several of the paintings. On the basis of this evidence Van Meegeren was convicted of forgery, on October 12, 1947 and sentenced one year in prison. While in prison, he suffered a heart attack and died on December 30, 1947. However, even following the evidence gathered by the panel of experts, many people still refused to believe that the famed “Disciples at Emmaus” was forged by Van Meegeren. Their contention was based on the fact that the other alleged forgeries and Van Meegeren’s nearly completed “Jesus Amongst the Doctors” were of a very inferior quality. Surely, they said, the creator of the beautiful “Disciples at Emmaus” could not produce such inferior pictures. Indeed, the “Disciples at Emmaus” was certified as an authentic Vermeer by the noted art historian A. Bredius and was bought by the Rembrandt Society for$170,000. The answer of the panel to these skeptics was that because Van Meegeren was keenly disappointed by his lack of status in the art world, he worked on the “Disciples at Emmaus” with the fierce determination of proving that he was better than a third rate painter. After producing such a masterpiece, his determination was gone. Moreover, after seeing how easy it was to dispose of the “Disciples at Emmaus” he devoted less effort to his  subsequent forgeries. This explanation failed to satisfy the skeptics. They demanded a thoroughly scientific and conclusive proof that the “Disciples at Emmaus” was indeed a forgery. This was done in 1967 by scientists at Carnegie Mellon University, and we would now like to describe that work.

The key to the dating of paintings and other materials such as rocks and fossils lies in the phenomenon of radioactivity discovered at the turn of the century. The physicist Rutherford and his colleagues showed that the atoms of certain “radioactive” elements are unstable and that within a given time period a fixed proportion of the atoms spontaneously disintegrates to form atoms of a new element. Because radioactivity is a property of  the atom, Rutherford showed that the radioactivity of a substance is directly proportional to  the number of atoms of the substance present. Thus, if $N(t)$ denotes the number of atoms present at time t, then

$\frac {dN}{dt}=-\lambda N$ equation I

The constant $\lambda$ which is positive is known as the decay constant of the substance. The larger

$\lambda$ is, of course, the faster the substance decays. One measure of the rate of disintegration of a substance is its half-life which is defined as the time required for half of a given quantity of radioactive atoms to decay. To compute the half-life of a substance in terms of $\lambda$, assume that at time $t_{0}$, $N(t_{0})=N_{0}$. Then, the solution of the initial-value problem

$\frac {dN}{dt}=-\lambda N_{0}$ if $N(t_{0})=N_{0}$ is

$N(t) =N_{0}exp(-\lambda \int_{t_{0}}^{t}ds)=N_{0}e^{-\lambda (t-t_{0})}$

or $\frac {N}{N_{0}} = exp (-\lambda (t-t_{0}))$.

Taking logarithm of both sides we obtain that

$-\lambda (t-t_{0})= ln \frac {N}{N_{0}}$ equation II

Now, if $\frac {N}{N_{0}}=1/2$ then $-\lambda (t-t_{0})= ln (1/2)$ so that

$(t-t_{0})= \frac {ln 2}{\lambda} = 0.6931/(\lambda)$ equation III

Thus, the half-life of a substance is $ln 2$ divided by the decay constant $\lambda$. The dimension of $\lambda$, which we suppress for  simplicity of  writing is reciprocal time. If t is measured in years, then $\lambda$ has the dimension of reciprocal years, and if t is measured in minutes, then $\lambda$ has the dimension of reciprocal minutes. The half-life of many substances have been determined and recorded. For example, the half-life of carbon-14 is 5568 years and the half-life of uranium-238 is 4.5 billion years.}

Now, the basis of “radioactive dating” is essentially the following. From equation II, we can solve for

$(t-t_{0})=(1/\lambda) ln (\frac {N}{N_{0}})$.

If $t_{0}$ is the time the substance was initially formed or  manufactured, then the age of the substance is

$(1/{\lambda}) ln (\frac {N_{0}}{N})$.

The decay constant $\lambda$ is known or can be computed in most instances. Moreover, we can usually evaluate N quite easily. Thus, if we knew $N_{0}$ we could determine the age of the substance. But, this is the real difficulty of course since we usually do not know $N_{0}$. In some instances though, we can either determine $N_{0}$ and such is the case for the  forgeries of Van Meegeren.

We begin with the following well-known facts of elementary chemistry. Almost all rocks in the earth’s crust contain a small quantity of uranium. The  uranium in the rock decays to another radioactive element, and that one decays to  another and another, and so forth in a series of elements that results in lead, which is not radioactive. The uranium (whose half-life is over four billion years) keeps feeding the elements following it in the series, so that as fast as the they decay, they are replaced by the elements before them.

Now, all paintings contain a small amount of the radioactive element lead-210 ($Pb^{210}$), and even a smaller amount of radium-226 ($Ra^{226}$), since these elements are contained in white lead (lead oxide), which is a pigment that artists have  used for over 2000 years. For the analysis which follows, in turn, is extracted from a rock called lead ore, in a process called smelting. In this process, the lead-210 in the ore goes along with the lead metal. However, 90-95% of  the radium and its descendants are removed with other waste products in a material called slag. Thus, most of  the supply of lead-210 is cut off and it begins to decay very rapidly, with a half-life of 22 years. This process continues until the lead-210 in the white lead is once more in radioactive equilibrium with the small amount of  radium present, that is, the disintegration of lead-210 is exactly balanced by the  disintegration of  the radium.

Let us now  use this information to compute the amount of lead-210 present in a sample in terms of the amount originally present at the  time of manufacture. Let $y(t)$ be the amount of  lead-210 per grain of white lead at time t, and $r(t)$, the number of disintegrations of radium-226 per minute per gram of  white lead at time t. If $\lambda$ is the decay constant for lead-210, then

$\frac {dy}{dt}=-\lambda y +r(t)$ with $y(t_{0})=y_{0}$ equation IV

Since we are only interested in a time period of at most 300 years, we may  assume that the radium-226, whose half-life is 1600 years, remains constant, so that $r(t)$ is a constant r. Multiplying both sides of the differential equation by the  integrating factor

$\mu (t)=e^{\lambda t}$ we obtain that

$\frac {d}{dt} e^{\lambda t}(y)=re^{\lambda t}$

Hence,

or $y(t)= (r/\lambda)(1- e^{-\lambda (t-t_{0})})+y_{0}e^{-\lambda (t-t_{0})})$ equation V

Now, $y(t)$ and r can be easily measured. Thus, if we knew $y_{0}$ we could use equation V to compute

$(t-t_{0})$ and consequently, we could determine the age of the painting. As we pointed out, though, we cannot measure

$y_{0}$ directly. One possible way out of this difficulty is to  use the fact that original quantity of lead-210 was in radioactive equilibrium with the larger amount of radium-226 in the ore from which the metal was extracted. Let us, therefore, take samples of different ores and count the number of disintegrations of  the radium-226 in the ores. This was done for a variety of ores and the results were tabulated. These numbers vary from 0.18 to 140. Consequently, the number of disintegrations of  the lead-210 per minute per gram of white lead at the time of manufacture will vary from 0.18 to 140. This implies that $y_{0}$ will also vary over a very large interval, since the number of disintegrations of lead-210 is  proportional to the amount present. Thus, we cannot use Equation V to obtain an accurate, or even a crude estimate of the age of  a painting.

To be continued in the next part,

— Nalin

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