4.3 Partial derivatives  (Page 6/11)

Lord kelvin and the age of earth

During the late 1800s, the scientists of the new field of geology were coming to the conclusion that Earth must be “millions and millions” of years old. At about the same time, Charles Darwin had published his treatise on evolution. Darwin’s view was that evolution needed many millions of years to take place, and he made a bold claim that the Weald chalk fields, where important fossils were found, were the result of $300$ million years of erosion.

At that time, eminent physicist William Thomson (Lord Kelvin) used an important partial differential equation, known as the heat diffusion equation , to estimate the age of Earth by determining how long it would take Earth to cool from molten rock to what we had at that time. His conclusion was a range of $20\text{to}400$ million years, but most likely about $50$ million years. For many decades, the proclamations of this irrefutable icon of science did not sit well with geologists or with Darwin.

Kelvin made reasonable assumptions based on what was known in his time, but he also made several assumptions that turned out to be wrong. One incorrect assumption was that Earth is solid and that the cooling was therefore via conduction only, hence justifying the use of the diffusion equation. But the most serious error was a forgivable one—omission of the fact that Earth contains radioactive elements that continually supply heat beneath Earth’s mantle. The discovery of radioactivity came near the end of Kelvin’s life and he acknowledged that his calculation would have to be modified.

Kelvin used the simple one-dimensional model applied only to Earth’s outer shell, and derived the age from graphs and the roughly known temperature gradient near Earth’s surface. Let’s take a look at a more appropriate version of the diffusion equation in radial coordinates, which has the form

Here, $T\left(r,t\right)$ is temperature as a function of $r$ (measured from the center of Earth) and time $t.$ $K$ is the heat conductivity—for molten rock, in this case. The standard method of solving such a partial differential equation is by separation of variables, where we express the solution as the product of functions containing each variable separately. In this case, we would write the temperature as

1. Substitute this form into [link] and, noting that $f(t)$ is constant with respect to distance $\left(r\right)$ and $R(r)$ is constant with respect to time $\left(t\right),$ show that
   
   $\frac{1}{f}\frac{\partial f}{\partial t}=\frac{K}{R}\left[\frac{{\partial }^{2}R}{\partial r^2}+\frac{2}{r}\frac{\partial R}{\partial r}\right].$
2. This equation represents the separation of variables we want. The left-hand side is only a function of $t$ and the right-hand side is only a function of $r,$ and they must be equal for all values of $r\text{and}t.$ Therefore, they both must be equal to a constant. Let’s call that constant $\text{−}{\lambda }^2.$ (The convenience of this choice is seen on substitution.) So, we have
   
   $\frac{1}{f}\frac{\partial f}{\partial t}=\text{−}{\lambda }^2\text{and}\frac{K}{R}\left[\frac{{\partial }^{2}R}{\partial r^2}+\frac{2}{r}\frac{\partial R}{\partial r}\right]=\text{−}{\lambda }^2.$
   
   Now, we can verify through direct substitution for each equation that the solutions are $f(t)=A{e}^{\text{−}{\lambda }^{2}t}$ and $R(r)=B\left(\frac{\text{sin}\alpha r}{r}\right)+C\left(\frac{\text{cos}\alpha r}{r}\right),$ where $\alpha =\lambda \text{/}\sqrt{K}.$ Note that $f(t)=A{e}^{+\lambda n^{2}t}$ is also a valid solution, so we could have chosen $+{\lambda }^2$ for our constant. Can you see why it would not be valid for this case as time increases?
3. Let’s now apply boundary conditions.
   1. The temperature must be finite at the center of Earth, $r=0.$ Which of the two constants, $B$ or $C,$ must therefore be zero to keep $R$ finite at $r=0?$ (Recall that $\text{sin}(\alpha r)\text{/}r\to \alpha =$ as $r\to 0,$ but $\text{cos}(\alpha r)\text{/}r$ behaves very differently.)
   2. Kelvin argued that when magma reaches Earth’s surface, it cools very rapidly. A person can often touch the surface within weeks of the flow. Therefore, the surface reached a moderate temperature very early and remained nearly constant at a surface temperature $T_s.$ For simplicity, let’s set $T=0\text{at}r=R_E$ and find $\alpha$ such that this is the temperature there for all time $t.$ (Kelvin took the value to be $300\text{K}\approx 80\text{°}\text{F}.$ We can add this $300\text{K}$ constant to our solution later.) For this to be true, the sine argument must be zero at $r=R_E.$ Note that $\alpha$ has an infinite series of values that satisfies this condition. Each value of $\alpha$ represents a valid solution (each with its own value for $A).$ The total or general solution is the sum of all these solutions.
   3. At $t=0,$ we assume that all of Earth was at an initial hot temperature $T_0$ (Kelvin took this to be about $7000\text{K}.)$ The application of this boundary condition involves the more advanced application of Fourier coefficients. As noted in part b. each value of ${\alpha }_n$ represents a valid solution, and the general solution is a sum of all these solutions. This results in a series solution:
      
      $T(r,t)=\left(\frac{T_0{R}_{\text{E}}}{\pi }\right){\displaystyle \sum _n\frac{{(\mathrm{-1})}^{n-1}}{n}{e}^{\text{−}\lambda n^{2}t}\frac{\text{sin}({\alpha }_{n}r)}{r},}\text{where}{\alpha }_n=n\pi \text{/}{R}_{\text{E}}.$

Note how the values of ${\alpha }_n$ come from the boundary condition applied in part b. The term $\frac{\text{−}{1}^{n-1}}{n}$ is the constant $A_n$ for each term in the series, determined from applying the Fourier method. Letting $\beta =\frac{\pi }{R_{\text{E}}},$ examine the first few terms of this solution shown here and note how ${\lambda }^2$ in the exponential causes the higher terms to decrease quickly as time progresses:

Near time $t=0,$ many terms of the solution are needed for accuracy. Inserting values for the conductivity $K$ and $\beta =\pi \text{/}{R}_{\text{E}}$ for time approaching merely thousands of years, only the first few terms make a significant contribution. Kelvin only needed to look at the solution near Earth’s surface ( [link] ) and, after a long time, determine what time best yielded the estimated temperature gradient known during his era $(1\text{°}\text{F}$ increase per $50\text{ft}).$ He simply chose a range of times with a gradient close to this value. In [link] , the solutions are plotted and scaled, with the $300-\text{K}$ surface temperature added. Note that the center of Earth would be relatively cool. At the time, it was thought Earth must be solid.

Epilog

On May $20,1904,$ physicist Ernest Rutherford spoke at the Royal Institution to announce a revised calculation that included the contribution of radioactivity as a source of Earth’s heat. In Rutherford’s own words:

“I came into the room, which was half-dark, and presently spotted Lord Kelvin in the audience, and realised that I was in for trouble at the last part of my speech dealing with the age of the Earth, where my views conflicted with his. To my relief, Kelvin fell fast asleep, but as I came to the important point, I saw the old bird sit up, open an eye and cock a baleful glance at me.

Then a sudden inspiration came, and I said Lord Kelvin had limited the age of the Earth, provided no new source [ of heat ] was discovered . That prophetic utterance referred to what we are now considering tonight, radium! Behold! The old boy beamed upon me.”

Rutherford calculated an age for Earth of about $500$ million years. Today’s accepted value of Earth’s age is about $4.6$ billion years.