T his paper is an addition to the series of papers on the exponential function begun by Albert Bartlett. 1 In particular, we ask how the graph of the exponential function y = e -t/t would appear if y were plotted versus ln t rather than the normal practice of plotting ln y versus t. In answering this question, we find a new way to interpret the mean life (or time constant) t using such a linear-log graph.Physics teachers are familiar with the graph of the exponential function. For example, the number of radioactive nuclei as a function of time N(t) is given by the following equation, where N 0 is the original number of nuclei.(1) Figure 1 shows the graph of versus t. We have used a mean life t of 4.0 s. We express this time as 4000 ms because it makes comparison of all the graphs presented in this paper easier.A physicist is often interested in the mean life τ (= ln 2 3 the half-life). To give τ a geometrical meaning, plot the natural logarithm of versus t as shown in Fig. 2.This graph, described by Eq. (2), is a straight line whose slope is .(2)In this case, the mean life τ is 4000 ms (or 4.0 s), so the slope of the graph in Fig. 2 is -0.25 s -1 . We want our students to develop the habit of asking "what if…?" questions (for example, see Ref.2). In this spirit, ask the following: "What shape would we find if we plotted versus ln t rather than ln versus t?" Figure 3 is such a graph.We can estimate the mean life from the preceding three graphs with varying degrees of difficulty. Using the linear plot in Fig. 1, the mean life is the elapsed time required for the ordinate to decrease to approximately 1/3 of its original value. (One can also see this on the log-linear graph in Fig. 2, but care must be exercised when interpolating on a logarithmic scale.) In Fig. 2, the mean life is the reciprocal of the magnitude of the log-linear graph's slope; one cannot tell at a glance what the slope is, let alone its reciprocal. In contrast, the unique and most interesting feature of Fig. 3 is that one can see the mean life at a glance: it is the time at which the graph's concavity changes. For the reader's convenience, the mathematics behind this change of concavity at the mean life is presented below using From elementary calculus we know that the concavity of the graph of y versus ln t will change when the second derivaExponential Decay

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fraction of the Earth solidifies from the outside in.) Therefore, only radioactive minerals in this relatively thin layer contribute to the energy flux at the Earth's surface, since energy released from decays deeper than has not had time to be conducted to the surface. With Kelvin's estimate of the Earth's age (10 8 yr, or approximately 3310 15 s), and of the thermal diffusivity (k < 10 -6 m 2 s -1 ), the layer's thickness is approximately 100 km. If radioactive minerals are uniformly distributed throughout the Earth, radioactivity in this layer contributes only a few mW/m 2 to the energy flux at Earth's surface, which itself is about 40 mW/m 2 . Had Kelvin known about radioactivity, he might therefore have used a temperature gradient about 5% smaller than the measured value (since the energy flux due only to the Earth cooling is slightly less than the total energy flux). He would thus have calculated a cooling time only about 10% larger [based on Eq. (3)] than his published value. Why was Perry's model consistent witha large age for the Earth? Why is Perry's model an exponential process? Perry hypothesized that thermal energy could be convected from Earth's interior (rather than Kelvin's idea that energy was conducted). Convection would have transported energy more rapidly than conduction; consequently, the temperature of a large part of the Earth's molten interior would be (spatially) uniform.Perry's argument uses physics accessible to first-year students. The rate at which energy passes though the surface of the Earth must equal the rate at which the molten interior loses energy as it cools. The first of these rates is (4) where DT is the difference between the temperatures on either side of the crust, L is the crust's thickness, r E is the Earth's radius, 4pr E 2 is the cross-sectional area through which thermal energy is conducted, and K is the thermal conductivity of the crust. The difference between the internal temperature T and the temperature outside the Earth is approximately equal to the internal temperature as long as the internal temperature is relatively large. The rate at which thermal energy is conducted through the Earth's crust is therefore given by the following equation:The second rate (at which the Earth's interior loses energy as it cools) is given by the following equation: T his is a companion to our previous paper 1 in which we give a published example, based primarily on Perry's work, 2,3 of a graph of ln y versus t when y is an exponential function of t. This work led us to the idea that Lord Kelvin's (William Thomson's) estimate of the Earth's age was wrong not because he did not account for radioactivity, as is commonly believed, 4 but because he used the wrong model for Earth's heat loss. We feel this idea is worth spreading. To this end (following England et al. 2,3 ), we examine two questions, the first about the radioactivity part and the second about Perry's alternate model for Earth's heat loss.1. Why would the inclusion of radioactivity not have affected Kelvin's estimate of ...