(b)The Die-away Curve

If we were to take $p$ as a proper fraction (less than unity), the curve would obviously tend to sink downwards, as in Figure 42, where each successive ordinate is $\frac{3}{4}$ of the height of the preceding one.

The equation is still \[ y=bp^x; \] but since $p$ is less than one, $\log_\epsilon p$ will be a negative quantity, and may be written $-a$; so that $p=\epsilon^{-a}$, and now our equation for the curve takes the form \[ y=b\epsilon^{-ax}. \]

The importance of this expression is that, in the case where the independent variable is time, the equation represents the course of a great many physical processes in which something is gradually dying away. Thus, the cooling of a hot body is represented (in Newton's celebrated “law of cooling”) by the equation \[ \theta_t=\theta_0 \epsilon^{-at}; \] where $\theta_0$ is the original excess of temperature of a hot body over that of its surroundings, $\theta_t$ the excess of temperature at the end of time $t$, and $a$ is a constant–namely, the constant of decrement, depending on the amount of surface exposed by the body, and on its coefficients of conductivity and emissivity, etc.

A similar formula, \[ Q_t=Q_0 \epsilon^{-at}, \] is used to express the charge of an electrified body, originally having a charge $Q_0$, which is leaking away with a constant of decrement $a$; which constant depends in this case on the capacity of the body and on the resistance of the leakage-path.

Oscillations given to a flexible spring die out after a time; and the dying-out of the amplitude of the motion may be expressed in a similar way.

In fact $\epsilon^{-at}$ serves as a die-away factor for all those phenomena in which the rate of decrease is proportional to the magnitude of that which is decreasing; or where, in our usual symbols, $\dfrac{dy}{dt}$ is proportional at every moment to the value that $y$ has at that moment. For we have only to inspect the curve, Figure 42 above, to see that, at every part of it, the slope $\dfrac{dy}{dx}$ is proportional to the height $y$; the curve becoming flatter as $y$ grows smaller. In symbols, thus $y=b\epsilon^{-ax}$ or \[ \log_\epsilon y = \log_\epsilon b - ax \log_\epsilon \epsilon = \log_\epsilon b - ax,\\ \text{and, differentiating,}\; \frac{1}{y}\, \frac{dy}{dx} = -a;\\ \text{hence}\; \frac{dy}{dx} = b\epsilon^{-ax} × (-a) = -ay; \] or, in words, the slope of the curve is downward, and proportional to $y$ and to the constant $a$.

We should have got the same result if we had taken the equation in the form \begin{align*} y &= bp^x; \\ \text{for then}\; \frac{dy}{dx} &= bp^x × \log_\epsilon p. \\ \text{But}\; \log_\epsilon p &= -a; \\ \text{giving us}\; \frac{dy}{dx} &= y × (-a) = -ay, \end{align*} as before.

The Time-constant. In the expression for the “die-away factor” $\epsilon^{-at}$, the quantity $a$ is the reciprocal of another quantity known as “the time-constant,” which we may denote by the symbol $T$. Then the die-away factor will be written $\epsilon^{-\frac{t}{T}}$; and it will be seen, by making $t = T$ that the meaning of $T$ $\left(\text{or of} \dfrac{1}{a}\right)$ is that this is the length of time which it takes for the original quantity (called $\theta_0$ or $Q_0$ in the preceding instances) to die away $\dfrac{1}{\epsilon}$th part–that is to $0.3678$–of its original value.

The values of $\epsilon^x$ and $\epsilon^{-x}$ are continually required in different branches of physics, and as they are given in very few sets of mathematical tables, some of the values are tabulated here for convenience.

As an example of the use of this table, suppose there is a hot body cooling, and that at the beginning of the experiment (i.e.: when $t = 0$) it is $72°$ hotter than the surrounding objects, and if the time-constant of its cooling is $20$ minutes (that is, if it takes $20$ minutes for its excess of temperature to fall to $\dfrac{1}{\epsilon}$ part of $72°$), then we can calculate to what it will have fallen in any given time $t$. For instance, let $t$ be $60$ minutes. Then $\dfrac{t}{T} = 60 ÷ 20 = 3$, and we shall have to find the value of $\epsilon^{-3}$, and then multiply the original $72°$ by this. The table shows that $\epsilon^{-3}$ is $0.0498$. So that at the end of $60$ minutes the excess of temperature will have fallen to $72° × 0.0498 = 3.586°$.

Further Examples.

(1) The strength of an electric current in a conductor at a time $t$ secs. after the application of the electromotive force producing it is given by the expression $C = \dfrac{E}{R}\left\{1 - \epsilon^{-\frac{Rt}{L}}\right\}$.

The time constant is $\dfrac{L}{R}$.

If $E = 10$, $R =1$, $L = 0.01$; then when $t$ is very large the term $\epsilon^{-\frac{Rt}{L}}$ becomes $1$, and $C = \dfrac{E}{R} = 10$; also \[ \frac{L}{R} = T = 0.01. \]

Its value at any time may be written: \[ C = 10 - 10\epsilon^{-\frac{t}{0.01}}, \] the time-constant being $0.01$. This means that it takes $0.01$ sec. for the variable term to fall by $\dfrac{1}{\epsilon} = 0.3678$ of its initial value $10\epsilon^{-\frac{0}{0.01}} = 10$.

To find the value of the current when $t = 0.001 \text{sec.}$, say, $\dfrac{t}{T} = 0.1$, $\epsilon^{-0.1} = 0.9048$ (from table).

It follows that, after $0.001$ sec., the variable term is $0.9048 × 10 = 9.048$, and the actual current is $10 - 9.048 = 0.952$.

Similarly, at the end of $0.1$ sec., \[ \frac{t}{T} = 10;\quad \epsilon^{-10} = 0.000045; \] the variable term is $10 × 0.000045 = 0.00045$, the current being $9.9995$.

(2) The intensity $I$ of a beam of light which has passed through a thickness $l$ cm. of some transparent medium is $I = I_0\epsilon^{-Kl}$, where $I_0$ is the initial intensity of the beam and $K$ is a “constant of absorption.”

This constant is usually found by experiments. If it be found, for instance, that a beam of light has its intensity diminished by 18% in passing through $10$ cms. of a certain transparent medium, this means that $82 = 100 × \epsilon^{-K×10}$ or $\epsilon^{-10K} = 0.82$, and from the table one sees that $10K = 0.20$ very nearly; hence $K = 0.02$.

To find the thickness that will reduce the intensity to half its value, one must find the value of $l$ which satisfies the equality $50 = 100 × \epsilon^{-0.02l}$, or $0.5 = \epsilon^{-0.02l}$. It is found by putting this equation in its logarithmic form, namely, \[ \log 0.5 = -0.02 × l × \log \epsilon, \] which gives \[ l = \frac{-0.3010}{-0.02 × 0.4343} = 34.7 \text{centimetres nearly}. \]

(3) The quantity $Q$ of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity $Q_0$ of the substance by the relation $Q = Q_0 \epsilon^{-\lambda t}$, where $\lambda$ is a constant and $t$ the time in seconds elapsed since the transformation began.

For “Radium $A$,” if time is expressed in seconds, experiment shows that $\lambda = 3.85 × 10^{-3}$. Find the time required for transforming half the substance. (This time is called the “mean life” of the substance.)

We have $0.5 = \epsilon^{-0.00385t}$. \begin{align*} \log 0.5 &= -0.00385t × \log \epsilon; \\ \text{and}\; t &= 3\text{ minutes very nearly}. \end{align*}

Exercises XIII