Drop a rubber ball on a hard floor from \(1.25\,\text{m}\). It rises to \(0.80\,\text{m}\), then to \(0.51\,\text{m}\), then to \(0.33\,\text{m}\). The ball loses less height at every bounce. Yet something stays the same from bounce to bounce. You may read this lesson on its own, whenever a physics lesson sends you here.
The same factor at every step
Divide each height by the one before it. \(0.80/1.25 = 0.64\), \(0.51/0.80 = 0.64\) and \(0.33/0.51 = 0.64\). The ball keeps the same fraction of its height at every bounce. The chapter Collisions shows why: the floor returns a fixed fraction of the speed, so it returns the square of that fraction of the height.
Call the fixed factor \(r\), and call the starting value \(y_0\). After one step the value is \(y_0r\). After two steps it is \(y_0r \times r = y_0r^2\). After \(n\) steps it is \(y_0r^n\). The number of steps sits in the power, so this kind of change is called exponential.
A quantity changes by a constant factor when it is multiplied by the same number \(r\) in every equal step of time, distance or count. If \(r > 1\) the quantity grows; if \(0 < r < 1\) it decays. Equal steps then give equal ratios, never equal differences.
Set \(r = 0.64\) in the figure above. The first bar falls by \(0.45\,\text{m}\) to the next, the next by only \(0.29\,\text{m}\). The falls shrink, but the ratio never changes. Now set \(r = 0.95\). The bars shrink so slowly that the ball bounces for a long time. The factor \(r\) alone decides the shape.
The sum of all the steps
The ball of the figure travels \(1.25\,\text{m}\) down, then \(0.80\,\text{m}\) up and \(0.80\,\text{m}\) down, and so on for ever. The bounces never quite stop, yet the total distance is finite. A sum of terms that shrink by a constant factor is a geometric series, and it has a short sum.
\[ S = a + ar + ar^2 + ar^3 + \dots \]
Why. the sum of all the terms, each \(r\) times the one before; \(a\) is the first term.
\[ rS = ar + ar^2 + ar^3 + \dots \]
Why. multiply every term by \(r\). Each term moves one place along the list.
\[ S - rS = a \]
Why. subtract the second line from the first. Every term after \(a\) cancels.
\[ S = \dfrac{a}{1 - r} \]
Read it. write the left side as \(S(1 - r)\) and divide by \(1 - r\). This needs \(0 < r < 1\), so that the terms really shrink.
For the ball, the distance after the first fall is twice the sum of the bounce heights. The first bounce height is \(1.25 \times 0.64 = 0.80\,\text{m}\), so the sum of all bounce heights is \(0.80/(1 - 0.64) = 2.22\,\text{m}\). The total distance is \(1.25 + 2 \times 2.22 = 5.69\,\text{m}\). The slider readout of the figure agrees.
A ball dropped from \(1.25\,\text{m}\) keeps \(64\,\%\) of its height at every bounce. How high does it rise after the third bounce, in metres?
Which list of readings changes by a constant factor?
From steps to a smooth change
A bounce is a natural step. A radioactive nucleus has no such step. The chapter Radioactivity shows that each nucleus has a fixed chance \(\lambda\) of decaying in each unit of time, at every instant. So the steps must be made as small as we like. Suppose \(\lambda t = 1\). Split the time into \(n\) equal steps, and let a fraction \(1/n\) of what is left decay in each step.
After the \(n\) steps the fraction left is \((1 - 1/n)^n\). With one step it is \(0\). With two steps it is \(0.25\), with four \(0.316\), with twelve \(0.352\) and with a hundred \(0.366\). The answer tends to \(0.368\). That number is \(1/e\), where \(e = 2.71828\) is a new constant of nature and of mathematics.
Growth behaves the same way. With a fraction \(1/n\) added in each of \(n\) steps, the value becomes \((1 + 1/n)^n\) times the start. This is \(2\) for one step, \(2.25\) for two, \(2.61\) for twelve and \(2.7146\) for 365. It tends to \(e\). So a continuous change at a steady fractional rate multiplies a quantity by \(e\) or by \(1/e\), not by \(2\) or by \(0\).
The number \(e\) is the limit of \((1 + 1/n)^n\) as \(n\) grows without end. Its value is \(e = 2.71828\ldots\). The limit of \((1 - 1/n)^n\) is \(1/e = 0.36788\ldots\). The function \(e^x\) is the exponential function.
The slope that equals the height
Why does physics choose this odd number? Consider the slope of the curve \(y = b^x\). Near any point, the slope is \((b^{x + h} - b^x)/h = b^x \times (b^h - 1)/h\) for a small \(h\). So the slope is the height \(b^x\) times a number that belongs to the base alone. Put \(h = 0.001\). For \(b = 2\) that number is \(0.693\). For \(b = 3\) it is \(1.099\). For \(b = e\) it is \(1.000\).
So \(e\) is the one base for which the slope of the curve equals its height, \(d(e^x)/dx = e^x\). The factor \(0.693\) for base 2 is not new either. It is the natural logarithm of 2, as the lesson on logarithms of this chapter shows. The derivative of \(e^x\) is derived in the chapter Differentiation.
What is the slope of the curve \(y = e^x\) at \(x = 2\)?
Find \((1 + 1/10)^{10}\), to three decimal places.
A rate proportional to the amount
Now the slope property proves its worth. Suppose a quantity changes at a rate proportional to itself, \(dN/dt = kN\). The function \(N = N_0e^{kt}\) has the slope \(kN_0e^{kt} = kN\), so it fits the law. At \(t = 0\) it gives \(N_0\). This is the solution; the chapter Integration derives it by separating the variables. A decaying sample has \(k = -\lambda\), so \(N = N_0e^{-\lambda t}\).
The smooth law still multiplies by a constant factor in equal steps. Over any interval \(T\) the factor is \(e^{kT}\), whatever the time at which the interval starts. With \(\lambda = 0.1\) per hour, a sample keeps \(e^{-0.1} = 0.905\) of its nuclei in each hour. It does not keep \(0.9\), because the chance acts on a shrinking sample through the hour.
A nuclide has a decay constant \(\lambda = 0.1\) per hour. What fraction of a sample is left after one hour? Give three decimal places.
A ball is dropped from \(5\,\text{m}\) onto a floor with coefficient of restitution \(e = 0.5\). Take \(g = 10\,\text{m/s}^2\). Find the height of the first bounce, the total distance the ball travels before it comes to rest, and the total time.
The floor returns the fraction \(0.5\) of the speed, so each bounce keeps \(r = 0.5^2 = 0.25\) of the height. (a) \(h_1 = 0.25 \times 5 = 1.25\,\text{m}\).
(b) The bounce heights form a geometric series with first term \(1.25\,\text{m}\) and factor \(0.25\). Their sum is \(1.25/(1 - 0.25) = 1.667\,\text{m}\). Each is travelled up and down, so the distance is \(5 + 2 \times 1.667 = 8.33\,\text{m}\).
(c) The first fall takes \(\sqrt{2 \times 5/10} = 1\,\text{s}\). Each flight after a bounce takes \(0.5\) times the one before, starting at \(2 \times 0.5 = 1\,\text{s}\). Those flights sum to \(1/(1 - 0.5) = 2\,\text{s}\), so the total time is \(3\,\text{s}\).
The letter \(e\) here is the coefficient of restitution, not the number \(2.718\).
The phenomenon of the lesson is now answered. From one factor, \(0.64\), the heights of every bounce follow as \(1.25 \times 0.64^n\). The geometric series gives the whole distance, \(5.69\,\text{m}\), although the bounces never end.
A chain reaction starts with 10 fissions, and each generation has \(1.5\) times as many fissions as the one before. How many fissions are there in generation 5, to the nearest whole number?
The law \(y_0r^n\) needs the same factor at every step. A real ball keeps less of its height when it hits hard at high speed, and air drag grows with speed. So the factor drifts a little from bounce to bounce. The sum \(a/(1 - r)\) also fails when \(r \ge 1\): the terms never shrink, and the sum grows without end.
The same idea appears throughout the book. The heights of a bouncing ball in the chapter Collisions fall by \(e^2\) at each bounce. The fissions of a chain reaction in the chapter on the nucleus grow by the factor \(k\) in each generation. A radioactive sample in the chapter Radioactivity halves in each half-life. The maxima of a damped oscillator shrink by a fixed ratio in each period. In each case, one factor per step describes the whole story.
- Subtracting the same amount at each step.Exponential change multiplies by the same factor. Equal steps give equal ratios, so the differences shrink or grow.
- Writing \(e^{-0.1} = 0.9\).\(e^{-x} \approx 1 - x\) only for very small \(x\). Here \(e^{-0.1} = 0.905\).
- Using the power rule to differentiate \(e^x\).The power rule is for \(x^n\). The slope of \(e^x\) is \(e^x\) itself.