# Exponential change

An exponential describes change by equal factors: a quantity whose rate of change is proportional to its own size. Whatever sits in the exponent must be a pure number, with no units.

Written for this edition, not translated from Einstein.

## What kind of change does an exponential describe?

Suppose a quantity changes at a rate proportional to its own size. A radioactive sample loses the same fraction of its atoms every hour; a beam of light loses the same fraction of its intensity in every millimetre of absorbing glass. Take 1,000 atoms that lose half their number each hour:

1. After one hour 500 remain, after two 250, after three 125.
2. The losses shrink as the sample does: 500 in the first hour, 250 in the second, 125 in the third.
3. Each hour multiplies the count by the same factor, one half. Equal steps, equal factors: that is exponential change.

The function e to the power x, with e about 2.718, is the one whose rate of change equals its own value everywhere, which is why it describes this kind of change. In symbols, with N₀ atoms at the start and a constant k that sets how fast they decay (for halving every hour, k is about 0.693 per hour):

$$
\begin{gathered} \frac{dN}{dt} = -kN, \\ \text{so}\quad N(t) = N_0\,e^{-kt} \end{gathered}
$$

If the rate of change of N is minus k times N, then N at time t is N nought times e to the minus k t.

Whatever sits in the exponent must be a pure number with no units. The reason is the series that defines e to the power x:

$$
e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \ldots
$$

e to the x equals one plus x plus x squared over two plus x cubed over six, and so on.

If x were a length, the series would add a length to an area to a volume, which has no meaning. In the decay law, k is a rate per hour and t is in hours, so kt is a pure number.

Two exponentials from the 1905 papers pass that check. Wien's radiation law, which the light-quanta paper uses from §4 on, contains βν/T: β times a frequency is a temperature, so dividing by the temperature T leaves a pure number. The formula below writes e to the power −βν/T as exp(−βν/T). exp(u) is another way of writing e to the power u, used when the power is long.

$$
\rho = \alpha\,\nu^{3} e^{-\beta \nu / T}
$$

Rho equals alpha nu cubed times e to the minus beta nu over T.

The Brownian paper's §4 solution contains x²/4Dt: x² is an area, and a diffusion coefficient D in square metres per second times a time t in seconds is also an area, so their ratio is a pure number.

$$
f(x,t) = \frac{n}{\sqrt{4\pi D}}\,\frac{e^{-x^{2}/4Dt}}{\sqrt{t}}
$$

f of x and t equals n over the square root of four pi D, times e to the minus x squared over four D t, divided by the square root of t.

## Worked example: The bell curve at its centre and one width out

The Brownian curve's shape comes from its exponential factor, e raised to −x²/4Dt. Follow that factor outward from the centre:

1. At the centre, x = 0, the exponent is 0 and e⁰ = 1: the curve is at its peak.
2. At x = √(2Dt), one root-mean-square width out, the exponent is −2Dt/4Dt = −1/2, and e raised to −1/2 is about 0.607 of the peak.
3. At x = √(4Dt), the exponent is −1, and e raised to −1 is about 0.368 of the peak.
4. Doubling the time doubles 4Dt, so each of these heights is reached √2 times farther from the centre.

None of these steps depends on the units chosen for x, t and D, as long as they agree with one another. That is what a pure-number exponent guarantees.

## Where this lesson stops

Whatever sits in an exponent must be a pure number. Check it by cancelling the units, as with βν/T and x²/4Dt.

## This lesson builds on

- [Rates of change and derivatives](/foundations/derivatives/)
- [Logarithms: turning products into sums](/foundations/logarithms/)