Foundation lesson

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:

After one hour 500 remain, after two 250, after three 125.
The losses shrink as the sample does: 500 in the first hour, 250 in the second, 125 in the third.
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}\,\exp\left(-\frac{\beta\,\nu}{T}\right)

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:

At the centre, x = 0, the exponent is 0 and e⁰ = 1: the curve is at its peak.
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.
At x = √(4Dt), the exponent is −1, and e raised to −1 is about 0.368 of the peak.
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.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: the same fraction lost every step

In linear change, an equal amount is added in every equal interval: y = y₀ + mt. In exponential change, the quantity is multiplied by an equal factor in every interval: y = y₀ · rⁿ after n intervals. When the factor is less than one, as here, every interval takes away the same fraction of what is left. Each 10-second step multiplies it by e^−0.5, about 0.607.

Select a step index n to inspect compounding:

Inspection at step n = 1

Formula: f(1) = (e^−0.5)¹ = e^−0.5

Fraction remaining: 0.606531 (60.65%)

Each 10-second interval scales the previous value by exactly e^−0.5 ≈ 0.606531.

Compounding decay steps with constant multiplier e⁻⁰·⁵
Step n	Elapsed time t (s)	Step multiplier	Fraction remaining	Percent
0	0	1.000000	1.000000	100.00%
1	10	e⁻⁰·⁵ ≈ 0.606531	0.606531	60.65%
2	20	e⁻⁰·⁵ ≈ 0.606531	0.367879	36.79%
3	30	e⁻⁰·⁵ ≈ 0.606531	0.223130	22.31%
4	40	e⁻⁰·⁵ ≈ 0.606531	0.135335	13.53%
5	50	e⁻⁰·⁵ ≈ 0.606531	0.082085	8.21%

Why an exponent has no units

There is no such number as e raised to three metres or five seconds: the series e^u = 1 + u + u²/2! + … would add metres to square metres. In a physical law the units inside an exponent cancel:

Wien’s law (Light §4): e^−βν/T. The frequency ν (s⁻¹) and temperature T (K) are balanced by β = h/k (s·K), making βν/T dimensionless.
Brownian diffusion Gaussian (Brownian §4): e^−x²/(4Dt). The numerator x² has units m²; the denominator 4Dt has units (m²/s · s) = m², so x²/(4Dt) is purely dimensionless.

What it shows, in words

Equal steps in time multiply what is left by equal factors: e^−0.5 per 10 seconds here, so the quantity falls to 1/e, about 0.368, every 20 seconds. That 20 seconds is the scale in the exponent: the fraction left after a time t is e raised to −t/(20 s), and t/(20 s) is a pure number.

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.

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