Foundation lesson

Rates of change and derivatives

Work out the average rate over shorter and shorter intervals. The value those averages settle on is the rate at the instant, and it is called the derivative.

Written for this edition, not translated from Einstein.

How can a body have a speed at a single instant?

A speed needs two readings of position and the time between them: divide the distance covered by the time taken, and you have the average speed over that interval. An instant has no duration, so that recipe cannot be used at an instant as it stands.

So shorten the interval and watch what happens. A ball rolls so that after t seconds it has gone 3t² metres: 3 m after one second, 12 m after two. Start at the one-second mark and average over shorter and shorter stretches:

Over the next second it goes from 3 m to 12 m: 9 m in 1 s, an average of 9 m/s.
Over the next tenth of a second it reaches 3.63 m: 0.63 m in 0.1 s, an average of 6.3 m/s.
Over the next hundredth it reaches 3.0603 m: 0.0603 m in 0.01 s, an average of 6.03 m/s.
Over the next thousandth it reaches 3.006003 m: 0.006003 m in 0.001 s, an average of 6.003 m/s.

The averages close in on 6 m/s. That number is the derivative: the ball's speed at the instant t = 1 s. The interval is never set to zero, which would mean dividing zero distance by zero time. It only shrinks, and the averages settle.

In symbols, with Δt for the short interval, the same recipe reads:

\frac{\mathrm{d} x}{\mathrm{d} t} = \lim_{\Delta t \to 0} \frac{x\left(t + \Delta t\right) - x\left(t\right)}{\Delta t}

d x by d t is the value that the change in position divided by the change in time settles on as delta t shrinks towards zero.

A derivative carries units: those of the quantity that changes, divided by those of the quantity you vary. Position in metres, varied over time in seconds, gives metres per second. On a graph of position against time, the derivative at a point is the steepness of the curve there.

Einstein's Brownian paper needs rates like this for a concentration of particles, which changes along a tube and in time at once. With two things changing, a rate has to say which one it follows, so the paper writes a curly ∂ and holds the other fixed:

\frac{\partial f}{\partial t} = D\,\frac{\partial^{2} f}{\partial x^{2}}

The partial derivative of f with respect to t equals D times the second partial derivative of f with respect to x.

The lesson on partial derivatives, listed at the end of this page, reads that equation one piece at a time.

Worked example: Why the averages settle on 6 m/s

The list of averages can be explained, not only watched. Start at any time t and let the interval be Δt. In that interval the ball moves 3(t + Δt)² − 3t² metres. Since (t + Δt)² is t² + 2tΔt + (Δt)², that distance is:

3\,\left(t + \Delta t\right)^{2} - 3\,t^{2} = 6\,t\,\Delta t + 3\,\left(\Delta t\right)^{2}

Three times t plus delta t, squared, minus three t squared, equals six t delta t plus three delta t squared.

Divide that distance by the time Δt to get the average speed over the interval:

\frac{6\,t\,\Delta t + 3\,\left(\Delta t\right)^{2}}{\Delta t} = 6\,t + 3\,\Delta t

Six t delta t plus three delta t squared, divided by delta t, equals six t plus three delta t.

The 3Δt term shrinks with the interval and leaves 6t metres per second. At t = 1 s that is 6 m/s. The extra 3, 0.3, 0.03 and 0.003 in the list above are that 3Δt term, for Δt of 1, 0.1, 0.01 and 0.001 seconds.

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.

The chord, and what it settles onto

The ball's position against time, with the chord over a whole second and the tangent at 1 second drawn through the same point.

The broken chord climbs 9 metres a second; the tangent climbs 6. The chord over a tenth of a second climbs 6.3, which at this size is already the tangent: no drawing that fits a page can separate them.

So the settling is plotted instead, one point for each interval the lesson averages over.

What it shows, in words

The first drawing is the ball's position against time, a curve rising more and more steeply. A point is marked at 1 second, 3 metres. The broken line through it meets the curve again a second later, at 12 metres, so it climbs 9 metres a second. The solid line through the same point is less steep and climbs 6: that is the tangent, and its steepness is the speed at the instant.

The second drawing plots the four averages against the interval each was taken over: 9, 6.3, 6.03, 6.003 metres per second for intervals of 1, 0.1, 0.01, 0.001 of a second. The first point sits well above the line at 6 and the other three lie on it. The interval is never set to zero; it only shrinks, and the averages settle.

What it does not show: the averages settle here because this curve is smooth at this instant. A curve with a corner would give one answer from the left and another from the right, and no tangent to draw, and nothing in either drawing rules that out. The lesson's worked example is what shows why these particular averages settle, by cancelling the 3Δt that the shrinking interval leaves behind.

Try it: nudge the frequency, watch the stopping voltage

In §8 of the light-quanta paper, Einstein predicts that the potential V needed to stop the electrons light frees from a metal rises in a straight line with the light's frequency ν, with the same slope for every metal. The derivative dV/dν is that slope: how many volts the stopping potential rises for each hertz of frequency.

Choose a frequency step Δν:

What the model computes

Baseline frequency (ν₀):: 6.00 × 10¹⁴ Hz
Frequency nudge (Δν):: +1.00 × 10¹³ Hz
Potential change (ΔV):: +0.04135668 V
Sensitivity ratio (ΔV / Δν):: 4.135667696 × 10⁻¹⁵ V·s (or V/Hz)
Modern h/e (2019 SI):: 4.135667696 × 10⁻¹⁵ V·s

Summary of frequency steps and sensitivity ratio
Nudge size	Δν (Hz)	ΔV (V)	Ratio ΔV / Δν (V·s)
+1.0 × 10¹⁴ Hz	1.0 × 10¹⁴	0.41357	4.135668 × 10⁻¹⁵
+5.0 × 10¹³ Hz	5.0 × 10¹³	0.20678	4.135668 × 10⁻¹⁵
+1.0 × 10¹³ Hz	1.0 × 10¹³	0.041357	4.135668 × 10⁻¹⁵
+2.0 × 10¹² Hz	2.0 × 10¹²	0.0082713	4.135668 × 10⁻¹⁵

What it shows, in words

A derivative carries units: volts divided by hertz gives volt-seconds. This model is Einstein's §8 straight line, with its slope set to the modern value of h/e, about 4.14 × 10⁻¹⁵ V·s. On a straight line every step size gives the same ratio ΔV / Δν, which is why the table repeats one number four times; on a curve the ratio would change with the step and settle only as Δν shrinks. The model was built with one slope for every metal, so it cannot show that real metals share one. Millikan measured that in 1916.

Where this lesson stops

A derivative is the value an average rate settles on as the interval shrinks, in units of one quantity per unit of the other. The interval shrinks but is never set to zero, so nothing is ever divided by zero.

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