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University physics · Feynman Vol. I · Ch. 5

Time and Distance

This lesson follows chapter 5 of the Feynman Lectures, which deals with the two quantities on which the whole of mechanics rests, time and distance. The thread running through the chapter is that both are defined by the way we measure them, and measuring, nearly always, means counting something that repeats. We go from Galileo's pulse to the size of the atomic nucleus, by way of the day, the pendulum, atomic clocks and triangulation. The original chapter can be read free of charge on the Caltech website.

  1. 1Galileo and the pulse
  2. 2The day
  3. 3The pendulum
  4. 4Oscillators
  5. 5Decay
  6. 6The standard clock
  7. 7Triangulation
  8. 8The nucleus
STEP 1

How did Galileo time a ball without a clock?

The first bench has a 4 m track, tilted just enough for the ball to gain 0.5 m/s of speed every second, and no clock at all. The only way to know when the ball passes each point is to use one's own heart as the clock and scratch the track, wherever the ball happens to be, on every beat. The example comes from Feynman, who rebuilds Galileo's first inclined-plane experiments in this way, and the numbers are ours.

With a 0.8 s pulse, the marks land 0.16 m, 0.64 m, 1.44 m and 2.56 m from the start, and the fifth falls on the end of the track, at 4 m. Divided by the first, these distances give 1, 4, 9, 16 and 25, the squares of the beat number. The gaps between neighbouring marks grow as 1, 3, 5, 7, so that on each beat the ball covers a little more track than on the one before. Today we write this as \(d \propto t^2\), and the missing constant is half the acceleration.

Nobody's heart keeps metronome time, and the bench lets the pulse vary from one beat to the next. The marks scatter, and the first suffers most, because it rests on a single beat and serves as the yardstick for all the others. Even so, release after release, the mean ratio hovers around 1, and the law of squares is still there, only buried in the noise.

On the bench, distance is the easy part, since a tape measure laid along the track will do. Time is the hard part, because the ball takes 4 s to come down and the mechanical clocks of the period drifted by minutes over a day. In Feynman's account, Galileo's way out was his own body, and this improvised stopwatch is enough to replace treatises in the Aristotelian mould with a table of numbers.

It is measurement that turns talk about motion into physics, since a relation such as 1, 4, 9, 16 only shows up once there are numbers. From here on, describing a motion will mean answering two questions, which Feynman reduces to where and when, and the rest of the chapter is about answering them with ever greater precision.

\(d \propto t^2\)\(d = \tfrac12\,a\,t^2\)\(d\) is the distance along the track, measured from the starting point, \(t\) is the time since release and \(a\) is the acceleration along the track, 0.5 m/s² on the bench.

In Feynman: §5-1 Motion ↗

Let's discuss

  • Release the ball with the pulse at 0.80 s and no irregularity. On which beat does it reach the end of the track, and what is \(d/d_1\) at that mark?
  • Change the pulse to 0.50 s and release again. The distances change, but what about the \(d/d_1\) column?
  • Raise the irregularity to 20% and release several times. Why does the first mark spoil the table more than the others?
  • Galileo did not know how long a beat lasted in seconds. Did he need to know in order to find the law?
Marks on the track
Beat kd (m)d/d₁
Mean ratio d/(d₁k²)
STEP 2

How can we tell whether one day lasts as long as the next?

The two graphs on the bench cover a whole year in São Paulo, at 23° south. The upper one shows by how much a day counted from one sunrise to the next exceeds 86,400 s, and the lower one does the same sum from one noon to the next. By noon we mean the Sun crossing its highest point, not twelve o'clock on the clock face. In late December both sums give about 29 s extra, and in mid-September the first drops to 63 s short, while the second stays at 22 s short.

The attentive reader will spot a cheat, because speaking of 86,400 s presupposes a clock, and a clock is precisely what we do not yet have. To break the circle, time comes to be defined by the way we measure it, by picking something that repeats and counting the repetitions. The day is the natural candidate, and it is the day that the bench puts to the test.

For that, the bench relies on a second clock that does not depend on the Sun, a one-hour hourglass, the same device Feynman proposes for the job. What it counts are the turns accumulated between two events of the Sun, which assumes someone keeping watch over the glass without a break.

Counted from sunrise to sunrise, the number of turns does not hold steady. In São Paulo it ranges from 23.98 in September to 24.01 in January, and at 50° of latitude the range widens to 23.96 to 24.03. Which of the two clocks is wrong? With only these two, there is no way to tell. What changes at sunrise is the length of daylight, which grows or shrinks with the seasons and brings sunrise forward or back from one morning to the next. At the equator daylight lasts practically 12 hours all year round, and the two curves on the bench become one.

From noon to noon, the count stays between 23.99 and 24.01 at any latitude, a swing smaller than the error of a real hourglass, which would read it as a round 24 turns. This does not prove that the sand always runs at the same rate, nor that the Sun returns to its highest point at identical intervals, and Feynman takes care to point it out. What the bench shows is that the two clocks agree to within ±0.01 turn a day, and it is this agreement, not a proof, that we call regular time. With better clocks, which Zoom on noon imitates, noon itself starts to wander, by up to about 30 s from one day to the next. This is the equation of time. It has two causes, the Earth's elliptical orbit, travelled fastest in early January, and the tilt of the axis, and, built up over the months, it leaves a sundial up to about a quarter of an hour fast or slow.

\(\Delta t_{\text{sunrise}} = 86{,}400\ \text{s}\)\(+\ \delta_{\text{sunrise}}(n)\)\(\Delta t_{\text{noon}} = 86{,}400\ \text{s}\)\(+\ \delta_{\text{noon}}(n)\)\(\Delta t_{\text{sunrise}}\) is the interval between sunrise on day \(n\) and sunrise on the following day, \(\Delta t_{\text{noon}}\) is the interval between two solar noons, and \(n\) is the day of the year, counted from 1 January. The leftovers \(\delta\) are what the bench's graphs show, in seconds.

In Feynman: §5-2 Time ↗

Let's discuss

  • With the latitude at −23°, go through the year. In which months does the sunrise-to-sunrise day exceed 86,400 s, and what is daylight doing at that time?
  • Take the latitude to 0°. What is left of the difference between the two curves?
  • Swap −23° for +23°. In the sunrise curve, the part that comes from the length of daylight changes sign, and the part that comes from the equation of time stays put. Why does only the first depend on the hemisphere?
  • Switch on Zoom on noon. At what times of year is the solar day longest, and by how much?
Excess, sunrise to sunrise
Excess, noon to noon
Hourglass turns
STEP 3

Does a pendulum always keep the same time?

The bench releases two identical 0.994 m pendulums at the same moment, the left one pulled out to 5° and the right one to 60°. The first takes 2.001 s to go out and come back, the second takes 2.146 s, and the 0.145 s difference piles up with every oscillation. After some seven round trips, one is heading right while the other heads left.

The time of one full oscillation, the period, therefore depends on the size of the swing. For small swings the dependence is weak, and \(T \approx T_0(1 + \theta_0^2/16)\) holds, but it grows quickly with the amplitude. At 90° the period is already 18% longer than \(T_0\), and the approximation, which gives 15%, begins to lag behind. The exact calculation uses the arithmetic–geometric mean, which converges in a few iterations and draws the solid curve on the graph.

Between 5° and 10°, on the other hand, the period changes by less than 0.15%. In the limit of small swings it tends to \(T_0 = 2\pi\sqrt{L/g}\), which depends only on the length of the string and the acceleration due to gravity, and it is this limit that makes the pendulum a clock.

In Feynman's account, it was Galileo who turned this limit into an instrument, treating the short-swinging pendulum as a marker of equal intervals. The hourglass from the previous step lets us check the idea. The 0.994 m pendulum completes about 1,800 oscillations in each hour of sand, and trading a 3° swing for an 8° one changes that count by about two oscillations, probably less than a real hourglass can resolve. Since the day has already passed the test in step 2, the pendulum lets us split the day far more finely than into hours.

The second is the 86,400th part of the mean day, \(24 \times 60 \times 60\) pieces, and the 0.994 m pendulum is called a seconds pendulum because each one-way swing, from one extreme to the other, lasts 1 s. The period, which counts there and back, is 2 s, and it is easy to confuse the two.

In a pendulum clock, an escapement gives a small push on each swing and restores what friction takes away, so that the amplitude stays the same. The dependence on amplitude is still there, as a fixed factor that the clockmaker compensates for by adjusting the length. What would spoil the rate is a change of amplitude, because going from 5° to 10° slows the pendulum by about 5 s an hour, or 2 minutes a day.

\(T_0 = 2\pi\sqrt{L/g}\)\(T \approx T_0\left(1 + \tfrac{\theta_0^2}{16}\right)\)\(T = \dfrac{T_0}{\operatorname{AGM}(1, \cos\tfrac{\theta_0}{2})}\)\(T_0\) is the period for small swings, \(L\) is the length of the string, \(g\) is the acceleration due to gravity, 9.81 m/s² on the bench, \(\theta_0\) is the amplitude in radians and \(T\) is the period, there and back, at that amplitude. AGM is the arithmetic–geometric mean, the common limit reached by repeatedly taking the arithmetic and the geometric mean of two numbers.

In Feynman: §5-3 Short times ↗

Let's discuss

  • With the amplitude at 30°, release both together and follow the counts. After how many oscillations do they swing in opposite directions, and how long does that take?
  • Take the amplitude to 90°. How far off is the approximation \(1 + \theta_0^2/16\), compared with the exact curve?
  • Change the length to 0.25 m. What happens to \(T_0\), and what happens to the ratio \(T/T_0\)?
  • A pendulum clock swings at 10° or 15°, far from the small-swing limit. Why does it keep good time all the same?
T₀, small swings
T at the chosen amplitude
Delay per hour, against the 5° one
Full oscillations (5° / right)
STEP 4

How do we measure a nanosecond?

The bench's screen is that of an oscilloscope, an instrument that traces the current in a circuit against time and freezes, in a single picture, a run of oscillations far too quick to count by eye. At the top is a slow oscillator, in a window spanning exactly one of its periods, and at the bottom is the next rung of a ladder of oscillators, ten times faster. Counting the lower peaks inside the window gives 10, and that count is all it takes to calibrate the fast oscillator against the slow one.

Nine rungs take us from 1 s to 1 ns, and the first can be a 25 cm pendulum, checked against the seconds pendulum of the previous step. A 0.1 s pendulum would already call for a 2.5 mm string, and a 1 ms one for a string a quarter of a micrometre long, so the rungs that follow have to be electrical circuits, in which the current reverses direction every half period.

Each rung is calibrated only if the one above it already is. The count to 10 is repeated nine times, always against a slower oscillator that passed the same test, and the pendulum's second, at the top of the chain, ends up as the reference for every rung below it.

The ruler under the screen is logarithmic, and each step along it multiplies time by 10. A single line holds the life of a resonance, about \(10^{-23}\) s, the period of visible light, some \(2\cdot10^{-15}\) s, the heartbeat, the day, the year, a human life, the age of the Earth, \(1.4\cdot10^{17}\) s, and that of the universe, \(4.4\cdot10^{17}\) s. That makes some 41 orders of magnitude, and the ladder of oscillators covers only nine of them, those between the heartbeat and the nanosecond.

\(T_{k+1} = T_k/10\)\(N = T_{\text{slow}}/T_{\text{fast}}\)\(T_k\) is the period of rung \(k\) of the ladder, with \(T_0 = 1\) s, and \(N\) is the number of oscillations of the fast one that fit into one period of the slow one, 10 on every rung.

In Feynman: §5-3 Short times ↗

Let's discuss

  • On rung 1, count the lower peaks inside the window. Does the count change on rung 9?
  • How many oscillations of the fast one fit into 1 s on rung 6? And how do we reach that number with counts that never go beyond 10?
  • On the ruler, how many steps separate the day from the year? And a human life from the age of the Earth?
  • A 25 cm pendulum has a period of 1 s. How long would it have to be to oscillate with a period of 1 µs, and how does that compare with the size of an atom?
Period of the slow one
Period of the fast one
Fast peaks per slow period
Fast oscillations in 1 s
STEP 5

How do we date a piece of wood?

The grid on the bench holds 400 nuclei of a single radioactive isotope, each lit for as long as it has not decayed. None of them has an appointed hour, and at every instant the bench draws lots to decide which go out. Even so, the count passes close to 200 after one half-life, \(T\), close to 100 after \(2T\) and close to 50 after \(3T\), always hugging the theoretical exponential. The fluctuations around the curve grow, in relative terms, as fewer nuclei remain.

Every nucleus on the grid has the same chance of decaying per unit time, whatever its age, and so the sample as a whole loses, in each half-life, half of what it still has. Read backwards, the fraction remaining gives the time elapsed. With an eighth of the grid lit, three half-lives have gone by, with a thousandth, close to ten, and in general the ratio \(B/A\) between what is left and what there was at the start equals \(\left(\tfrac12\right)^{t/T}\), which is enough to read off the age \(t\).

This clock has no tick. There is nothing to count, only a fraction to measure, and it keeps running long after any human counter has gone home, which makes it useful over thousands or billions of years.

The practical difficulty is knowing how much there was at the start, and for carbon the atmosphere supplies the answer. Cosmic rays make carbon-14 out of nitrogen high in the atmosphere, and decay removes it, so the fraction of carbon-14 in the carbon dioxide of the air tends to stay roughly constant. A living tree exchanges carbon with the air and keeps that fraction, and once it dies the exchange stops and the clock starts to run. The half-life is 5,730 years, a value measured with more care than Feynman's round 5000, and the small variations of the fraction in the air over the millennia are corrected with calibration curves. These curves come from natural archives in which every year leaves its own trace, such as tree rings and the yearly layers of sediment at the bottom of certain lakes, counted one by one and also dated by carbon-14.

What limits the dating is the count. Each decay is a click in the detector, \(N\) clicks carry a statistical uncertainty of about \(\sqrt N\), and that turns into an age uncertainty of \(T/(\ln 2\,\sqrt N)\). A 12,000-year-old sample keeps 23% of its original carbon-14, and a detector that would register 10,000 decays from a fresh sample registers 2,342 in the same time, which gives ±171 years. At 60,000 years only 0.07% is left, the same measurement counts some 7 decays and is off by more than 3,000 years. In practice, background radiation and contamination by recent carbon put the limit at around 50,000 years, and longer times call for isotopes with longer half-lives, such as uranium-238, which dates rocks and the Earth itself.

\(\dfrac{B}{A} = \left(\tfrac12\right)^{t/T}\)\(t = T\,\log_2\dfrac{A}{B}\)\(\delta t \approx \dfrac{T}{\ln 2}\,\dfrac{1}{\sqrt{N}}\)\(A\) is the amount of radioactive material when the sample formed, \(B\) is the amount today, \(t\) is the age, \(T\) is the half-life, 5,730 years for carbon-14, \(N\) is the number of decays counted in the sample and \(\delta t\) is the age uncertainty that comes from the count alone.

In Feynman: §5-4 Long times ↗

Let's discuss

  • In the Half-life scenario, restart a few times and note how many nuclei remain at \(T\), \(2T\) and \(3T\). Which of the three numbers varies most from one run to the next, in proportion?
  • Decay serves as a clock without anything repeating. What plays the part of the repetition?
  • In the Dating scenario, set the age to 5,730 years and then to 11,460. What is \(B/A\) in each case?
  • Take the age to 20,000 years. From how many decays counted in a fresh sample does the uncertainty drop below 10% of the age?
B/A
Estimated age
Uncertainty
STEP 6

Which clock is the standard?

The bench follows six clocks for 100 years and plots how far each has run ahead of a chosen reference. Three are quartz clocks, with rate errors of order \(10^{-8}\), about 1 ms a day, two are atomic, with errors of \(10^{-13}\) and \(-2\cdot10^{-13}\), and the sixth is the Earth itself, which counts days. With the Earth as reference, the other five all take on the same curve, added to the straight line of each one's own error, and that curve reaches 31 s by the end of the century.

With the mean of the two atomic clocks as reference, it becomes clear where that curve comes from. The atomic clocks disagree by less than 1 ms in 100 years, each quartz follows a straight line with the slope of its error, and only the Earth bends, with the same 31 s lag by the end of the century. The lag is a tiny daily excess that piles up. On the bench, which starts with a day of exactly 86,400 s, that excess averages 0.85 ms over the century, and 36,525 days add up to 31 s.

The excess has two parts. The day grows slowly, by about 1.7 ms per century, mainly because tidal friction brakes the rotation, and it would grow by some 2.3 ms if the crust, which is still settling since the end of the last glaciation, did not offset part of it. On top of that trend, the day wobbles over the year by about 1 ms, mainly because the winds trade angular momentum with the solid part of the planet.

It remains to decide who is running slow, the Earth or everyone else, and there is no perfect clock to act as referee. The criterion, spelt out in Feynman, is to trust the clocks that agree with one another, and by it the one running slow is the Earth, the only one drifting away from a group that keeps together. An atomic clock counts the oscillations tied to a transition between two energy levels of an atom, a frequency set by the structure of the atom, which hardly changes with the surroundings of the instrument and is the same in every caesium-133 atom.

Before that, and for a long time, the standard was the Earth. Pulse, hourglass and wall clock disagree with one another with no clear reason to prefer any of them, the Earth's rotation is the same for everybody, and the second was defined as 1/86,400 of the mean solar day, an average over the year that wipes out the equation of time from step 2. Since 1967, the second has been defined by caesium-133, as 9,192,631,770 periods of the radiation emitted in the transition between the two hyperfine levels of its ground state. As the mean day is still slightly longer than 86,400 of these seconds, leap seconds, inserted since 1972, keep civil time close to the Earth's rotation, and the decision taken in 2022 is to abandon them by 2035.

\(1\ \text{s} = 9{,}192{,}631{,}770\)\(\text{periods of caesium-133}\)\(\Delta t_{\text{Earth}}(N\ \text{days}) = \sum_{d} \delta(d)\)\(\Delta t_{\text{Earth}}\) is how far the Earth lags, after \(N\) days, behind an ideal clock, and \(\delta(d)\) is how much day \(d\) exceeds 86,400 s, with one part that grows by 1.7 ms per century and another that oscillates over the year.

In Feynman: §5-5 Units and standards of time ↗

Let's discuss

  • With the Earth as reference, compare the curves of the two atomic clocks with those of the quartz clocks. What do they all have in common?
  • Switch to the mean of the atomic clocks. How far does the Earth lag in 100 years, and why is its curve not a straight line?
  • Switch on the zoom. What is the amplitude of the Earth's yearly wobble, and would it show up against a quartz of \(10^{-8}\)?
  • Choose quartz 1 as reference and raise the error to \(10^{-6}\). What would someone conclude who had only that clock?
Earth's lag in 100 years
Atomic disagreement in 100 years
STEP 7

How far away is a star?

In the Star scenario, the bench starts from a parallax of 0.77″, the value measured for Proxima Centauri, the smallest of the three stars in the Alpha Centauri system and the closest to the Sun. Seen six months apart, it shifts slightly against the far more distant stars of the background, because the Earth has moved to the other side of its orbit. The base of the triangle is the diameter of the orbit, 2 AU, about 300 million km, and the parallax is half the angle at the apex. With 0.77″, the star lies at 1.30 pc, or 4.2 light-years, some 270 thousand times the distance from the Earth to the Sun.

The parsec is the distance at which one astronomical unit, set perpendicular to the line of sight, spans one arcsecond, and it was chosen so that the distance in parsecs is the inverse of the parallax in arcseconds. What this simple sum hides is the error. A telescope measures angles with a more or less fixed error, whatever the distance of the star, and the relative error of the distance is that error divided by the parallax. With 0.01″ of error, Proxima comes out with 1.3%, a star at 10 pc with 10%, and at 100 pc, where the parallax is as large as the error itself, the measurement no longer says anything about the distance.

From the ground, through the atmosphere, getting down to 0.01″ is already hard work. The Hipparcos satellite, in the 1990s, measured parallaxes with errors near 0.001″. Gaia, which observed the sky from 2014 to 2025, reaches a few hundredths of a thousandth of an arcsecond on the brightest stars, which takes triangulation out to thousands of parsecs, still within our galaxy.

In the Satellite scenario, 1° of error in each sighting becomes a 3.8% error in the height, some 19 km out of 500. The same triangle shrunk to the size of a courtyard, with a 10 m base and a point 5 m up, can be checked with a tape measure, and the two readings coincide within those same 3.8%, or within whatever the tape and the protractor allow. This is what, for us, makes triangulation a measurement of distance, and not merely a calculation with angles: wherever a tape can be laid alongside the triangle, the two readings match. That match is a result of measurement, and it tells us that the angles of the triangles we are able to measure add up to 180°, a point Feynman also stresses.

The Satellite and Moon scenarios use two observers on the Earth's surface. A satellite 500 km up, seen from two stations 1,000 km apart, appears at wide angles, and even 1° of error in each sighting costs less than 4% in the height. For the Moon, at 384,000 km, a 10,000 km base leaves an apex angle of only 1.5°. An error of 0.1° already costs 14%, and from about 0.75° upwards the triangle may not even close.

The Sun stays off the bench because of a number. With the Earth's diameter as the base, the apex angle would be some 18″, on a disc half a degree wide with no sharp edges or fixed features to aim at. Today the distance to it comes from a clock rather than an angle: a radar pulse sent to Venus returns within a few minutes, and the round-trip time, measured with the oscillators and atomic clocks of steps 4 to 6 and multiplied by the speed of light, gives the distance in kilometres. That one distance is enough, because the positions of the planets and Kepler's laws already fix the proportions of the whole system, and before radar the same role fell to triangulating nearby asteroids, such as Eros, the two examples Feynman cites.

Beyond the reach of parallax, the way forward is comparison. If a star has the same colour and the same spectrum as others whose distance parallax has already given, it seems reasonable to assume that it emits the same power, and the flux reaching us falls with the square of the distance. The Standard candle scenario does this with a star just like the Sun. The error now comes from how closely the star resembles the model, and a 20% uncertainty in the power becomes 10% in the distance, near or far. With ever brighter candles, the same idea reaches the galaxies, and the most distant ones yet observed lie at distances of order \(10^{26}\) m.

\(d = \dfrac{b\,\sin\alpha\,\sin\beta}{\sin(\alpha+\beta)}\)\(d\,[\text{pc}] = \dfrac{1}{p\,['']}\)\(\dfrac{\delta d}{d} = \dfrac{\delta p}{p}\)\(F = \dfrac{L}{4\pi r^2}\)\(d\) is the distance from the target to the base, \(b\) is the length of the base, \(\alpha\) and \(\beta\) are the angles the two sightings make with it, \(p\) is the parallax in arcseconds, \(\delta p\) is the error in the angle and \(\delta d\) that in the distance. \(F\) is the flux arriving at a distance \(r\) from the star, in W/m², and \(L\) is the power it emits, \(3.8\cdot10^{26}\) W for the Sun.

In Feynman: §5-6 Large distances ↗

Let's discuss

  • In the Star scenario, with the parallax at 0.77″, what is the distance in parsecs and in light-years? And with a parallax ten times smaller?
  • With the angular error at 0.01″, from what parallax does the relative error exceed 100%, and what distance does that correspond to?
  • In the Moon scenario, raise the error until triangulation stops working. How does that error compare with the apex angle?
  • In the Standard candle scenario, divide the flux by 4. What happens to the distance, and to the relative error?
Distance
Apex angle
Relative error
STEP 8

How do we measure the size of a nucleus?

The bench shows a carbon plate face on, as a particle in the beam would meet it, with each nucleus drawn as a disc. The electrons are left out, since they are too light to stop a very fast particle, and what removes a particle from the beam is a collision with a nucleus. At true scale, the nuclei in 1 cm of carbon block only 2.5% of the area and would barely show. That is why the figure multiplies this area by a factor, ×20 with the initial values, which amounts to looking at a 20 cm plate. Each shot lands at a random point, and the counts refer to this equivalent plate, with the factor taken out when the radius is worked out.

The area a target blocks off, as the beam sees it, is called its cross section, \(\sigma\). In a thin layer, of thickness \(dx\), with \(n\) nuclei per unit volume, the discs cover a fraction \(n\sigma\,dx\) of the area, and that is the fraction of the beam the layer removes. Layer upon layer, the fraction that makes it through a thickness \(x\) falls as \(e^{-n\sigma x}\), the same exponential as in step 5, with thickness in place of time.

For carbon, \(n \approx 1.1\cdot10^{23}\) nuclei per cm³. If 97.5% of the beam gets through 1 cm, then \(n\sigma x = \ln(1/0.975) \approx 0.025\), \(\sigma \approx 2.3\cdot10^{-25}\) cm² and, treating the nucleus as a sphere, \(r = \sqrt{\sigma/\pi} \approx 2.7\) fm. One fm, or fermi, is \(10^{-15}\) m, and the empirical rule \(r \approx 1.2\,A^{1/3}\) fm, where \(A\) is the number of protons and neutrons, gives practically the same value for the 12 of carbon.

One might expect that, in a whole centimetre of carbon, the nuclei would be hiding one another, and that is not what happens. With 2.5% of the area covered, a nucleus on the back face has at most some 2.5% chance of having another in front of it, and on average, across the thickness, half that, so the simple sum, stopped fraction equal to \(n\sigma x\), is only slightly off. In 10 cm the summed coverage goes past 25%, the shadows begin to overlap, and the simple sum, which gives 25% of particles stopped, already departs from the 22% of the exponential.

Like the ladder of oscillators in step 4, the measurement of distance comes down in rungs, and each stretch of the ruler under the bench has its own instrument, the tape measure down to the millimetre, the optical microscope to about a micrometre, the electron microscope down to atoms, X-rays for the spacing in crystals, close to \(10^{-10}\) m, and, at the scale of the nucleus, only the particle counting of this bench.

From atom to nucleus there is still a jump of \(10^5\), the same that separates a 7 m room from a hair, and no image spans that jump. The size of the atom had an independent check, the volume of a mole of crystal divided by Avogadro's number, whereas for the nucleus what we measure is how much it gets in the way of a beam. That is why the bench counts particles rather than looking.

The ruler under the bench is logarithmic, like the time ruler in step 4, and it runs from the nucleus to the most distant galaxy yet observed, from \(10^{-15}\) m to \(10^{26}\) m. That makes some 41 orders of magnitude, nearly as many as on the time ruler, and each stretch of it needed its own recipe for distance, one we accept only because it matches the previous recipe where the two overlap.

\(\dfrac{N}{N_0} = e^{-n\sigma x}\)\(\sigma = \pi r^2\)\(n_{\text{C}} \approx 1.1\cdot10^{23}\ \text{cm}^{-3}\)\(N_0\) is the number of particles reaching the plate and \(N\) the number leaving the other side, \(n\) is the number of nuclei per unit volume, \(\sigma\) is the cross section of a nucleus, \(x\) is the thickness of the plate and \(r\) is the radius of the nucleus. \(n_{\text{C}}\) is the value of \(n\) for carbon, with a density of 2.2 g/cm³.

In Feynman: §5-7 Short distances ↗

Let's discuss

  • Fire 1,000 particles with the initial values. Does the counted fraction agree with the theoretical one, within the uncertainty? And how much would get through a real 1 cm plate?
  • Double the nuclear radius, from 2.7 to 5.4 fm. By how much is \(n\sigma x\) multiplied, and what happens to the figure's exaggeration factor?
  • Take the thickness to 10 cm and compare the fraction of stopped particles with \(n\sigma x\). How far off is the simple sum?
  • Fire 1,000 particles and then 3,000 more. What happens to the uncertainty of the deduced radius?
Transmitted, counted
Transmitted, theoretical
On the real plate
Deduced radius

Where we go next

Steps 5 and 8 ran into the same limit: a count of \(N\) random events fluctuates by about \(\sqrt N\), and it is that fluctuation that decides how much can be claimed about the age of a piece of wood or the radius of a nucleus. Feynman's next chapter, Probability, deals with these fluctuations and with what can be concluded in spite of them. It is also on the Caltech website, and the lesson in this track that goes with it is Probability.