Einstein and Feynman: The Man Who Bent Spacetime and the Man Who Counted Light

0. Key points

• Starting from a single line — “the speed of light is the same for everyone” — Einstein derived the conclusion that time itself runs at different rates for different observers. In this article we follow that derivation all the way through using nothing beyond the Pythagorean theorem.
• $E = mc^2$ is not a formula for atomic bombs. It is the claim that energy and mass are the same thing, and it can be derived from a single thought experiment: throwing light across a box.
• General relativity rewrote gravity as the curvature of spacetime. The effect is not a fantasy: it is measured every day in the form of a $38\ \mu\mathrm{s}$ daily drift of the clocks aboard GPS satellites.
• Feynman organized quantum electrodynamics (QED) into a recipe: add up the “arrows” for every possible path. The theory’s prediction for the strength of the electron’s magnetism agrees with experiment to twelve digits. It is the most precisely tested theory in physics.
• Feynman cracked safes, played the bongos, and at the hearings on the Space Shuttle disaster demonstrated the cause with a glass of ice water and a piece of rubber. His anecdotes are not merely entertaining; they are a working model of the scientific ethic “do not fool yourself.”

1. Motivation: two “peculiar” geniuses

Ask someone to name two physicists and most people will say Albert Einstein (1879–1955) and Richard P. Feynman (1918–1988). The old man sticking out his tongue in the photograph, and the bongo-playing man with the New York accent. Both are remembered less as “geniuses” than as eccentrics.

But these two are not famous because of their oddities. They are the people who, each almost single-handedly, built two of the foundations of modern physics: relativity and quantum electrodynamics. And they share something. Neither of them ever once accepted “because everyone says so” as a reason.

What Einstein doubted was an assumption that nobody had questioned in the two hundred years since Newton: that time flows at the same rate everywhere in the universe. What Feynman doubted was an assumption written into the quantum theory textbooks of his day: that a particle travels along a single path. Both men, having thrown away the assumption they doubted, found a far simpler and more beautiful world on the other side.

In this article we enjoy the anecdotes, but we also follow what these two actually calculated, with the formulas. Formulas will appear, but all we need is the Pythagorean theorem and division.

Tip

Stories about geniuses “being born different” are not very useful. What is worth noticing here is a single shared trait: neither man let a point of confusion sit unresolved until he was satisfied. Einstein did not leave the question that struck him at sixteen — “if I ran alongside a light beam, would the light look frozen?” — untouched for ten years.

2. Preliminaries: the strange fact about the speed of light

Let us set the stage.

Definition 2.1（Inertial frame）

An inertial frame is a coordinate system in which a body free of external forces appears to remain at rest, or to keep moving in a straight line at constant speed. The interior of a train moving at constant velocity, and the surface of the stationary ground (ignoring the effects of rotation), are both inertial frames. The interior of an accelerating train is not.

Everyday velocities add. Throw a ball forward at 10 km/h inside a train moving at 100 km/h and, to someone on the ground, the ball travels at 110 km/h. Yet experiments at the end of the nineteenth century — the Michelson–Morley experiment being the famous one — showed that this addition fails for light. The Earth orbits the Sun at about 30 km/s, and yet no difference was found between the speed of light travelling along the direction of that motion and the speed of light travelling transverse to it.

Einstein made a bold move: rather than trying to explain this experimental result, he accepted it as a fact and placed it at the foundation.

Axiom 2.2（The two principles of special relativity）

1. Principle of relativity: the laws of physics take the same form in every inertial frame. No experiment can determine which inertial frame is “truly at rest”.
2. Principle of the constancy of the speed of light: the speed of light in vacuum, $c$, is independent of the motion of the source and of the motion of the observer, taking the same value $c = 2.99792458 \times 10^8\ \mathrm{m/s}$ in every inertial frame.

Read naively, these two look contradictory. Light emitted forward from a rocket moving at $0.9c$ ought to travel at $1.9c$ as seen from the ground. Axiom 2.2 forbids that. There is only one way out of the contradiction: give ground on the “time” in speed = distance ÷ time.

3. Einstein (1): time stretches

3.1. A tool called the light clock

Let us build a thought-experiment device. Place two mirrors facing each other a distance $L$ apart and bounce light between them. Think of it as a clock that ticks once per round trip of the light. We call this a light clock.

A light clock at rest (left) and a light clock moving to the right with speed v (right). Inside the moving clock, the light must travel a longer, slanted path.

Theorem 3.1（Time dilation）

Consider a clock in uniform straight-line motion with speed $v$ (where $0 \le v < c$) relative to an inertial frame $S$. Let $\Delta \tau$ be the time interval between two events as measured in the inertial frame moving with the clock (the proper time; this is the same quantity as proper time（Definition 4.1）［Physics in Everyday Life］). Then the interval $\Delta t$ between the same two events measured in $S$ satisfies

$$\Delta t = \gamma\, \Delta \tau, \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

Since $\gamma \ge 1$ when $0 \le v < c$, we have $\Delta t \ge \Delta \tau$. That is, a moving clock appears to run slow.

Proof（Theorem 3.1）

We argue with the light clock, orienting the mirror separation $L$ perpendicular to the direction of motion.

(i) The frame moving with the clock. In this frame the light clock is at rest. The light travels a distance $L$ up and $L$ down, so the round trip takes

$$\Delta \tau = \frac{2L}{c}$$

Here we have used the constancy of the speed of light from Axiom 2.2, granting that light moves at $c$ in this frame too.

(ii) The frame $S$. Let $\Delta t$ be the duration of the same round trip. During that time the whole clock moves sideways by $v\,\Delta t$, so on the upward leg the light travels $v\,\Delta t/2$ horizontally and $L$ vertically along a slanted path. By the Pythagorean theorem, the upward leg has length

$$\sqrt{L^2 + \left(\frac{v\,\Delta t}{2}\right)^2}$$

The return leg has the same length, so the round-trip path is twice this. Now we use Axiom 2.2 again: in $S$ as well the light travels at $c$, hence

$$c\,\Delta t = 2\sqrt{L^2 + \left(\frac{v\,\Delta t}{2}\right)^2}.$$

(iii) Solving. Squaring both sides gives $c^2 (\Delta t)^2 = 4L^2 + v^2 (\Delta t)^2$, and rearranging for $\Delta t$,

$$(\Delta t)^2 = \frac{4L^2}{c^2 - v^2}, \qquad \Delta t = \frac{2L}{\sqrt{c^2 - v^2}} = \frac{2L}{c}\cdot\frac{1}{\sqrt{1 - v^2/c^2}}.$$

(In (iii) we used $c^2 - v^2 > 0$, which follows from $v < c$, and took the positive square root.)

Substituting $\Delta \tau = 2L/c$ from (i) yields $\Delta t = \gamma\, \Delta \tau$.

(iv) Does the same hold for clocks other than light clocks? Yes — for a spring-driven watch, and for a heartbeat, alike. If only the light clock ran slow while other clocks did not, then merely comparing the two clocks would let us decide whether we are “really moving”, in violation of the principle of relativity in Axiom 2.2. What runs slow is not the clock as a device; it is time itself.

∎

Let us look at the size of $\gamma$. At the speed of a bullet train, $v = 300\ \mathrm{km/h} \approx 83\ \mathrm{m/s}$, we have $v/c \approx 2.8 \times 10^{-7}$ and $\gamma - 1 \approx 3.8 \times 10^{-14}$. Ride for a full year ($3.2\times 10^7$ seconds) and the discrepancy barely reaches a microsecond. No wonder we never notice it. But at $v = 0.99c$ we get $\gamma \approx 7.1$, and at $v = 0.999c$, $\gamma \approx 22.4$: the growth becomes explosive as $v$ approaches $c$.

3.2. Evidence falling from the sky

Example 3.2（Why muons reach the ground）

Cosmic rays create particles called muons in the upper atmosphere, at an altitude of about $15\ \mathrm{km}$. A muon at rest has a mean lifetime of only $\tau_0 = 2.2\ \mu\mathrm{s} = 2.2\times 10^{-6}\ \mathrm{s}$.

The distance light covers in that lifetime is

$$c\,\tau_0 = (3.0\times 10^8)\times(2.2\times 10^{-6}) \approx 6.6\times 10^{2}\ \mathrm{m} = 0.66\ \mathrm{km}$$

which is only 4.4 % of $15\ \mathrm{km}$. In Newtonian mechanics, muons should never reach the ground.

Yet they are observed at the surface in large numbers. Let us apply Theorem 3.1. For a muon with $v = 0.995c$,

$$\gamma = \frac{1}{\sqrt{1 - 0.995^2}} = \frac{1}{\sqrt{1 - 0.990025}} = \frac{1}{\sqrt{0.009975}} = \frac{1}{0.09987} \approx 10.0$$

The lifetime seen by a ground observer becomes $\gamma\tau_0 \approx 22\ \mu\mathrm{s}$, and the distance covered is

$$0.995 \times (3.0\times 10^8) \times (22\times 10^{-6}) \approx 6.6\times 10^{3}\ \mathrm{m} = 6.6\ \mathrm{km}$$

stretched by exactly the factor $\gamma$. Real muons often have $\gamma$ in the tens or hundreds, and those pass through $15\ \mathrm{km}$ with room to spare.

Rossi and Hall confirmed this around 1940 by counting muons at the summit and at the base of Mount Washington (altitude about $1.9\ \mathrm{km}$). The fraction of the particles counted at the summit that survived to the base was far too large to explain without time dilation.

Remark 3.3

“If a moving clock runs slow, then from the other party’s point of view mine runs slow. Isn’t that a contradiction?” This is a good instinct, and the answer is: there is no contradiction. Theorem 3.1 compares “the time difference between two events as measured in some inertial frame”, and which events count as simultaneous differs from frame to frame, so both observers can say “the other one is running slow” without any clash. We treat this relativity of simultaneity in the Appendix.

4. Einstein (2): deriving $E = mc^2$ inside a box

In 1905, a few months after publishing the paper on special relativity, Einstein wrote a three-page follow-up. It contained the most famous formula in physics. That formula is not a blueprint for an atomic bomb. It is the claim that mass and energy are not separate things, but one quantity measured in two different units.

Theorem 4.1（Equivalence of mass and energy）

If a body at rest emits energy $E$, its mass decreases by

$$\Delta m = \frac{E}{c^2}$$

Conversely, a body of mass $m$ at rest possesses, by virtue of its existence alone, the energy $E = mc^2$.

Proof（Theorem 4.1）

We reproduce the “box and light” argument Einstein gave in 1906. Only two facts are needed.

• Fact A: light of energy $E$ carries momentum $p = E/c$ (a classical result of Maxwell’s electromagnetism).
• Fact B: in a system with no external forces, the centre of mass does not move (the same content as conservation of momentum in Newtonian mechanics).

(i) Setup. A closed box of mass $M$ and length $L$ sits at rest in vacuum. At time 0 the left wall emits light of energy $E$ to the right, and the right wall absorbs it. No force acts from outside.

(ii) Recoil of the box. By Fact A the light carries momentum $E/c$ to the right. Since the total momentum must remain 0, the box carries momentum $E/c$ to the left. Its speed is

$$v = \frac{E}{Mc}$$

(assuming $v \ll c$ and writing the Newtonian approximation).

(iii) How far the box shifts. The time for the light to cross the box is approximately $t = L/c$ when $v$ is small. During this time the box moves left by

$$\Delta x = v t = \frac{E}{Mc}\cdot\frac{L}{c} = \frac{EL}{Mc^2}$$

When the light is absorbed by the right wall the momentum is cancelled and the box stops, coming to rest displaced $\Delta x$ to the left of where it began.

(iv) Removing the contradiction. This violates Fact B. With no external force, the whole box (mass $M$) has moved left by $\Delta x$, so the centre of mass has shifted left. The only way to keep the centre of mass fixed is to suppose that the light carried a mass $m$ from the left wall to the right wall, i.e. a distance $L$ to the right. The condition that the centre of mass not move is

$$m \cdot L = M \cdot \Delta x$$

Substituting $\Delta x$ from (iii),

$$m L = M \cdot \frac{EL}{Mc^2} = \frac{EL}{c^2}, \qquad\text{hence}\qquad m = \frac{E}{c^2}.$$

The left wall lost energy $E$ and with it a mass $E/c^2$; the right wall received both. Extract energy from a body and its mass drops by exactly that amount.

∎

Example 4.2（How many tonnes does the Sun shed per second?）

The power radiated by the Sun (its luminosity) is $L_\odot = 3.85 \times 10^{26}\ \mathrm{W}$. By Theorem 4.1, the mass the Sun loses each second is

$$\frac{L_\odot}{c^2} = \frac{3.85\times10^{26}\ \mathrm{J/s}}{(3.00\times10^{8}\ \mathrm{m/s})^2} = \frac{3.85\times10^{26}}{8.99\times10^{16}} \approx 4.3\times 10^{9}\ \mathrm{kg/s}$$

That is 4.3 million tonnes every second, or $1.4\times10^{17}\ \mathrm{kg}$ per year.

The number looks staggering, but set against the Sun’s mass of $2.0\times10^{30}\ \mathrm{kg}$, shining for ten billion years ($3.2\times10^{17}$ seconds) costs only

$$\frac{4.3\times10^{9}\times 3.2\times10^{17}}{2.0\times10^{30}} \approx 6.9\times10^{-4}$$

that is, a mere $0.07\ \%$. What powers the Sun is the tiny mass difference released when hydrogen fuses into helium.

5. Einstein (3): gravity as the curvature of spacetime

Special relativity had a hole in it: it could handle neither accelerated frames nor gravity. In 1907 Einstein arrived at what he would later call “the happiest thought of my life”.

Axiom 5.1（The equivalence principle）

A laboratory at rest in a uniform gravitational field and a laboratory accelerating at a constant rate $g$ in gravity-free space cannot be distinguished by any mechanical experiment performed inside.

While an elevator is falling, the people inside are weightless. Conversely, inside an accelerating rocket in space they are pressed to the floor. Axiom 5.1 declares that these situations are not merely similar — they are the same. Out of this grew general relativity (1915), in which gravity is not a force but the geometric curvature of spacetime. Bodies are not pulled onto curved trajectories by gravity; they travel “straight” through a curved spacetime.

The first dramatic test of the theory came with the solar eclipse of 1919. Eddington’s expeditions measured the bending of starlight grazing the edge of the Sun and obtained a result close to the general-relativistic value, roughly twice the Newtonian prediction. The next day, Einstein was a worldwide celebrity.

Proposition 5.2（Difference in clock rates due to gravity (weak-field approximation)）

A clock at rest at a location with gravitational potential $\Phi$ runs, relative to a clock at rest at $\Phi = 0$ (infinity), at the rate

$$\frac{d\tau}{dt} \approx 1 + \frac{\Phi}{c^2} \qquad (|\Phi| \ll c^2)$$

Near the Earth’s surface $\Phi = -GM/r$, so a clock placed higher up (larger $r$) runs faster. When the height difference $h$ is small compared with the Earth’s radius, the fractional difference in rate is $gh/c^2$.

Proof（Proposition 5.2）

Send light of frequency $\nu$ upward between two points separated by a height $h$.

By Theorem 4.1, light of energy $E = h_{\mathrm{P}}\nu$ (where $h_{\mathrm{P}}$ is Planck’s constant; see the Planck relation（Definition 2.1）［Physics in Everyday Life］) effectively carries a mass $m = E/c^2$. If this “mass” climbs a height $h$ against gravity, it should lose energy equal to the gain in potential energy:

$$\Delta E = m g h = \frac{E}{c^2}\, g h .$$

Hence on arrival at the top the energy is $E' = E(1 - gh/c^2)$ and the frequency has dropped to

$$\frac{\nu'}{\nu} = 1 - \frac{gh}{c^2}$$

(gravitational redshift).

Here is the point. The lower clock emits light oscillating $\nu$ times per second, but only $\nu'$ oscillations arrive per second at the top. To the upper observer, “the lower clock ticks only $\nu' < \nu$ times per second” — that is, the lower clock runs slow. Equivalently, the upper clock runs faster by the fraction $gh/c^2$. In a uniform field we may write $\Phi = gh$, which matches the $\Phi/c^2$ in the statement.

(This argument estimates the energy loss of the light classically. It is a shortcut, and although it gives the right coefficient, strictly the result should be derived from the metric of general relativity. The 1959 experiment of Pound and Rebka in a Harvard tower of height $22.5\ \mathrm{m}$ confirmed the formula to good accuracy.)

∎

Example 5.3（By how many microseconds per day do GPS clocks drift?）

GPS satellites move on circular orbits at $r_s = 2.66\times10^{7}\ \mathrm{m}$ from the Earth’s centre, while the surface is at $r_E = 6.37\times10^{6}\ \mathrm{m}$. We use $GM = 3.986\times10^{14}\ \mathrm{m^3/s^2}$ for the Earth and $c^2 = 8.988\times10^{16}\ \mathrm{m^2/s^2}$.

(a) The gravitational effect (Proposition 5.2). The satellite is higher than the surface, so its clock runs fast. The fractional difference is

$$\frac{GM}{c^2}\left(\frac{1}{r_E} - \frac{1}{r_s}\right) = 4.435\times10^{-3}\times\left(1.570\times10^{-7} - 3.765\times10^{-8}\right)$$$$= 4.435\times10^{-3}\times 1.193\times10^{-7} = 5.29\times10^{-10}.$$

Multiplying by one day $= 86400\ \mathrm{s}$ gives $4.57\times10^{-5}\ \mathrm{s} = 45.7\ \mu\mathrm{s}$ gained.

(b) The velocity effect (Theorem 3.1). The orbital speed is $v = \sqrt{GM/r_s} = \sqrt{1.501\times10^{7}} = 3.87\times10^{3}\ \mathrm{m/s}$. Since $v \ll c$ we may approximate $\gamma \approx 1 + v^2/(2c^2)$, so the fractional slowing is

$$\frac{v^2}{2c^2} = \frac{1.501\times10^{7}}{2\times 8.988\times10^{16}} = 8.35\times10^{-11}.$$

Over one day this is $7.2\ \mu\mathrm{s}$ lost.

(c) Total. $45.7 - 7.2 = 38.5\ \mu\mathrm{s}$: the satellite clock gains about $38\ \mu\mathrm{s}$ per day. Without correction, the positional error would reach

$$c \times 38.5\times10^{-6}\ \mathrm{s} = 3.00\times10^{8}\times3.85\times10^{-5} \approx 1.2\times10^{4}\ \mathrm{m} = 12\ \mathrm{km}$$

per day. Your car navigation would be 12 km off after one day. Real GPS satellites therefore carry clocks whose frequencies are offset by exactly this amount from the start. Relativity is at work inside your phone every day (see also the difference in rate between clocks in orbit and on the ground（Theorem 4.2）［Physics in Everyday Life］ and the GPS numbers（Example 4.3）［Physics in Everyday Life］ in Physics in Everyday Life).

6. Feynman (1): the strange theory of light and electrons

6.1. The war against infinity

We move the scene to the 1940s. What tormented physicists then was a disease: any attempt to compute the interaction of electrons with light (the electromagnetic field) accurately in quantum mechanics gave the answer infinity. Computing the effect of an electron interacting with the field it creates itself made the result diverge.

Three people solved this: Julian Schwinger, Sin-Itiro Tomonaga, and Richard Feynman. They shared the 1965 Nobel Prize in Physics. The method is called renormalization. Crudely put, it is a procedure for pushing the infinities into the “bare, unobservable mass and charge of the electron” and rewriting the theory purely in terms of quantities that can be measured.

Schwinger’s and Tomonaga’s methods were mathematically rigorous, but the calculations were penance. What Feynman contributed was a way of calculating by drawing pictures.

6.2. Summing over all paths

Feynman’s starting point is the path integral idea he formulated in 1948.

Definition 6.1（Probability amplitude and the sum over paths）

In quantum mechanics, an event (say, “an electron leaving point $A$ arrives at point $B$”) is assigned a complex number $z$ called the probability amplitude. The observed probability is $|z|^2$ (the Born rule（Definition 2.2）［Schrödinger's Cat］).

When the event can happen along several routes, the total amplitude is the sum of the amplitudes of the individual routes:

$$z_{\text{total}} = z_1 + z_2 + \cdots, \qquad P = |z_{\text{total}}|^2 .$$

Think of a complex number as an “arrow” with a length and a direction. Adding amplitudes means laying arrows end to end.

That single point — adding arrows rather than probabilities — changes everything, because arrows pointing in opposite directions cancel.

Example 6.2（Reading the double slit as a sum of arrows）

Suppose an electron passes through two slits and arrives at a point $P$ on the screen. Write the amplitudes for each slit alone, normalized to equal length, as $z_1 = e^{i\varphi_1}$ and $z_2 = e^{i\varphi_2}$.

With one slit open the probabilities are $|z_1|^2 = 1$ and $|z_2|^2 = 1$, whose naive sum is 2. But with both slits open, Definition 6.1 gives

$$P = |e^{i\varphi_1} + e^{i\varphi_2}|^2 = (e^{i\varphi_1} + e^{i\varphi_2})(e^{-i\varphi_1} + e^{-i\varphi_2}) = 2 + e^{i(\varphi_1-\varphi_2)} + e^{-i(\varphi_1-\varphi_2)}$$$$= 2 + 2\cos(\varphi_1 - \varphi_2).$$

The phase difference $\varphi_1 - \varphi_2$ is fixed by the difference in the distances from the slits to $P$. If that difference is an integer number of wavelengths, $\cos = 1$ and $P = 4$; if it is a half wavelength, $\cos = -1$ and $P = 0$.

There are places brighter than the probability sum 2 (bright fringes) and places where nothing arrives at all (dark fringes). These are the interference fringes. The famous fact that asking “which slit did the electron go through?” destroys the fringes leads to the measurement problem（Remark 5.1）［Schrödinger's Cat］, treated in the article Schrödinger’s Cat.

6.3. Feynman diagrams

Feynman organized the innumerable “routes” contained in the interaction of electrons and photons into pictures made of lines and points.

The simplest Feynman diagram: two electrons scatter by exchanging a single photon. Straight lines are electrons, the wavy line is a photon, and the filled dots where lines meet are the interaction points (vertices) at which an electron emits or absorbs a photon.

The rules are these. Assign a fixed factor to each element of the diagram; multiplying them together gives the amplitude for that route. Then add up all the diagrams (Definition 6.1). Each vertex contributes a factor proportional to the coupling constant $e$.

Definition 6.3（The fine-structure constant）

The dimensionless constant measuring the strength of the electromagnetic interaction,

$$\alpha = \frac{e^2}{4\pi\varepsilon_0 \hbar c} = \frac{1}{137.036\ldots} \approx 0.0072973$$

is called the fine-structure constant. Every two additional vertices shrink a diagram’s contribution to the amplitude by roughly a factor of $\alpha$.

The fact that $\alpha$ is comfortably smaller than 1 is what makes QED a computable theory. The more complicated the diagram, the smaller its contribution, so we can add diagrams in order of simplicity and truncate at whatever precision we need. Feynman repeatedly called this constant a “magic number” that no theory can derive. Why 137 remains unsolved to this day (see Unsolved Problems in Physics).

6.4. Agreement to twelve digits

Let us see how accurate QED is, with concrete numbers. An electron behaves as a small magnet, and the strength of that magnet is measured by a quantity called the $g$ factor.

Proposition 6.4（The anomalous magnetic moment of the electron）

The Dirac equation (the relativistic equation for the electron) by itself predicts $g = 2$, that is, $g/2 = 1$. In QED, however, the contribution of diagrams in which the electron emits a photon and reabsorbs it gives

$$\frac{g}{2} = 1 + \frac{\alpha}{2\pi} + O(\alpha^2)$$

The $\alpha/(2\pi)$ term was computed by Schwinger in 1948.

Remark 6.5

Proving Proposition 6.4 requires a perturbative QED calculation (renormalization of a one-loop integral), which is beyond the scope of this article. For Feynman-diagram calculations, see Feynman’s QED: The Strange Theory of Light and Matter and the references. Here we only check the numbers.

Example 6.6（The most accurately predicted number in human history）

Let us evaluate the first correction in Proposition 6.4. Using $\alpha = 1/137.036$ from Definition 6.3,

$$\frac{\alpha}{2\pi} = \frac{1}{137.036 \times 6.28319} = \frac{1}{861.02} = 0.00116141 .$$

Hence $g/2 \approx 1.00116141$.

The measured value reported by Fan and collaborators in 2023 is

$$\frac{g}{2} = 1.00115965218059 \pm 0.00000000000013$$

With Schwinger’s single term we get $0.00116141$ against $0.00115965$: the leading three digits already agree. Starting from Dirac’s $g/2 = 1$, everything in $1.00116$ from the third decimal place onward is precisely the effect of the electron constantly emitting and reabsorbing photons.

Carrying the calculation through the $\alpha^2$, $\alpha^3$, $\alpha^4$ and $\alpha^5$ terms — thousands of Feynman diagrams — gives a theoretical value agreeing with the experiment above to twelve significant figures. Feynman likened this precision to measuring the distance from Los Angeles to New York to within the thickness of a single human hair.

7. Feynman (2): safes, bongos and O-rings

What made Feynman’s name known to the public was not QED but the anecdote collection Surely You’re Joking, Mr. Feynman!. We give just four. In each of them, beneath the joke, his view of science shows through.

Cracking safes. During the Second World War, Feynman was at Los Alamos working on the atomic bomb. To amuse himself he opened one colleague’s classified filing cabinet after another. He had no special skill. He had simply observed and tested a few things: that many people left the dial on the factory default, that mathematicians tend to use digits of $\pi$ or $e$ as combinations, and that the numbers can be inferred from the position of the dial before locking. This was an early statement of a point still valid today: the weakest part of security is human habit.

Bongos and drawing. While in Brazil, Feynman became absorbed in percussion, to the point of marching in a carnival parade. In his fifties, declaring that he would “prove a scientist can understand art”, he traded skills with a painter friend and improved to the point of selling drawings. For him, physics, percussion and drawing were the same activity: try it and see whether it works.

The Nobel telephone call. In 1965 the call announcing the prize came in the small hours. Feynman is said to have replied first that they should call back at a more reasonable time. He believed prizes and titles got in the way of research, and for a while considered declining.

O-rings and ice water. In 1986 he joined, despite his illness, the presidential commission investigating the explosion of the Space Shuttle Challenger. The cause was that the rubber O-rings sealing the joints of the solid rocket boosters had lost their elasticity in the low temperature on the day of launch (about $-0.6\ ^\circ\mathrm{C}$). At a televised hearing he dropped a sample of O-ring material into a glass of ice water and pulled it out again. The rubber stayed squeezed flat and did not spring back. That one minute of demonstration convinced the world more than hundreds of pages of report.

At the end of the personal appendix he attached to the accident report he wrote, in substance: nature must take precedence over public relations. For a technology to succeed, reality must prevail over appearances, because nature cannot be fooled.

Note

A warning Feynman often gave in lectures: “The first principle is that you must not fool yourself — and you are the easiest person to fool” (from his 1974 Caltech commencement address, “Cargo Cult Science”). Both the safecracking and the O-ring can be read as this principle in practice.

8. What the two men shared

Einstein and Feynman differed in temperament, in era, and in subject matter. What they had in common was that their standard for “understanding” was extraordinarily high.

Einstein knew Maxwell’s equations were correct, but he could not bear being unable to answer “what would I see if I ran alongside the light?” Feynman knew how to apply the calculational rules of quantum mechanics, but he was not satisfied until he could restate why those rules are what they are, in his own words.

Their other shared trait was that they converted answers into predictions. Einstein did not stop at “spacetime curves”; he produced a number, that starlight would be displaced by $1.75$ arcseconds during an eclipse. Feynman did not stop at “the electron emits and reabsorbs photons”; he produced a number for $g/2$ to the twelfth decimal place. Once you produce a number, you can be destroyed by being wrong. Making that bet is what physics is.

One difference: Einstein never accepted, throughout his life, the probabilistic interpretation of the quantum mechanics he himself had helped found. The line “God does not play dice”, from a letter to Born, is famous. Feynman, by contrast, accepted the probabilistic interpretation as his starting point and built his calculational method on top of it. This clash over determinism connects to the article Laplace’s Demon and Determinism (Laplace's demon（Definition 3.1）［Laplace's Demon and Determinism］ and Kennard's uncertainty relation（Theorem 5.1）［Laplace's Demon and Determinism］) and to the relation between information and physics discussed in Maxwell’s Demon (Landauer's principle（Theorem 6.4）［Maxwell's Demon］).

9. Exercises

Exercise 9.1Easy

Find the speed $v$ at which $\gamma = 2$. At that speed, how many months elapse aboard a spacecraft while one year passes on the ground?

Solution

From $\gamma = 1/\sqrt{1 - v^2/c^2} = 2$ we get $\sqrt{1 - v^2/c^2} = 1/2$; squaring both sides, $1 - v^2/c^2 = 1/4$, so $v^2/c^2 = 3/4$. Hence

$$v = \frac{\sqrt{3}}{2}c \approx 0.866\,c \approx 2.60\times10^{8}\ \mathrm{m/s} .$$

By Theorem 3.1, $\Delta t = \gamma\,\Delta\tau$, so for $\Delta t = 12$ months on the ground the proper time aboard is $\Delta\tau = \Delta t/\gamma = 6$ months.

Exercise 9.2Standard

Suppose a mass of $1\ \mathrm{g}$ were converted entirely into energy. How many joules would that be? How many tonnes of TNT is that equivalent to? Take one tonne of TNT to be $4.18\times10^{9}\ \mathrm{J}$.

Solution

By Theorem 4.1,

$$E = mc^2 = (1.00\times10^{-3}\ \mathrm{kg})\times(3.00\times10^{8}\ \mathrm{m/s})^2 = 1.00\times10^{-3}\times 9.00\times10^{16} = 9.00\times10^{13}\ \mathrm{J}.$$

In TNT equivalent,

$$\frac{9.00\times10^{13}}{4.18\times10^{9}} = 2.15\times10^{4}\ \text{tonnes} = 21.5\ \text{kilotons}$$

which is the same order of magnitude as the roughly 15 kilotons estimated for the yield of the bomb dropped on Hiroshima.

Note that in actual fission only about $0.1\ \%$ of the fuel is converted into energy, so obtaining a mass defect of $1\ \mathrm{g}$ requires several kilograms of nuclear fuel. It is not the case that “prepare $1\ \mathrm{g}$ of matter and you can extract this energy”.

Exercise 9.3Standard

By how many nanoseconds per day does a clock on the observation deck of the Tokyo Skytree (height $h = 450\ \mathrm{m}$) gain on a clock at ground level? Take $g = 9.8\ \mathrm{m/s^2}$, $c = 3.00\times10^{8}\ \mathrm{m/s}$, and one day $= 86400\ \mathrm{s}$.

Solution

By Proposition 5.2, the fractional difference in rate is

$$\frac{gh}{c^2} = \frac{9.8 \times 450}{9.00\times10^{16}} = \frac{4410}{9.00\times10^{16}} = 4.90\times10^{-14}.$$

Over one day,

$$4.90\times10^{-14} \times 86400 = 4.23\times10^{-9}\ \mathrm{s} \approx 4.2\ \mathrm{ns}$$

This has actually been measured. In 2020 the group of Hidetoshi Katori placed two transportable optical lattice clocks at the ground floor and the observation deck of the Skytree and directly detected this difference of order $10^{-14}$. We now live in an era in which general relativity can be tested not in space but at a tourist attraction in Tokyo.

Exercise 9.4Hard

The next correction after Proposition 6.4 can be estimated to be roughly of order $(\alpha/2\pi)^2$. Compute this quantity and state around which decimal place of $g/2$ it becomes relevant. Also explain what difficulties would arise in QED calculations if $\alpha$ were a constant larger than $1$.

Solution

Squaring $\alpha/(2\pi) = 1.161\times10^{-3}$ from Example 6.6,

$$\left(\frac{\alpha}{2\pi}\right)^2 = (1.161\times10^{-3})^2 = 1.35\times10^{-6}$$

This affects the sixth decimal place of $g/2 = 1.00115965\ldots$, i.e. around the underlined digit in $1.00115\underline{9}65$. In fact the coefficient of the $\alpha^2$ term is not $1$ but about $-0.328$ (as the coefficient of $(\alpha/\pi)^2$), so the contribution is somewhat smaller than $10^{-6}$; still, the order-of-magnitude estimate is sound.

If $\alpha > 1$, each pair of additional vertices would make the contribution larger. As stated just after Definition 6.3, QED calculations rely on “more complicated diagrams contribute less”, adding diagrams in order of simplicity and truncating partway (perturbation theory). With $\alpha > 1$ this series diverges, and no number of computed terms yields an approximation. Indeed, in the strong interaction (quantum chromodynamics) the coupling reaches order $1$ or more at low energies, which is why adding a handful of Feynman diagrams does not work there and other methods, such as numerical computation on a lattice, are required.

References

• A. Einstein, “Zur Elektrodynamik bewegter Körper”, Annalen der Physik 17 (1905), 891–921. — The original paper on special relativity. The content of Theorem 3.1 is in §4.
• A. Einstein, “Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?”, Annalen der Physik 18 (1905), 639–641. — The three-page paper deriving $E = mc^2$.
• R. P. Feynman, QED: The Strange Theory of Light and Matter, Princeton University Press, 1985. — A popular lecture series explaining QED as the addition of arrows, with almost no formulas. (Japanese translation: Hikari to Busshitsu no Fushigi na Riron, trans. Tsuneyoshi Kamae and Masako Onuki, Iwanami Gendai Bunko, in Japanese.)
• R. P. Feynman, Surely You’re Joking, Mr. Feynman!. — The safecracking, the Brazilian bongos and the Nobel Prize anecdotes are in this book. (Japanese translation: Gojōdan Deshō, Feynman-san, trans. Masako Onuki, Iwanami Gendai Bunko, in Japanese.)
• R. P. Feynman, R. B. Leighton, M. Sands, The Feynman Lectures on Physics. — Volume I (mechanics) contains the light clock and relativity; Volume III (quantum mechanics) contains the discussion of the double slit.
• M. Takamoto, I. Ushijima, N. Ohmae, et al., “Test of general relativity by a pair of transportable optical lattice clocks”, Nature Photonics 14 (2020), 411–415. DOI: 10.1038/s41566-020-0619-8 — The test of the gravitational redshift at the Tokyo Skytree.
• X. Fan, T. G. Myers, B. A. D. Sukra, G. Gabrielse, “Measurement of the Electron Magnetic Moment”, Physical Review Letters 130 (2023), 071801. DOI: 10.1103/PhysRevLett.130.071801 — The source of the measured $g/2$ used in Example 6.6.

Appendix: The relativity of simultaneity

Let us look more carefully at why “moving clocks run slow” leads to no contradiction even though the statement is symmetric.

From the middle of a train of length $2\ell$, fire light simultaneously toward the front and toward the rear. For a passenger on the train, the light travels a distance $\ell$ in both directions, so it reaches the front wall and the rear wall simultaneously.

But to someone standing on the platform, the train is moving forward. The rear wall approaches the light while the front wall runs away from it. By Axiom 2.2, the light travels at $c$ in both directions for the platform observer too. Therefore the light arrives at the rear wall first.

The same two events (light reaches the front wall / light reaches the rear wall) are simultaneous in one inertial frame and not simultaneous in another. This is the relativity of simultaneity.

This fact resolves the question raised in Remark 3.3. To “compare A’s clock with B’s clock” one must read both at the same time. But the meaning of “at the same time” differs from frame to frame, so when A says “B’s clock is slow” and B says “A’s clock is slow”, the two are simply comparing different pairs of events, and no direct contradiction arises.

A contradiction would genuinely be at stake only if the two clocks were brought to the same place and set side by side. That is the twin paradox. Reunite the twin who stayed on Earth with the twin who made a round trip by rocket, and the rocket twin is indeed younger. The symmetry is broken because the rocket twin accelerates in order to turn around, and so does not remain in a single inertial frame throughout. Theorem 3.1 applies only to comparisons between inertial frames, and the origin of the asymmetry lies precisely in the fact that the rocket twin switches inertial frames at the turnaround.