The Light Clock and Time Dilation

Take a rod, put a mirror on each end, and send a flash of light bouncing between them. Count the round trips and you have a clock — the only clock in physics whose rate you can compute from first principles, because the one thing you know for certain about light is how fast it goes. Now slide the rod sideways. The far mirror moves on while the light is in flight, so the light no longer crosses straight: it zig-zags. And the one thing it cannot do is hurry. A longer road at a fixed speed takes longer, and the moving clock ticks slow. What turns this into physics is that the geometry is a right triangle. In half a tick the light must cover the gap $L$ across the rod while the clock carries it $v\,\Delta t/2$ along the track; its path is the hypotenuse, of length $c\,\Delta t/2$ , because light goes at $c$ for everybody. Three sides, one theorem, and out drops $\Delta t = \gamma\,\Delta t_0$ with $\gamma = 1/\sqrt{1-\beta^2}$ — the Lorentz factor, from nothing but the constancy of $c$ and Pythagoras. Take $L = 0.15\ \mathrm{m}$ and $\beta = 0.8$ and the triangle is literally the 3-4-5 of school geometry: legs $0.15$ and $0.20$ metres, hypotenuse $0.25$ , and a tick stretched from $1.0007$ to $1.6678$ nanoseconds. This article owns that derivation, and the step almost every textbook skips: the proof that the gap $L$ is the same in both frames, without which the argument is circular. The general statement of the effect and the experimental record belong to Time Dilation; the behaviour of $\gamma$ as a function belongs to The Lorentz Factor Gamma.

Key takeaways

A light clock is a photon bouncing between mirrors a gap $L$ apart. One tick is one round trip, lasting $\Delta t_0 = 2L/c$ in the clock's own frame — the proper time between "photon leaves the mirror" and "photon returns to it".
Watch it glide past at $v$ and the photon zig-zags. Each half-tick is a right triangle: legs $L$ and $v\,\Delta t/2$ , hypotenuse $c\,\Delta t/2$ — because the second postulate refuses to let light go faster.
Pythagoras gives $\Delta t = 2L/\sqrt{c^2-v^2} = \gamma\,\Delta t_0$ , with the lorentz factor $\gamma = 1/\sqrt{1-\beta^2}$ , $\beta = v/c$ . The gap $L$ cancels: the result cannot be about clocks.
The derivation needs one lemma — the transverse gap is unchanged. That is a theorem, not an assumption: a rod either scratches a ring or it does not, and both frames must agree.
At $\beta = 0.8$ a $15\ \mathrm{cm}$ clock has $\gamma = 5/3$ , an exact 3-4-5 triangle, and a tick of $1.6678\ \mathrm{ns}$ against $1.0007\ \mathrm{ns}$ at rest. A muon at $\beta = 0.995$ has $\gamma = 10.0125$ : its $2.197\ \mu\mathrm{s}$ lifetime reads $22.0\ \mu\mathrm{s}$ in the lab.
Key idea: light travels at $c$ in every frame, so when a light clock moves the photon's road grows while its speed does not — and the tick stretches by exactly the ratio of hypotenuse to vertical leg, which is $\gamma$ .

The core insight

The photon in a moving light clock must travel further per tick than the photon in a stationary one — a hypotenuse instead of a leg — and it is forbidden to compensate by going faster. The extra path is paid for in extra time, and the payment is the ratio of hypotenuse to leg, which the triangle fixes at $1/\sqrt{1-\beta^2} = \gamma$ . Nothing about mirrors or photons survives: $L$ cancels, and what is left is a statement about time.

Where this lives today

The light clock stopped being a thought experiment around 1960. An optical atomic clock is, functionally, the device in the punchline figure: a fixed length, a wave crossing it at $c$ , a count of cycles. The best of them — strontium lattice and aluminium-ion clocks — reach fractional frequency uncertainties near $10^{-18}$ , so they resolve $\gamma - 1 \approx 10^{-18}$ , corresponding to $\beta \approx 1.4\times10^{-9}$ : about $0.4\ \mathrm{m/s}$ , or walking speed. The same geometry runs unnoticed inside every accelerator — muons stored in the CERN $g\!-\!2$ ring at $\gamma \approx 29$ survive long enough to be measured only because their internal clock runs at one twenty-ninth of the laboratory's — and overhead, where each GPS satellite's atomic clock loses $7.2\ \mu\mathrm{s}$ per day to its orbital motion, computed from the same $\gamma$ that came out of a 3-4-5 triangle.

The physical picture

Everyone has thrown a ball straight up inside a moving car. From the back seat it goes up and comes down; from the roadside it traces a parabola — a longer path in the same time. No mystery: the ball picked up the car's forward velocity, so it travels further and faster, and the two extras cancel exactly. Now do it with light. The photon must still travel further, but it is flatly forbidden to travel faster: the second postulate says every inertial observer, whatever their motion, measures the same $c$ . The compensation that rescues the ball is unavailable, and the debt is settled in the only currency left.

Interactive Simulation

The Light Clock — a moving clock's photon travels a longer diagonal path, so it ticks slower (Δt = γΔτ). Set the clock's speed and compare tick counts.

Figure

The same clock twice. LEFT: at rest, the photon bounces straight up and back across the gap L, so a tick lasts Δt₀ = 2L/c — the proper time. RIGHT: the clock glides at v, sliding vΔt in one tick while the photon zig-zags. Half a tick is a right triangle — legs L and vΔt/2, hypotenuse cΔt/2 — because the photon still travels at exactly c.

Put a number on it. Take $L = 0.15\ \mathrm{m}$ — the clock then ticks once per nanosecond — and set it moving at $\beta = 0.8$ .

Figure

PUNCHLINE — the triangle with real numbers. A clock of gap L = 0.15 m ticks in Δt₀ = 1.0007 ns at rest. At β = 0.8 the half-tick triangle has legs 0.15 m and 0.20 m and hypotenuse 0.25 m: a 3-4-5 triangle, drawn to scale. The photon covers 0.50 m of zig-zag against 0.30 m of bounce — a ratio of 5/3 — so the tick stretches to 1.6678 ns. That 5/3 is γ.

Notice what the figure does not contain: any property of a mirror, a photon or a rod. The inputs are a length, a speed, and the refusal of light to be hurried.

The core equations

Standard configuration: $S'$ carries the clock and slides along the $x$ -axis of $S$ at speed $v$ , with $\beta \equiv v/c$ . The mirrors are separated along $y$ , perpendicular to the motion.

\Delta t_0 = \frac{2L}{c} \qquad (1)

where: $\Delta t_0$ — one tick as read by the clock itself; $L$ — the mirror gap; $c = 299\,792\,458\ \mathrm{m/s}$ . Physical reading: both tick events happen at the same place in the clock's own frame, so $\Delta t_0$ is a proper time — read off one clock, not inferred from two. In $S$ the clock slides $v\,\Delta t$ per tick, so the photon travels diagonally, and half a tick is a right triangle whose vertical leg is the gap and whose horizontal leg is half the slide.

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

where: $\Delta t$ — the same tick measured in $S$ ; $c\,\Delta t/2$ — the hypotenuse, the distance the photon covers in half a tick. Physical reading: this is where the second postulate enters, and it enters exactly once. The hypotenuse is $c\,\Delta t/2$ , not $\sqrt{c^2+v^2}\,\Delta t/2$ — the photon does not inherit the clock's sideways velocity.

\Delta t = \frac{2L}{\sqrt{c^{2}-v^{2}}} = \frac{2L/c}{\sqrt{1-\beta^{2}}} = \gamma\,\Delta t_0, \qquad \gamma \equiv \frac{1}{\sqrt{1-\beta^{2}}} \qquad (3)

where: $\gamma$ — the Lorentz factor, $\gamma \ge 1$ , with equality only at $\beta = 0$ . Physical reading: the gap $L$ has cancelled, so whatever the clock is made of, the moving tick exceeds the rest tick by the same universal factor — the first sign that this is a fact about time, not about clocks. One silent input was used, and it is a theorem, not a convention.

L'_{\perp} = L_{\perp} \qquad\text{(lengths perpendicular to the motion are unchanged)} \qquad (4)

where: $L_\perp,\ L'_\perp$ — a transverse length as measured in the two frames. Physical reading: without this, (2) has no right to put $L$ on the vertical leg — the lab's gap could be some unknown $f(\beta)L$ , and $f$ would contaminate the answer. Derivation B proves $f = 1$ . Finally, the same tick read as a spacetime interval; the two events are separated by $\Delta x = v\,\Delta t$ in $S$ and by nothing at all in the clock's frame.

c^{2}\Delta\tau^{2} = c^{2}\Delta t^{2} - \Delta x^{2} \quad\Longrightarrow\quad \Delta\tau = \frac{\Delta t}{\gamma} \qquad (5)

where: $\Delta\tau$ — the proper time between the events, here equal to $\Delta t_0$ . Physical reading: dilation is what the invariance of the interval looks like when one frame sits still with the clock. The Invariant Spacetime Interval develops it properly.

Where the results come from

Derivation AThe tick of a moving light clock — Pythagoras and nothing else

The goal is to obtain the Lorentz factor from the ground up, using only the constancy of $c$ and elementary geometry: set the clock at rest, compute its tick, then view the identical clock from a frame in which it moves transversely to its mirror axis. What emerges is not merely a formula for light clocks. The gap $L$ cancels, so the answer cannot depend on the clock's construction — and the last step turns that cancellation into an argument about every clock there is.

Step 1. In the clock's rest frame the photon goes straight up the gap $L$ and straight back, covering $2L$ at speed $c$ , so one tick lasts $\Delta t_0 = 2L/c$ . With $L = 0.15\ \mathrm{m}$ that is $1.0007\ \mathrm{ns}$ . Why this is a proper time and not merely "a time": the photon departs from and returns to the same mirror, so one clock is present at both events — which is the definition.

Step 2. Let the clock move at $v$ along $x$ . In one tick it slides $v\,\Delta t$ , so in the first half-tick the photon must reach a top mirror that has moved $v\,\Delta t/2$ along $x$ while staying $L$ above the bottom one. Its path is the hypotenuse of a right triangle with legs $L$ and $v\,\Delta t/2$ , and that hypotenuse is $c\,\Delta t/2$ long, because the photon travels at $c$ here too: $(c\Delta t/2)^2 = L^2 + (v\Delta t/2)^2$ . Why this is the entire physics: a projectile fired upward from the moving clock would trace a longer path too, but it would move faster — at $\sqrt{c^2+v^2}$ — and the effects would cancel. The postulate forbids that compensation for light.

Step 3. Collect terms: $\Delta t^2(c^2 - v^2) = 4L^2$ , so $\Delta t = 2L/\sqrt{c^2-v^2}$ , and dividing by $c$ exposes the rest tick, $\Delta t = (2L/c)/\sqrt{1-\beta^2} = \gamma\,\Delta t_0$ . Why only the positive root: $\Delta t$ runs from an emission to a later absorption.

Step 4. Read off what vanished. The gap $L$ cancels, so $\Delta t/\Delta t_0 = \gamma$ for every $L$ , and nothing about mirrors or reflectivity survives either. So if some other clock — a pendulum, a decaying muon, a beating heart — failed to dilate by the same $\gamma$ , you could strap it to a light clock and read your absolute velocity off the drift between them. Why this closes the argument: the principle of relativity forbids exactly that, so the light clock is a scaffold, not a mechanism — used because its rate is calculable, discarded because the answer proves independent of it. *Verified in verify_aq-sr-light-clock-time-dilation.py: sympy solves $(c\Delta t/2)^2 = L^2 + (v\Delta t/2)^2$ for $\Delta t$ , returns the single positive root $2L/\sqrt{c^2-v^2}$ , and confirms it equals $\gamma\cdot 2L/c$ identically in $v$ and $L$ .*

Go deeper · Derivation B · 2 minThe lemma everyone skips — why the transverse gap L does not change

Derivation BThe lemma everyone skips — why the transverse gap L does not change

Every textbook light clock puts $L$ on the vertical leg without comment. It has no right to. In the lab the gap is $f(\beta)L$ for some unknown $f$ , and carrying $f$ through turns equation (3) into $\Delta t = \gamma f(\beta)\Delta t_0$ — so the whole result hangs on $f = 1$ . You cannot borrow that from "contraction acts only along the motion", because contraction is normally derived using time dilation. That is a circle. The escape uses the principle of relativity plus one fact no observer can dispute.

Step 1. Frame $S$ holds a rigid ring of radius $L$ , centred on the $x$ -axis in the $y$ – $z$ plane, so the motion goes straight through the hole. Frame $S'$ , moving at $+v$ along $x$ , carries a rod of proper length $2L$ held perpendicular to the motion and aimed at the hole; each end is tipped with a scriber, and the rim is polished. Why a ring and a rod: the pairing converts a question about lengths — which is frame-relative — into a question about contact, which is not.

Step 2. Either the scribers touch the rim and leave a scratch, or they do not. A scratch is a mark on a physical object: afterwards it is there, for everyone, in every frame, forever. Why this is the pivot: relativity lets observers disagree about lengths, times and simultaneity, but never about the intersection of two worldlines.

Step 3. Suppose transverse lengths shrink, $f(\beta) < 1$ . Then $S$ says the moving rod is shorter than $2L$ and passes cleanly: no scratch. Now stand in $S'$ . There it is the ring that moves, at $-v$ , and relativity insists the same law applies — the same $f$ , since $f$ can depend only on $|\beta|$ . So $S'$ finds the ring's radius cut to $fL < L$ while the rod keeps its full proper half-length $L$ : the rod is too big for the hole, and it scratches. Why this finishes it: $S$ predicts no scratch, $S'$ predicts a scratch, and the scratch is absolute — so $f < 1$ is dead.

Step 4. Repeat with $f > 1$ and the identical contradiction appears with the roles swapped. The only survivor is $f(\beta) = 1$ : transverse lengths are identical in both frames, exactly. Equivalently, reciprocity demands $f = 1/f$ , whose only positive root is $1$ . Why this must come before the light clock: with $f = 1$ established independently, Step 2 of Derivation A is legitimate and $\gamma$ is no longer derived in a circle. Lengths <strong>along</strong> the motion get a different answer — that is Length Contraction's business. *Verified in verify_aq-sr-light-clock-time-dilation.py: sympy solves $f = 1/f$ over the positive reals and returns the unique root $f = 1$ ; the light-clock algebra is re-run with a general transverse factor $f$ , returning $\Delta t = \gamma f\,\Delta t_0$ .*

Go deeper · Derivation C · 1 minThe same tick from the invariant interval — a route with no light in it

Derivation CThe same tick from the invariant interval — a route with no light in it

A derivation that leans on a photon invites the suspicion that the answer is somehow about photons. It is not, and the cleanest way to show that is to reach the identical result along a road that never mentions light: the invariance of the spacetime interval $s^2 = c^2\Delta t^2 - \Delta x^2$ , which belongs to The Invariant Spacetime Interval. We borrow only its defining property — every inertial observer computes the same value for it.

Step 1. In the clock's own frame the two tick events are co-located, so $\Delta x' = 0$ and $s^2 = c^2\Delta\tau^2$ , with $\Delta\tau$ the clock's own reading. Why this frame is special without being privileged: it is the unique frame in which the events coincide in space — but the $s^2$ it computes is the value every other frame computes.

Step 2. In $S$ the events lie on the clock's worldline and the clock moves at constant $v$ , so $\Delta x = v\,\Delta t$ and $s^2 = c^2\Delta t^2(1-\beta^2)$ . Why the displacement is forced: given the velocity and the elapsed coordinate time, the spatial separation is not free.

Step 3. Equate the two — that is what invariant means — to get $\Delta t = \gamma\,\Delta\tau$ : the light clock's answer, with no mirrors anywhere. Why this is worth having: both routes draw on the postulates through different pipes, and their agreement establishes that the light clock smuggled in nothing optical. *Verified in verify_aq-sr-light-clock-time-dilation.py: sympy substitutes $\Delta x = v\Delta t$ into $c^2\Delta t^2 - \Delta x^2$ , equates it to $c^2\Delta\tau^2$ , solves for $\Delta t$ and returns $\gamma\,\Delta\tau$ .*

Limiting cases

For the $L = 0.15\ \mathrm{m}$ clock, whose proper tick is $\Delta t_0 = 1.0007\ \mathrm{ns}$ :

Case	Result	Meaning
At rest, $\beta = 0$	$\gamma = 1$ ; $\Delta t = 1.0007\ \mathrm{ns}$	the zig-zag degenerates to a straight line
Earth's orbit, $\beta = 9.94\times10^{-5}$	$\gamma - 1 = 4.9\times10^{-9}$ ; tick lengthens by $4.9\ \mathrm{as}$	dilation is quadratic in $\beta$ — five attoseconds on a nanosecond. This is why Newton lasted two centuries
$\beta = 0.6$	$\gamma = 1.25$ ; $\Delta t = 1.2509\ \mathrm{ns}$	the photon's road is $25\%$ longer
$\beta = 0.8$	$\gamma = 5/3$ ; $\Delta t = 1.6678\ \mathrm{ns}$	the 3-4-5 triangle: legs $0.15$ and $0.20\ \mathrm{m}$ , hypotenuse $0.25\ \mathrm{m}$
$\beta = 0.995$ (muon)	$\gamma = 10.0125$ ; $\Delta t = 10.019\ \mathrm{ns}$	the photon travels almost sideways; it barely climbs
$\beta \to 1$	$\gamma \to \infty$	the triangle grows infinitely long and infinitesimally tall; the clock never ticks

Common trap“It’s an artefact of using light to measure time — a mechanical clock would disagree, and then we’d know who is really moving.”

This objection is almost right about what the derivation shows and completely wrong about what it means. The argument as given establishes only that a light clock dilates. What rules out a wristwatch keeping stubbornly correct time is not a longer calculation but the principle of relativity. Suppose the wristwatch did not dilate: bolt it to the light clock, seal both in a windowless laboratory, and watch the readings drift apart. The drift rate would give you $\gamma$ , hence $|\beta|$ — your laboratory's absolute velocity, measured from the inside with the shutters closed, which is the one thing a century of experiment says cannot be done. So either every clock dilates by precisely the same $\gamma$ , or the principle of relativity is false. There is no third option. The light clock does not cause time dilation; it is merely the one clock whose rate can be computed in advance.

Worked examples

1 · The 3-4-5 light clock

A light clock has mirrors $L = 0.15\ \mathrm{m}$ apart, bouncing the photon along $y$ . It is carried past the laboratory at $\beta = 0.8$ along $x$ . Find the proper tick, the laboratory tick, the sides of the half-tick triangle, and the distance the photon actually travels in one tick.

Step 1. The proper tick, from (1): the photon covers $2L = 0.30\ \mathrm{m}$ at $c$ , so $\Delta t_0 = 0.30/(2.99792458\times10^{8}) = \boxed{1.0007\ \mathrm{ns}}$ — a clock ticking at very nearly one gigahertz.

Step 2. The Lorentz factor: $\gamma = 1/\sqrt{1 - 0.8^2} = 1/\sqrt{0.36} = \boxed{5/3 \approx 1.6667}$ . The laboratory tick, from (3), is $\Delta t = \gamma\,\Delta t_0 = \boxed{1.6678\ \mathrm{ns}}$ — a $67\%$ stretch.

Step 3. The triangle. Vertical leg $L = 0.15\ \mathrm{m}$ ; hypotenuse $c\,\Delta t/2 = \gamma L = \boxed{0.25\ \mathrm{m}}$ ; horizontal leg $v\,\Delta t/2 = \beta \times 0.25 = \boxed{0.20\ \mathrm{m}}$ . Check: $0.15^2 + 0.20^2 = 0.0625 = 0.25^2$ — exactly the 3-4-5 triangle, and not by accident: $\beta = 4/5$ makes the horizontal leg four-fifths of the hypotenuse, while $1/\gamma = 3/5$ makes the vertical leg three-fifths of it.

Step 4. The road travelled. In the laboratory the photon covers the whole zig-zag, $c\,\Delta t = \boxed{0.50\ \mathrm{m}}$ ; at rest, only the bounce, $2L = 0.30\ \mathrm{m}$ . The ratio is $0.50/0.30 = 5/3 = \gamma$ — necessarily, since both journeys are made at the same speed. The clock meanwhile slides $v\,\Delta t = \boxed{0.40\ \mathrm{m}}$ . Time dilation is the ratio of two path lengths.

2 · The muon as a light clock

A cosmic-ray muon has mean lifetime $\tau_0 = 2.197\ \mu\mathrm{s}$ in its own frame and crosses the laboratory at $\beta = 0.995$ . Treat it as a clock — an unstable particle is exactly that — and find its laboratory lifetime, the gap of the equivalent light clock, and how far it gets before decaying.

Step 1. $\gamma = 1/\sqrt{1 - 0.995^2} = 1/\sqrt{0.009975} = \boxed{10.0125}$ . Note how violently $\gamma$ responds near $\beta = 1$ : a mere $0.995$ already buys a factor of ten.

Step 2. Decay is a process at rest in the muon, so $\tau_0$ is a proper time and (3) applies directly: $\Delta t = \gamma\tau_0 = \boxed{22.0\ \mu\mathrm{s}}$ . In the laboratory this clock runs for twenty-two microseconds; on its own wristwatch, for two.

Step 3. The equivalent light clock. A light clock ticking once per $2.197\ \mu\mathrm{s}$ needs a gap $L = c\tau_0/2 = \boxed{329\ \mathrm{m}}$ . To relativity, a decaying muon is a pair of mirrors a third of a kilometre apart — and it dilates for the same reason, by the same factor.

Step 4. The distance covered. Without dilation the muon travels $v\tau_0 = \boxed{655\ \mathrm{m}}$ — barely off the top of a tall mountain. With dilation it travels $v\gamma\tau_0 = \boxed{6.56\ \mathrm{km}}$ , ten times as far, which is why cosmic-ray muons reach sea level in numbers no non-relativistic calculation can account for. That count has actually been made, on a mountain, and it belongs to Time Dilation.

Physics of the portrait

The light clock has a confusing pedigree. It is not in Einstein's 1905 paper — there the moving clock's slowing appears in §4, read straight off the transformation, with no mirrors in sight. It was invented four years later by two Americans, and made canonical half a century after that by a chapter of an introductory course at Caltech.

Origins · 2 minRichard Feynman and how this was discovered

Richard Feynman

1918–1988 · the light clock's canonical form, FLP I §15-4 (1963)

Feynman did not invent the light clock, and this card would be worthless if it said he had. Gilbert N. Lewis and Richard C. Tolman published the construction in 1909 (Proc. Amer. Acad. Arts Sci. 44, 711–724), reasoning about a light signal bouncing between mirrors in a moving system and reading the ratio $1/\sqrt{1-\beta^2}$ off the triangle; Langevin used the same device in 1911. What Feynman supplied, in The Feynman Lectures on Physics Vol. I, Ch. 15 §15-4 — "Transformation of time", published in 1963 — was its canonical form and its permanent place in the curriculum: a rod with a mirror at each end, a light signal ticking up and down it, and then the rod set moving so that the light zig-zags. The placement is itself an argument: §15-3 of the same chapter is the Michelson–Morley experiment, whose transverse arm is timed with this very triangle, so the reader meets the zig-zag first as an interferometer and then, a few pages later, as a clock — and the null result and time dilation stand revealed as two readings of one piece of geometry. Feynman shared the 1965 Nobel Prize in Physics for quantum electrodynamics. The lesson of §15-4 outlives the physics: when a principle is hard to believe, build the simplest machine that must obey it, and compute.

History

1887

— Michelson and Morley time light along an arm perpendicular to the ether wind: the same zig-zag, and the same right triangle, from which the light clock will later be built.

1897

— Larmor, treating electrons circling in a moving system, writes that they describe corresponding parts of their orbits in times shorter by the ratio √(1 − v²/c²) — the first appearance of time dilation in print. He repeats it in Aether and Matter (1900).

1905

— Einstein, "Zur Elektrodynamik bewegter Körper", Annalen der Physik 17, 891–921 (received 30 June, published 26 September). The slowing of a moving clock is derived in §4, from the transformation. There is no light clock in the paper.

1909

— Lewis and Tolman, "The Principle of Relativity, and Non-Newtonian Mechanics", Proc. Amer. Acad. Arts Sci. 44, 711–724: a light signal reflected between mirrors in a moving system, with 1/√(1 − β²) read off the geometry. The light clock's first appearance.

1911

— Langevin uses the same reflecting-mirror construction.

1941

— Rossi and Hall measure muon decay at Echo Lake (3240 m) and Denver (1616 m), Colorado: time dilation observed acting on a real clock.

1963

— Feynman, The Feynman Lectures on Physics Vol. I, Ch. 15 §15-4, "Transformation of time": the rod with a mirror at each end, and the light clock's arrival as the standard derivation.

1966

— Taylor and Wheeler, Spacetime Physics, make the interval and proper time the spine of the subject, with the light clock as the picture that motivates them.

Going further

Texts. Feynman, Leighton and Sands, The Feynman Lectures on Physics, Vol. I, Ch. 15 §15-4 · Taylor and Wheeler, Spacetime Physics, 2nd ed., Ch. 1 and 3 · Rindler, Introduction to Special Relativity, 2nd ed., §2 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.1.

Provenance. The Pythagorean derivation is standard (Taylor and Wheeler Ch. 1; Griffiths §12.1.2; Feynman I §15-4); the transverse lemma follows the ring-and-rod argument of Rindler §2. Every relation is re-derived and every number — ticks, triangle, muon figures, limiting cases, figure captions — recomputed in `verify_aq-sr-light-clock-time-dilation.py`, from $c = 299\,792\,458\ \mathrm{m/s}$ and $\tau_0 = 2.197\ \mu\mathrm{s}$ . Base note: the dry base gave the light clock no attribution at all, leaving the impression the construction is Einstein's — it is Lewis and Tolman's (1909) — and closed each derivation with a "cross-checked against [textbook]" line recording no computation. Both are replaced.

Builds on. The Two Postulates of Special Relativity · Einstein Clock Synchronization · The Michelson–Morley Experiment. Leads to. Time Dilation · The Lorentz Factor Gamma · Proper Time · Length Contraction · The Invariant Spacetime Interval.

If the photon must keep up with the moving mirrors, doesn't it need a sideways velocity — and doesn't that push its speed above c?

It does acquire a sideways component, and no, its total speed is still exactly $c$ — that is the whole content of the second postulate. Decompose the photon's lab-frame velocity along the hypotenuse: a component $v$ along $x$ , matching the clock, and a component $\sqrt{c^2 - v^2}$ along $y$ . Those add in quadrature to $c$ , not to something bigger. A thrown ball keeps its full upward speed and gains a sideways one; the photon cannot. It buys its sideways component by surrendering vertical speed — its climb rate falls from $c$ to $c/\gamma$ — and that reduced climb rate across the fixed gap $L$ is the dilation.

Isn't this circular? You use the constancy of c to derive time dilation, but "speed" already involves time.

It would be circular if the derivation had to establish what a metre and a second are. It does not. Each frame comes equipped in advance with rulers at rest in it and clocks at rest in it, synchronized within that frame by the Einstein convention — that is what an inertial frame is, and it is the business of Einstein Clock Synchronization. The second postulate is then a well-posed empirical claim: measured with those rulers and clocks, light goes at $c$ in every frame. The light clock takes that as input and derives a relation between the readings of two different frames' clocks.

Which clock is "really" slow? It is moving relative to me, but I am moving relative to it.

Both statements are true and neither is privileged. To say "your clock is slow" I must compare your single clock against two of mine — one where the tick starts, one where it ends — because your clock moved between the events. You do the same to me, using two of your clocks. Different pairs of events, different sets of clocks, no contradiction. What makes the books balance is that we disagree about whether the other's two clocks are synchronized. Time Dilation settles it with the nanoseconds accounted for on both sides.

What if I build the clock with the mirrors along the direction of motion instead of across it?

You get the same answer — but only if you also let the rod contract by $1/\gamma$ , which is exactly why the transverse clock is used for the derivation. Run the longitudinal case with an uncontracted rod of length $L$ : the round trip takes $(2L/c)/(1-\beta^2) = \gamma^2\Delta t_0$ , too big by a factor $\gamma$ . Let the rod shorten to $L/\gamma$ and it collapses to $\gamma\Delta t_0$ . So the longitudinal clock does not hand you $\gamma$ for free; it hands you a consistency condition between dilation and contraction, which belongs to Length Contraction.

Does the photon experience anything during all this? People say light "experiences no time".

Along a light ray the interval vanishes — $c^2\Delta t^2 - \Delta x^2 = 0$ — so equation (5) gives $\Delta\tau = 0$ , and the slogan says exactly that. It is harmless provided you do not promote it into a frame. There is no inertial frame moving at $c$ : setting $\beta = 1$ makes $\gamma$ infinite and the transformation singular, so "the photon's point of view" is not a place you can stand and ask questions from.

The effect is enormous for muons and invisible for aeroplanes. Where does it switch on?

It never switches on; it is always present, and it is quadratic. For small $\beta$ , $\gamma \approx 1 + \tfrac12\beta^2$ , so the fractional slowing is $\beta^2/2$ — half the square of a small number. At Earth's orbital speed ( $\beta \approx 10^{-4}$ ) that is $5\times10^{-9}$ : our $1.0007\ \mathrm{ns}$ clock loses five attoseconds a tick. At an airliner's $250\ \mathrm{m/s}$ it is $3.5\times10^{-13}$ ; at $\beta = 0.995$ it is a factor of ten. No threshold, only a curve — the subject of The Lorentz Factor Gamma.