Time Dilation

In 1963 two MIT physicists carried a scintillation detector to the summit of Mount Washington — $1917\ \mathrm{m}$ above sea level — counted muons for six runs, and found $563$ of them an hour. Then they drove the apparatus down to Cambridge, Massachusetts, a drop of $\Delta h = 1907\ \mathrm{m}$ , and counted again. These muons travel at $\beta \approx 0.995$ and live, on average, $2.197\ \mu\mathrm{s}$ ; the fall takes $6.393\ \mu\mathrm{s}$ , which is $2.91$ lifetimes. Nineteen out of twenty should have died on the way. About $412$ per hour arrived. They arrive because a moving clock runs slow — and a muon is a clock. Between "muon is created" and "muon decays" the muon is present at both events: it carries the proper time $\Delta\tau$ , and its own clock ticks off just $0.6385\ \mu\mathrm{s}$ during a fall the mountain says took $6.393$ . The relation is $\Delta t = \gamma\,\Delta\tau$ with $\gamma = 10.0125$ , and it follows in two lines from the Lorentz transformation: set $\Delta x' = 0$ — both events at the same place in the muon's frame — and the factor drops out. The Light Clock and Time Dilation gets the same $\gamma$ from a triangle; that article owns the derivation and we do not repeat it. Two statements about proper time are routinely muddled, and they are different. First: between two events on a single clock's worldline, that clock's own reading is the smallest elapsed time any inertial frame assigns to the pair. Second: among all timelike worldlines joining two fixed events, the inertial one carries the largest proper time. The first is this article's business; the second is the maximal-proper-time theorem, and it belongs to Proper Time. This article owns $\Delta t = \gamma\,\Delta\tau$ from the transformation, the resolution of the reciprocity puzzle, and the experimental record.

Key takeaways

Time dilation is $\Delta t = \gamma\,\Delta\tau$ : the proper time $\Delta\tau$ read on a single clock present at both events is the smallest elapsed time any inertial frame assigns to that pair.
It falls out of the inverse Lorentz transformation in two lines. Put both events at the same place in the clock's frame, $\Delta x' = 0$ , and $\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)$ collapses to $\Delta t = \gamma\,\Delta\tau$ . No light needed.
Reciprocity is not a contradiction. Each frame finds the other's clock slow by the same $\gamma$ , because the comparison is not symmetric: one side uses one clock, the other uses two — and they disagree about whether that pair is synchronized.
The muon is the classic clock. Frisch and Smith (1963): $563$ /hr at the Mount Washington summit, $\Delta h = 1907\ \mathrm{m}$ , $\beta = 0.995$ , $\gamma = 10.0125$ . Without dilation $\sim\!31$ /hr should survive; with it, $\sim\!421$ /hr. About $412$ /hr were counted.
The effect is engineering, not philosophy: an atomic clock on a GPS satellite ( $v = 3.874\ \mathrm{km/s}$ ) loses $7.2\ \mu\mathrm{s}$ per day to its motion.
Key idea: proper time is what a clock reads along its own worldline, and it is frame-independent; coordinate time is what a lattice of synchronized clocks reads, and it is not. Time dilation is the exchange rate between them, $\Delta t = \gamma\,\Delta\tau$ .

The core insight

Two events happen at the same place in one frame — the muon is born, the muon decays, both on the muon. That frame is special not because it is preferred but because a single clock is present at both events, and its reading is the proper time $\Delta\tau$ . Every other frame must use two separated clocks to time the same pair, and gets a larger number, $\Delta t = \gamma\,\Delta\tau$ . The asymmetry between one clock and two is not an artefact of the apparatus: it is time dilation.

Where this lives today

The muon is still working. Its dilated lifetime is what makes muon storage rings possible at all — the $g\!-\!2$ measurements at CERN and Fermilab run muons at $\gamma \approx 29$ , so that $2.197\ \mu\mathrm{s}$ becomes $64\ \mu\mathrm{s}$ of laboratory time, long enough to watch the spin precess thousands of times. Every collider's beam-lifetime budget is written in $\gamma\tau_0$ . Overhead, the correction is continuous. Each GPS satellite's caesium standard is deliberately detuned before launch — set to $10.229\,999\,995\,43\ \mathrm{MHz}$ rather than $10.23$ , a fractional offset of $-4.4647\times10^{-10}$ — so that in orbit, with $-7.2\ \mu\mathrm{s/day}$ of motion and $+45.8$ of gravity applied, it ticks at the rate the ground expects. In the laboratory the bar keeps rising: optical clock comparisons now resolve fractional frequency differences near $10^{-18}$ , enough to see the dilation of a clock moving at walking pace across a bench.

The physical picture

Ask what it takes to time a moving object. If the object carries a clock and both events happen on it, one clock suffices and you read its face twice. If the object is moving past you, the two events happen at different places in your laboratory, so you need a clock at each place — and those two clocks must be synchronized before the measurement means anything. That is where the asymmetry hides. Synchronization is not frame-independent: the moving object's frame will insist your two clocks were never synchronized, and the amount by which it says they are out is exactly the amount by which its own clock appears to lag. The books balance, in both directions, to the last nanosecond.

Figure

One clock or two — the asymmetry that IS time dilation. LEFT: in the clock's own frame S′ both events happen at the same place, so ONE clock reads both and gives the proper time Δτ. RIGHT: in the lab S the clock has moved, so the events must be timed by TWO separated clocks — and S′ says that pair is out of synchrony by vL/c². The lab time is Δt = γΔτ.

Now watch it happen. The muon experiment is the cleanest demonstration in physics, because the classical prediction and the relativistic one differ not by a percent but by more than a factor of thirteen.

Figure

PUNCHLINE — Frisch and Smith, 1963. 563 muons per hour at the Mount Washington summit. The 1907 m fall takes 6.393 μs in the lab — 2.91 muon lifetimes — so with NO dilation only e^(−2.91) = 5.4 % survive, about 31 per hour. With dilation the muon's own clock advances just 0.6385 μs = 0.291 lifetimes, so 74.8 % survive: about 421 per hour. About 412 were counted. Without dilation 95 % of these muons should have died on the way down. They didn't.

The core equations

Standard configuration: $S'$ moves at $+v$ along the $x$ -axis of $S$ , $\beta \equiv v/c$ , $\gamma \equiv 1/\sqrt{1-\beta^2}$ . The clock of interest is at rest in $S'$ .

t = \gamma\left(t' + \frac{v x'}{c^{2}}\right), \qquad x = \gamma\left(x' + v t'\right) \qquad (1)

where: $(t', x')$ — an event's coordinates in the clock's frame; $(t, x)$ — the same event in the lab. Physical reading: this is the inverse Lorentz transformation, the $v \to -v$ image of the forward map. Everything below is read off it.

\Delta t = \gamma\,\Delta\tau, \qquad \gamma = \frac{1}{\sqrt{1-\beta^{2}}} \ge 1 \qquad (2)

where: $\Delta\tau$ — the proper time, the interval read on a single clock present at both events; $\Delta t$ — the coordinate time between them in a frame where that clock moves at $v$ . Physical reading: the condition $\Delta x' = 0$ is not decoration — it is the definition of $\Delta\tau$ , and (2) is false without it. If the events are at different places in the primed frame you must use the full transformation (1), which mixes space into time.

\mathrm{d}\tau = \frac{\mathrm{d}t}{\gamma}, \qquad \tau = \int \sqrt{1 - \beta^{2}(t)}\ \mathrm{d}t \qquad (3)

where: $\beta(t)$ — the clock's instantaneous speed, which need not be constant. Physical reading: proper time is an integral along a worldline — a property of the path, not of the endpoints. That integral, and the theorem that the straight path maximizes it, belong to Proper Time.

N(t) = N_0\,e^{-t/(\gamma\tau_0)} = N_0\,e^{-\Delta\tau/\tau_0} \qquad (4)

where: $\tau_0$ — the mean lifetime in the particle's rest frame ( $2.197\ \mu\mathrm{s}$ for the muon); $N_0$ — the number at the start. Physical reading: the two forms are one statement. Decay runs on the particle's own clock, so survival depends only on the proper time elapsed — which is what makes an unstable particle a usable clock and the muon experiment a clean measurement of $\gamma$ .

\Delta T = \frac{v L_0}{c^{2}} \qquad\text{(leading clocks LAG)} \qquad (5)

where: $L_0$ — the proper distance between two clocks that are synchronized in their own frame; $\Delta T$ — the offset another frame attributes to them. Physical reading: there is no $\gamma$ here — the $\gamma$ from the time transformation and the $1/\gamma$ from the contracted separation cancel. This offset is the bookkeeping entry that makes reciprocity consistent; its full treatment belongs to Relativity of Simultaneity.

Where the results come from

Derivation AΔt = γΔτ, straight from the transformation

The light-clock route — The Light Clock and Time Dilation — obtains $\gamma$ from a right triangle and the constancy of $c$ . This card obtains the identical result from the Lorentz transformation, with no photons anywhere. The point is not economy but generality: the light clock proves the effect for one particular device, whereas the transformation is a statement about coordinates, so what follows applies to any clock whatever and to any pair of events on one worldline. The whole derivation turns on one condition: $\Delta x' = 0$ .

Step 1. Let the clock sit at rest in $S'$ and let the two events be two readings of its face — a tick and the next tick, or a creation and a decay. Because the clock does not move in $S'$ , the events occur at the same $x'$ : $\Delta x' = 0$ , and the elapsed $\Delta t'$ is the proper time $\Delta\tau$ . Why this step carries all the physics: "proper time" is not a new quantity but a coordinate time evaluated in the one frame where the events are co-located — and that frame exists precisely because both lie on a single timelike worldline.

Step 2. Take differences of the first equation in (1): $\Delta t = \gamma\left(\Delta t' + v\,\Delta x'/c^2\right)$ . The transformation is linear, so it acts on differences exactly as on coordinates. Why linearity matters: were the map non-linear, coordinate differences would depend on where you put the origin, and $\Delta t$ would not be well defined.

Step 3. Impose $\Delta x' = 0$ . The second term dies and $\Delta t = \gamma\,\Delta t' = \gamma\,\Delta\tau$ , with $\gamma \ge 1$ . The lab's elapsed time is the larger number; the clock's own reading is the smaller. Why the inequality runs this way: $\gamma \ge 1$ always, with equality only at $\beta = 0$ , so no inertial frame ever assigns the pair a time smaller than the moving clock's own. Between two events on one worldline, proper time is the minimum. *The algebra can be checked directly: applying the inverse Lorentz transformation to a coordinate difference and substituting $\Delta x' = 0$ returns $\Delta t = \gamma\Delta t'$ identically in $\beta$ ; the inequality $\gamma \ge 1$ holds with equality if and only if $\beta = 0$ , and the same $\gamma$ emerges from the light-clock triangle by a wholly independent route.*

Go deeper · Derivation B · 2 minThe reciprocity theorem — why "each finds the other slow" is not a contradiction

Derivation BThe reciprocity theorem — why "each finds the other slow" is not a contradiction

Here is the objection that has felled generations of students. If $A$ finds $B$ 's clock slow, and $B$ finds $A$ 's slow by the same factor, surely one of them is wrong. Neither is, and the resolution is not a verbal trick: the two statements refer to different measurements, and done honestly the two accounts of the same physical facts agree to the last nanosecond. Comparing clocks across a distance requires a synchronization convention, and the two frames do not share one.

Step 1. $A$ (frame $S$ ) owns two clocks at rest, at $x = 0$ and $x = L$ , synchronized in $S$ . $B$ carries one clock and flies past at $+v$ . Event 1: $B$ passes $A$ 's near clock; set all readings to zero. Event 2: $B$ passes $A$ 's far clock. Why one clock against two: this is forced, not chosen — $B$ 's clock is at two different places in $S$ , so $S$ has no alternative but to use a clock at each.

Step 2. $A$ 's account. The far clock is reached after $\Delta t = L/v$ ; by Derivation A, $B$ 's single clock reads $\Delta\tau = \Delta t/\gamma$ . So $A$ says: " $B$ 's clock ran slow by $\gamma$ ." With $L = 100\ \mathrm{m}$ , $\beta = 0.6$ : $\Delta t = 555.94\ \mathrm{ns}$ , $\Delta\tau = 444.75\ \mathrm{ns}$ . Why this is a legitimate measurement: both readings are direct local comparisons — no signals, no delays.

Step 3. $B$ 's account of the same two events. In $B$ 's frame $A$ 's clocks are the moving ones, so each runs slow by $\gamma$ : over $B$ 's own $444.75\ \mathrm{ns}$ the far clock advances only $444.75/1.25 = 355.80\ \mathrm{ns}$ . But $B$ also says $A$ 's two clocks were never synchronized — the far one was already ahead by $vL/c^2 = 200.14\ \mathrm{ns}$ at event 1 (equation 5). Why the offset has this sign: $A$ 's clocks move in the $-x'$ direction in $B$ 's frame, so the rearmost of them in $A$ 's own motion — the one at $x = L$ — leads.

Step 4. Add up $B$ 's books. The far clock's reading at event 2 is $200.14 + 355.80 = 555.94\ \mathrm{ns}$ — precisely the $\Delta t$ that $A$ reported. Symbolically, $\dfrac{vL}{c^{2}} + \dfrac{\Delta\tau}{\gamma} = \dfrac{vL}{c^{2}} + \dfrac{L}{v}\left(1-\beta^{2}\right) = \dfrac{L}{v} = \Delta t$ , exactly and for every $\beta$ . Why this settles it: both observers agree about every local clock reading — the only things anyone actually observes — while disagreeing about rates and about synchrony, and those two disagreements are exactly the same size and cancel. The comparison was never symmetric: one side used one clock, the other used two. *The identity $vL/c^2 + (L/v)(1-\beta^2) = L/v$ holds for all $\beta$ , and the $100\ \mathrm{m}$ , $\beta = 0.6$ ledger confirms it numerically: $\Delta t = 555.94\ \mathrm{ns}$ , $\Delta\tau = 444.75\ \mathrm{ns}$ , offset $200.14\ \mathrm{ns}$ , far-clock advance $355.80\ \mathrm{ns}$ , and the sum $200.14 + 355.80 = 555.94\ \mathrm{ns}$ — balancing in both directions.*

Go deeper · Derivation C · 1 minWhy a decaying particle is a clock — survival depends only on proper time

Derivation CWhy a decaying particle is a clock — survival depends only on proper time

The muon experiment is usually told as though the dilated lifetime $\gamma\tau_0$ were the whole story. The missing step is worth a card, because it licenses the entire experimental record. Decay is a Poisson process with a rest-frame mean life $\tau_0$ , so survival after a proper time $\Delta\tau$ is $e^{-\Delta\tau/\tau_0}$ . What must be shown is that this — and not the coordinate time — governs the counts, so that the ratio of the two counts measures $\gamma$ .

Step 1. In the particle's rest frame nothing is moving and there is no relativity to do: survival after its own elapsed time $\Delta\tau$ is $S = e^{-\Delta\tau/\tau_0}$ , with $\tau_0 = 2.197\ \mu\mathrm{s}$ for a muon. Why state the decay law there: $\tau_0$ is a property of the muon — of the weak interaction that destroys it — not of anybody's coordinates.

Step 2. Translate to the lab. The flight lasts $\Delta t$ there, and by Derivation A the particle's own clock advances only $\Delta\tau = \Delta t/\gamma$ , so $S = e^{-\Delta t/(\gamma\tau_0)}$ — equation (4), and the statement that the lab-frame mean lifetime is $\gamma\tau_0$ . Why the two forms are one: they are literally the same number, written in the two frames' variables.

Step 3. Read off the experiment. Count $N_0$ at the top and $N$ at the bottom; then $\gamma = \dfrac{\Delta t/\tau_0}{\ln(N_0/N)}$ . Three measurable numbers deliver $\gamma$ , which can then be checked against $1/\sqrt{1-\beta^2}$ from the measured speed. Why this is the cleanest possible test: nothing about the particle's internal machinery enters — the muon is used purely as a clock whose rate is known. *The two survival expressions $e^{-\Delta\tau/\tau_0}$ and $e^{-\Delta t/(\gamma\tau_0)}$ are identical under $\Delta\tau = \Delta t/\gamma$ , and the relation inverts to $\gamma = (\Delta t/\tau_0)/\ln(N_0/N)$ . Numerically, inverting the published counts ( $563 \to 412$ ) returns $\gamma \approx 9.3$ — above Frisch and Smith's measured $8.8 \pm 0.8$ but below the single-speed $10.01$ , which is exactly the signature of a beam that decelerates on the way down.*

Limiting cases

Case	Result	Meaning
$\beta \to 0$	$\gamma \to 1$ , $\Delta t \to \Delta\tau$	time becomes absolute; the Galilean world, where $t' = t$
GPS satellite, $\beta = 1.29\times10^{-5}$	$\gamma - 1 = 8.35\times10^{-11}$	$7.2\ \mu\mathrm{s}$ per day — invisible to a wristwatch, fatal to navigation
$\beta = 0.6$	$\gamma = 1.25$	a $100\ \mathrm{m}$ frame's clocks are out of sync by $200\ \mathrm{ns}$ as judged from outside
$\beta = 0.995$ (Frisch–Smith)	$\gamma = 10.0125$	a $2.197\ \mu\mathrm{s}$ lifetime reads $22.0\ \mu\mathrm{s}$ in the lab
$\beta \to 1$	$\gamma \to \infty$ , $\Delta\tau \to 0$	the particle's own clock nearly stops; an LHC proton has $\gamma \approx 7460$
$\beta = 1$ (light)	$\Delta\tau = 0$	a null worldline carries zero proper time — and admits no rest frame at all

Common trap“If each of them finds the other's clock slow, then one of them has to be wrong.”

Neither is wrong, and the reason is arithmetic, not a verbal manoeuvre — Derivation B does it. The trap is the phrase "the other's clock", which sounds like a single well-defined comparison and is not. When $A$ says " $B$ 's clock is slow", $A$ is comparing one of $B$ 's clocks against two of $A$ 's, read at two different places; when $B$ says the same of $A$ , $B$ does the mirror image. Those are different experiments on different pairs of events, and no principle of logic requires them to give reciprocal answers — though they do. What makes the accounts consistent is that each frame regards the other's pair of clocks as out of synchrony by exactly $vL/c^2$ . Run the ledger at $L = 100\ \mathrm{m}$ , $\beta = 0.6$ : the $200\ \mathrm{ns}$ offset plus the $355.80\ \mathrm{ns}$ that $B$ says $A$ 's clock advanced comes to $555.94\ \mathrm{ns}$ — exactly what $A$ says elapsed. A genuine, frame-independent asymmetry appears only when the clocks are brought back together at one place, which requires acceleration: that is The Twin Paradox.

Worked examples

1 · Frisch and Smith on Mount Washington (1963)

Muons selected at the Mount Washington summit ( $1917\ \mathrm{m}$ above sea level) have $\beta = 0.995$ ; the Cambridge detector sits $\Delta h = 1907\ \mathrm{m}$ lower. The muon mean lifetime is $\tau_0 = 2.197\ \mu\mathrm{s}$ , and $563$ muons per hour were counted at the summit. How many should reach sea level with and without time dilation?

Step 1. The Lorentz factor: $\gamma = 1/\sqrt{1 - 0.995^2} = 1/\sqrt{0.009975} = \boxed{10.0125}$ .

Step 2. The lab-frame flight time: $\Delta t = \Delta h/(\beta c) = 1907/(0.995 \times 2.998\times10^{8}) = \boxed{6.393\ \mu\mathrm{s}}$ , which is $6.393/2.197 = \boxed{2.91}$ muon lifetimes.

Step 3. Without dilation, the muon's own clock would also run for $6.393\ \mu\mathrm{s}$ , and the survival fraction would be $e^{-2.91} = \boxed{0.0545}$ — about $5.4\%$ . Of $563$ per hour, that leaves $563 \times 0.0545 = \boxed{31\ \text{per hour}}$ . Ninety-five per cent of the muons should have died on the way down.

Step 4. With dilation the muon's own clock advances only $\Delta\tau = \Delta t/\gamma = 6.393/10.0125 = \boxed{0.6385\ \mu\mathrm{s}}$ , which is $\Delta\tau/\tau_0 = \boxed{0.291}$ lifetimes. By Derivation C the survival fraction is $e^{-0.291} = \boxed{0.748}$ , so $563 \times 0.748 = \boxed{421\ \text{per hour}}$ should arrive. Frisch and Smith counted about $412$ .

Step 5. Be honest about the residuals — and about why they are there, because the reason is real physics and not a fudge. Our clean model gives $31$ and $421$ ; the published figures are $27$ expected classically and about $412$ observed (the paper's own abstract quotes $565 \pm 10$ decays per hour at the summit and $409 \pm 9$ at sea level). And Frisch and Smith's measured dilation factor is $8.8 \pm 0.8$ — not $10.01$ . The clean model has assumed a single, constant speed, and that is exactly what the experiment does not have. The selected muons span a band, $0.9950c$ to $0.9954c$ ( $\gamma$ from $10.01$ to $10.44$ , averaging about $10.2$ ); more importantly, they lose energy to ionization on the way down, so by Cambridge their $\gamma$ has fallen to roughly $6.8$ . The factor to compare against is therefore not $10.01$ but what the paper's abstract calls "the effective time dilation factor calculated for mesons of these speeds in our detection geometry" — and they give it as $\boxed{8.4 \pm 2}$ , against which their measured $8.8 \pm 0.8$ agrees comfortably. None of this touches the verdict: $412$ counted against $27$ predicted classically is a factor of about fifteen, and no amount of deceleration bookkeeping closes a gap like that.

2 · The GPS constellation

A GPS satellite orbits at $v = 3.874\ \mathrm{km/s}$ . By how much does its atomic clock lose to a clock on the ground each day through motion alone, and what would that do to a position fix?

Step 1. $\beta = 3874/(2.998\times10^{8}) = 1.29\times10^{-5}$ — tiny, so use the series: $\gamma - 1 \approx \tfrac12\beta^2 = \boxed{8.35\times10^{-11}}$ .

Step 2. Over one day the satellite clock loses $(\gamma - 1) \times 86\,400\ \mathrm{s} = \boxed{7.2\ \mu\mathrm{s/day}}$ relative to the ground — the special-relativistic term, and it makes the orbiting clock run slow.

Step 3. Gravity pushes the other way and harder: the satellite clock gains about $45.8\ \mu\mathrm{s/day}$ from the weaker potential at orbital altitude, so the net offset built into every GPS clock before launch is $+38.6\ \mu\mathrm{s/day}$ . (That budget is general-relativistic and belongs elsewhere; the $7.2$ is the piece this article owns.) Converted to distance, $38.6\ \mu\mathrm{s} \times c = \boxed{11.6\ \mathrm{km/day}}$ of ranging error if the correction were dropped. GPS is not a delicate test of relativity — it is a system that does not function without it.

Physics of the portrait

Time dilation has no single discoverer, and that is the point of the card: it was written down before anyone believed it, derived before anyone measured it, and measured by people who were mostly trying to do something else. The concept, not a person, is the portrait.

Origins · 2 minTime dilation and how this was discovered

Time dilation

1897–2014 · Larmor · Einstein · Ives and Stilwell · Rossi and Hall · Frisch and Smith

The first statement in print belongs to Joseph Larmor, who wrote in 1897 that electrons circling in a moving system "describe corresponding parts of their orbits in times shorter" by the ratio $\sqrt{1-v^2/c^2}$ — the effect exactly, though embedded in an ether theory he never abandoned; he repeated it in Aether and Matter (1900). Einstein derived it from the postulates in §4 of the 1905 paper. For thirty-three years it stayed arithmetic. Then in 1938 Herbert Ives and G. R. Stilwell, at Bell Telephone Laboratories, measured the light emitted forwards and backwards by fast hydrogen canal rays and found the mean of the two wavelengths shifted from the rest value by exactly the second-order factor — the transverse Doppler effect, and the first laboratory sighting of a moving clock running slow. (Ives believed to the end that he was confirming an ether theory.) Bruno Rossi and D. B. Hall watched cosmic-ray muons decay at Echo Lake and Denver in 1940; David Frisch and James Smith turned that into the definitive experiment on Mount Washington in 1963. Today the effect is confirmed to parts in $10^9$ and is a design constraint in orbit.

History

1897

— Larmor writes that electrons in a moving system describe corresponding parts of their orbits in times shorter by the ratio √(1 − v²/c²): the first appearance of time dilation. Repeated in Aether and Matter (1900).

1905

— Einstein, "Zur Elektrodynamik bewegter Körper", Annalen der Physik 17, 891–921 (received 30 June, published 26 September): the moving clock runs slow, derived in §4 from the transformation.

1938

— Ives and Stilwell, "An Experimental Study of the Rate of a Moving Atomic Clock", Journal of the Optical Society of America 28, 215–226: the transverse Doppler shift of hydrogen canal rays. The first laboratory measurement.

1941

— Rossi and Hall, Physical Review 59, 223: muon decay measured at Echo Lake (3240 m) and Denver (1616 m), Colorado. A real clock, dilated.

1963

— Frisch and Smith, "Measurement of the Relativistic Time Dilation Using μ-Mesons", American Journal of Physics 31, 342–355: 563 muons per hour at the Mount Washington summit (1917 m), about 412 in Cambridge after a 1907 m fall, 27 expected without dilation. Measured dilation factor 8.8 ± 0.8, against a predicted 8.4 ± 2 averaged over the decelerating descent.

1971

— Hafele and Keating fly caesium clocks east and west around the world in October. Predicted −40 ± 23 ns eastward and +275 ± 21 ns westward; observed −59 ± 10 ns and +273 ± 7 ns.

2014

— Botermann and colleagues, Physical Review Letters 113, 120405: lithium-7 ions stored at β = 0.338 in the ESR ring at GSI Darmstadt confirm the relation between γ and β to 2.3 parts per billion.

Going further

Texts. 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 · D. H. Frisch and J. H. Smith, American Journal of Physics 31 (1963) 342 · N. Ashby, "Relativity in the Global Positioning System", Living Reviews in Relativity 6 (2003) 1. The anchor values used throughout are $c = 299\,792\,458\ \mathrm{m/s}$ , $\tau_0 = 2.197\ \mu\mathrm{s}$ , $\Delta h = 1907\ \mathrm{m}$ , $\beta = 0.995$ . The reciprocity identity $vL/c^2 + (L/v)(1-\beta^2) = L/v$ holds for all $\beta$ , and $e^{-\Delta\tau/\tau_0} = e^{-\Delta t/(\gamma\tau_0)}$ under $\Delta\tau = \Delta t/\gamma$ . The clean single-speed model gives $31$ /hr and $421$ /hr; Frisch and Smith report $27$ and about $412$ (their abstract: $565 \pm 10$ and $409 \pm 9$ decays/hr), with a measured dilation factor $8.8 \pm 0.8$ against a predicted $8.4 \pm 2$ . The difference from the idealized $\gamma = 10.01$ is physical, not cosmetic: the muons span $0.9950c$ – $0.9954c$ and decelerate by ionization loss during the fall, so their $\gamma$ drops from about $10.2$ at the summit to about $6.8$ at Cambridge, and the factor to compare against is the average over the descent. A geometry note, because these two numbers are easy to confuse: the Mount Washington summit stands $1917\ \mathrm{m}$ above sea level, and the Cambridge detector about $10\ \mathrm{m}$ above it — so the muons' flight path, which is the drop between the two counting stations, is $\Delta h = 1907\ \mathrm{m}$ . The flight path is the drop, not the summit elevation.

Builds on. The Two Postulates of Special Relativity · The Light Clock and Time Dilation · The Lorentz Factor Gamma · Einstein Clock Synchronization. Leads to. Relativity of Simultaneity · Length Contraction · Proper Time · The Invariant Spacetime Interval · The Lorentz Transformation Equations.

Is the moving clock really running slow, or does it just look that way because of the time light takes to reach me?

Really. Light-travel delay is a separate effect, and it is removed before anyone quotes a time dilation. The measurement in Derivation B involves no signals at all: $B$ 's clock face is compared locally, side by side, with $A$ 's near clock at one event and $A$ 's far clock at another. Nothing propagates, nothing is delayed, and the discrepancy survives. What you literally see through a telescope mixes dilation with the changing light-travel time — that is the relativistic Doppler effect, not $\Delta t = \gamma\Delta\tau$ .

If proper time is the smallest elapsed time, why is the travelling twin said to age the LEAST — isn't that the same as saying she has the longest proper time?

This is the confusion the base of this article fell into, so separate it cleanly. Statement one: between two events on a single clock's worldline, that clock's proper time is the smallest elapsed time any frame assigns — this article's $\Delta t = \gamma\Delta\tau$ . Statement two: among all timelike worldlines joining two fixed events, the inertial one carries the largest proper time. The first compares frames describing one worldline; the second compares different worldlines between the same endpoints. The travelling twin ages less because her worldline is not the inertial one. That theorem belongs to Proper Time, its dramatization to The Twin Paradox.

Does the muon "know" it is moving fast and slow its decay down?

No — and this is the cleanest way to see that dilation is not a mechanism acting on the clock. In the muon's own frame nothing is unusual: it is at rest, it has its ordinary $2.197\ \mu\mathrm{s}$ mean life, and it decays on schedule. What changes between frames is not the muon but the elapsed time between the two events: the mountain's frame assigns $6.393\ \mu\mathrm{s}$ to a flight the muon experiences as $0.6385\ \mu\mathrm{s}$ , and nothing in the weak interaction has been altered. (The muon has an equally good account of why it reaches the ground: from where it sits the mountain is length-contracted — the subject of Length Contraction.)

Why does the leading-clock offset vL/c² have no γ in it?

Because two factors of $\gamma$ cancel, and it is worth seeing where. Apply the time transformation to two clocks separated by $\Delta x$ in the lab and simultaneous there: $\Delta t' = -\gamma v \Delta x/c^2$ . But the separation of two clocks whose proper separation is $L_0$ is, in the lab, the contracted $\Delta x = L_0/\gamma$ . Substitute and the $\gamma$ from the transformation meets the $1/\gamma$ from the contraction: $\Delta t' = -v L_0/c^2$ , clean. This is why the offset is quoted with the proper length and no $\gamma$ , and it is the form used in the reciprocity ledger.

The Frisch–Smith paper says the dilation factor was 8.8 ± 0.8, but γ = 10.01 at 0.995c. Is that a discrepancy?

It is not a discrepancy, and the reason is worth knowing. The muons do not fall at a constant speed. They are selected in a band ( $0.9950c$ – $0.9954c$ , so $\gamma$ between $10.01$ and $10.44$ ), and on the way down they lose energy to ionization in the atmosphere, arriving in Cambridge with $\gamma$ around $6.8$ . The factor to compare against is therefore the one averaged over the descent, which Frisch and Smith give as $8.4 \pm 2$ — and their measured $8.8 \pm 0.8$ sits comfortably inside it. The idealized $\gamma = 10.01$ is the value at the top of the fall, not along it. What is decisive is the size of the effect, not its third digit: $27$ counts an hour predicted classically, about $412$ observed.

Does time dilation apply to gravity too? GPS seems to mix the two.

GPS does mix them, and they are genuinely different effects. The $-7.2\ \mu\mathrm{s/day}$ here is kinematic: it depends on the satellite's speed and follows from special relativity alone. The $+45.8\ \mu\mathrm{s/day}$ is gravitational: it depends on the potential difference between orbit and ground, and it is a prediction of general relativity, which special relativity cannot supply. They act in opposite senses and gravity wins, leaving the net $+38.6\ \mu\mathrm{s/day}$ dialled into the satellite clocks before launch. Only the first term is this article's.