---
title: Time Dilation, Length Contraction, and Paradoxes
module: Foundations of Relativity
moduleNumber: 1
lessonNumber: 3
order: 103
summary: >
  A light clock and the constancy of c give the two headline effects directly:
  a moving clock runs slow by gamma, and a moving rod is short by the same
  factor. Cosmic-ray muons reaching sea level are the standing experimental
  proof. The relativistic Doppler effect adds the time-dilation factor to the
  classical shift, and the twin and pole-barn paradoxes dissolve once the
  relativity of simultaneity is taken seriously.
topics: [Foundations of Relativity]
draft: false
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 1 — Relativity I; §1-4 Time Dilation and Length Contraction"
  - book: Tipler & Llewellyn
    ref: "§1-5 The Doppler Effect; §1-6 The Twin Paradox and Other Surprises"
---

The [Lorentz
transformation](/relativity/foundations/lorentz-transformation-spacetime)
carries two consequences that no classical intuition survives: a moving clock
ticks slowly and a moving rod is short, each by the factor $\gamma =
(1 - \beta^2)^{-1/2}$. Both follow from a single geometric argument built out of
the constancy of $c$, and both are confirmed daily by particles that reach the
ground only because the effects are real.[^tl-dil]

## The light clock and proper time

A light clock is a pulse bouncing between a source and a mirror a fixed distance
$D$ apart. In the clock's rest frame $S'$ the round trip takes

$$
\Delta \tau = \frac{2D}{c}.
$$

Now view the same clock from a frame $S$ in which it moves at speed $v$
perpendicular to the pulse's path. Between emission and return the clock
advances a horizontal distance $v\,\Delta t$, so the pulse travels a longer,
slanted path. Because $c$ is the same in $S$, the longer path takes longer. The
geometry is a right triangle: half the light path is the hypotenuse, $D$ the
vertical leg, and $v\,\Delta t/2$ the horizontal leg.

$$
% caption: The light clock seen from a frame in which it moves right at v. The
% pulse follows the slanted path; half of it is the hypotenuse of a right
% triangle with legs D and v times delta-t over two.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % base line (motion)
  \draw[black] (-0.3,0) -- (6.3,0);
  \draw[->, black] (5.0,-0.5) -- (6.0,-0.5) node[right] {v};
  % positions of source: start (0,0), mirror at top, return at (4,0)
  \coordinate (s0) at (0,0);
  \coordinate (mir) at (2,2.4);
  \coordinate (s1) at (4,0);
  % slanted light path
  \draw[acc, thick, ->] (s0) -- (mir);
  \draw[acc, thick, ->] (mir) -- (s1);
  % vertical D
  \draw[black, dashed] (2,0) -- (mir) node[midway, right] {D};
  % half base
  \draw[black] (0,0) -- (2,0);
  \node[anchor=north] at (1,-0.25) {v times delta t over 2};
  % hypotenuse label
  \node[acc, anchor=south east] at (1.05,1.25) {c times delta t over 2};
  % right angle mark
  \draw[black] (1.8,0) -- (1.8,0.2) -- (2,0.2);
  % dots
  \fill[black] (s0) circle (1.6pt); \fill[black] (s1) circle (1.6pt);
  \node[anchor=north west] at (4.1,-0.05) {return};
\end{tikzpicture}
$$

Applying the Pythagorean theorem to half the trip,

$$
\parens{\frac{c\,\Delta t}{2}}^2 = D^2 + \parens{\frac{v\,\Delta t}{2}}^2,
$$

solving for $\Delta t$, and using $\Delta \tau = 2D/c$ gives

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

Since $\gamma \ge 1$, the moving-frame interval $\Delta t$ is always the longer
one. The special interval $\Delta \tau$, read on a single clock present at both
events, is the **proper time**.

> **Definition (Proper time).** The time interval $\Delta \tau$ between two
> events measured by a single clock that is present at both, so that they occur
> at the same place in that clock's frame. It is the shortest interval any
> inertial observer measures between the two events; every other frame reads
> $\Delta t = \gamma\,\Delta\tau$. A moving clock runs slow.

The same result comes from the transformation directly. Writing $\Delta t =
\gamma(\Delta t' + v\,\Delta x'/c^2)$ and setting $\Delta x' = 0$ (one clock, one
place) leaves $\Delta t = \gamma\,\Delta t'$. The condition $\Delta x' = 0$ is
what makes $\Delta t'$ a proper time; if the two events happen at different
places in every frame, no clock reads a proper interval between them.

## Muon decay

Time dilation is not a thought experiment. Muons are created by cosmic rays high
in the atmosphere and decay with a mean lifetime $\tau \approx 2\ \mu\mathrm{s}$
in their rest frame, following $N(t) = N_0 e^{-t/\tau}$. A muon moving at
$0.998c$ would travel only $v\tau \approx 600\ \mathrm{m}$ in one rest-frame
lifetime, so almost none should survive the several-kilometer fall to sea level.
Yet many arrive.

In Earth's frame the lifetime is dilated by $\gamma = 1/\sqrt{1 - 0.998^2}
\approx 15$, giving a mean lifetime of $30\ \mu\mathrm{s}$ and a mean travel
distance of $9000\ \mathrm{m}$. Of $10^8$ muons at $9000\ \mathrm{m}$, the
classical count reaching the ground after $15$ rest-frame lifetimes would be
$N = 10^8 e^{-15} \approx 31$; the relativistic count, using one dilated
lifetime for the trip, is $N = 10^8 e^{-1} \approx 3.7 \times 10^7$. The
observed flux matches the relativistic prediction.

$$
% caption: Muons surviving the fall to sea level. The classical no-dilation
% prediction leaves about thirty of a hundred million; time dilation leaves
% about thirty-seven million, which is what is observed.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (0,3.6) node[above] {count (log scale)};
  \draw[black] (0,0) -- (6,0);
  % start bar (10^8)
  \fill[black] (0.5,0) rectangle (1.5,3.2);
  \draw[black] (0.5,0) rectangle (1.5,3.2);
  \node[anchor=north, align=center] at (1.0,-0.1) {created\\100 million};
  % relativistic (3.7e7) ~ log close to top
  \fill[acc!14] (2.6,0) rectangle (3.6,2.85);
  \draw[acc] (2.6,0) rectangle (3.6,2.85);
  \node[anchor=north, align=center] at (3.1,-0.1) {with dilation\\37 million};
  % classical (31) tiny
  \fill[black] (4.7,0) rectangle (5.7,0.7);
  \draw[black] (4.7,0) rectangle (5.7,0.7);
  \node[anchor=north, align=center] at (5.2,-0.1) {no dilation\\about 31};
\end{tikzpicture}
$$

From the muon's own frame nothing decays faster than usual; instead the
atmosphere rushes past at $0.998c$ and the $9000\ \mathrm{m}$ of air is
contracted to $600\ \mathrm{m}$, short enough to cross within $2\ \mu\mathrm{s}$.
The two frames describe the survival with different mechanisms, dilation in one
and contraction in the other, and agree on the count.

## Length contraction

To measure a moving rod's length, an observer must mark the positions of both
ends _at the same instant_ in the measuring frame. Let the rod lie at rest in
$S'$ with proper length $L_p = x'_2 - x'_1$. In $S$ the ends are located
simultaneously, $t_2 = t_1$. Using $x'_2 - x'_1 = \gamma[(x_2 - x_1) - v(t_2 -
t_1)]$ with $t_2 = t_1$ gives $L_p = \gamma\,L$, so

$$
L = \frac{L_p}{\gamma} = L_p\sqrt{1 - \frac{v^2}{c^2}}.
$$

> **Definition (Proper length).** The length $L_p$ of an object measured in the
> frame where it is at rest. In any frame in which it moves at speed $v$, its
> length along the direction of motion is $L = L_p/\gamma$, shorter than the
> proper length. Dimensions perpendicular to the motion are unchanged.

Contraction acts only along the motion. A square carried at high speed becomes a
foreshortened parallelogram, not a smaller square, because the transverse sides
keep their length while the parallel sides shrink. The contraction is
appreciable only near $c$: $L/L_p$ stays close to $1$ until $\beta \gtrsim 0.5$
and drops to zero as $\beta \to 1$.

$$
% caption: The contraction ratio L over L-p versus beta. It departs from 1 only
% at high speed and falls to zero as the speed approaches c.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.4,0) node[right] {beta};
  \draw[->, black] (0,0) -- (0,3.6) node[above] {length ratio};
  \foreach \x/\l in {0/0, 3/0.5, 6/1} \draw[black] (\x,0.05) -- (\x,-0.05) node[below] {\l};
  \draw[black] (-0.05,3) -- (0.05,3) node[left] {1};
  % L/Lp = sqrt(1-x^2), x from 0..1 mapped to 0..6, y scaled by 3
  \draw[acc, very thick]
    (0,3.0) .. controls (2.0,2.9) and (3.4,2.6) ..
    (4.2,2.2) .. controls (4.9,1.8) and (5.5,1.15) ..
    (5.8,0.75) .. controls (5.92,0.5) and (5.98,0.2) .. (6,0);
\end{tikzpicture}
$$

Lorentz and FitzGerald had proposed exactly this contraction, in the direction
of motion, to explain the Michelson-Morley null result, before Einstein derived
it from the postulates. In relativity it is not a mechanical compression of the
rod but a property of how the two frames slice spacetime into "space at one
instant."

## The relativistic Doppler effect

A source emitting $N$ waves of proper frequency $f_0$ over a proper time
interval spreads them over a distance that depends on whether it approaches or
recedes, and the emission interval is itself time-dilated. Combining the two
effects, an observer and source approaching each other measure

$$
f = \sqrt{\frac{1 + \beta}{1 - \beta}}\; f_0 \qquad \text{(approaching, blueshift)},
$$

and receding,

$$
f = \sqrt{\frac{1 - \beta}{1 + \beta}}\; f_0 \qquad \text{(receding, redshift)}.
$$

Unlike the classical Doppler effect for sound, the relativistic formula depends
only on the relative speed $\beta$, not on which of source or observer moves,
because there is no medium to move against. For $\beta \ll 1$ both reduce to
$f/f_0 \approx 1 \pm \beta$, so $|\Delta f / f_0| \approx \beta$.

$$
% caption: Received frequency over proper frequency versus beta, for approach
% (upper branch, blueshift) and recession (lower branch, redshift). Both meet at
% 1 when the relative speed is zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (0,4.0) node[above] {frequency ratio};
  \draw[->, black] (0,1.8) -- (6.3,1.8) node[right] {beta};
  \draw[black] (-0.05,1.8) -- (0.05,1.8) node[left] {1};
  \foreach \x/\l in {2/0.3, 4/0.6, 5.7/0.9} \draw[black] (\x,1.75) -- (\x,1.85) node[below] {\l};
  % approaching: sqrt((1+b)/(1-b)) rising
  \draw[acc, very thick]
    (0,1.8) .. controls (2.0,2.3) and (3.6,2.9) .. (4.6,3.4)
    .. controls (5.2,3.7) and (5.6,3.85) .. (5.9,3.95);
  \node[acc, anchor=south] at (5.0,3.5) {approaching};
  % receding: sqrt((1-b)/(1+b)) falling toward 0
  \draw[black, very thick]
    (0,1.8) .. controls (2.0,1.35) and (3.6,0.9) .. (4.6,0.6)
    .. controls (5.2,0.45) and (5.6,0.32) .. (5.9,0.25);
  \node[black, anchor=north] at (5.0,0.55) {receding};
\end{tikzpicture}
$$

> **Example (Redshift of a distant galaxy).** Hydrogen's Balmer line at
> $\lambda_0 = 656\ \mathrm{nm}$ is observed at $\lambda = 1458\ \mathrm{nm}$ in
> a galaxy's light. Since $f \propto 1/\lambda$, the receding formula in
> wavelength form is $\lambda_0/\lambda = \sqrt{(1-\beta)/(1+\beta)}$, so
> $(\lambda_0/\lambda)^2 = (656/1458)^2 = 0.202 = (1-\beta)/(1+\beta)$. Solving,
> $\beta = 0.664$: the galaxy recedes at $0.664c$.

## The twin paradox

Homer stays on Earth while his twin Ulysses travels to a star at $\beta = 0.8$
($\gamma = 5/3$) and returns. Each twin, naively, could regard the other as the
one who moved and expect the other to age less. Both cannot be right. The
resolution is that the situations are not symmetric: Homer stays in one inertial
frame the whole time, while Ulysses switches frames at the turnaround, and only
Ulysses feels the acceleration.

In Homer's frame the outbound leg takes $\Delta t = 5\ \mathrm{y}$ and the return
another $5\ \mathrm{y}$, so Homer ages $10\ \mathrm{y}$. Ulysses' clock reads the
proper time along each leg, $\Delta\tau = \Delta t/\gamma = 5/(5/3) = 3\
\mathrm{y}$ each way, so Ulysses ages $6\ \mathrm{y}$ and returns four years
younger. The invariant interval gives the same answer frame-independently:
Ulysses' worldline has $\Delta x' = 0$ on each leg, so $\Delta\tau = \Delta s/c$,
while Homer's straight worldline accumulates more proper time.

$$
% caption: Ulysses' worldline in Homer's frame. The outbound and return legs
% tilt oppositely and meet at the turnaround; Homer's worldline is the straight
% vertical. The bend, absent from Homer's line, breaks the symmetry.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.3,0) -- (4.4,0) node[right] {x};
  \draw[->, black] (0,-0.3) -- (0,5.2) node[above] {ct};
  % Homer worldline (vertical)
  \draw[black!70, very thick] (0,0) -- (0,5.0);
  \node[black!70, anchor=east] at (0,3.6) {Homer};
  % Ulysses out and back, turnaround at (3, 2.5)
  \draw[acc, very thick] (0,0) -- (3,2.5);
  \draw[acc, very thick] (3,2.5) -- (0,5.0);
  \node[acc, anchor=west] at (1.6,1.0) {Ulysses out};
  \node[acc, anchor=west] at (1.6,3.9) {Ulysses back};
  \fill[acc] (3,2.5) circle (2pt);
  \node[acc, anchor=west] at (3.05,2.5) {turnaround};
  \fill[black] (0,0) circle (1.8pt); \fill[black] (0,5.0) circle (1.8pt);
  \node[anchor=east] at (0,0.15) {depart};
  \node[anchor=east] at (0,4.85) {reunite};
\end{tikzpicture}
$$

Special relativity handles the accelerated turnaround perfectly well as long as
the analysis is done from an inertial frame such as Homer's; the paradox comes
only from wrongly assuming the two frames are interchangeable. The age
difference traces to the relativity of simultaneity: at the turnaround, Ulysses'
notion of "now on Earth" jumps forward, skipping over years of Homer's life that
Ulysses never shares a frame with.

## The pole and barn

A runner carries a $10\ \mathrm{m}$ pole toward a $5\ \mathrm{m}$ barn at
$\beta = 0.866$ ($\gamma = 2$). In the barn's frame the pole is contracted to
$L = 10/2 = 5\ \mathrm{m}$ and fits exactly; the farmer can briefly close both
doors with the pole inside. In the runner's frame the pole is $10\ \mathrm{m}$
and the barn is contracted to $2.5\ \mathrm{m}$, so the pole cannot possibly fit.
Both are correct, because "both doors shut with the pole inside" means _two
events are simultaneous_, and simultaneity is frame-dependent. In the barn frame
the back door shuts and the front door shuts at the same instant; in the runner's
frame the back door shuts (pole tip reaching it) well before the front door
shuts, and there is never an instant with the whole pole enclosed.

| Frame | Pole length | Barn length | Both doors shut with pole inside? |
| --- | --- | --- | --- |
| Barn (farmer) | $5\ \mathrm{m}$ (contracted) | $5\ \mathrm{m}$ (proper) | yes, the two shuttings are simultaneous |
| Runner (pole) | $10\ \mathrm{m}$ (proper) | $2.5\ \mathrm{m}$ (contracted) | no, the shuttings happen at different times |

Neither paradox needs new physics. Each is the relativity of simultaneity read
back into ordinary language, which quietly assumes a universal "now" that
relativity has removed. The dynamics built on these kinematics, momentum and
energy that stay conserved across frames, are the subject of the
[next lesson](/relativity/foundations/relativistic-momentum-energy).

[^tl-dil]: **Tipler & Llewellyn**, _Modern Physics_, §1-4 to §1-6 — Time Dilation and Length Contraction (light-clock derivation, muon decay, proper time and proper length), the relativistic Doppler Effect, and the Twin Paradox and pole-barn paradox resolved through the relativity of simultaneity.