Time Dilation, Length Contraction, and Paradoxes

The Lorentz transformation 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.¹

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 $S^{\prime }$ the round trip takes

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.

Applying the Pythagorean theorem to half the trip,

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

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.

The same result comes from the transformation directly. Writing $\Delta t=\gamma (\Delta t^{\prime }+v\Delta x^{\prime }/c^2)$ and setting $\Delta x^{\prime }=0$ (one clock, one place) leaves $\Delta t=\gamma \Delta t^{\prime }$. The condition $\Delta x^{\prime }=0$ is what makes $\Delta t^{\prime }$ 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.

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 $S^{\prime }$ with proper length $L_p=x_2^{\prime }-x_1^{\prime }$. In $S$ the ends are located simultaneously, $t_2=t_1$. Using $x_2^{\prime }-x_1^{\prime }=\gamma [(x_2-x_1)-v(t_2-t_1)]$ with $t_2=t_1$ gives $L_p=\gamma L$, so

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 $\beta \to 1$.

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

and receding,

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 $|\Delta f/f_0|\approx \beta$.

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 $\Delta x^{\prime }=0$ on each leg, so $\Delta \tau =\Delta s/c$, while Homer's straight worldline accumulates more proper time.

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.

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.