Length contraction — derivation: Worked examples

Before anything, recall the two lengths, in plain words:

==Proper length $L_{0}$ == — the length in the frame where the object sits still. The biggest value anyone measures.
==Measured length $L$ == — the length in a frame where the object whizzes past at speed $v$ . Always $L \leq L_{0}$ .
== $γ$ (gamma)== — the "stretch/shrink dial," $γ = 1/ 1 - v^{2} / c^{2}$ . It is always $\geq 1$ . (See Lorentz factor gamma.)

Everything below is just these three quantities and the one relation $L = L_{0} / γ$ turned to face different questions.

The scenario matrix

Length contraction has no "quadrants" like an angle does — but it has its own family of distinct cases. Here is every cell we must cover:

#	Cell (case class)	What varies	Covered by
A	Forward direction — given $L_{0}$ and $v$ , find $L$	shrink a known rod	Ex 1
B	Inverse direction — given $L$ and $L_{0}$ , find $v$	solve for speed	Ex 2
C	Zero-speed limit ( $v = 0$ )	degenerate: $γ = 1$	Ex 3
D	Light-speed limit ( $v \to c$ )	limiting behaviour: $L \to 0$	Ex 3
E	Slow / everyday speed ( $v ≪ c$ )	why we never notice it	Ex 4
F	Symmetry / "who sees whom" — both frames	reciprocity, no paradox	Ex 5
G	Perpendicular dimension	direction that does not contract	Ex 6
H	Real-world word problem	muon / astronomy	Ex 7
I	Exam twist — mixed with time, or "contracted distance travelled"	combine effects	Ex 8
J	Diagonal object — motion not along the rod's length	only the $x$ -part shrinks	Ex 9

The figures below are drawn in a high-contrast style: black lines on white, with the one red object being whatever we are actually measuring in that figure.

Cell A — Forward direction

Worked example Example 1 — Shrink a known rod (Cell A)

A metre-scale rod has proper length $L_{0} = 10$ m and flies past the lab at $v = 0.8 c$ . How long does the lab measure it?

Forecast: $0.8 c$ is fast, so expect a noticeable shrink — guess somewhere between 5 and 7 m before reading on.

In the figure, the upper black bar is the rest-frame rod (full $10$ m, its end-caps at $0$ and $10$ ); the lower red bar with the little $v$ -arrow is the same rod as the lab measures it — deliberately drawn stopping short at $6$ so you can see the missing $4$ m. Keep your eye on that red bar as we compute it.

Step 1 — Build $γ$ . $γ = \frac{1}{1 - 0. 8 ^{2}} = \frac{1}{1 - 0.64} = \frac{1}{0.36} = \frac{1}{0.6} = \frac{5}{3} \approx 1.667.$ Why this step? $γ$ is the dial that tells us how much shorter. We cannot shrink anything until we know it.

Step 2 — Divide (never multiply). $L = \frac{L _{0}}{γ} = \frac{10}{1.667} = 6.0 m .$ Why divide? The lab watches the rod move, so the lab measures the contracted (smaller) length. Proper length $L_{0}$ is the maximum, so it must sit on top of the fraction. That $6$ m is exactly where the red bar ends in the figure.

Verify: $6.0 < 10$ ✔ — shorter, as required. Units: metres ÷ (dimensionless $γ$ ) = metres ✔.

Cell B — Inverse direction

Worked example Example 2 — Find the speed (Cell B)

A rod is known to be $1.00$ m long at rest, but the lab measures it as $0.60$ m. How fast is it moving?

Forecast: It shrank to 60% — that's a big shrink, so $v$ should be a large fraction of $c$ . Guess above $0.7 c$ .

Step 1 — Write the ratio form. $\frac{L}{L _{0}} = \frac{1}{γ} = 1 - \frac{v ^{2}}{c ^{2}} \Rightarrow 0.60 = 1 - \frac{v ^{2}}{c ^{2}} .$ Why this step? We're given the ratio $L / L_{0}$ , and the ratio equals exactly the square-root factor. No need to compute $γ$ separately.

Step 2 — Square both sides to peel off the root. $0.36 = 1 - \frac{v ^{2}}{c ^{2}} \Rightarrow \frac{v ^{2}}{c ^{2}} = 0.64.$ Why square? The unknown $v$ is trapped under a square root; squaring is the tool that undoes it.

Step 3 — Take the root. $\frac{v}{c} = 0.80 \Rightarrow v = 0.8 c .$ Why the positive root? Speed is a magnitude — never negative.

Verify: Plug back: $γ = 1/ 1 - 0.64 = 5/3$ , so $L = 1.00 \times 0.6 = 0.60$ m ✔. And $0.8 c > 0.7 c$ , matching the forecast.

Cells C & D — the two extremes

Worked example Example 3 — Zero speed and light speed (Cells C, D)

Take the same $L_{0} = 10$ m rod. What length does the lab measure (a) when $v = 0$ , and (b) as $v \to c$ ?

Forecast: At rest, nothing moves — no shrink at all. Near light speed, the shrink should run wild — guess it goes toward zero.

Part (a) — $v = 0$ (Cell C, degenerate input). $γ = \frac{1}{1 - 0} = 1, L = \frac{10}{1} = 10 m .$ Why this matters: When the rod is at rest in the lab too, the lab frame is the rest frame, so $L = L_{0}$ . The formula correctly gives back the proper length — a vital sanity check on any relativity formula.

Part (b) — $v \to c$ (Cell D, limiting behaviour). As $v \to c$ , the term $v^{2} / c^{2} \to 1$ , so $1 - \frac{v ^{2}}{c ^{2}} \to 0 = 0 \Rightarrow L \to 0.$ Why this matters: Approaching light speed, the rod is squashed toward zero length in the lab. Look at the red curve in the figure — flat and lazy for small $v$ , then plunging off a cliff near $c$ . This is why "everyday" motion shows no effect, but light-speed motion is dramatic.

Verify (numerically): at $v = 0.99 c$ , $γ = 1/ 1 - 0.9801 \approx 7.09$ , $L \approx 10/7.09 \approx 1.41$ m — already tiny, confirming the plunge. ✔

Cell E — everyday speeds

Worked example Example 4 — Why you never notice it (Cell E)

A $100$ m train travels at $v = 300$ km/h $\approx 83.3$ m/s. By how much does the lab see it contract? (Use $c = 3 \times 1 0^{8}$ m/s.)

Forecast: Trains obviously don't look shorter. So the effect must be absurdly tiny — guess far less than the width of an atom.

Step 1 — Compute $v / c$ . $\frac{v}{c} = \frac{83.3}{3 \times 1 0 ^{8}} \approx 2.78 \times 1 0^{- 7} .$ Why this step? Contraction depends only on the ratio $v / c$ ; a train's speed is a whisper compared to light.

Step 2 — Use the tiny- $x$ shortcut. For a very small number $x = v^{2} / c^{2}$ , the exact $1 - x$ is almost $1 - \frac{1}{2} x$ . So the amount lost is $L_{0} - L \approx L_{0} \cdot \frac{1}{2} \frac{v ^{2}}{c ^{2}} = 100 \times \frac{1}{2} (2.78 \times 1 0^{- 7})^{2} \approx 3.86 \times 1 0^{- 12} m .$ Why this tool (the approximation)? Plugging $2.78 \times 1 0^{- 7}$ into a calculator's square root just returns $1.000000$ — the change hides below its precision. The linear approximation is the only way to see how small the loss really is.

Verify: $3.86 \times 1 0^{- 12}$ m is roughly one forty-thousandth the width of an atom ( $\sim 1 0^{- 10}$ m) — utterly unmeasurable ✔. This is exactly why length contraction stayed hidden until we studied things moving near $c$ .

Cell F — symmetry, "who sees whom"

Worked example Example 5 — Both see the other shorter (Cell F)

Two identical rods, each $L_{0} = 12$ m in its own frame, fly past each other at $v = 0.6 c$ . Ann rides rod A; Bob rides rod B. How long does Ann measure Bob's rod, and how long does Bob measure Ann's? Is it a contradiction?

Forecast: By the principle of relativity, neither is "specially moving." So expect a symmetric answer — each measures the other equally short.

Step 1 — Compute the shared $γ$ . $γ = \frac{1}{1 - 0.36} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25.$ Why one $γ$ ? The relative speed between the frames is the same $0.6 c$ either way, so both use the identical factor.

Step 2 — Ann measures Bob's moving rod. $L_{Bob, by Ann} = \frac{12}{1.25} = 9.6 m .$

Step 3 — Bob measures Ann's moving rod. $L_{Ann, by Bob} = \frac{12}{1.25} = 9.6 m .$ Why identical? From Ann's seat, Bob's rod moves; from Bob's seat, Ann's rod moves. Each applies the same formula to the other's rod.

Verify — no contradiction: They are not comparing the same pair of events. Ann marks Bob's two ends simultaneously in Ann's time; Bob uses his own "same time." Because simultaneity is frame-dependent, those are literally different measurements. Both being $9.6$ m is perfectly consistent ✔.

Cell G — the direction that does not shrink

Worked example Example 6 — Perpendicular dimension (Cell G)

A rectangular flag is $2.0$ m long (along the motion) and $1.0$ m tall (perpendicular). It flies at $v = 0.8 c$ along its length. Give both measured dimensions in the lab.

Forecast: Only the length aligned with $v$ should shrink. The height should stay put. Guess: length $\to 1.2$ m, height $\to 1.0$ m.

Step 1 — Contract the along-motion length. With $γ = 5/3$ : $L_{∥} = \frac{2.0}{1.667} = 1.2 m .$ Why only this one? The derivation used $x^{'} = γ (x - v t)$ — an equation about the $x$ -axis only, the direction of motion. Nothing in it touches $y$ or $z$ .

Step 2 — Leave the height alone. $L_{⊥} = 1.0 m (unchanged) .$ Why unchanged — the ring argument: Imagine a ring moving through a hole it exactly fits at rest. If the perpendicular dimension shrank, one frame would say "ring slips through freely" and another "ring jams" — an impossible physical disagreement. So perpendicular lengths must be frame-independent.

Verify: Final flag = $1.2$ m $\times$ $1.0$ m. Only the red (motion-aligned) side changed ✔.

Cell H — real-world word problem

Worked example Example 7 — The muon's thin atmosphere (Cell H)

A muon streaks toward the ground at $v = 0.98 c$ . In Earth's frame the atmosphere it must cross is $L_{0} = 10$ km thick. In the muon's own frame, how thick is that layer?

Forecast: From the muon's seat, the atmosphere rushes past it. So the muon sees a contracted, thinner layer — guess a couple of km.

Step 1 — Identify whose proper length is whose. The $10$ km is measured by Earth, at rest relative to the atmosphere — so $10$ km is the atmosphere's proper length $L_{0}$ . Why care? Getting proper vs. contracted backwards is the #1 way to divide when you should… still divide, but with the wrong number.

Step 2 — Compute $γ$ . $γ = \frac{1}{1 - 0.9 8 ^{2}} = \frac{1}{1 - 0.9604} = \frac{1}{0.0396} \approx 5.025.$

Step 3 — Contract into the muon's frame. $L = \frac{10}{5.025} \approx 1.99 km .$ Why divide? In the muon's frame the atmosphere moves, so the muon sees the contracted value.

Verify: $1.99$ km $< 10$ km ✔. This thinner layer is exactly why muons — which decay quickly — still reach the ground: they only have $\sim 2$ km to cross in their own frame. See the Muon decay experiment. ✔

Cell I — exam twist (combine with time / distance)

Worked example Example 8 — Contracted distance vs. dilated time (Cell I)

A spaceship flies from Earth to a star $L_{0} = 8.0$ light-years away (Earth-frame distance) at $v = 0.8 c$ . (a) How far is the trip in the ship's frame? (b) How long does the trip take by the ship's own clock?

Forecast: The ship sees a shorter distance (contraction) and a shorter time (its clock ticks less). Both smaller.

Step 1 — $γ$ again. $γ = 5/3 \approx 1.667$ (same as Ex 1, since $v = 0.8 c$ ).

Step 2 — (a) Contract the distance. The $8.0$ ly is Earth's proper distance: $L = \frac{8.0}{1.667} = 4.8 ly .$ Why divide? In the ship's frame the star rushes toward it, so the gap is contracted.

Step 3 — (b) Ship's travel time. In the ship's frame it covers the contracted distance $4.8$ ly at speed $0.8 c$ : $t_{ship} = \frac{L}{v} = \frac{4.8 ly}{0.8 c} = 6.0 years .$ Why use contracted $L$ here? The ship must use distances as it measures them with its own clock — mixing Earth's distance with the ship's clock is the classic trap.

Verify (cross-check via time dilation): Earth-frame travel time $= 8.0 ly /0.8 c = 10$ yr. The ship's clock (proper time) should read $10/ γ = 10/1.667 = 6.0$ yr ✔ — both methods agree, confirming length contraction and Time dilation are two faces of the same Lorentz transformation.

Cell J — diagonal object

Worked example Example 9 — A rod tilted to the motion (Cell J)

In its rest frame a rod has length $L_{0} = 5.0$ m and points at $3 7^{\circ}$ above the direction it will later move. It then flies at $v = 0.8 c$ along the $x$ -axis. What is its measured length in the lab? ( $cos 3 7^{\circ} = 0.8$ , $sin 3 7^{\circ} = 0.6$ .)

Forecast: Only the along-motion part shrinks; the up part survives. So the rod gets shorter and appears steeper. Guess a total a bit under 5 m.

Step 1 — Split the rod into $x$ and $y$ pieces. $L_{0 x} = 5.0 cos 3 7^{\circ} = 4.0 m, L_{0 y} = 5.0 sin 3 7^{\circ} = 3.0 m .$ Why split? Contraction acts only along $x$ . Breaking the rod into an along-part and an across-part lets us apply the rule to each separately.

Step 2 — Contract only the $x$ -piece ( $γ = 5/3$ ): $L_{x} = \frac{4.0}{1.667} = 2.4 m, L_{y} = 3.0 m (unchanged) .$ Why only $x$ ? By the very rule proved in Cell G, the perpendicular ( $y$ ) part is untouched; only the along-motion ( $x$ ) part feels $γ$ .

Step 3 — Rebuild the length with the Pythagorean theorem: $L = L_{x}^{2} + L_{y}^{2} = 2. 4^{2} + 3. 0^{2} = 5.76 + 9.0 = 14.76 \approx 3.84 m .$ Why Pythagoras? The two surviving pieces $L_{x}$ and $L_{y}$ meet at a right angle, so the true rod length is the hypotenuse of that little triangle — Pythagoras is exactly the tool that turns two perpendicular sides into one straight length.

Verify: $3.84 < 5.0$ ✔ (shorter, matching the forecast). The new angle is $arctan (3.0/2.4) = arctan (1.25) \approx 51. 3^{\circ}$ — steeper than $3 7^{\circ}$ , exactly as forecast, because only the horizontal side shrank. ✔

Active recall

Recall Self-test on the cells

Given $L_{0}$ and $v$ , do you multiply or divide by $γ$ ? ::: Divide — the lab sees the contracted (smaller) length. A rod shrinks to $0.60$ of its rest length; find $v$ . ::: $0.6 = 1 - v^{2} / c^{2} \Rightarrow v = 0.8 c$ . At $v = 0$ what does the formula give? ::: $L = L_{0}$ — no contraction, the sanity check. As $v \to c$ what happens to $L$ ? ::: $L \to 0$ — squashed toward zero length. Does the height of a moving flag change? ::: No — only the dimension along the motion contracts. If two ships each measure the other's rod as shorter, is that a paradox? ::: No — they use different notions of "same time" (relativity of simultaneity). For a tilted rod, which component contracts? ::: Only the component along the direction of motion; recombine with Pythagoras. Ship crosses an 8 ly gap at $0.8 c$ — its own clock reads? ::: $6.0$ years (uses contracted $4.8$ ly ÷ $0.8 c$ ).

Connections

Length contraction — derivation — the parent this page drills.
Lorentz transformation — where every step ultimately comes from.
Time dilation — the partner effect, cross-checked in Ex 8.
Relativity of simultaneity — resolves the Ex 5 "paradox."
Lorentz factor gamma — the $γ$ dial used in every example.
Muon decay experiment — the real-world Ex 7.
Spacetime interval — the invariant behind the Ex 8 agreement.

Worked examples — Length contraction — derivation