Length Contraction — Moving Objects Shrink
Length Contraction
Just as time stretches for moving clocks, space itself contracts for moving objects. A meter stick flying past you at high speed is physically shorter in the direction of its motion than an identical meter stick at rest. This is length contraction, and it is the spatial counterpart of time dilation.
The mathematics is equally simple. If an object has a rest length of L-zero — called the proper length, measured in the frame where the object is stationary — then an observer who sees the object moving at speed v measures a contracted length L equals L-zero divided by gamma, or equivalently L equals L-zero times the square root of one minus v-squared over c-squared.
Because gamma is always at least one, the contracted length is always less than or equal to the proper length. At low speeds gamma is essentially one and the contraction is undetectable. At 87 percent of c, gamma is two and the object appears half its rest length. At 99.5 percent of c, gamma is ten and the object is contracted to one-tenth of its rest length.
Length contraction only occurs along the direction of motion. Dimensions perpendicular to the velocity are unaffected. A rocket flying past you would appear flattened like a pancake in its direction of travel, but its height and width would look normal.
This effect has been experimentally verified in particle accelerators, where fast-moving nuclei appear as flattened disks to laboratory detectors. It also resolves famous thought experiments like the ladder-in-the-barn paradox, where a relativistically contracted ladder can momentarily fit inside a barn that is shorter than the ladder's rest length.
If $L_0$ is the proper length (rest length), the contracted length observed by a moving observer is:
$$L = \frac{L_0}{\gamma} = L_0 \sqrt{1 - \frac{v^2}{c^2}}$$
Since $\gamma \ge 1$, we always have $L \le L_0$: the moving object appears shorter.
The proper length $L_0$ is the length of an object measured in the frame where the object is at rest. It is always the longest length anyone can measure for that object.
A spaceship has a rest length of 100 m and flies past Earth at $v = 0.8c$. What length do Earth observers measure?
Step 1: $\gamma = 1/\sqrt{1 - 0.64} = 1/0.6 = 5/3$
Step 2: $L = L_0/\gamma = 100/(5/3) = 100 \times 3/5 = 60$ m
Earth observers see the 100 m ship contracted to just 60 m.
A ruler appears to be 0.5 m long when moving at $0.6c$. What is its proper length?
Step 1: $\gamma = 1/\sqrt{1 - 0.36} = 1/\sqrt{0.64} = 1/0.8 = 1.25$
Step 2: $L_0 = \gamma \cdot L = 1.25 \times 0.5 = 0.625$ m
At what speed does a 200 m train appear to be 100 m long?
Step 1: $L = L_0/\gamma$, so $\gamma = L_0/L = 200/100 = 2$.
Step 2: $\gamma = 2 \implies v = c\sqrt{1 - 1/4} = c\sqrt{3}/2 \approx 0.866c$.
The distance from Earth to a star is 10 light-years. A ship travels at $0.99c$. What distance does the ship's crew measure?
Step 1: $\gamma = 1/\sqrt{1-0.9801} = 1/\sqrt{0.0199} \approx 7.089$
Step 2: $L = 10/7.089 \approx 1.41$ light-years
The crew sees the trip as only 1.41 light-years — the universe literally shrinks in front of them.
Practice Problems
Show Answer Key
1. $\gamma = 1.25$. $L = 50/1.25 = 40$ m.
2. $\gamma \approx 2.294$. $L_0 = 2.294 \times 2 \approx 4.59$ m.
3. $\gamma = 10/5 = 2$. $v = c\sqrt{3}/2 \approx 0.866c$.
4. $\gamma \approx 70.71$. $L \approx 100000/70.71 \approx 1414$ light-years.
5. $\gamma \approx 1.400$. $L \approx 100/1.400 \approx 71.4$ yards.
6. No. Contraction only occurs along the direction of motion.
7. $\gamma = 2$. Ladder contracts to $20/2 = 10$ m. It just fits.
8. $L = L_0/5 = 20\%$ of its rest length.
9. $\gamma \approx 7.089$. $L \approx 1/7.089 \approx 0.141$ fm.
10. $\gamma \approx 3.203$. $L \approx 4.37/3.203 \approx 1.36$ light-years.