Understanding Length Contraction Revision Notes | A Level AQA Physics

1. The Puzzle: Why Does Motion Shrink Objects?

Imagine a spacecraft travelling at 87% of the speed of light. To an observer watching from Earth, that spacecraft appears physically shorter than when it was sitting on the launch pad. Not an optical illusion — an actual, measurable reduction in length. This is length contraction, one of the most striking predictions of Einstein's special relativity.

Length contraction follows directly from two postulates that Einstein proposed in 1905: the laws of physics are identical in all inertial frames, and the speed of light in a vacuum is the same for all observers regardless of their motion. Once you accept those two ideas, length contraction is not optional — it is a mathematical consequence.

At everyday speeds — a car at 30 m/s, even a jet at 300 m/s — the contraction is so vanishingly small that no instrument could detect it. The effect only becomes significant when an object's speed becomes an appreciable fraction of $c = 3.0 \times 10^8$ m/s. That is why this topic sits firmly in the 'turning points in physics' section of the AQA specification: it marks the moment classical mechanics broke down.

2. Proper Length and the Key Formula

Before applying the formula, you need to be precise about what you are measuring. The proper length $l_0$ is the length of an object measured by an observer who is at rest relative to that object — in other words, measured in the object's own rest frame. Think of it as the length stamped on the object's passport: the value it has when nobody is chasing it.

The term inside the square root, $\left(1 - \frac{v^2}{c^2}\right)$, is always between 0 and 1 for any real speed below $c$. That means $\sqrt{1 - \frac{v^2}{c^2}}$ is always less than or equal to 1, so $l \leq l_0$. The contracted length can never exceed the proper length — and it can never be negative.

Notice that the formula contains $\frac{v^2}{c^2}$, which is the square of the ratio of the object's speed to the speed of light. This ratio is sometimes written as $\beta^2$ in more advanced texts, but AQA requires you to work directly with $v$ and $c$. Keep $c = 3.0 \times 10^8$ m/s unless a question specifies otherwise.

3. Applying the Formula: Worked Calculations

Theory clicks into place when you work through a real calculation. The most common exam error is mishandling the arithmetic inside the square root — particularly squaring the speed and the speed of light separately before dividing. Work through each step deliberately.

Worked example: Calculating the contracted length of a spacecraft

A spacecraft has a proper length of 200 m. It travels at $v = 2.4 \times 10^8$ m/s relative to an observer on Earth. Calculate the length of the spacecraft as measured by the Earth observer. [3 marks]

1. $l = l_0\sqrt{1 - \dfrac{v^2}{c^2}}$ — [M1] Stating the correct formula earns the method mark. Always write it out in full before substituting.
2. $v^2 = (2.4 \times 10^8)^2 = 2.4 \times 2.4 \times 10^{16} = 5.76 \times 10^{16}$ m$^2$s$^{-2}$ $c^2 = (3.0 \times 10^8)^2 = 3.0 \times 3.0 \times 10^{16} = 9.0 \times 10^{16}$ m$^2$s$^{-2}$ — Square each value explicitly. A very common error is writing $(3.0 \times 10^8)^2 = 3.0 \times 10^{16}$ — the coefficient must also be squared: $3.0^2 = 9.0$.
3. $\dfrac{v^2}{c^2} = \dfrac{5.76 \times 10^{16}}{9.0 \times 10^{16}} = \dfrac{5.76}{9.0} = 0.64$ — The powers of ten cancel, leaving a dimensionless ratio. This ratio is always between 0 and 1 for any sub-light speed.
4. $\sqrt{1 - 0.64} = \sqrt{0.36} = 0.60$ — Check: $0.60 \times 0.60 = 0.36$. The Lorentz factor here is 0.60, meaning the spacecraft appears 60% of its proper length.
5. $l = 200 \times 0.60 = 120$ m — [A1] The contracted length is 120 m. Include units. A sanity check: $l < l_0$ (120 m < 200 m), which is physically correct.

Answer: $l = 120$ m

That sanity check in the final step is worth building into every answer: the contracted length must always be strictly less than the proper length (unless $v = 0$). If your answer gives $l > l_0$, something has gone wrong in the arithmetic.

4. Direction of Contraction and Physical Interpretation

Length contraction is not a uniform shrinkage in all directions. It acts only along the direction of motion. A spacecraft moving horizontally contracts in its horizontal dimension; its height and width, which are perpendicular to the motion, remain unchanged as measured by the stationary observer.

This is a direct consequence of the mathematics: the Lorentz transformation only modifies the spatial coordinate parallel to the relative velocity. Dimensions perpendicular to motion transform identically between frames. So if a spacecraft is 200 m long and 20 m wide, and it moves along its length, the observer measures a shorter length but the same 20 m width.

A subtler point worth appreciating: length contraction is symmetric. If observer A sees observer B's ruler contracted, then observer B equally sees observer A's ruler contracted by the same factor. Neither frame is 'really' moving — both are inertial, and both are equally valid. This symmetry is what makes special relativity genuinely different from classical mechanics, where there was an assumed absolute space.

5. Why Length Contraction Happens: The Conceptual Argument

Length contraction and time dilation are two sides of the same coin. Recall from the time dilation result: a moving clock runs slow by the factor $\sqrt{1 - v^2/c^2}$. Now consider measuring the length of a moving object. To measure a length, you must record the positions of both ends simultaneously. But simultaneity is relative — two events that are simultaneous in one frame are not simultaneous in another frame moving relative to it.

Because the moving object's clock runs slow, and because simultaneity shifts between frames, the stationary observer ends up recording the two end-positions at moments that, from the object's perspective, are not simultaneous. The net result is that the stationary observer measures a shorter distance between the ends. The mathematics works out to exactly the same Lorentz factor that appears in time dilation.

A useful way to see the consistency: a muon created in the upper atmosphere at 15 km altitude reaches Earth's surface even though, at its speed, classical physics predicts it should decay first. From Earth's frame, the muon's clock runs slow (time dilation). From the muon's frame, the atmosphere is length-contracted to a much shorter distance. Both frames agree on the outcome — the muon reaches the surface — but they explain it differently. Length contraction and time dilation are not competing effects; they are the same physical reality described from different frames.

6. The Negligible Effect at Everyday Speeds

One of the most important things to appreciate about length contraction is how utterly undetectable it is at human-scale speeds. Consider a car travelling at 30 m/s — fast by road standards, but a tiny fraction of $c$.

Worked example: Estimating length contraction at a non-relativistic speed

A car of proper length 4.0 m travels at 30 m/s. Calculate the contracted length as measured by a stationary observer, and comment on whether the effect is detectable.

1. $\dfrac{v^2}{c^2} = \dfrac{30^2}{(3.0 \times 10^8)^2} = \dfrac{900}{9.0 \times 10^{16}} = 1.0 \times 10^{-14}$ — $30^2 = 30 \times 30 = 900$. The ratio is extraordinarily small — fourteen orders of magnitude below 1.
2. $\sqrt{1 - 1.0 \times 10^{-14}} \approx 1 - 5.0 \times 10^{-15}$ — For very small $x$, $\sqrt{1-x} \approx 1 - x/2$. The Lorentz factor is indistinguishable from 1 at this speed.
3. $l = 4.0 \times (1 - 5.0 \times 10^{-15}) \approx 4.0$ m\n(Difference $\approx 2.0 \times 10^{-14}$ m) — The contraction is about $2 \times 10^{-14}$ m — far smaller than an atom (roughly $10^{-10}$ m in diameter). Completely undetectable.

Answer: Contracted length $\approx 4.0$ m; the contraction is of order $10^{-14}$ m and is entirely undetectable.

This calculation explains why Newton's mechanics worked perfectly well for centuries. At speeds far below $c$, the Lorentz factor is so close to 1 that classical mechanics is an excellent approximation. Special relativity only becomes essential when $v$ is a significant fraction of $c$ — typically above about 10% of $c$, where the contraction exceeds 0.5%.

7. Exam Technique and Common Mistakes

Length contraction questions on AQA papers tend to follow a predictable pattern: you are given a proper length and a speed, and asked to calculate the contracted length. Occasionally the question is inverted — you are given $l$ and asked to find $v$. Either way, the same formula applies; you are rearranging for a different variable.

Worked example: Rearranging for speed given contracted length

A rod has a proper length of 3.0 m. An observer measures its length as 2.4 m. Calculate the speed of the rod relative to the observer. [4 marks]

1. $l = l_0\sqrt{1 - \dfrac{v^2}{c^2}}$ $2.4 = 3.0\sqrt{1 - \dfrac{v^2}{c^2}}$ — [M1] Correct formula with substitution. Identify $l = 2.4$ m and $l_0 = 3.0$ m.
2. $\sqrt{1 - \dfrac{v^2}{c^2}} = \dfrac{2.4}{3.0} = 0.80$ — [M1] Divide both sides by $l_0 = 3.0$. The Lorentz factor is 0.80.
3. $1 - \dfrac{v^2}{c^2} = 0.80^2 = 0.80 \times 0.80 = 0.64$ — Squaring removes the square root. Check: $0.80 \times 0.80 = 0.64$.
4. $\dfrac{v^2}{c^2} = 1 - 0.64 = 0.36$ — Rearrange by subtracting from 1.
5. $v^2 = 0.36 \times c^2 = 0.36 \times (3.0 \times 10^8)^2 = 0.36 \times 9.0 \times 10^{16} = 3.24 \times 10^{16}$ m$^2$s$^{-2}$ $v = \sqrt{3.24 \times 10^{16}} = \sqrt{3.24} \times 10^8$ $\sqrt{3.24} \approx 1.80$ (since $1.80 \times 1.80 = 3.24$) $v = 1.80 \times 10^8$ m/s — [A1] Alternatively, note $v = 0.60c = 0.60 \times 3.0 \times 10^8 = 1.8 \times 10^8$ m/s. Sanity check: $v < c$, as required.

Answer: $v = 1.8 \times 10^8$ m/s (or equivalently $0.60c$)

When a question uses the command word calculate, you must show the formula, the substitution with values and units, and the final answer with units. Omitting any of these steps risks losing method marks even if your numerical answer is correct. When the command word is explain, you need cause-and-effect language: state what happens and why, linking back to the postulates of special relativity or the Lorentz factor explicitly.

Length contraction, described by $l = l_0\sqrt{1 - v^2/c^2}$, is a real, measurable reduction in the length of a moving object along its direction of motion, arising directly from the postulates of special relativity — and it is one of the clearest signs that space and time are not the absolute, fixed backdrop that Newton assumed.