Year 1 Sem 2/LC Classical Mechanics and Relativity 2/Lecture 09

Lecture 9: Special Relativity II: Length Contraction
Thu 19 Feb 2026 15:00

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Proving Length Contraction

Consider a horizontal laser cavity. We fire a laser from the left of the cavity (point A) to the right of the cavity (point B). The laser bounces off a mirror at point B and bounces back.

The cavity has length $L_0$ as measured in the rest frame $\Sigma_0$. In this frame:

$$t_0 - \frac{2L_0}{c}, \qquad \text{Where $t_0$ is the time taken for the A->B->A journey in $\Sigma_0$}$$

Now suppose the cavity moves at speed $v$ relative to a stationary observer in a different frame ($\Sigma$).

The observer sees the cavity as having length $L$, how long does the A->B->A journey take as observed by a stationary observer in $\Sigma$.

$$t = t_{A \to B} + t_{B \to A}$$

The cavity is moving away from the observer, so the laser beam doesn’t just travel $L$ when going from $A \to B$. It must travel $L$ plus the distance the whole cavity has moved away in this time (as $B$ is moving away from the pulse). This extra distance is $vt_{A \to B}$, hence the distanced travelled when going from A to B is $L+ vt_{A \to B}$.

As $c$ is constant in all frames, we can say:

$$c = \frac{L + vt_{A \to B}}{t_{A \to B}}$$

$$t_{A \to B} (c - v) = L$$

$$t_{A \to B} = \frac{L}{c - v}$$

And now on the return journey from $B \to A$, as it travels back towards $A$, the cavity is moving in the same direction as the light, so $A$ “catches up” to the pulse and results in a smaller required distance to be travelled, $L - vt_{B \to A}$

$$c = \frac{L - vt_{B \to A}}{t_{B \to A}} \implies t_{B \to A} = \frac{L}{c+v}$$

And returning to the total time:

$$t = t_{A \to B} + t_{B \to A}$$

$$t = \frac{L}{c - v} + \frac{L}{c + v}$$

$$= L \left(\frac{1}{c - v} + \frac{1}{c + v}\right)$$

$$= L \left(\frac{c + v + c - v}{(c-v)(c+v)}\right)$$

$$= L \frac{2c}{c^2 - v^2}$$

$$= \frac{2L}{c} \frac{c^2}{c^2 - v^2}$$

$$= \frac{2L}{c} \frac{1}{1 - \frac{v^2}{c^2}}$$

$$= \frac{2L}{c} \gamma^2$$

We now apply time dilation, which says that $t = \gamma t_0$, hence:

$$\gamma t_0 = \frac{2L}{c} \gamma^2$$

$$t_0 = \frac{2L}{c} \gamma$$

From the rest frame, we know that $t_0 = 2L_0 / c$, so:

$$\frac{2L_0}{c} = \frac{2L \gamma}{c}$$

$$L_0 = L \gamma$$

As $\gamma > 1, \, \forall (u < c)$, the proper length measured from the rest frame will always be greater than the length measured by an observer, hence length is contracted for a moving object.

Example

In a particle accelerator, protons are accelerated to a measly $0.9c$. These protons pass through a tunnel of length $2 \mathrm{km}$ as viewed from the a laboratory rest frame at the accelerator.

Recall that:

$$t = \gamma t_0$$

$$L = \frac{L_0}{\gamma}$$

$$\gamma = \frac{1}{\sqrt{1 - \beta^2}}, \qquad \text{where: } \beta = u/c$$

How long would the journey take, according to the rest frame?

Viewed in the lab frame, and $v = d/t$ so:

$$t = \frac{2 \times 10^3 \mathrm{m}}{0.9 \times 3 \times 10^8} = 7.4 \times 10^{-6}\mathrm{s} = 7.4 \mathrm{\mu\, s}$$

No relativity needed!

How long would the journey take, according to the proton’s frame?

We do need to consider relativity here, and use time dilation:

$$t = \gamma t_0$$

$$\gamma = \frac{1}{\sqrt{1 - 0.9^2}} = 2.3$$

$$t = \gamma t_0 \implies t_0 = \frac{t}{\gamma} = \frac{7.4 \mathrm{\mu\, s}}{2.3} = 3.2 \mathrm{\mu\,s}$$

How long is the tunnel, according to the proton’s frame?

The rest frame now corresponds to the tunnel. It has proper length in its rest frame of $2 \mathrm{k\, m}$.

As viewed by the protons in their rest frame, it is the tunnel that is moving and is rushing towards them at $0.9c$. This is therefore a length contraction problem.

$$L = \frac{L_0}{\gamma} = \frac{2 \mathrm{km}}{2.3} = 870 \mathrm{m}$$

Checking Values

To check, we can reuse $v = d/t$ but this time in the proton rest frame.

$$d = 870\mathrm{m}, \quad t = 3.2 \times 10^{-6} \mathrm{s}$$

Hence

$$v = \frac{870 \mathrm{m}}{3.2 \times 10^{-6} \mathrm{s}} = 2.7 \times 10^8 \mathrm{ms^{-1}} = 0.9c \text{ as required!}$$

Example II: Cosmic Rays

The highest energy cosmic rays are protons with massive energies $E \sim 10^{20} \mathrm{eV}$, compared to the LHC with $E \sim 10^{12} \mathrm{eV}$

This corresponds to a $\gamma = 10^{11}$, and our galaxy is $\sim 10^{20}$m across. ¹

As viewed on earth, these cosmic protons travel at $v \approx c$, hence,

$$t \approx \frac{d}{v} \approx \frac{10^{20}}{3 \times 10^8} = 3 \times 10^{11} \mathrm{s} \approx 10^4 \mathrm{ years}$$

As viewed by the protons:

$$t_0 = \frac{t}{\gamma} = \frac{3 \times 10^{11}}{10^{11}} \approx 3 \mathrm{s}$$

Which is starkly different!

Lorentz Transformations

In general, transforming between coordinates in two different frames can get extremely messy, more than can easily be handed in simple applications of length contraction or time dilation.

The Lorentz Transformations provide a general set of coordinate transformations between $\Sigma$ and $\Sigma^\prime$, i.e. $(x, y, z, t) \to (x^\prime, y^\prime, z^\prime, t^\prime)$.

These transformations must be:

Symmetric about a change in sign of $u$, i.e. the transformation from $\Sigma \to \Sigma^\prime$ with $u$ and the transformation from $\Sigma^\prime \to \Sigma$ with $-u$ must be the same.
They must be linear as otherwise a constant speed in one would not be constant in the other (and therefore not inertial frames).
If relative velocity is only in the $x$ direction, the transformations must be independent of $y$ and $z$.

These transformations are (assuming motion only in $x$):

$$x = \gamma (x^\prime + v t^\prime)$$

$$y = y^\prime$$

$$z = z^\prime$$

$$t = \gamma \left(t^\prime + \frac{x^\prime v}{c^2}\right)$$

And:

$$x^\prime = \gamma (x- v t)$$

$$y^\prime = y$$

$$z^\prime = z$$

$$t^\prime = \gamma \left(t - \frac{x v}{c^2}\right)$$

We do not need to know the derivations, but do need to know the results.

Footnotes

1.

$\gamma = E/mc^2$ - will be expanded on in later lectures.

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