Which Length Contracts? Moving Rods in HSC Physics

A 5.0 m rod is travelling to the right at \(0.60c\), but it sits diagonally across its spacecraft. Should an observer standing beside the track measure its length as 4.0 m, since length contraction reduces lengths by 20% at that speed?

Not necessarily. Only the component of length parallel to the relative motion contracts. The perpendicular component stays unchanged. If the rod is diagonal, you must contract its parallel component first, then calculate its new total length.

That gives us three different cases: a rod pointing along the motion gets shorter by the full contraction factor, a perpendicular rod keeps its length, and a diagonal rod changes both its measured length and its measured angle.

01Which length is actually being measured?

Imagine a spacecraft carrying a straight antenna. We can describe its orientation using two directions:

• The \(x\)-direction runs horizontally, parallel to the spacecraft’s motion.
• The \(y\)-direction runs vertically across the spacecraft, perpendicular to its motion.

We’ll take motion to the right as positive \(x\). Assume the antenna is straight, doesn’t rotate relative to the spacecraft, and moves at constant velocity. We’re comparing inertial reference frames, not studying acceleration or physical bending.

Imagine the antenna casting two imaginary shadows onto rulers: one along \(x\), and one along \(y\). The shadows represent its horizontal and vertical components.

Special relativity changes the horizontal component measured by the laboratory but leaves the vertical component unchanged.

The shadow idea is only a way to visualise the geometry. An actual photograph involves light-travel times, which introduce another complication.

Start with the rest frame

The antenna’s rest frame is the reference frame in which both its endpoints are stationary.

In that frame, we define:

• \(L_0\): the antenna’s proper length, or rest length, in metres (m).
• \(\Delta x_0\): its endpoint separation along \(x\), in metres.
• \(\Delta y_0\): its endpoint separation along \(y\), in metres.

For a straight antenna, Pythagoras gives:

\[
L_0=\sqrt{\Delta x_0^2+\Delta y_0^2}
\]

Now consider a laboratory observer who sees the antenna travelling along \(+x\).

The laboratory measures different endpoint separations, \(\Delta x\) and \(\Delta y\). These must be measured at the same laboratory time.

The important distinction is:

Here, \(v\) is the relative speed, and \(\gamma\) (gamma) is the Lorentz factor. We’ll calculate it shortly.

Notice that the laboratory does not automatically measure \(L_0/\gamma\) for a diagonal rod. That shortcut works only when the rod is parallel to the motion.

02How could an experiment distinguish these possibilities?

Suppose you wanted to measure a diagonal antenna moving past a laboratory at \(0.60c\).

You know that the antenna is 5.00 m long in its own rest frame, with a horizontal endpoint separation of 3.00 m and a vertical separation of 4.00 m.

How would you find out whether its entire length contracts or only its horizontal component?

First design: photograph the moving antenna

You could place a camera beside the track and photograph the antenna as it passes.

The naive prediction is that you can simply measure the antenna’s endpoints in the image.

But there’s a problem.

Light from the two endpoints can travel different distances to the camera. Photons arriving together at the camera may have left the endpoints at different times.

At relativistic speeds, that difference matters. A photograph’s apparent shape is not necessarily the same as the object’s length measured between simultaneous positions in the laboratory.

The first design hasn’t isolated the quantity we’re trying to measure.

Revised design: synchronised position detectors

Instead, imagine an idealised laboratory equipped with a two-dimensional grid of position sensors.

Each sensor has a known position and access to clocks synchronised in the laboratory frame.

At one chosen laboratory time, the sensors record the locations of both antenna endpoints. Their readings can be transmitted and compared afterwards.

We can then calculate the horizontal and vertical separations without confusing differences in observation time with differences in position.

This is a thought experiment, not a report of measurements from a real laboratory.

For our 3-4-5 m antenna, three competing models make different predictions:

The decisive observation would be the vertical separation.

Both the incorrect whole-antenna model and special relativity predict a horizontal separation of 2.40 m. Only special relativity predicts that the vertical separation remains 4.00 m.

That’s why measuring both components is more informative than recording one total length.

03The equations: contract the component, not the entire rod

The Lorentz factor is

\[
\gamma=\frac{1}{\sqrt{1-v^2/c^2}}
\]

where:

• \(\gamma\) is a dimensionless number.
• \(v\) is the relative speed between the laboratory and rod, in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

For a rod moving along \(x\), its components transform as:

\[
\begin{aligned}
\Delta x&=\frac{\Delta x_0}{\gamma}\\
\Delta y&=\Delta y_0
\end{aligned}
\]

The laboratory-frame total length is then

\[
L=\sqrt{\Delta x^2+\Delta y^2}
\]

where \(L\) is the simultaneous end-to-end length measured by the laboratory observer.

If the rod is entirely horizontal, \(\Delta y_0=0\), so \(L=L_0/\gamma\).

If it is entirely vertical, \(\Delta x_0=0\), so \(L=L_0\).

For any other orientation, the laboratory length lies between these two extremes.

04Does the rod’s mass contract too?

No. Length and mass must be treated differently.

In modern special relativity, the mass usually written \(m_0\) is the object’s invariant mass, also called its rest mass. Every inertial observer agrees on its value.

What changes between frames is the object’s total energy.

For an isolated rod with invariant mass \(m_0\), its rest energy is

\[
E_0=m_0c^2
\]

In a frame where the rod moves at speed \(v\), its total energy is

\[
E=\gamma m_0c^2
\]

Here, \(E_0\) and \(E\) are measured in joules (J), and \(m_0\) is measured in kilograms (kg).

The difference is kinetic energy:

\[
K=E-E_0
\]

Suppose our idealised antenna has an invariant mass of 1.00 kg and moves at \(0.60c\).

Its rest energy is \(9.00\times10^{16}\ \mathrm{J}\), while its laboratory total energy is \(1.125\times10^{17}\ \mathrm{J}\).

Its invariant mass is still 1.00 kg.

You might encounter the older term relativistic mass, defined using \(\gamma m_0\). It is less useful here because it can make a change in energy sound like a change in invariant mass.

For the 2026 HSC Physics course, keep the distinction clear: relative motion changes measured length and total energy, but it does not change an object’s invariant mass.

This comparison assumes the rod remains intact and its internal energy doesn’t change. Heating, deforming, or breaking an object can change the invariant mass of the system being considered. Simply viewing the same unchanged object from another inertial frame cannot.

05A quick decision method for moving-rod questions

Before substituting into a length-contraction equation, make four decisions.

There are three useful warning signs:

1. If you’ve contracted a perpendicular separation, you’ve applied the wrong rule.
2. If a valid laboratory-frame length for an unchanged rod is greater than its proper length, check your frames, timing, and calculations.
3. If you’ve changed the rod’s invariant mass simply because its speed changed, you’ve confused mass with total energy.

The second warning assumes you’re comparing simultaneous endpoint separations for a straight, unchanged rod moving inertially. A distance between endpoint positions recorded at different times is not necessarily a rod length.

06Questions and solutions

The questions become progressively harder. The final problem is a conceptual extension beyond a routine HSC length-contraction calculation.

Question 1

A straight 3.60 m rod moves at \(0.80c\) relative to a laboratory, with its length parallel to its motion.

Calculate its laboratory-frame length.

If the same rod were instead oriented perpendicular to the direction of motion, what laboratory-frame length would be measured?

Solution 1

The laboratory measures 2.16 m when the rod is parallel to the motion, and 3.60 m when it is perpendicular.

For the parallel rod, apply length contraction:

\[
\begin{aligned}
L&=L_0\sqrt{1-v^2/c^2}\\
&=3.60\sqrt{1-0.80^2}\\
&=3.60(0.60)\\
&=2.16\ \mathrm{m}
\end{aligned}
\]

Here, \(L_0=3.60\ \mathrm{m}\) is the rest length, and \(L\) is the laboratory length.

For the perpendicular rod, none of its length lies along the relative motion. Its full 3.60 m length is transverse and remains unchanged.

The tempting mistake is to apply the factor 0.60 to both orientations. The speed is the same, but the relevant direction of measurement is different.

Question 2

A laboratory’s synchronised sensors measure the simultaneous endpoint separations of a moving diagonal rod as:

• Horizontal separation: 3.60 m.
• Vertical separation: 8.00 m.

The rod moves along \(+x\) at \(0.80c\) and has no rotation relative to its own rest frame.

Calculate:

a) The rod’s proper length.

b) The angle it makes with the positive \(x\)-axis in its rest frame.

c) The total length and angle measured in the laboratory.

A student suggests finding the proper length by dividing the laboratory’s total length by 0.60. Explain whether that method is valid.

Solution 2

The rod’s proper length is 10.0 m, with a rest-frame angle of \(53.1^\circ\). In the laboratory, its length is approximately 8.77 m, and its angle is \(65.8^\circ\).

The student’s method is invalid because the rod is diagonal.

Part a: Recover the proper length.

At \(0.80c\),

\[
\gamma=\frac{1}{\sqrt{1-0.80^2}}
=\frac{1}{0.60}
\]

The horizontal rest-frame component must be larger than the laboratory component:

\[
\begin{aligned}
\Delta x_0&=\gamma\Delta x\\
&=\frac{3.60}{0.60}\\
&=6.00\ \mathrm{m}
\end{aligned}
\]

The vertical component is unchanged:

\[
\Delta y_0=8.00\ \mathrm{m}
\]

Now apply Pythagoras:

\[
\begin{aligned}
L_0&=\sqrt{6.00^2+8.00^2}\\
&=\sqrt{100}\\
&=10.0\ \mathrm{m}
\end{aligned}
\]

Part b: Find the rest-frame angle.

\[
\begin{aligned}
\tan\theta_0&=\frac{8.00}{6.00}\\
\theta_0&=53.1^\circ
\end{aligned}
\]

Part c: Find the laboratory length and angle.

\[
\begin{aligned}
L&=\sqrt{3.60^2+8.00^2}\\
&=\sqrt{76.96}\\
&=8.77\ \mathrm{m}
\end{aligned}
\]

And

\[
\begin{aligned}
\theta&=\tan^{-1}\left(\frac{8.00}{3.60}\right)\\
&=65.8^\circ
\end{aligned}
\]

The rod is steeper and shorter in the laboratory.

Why the student’s shortcut fails:

Dividing the full laboratory length by 0.60 would give approximately 14.6 m.

That assumes the entire end-to-end separation was reduced by the same factor. But only the horizontal component contracted; the vertical component remained 8.00 m.

The correct method is to reconstruct the rest-frame components separately, then combine them.

Question 3

A diagonal rod moves at a constant \(0.60c\) along \(+x\).

Two laboratory sensors report the following endpoint positions:

Here, \(1\ \mathrm{ns}=10^{-9}\ \mathrm{s}\).

The rod has no vertical motion, doesn’t rotate, and moves inertially throughout the measurements.

A student calculates the distance between the two recorded positions and claims this is the rod’s laboratory length.

a) Explain why that calculation is not a valid length measurement.

b) Calculate the rod’s actual simultaneous laboratory-frame length.

c) Determine its proper length.

Solution 3

The actual laboratory-frame length is 5.66 m, and the proper length is 6.40 m. Simply subtracting the recorded endpoint positions is invalid because they were measured 15.0 ns apart.

Part a: Identify the timing error.

The rod continues moving between the two detections.

A direct laboratory length requires the positions of both endpoints at one laboratory time. The readings supplied refer to two different times.

Their horizontal coordinate difference is 6.70 m, but that difference includes the rod’s movement during the 15.0 ns interval.

Part b: Reconstruct simultaneous endpoint positions.

The rod’s speed is

\[
v=0.60(3.00\times10^8)
=1.80\times10^8\ \mathrm{m\,s^{-1}}
\]

During 15.0 ns, it moves

\[
\begin{aligned}
d&=v\Delta t\\
&=(1.80\times10^8)(15.0\times10^{-9})\\
&=2.70\ \mathrm{m}
\end{aligned}
\]

Here, \(d\) is the horizontal distance moved between detections, and \(\Delta t\) is their time difference.

So the front endpoint’s horizontal position at \(t=0\) was

\[
x_{\mathrm{front}}=8.70-2.70
=6.00\ \mathrm{m}
\]

At that same time, the rear was at \(x=2.00\ \mathrm{m}\).

Therefore,

\[
\Delta x=6.00-2.00
=4.00\ \mathrm{m}
\]

The vertical separation remains \(\Delta y=4.00\ \mathrm{m}\), since the rod has no vertical motion.

The actual laboratory length is

\[
\begin{aligned}
L&=\sqrt{4.00^2+4.00^2}\\
&=\sqrt{32.0}\\
&=5.66\ \mathrm{m}
\end{aligned}
\]

Part c: Recover the proper length.

At \(0.60c\), the contraction factor is 0.80.

\[
\begin{aligned}
\Delta x_0&=\frac{4.00}{0.80}\\
&=5.00\ \mathrm{m}
\end{aligned}
\]

The vertical component remains 4.00 m.

\[
\begin{aligned}
L_0&=\sqrt{5.00^2+4.00^2}\\
&=\sqrt{41.0}\\
&=6.40\ \mathrm{m}
\end{aligned}
\]

The tempting route would give

\[
\sqrt{6.70^2+4.00^2}=7.80\ \mathrm{m}
\]

That value is even greater than the proper length.

This doesn’t disprove length contraction. It tells us that the 7.80 m calculation was never a valid simultaneous length measurement.

Question 4

Two straight rods, A and B, each have a proper length of 10.0 m and an invariant mass of 0.500 kg.

They move to the right relative to the same laboratory.

• Rod A is parallel to the direction of motion and travels at \(0.60c\).
• Rod B travels faster, at \(0.80c\). In its own rest frame, its endpoint separations are 6.00 m horizontally and 8.00 m vertically.

Assume the rods are unchanged internally, and the given masses are their total invariant masses.

a) Which rod has the greater laboratory-frame end-to-end length?

b) Which rod has the greater total laboratory-frame energy?

c) Explain why a laboratory observer cannot rank the rods’ speeds using their measured total lengths alone.

Solution 4

Rod B has both the greater laboratory-frame length and the greater laboratory-frame total energy, even though it moves faster. Both rods have the same invariant mass of 0.500 kg.

The key is that the rods have different orientations.

Part a: Compare their laboratory lengths.

For Rod A, the contraction factor is 0.80:

\[
\begin{aligned}
L_A&=10.0(0.80)\\
&=8.00\ \mathrm{m}
\end{aligned}
\]

For Rod B, the contraction factor is 0.60.

Its horizontal component becomes

\[
\Delta x_B=6.00(0.60)=3.60\ \mathrm{m}
\]

Its vertical component remains

\[
\Delta y_B=8.00\ \mathrm{m}
\]

Therefore,

\[
\begin{aligned}
L_B&=\sqrt{3.60^2+8.00^2}\\
&=8.77\ \mathrm{m}
\end{aligned}
\]

Rod B is approximately 0.77 m longer than Rod A according to simultaneous laboratory measurements.

Part b: Compare their total energies.

Both rods have the same rest energy:

\[
\begin{aligned}
E_0&=m_0c^2\\
&=0.500(3.00\times10^8)^2\\
&=4.50\times10^{16}\ \mathrm{J}
\end{aligned}
\]

Rod A has \(\gamma_A=1.25\):

\[
\begin{aligned}
E_A&=1.25E_0\\
&=5.63\times10^{16}\ \mathrm{J}
\end{aligned}
\]

Rod B has \(\gamma_B=1/0.60\):

\[
\begin{aligned}
E_B&=\frac{E_0}{0.60}\\
&=7.50\times10^{16}\ \mathrm{J}
\end{aligned}
\]

Rod B has greater total laboratory energy because its speed is greater.

Its invariant mass is not greater.

Part c: Explain the apparently reversed length comparison.

The tempting prediction is that Rod B must be shorter because it travels faster.

That prediction would be correct if both rods had the same proper length and both lay parallel to the motion.

But Rod B has an 8.00 m transverse component that doesn’t contract. Rod A has no transverse component at all.

A faster rod can therefore have a greater measured total length than a slower rod when their orientations differ.

Meanwhile, for rods with the same invariant mass, laboratory total energy increases with \(\gamma\), independently of their orientation.

The two comparisons answer different questions. Length depends on both orientation and speed; total energy here depends on invariant mass and speed.

Question 5

A thin, straight 5.00 m rod moves right at \(0.60c\) through an idealised rectangular laboratory chamber.

In the rod’s rest frame, its endpoint separations are:

• 3.00 m horizontally, parallel to the motion.
• 4.00 m vertically, perpendicular to the motion.

The chamber has two vertical shutters, one at \(x=0\) and the other at \(x=2.50\ \mathrm{m}\). It is tall enough for the rod to fit vertically.

At laboratory time \(t=0\), the rod’s rear endpoint is at \(x=0.050\ \mathrm{m}\), and its front endpoint is at \(x=2.450\ \mathrm{m}\).

Both shutters close simultaneously according to synchronised laboratory clocks. They remain closed for \(0.20\ \mathrm{ns}\), then reopen.

Assume the shutters close and reopen instantaneously, the rod has negligible thickness, and there are no collisions except those caused by contact with a closed shutter.

A student argues:

“The laboratory says the rod fits inside, but the rod’s rest frame says its horizontal length is 3.00 m while the chamber is contracted to only 2.00 m. So the rod must hit a shutter.”

A second student argues:

“If both shutters are closed together in the laboratory, they must also be closed together in the rod’s frame. The rod is trapped.”

Determine whether either argument is correct.

Your explanation must establish:

a) Whether the rod collides with a shutter during the 0.20 ns closure, using a numerical calculation.

b) How the rod and chamber can have apparently incompatible lengths in the two reference frames without contradicting each other.

c) Which shutter closes first in the rod’s rest frame, and why.

d) What would change if the shutters instead remained closed for \(0.40\ \mathrm{ns}\).

Solution 5

Neither student is correct. The rod passes through without a collision when the shutters remain closed for 0.20 ns. A 0.40 ns closure, however, causes the rod’s front endpoint to hit the right shutter.

The apparent contradiction comes from confusing simultaneous events in different reference frames. A change in measured length doesn’t create a disagreement about whether a physical collision occurs.

Part a: Test the laboratory clearance.

At \(0.60c\), the rod’s horizontal component contracts by a factor of 0.80:

\[
\begin{aligned}
\Delta x&=3.00(0.80)\\
&=2.40\ \mathrm{m}
\end{aligned}
\]

Its vertical component remains 4.00 m.

At \(t=0\), both endpoints are between the laboratory shutters, as specified.

The rod travels at

\[
v=1.80\times10^8\ \mathrm{m\,s^{-1}}
\]

During 0.20 ns, it moves

\[
\begin{aligned}
d&=v\Delta t\\
&=(1.80\times10^8)(0.20\times10^{-9})\\
&=0.036\ \mathrm{m}
\end{aligned}
\]

The front endpoint moves from \(x=2.450\ \mathrm{m}\) to

\[
x=2.450+0.036=2.486\ \mathrm{m}
\]

The right shutter is at \(x=2.500\ \mathrm{m}\).

The remaining clearance is

\[
2.500-2.486=0.014\ \mathrm{m}
\]

So the front endpoint remains 1.4 cm away from the right shutter when it reopens.

The rear endpoint moves right, away from the left shutter.

There is no collision.

Part b: Explain why the rod doesn’t fit in its own frame.

In the rod’s rest frame, the horizontal endpoint separation is 3.00 m.

The chamber is moving left at \(0.60c\), so its horizontal separation between shutters is

\[
\begin{aligned}
D’&=2.50(0.80)\\
&=2.00\ \mathrm{m}
\end{aligned}
\]

Here, \(D’\) is the chamber’s shutter separation measured simultaneously in the rod’s frame.

The rod’s 3.00 m horizontal extent is indeed larger than the chamber’s 2.00 m extent.

So there is no instant in the rod’s frame when the entire rod lies between both shutters.

But that doesn’t contradict the laboratory result.

The laboratory’s statement that the entire rod is inside the chamber refers to the positions of both endpoints at one laboratory time.

The rod’s frame does not assign the same time to those separated endpoint-position events.

A simultaneous laboratory snapshot isn’t a simultaneous rod-frame snapshot.

Part c: Determine the shutter order.

The right shutter, at larger \(x\), closes before the left shutter according to clocks synchronised in the rod’s frame.

The closures are simultaneous only in the laboratory frame.

This is the relativity of simultaneity: two events separated along the direction of relative motion can be simultaneous in one inertial frame but occur at different times in another.

The same applies to the reopening events.

For an optional quantitative extension, the time-coordinate relationship between these frames gives

\[
\Delta t’=-\gamma\frac{v\Delta x}{c^2}
\]

where \(\Delta t’\) is the right shutter’s closing time minus the left shutter’s closing time in the rod’s frame, and \(\Delta x=2.50\ \mathrm{m}\) is their laboratory separation.

Substituting \(\gamma=1.25\):

\[
\begin{aligned}
\Delta t’&=-1.25\frac{(1.80\times10^8)(2.50)}{c^2}\\
&=-6.25\ \mathrm{ns}
\end{aligned}
\]

The negative sign means the right shutter closes 6.25 ns earlier.

Each shutter’s 0.20 ns closure lasts 0.25 ns in the rod’s frame because of time dilation. The intervals therefore don’t overlap in that frame.

The rod-frame observer never sees both shutters closed at once.

Part d: Extend the closure to 0.40 ns.

During 0.40 ns, the rod would travel

\[
\begin{aligned}
d&=(1.80\times10^8)(0.40\times10^{-9})\\
&=0.072\ \mathrm{m}
\end{aligned}
\]

But the front endpoint begins only 0.050 m from the right shutter.

It reaches that shutter after

\[
\begin{aligned}
t&=\frac{d}{v}\\
&=\frac{0.050}{1.80\times10^8}\\
&=2.78\times10^{-10}\ \mathrm{s}\\
&=0.278\ \mathrm{ns}
\end{aligned}
\]

Because the right shutter remains closed for 0.40 ns, the front endpoint strikes it before it reopens.

Both reference frames must agree that this collision occurs. They may disagree about the times and distances involved, but not about whether the same physical collision event happened.

That’s the deeper constraint: relativity changes how observers divide events into space and time, not which physical interactions occur.

07What to carry into the next relativity problem

The calculation rule is simple: identify the rod’s rest frame, resolve its length into components, and contract only the component parallel to relative motion.

The harder judgement is deciding whether the measurements actually represent a length. Endpoint positions measured at different times cannot simply be subtracted, and simultaneous events in one reference frame need not be simultaneous in another.

For the foundational proper-length equation, see Length Contraction in HSC Physics: Proper Length and Calculations.

And when a problem changes from the rod’s geometry to its momentum or energy, carry the same frame discipline with you. The rod’s invariant mass stays fixed, even as its total energy depends on the observer. That distinction is developed in Rest Mass vs Relativistic Energy: HSC Physics Explained.

The next time a moving-rod question seems to produce two incompatible lengths, don’t immediately change the equation. Check the direction of motion, the frame in which the endpoints are at rest, and which events were measured simultaneously.