§-Syllabus dot point

NSWPhysicsSyllabus dot point

Inquiry Question 3: What evidence supports the relativistic model of the universe?

Analyse the Michelson-Morley experiment, state Einstein's two postulates of special relativity, and apply the consequences of time dilation, length contraction and relativity of simultaneity

A focused answer to the HSC Physics Module 7 dot point on light and special relativity. The Michelson-Morley null result, Einstein's two postulates, and quantitative application of time dilation t = gamma t_0, length contraction L = L_0 / gamma and relativity of simultaneity.

Generated by Claude Opus 4.814 min answerUpdated 2026-07-01

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context

What this dot point is asking

NESA wants you to summarise the Michelson-Morley experiment, state Einstein's two postulates, and apply time dilation and length contraction quantitatively. You should also be able to describe the relativity of simultaneity qualitatively.

The answer

The aether problem and Michelson-Morley

By the late nineteenth century, light was understood as a wave. Waves needed a medium, so physicists postulated the luminiferous aether: a hypothetical, all-pervading substance through which light propagated. The aether was assumed stationary (or close to it), so the Earth must move through it at orbital speed (about $30$ km/s).

Michelson and Morley's interferometer aimed to detect this motion. A beam of monochromatic light was split into two perpendicular paths by a half-silvered mirror, reflected off mirrors at equal distance, and recombined. Any difference in transit time between the two paths would produce interference fringes. Rotating the apparatus by $90^\circ$ would swap the "along-aether" and "across-aether" paths and should shift the fringes by a predictable amount (about $0.4$ of a fringe for the 1887 setup).

The result was null: no significant fringe shift was observed in any orientation, season or location. The experiment was repeated many times with increasing precision, always null. The simplest interpretation: there is no aether.

Einstein's two postulates (1905)

Special relativity rests on just two postulates:

Principle of relativity. The laws of physics are the same in all inertial (non-accelerating) reference frames. No experiment can identify an absolute rest frame.
Constancy of the speed of light. The speed of light in vacuum, $c$ , is the same in all inertial frames, regardless of the motion of the source or the observer.

The second postulate is the radical one. It immediately removes the need for an aether and makes the Michelson-Morley null result automatic. But it forces strange consequences for space and time.

The Lorentz factor

Most relativistic formulas use:

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

At everyday speeds, $\gamma \approx 1$ and relativistic effects are negligible. At $v = 0.5c$ , $\gamma \approx 1.155$ ; at $v = 0.9c$ , $\gamma \approx 2.29$ ; at $v = 0.99c$ , $\gamma \approx 7.09$ .

Time dilation

A clock moving with speed $v$ relative to an observer ticks slowly compared to a clock at rest with that observer. If the moving clock measures proper time $t_0$ (time between two events at the same place in its own frame), the time interval measured by the stationary observer is:

t = \gamma t_0

The classic thought experiment: a light clock bounces a photon between two parallel mirrors a distance $L_0$ apart. In its rest frame the round trip is $t_0 = 2 L_0 / c$ . In a frame where the clock moves sideways at speed $v$ , the photon traces a longer zig-zag path, but still travels at $c$ (postulate 2). Equating distances gives $t = \gamma t_0$ .

Time dilation has been confirmed by atomic clocks flown on aircraft (Hafele-Keating, 1971) and by the increased lifetime of fast-moving muons (covered in evidence-for-special-relativity).

Length contraction

An object moving at speed $v$ along its length is measured to be shorter than its proper length $L_0$ (the length in its rest frame) by:

L = \frac{L_0}{\gamma}

Contraction is only along the direction of motion; perpendicular dimensions are unchanged. Like time dilation, it is real in the sense that any measurement in the observer's frame, made with synchronised rulers, gives the contracted value. There is no internal stress in the object; it is the geometry of spacetime that differs between frames.

Relativity of simultaneity

Two events that are simultaneous in one inertial frame are generally not simultaneous in another moving relative to the first.

Einstein's train thought experiment: lightning strikes both ends of a train simultaneously according to an observer on the embankment. The flashes reach the embankment observer at the same time. But an observer at the centre of the moving train is travelling toward the front flash and away from the back flash, so the front flash reaches them first. Because the speed of light is the same in both frames (postulate 2), the train observer must conclude the front strike happened earlier than the back strike. The two observers disagree on which events were simultaneous.

This is the deepest consequence of the postulates: there is no universal "now". Time ordering of causally connected events (cause before effect) is preserved, but ordering of spacelike-separated events depends on the frame.

Worked examples

Model 1 - muon trip (time dilation and length contraction agree). A muon created at the top of the atmosphere ( $15$ km altitude) travels at $0.998c$ toward the ground. Its proper lifetime is $2.2\ \mu\text{s}$ . Show that time dilation lets it reach the ground, and confirm the same conclusion using length contraction.

In the Earth frame: $\gamma = \dfrac{1}{\sqrt{1 - 0.998^2}} = \dfrac{1}{\sqrt{0.003996}} = 15.8$ .

Lifetime as measured from Earth: $t = \gamma t_0 = 15.8 \times 2.2 \times 10^{-6}\ \text{s} = 3.48 \times 10^{-5}\ \text{s}$ .

Distance the muon can cover: $d = vt = (0.998 \times 3.00 \times 10^8\ \text{m s}^{-1})(3.48 \times 10^{-5}\ \text{s}) = 1.04 \times 10^4\ \text{m} = 10.4\ \text{km}$ .

Without dilation, the muon would only travel $v t_0 = 660\ \text{m}$ in $2.2\ \mu\text{s}$ and almost none would reach the surface, so time dilation explains why many do.

In the muon's own frame, its lifetime is the proper time $2.2\ \mu\text{s}$ , but the atmosphere is length-contracted: $L = \dfrac{15\ \text{km}}{15.8} = 0.95\ \text{km}$ , so it only has to cover this much-shorter distance in its proper lifetime. Both frames agree that the muon reaches the ground, via different mechanisms.

Marker's note: state both mechanisms explicitly - the Earth observer explains survival via the muon's dilated lifetime, the muon's own frame explains it via a contracted atmosphere. Giving only one mechanism when the question asks for both loses marks.

Model 2 - length contraction of a spacecraft. A spacecraft has a proper length of $90$ m. Calculate its length as measured by a space-station observer past whom it flies at $v = 0.60c$ .

$\gamma = \dfrac{1}{\sqrt{1 - 0.60^2}} = \dfrac{1}{\sqrt{0.64}} = 1.25$ .

$L = \dfrac{L_0}{\gamma} = \dfrac{90\ \text{m}}{1.25} = 72\ \text{m}$ .

Marker's note: identify which observer measures the proper length before substituting ( $L_0 = 90$ m is measured in the ship's own rest frame); the station observer, who sees the ship moving, always measures the shorter value.

Common traps

Applying time dilation backwards: The moving clock measures the proper time $t_0$ (the shorter interval). The observer who sees the clock moving measures $t = \gamma t_0$ (longer interval).
Applying length contraction backwards: The object at rest in some frame has the proper length $L_0$ in that frame. Any frame that sees the object moving measures a shorter length.
Confusing proper time with elapsed coordinate time: Proper time is between two events at the same place in some frame. Coordinate time depends on the synchronisation of clocks at different places.
Treating $\gamma$ as something that can be smaller than 1: $\gamma \geq 1$ always, equal to 1 only when $v = 0$ .

Saying "moving clocks run slow because of physical effects on their mechanism". Any clock (mechanical, atomic, biological) is dilated equally because it is a property of time itself, not of the clock.

Exam technique

Match the command word: "outline/describe" wants the Michelson-Morley set-up and its null result; "state" wants the two postulates verbatim; "calculate" wants $\gamma$ shown explicitly before it is used in $t = \gamma t_0$ or $L = L_0/\gamma$ ; "explain/analyse" wants the causal chain from postulate 2 to the physical consequence.

Always compute $\gamma$ as a first, separate step and quote it to at least three significant figures before using it in a further calculation - a rounding error in $\gamma$ propagates into every later answer. Use the exact NESA value $c = 3.00 \times 10^8\ \text{m s}^{-1}$ , keep $v$ as a multiple of $c$ where possible (it cancels cleanly), and give final answers to two or three significant figures with the correct unit.

On a $\gamma$ -versus- $v/c$ graph, read the value straight up from the given speed on the horizontal axis to the curve, then across to the vertical axis; the curve's steepening slope near $v = c$ is itself worth describing in an "analyse the graph" question, not just the numerical reading.

In plain English

Imagine you are on a train with a laser pointer, bouncing a laser pulse straight up to a mirror on the ceiling and back down, like a very fast game of catch with light. To you, on the train, the light just goes straight up and straight down. But to someone standing on the platform watching the train zoom past, the light does not go straight up and down - because the train (and the mirror) is moving sideways the whole time, so from the platform the light traces a diagonal path, up and across, then down and across again. That diagonal path is longer than the straight up-and-down path. Here is the strange part: light always travels at the exact same speed for both of you, no exceptions. If the platform observer's light has to cover a longer path at the same speed, it must take a longer time to do it. So the platform observer sees your light-clock's "tick" take longer than you do, on your own clock - your clock, from their point of view, is running slow. This is time dilation, and it is not a trick of eyesight or delay: every observer moving relative to you would genuinely measure your clock running slower than their own, and you would measure theirs running slow right back. Nobody is "wrong" - there is simply no single, universal clock that everyone shares, because the only thing every observer agrees on is the speed of light itself.

Examples in context

Example 1. GPS satellite clocks over Sydney. GPS satellites orbit at $v = 3.87 \text{ km/s}$ , so $\beta = v/c = 1.29 \times 10^{-5}$ and $\gamma \approx 1 + \beta^2/2 = 1 + 8.3 \times 10^{-11}$ . Per day ( $86{,}400 \text{ s}$ ), special-relativistic time dilation slows the satellite clock by $\Delta t = (\gamma - 1) \times 86{,}400 = 7.2 \mu\text{s}$ relative to a ground clock. (General relativity adds $+45.9 \mu\text{s}$ for altitude, giving net $+38.7 \mu\text{s}/\text{day}$ ). Without applying this correction, GPS positions over Sydney would drift by $\sim 11 \text{ km}/\text{day}$ . The satellite oscillators are tuned to $10.22999999543 \text{ MHz}$ instead of $10.23 \text{ MHz}$ at launch to compensate.

Example 2. Cosmic muon flux measured at sea level in NSW. Muons created at $h = 15 \text{ km}$ altitude live (in their rest frame) $\tau_0 = 2.2 \mu\text{s}$ before decaying. At classical speed $v = 0.998c$ , distance travelled in one lifetime is $v \tau_0 = 660 \text{ m}$ , far short of $15 \text{ km}$ . Yet many muons reach sea level. In our frame, time dilates: $\tau = \gamma \tau_0 = 15.8 \times 2.2 \mu\text{s} = 34.8 \mu\text{s}$ , allowing $v \tau = 0.998 c \times 34.8 \mu\text{s} = 10{,}400 \text{ m}$ of travel. In the muon's frame, the atmosphere is length-contracted to $h/\gamma = 950 \text{ m}$ . Both viewpoints agree on the observed sea-level flux.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC5 marksA spacecraft travels at 0.80c relative to Earth. Its proper length (measured in its own rest frame) is 50 m and a clock on board ticks off 1.0 hour during a journey segment. Calculate the length of the spacecraft and the elapsed time as measured by an Earth observer, and explain the meaning of 'proper time' and 'proper length'.

Show worked answer →

Lorentz factor at $v = 0.80c$ :

$\gamma = 1 / \sqrt{1 - v^2/c^2} = 1 / \sqrt{1 - 0.64} = 1 / \sqrt{0.36} = 1 / 0.6 = 5/3 \approx 1.667$ .

Length contraction (Earth observer sees the moving ship as contracted along its direction of motion):

$L = L_0 / \gamma = 50 / 1.667 = 30$ m.

Time dilation (Earth observer sees the moving clock running slow, so the same on-board interval corresponds to a longer Earth interval):

$t = \gamma t_0 = 1.667 \times 1.0 = 1.67$ h.

Proper time $t_0$ : the time interval between two events that occur at the same location in some frame, measured by a clock at rest in that frame. The on-board clock measures proper time for events happening on the spacecraft.

Proper length $L_0$ : the length of an object measured in the rest frame of the object. The on-board crew measures proper length.

Markers reward correct $\gamma$ , both calculations, and clear definitions linking proper time and length to the rest frame of the observed system. A common error is to apply the contraction the wrong way (Earth observer sees a shorter ship, not a longer one).

2020 HSC3 marksOutline the Michelson-Morley experiment and explain how its null result was a key piece of evidence for special relativity.

Show worked answer →

A Michelson interferometer splits a beam of monochromatic light into two perpendicular paths, reflects each off a mirror, and recombines them to form an interference pattern. Michelson and Morley (1887) aimed to detect the Earth's motion through the hypothesised "luminiferous aether", a medium thought necessary for light to propagate.

Reasoning: if the Earth moves through a stationary aether at speed $v$ , light travelling parallel to that motion takes a different round-trip time from light travelling perpendicular to it. Rotating the apparatus would change which path was "along the wind" and shift the interference fringes by a predictable amount.

Result: no fringe shift was observed (or only one far smaller than predicted) regardless of orientation or time of year. This null result implied no aether wind exists, which conflicted with the wave model that required a medium.

Significance for special relativity: Einstein's second postulate, that the speed of light in vacuum is the same in all inertial frames, makes the Michelson-Morley null result automatic; there is no aether and no preferred frame to detect.

Markers reward correct description of the apparatus, the expected versus observed result, and the link to Einstein's postulate.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksA spacecraft clock runs for a proper time of

4.0

s while the ship travels at

v = 0.60c

relative to Earth. Calculate the Lorentz factor

\gamma

and the time an Earth observer measures for the same interval.

Show worked solution →

$\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}} = \dfrac{1}{\sqrt{1 - 0.60^2}} = \dfrac{1}{\sqrt{0.64}} = 1.25$ .

The ship's clock measures the proper time $t_0 = 4.0$ s (it is present at both events), so the Earth observer measures $t = \gamma t_0 = 1.25 \times 4.0 = 5.0$ s.

Marks: one for $\gamma = 1.25$ correctly evaluated, one for $t = 5.0\ \text{s}$ with the reasoning that the Earth clock reads the longer, dilated time.

foundation2 marksA rod has a proper length of

120

m. Calculate its length as measured by an observer relative to whom it moves at

v = 0.80c

Show worked solution →

$\gamma = \dfrac{1}{\sqrt{1 - 0.80^2}} = \dfrac{1}{\sqrt{0.36}} = \dfrac{1}{0.60} = 1.667$ .

$L = \dfrac{L_0}{\gamma} = \dfrac{120}{1.667} = 72\ \text{m}$ .

Marks: one for the correct $\gamma$ , one for $L = 72\ \text{m}$ stated to two significant figures with the reasoning that the moving observer measures a contraction along the direction of motion only.

foundation2 marksState Einstein's two postulates of special relativity, and identify which one directly explains the Michelson-Morley null result.

Show worked solution →

Postulate 1 (relativity): the laws of physics are the same in all inertial reference frames.

Postulate 2 (constancy of $c$ ): the speed of light in vacuum is the same in all inertial frames, independent of the motion of the source or observer.

Postulate 2 explains the null result: if $c$ is the same in every direction and in every frame, there is no "aether wind" for light to travel faster or slower against, so no fringe shift can ever appear regardless of the interferometer's orientation.

Marks: one for both postulates stated accurately, one for correctly identifying postulate 2 as the explanation of the null result.

core4 marksThe graph shows the Lorentz factor

\gamma

against speed

v

(as a fraction of

c

) for the readings marked. **(a)** Describe the shape of the curve as

v \to c

. **(b)** Using the marked point at

v = 0.80c

, read off

\gamma

and verify it by calculation. **(c)** A spaceship's on-board clock records a proper time of

10.0

s at

v = 0.80c

. Use your value of

\gamma

to find the Earth-frame time.

Show worked solution →

(a) The curve starts near $\gamma = 1$ at low speed and rises slowly at first, then curves upward increasingly steeply, diverging toward infinity as $v \to c$ . No massive object can reach $\gamma = \infty$ , consistent with $c$ being an unreachable limit.

(b) Reading the marked point at $v = 0.80c$ gives $\gamma \approx 1.7$ . By calculation: $\gamma = \dfrac{1}{\sqrt{1 - 0.80^2}} = \dfrac{1}{\sqrt{0.36}} = 1.667$ , which matches the graph.

(c) $t = \gamma t_0 = 1.667 \times 10.0 = 16.7\ \text{s}$ .

Marks: one for describing the increasingly steep, diverging shape as $v \to c$ , one for correctly reading $\gamma \approx 1.7$ from the graph, one for the calculated $\gamma = 1.667$ matching the reading, one for $t = 16.7\ \text{s}$ .

core4 marksA muon is created with proper lifetime

\tau_0 = 1.5\ \mu\text{s}

and travels at

v = 0.95c

toward the ground. Calculate (a) the Lorentz factor, (b) the muon's lifetime as measured in the Earth frame, and (c) the distance it can travel (Earth frame) before decaying.

Show worked solution →

(a) $\gamma = \dfrac{1}{\sqrt{1 - 0.95^2}} = \dfrac{1}{\sqrt{0.0975}} = 3.20$ .

(b) $\tau = \gamma \tau_0 = 3.20 \times 1.5 \times 10^{-6} = 4.8 \times 10^{-6}\ \text{s}$ .

(c) $v = 0.95 \times 3.00 \times 10^8 = 2.85 \times 10^8\ \text{m s}^{-1}$ , so $d = v\tau = 2.85 \times 10^8 \times 4.8 \times 10^{-6} = 1.4 \times 10^3\ \text{m}$ .

Marks: one for $\gamma = 3.20$ , one for $\tau = 4.8 \times 10^{-6}\ \text{s}$ , one for the correct speed in m/s, one for $d = 1.4 \times 10^3\ \text{m}$ (about $1.4$ km) with the unit.

core3 marksAn observer measures a spacecraft's length as

45

m while it passes at

v = 0.60c

. Calculate the spacecraft's proper length, and state which observer (the one on the ground or the one on the spacecraft) measures the proper length.

Show worked solution →

$\gamma = \dfrac{1}{\sqrt{1 - 0.60^2}} = 1.25$ .

Rearranging $L = L_0 / \gamma$ gives $L_0 = \gamma L = 1.25 \times 45 = 56\ \text{m}$ .

The proper length is measured by an observer at rest relative to the object, so it is the spacecraft's own crew (on board the ship) who measure the proper length of $56$ m; the ground observer, for whom the ship is moving, measures the contracted $45$ m.

Marks: one for $\gamma = 1.25$ , one for $L_0 = 56\ \text{m}$ , one for correctly identifying the on-board observer as measuring the proper length.

exam6 marksAnalyse how the Michelson-Morley experiment and Einstein's postulates of special relativity together overturned the nineteenth-century model of light propagating through a stationary aether.

Show worked solution →

Band-6 plan. (1) State the aether model and what it predicted for the interferometer. (2) Describe the experiment and its null result. (3) State how classical physicists tried (and failed) to rescue the aether. (4) State Einstein's two postulates and show postulate 2 makes the null result automatic, removing the need for an aether. (5) Conclude with the significance: a change in the model of space and time itself, not just a failed detection.

Model answer. Nineteenth-century physics treated light as a wave requiring a medium, the luminiferous aether, assumed to be a fixed background against which absolute motion could be measured. Because the Earth orbits the Sun at about $30\ \text{km s}^{-1}$ , it should move through this aether, so light travelling parallel to the Earth's motion should take a measurably different round-trip time from light travelling perpendicular to it.

Michelson and Morley's interferometer split a beam into two perpendicular paths, reflected each off a mirror, and recombined them to look for the resulting interference-fringe shift as the apparatus was rotated. Despite using an apparatus sensitive enough to detect a small fraction of the predicted shift, no significant fringe shift was seen in any orientation, season or location. The result was resoundingly null, and was reproduced many times with increasing precision.

Attempts to rescue the aether, such as an aether "dragged along" by the Earth or an aether-induced contraction of the apparatus itself (the Lorentz-FitzGerald contraction, introduced ad hoc), could each explain the null result in isolation but conflicted with other observations (such as stellar aberration) or amounted to unfalsifiable patchwork rather than a coherent theory.

Einstein instead removed the aether from physics altogether. His two postulates - that the laws of physics are identical in every inertial frame, and that the speed of light in vacuum, $c$ , is the same for every inertial observer regardless of the motion of the source - make the Michelson-Morley null result an automatic, necessary consequence: since $c$ is the same in every direction in every frame, there is no aether wind to detect and no fringe shift is possible at any orientation, with no special assumptions required.

This was a change in the model of space and time, not merely a discarded medium: the postulates force time dilation, length contraction and the relativity of simultaneity as their logical consequences. The Michelson-Morley result is therefore best understood not as evidence collected and later explained, but as the deepest early clue that the aether concept, and with it absolute space and time, could not be right.

Marker's note: the top band explains WHY the null result is a problem for the aether model (not just "no fringes were seen"), names the failed rescue attempts, and shows the logical link from postulate 2 to the automatic explanation of the null result. A response that only describes the apparatus without this causal chain caps in the middle band.

exam7 marksEvaluate the claim that time dilation and length contraction are merely 'measurement illusions' caused by the finite speed of light, rather than real physical effects.

Show worked solution →

Band-6 plan. (1) State the claim and what "illusion" would mean (an artefact of signal delay, correctable to reveal a single true time/length). (2) Distinguish this from relativistic effects, which persist after signal-delay is removed. (3) Give at least one piece of independent evidence (muon decay or atomic clocks) that only makes sense if the effects are physically real. (4) Explain WHY they are real: a consequence of the invariance of $c$ , not of light taking time to reach an observer. (5) Reach an explicit judgement rejecting the "illusion" framing.

Model answer. The claim conflates two different things: the ordinary delay caused by light taking time to travel from an event to an observer (which IS correctable, and which physicists always subtract out first), and the relativistic effects of time dilation and length contraction, which remain after that correction and cannot be removed by any choice of viewing position.

If dilation were only a signal-delay illusion, an observer co-located with the moving clock (removing all travel-time effects) would agree with the "stationary" observer on the elapsed time. This is not what happens: two observers in relative motion, each correcting fully for light travel time, still calculate different proper times between the same two events, and both calculations are internally consistent and correct in their own frame.

Independent evidence confirms the effect is physically real rather than an artefact of observation. Fast-moving muons created in the upper atmosphere have a proper lifetime of only $2.2\ \mu\text{s}$ , far too short at classical speeds to reach sea level; yet a large flux is detected at ground level, because in the Earth's frame $\tau = \gamma\tau_0$ genuinely extends how long the muon exists before it decays. Caesium atomic clocks flown on aircraft (the Hafele-Keating experiment) were subsequently found to read a measurably different elapsed time from clocks left on the ground, entirely consistent with the predicted dilation, and this discrepancy is used operationally to correct GPS satellite clocks today.

The underlying reason the effects are real is that they follow directly from Einstein's second postulate: because every inertial observer measures the same value of $c$ , the geometry of space and time itself, not the observer's viewpoint, must differ between frames for events to be timed and measured consistently. Time dilation and length contraction are properties of spacetime intervals, affecting every physical process (clocks, muon decay, biological ageing) identically, which rules out an explanation based on any one measuring instrument or viewing delay.

Overall, the "measurement illusion" framing should be rejected: the effects persist after all signal-delay corrections, are confirmed by independent physical processes (muon flux, atomic clocks), and follow necessarily from the invariance of $c$ . They are real, frame-dependent physical effects, not an artefact of how or when information reaches an observer.

Marker's note: the top band explicitly distinguishes signal delay from relativistic dilation/contraction, cites at least one concrete confirming observation (muon lifetime or Hafele-Keating/GPS), and explains the postulate-based reason the effects are real, ending in an explicit judgement. A response that just restates the formulas without addressing the "illusion" claim caps in the middle band.