Galilean and special relativity

Topic A: Space, Time and Motion

Lesson A.5 Galilean and Special Relativity | IB Physics HL

How can observers moving relative to one another measure different times and lengths while agreeing on the laws of physics? This lesson connects the everyday Galilean description of motion to special relativity. Careful identification of reference frames, events and measurement procedures makes the apparently surprising results consistent.

Learning objectives

Distinguish inertial and non-inertial reference frames.
Apply Galilean coordinate and velocity transformations.
State the two postulates of special relativity.
Calculate and interpret the Lorentz factor.
Apply Lorentz coordinate and velocity transformations.
Identify proper time and proper length before calculating dilation or contraction.
Explain relativity of simultaneity.
Calculate the invariant spacetime interval and identify causal relationships.
Interpret worldlines, light rays and moving axes on spacetime diagrams.
Explain muon evidence and compare the Earth and muon descriptions.
Evaluate a data-based investigation of relativistic decay.

Notation and the standard frame arrangement

Let S′ move at constant velocity +v along the x-axis of S. The axes are parallel, and the origins coincide when $t=t'=0$ . An event has coordinates (x, t) in S and (x′, t′) in S′. An object’s velocity is u in S and u′ in S′. The frame velocity v and object velocity u are different quantities.

Use the IB notation $\Delta t_0$ for proper time and $L_0$ for proper length. Unless otherwise stated, consider one spatial dimension and inertial motion. Primes identify another frame, rather than a later time.

$c=3.00\times10^8\,\mathrm{m\,s^{-1}},\quad\beta=\frac{v}{c}$

A.5.1 — Reference frames and events

A reference frame consists of spatial coordinates and a system of clocks used to assign positions and times. An event is an occurrence at a particular position and time, such as a flash emitted or a particle decaying.

An inertial frame is one in which a body with no resultant force moves at constant velocity. Frames moving at constant velocity relative to an inertial frame are also inertial. No inertial frame is an absolute state of rest.

A non-inertial frame accelerates or rotates relative to an inertial frame. Applying Newton’s laws within it requires additional inertial, or fictitious, forces. A car accelerating or a rotating platform provides an example.

In classical Galilean relativity, the laws of mechanics have the same form in all inertial frames, and time is absolute. Positions and velocities can differ between frames; distances measured between simultaneous endpoints are unchanged by the Galilean transformation.

A.5.2 — Galilean transformations

For the standard frame arrangement:

$x'=x-vt,\quad t'=t$

$y'=y,\quad z'=z$

The origin of S′ is at $x=vt$ in S. Subtract that displacement to obtain an event’s position relative to S′. For differences between events:

$\Delta x'=\Delta x-v\Delta t,\quad\Delta t'=\Delta t$

Galilean velocity transformation

$u'=u-v,\quad u=u'+v$

These are signed one-dimensional velocity equations. Acceleration is unchanged between these frames because v is constant. Galilean mechanics is an excellent approximation in ordinary situations where relativistic corrections are negligible.

Worked example 1 — A passenger on a moving train

A train moves right at 20.0 m/s relative to the ground. A passenger walks forward at 1.50 m/s relative to the train. Find their ground-frame velocity. What if they walk backward at the same speed?

Show worked answer

Given: Train velocity +20.0 m/s. Take right as positive; the train is S′ and the ground is S.

Step 1 — Forward walking.

$u=u'+v=1.50+20.0=21.5\,\mathrm{m\,s^{-1}}$

Step 2 — Backward walking. The train-frame velocity is negative.

$u=-1.50+20.0=18.5\,\mathrm{m\,s^{-1}}$

Final answers: Forward walking: 21.5 m/s right. Backward walking: 18.5 m/s right.

Check: Walking backward relative to the train does not necessarily mean moving left relative to the ground.

Worked example 2 — Transforming an everyday event

S′ moves at +20.0 m/s relative to S. An event occurs at x = 150 m and t = 5.00 s in S. Use Galilean transformations to find its coordinates in S′.

Show worked answer

Given: Origins coincide at zero time; v = +20.0 m/s, x = 150 m and t = 5.00 s.

Step 1 — Transform position.

$x'=x-vt=150-(20.0)(5.00)=50.0\,\mathrm{m}$

Step 2 — Transform time.

$t'=t=5.00\,\mathrm{s}$

Final answers: Coordinates in S′: x′ = 50.0 m and t′ = 5.00 s.

Check: The moving origin is 100 m from the S origin at this instant.

A.5.3 — Why special relativity is needed

Applying $u'=u-v$ to a light ray with $u=c$ predicts $u'=c-v$ . This conflicts with the invariant speed of light in vacuum. The Galilean transformation therefore cannot describe all physics at relativistic speeds.

Einstein’s two postulates

Principle of relativity: The laws of physics are the same in all inertial reference frames.
Invariant light speed: Light in vacuum has speed c in every inertial frame, independent of the motion of its source.

The consequences include frame-dependent simultaneity, time intervals and measured lengths. They concern measurements assigned using coordinates and synchronized clocks, rather than uncorrected arrival times of light signals.

A.5.4 — The Lorentz factor

$\gamma=\frac1{\sqrt{1-\frac{v^2}{c^2}}}=\frac1{\sqrt{1-\beta^2}}$

The Lorentz factor is dimensionless, depends on the magnitude of relative velocity and is at least one for $|v|< c$ . It approaches one at low speeds and increases without bound as speed approaches c. A massive object has no inertial rest frame moving at c.

Useful Lorentz factors
Speed	γ
0	1
0.600c	1.25
0.800c	1.67
0.900c	2.29
0.980c	5.03

Keep the unrounded γ during working. In particular, $\gamma=5/3$ at 0.800c; using 1.67 too early introduces avoidable rounding error.

A.5.5 — Proper time and time dilation

The proper time interval $\Delta t_0$ between two events is measured by a clock present at both events. For the constant-velocity problems here, it is measured in the inertial frame where the events occur at the same position.

Another inertial frame sees that clock move between the two events and assigns a longer coordinate time interval:

$\Delta t=\gamma\Delta t_0$

Each inertial observer can describe the other’s clock as running slowly after correcting for signal travel time. This reciprocity is consistent because their synchronized clocks define different simultaneity slices. Do not identify proper time simply by deciding that Earth is “stationary”.

Worked example 3 — A process on a spaceship

A spaceship moves at 0.800c relative to Earth. A clock fixed on the spaceship records 5.00 s between two events at that clock. Find the Earth-frame interval.

Show worked answer

Given: The spaceship clock is present at both events, so its interval is the proper time, 5.00 s.

Step 1 — Calculate γ.

$\gamma=\frac1{\sqrt{1-0.800^2}}=\frac53$

Step 2 — Apply time dilation using unrounded γ.

$\Delta t=\gamma\Delta t_0=\frac53(5.00)=8.33\,\mathrm{s}$

Final answers: Earth-frame time interval 8.33 s.

Check: The Earth interval exceeds the spaceship interval. Keep γ unrounded until the final answer.

Show a light-clock derivation

Consider a clock with two mirrors separated by d perpendicular to its motion. A light pulse travels to the other mirror and returns. In the clock’s rest frame:

$c\Delta t_0=2d$

In a frame where the clock moves horizontally at speed v, the light follows diagonal paths. For half the round trip, Pythagoras gives:

$\left(\frac{c\Delta t}{2}\right)^2=d^2+\left(\frac{v\Delta t}{2}\right)^2$

Substitute $d=c\Delta t_0/2$ and rearrange:

$(c^2-v^2)(\Delta t)^2=c^2(\Delta t_0)^2$

$\Delta t=\frac{\Delta t_0}{\sqrt{1-v^2/c^2}}=\gamma\Delta t_0$

The same vacuum light speed c is used in both frames.

A.5.6 — Proper length and length contraction

The proper length $L_0$ is measured in the object’s rest frame. A frame in which the object moves measures a shorter length parallel to the relative motion:

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

A moving length is measured by recording the positions of both endpoints at the same time in the measuring frame. Measurements of endpoints at different times do not directly give the object’s length in that frame.

Dimensions perpendicular to the relative motion do not contract. Length contraction describes a coordinate measurement, not a physical crushing in the object’s own rest frame and not necessarily its photographed appearance.

Worked example 4 — The measured length of a spaceship

A spaceship has proper length 100 m along its motion and travels at 0.800c relative to Earth. Find its Earth-frame length.

Show worked answer

Given: Rest length 100 m and relative speed 0.800c. Earth measures both endpoints simultaneously in the Earth frame.

Step 1 — Use the Lorentz factor.

$\gamma=\frac53$

Step 2 — Apply length contraction.

$L=\frac{L_0}{\gamma}=\frac{100}{5/3}=60.0\,\mathrm{m}$

Final answers: Earth-frame length 60.0 m.

Check: The spaceship’s rest length stays 100 m. Its width perpendicular to motion is unchanged.

A.5.7 — Lorentz transformations

For S′ moving at +v relative to S, with coincident origins at zero time, the IB Lorentz transformations are:

$x'=\gamma(x-vt)$

$t'=\gamma\left(t-\frac{vx}{c^2}\right)$

$y'=y,\quad z'=z$

Space and time coordinates mix. These transform the coordinates of the same event. They are not merely time-dilation and length-contraction formulas.

For differences between two events, the same linear relationships apply:

$\Delta x'=\gamma(\Delta x-v\Delta t)$

$\Delta t'=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)$

The inverse transformations follow by exchanging primed and unprimed coordinates and replacing v by −v:

$x=\gamma(x'+vt'),\quad t=\gamma\left(t'+\frac{vx'}{c^2}\right)$

Use the full transformation: $\Delta t=\gamma\Delta t_0$ is a special case for the appropriate co-located events. It is not a general transformation between arbitrary event times.

Worked example 5 — Transforming a relativistic event

S′ moves at +0.600c relative to S. An event has x = 900 m and t = 5.00 μs in S. Find x′ and t′.

Show worked answer

Given: v = 0.600c, γ = 1.25, x = 900 m and t = 5.00 × 10⁻⁶ s. The origins coincide at zero time.

Step 1 — Transform position.

$x'=1.25[900-(0.600)(3.00\times10^8)(5.00\times10^{-6})]=0\,\mathrm{m}$

Step 2 — Calculate the time correction.

$\frac{vx}{c^2}=\frac{0.600(900)}{3.00\times10^8}=1.80\times10^{-6}\,\mathrm{s}$

Step 3 — Transform time.

$t'=1.25(5.00-1.80)\,\mu\mathrm{s}=4.00\,\mu\mathrm{s}$

Final answers: x′ = 0 m; t′ = 4.00 μs.

Check: The event lies on the worldline of the moving origin. Relative to the shared origin event, the S′ interval is proper time.

A.5.8 — Relativity of simultaneity

Two events are simultaneous in a frame when their time coordinates are equal. If events at different positions are simultaneous in S, then $\Delta t=0$ , but:

$\Delta t'=-\gamma\frac{v\Delta x}{c^2}$

They are generally not simultaneous in S′. For positive v and positive $\Delta x$ , the event farther along +x occurs earlier in S′. This is a difference in assigned event times after correcting signal delays.

When transforming two endpoint events simultaneous in S, they are generally not simultaneous in S′. Their transformed position difference is therefore not automatically the object’s length in S′.

Worked example 6 — Two flashes simultaneous in one frame

Flash A occurs at x = 0 and flash B at x = 600 m, both at t = 0 in S. S′ moves at +0.600c. Find the time difference t′B − t′A.

Show worked answer

Given: Δt = 0, Δx = +600 m and γ = 1.25.

Step 1 — Use the difference transformation.

$\Delta t'=\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right)$

Step 2 — Substitute.

$\Delta t'=-1.25\frac{0.600(600)}{3.00\times10^8}=-1.50\times10^{-6}\,\mathrm{s}$

Final answers: Δt′ = −1.50 μs: B occurs 1.50 μs before A in S′.

Check: These separated flashes are spacelike separated, so this change in time order does not reverse a causal relationship.

A.5.9 — Relativistic velocity transformation

For collinear motion, the IB velocity transformation is:

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Its inverse is:

$u=\frac{u'+v}{1+\frac{u'v}{c^2}}$

Use signed velocities. If speeds are much smaller than c, the denominator is approximately one and the Galilean result is recovered. At relativistic speeds, adding two subluminal velocities in the same direction gives a result smaller than c.

Show the derivation and light-speed check

For a short displacement of the moving object, divide the Lorentz differences:

$u'=\frac{\Delta x'}{\Delta t'}=\frac{\Delta x-v\Delta t}{\Delta t-v\Delta x/c^2}$

Divide numerator and denominator by $\Delta t$ , using $u=\Delta x/\Delta t$ for uniform motion:

$u'=\frac{u-v}{1-uv/c^2}$

For a light ray moving along +x, substitute $u=c$ :

$u'=\frac{c-v}{1-v/c}=c$

A light ray along −x similarly gives $u'=-c$ .

Worked example 7 — Launching a probe from a fast spaceship

A spaceship moves at +0.600c relative to Earth. It launches a probe at +0.800c relative to itself. Find the probe’s Earth-frame velocity. Also find the Earth-frame velocity if the probe is launched backward at −0.800c.

Show worked answer

Given: Frame velocity v = +0.600c. The probe’s spaceship-frame velocity is u′.

Step 1 — Forward launch. Use the inverse velocity transformation.

$u=\frac{0.800c+0.600c}{1+(0.800)(0.600)}=\frac{1.400}{1.480}c=0.946c$

Step 2 — Backward launch. Keep the negative sign.

$u=\frac{-0.800c+0.600c}{1+(-0.800)(0.600)}=\frac{-0.200}{0.520}c=-0.385c$

Final answers: Forward probe: +0.946c. Backward probe: −0.385c relative to Earth.

Check: The classical forward result 1.40c is incompatible with special relativity.

A.5.10 — The invariant spacetime interval

Different inertial frames can assign different time and position separations to the same pair of events. They agree on the squared spacetime interval. For one-dimensional motion, the IB expression is:

$(\Delta s)^2=(c\Delta t)^2-(\Delta x)^2$

$(c\Delta t')^2-(\Delta x')^2=(c\Delta t)^2-(\Delta x)^2$

The squared interval has units m². This is a signed spacetime quantity, not an ordinary Euclidean distance squared; it can be negative.

Event separations and causality
Type	Squared interval	Meaning
Timelike	$(\Delta s)^2>0$	A frame exists where the events share a position. A subluminal signal can connect them; their time order is preserved.
Lightlike	$(\Delta s)^2=0$	A light signal can connect them. No inertial rest frame of light exists.
Spacelike	$(\Delta s)^2<0$	No signal at or below c can connect them. A frame exists where they are simultaneous; their time order can differ between frames.

For timelike events, the frame where $\Delta x'=0$ gives:

$(\Delta s)^2=(c\Delta t_0)^2$

For non-zero spacelike separation, no co-located-event frame and no proper time between those events exist.

Worked example 8 — Interval and proper time

Two events in S have Δt = 5.00 μs and Δx = 900 m. Find the squared interval, classify the separation and calculate proper time.

Show worked answer

Given: Time separation 5.00 × 10⁻⁶ s and position separation 900 m.

Step 1 — Calculate cΔt.

$c\Delta t=(3.00\times10^8)(5.00\times10^{-6})=1500\,\mathrm{m}$

Step 2 — Calculate the squared interval.

$(\Delta s)^2=(1500)^2-(900)^2=1.44\times10^6\,\mathrm{m^2}$

Step 3 — Use the positive interval to find proper time.

$\Delta t_0=\frac{\sqrt{1.44\times10^6}}{3.00\times10^8}=4.00\,\mu\mathrm{s}$

Final answers: Squared interval +1.44 × 10⁶ m²; timelike separation; proper time 4.00 μs.

Check: The co-located-event frame moves at Δx/Δt = 0.600c, consistent with worked example 5.

A.5.11 — Spacetime diagrams

A spacetime diagram places x horizontally and ct vertically. An event is a point. A worldline is the sequence of events occupied by an object.

A stationary object has a vertical worldline.
Uniform velocity gives a straight worldline.
Changing velocity gives a worldline with changing direction.
Light rays obey $x=\pm ct$ . They appear at 45° only when ct and x use equal scale lengths.
A massive object’s worldline lies inside the light cone; light lies on its boundary.

The light cone through an event separates possible causal connections from spacelike-separated events. The future cone contains events a signal could reach; the past cone contains events that could have influenced the chosen event.

Worldline angle and speed

On an equally scaled ct–x diagram, a uniform-velocity worldline makes angle θ with the vertical time axis:

$\tan\theta=\frac{v}{c}$

For a line through the origin, $\tan\theta=x/(ct)$ . Otherwise, use coordinate differences along the line. The gradient of ct plotted against x is $c/v$ , rather than v.

Moving axes and simultaneity

The ct′ axis is the worldline of the moving origin, where $x'=0$ . The x′ axis is the set of events with $t'=0$ :

$ct'\text{ axis:}\quad x=vt,\quad\frac{x}{ct}=\beta$

$x'\text{ axis:}\quad ct=\beta x$

Each moving axis tilts toward the nearest light line. Lines of constant t are horizontal; lines of constant t′ are parallel to the x′ axis. The primed axes are not perpendicular in the Euclidean drawing.

Read primed coordinates using lines parallel to the other primed axis and correctly calibrated scale marks. Do not use ordinary perpendicular projection or assume equal Euclidean spacing of the two frames’ unit marks.

Diagram measurements: Measure angles only when consistent scales are supplied. On schematic drawings, use the equations instead.

Worked example 9 — A spacetime worldline angle

An object moves at +0.600c. Find its worldline angle to the ct axis and its ct-against-x gradient on an equally scaled diagram.

Show worked answer

Given: Uniform velocity +0.600c and equal ct and x scales.

Step 1 — Find the angle from the time axis.

$\theta=\tan^{-1}(0.600)=31.0^\circ$

Step 2 — Find the plotted gradient.

$\frac{\Delta(ct)}{\Delta x}=\frac{c}{v}=\frac1{0.600}=1.67$

Final answers: Angle 31.0° from the ct axis; worldline gradient 1.67.

Check: The angle is smaller than 45°, so the speed is below c. The x′ axis tilts 31.0° above x for the same frame velocity.

A.5.12 — Muon decay as evidence

Muons produced by cosmic-ray interactions in the atmosphere have a short proper mean lifetime. Decay is random: a mean lifetime is not a fixed deadline at which every muon decays.

$\Delta t_0=2.20\,\mu\mathrm{s}$

In the Earth frame, relativistic muons have a longer mean lifetime and a characteristic mean decay distance:

$\Delta t=\gamma\Delta t_0,\quad d_{\mathrm{mean}}=v\gamma\Delta t_0$

In a muon’s rest frame, its lifetime remains the proper lifetime while the atmosphere’s path length contracts. Both frames predict the same survival probability for the same creation and arrival events.

Experimental lifetime measurements, including the CERN muon storage-ring measurements, support time dilation. See the primary reference Measurements of relativistic time dilatation for positive and negative muons in a circular orbit.

Worked example 10 — Muon motion in two frames

A muon moves at 0.980c with proper mean lifetime 2.20 μs. Calculate its Earth-frame mean lifetime and characteristic decay distance. For a 3.00 km atmospheric path, find the path length and travel time in the muon frame.

Show worked answer

Given: Speed 0.980c, proper mean lifetime 2.20 μs and Earth-rest path 3000 m.

Step 1 — Find γ and the Earth-frame mean lifetime.

$\gamma=\frac1{\sqrt{1-0.980^2}}=5.02519$

$\Delta t=\gamma\Delta t_0=11.1\,\mu\mathrm{s}$

Step 2 — Find characteristic distance using unrounded values.

$d_{\mathrm{mean}}=v\gamma\Delta t_0=3.25\times10^3\,\mathrm{m}$

Step 3 — Contract the atmospheric path.

$L=\frac{3000}{5.02519}=597\,\mathrm{m}$

Step 4 — Calculate travel time in the muon frame. The ground approaches at 0.980c.

$t_{\mathrm{travel},\mu}=\frac{L}{v}=2.03\,\mu\mathrm{s}$

Final answers: Earth mean lifetime 11.1 μs; characteristic decay distance 3.25 km. Muon-frame path 597 m and travel time 2.03 μs.

Check: Without dilation, vΔt₀ is only 647 m. Some muons decay earlier and others travel farther than the mean distance.

Worked example 11 — Speed from clock intervals

A moving clock records 4.00 s between events at the clock. The laboratory assigns 8.00 s between the same events. Find relative speed.

Show worked answer

Given: Proper time 4.00 s and laboratory interval 8.00 s.

Step 1 — Determine γ.

$\gamma=\frac{\Delta t}{\Delta t_0}=\frac{8.00}{4.00}=2.00$

Step 2 — Rearrange the Lorentz factor.

$\frac{v}{c}=\sqrt{1-\frac1{\gamma^2}}=\sqrt{1-\frac14}=0.866$

Final answers: Relative speed 0.866c, or 2.60 × 10⁸ m/s.

Check: The intervals determine the speed magnitude, not its direction.

A.5.13 — A return journey and the twin question

A traveller who turns around and returns follows a different spacetime path from the stay-at-home observer. The traveller changes velocity and cannot describe the whole trip using one unchanged inertial frame. Reciprocal clock-rate statements for inertial motion therefore do not imply equal ages at reunion.

Calculate elapsed proper time along each inertial leg. For an idealized brief turnaround, add the leg intervals; a realistic calculation includes acceleration periods as well.

Worked example 12 — An idealized round trip

A destination is 3.00 light-years from Earth in the Earth frame. A traveller goes there and back at 0.600c with a negligibly brief turnaround. Neglect gravity. Compare Earth and traveller elapsed times.

Show worked answer

Given: One-way Earth distance 3.00 light-years and speed 0.600 light-years per year; γ = 1.25.

Step 1 — Find Earth time for the two legs.

$t_{\mathrm{Earth,one-way}}=\frac{3.00}{0.600}=5.00\,\mathrm{yr}$

$t_{\mathrm{Earth,total}}=10.0\,\mathrm{yr}$

Step 2 — Find proper time on each leg and add.

$\Delta t_{0,\mathrm{one-way}}=\frac{5.00}{1.25}=4.00\,\mathrm{yr}$

$t_{\mathrm{traveller,total}}=8.00\,\mathrm{yr}$

Final answers: Earth elapsed time 10.0 years; traveller elapsed time 8.00 years.

Check: Time dilation is applied separately to the constant-velocity legs. The turnaround breaks the symmetry of the complete journeys.

A.5.14 — Common misconceptions and exam technique

“Earth always measures proper time.” Identify the clock present at both events.
“Time dilation applies to any event pair.” Use Lorentz transformations unless proper-time conditions apply.
“Length contraction physically crushes an object.” It is a measurement using simultaneous endpoints.
“All dimensions contract.” Only the dimension parallel to relative motion contracts.
“A muon must decay at its mean lifetime.” Individual decay is random.
“Light has a rest frame.” No inertial frame moves at c.
“Any event order can reverse.” The order of causally connected timelike or lightlike events is preserved.
“Light lines always look like 45° lines.” This requires equal ct and x scales.

A reliable solution sequence

Name S and S′ and state which moves at +v.
Identify whether the question concerns coordinates, velocity, a clock interval or a length.
Identify proper time or proper length.
Calculate γ and retain extra digits.
Use signed velocities and consistent units.
State the frame of each result and interpret negative signs.
Check speeds remain below c and the interval is invariant.

A.5.15 — IB-style multiple-choice practice

These are original practice questions. Choose one answer before expanding the explanation.

Multiple choice 1 [1 mark]

Which is an inertial frame in the idealized situation described?

A car accelerating along a straight road.
A spacecraft at constant velocity relative to an inertial laboratory.
A rotating platform.
A car at constant speed around a circle.

Show answer and explanation

Answer: B. Constant velocity relative to an inertial frame defines another inertial frame. Rotation or changing velocity requires acceleration.

Multiple choice 2 [1 mark]

Why is the Galilean velocity transformation inadequate for light?

It predicts no acceleration for free motion.
It uses signed velocities.
It predicts a different light speed for a relatively moving observer.
It changes the source mass.

Show answer and explanation

Answer: C. Using $u'=u-v$ with $u=c$ gives $c-v$ , contrary to invariant light speed.

Multiple choice 3 [1 mark]

Where is proper time measured in the constant-velocity problems of this lesson?

In the frame where both events occur at the same position.
Always on Earth.
In the frame where the events are simultaneous.
Always by clocks at opposite ends of a rod.

Show answer and explanation

Answer: A. The clock present at both events measures proper time; the events are co-located in its inertial rest frame.

Multiple choice 4 [1 mark]

How must a laboratory measure the length of a rod moving along its length?

Record its front now and rear later.
Use a photograph without correcting signal travel time.
Assume its rest length.
Record both endpoints at the same laboratory time.

Show answer and explanation

Answer: D. Moving length requires simultaneous endpoint positions in the measuring frame.

Multiple choice 5 [1 mark]

A spacecraft moves at 0.600c and launches a probe forward at 0.800c relative to itself. What is the Earth-frame probe speed?

1.40c.
0.946c.
0.200c.
0.800c.

Show answer and explanation

Answer: B. $u/c=(0.800+0.600)/(1+0.800\times0.600)=0.946$ .

Multiple choice 6 [1 mark]

Two events have a negative squared spacetime interval. Which statement is correct?

A massive particle can connect them.
Every frame assigns the same time separation.
They are spacelike separated and cannot be connected by a signal at or below c.
They share a position in some inertial frame.

Show answer and explanation

Answer: C. Their spatial separation exceeds c times their time separation, preventing causal connection at or below c.

A.5.16 — Extended-response practice

Show working and state the reference frame of each answer. Mark allocations are suggested guides.

Practice 1 — Galilean coordinates and velocity [4 marks]

S′ moves at +15.0 m/s relative to S. An event has x = 80.0 m and t = 4.00 s in S. An object has velocity +18.0 m/s in S. Find the event coordinates and object velocity in S′.

Show worked answer

Use the Galilean coordinate transformation. [1]

$x'=80.0-(15.0)(4.00)=20.0\,\mathrm{m}$

Correct position. [1]

$t'=4.00\,\mathrm{s}$

Correct time. [1]

$u'=18.0-15.0=3.00\,\mathrm{m\,s^{-1}}$

Correct signed velocity. [1]

Practice 2 — Postulates and classical limitations [4 marks]

State the two postulates of special relativity and explain why the Galilean velocity transformation cannot satisfy them for light.

Show worked answer

The laws of physics are the same in all inertial frames. [1] Light in vacuum has speed c in every inertial frame, independent of source motion. [1]

The Galilean result $u'=u-v$ gives $u'=c-v$ for a light ray. [1] This predicts frame-dependent light speed and conflicts with the second postulate. [1]

Practice 3 — Identifying proper time [4 marks]

A clock moves at 0.600c relative to a laboratory and records 8.00 s between events at its position. Identify proper time and calculate the laboratory interval.

Show worked answer

The moving clock is present at both events, so 8.00 s is proper time. [1]

$\gamma=\frac1{\sqrt{1-0.600^2}}=1.25$

Correct γ. [1]

Use $\Delta t=\gamma\Delta t_0$ . [1]

$\Delta t=(1.25)(8.00)=10.0\,\mathrm{s}$

Correct laboratory interval. [1]

Practice 4 — Length and simultaneity [4 marks]

A rod has proper length 240 m and moves parallel to its length at 0.800c. Find its laboratory length. State how its endpoints must be measured and whether its perpendicular width contracts.

Show worked answer

Use $\gamma=5/3$ and $L=L_0/\gamma$ . [1]

$L=\frac{240}{5/3}=144\,\mathrm{m}$

Correct length. [1]

Measure both endpoints at the same laboratory-frame time. [1] Its perpendicular width does not contract. [1]

Practice 5 — Lorentz coordinates [5 marks]

S′ moves at +0.800c relative to S. The origins coincide at zero time. An event occurs at x = 1500 m and t = 8.00 μs in S. Determine x′ and t′.

Show worked answer

$\gamma=\frac53$

Correct γ. [1]

Use the Lorentz position transformation. [1]

$x'=\frac53[1500-(0.800)(3.00\times10^8)(8.00\times10^{-6})]=-700\,\mathrm{m}$

Correct signed position. [1]

Use the Lorentz time transformation. [1]

$t'=\frac53\left(8.00\times10^{-6}-\frac{0.800(1500)}{3.00\times10^8}\right)=6.67\,\mu\mathrm{s}$

Correct time. [1]

Practice 6 — Simultaneous endpoint events [5 marks]

Two events are simultaneous in S and separated by Δx = +900 m. S′ moves at +0.800c. Find Δt′ and Δx′. Explain why the transformed separation is not automatically a rod length in S′.

Show worked answer

Use $\Delta t=0$ and $\gamma=5/3$ . [1]

$\Delta t'=-\frac53\frac{0.800(900)}{3.00\times10^8}=-4.00\,\mu\mathrm{s}$

Correct time difference and interpretation: the +x event is earlier in S′. [2]

$\Delta x'=\frac53(900)=1500\,\mathrm{m}$

Correct event-position difference. [1]

The transformed events are not simultaneous in S′, so they do not directly specify a length measured in that frame. [1]

Practice 7 — Relativistic relative velocity [4 marks]

An object moves at +0.850c in S. S′ moves at +0.500c relative to S. Find u′ and compare it with the classical result.

Show worked answer

Use the signed relativistic velocity transformation. [1]

$u'=\frac{0.850c-0.500c}{1-(0.850)(0.500)}=\frac{0.350}{0.575}c=0.609c$

Correct value. [1] The object moves in the positive direction in S′. [1] The Galilean result would be 0.350c; the relativistic denominator is essential at these speeds. [1]

Practice 8 — Interval, causality and proper time [6 marks]

Events in S have Δt = 4.00 μs and Δx = 900 m. Calculate the squared interval, identify whether a subluminal signal could connect them, find proper time and determine the velocity of a frame where they share a position.

Show worked answer

Convert time to SI units and calculate cΔt = 1200 m. [1]

$(\Delta s)^2=(1200)^2-(900)^2=6.30\times10^5\,\mathrm{m^2}$

Correct interval. [1] The interval is timelike, so a subluminal signal could connect the events. [1]

Use $\Delta t_0=\sqrt{(\Delta s)^2}/c$ . [1]

$\Delta t_0=\frac{\sqrt{6.30\times10^5}}{3.00\times10^8}=2.65\,\mu\mathrm{s}$

Correct proper time. [1]

$v=\frac{\Delta x}{\Delta t}=\frac{900}{4.00\times10^{-6}}=0.750c$

Correct co-located-event frame velocity. [1]

A.5.17 — Lab-style investigation: muon survival

Research question

Does the decay of a relativistic muon beam agree with time dilation rather than an unchanged laboratory lifetime?

This is a simulated data-analysis investigation, not a claim that these measurements were collected in a school laboratory. Consider a collimated beam with speed 0.980c and proper mean lifetime 2.20 μs. Equal-exposure measurements are made at several distances. Detector efficiency and geometric losses are assumed corrected; the mean background contribution is 40 counts at each distance.

Illustrative muon counts over equal exposures
Distance h / m	Measured counts	Background counts
0	10040	40
1000	7390	40
2000	5450	40
3000	4010	40
4000	2970	40

For this investigation, the exponential survival law is supplied:

$N=N_0e^{-t/\overline{T}}$

N is the corrected count, N₀ is its value at h = 0 and $\overline{T}$ is the laboratory mean lifetime. With $t=h/v$ :

$\ln\left(\frac{N}{N_0}\right)=-\frac{h}{v\overline{T}}$

Correct the counts for background.
State the linear graph and the physical meaning of its gradient.
Use the first and last rows to estimate the laboratory mean lifetime and γ.
Compare with special relativity and with a no-dilation prediction.
Identify limitations, improvements and a method for estimating uncertainty.

Show investigation answers

(a) Subtract 40 counts from each measurement. Corrected counts are 10000, 7350, 5410, 3970 and 2930.

(b) Plot $\ln(N/N_0)$ vertically against h horizontally. The ideal gradient is $b=-1/(v\overline{T})$ and the intercept is zero.

(c) An endpoint estimate gives:

$b=\frac{\ln(2930/10000)-0}{4000-0}=-3.07\times10^{-4}\,\mathrm{m^{-1}}$

$\overline{T}=-\frac1{vb}=\frac1{(0.980)(3.00\times10^8)|b|}=11.1\,\mu\mathrm{s}$

$\gamma_{\mathrm{estimated}}=\frac{\overline{T}}{\Delta t_0}=5.04$

Keep unrounded intermediate values. A best-fit line through all points is preferable in a complete analysis.

(d) Special relativity predicts $\gamma=5.03$ and a mean laboratory lifetime of 11.1 μs, consistent with the simulated data. Without dilation:

$N(4000)=10000e^{-4000/[v(2.20\times10^{-6})]}\approx20.6$

This is far smaller than 2930 corrected counts. The comparison tests models for an ensemble, not the fate of a particular muon.

(e) Real measurements require corrections for detector efficiency, acceptance, background and the beam’s spread in speed. Repeat equal-exposure measurements and calibrate the detectors. Propagate counting and background uncertainties to the logarithmic values, including uncertainty in N₀; normalization creates correlated uncertainties. Estimate the slope range or fit uncertainty, and include velocity uncertainty when calculating lifetime. The approximate counting uncertainty $\sqrt{N_{\mathrm{measured}}}$ applies to Poisson counts, with background uncertainty added separately.

A.5.18 — Trusting the eternal God

Scripture — Psalm 90:2, KJV

“Before the mountains were brought forth, or ever thou hadst formed the earth and the world, even from everlasting to everlasting, thou art God.”

Creation: Christians believe that God created and sustains the universe. Studying space and time invites wonder at creation and gratitude for our ability to investigate it. Relativity describes physical measurements; the Bible expresses our faith in the eternal Creator.

The Fall: Our lives are limited, and sin damages our relationship with God and with one another. Knowledge alone cannot meet our deepest need for forgiveness. Humility, honesty and care for others matter as much as academic success.

Redemption: Jesus Christ entered human history, died for our sins and rose again. Through Him, God offers forgiveness and eternal life. You are invited to trust Jesus as your personal Saviour and Lord, placing your hope in His grace.

Restoration: Christian hope looks toward God’s renewal of creation. Use your time and abilities to serve others. Read and study the Bible, spend time in prayer and connect with other believers to grow in faith and knowledge of God.

Reflection: How are you using the time entrusted to you? What step can you take today to grow in faith and help another person?

Prayer: Eternal God, thank You for creation and for the new life offered through Jesus. Help me learn with humility, use my time wisely and serve others with love. Guide my studies and my life in Your truth. Amen.

Lesson summary and readiness checklist

Galilean transformations use $x'=x-vt$ , $t'=t$ and $u'=u-v$ .
Special relativity preserves the laws of physics and the vacuum light speed in all inertial frames.
Use $\gamma=1/\sqrt{1-v^2/c^2}$ .
Lorentz transformations mix space and time coordinates of the same event.
Use $u'=(u-v)/(1-uv/c^2)$ at relativistic speeds.
Identify proper time before applying $\Delta t=\gamma\Delta t_0$ .
Measure moving lengths with simultaneous endpoints and use $L=L_0/\gamma$ .
The squared interval $(\Delta s)^2=(c\Delta t)^2-(\Delta x)^2$ is invariant.
On equally scaled ct–x diagrams, $\tan\theta=v/c$ for the angle from the time axis.
Muon survival has consistent Earth-frame and muon-frame explanations.

Topic A complete: Review kinematics, forces and momentum, work and energy, rigid body mechanics and relativity. In each problem, identify the physical system, reference frame and assumptions before choosing an equation.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Galilean and special relativity

Topic A: Space, Time and Motion

Learning objectives

Notation and the standard frame arrangement

A.5.1 — Reference frames and events

A.5.2 — Galilean transformations

Galilean velocity transformation

Worked example 1 — A passenger on a moving train

Worked example 2 — Transforming an everyday event

A.5.3 — Why special relativity is needed

Einstein’s two postulates

A.5.4 — The Lorentz factor

A.5.5 — Proper time and time dilation

Worked example 3 — A process on a spaceship

A.5.6 — Proper length and length contraction

Worked example 4 — The measured length of a spaceship

A.5.7 — Lorentz transformations

Worked example 5 — Transforming a relativistic event

A.5.8 — Relativity of simultaneity

Worked example 6 — Two flashes simultaneous in one frame

A.5.9 — Relativistic velocity transformation

Worked example 7 — Launching a probe from a fast spaceship

A.5.10 — The invariant spacetime interval

Worked example 8 — Interval and proper time

A.5.11 — Spacetime diagrams

Worldline angle and speed

Moving axes and simultaneity

Worked example 9 — A spacetime worldline angle

A.5.12 — Muon decay as evidence

Worked example 10 — Muon motion in two frames

Worked example 11 — Speed from clock intervals

A.5.13 — A return journey and the twin question

Worked example 12 — An idealized round trip

A.5.14 — Common misconceptions and exam technique

A reliable solution sequence

A.5.15 — IB-style multiple-choice practice

Multiple choice 1 [1 mark]

Multiple choice 2 [1 mark]

Multiple choice 3 [1 mark]

Multiple choice 4 [1 mark]

Multiple choice 5 [1 mark]

Multiple choice 6 [1 mark]

A.5.16 — Extended-response practice

Practice 1 — Galilean coordinates and velocity [4 marks]

Practice 2 — Postulates and classical limitations [4 marks]

Practice 3 — Identifying proper time [4 marks]

Practice 4 — Length and simultaneity [4 marks]

Practice 5 — Lorentz coordinates [5 marks]

Practice 6 — Simultaneous endpoint events [5 marks]

Practice 7 — Relativistic relative velocity [4 marks]

Practice 8 — Interval, causality and proper time [6 marks]

A.5.17 — Lab-style investigation: muon survival

Research question

A.5.18 — Trusting the eternal God

Scripture — Psalm 90:2, KJV

Lesson summary and readiness checklist

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