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Relativity · 27.1

Frames & Galilean Relativity

Pour coffee on a smoothly cruising train and it falls straight into the cup, exactly as it would at home. Galileo noticed this on ships four centuries ago: below deck, with the sea calm, no experiment can tell you whether the ship is moving. That innocent observation is the first relativity principle.

01

Build the model

Connect the measurement to the mechanism.

A reference frame is a coordinate grid plus a clock; an inertial frame is one in which Newton’s first law holds — no fictitious pushes appear. The Galilean principle of relativity says the laws of mechanics are identical in every inertial frame: uniform velocity is undetectable from inside. Translating between frames is bookkeeping — the Galilean transformations x′ = x − vt, t′ = t — and differentiating gives velocity addition u′ = u − v, then a′ = a: both frames agree on accelerations, hence on forces, hence on all of mechanics.

The picture cracked when Maxwell’s equations predicted a definite speed for light, c = 3.00 × 10⁸ m s⁻¹, with no frame attached. Galileo insists a moving observer must measure c − v; Maxwell’s equations contain no v. One of them had to give.

Simple definition
An inertial frame is one moving at constant velocity in which Newton’s laws hold; Galilean relativity says mechanics is identical in all of them, with frames related by x′ = x − vt, t′ = t and velocities by u′ = u − v.
Example
A flight attendant walks forward at 1 m/s in a plane cruising at 250 m/s: the ground measures 251 m/s, the plane measures 1 m/s — and the coffee pours identically for both observers.
Galilean transformationx′ = x − vt · t′ = t

The moving observer’s origin has slid a distance vt away, so subtract it. The second equation is the hidden assumption of all pre-1905 physics: one universal clock for everybody.

v = relative speed of the frames along x

Velocity additionu′ = u − v

Speeds simply add and subtract: a ball thrown at 8 m/s inside a 30 m/s train does 38 m/s past the platform. Utterly reliable at everyday speeds — and, it turns out, only approximately true.

Differentiate x′ = x − vt with t′ = t

Acceleration is invarianta′ = a, so F = ma in both frames

Differentiating u′ = u − v kills the constant v. Both frames agree on every acceleration and every force — which is exactly why no mechanics experiment can reveal uniform motion.

v is constant, so it dies under d/dt

The problem with lightGalileo: c′ = c − v · Maxwell: c′ = c

Maxwell’s equations fix the speed of light from constants of electricity and magnetism, without asking who is measuring. Galilean addition says different observers must disagree. Both cannot be right.

c = 1/√(μ₀ε₀) — no frame appears

01

What counts as inertial

A frame gliding at constant velocity is inertial: a puck released at rest stays put. A braking car or a rotating carousel is not — loose objects accelerate with no force in sight, and you must invent fictitious forces to patch Newton up. The Earth’s surface spins and orbits, but the accelerations are so small that it passes as inertial for laboratory physics.

02

Relativity is about symmetry, not motion

The deep claim is not that things move — it is that the laws cannot tell inertial frames apart. Drop a ball on the train and it lands at your feet, because it shares the train’s velocity and gains none horizontally. Every mechanics experiment comes out identical, so the question ‘are we really moving?’ has no mechanical answer. Only relative velocity between frames is measurable.

03

The crack Maxwell opened

By 1865 light was an electromagnetic wave with speed c set by μ₀ and ε₀. Waves usually have a medium — sound needs air — so physicists invented the ‘aether’ and expected light’s measured speed to vary with the observer’s motion through it, by u′ = c − v. The Earth orbits at 30 km/s, so the effect had to be there at 1 part in 10⁴. The next lesson is the experiment that went looking — and found nothing.

02

Change one variable at a time

Make the relationship visible.

Galileo’s whole law is a downward shift of the line u′ = u by v. It works flawlessly for balls and passengers — then try it on light: the train observer should measure c − v, a different light speed in every frame. Maxwell’s equations, and lesson 27.2’s experiment, say that never happens.

u′ = u (v = 0)u′ = u − vuu′

Train-frame speed u′ = u − v8 m/s

Ground-frame speed u20 m/s

Same rule applied to lightc − v = 299,792,446 m/s

What experiment finds for lightc in both frames — Galileo fails here

03

Catch the common trap

Explain before calculating.

You are sealed in a windowless cabin moving at constant velocity. Which experiment can reveal your speed?

Choose an answer to test the model.

04

Worked examples

State the rule, substitute, then check units.

EasyA passenger walks forward at 2.0 m/s along a train moving at 30.0 m/s. Find her velocity in the ground frame, and her velocity if she walks toward the rear instead.
  1. Ground frame: u = u′ + v = 2.0 + 30.0 = 32.0 m/s.
  2. Walking rearward, u′ = −2.0 m/s, so u = −2.0 + 30.0 = 28.0 m/s.
  3. Both answers come from the same rule u′ = u − v, rearranged — velocities add like ordinary vectors in Galilean physics.

Answer32.0 m/s forward; 28.0 m/s walking rearward

MediumA ball is thrown at 8.0 m/s (train frame) inside a carriage doing 30.0 m/s. Where is the ball after 3.0 s in each frame? Verify the Galilean transformation connects the answers.
  1. Train frame: x′ = u′t = 8.0 × 3.0 = 24.0 m from the thrower.
  2. Ground frame: u = 38.0 m/s, so x = 38.0 × 3.0 = 114.0 m.
  3. Check: x′ = x − vt = 114.0 − 30.0 × 3.0 = 24.0 m ✓ — same event, two coordinate grids.

Answer24.0 m (train frame), 114.0 m (ground frame), linked by x′ = x − vt

HardMaxwell’s equations give c = 1/√(μ₀ε₀) = 3.00 × 10⁸ m/s with no reference frame specified. The Earth orbits the Sun at 30 km/s. What fractional change in the measured speed of light did Galilean physics predict across the year, and why was that a crisis?
  1. Galilean addition: measured speed should be c ± v depending on the direction of travel through the supposed aether.
  2. Fractional shift: v/c = 3.0 × 10⁴ ÷ 3.0 × 10⁸ = 1.0 × 10⁻⁴ — one part in ten thousand, reversing over six months.
  3. But Maxwell’s c is built only from μ₀ and ε₀, constants with no v in them. Either Maxwell’s equations were wrong in moving frames, or the Galilean transformation was. Interferometers were precise enough to referee — see 27.2.

AnswerA ±1 × 10⁻⁴ seasonal shift was predicted; Maxwell’s frame-free c said there should be none

ChallengingStarting from u′ = u − v with v constant, show that both frames agree on accelerations and hence on Newton’s second law — and identify the single assumption that special relativity will later abandon.
  1. Differentiate with respect to time: du′/dt = du/dt − 0, so a′ = a — the constant v vanishes.
  2. With identical masses and forces, F = ma holds in both frames: no mechanics experiment can distinguish them. Galilean relativity is proved, not assumed, for mechanics.
  3. The derivation silently used t′ = t — the same clock for all observers, which let us differentiate both sides ‘with respect to time’ as if time were shared. That is the assumption Einstein drops: once simultaneity is frame-dependent, u′ = u − v must be rebuilt (27.4).

Answera′ = a follows, so mechanics is frame-blind; the buried assumption is universal time t′ = t

Exam diagrams for this topic1 figure to inspect and practiseQuestions, hints and marking points in one compact subsection.

See it. Read it. Work it.

These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.

  1. 01

    InspectRead the figure comment.

  2. 02

    TraceFollow labels, arrows and axes.

  3. 03

    AnswerWork one part at a time.

  4. 04

    CheckReveal hints and marking points.

IB

01Figure 5Rigid body mechanics · Galilean and special relativityIB
Solid cylinder released from rest on a sloperadius 0.15 msolid cylinder, 2.0 kg, released from rest1.2 mbottom of the slope

Figure comment

Figure 5Side view. A straight hatched slope runs down from the upper left to a horizontal surface at the lower right. A cylinder is drawn end-on resting on the slope near the top, labelled solid cylinder, 2.0 kg, released from rest, with a line from its centre to the rim labelled radius 0.15 m. A dashed horizontal line runs to the right from the level of the cylinder's centre, and a dimension line marks 1.2 m between that level and the horizontal surface at the bottom of the slope.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

Guided questions 5 parts

Reading cue. The 1.2 m is measured from the dashed line through the axis, so it is the drop of the centre of mass and not the length of the slope; 0.15 m is a radius, not a diameter.

  1. aState State the relationship between the translational speed of the cylinder's centre and its angular speed while it rolls without slipping, using the dimension marked on the drawing.

    recall2 marks

    Check answer 2 marks
    1. v = ωR
    2. with R = 0.15 m from the figure, v = 0.15ω
  2. bDetermine The cylinder reaches the bottom of the slope with a translational speed of 3.96 m s⁻¹. Determine its angular speed there.

    routine3 marks

    Check answer 3 marks
    1. rolling without slipping, so ω = v / R
    2. ω = 3.96 / 0.15
    3. ω = 26 rad s⁻¹
  3. cDetermine The slope is inclined at 25° to the horizontal. Determine the acceleration of the cylinder's centre down the slope, and the time it takes to reach the bottom from rest.

    demanding4 marks

    Check answer 4 marks
    1. for a solid cylinder a = g sin θ / (1 + I/MR²) = g sin θ / 1.5 = (2/3)g sin θ
    2. a = (2/3) × 9.81 × sin 25° = 2.76 m s⁻²
    3. the distance travelled is along the slope, 1.2 / sin 25° = 2.84 m, not 1.2 m
    4. t = √(2 × 2.84 / 2.76) = 1.43 s, which checks against v = at = 2.76 × 1.43 = 3.96 m s⁻¹
  4. dShow (that) Friction is the only force exerting a torque about the axis. Show that the coefficient of static friction must be at least (tan θ)/3 for the cylinder to roll without slipping, and evaluate this for the 25° slope.

    top of the paper4 marks

    Check answer 4 marks
    1. torque about the axis: fR = Iα = ½MR²(a/R), so f = ½Ma
    2. substituting a = (2/3)g sin θ gives f = (1/3)Mg sin θ
    3. the normal force is N = Mg cos θ, so μ_min = f/N = (tan θ)/3
    4. for θ = 25°, μ_min = 0.466 / 3 = 0.16

Transfer challenge

A block of mass 1.5 kg hangs from a light string wound round the rim of a solid cylindrical pulley of mass 2.0 kg and radius 0.15 m, free to turn about a fixed horizontal axis. Determine the acceleration of the block and the tension in the string.

Check answer 4 marks
  1. pulley: TR = Iα = ½MR²(a/R), so T = ½Ma = 1.0a
  2. block: mg − T = ma, so 1.5 × 9.81 − 1.0a = 1.5a
  3. a = 14.7 / 2.5 = 5.9 m s⁻², and T = 1.0 × 5.9 = 5.9 N
  4. the acceleration is below g because part of the released potential energy goes into spinning the pulley, exactly as it went into spinning the cylinder on the slope