Galilean Relativity

Whenever something moves, the answer to "how fast?" depends on who is asking. A ball rolling across the floor of a train looks completely different to a passenger sitting still on the train and to someone standing on the platform watching the train go by. This workbook introduces reference frames — the coordinate systems we use to describe motion — and the simple equations that let us convert a measurement made in one reference frame into a measurement made in another. These equations were worked out by Galileo and refined by Newton, and the whole set of ideas is known as Galilean relativity.

By the end of this workbook you should be able to:

explain what is meant by a reference frame, an event and an inertial reference frame
state that Newton's laws of motion are the same in every inertial reference frame — this is Galilean relativity
use the Galilean transformation equations, x′ = x − vt and t′ = t, to relate the position and time of an event as measured in two reference frames
use the Galilean velocity addition equation, u′ = u − v, to relate the velocity of an object as measured in two reference frames

Name

1. Reference frames

Imagine you are sitting on a train, reading a book. To you, the book is not moving at all. But to someone standing on the platform watching the train go past, both you and the book are moving very quickly. Neither person is wrong — they are simply describing the same situation from a different reference frame.

A reference frame is a coordinate system, with a set of axes and a clock, that we use to assign a position and a time to something that happens.

A single, precisely located and timed happening — a flash of light, a ball bouncing, two cars passing each other — is called an event.

Physicists often imagine a hypothetical person, called an observer, taking the measurements in a particular reference frame. An observer is always at rest relative to their own reference frame — the observer on the train is at rest in the train's reference frame, and the observer on the platform is at rest in the platform's reference frame.

Figure 1. Observer A throws a ball straight up while riding on a moving cart. A, who moves with the cart, sees the ball go straight up and straight back down. B, standing still on the ground, sees the same ball follow a curved path, because the cart (and the ball) are also moving sideways in B's reference frame.

Both descriptions in Figure 1 are correct — they are just measurements taken in different reference frames. This is the central idea behind everything in this workbook: all reference frames are equally valid. There is no single "correct" point of view.

1 In your own words, explain what is meant by an event in physics, and give an example of your own (not one already used in this workbook).

Think about something that happens at one specific place and one specific instant — not something that lasts a long time or covers a large distance.

Model answer

An event is a single, precisely located happening — it occurs at one specific point in space and at one specific instant in time. Example: a firework exploding, a car's indicator light flashing once, or two runners crossing the finish line at the same moment.

2 Using Figure 1, explain why observer A records a straight-line path for the ball, while observer B records a curved path — even though it is the same ball, thrown in the same way.

A is moving sideways at the same speed as the ball's sideways motion the whole time. B is not.

Model answer

In A's reference frame, A and the ball are both moving sideways with the cart at the same speed, so A only ever sees the ball's up-and-down motion — a straight vertical line. In B's reference frame, the ball has both an up-and-down motion (from being thrown) and a sideways motion (because it is moving with the cart), so B sees the combination of the two as a curved path.

3 A skydiver jumps from a plane and, a few seconds later, pulls their parachute cord, which slows their fall sharply. Describe how this event (the parachute opening) would be described by (a) a camera that is falling alongside the skydiver at the same speed, and (b) an observer standing on the ground.

Ask yourself: in each reference frame, who (or what) is treated as "not moving", and what does everything else appear to do relative to that?

Model answer

(a) In the falling camera's reference frame, the camera treats itself as stationary, so before the cord is pulled the skydiver appears to stay still relative to the camera. The moment the cord is pulled, the skydiver decelerates and appears to move away from the camera (upwards, relative to the camera). (b) In the ground observer's reference frame, the skydiver is falling quickly downwards the whole time; when the cord is pulled, the observer simply records the skydiver's downward speed decreasing.

2. Inertial frames and Galilean relativity

Some reference frames are more useful than others for applying Newton's laws of motion. Picture an observer floating in deep space, far from any planets, inside a spacecraft with no windows. If the observer very carefully releases a small object in mid-air, what happens to it?

If the spacecraft is moving at a constant velocity (in a straight line, at a steady speed) — or is completely at rest — the released object simply stays exactly where it was let go. It does not drift towards any wall. But if the spacecraft is accelerating (its engines are firing, speeding it up), the object appears to drift backwards, towards the rear wall, even though no force is acting on it. To the observer, it would look as though Newton's first law had been broken.

An inertial reference frame is one that is not accelerating. Newton's laws of motion apply correctly, exactly as we normally state them, only in inertial reference frames.

This idea — that the same laws of motion apply, unchanged, in every inertial reference frame, no matter how fast that frame is moving (as long as it isn't accelerating) — is the heart of Galilean relativity.

Key idea Newton's laws of motion are the same in all inertial reference frames. This is known as Galilean relativity. There is no single reference frame that is "correct" or "truly stationary" — every inertial observer's point of view is equally valid.

Worth remembering

It's tempting to assume the ground beneath your feet is "properly" stationary. It isn't! The Earth spins on its axis once roughly every 24 hours, and its radius is about 6400 km. A point on the equator is therefore moving at roughly:

v = 2πR / T = (2 × π × 6.4 × 10⁶) / (24 × 3600) ≈ 465 m s⁻¹

That's roughly 1700 km/h — faster than a jet airliner — yet you don't feel it, because you, the ground, and the air around you are all moving together at (very nearly) constant velocity. For everyday physics we still treat the Earth's surface as a good inertial reference frame, even though it is moving.

4 State, in your own words, what is meant by an inertial reference frame.

Model answer

An inertial reference frame is a reference frame that is not accelerating (it is either at rest or moving with constant velocity). Newton's laws of motion apply correctly within such a frame.

5 For each situation below, state whether the reference frame described is inertial or non-inertial, and give a reason.
(a) a lift (elevator) moving upwards at a constant 1.5 m s⁻¹
(b) a car braking sharply as it approaches a red light
(c) a satellite travelling in a circular orbit around the Earth at constant speed
(d) a spacecraft drifting through deep space with its engines switched off

A reference frame is non-inertial if its speed or direction of travel is changing. Moving in a circle at constant speed still counts as accelerating, because the direction of motion is constantly changing.

Model answer

(a) Inertial — constant velocity, so no acceleration. (b) Non-inertial — the car's speed is decreasing, so it is accelerating (decelerating). (c) Non-inertial — even though the speed is constant, the direction keeps changing, so the velocity (and therefore the frame) is accelerating. (d) Inertial — no engines firing, so no resultant force and no acceleration.

6 Explain how an observer floating inside a windowless spacecraft, with no view of the outside world, could use the "release an object in mid-air" thought experiment to test whether their spacecraft is an inertial reference frame.

Model answer

The observer releases a small object gently in mid-air, with no force applied to it. If the object stays exactly where it was released, no resultant force is acting on it, so the spacecraft is not accelerating — it is an inertial reference frame. If the object appears to drift towards one wall on its own, the spacecraft itself must be accelerating in the opposite direction, so it is a non-inertial frame.

3. The Galilean transformation equations

If two observers are in different (inertial) reference frames, how do we convert a measurement made by one of them into the equivalent measurement made by the other? We call this process a transformation.

Suppose reference frame S is stationary, and reference frame S′ (say, S′ is a camera) moves at a constant velocity v relative to S, along the x-direction. We will always assume that the origins of S and S′ are in the same place at t = 0 — the moment the clocks in both frames are started.

An event that happens somewhere along the x-axis can be given coordinates (x, t) in frame S, and different coordinates (x′, t′) in frame S′. In Newton's model of the Universe, it was assumed that time passes at the same rate for every observer, however they are moving. That single assumption leads directly to the Galilean transformation equations.

x′ = x − vt
t′ = t where v is the velocity of frame S′ relative to frame S, and the origins of S and S′ coincide at t = t′ = 0

The second equation, t′ = t, looks almost too simple to be worth writing down — but it is a genuine assumption, not something that can be proved. Galileo and Newton assumed that every observer, however fast they are moving, would agree on the time of any event and on the time interval between two events. This assumption works extremely well at everyday speeds.

Exploring the transformation

The simulation below matches the scenario of a stationary object on a sports field, 88 m from a fixed starting point, with a camera moving away from that starting point at a constant velocity v. Use the time slider to see how x′ — the object's distance from the moving camera — changes, while x itself (the object's distance from the stationary starting point) stays fixed.

Simulation 1 — Galilean reference frames

Interactive simulation — view this workbook on a computer or tablet to use it.

Camera velocity, v = 2.0 m s⁻¹

Object's distance in S, x = 88 m

Time, t = 0.0 s

t = 0.0 s x = 88.0 m vt = 0.0 m x′ = x − vt = 88.0 m

7 Using the default settings on Simulation 1 (x = 88 m, v = 2.0 m s⁻¹), read off the value of x′ at t = 2 s, t = 4 s and t = 6 s, and note them below. Then describe, in words, the pattern you see as t increases.

Model answer

t = 2 s: x′ = 88 − (2 × 2) = 84 m. t = 4 s: x′ = 88 − (2 × 4) = 80 m. t = 6 s: x′ = 88 − (2 × 6) = 76 m. As t increases, x′ decreases at a constant rate of 2 m per second — because the camera is catching up to the object at 2 m s⁻¹, so the object's distance from the camera shrinks by 2 m every second, while its distance from the fixed starting point, x, does not change at all.

Worked example

A train passes a platform, travelling at a constant velocity of 20 m s⁻¹. The observer on the platform (frame S) and the observer on the train (frame S′) both start their clocks at the instant the train passes. After 5.0 s, the observer on the platform records a flash of light (the event) at x = 120 m, t = 5.0 s. Determine the position, x′, of this event according to the observer on the train.

x′ = x − vt

x′ = 120 − (20 × 5.0)

x′ = 120 − 100 = complete this step yourself

(Since t′ = t, the time of the event is the same for both observers: t′ = 5.0 s.)

8 A cyclist rides past a lamppost at a constant velocity of 8.0 m s⁻¹, starting her stopwatch at the moment she passes it. 6.0 s later, she notices a dog barking at a point 70 m from the lamppost (measured by someone standing still next to the lamppost). Calculate the position of the barking dog, x′, according to the cyclist.

Let the person by the lamppost be reference frame S, and the cyclist be reference frame S′. Use x′ = x − vt.

Model answer

x′ = x − vt

x′ = 70 − (8.0 × 6.0)

x′ = 70 − 48 = 22 (finish evaluating this yourself, including units)

9 Explain why t′ = t is described as an assumption of Galilean relativity, rather than something that has been proved to be true.

Model answer

Newton and Galileo simply assumed that time passes at the same rate for every observer, no matter how they are moving, because this matches everyday experience — at ordinary speeds, no one has ever noticed two clocks disagreeing because one was moving. It was accepted as a sensible starting point (a postulate), not something that had been directly tested at very high speeds.

10 A drone hovers at a fixed point, x = −40 m (that is, 40 m behind the origin of frame S). A second drone, representing the origin of frame S′, flies past the origin of S at t = 0 and travels at a constant velocity of v = −5.0 m s⁻¹ (moving in the negative x-direction). Calculate x′, the position of the first drone according to the second drone, at t = 3.0 s.

Substitute the values directly into x′ = x − vt, being careful with the negative signs.

Model answer

x′ = x − vt

x′ = (−40) − (−5.0 × 3.0)

x′ = −40 − (−15) = −40 + 15 = (finish evaluating this yourself, including sign and units)

4. The velocity addition equation

We often need to relate the velocity of a moving object, as measured by two different observers. Suppose an object has velocity u in the stationary frame S, and velocity u′ in frame S′, which is itself moving at velocity v relative to S. Because x′ = x − vt, working out how the object's position changes with time in each frame leads to a very similar-looking equation for velocity.

u′ = u − v the Galilean velocity addition equation, where u is the object's velocity in frame S and u′ is its velocity in frame S′, which moves at velocity v relative to S

This equation is often rearranged as u = u′ + v when u′ (the velocity in the moving frame) is the known quantity and u (the velocity in the stationary, "ground" frame) is what needs to be found.

Simulation 2 — velocity addition

Drag the sliders to set the train's velocity, v, and the ball's velocity relative to the train, u′. Watch how the ball moves relative to the ground.

Interactive simulation — view this workbook on a computer or tablet to use it.

Train velocity, v = 4.0 m s⁻¹

Ball's velocity relative to train, u′ = 2.0 m s⁻¹

From the reference frame of the ground observer: v = 4.0 m s⁻¹ u′ = 2.0 m s⁻¹ u = u′ + v = 6.0 m s⁻¹

Worked example

A train is moving with a constant velocity of 16.0 m s⁻¹. A ball rolls along the floor of the train, in the direction of travel, with a constant velocity of 3.0 m s⁻¹ relative to the train.

(a) Calculate the velocity of the ball as recorded by an observer standing on the ground outside.

u′ = u − v

3.0 = u − 16.0

u = 3.0 + 16.0 = complete this step yourself

(b) Now suppose instead the ball rolls towards the back of the train with the same speed. Set up the equation for this case (do not evaluate the final answer).

u′ = u − v

−3.0 = u − 16.0

u = −3.0 + 16.0 = complete this step yourself

11 Using the same train from the worked example (v = 16.0 m s⁻¹), now suppose the whole train is travelling in the opposite direction, so v = −16.0 m s⁻¹, while the ball still rolls towards the front of the train at u′ = 3.0 m s⁻¹ relative to the train. Calculate u, the ball's velocity according to the ground observer.

u′ = u − v, so u = u′ + v. Substitute the values carefully, including the negative sign on v.

Model answer

u′ = u − v

3.0 = u − (−16.0)

u = 3.0 − 16.0 = (finish evaluating this yourself, including sign and units)

12 A skateboarder, D, moves at a constant 4.0 m s⁻¹ across a playground. While moving, D throws a ball forwards so that, relative to the skateboard, the ball travels at 2.0 m s⁻¹. Calculate the velocity of the ball, u, as recorded by a stationary observer, C, standing on the ground.

The skateboard is frame S′, moving at v = 4.0 m s⁻¹. The ball's velocity relative to the skateboard is u′ = 2.0 m s⁻¹.

Model answer

u′ = u − v

2.0 = u − 4.0

u = 2.0 + 4.0 = (finish evaluating this yourself, including units)

13 Newton assumed that distance and time were the same for every observer (x and t could be transformed simply, and t′ = t exactly). Using the equation u′ = u − v, explain why velocity is different: why can two observers, in different inertial reference frames, never agree on the velocity of a moving object (unless v = 0)?

Model answer

Because u′ = u − v, the velocity measured in frame S′ always differs from the velocity measured in frame S by exactly v, the relative velocity between the two frames. Unless the two frames happen to be at rest relative to each other (v = 0), the two observers will always calculate a different value for the object's velocity — velocity is not invariant (the same for everyone); it depends on the observer's own reference frame.

Glossary

Reference frame: A coordinate system, with axes and a clock, used to assign a position and a time to an event.
Event: A single incident that occurs at one exact point in space and one exact instant in time.
Observer: A hypothetical person who takes measurements from a single, specific reference frame, and is always at rest relative to that frame.
Inertial reference frame: A reference frame that is not accelerating, in which Newton's laws of motion apply correctly.
Galilean relativity: The idea that Newton's laws of motion are the same in every inertial reference frame — there is no single "correct" reference frame.
Galilean transformation: The pair of equations, x′ = x − vt and t′ = t, used to convert the position and time of an event from one inertial reference frame to another.
Velocity addition equation: The equation u′ = u − v, which relates the velocity of an object as measured in two different inertial reference frames moving at relative velocity v.