You already know that motion looks different from different reference frames. Galilean relativity is the toolkit that lets you convert between those frames — to take a position, a time, or a velocity measured by one observer and work out exactly what a second, moving observer would measure. It’s the everyday, low-speed version of relativity, and it’s beautifully simple: you just add and subtract.
📘 What you need to know
The laws of physics are the same in all inertial frames — this is Galilean relativity
Newton treated space and time as fixed and absolute (the same for everyone)
Galilean transformations convert position and time between frames S and S′
Position: x = x′ + vt (and x′ = x − vt)
Time is unchanged: t = t′, and sideways coordinates are unchanged: y = y′
Velocity addition: same direction u = u′ + v; opposite direction subtracts
These work only when speeds are much less than the speed of light
The same laws in every frame
We use inertial frames for one powerful reason: Newton’s laws of motion work identically in all of them. That principle has a name — Galilean relativity. An object left alone keeps moving in a straight line at constant velocity (Newton’s first law) whether you watch it from a platform or from a train gliding past at steady speed.
So the same laws of physics apply no matter which inertial frame you’re in, as long as the frames move in straight lines at constant velocity relative to each other. An object with constant velocity in one frame still has a constant — though possibly different — velocity in another.
Drop a ball on a smoothly moving train and it lands right under your hand — exactly as on the platform. The laws of physics don’t change between inertial frames. That’s Galilean relativity.
Newton went one step further and assumed space and time were fixed and absolute — that a one-second interval or a one-metre length is the same for every observer, everywhere. At everyday speeds this works perfectly. (Later, near the speed of light, we’ll see it breaks down — but that’s a story for special relativity.)
Here’s the mental picture: Galilean relativity says everyone agrees on clocks and rulers — a second is a second, a metre is a metre — they just disagree on positions and velocities because they’re measuring from different starting points that are sliding past each other. That single assumption is what makes the maths so easy.
Galilean transformation equations
Now for the tool itself. We have our stationary frame S with coordinates (x, y, t) and a moving frame S′ with coordinates (x′, y′, t′) sliding along the x-direction at velocity v. A Galilean transformation lets us swap a measurement from one frame into the other.
Imagine both observers time an event — say a balloon popping. Observer S′ is moving, so by the time the balloon pops their whole frame has slid forward by a distance vt. So the position S measures is the position S′ measures, plus that shift:
Frame S′ has slid forward by vt. So the distance S measures to the event is the moving-frame distance x′ plus that shift: x = x′ + vt.
That gives the full set of Galilean transformation equations. Notice how little actually changes:
Galilean transformations (S ↔ S′)x = x′ + vt or x′ = x − vty = y′z = z′t = t′
The y, z and t coordinates are identical in both frames, because the relative motion is only along x and time is absolute. You really only need to remember one equation, x = x′ + vt — the other is just it rearranged.
WE 1
A bus passes a stop at a constant 12 m s−1. A passenger walks forward along the bus at 1.2 m s−1 for 8.0 s. How far does the passenger travel according to (a) themselves, and (b) a person at the stop?
(a) Passenger’s own frame (S′)
they’re at rest relative to the bus, so use x′ = u′t
x′ = 1.2 × 8.0 = 9.6 m(b) Person at the stop (S) — Galilean transform
x = x′ + vt
x = 9.6 + (12 × 8.0) = 9.6 + 96x = 105.6 mThe passenger only “walks” 9.6 m, but the bus carries them 96 m more past the stop.
The golden rule for these: if the event happens inside your own frame, use plain old distance = speed × time — no transformation needed. You only reach for x = x′ + vt when you’re measuring something that’s happening in a frame moving relative to you.
Velocity addition
The same idea transforms velocities. If several things are moving, we just combine their velocities as vectors. Let’s fix the labels — they trip people up more than the maths does:
u′ in moving frame S′
+ v
v speed of frame S′
=
u in stationary frame S
So u is the object’s velocity in the stationary frame, u′ is its velocity in the moving frame, and v is the speed of the moving frame itself. When the object moves in the same direction as the frame, the velocities add:
Velocity additionu = u′ + v or u′ = u − v
When the object moves in the opposite direction to the frame, its velocity is negative — so the same equation naturally subtracts. The key is choosing a positive direction first (usually the direction of v) and sticking to it.
Same direction: velocities add. Opposite direction: u′ is negative, so the equation subtracts. Pick a positive direction first and the signs take care of themselves.
WE 2
A river flows at 2.5 m s−1. A swimmer swims downstream at 1.3 m s−1 relative to the water. How fast does someone on the bank see the swimmer moving?
Step 1 — identify the frames
water = moving frame (v = 2.5), swimmer measured in water (u′ = 1.3)
Step 2 — same direction, so add
u = u′ + v
u = 1.3 + 2.5u = 3.8 m s⁻¹ downstreamThe current carries the swimmer along, so the bank sees them faster than they swim.
WE 3
In the same 2.5 m s−1 river, the swimmer now swims upstream at 1.3 m s−1 relative to the water. What speed does the bank observer measure now?
Step 1 — set downstream as positive
upstream swim is negative: u′ = −1.3
Step 2 — add (with the sign)
u = u′ + v = −1.3 + 2.5
u = 1.2 m s⁻¹ downstreamEven swimming against it, the current wins — the bank still sees them drift downstream.
WE 4
Two cars drive along a straight road in opposite directions. Car P travels east at 25 m s−1 and Car Q travels west at 19 m s−1, both relative to the ground. What is the velocity of Car P relative to Car Q?
Step 1 — set east as positive
P: u = +25 (from ground). Q is the frame: v = −19
Step 2 — velocity of P in Q’s frame
u′ = u − v
u′ = 25 − (−19) = 25 + 19u′ = 44 m s⁻¹ eastClosing on each other, so the relative speed is the sum. “Relative to Q” means: what does Q measure?
WE 5
An escalator moves upward at 0.7 m s−1. (a) A person walks up it at 0.9 m s−1 relative to the steps — how fast do they rise relative to the ground? (b) Another person walks down the up-escalator at 0.9 m s−1 relative to the steps — what is their velocity relative to the ground?
(a) Walking up (same direction)
u = u′ + v = 0.9 + 0.7
u = 1.6 m s⁻¹ up(b) Walking down (opposite), so u′ = −0.9
u = u′ + v = −0.9 + 0.7
u = −0.2 m s⁻¹ (0.2 m s⁻¹ down)They walk faster than the escalator lifts them, so they slowly lose the battle — net motion is gently downward.
🛠️ Solving a Galilean relativity problem
Label the frames. Which is stationary (S) and which is moving (S′)? What is v?
Pick a positive direction — usually the direction of v — and give opposite motions a minus sign.
Same frame as the event? Just use distance = speed × time. No transform needed.
Position across frames? Use x = x′ + vt (or rearrange for x′).
Velocity across frames? Use u = u′ + v, keeping every sign.
💡 Top tips
One equation to remember.x = x′ + vt rearranges to x′ = x − vt — same equation.
Direction beats formula. Decide the positive direction first, then let the signs do the work.
“Relative to X” = X’s frame. The observer named after “relative to” is the one measuring.
No transform inside your own frame. Measure an event in your frame with ordinary physics.
Only for low speeds. Galilean rules break down near the speed of light — that needs Lorentz.
Quick recap: Galilean relativity says the laws of physics are the same in every inertial frame, with space and time absolute. Convert positions with x = x′ + vt and velocities with u = u′ + v, watching directions — but only when speeds are far below c.
⚠ Common mistakes
Ignoring direction — velocity is a vector, so opposite motion needs a minus sign
Applying a transformation to an event measured within your own frame
Mixing up u (stationary frame) and u′ (moving frame), or losing track of v
Forgetting that the positive direction is set by the frame’s velocity v
Using Galilean rules at relativistic speeds — they only hold when v « c
You’ve now got the machinery to move measurements between frames — and it all rested on one quiet assumption: that space and time are the same for everyone. That assumption is about to be challenged. Next we meet the postulates of special relativity, where Einstein keeps the “same laws in every frame” idea but adds a shocking twist about the speed of light — and simple addition stops working.
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