Visual walkthrough — Reference frames — Galilean transformations

We use only two ideas the whole way: (1) an arrow that points from one place to another (a "vector"), and (2) the head-to-tail rule for adding two arrows. Both are built in Vectors — addition and components — glance there if arrows are new.

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Step 1 — Two observers, two rulers

WHAT. We draw two observers. Call the first one $S$ (think: someone standing on the ground). Call the second one $S^{\prime }$ — read aloud as "S-prime", the little tick mark just means "the second one" (think: someone riding a train). Each carries a ruler laid along an $x$-direction and a $y$-direction, plus a clock.

WHY. A measurement is meaningless until we say who measured it. Position and time are only numbers relative to somebody's ruler and clock. So before any formula, we must nail down the two rulers. This is the whole point of a reference frame.

PICTURE. The ground observer's origin is $O$ (magenta). The train observer's origin is $O^{\prime }$ (violet). At the instant $t=0$ we deliberately slide them so the two origins sit on top of each other — this just saves us a constant and costs nothing.

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Step 2 — The train's origin drifts away

WHAT. The train moves at a constant velocity, an arrow we call $\vec{V}$ (orange). "Constant" means the arrow never changes length or direction. After a time $t$ has ticked by, where has the train's origin $O^{\prime }$ slid to, as measured by the ground?

WHY. Everything downstream is bookkeeping about the gap between the two origins. That gap is the only difference between the two observers, so we compute it first. And the rule for it is the most basic one in kinematics: distance travelled = velocity × time.

PICTURE. Starting from $O$, the origin $O^{\prime }$ has slid to the tip of the arrow $\vec{V}t$.

$\underset{\text{gap between origins}}{\underbrace{\vec{V}t}}=\underset{\begin{array}{c}\text{train’s velocity}\\ \text{(fixed arrow)}\end{array}}{\underbrace{\vec{V}}}\times \underset{\begin{array}{c}\text{time on}\\ \text{the clock}\end{array}}{\underbrace{t}}$

Reading it term by term: $\vec{V}$ sets the direction and speed of the drift; multiplying by $t$ stretches that arrow longer the more time passes. At $t=0$ the arrow has zero length — the origins coincide, exactly as we arranged in Step 1.

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Step 3 — One event, seen two ways

WHAT. Now something happens — a firecracker pops. That is an event: a definite place at a definite time. The ground calls its position $\vec{r}$; the train calls the same pop's position $\vec{r}{}^{\prime }$. We want the relationship between these two arrows.

WHY. This is the heart of the derivation, and it needs no physics at all — it is pure geometry. To reach the event from $O$, you can either go straight there ($\vec{r}$), or take a detour: first walk to the train's origin ($\vec{V}t$), then walk from there to the event ($\vec{r}{}^{\prime }$). Both routes end at the same spot, so the arrows must match up.

PICTURE. Look at the triangle of three arrows. The head-to-tail rule says the straight arrow equals the two-leg path:

$\underset{\begin{array}{c}\text{event as seen}\\ \text{from ground }O\end{array}}{\underbrace{\vec{r}}}=\underset{\begin{array}{c}O\to O^{\prime }\\ \text{(from Step 2)}\end{array}}{\underbrace{\vec{V}t}}+\underset{\begin{array}{c}{O}^{\prime }\to \text{event}\\ \text{(train’s answer)}\end{array}}{\underbrace{\vec{r}{}^{\prime }}}$

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Step 4 — Rearrange for the train's coordinates

WHAT. We now solve the triangle equation for $\vec{r}{}^{\prime }$ — the primed observer's answer in terms of quantities the ground can measure.

WHY. In practice we usually know the ground numbers ($\vec{r}$, $t$, $\vec{V}$) and want the train's numbers. Moving $\vec{V}t$ to the other side of the equals sign does exactly that.

PICTURE. Subtracting $\vec{V}t$ means "walk back along the orange arrow" — the same triangle, read backwards.

$\boxed{\underset{\text{train’s position}}{\underbrace{\vec{r}{}^{\prime }}}=\underset{\text{ground’s position}}{\underbrace{\vec{r}}}-\underset{\begin{array}{c}\text{undo the}\\ \text{origin drift}\end{array}}{\underbrace{\vec{V}t}},\underset{\begin{array}{c}\text{clocks agree}\\ \text{(the Galilean bet)}\end{array}}{\underbrace{t^{\prime }=t}}}$

In components with $\vec{V}=V\hat{x}$ (train moving along $x$):

$x^{\prime }=x-Vt,y^{\prime }=y,z^{\prime }=z,t^{\prime }=t.$

Only the $x$ that lines up with $\vec{V}$ gets shifted; the sideways directions are untouched — read straight off the triangle.

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Step 5 — First derivative: how velocities transform

WHAT. Velocity is how fast a position arrow changes each second. We write it with $\frac{d}{dt}$, which is just shorthand for "rate of change per unit time". Apply it to the whole position rule.

WHY use a derivative here? Because velocity is defined as the time-rate-of-change of position. We are not choosing a fancy tool for style — the question "how do velocities compare?" is literally the question "how does the position relation change per second?", and $\frac{d}{dt}$ is the operation that answers exactly that. No other tool measures a per-second rate.

PICTURE. Take $\frac{d}{dt}$ of $\vec{r}{}^{\prime }=\vec{r}-\vec{V}t$. Because $t^{\prime }=t$, the two observers' clocks run at the same rate, so differentiating "per second" means the same to both. The rate of change of $\vec{r}$ is the ground velocity $\vec{u}$; of $\vec{r}{}^{\prime }$ is the train velocity $\vec{u}{}^{\prime }$; and the rate of change of $\vec{V}t$ is just $\vec{V}$ (since $\vec{V}$ is a fixed arrow):

$\underset{\begin{array}{c}\text{event’s velocity}\\ \text{per train}\end{array}}{\underbrace{\vec{u}{}^{\prime }}}=\underset{\begin{array}{c}\text{event’s velocity}\\ \text{per ground}\end{array}}{\underbrace{\vec{u}}}-\underset{=\vec{V}}{\underbrace{\frac{d(\vec{V}t)}{dt}}}⟹\boxed{\vec{u}=\vec{u}{}^{\prime }+\vec{V}}$

Reading the box: the ground velocity is the train's own reading plus the train's velocity. The velocities differ by exactly one $\vec{V}$ — one power of $t$ has been "knocked off" the $\vec{V}t$ gap.

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Step 6 — Second derivative: acceleration is shared

WHAT. Acceleration is how fast the velocity arrow changes each second — apply $\frac{d}{dt}$ one more time, now to the velocity rule.

WHY. Same logic as Step 5, one level up: acceleration is defined as the rate of change of velocity, so we differentiate again. The crucial fact is that $\vec{V}$ is constant, and the rate of change of something that never changes is zero.

PICTURE. Differentiate $\vec{u}=\vec{u}{}^{\prime }+\vec{V}$:

$\underset{\text{ground’s accel.}}{\underbrace{\vec{a}}}=\underset{\text{train’s accel.}}{\underbrace{\vec{a}{}^{\prime }}}+\underset{=0\text{(constant!)}}{\underbrace{\frac{d\vec{V}}{dt}}}⟹\boxed{\vec{a}{}^{\prime }=\vec{a}}$

The extra shift has now vanished entirely. Both observers measure the identical acceleration. Since mass is the same for both, Newton's $\vec{F}=m\vec{a}$ reads the same in both frames — that is Galilean relativity.

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Step 7 — The degenerate cases (nothing left uncovered)

WHAT. Let us push every knob to an extreme and check the pictures still hold.

WHY. A rule you trust must survive its edge cases. If any special input broke the triangle, we would not really understand it.

PICTURE. Four collapses of the same triangle:

• $t=0$ — the gap $\vec{V}t$ has zero length. The triangle flattens to a point: $\vec{r}{}^{\prime }=\vec{r}$. The two observers agree at the starting instant, by our Step-1 setup.
• $\vec{V}=\vec{0}$ — the train never moves. Then $\vec{V}t=\vec{0}$ for all time: $\vec{r}{}^{\prime }=\vec{r}$, $\vec{u}{}^{\prime }=\vec{u}$. Same frame, same answers.
• Event at $O^{\prime }$ (firecracker on the train's origin) — then $\vec{r}{}^{\prime }=\vec{0}$ and $\vec{r}=\vec{V}t$: the event just tracks the drifting origin, recovering Step 2.
• $\vec{V}$ not constant (accelerating train) — now $\frac{d\vec{V}}{dt}=\vec{A}\ne 0$, so Step 6 gives $\vec{a}{}^{\prime }=\vec{a}-\vec{A}$. The frame is non-inertial and pseudo-forces appear. See Inertial and non-inertial frames — constant $\vec{V}$ was the magic ingredient.

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The one-picture summary

The entire derivation is one triangle differentiated twice. Position differs by the full drift arrow $\vec{V}t$; velocity differs by the constant arrow $\vec{V}$; acceleration differs by nothing.

Recall Feynman retelling — the whole walkthrough in plain words

Picture you and a friend. Your friend hops on a train that glides at a steady speed. At the whistle ($t=0$) you're standing shoulder to shoulder, so you agree on where everything is. As the train pulls away, the gap between you grows exactly by "train-speed times seconds" — that's the orange arrow. Now a firecracker pops somewhere. To find where you think it is, you can walk straight to it, or first walk over to your friend and then out to the firecracker — same trip, so those arrows must add up. That adding-up is the position rule. Ask "how fast does the firecracker move?" — that's just how fast its arrow changes each second. Because the gap grows steadily, your two answers differ by exactly the train's speed: velocities differ by one $\vec{V}$. Ask "how fast does it speed up?" — differences that were steady now contribute nothing, because a steady train-speed doesn't itself speed up. So you both get the same acceleration. That shared number is why the same laws of physics work on the train and on the ground — and it all fell out of one triangle, drawn once and read three times.

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Recall checkpoints

Recall

Why does the position rule need arrows (vectors) and not just number subtraction? ::: Because the event can lie sideways to the train's motion; arrows keep the sideways parts equal and shift only the along-$\vec{V}$ part automatically. In the chain $\vec{V}t\to \vec{V}\to 0$, what operation moves you one step right? ::: A time-derivative $d/dt$ — each one peels off a power of $t$. What single assumption makes the transformation Galilean rather than Lorentz? ::: Absolute time, $t^{\prime }=t$ (all clocks tick identically). If the train accelerates at $\vec{A}$, what does Step 6 become? ::: $\vec{a}{}^{\prime }=\vec{a}-\vec{A}$ — the frame is non-inertial and pseudo-forces appear. At $t=0$ why do the two observers agree on position? ::: The drift arrow $\vec{V}t$ has zero length, so $\vec{r}{}^{\prime }=\vec{r}$.

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Connections

• Vectors — addition and components — the head-to-tail triangle is the engine of Step 3.
• Relative velocity — Step 5 rearranged: ${\vec{u}}_{A/B}={\vec{u}}_A-{\vec{u}}_B$.
• Inertial and non-inertial frames — Step 7's accelerating-train case.
• Newton's laws of motion — invariant because of Step 6.
• Special relativity — Lorentz transformation — what replaces $t^{\prime }=t$ when $v\sim c$.