Foundations — Reference frames — Galilean transformations

Before you can read a single line of the parent note, you need to own the alphabet it speaks. Below, every symbol is built from nothing: plain words → a picture → why the topic needs it. Read top to bottom; each rung stands on the one below it.

1. A point in space, and its arrow

Everything starts with the idea of where something is. Not "how far" — where, meaning direction and distance from an agreed starting spot.

Picture a giant sheet of graph paper. Somebody stabs a pin into one crossing and says "this is $O$". Two rulers stretch out from that pin: one east ($x$), one north ($y$). That is a coordinate system.

Look at the figure above. The black dot at the crossing is $O$. The coral arrow runs from $O$ to the green object — that arrow is the position vector. The butter-yellow dashes show its two components: walk right along $x$, then up along $y$, and you land on the tip.

Why an arrow and not just a pair of numbers? Because an arrow shows the direction at a glance, and — crucially for this topic — arrows can be added tip-to-tail. That single property is what lets us combine "where the train is" with "where the ball is inside the train". You'll want Vectors — addition and components fresh in your mind.

In this second figure the four coloured arrows land in the four corners. Read off each tip: the top-right arrow is $(+,+)$ = east-and-north; the top-left is $(-,+)$ = west-and-north; bottom-left $(-,-)$ = west-and-south; bottom-right $(+,-)$ = east-and-south. You must be comfortable with both signs before you subtract vectors, because $\vec{r}-\vec{V}t$ will routinely produce negatives.

2. A moment in time, and the clock

Space alone is not enough. A firecracker "pops" — it has a where and a when.

Picture a flashbulb going off. Its position is an arrow $\vec{r}$; its time is a number $t$ frozen on the clock. The whole game of this topic is: two observers point at the same flash and write down different $\vec{r}$ but (in the Galilean world) the same $t$.

3. Velocity — how the arrow moves

Why do we need velocity as a vector, not just a speed? Because "5 m/s north" and "5 m/s east" are different motions even though the speed is identical — and in Example 2 of the parent (the river boat) the two directions must be added as arrows, giving a 5 m/s diagonal, not "3 + 4 = 7".

We use the derivative here and no other tool because "rate of change of position" is precisely what velocity means. The word and the symbol are the same idea.

In this figure, the lavender curve is position climbing with time. The straight coral line just touches it at the green dot — its steepness (slope) is the velocity $\vec{u}$ at that instant. Steeper coral line ⇒ faster motion. Keep this "velocity = slope" picture: we are about to differentiate a whole equation, and differentiating just means "take the slope of both sides".

4. Acceleration — how the velocity changes

Apply the derivative twice to position and you get acceleration; that stacked symbol $\frac{d^2}{d{t}^2}$ just means "derivative of the derivative". Why does the topic care so much about $\vec{a}$? Because it is the one quantity two inertial observers agree on — the anchor of the whole chapter and the reason Newton's laws of motion look the same to everyone.

5. The two frames $S$ and $S^{\prime }$, and the relative velocity $\vec{V}$

Now we name the two observers once and for all, so no symbol appears unannounced.

Read this figure carefully — it contains the whole topic. The coral star is one flash (one event) both observers watch. The ground observer $S$ draws the slate arrow $\vec{r}$ from $O$ to the star. The train observer $S^{\prime }$ draws the coral arrow $\vec{r}{}^{\prime }$ from $O^{\prime }$ to the same star. The mint arrow along the bottom is the drift $\vec{V}t$ — how far $O^{\prime }$ has slid. The three arrows form a tip-to-tail triangle.

6. Turning one observer's numbers into the other's — the derivation

Everything above was vocabulary. Here is the payoff: the dictionary that converts $S$'s numbers into $S^{\prime }$'s.

Step 1 — read the triangle. In the figure of §5, to reach the star from $O$ you can go straight ($\vec{r}$), or the long way: first drift to $O^{\prime }$ (that's $\vec{V}t$, the mint arrow), then from $O^{\prime }$ to the star (that's $\vec{r}{}^{\prime }$, the coral arrow). Same destination, so the two routes are equal: $\vec{r}=\vec{V}t+\vec{r}{}^{\prime }.$ Why this step? Pure tip-to-tail vector addition — no physics yet, just geometry of the triangle you can see.

Step 2 — solve for the primed position. Subtract $\vec{V}t$ from both sides to get what the train observer measures in terms of what the ground observer measures:

The line $t^{\prime }=t$ is the hidden universal-time assumption that makes this Galilean. This boxed pair is the "aha": hand me any event's $(\vec{r},t)$ in the ground frame and I hand you back the train's $(\vec{r}{}^{\prime },t^{\prime })$.

Step 3 — differentiate to get the velocity rule. Take the slope-in-time ($\frac{d}{dt}$) of both sides of $\vec{r}{}^{\prime }=\vec{r}-\vec{V}t$. Since $t^{\prime }=t$ the two clocks run together, and $\vec{V}$ is constant so the slope of $\vec{V}t$ is just $\vec{V}$: $\frac{d\vec{r}{}^{\prime }}{dt}=\frac{d\vec{r}}{dt}-\vec{V}⟹\vec{u}{}^{\prime }=\vec{u}-\vec{V}.$ Move $\vec{V}$ to the other side and you have the everyday form:

In this figure the same triangle idea appears for velocities: the coral arrow $\vec{u}{}^{\prime }$ (velocity in the train) placed tip-to-tail with the mint arrow $\vec{V}$ (train's velocity) gives the slate arrow $\vec{u}$ (velocity from the ground). Differentiating the position triangle simply produced a velocity triangle — that is why the addition law has exactly the same shape.

Step 4 — differentiate once more. The slope-in-time of the constant $\vec{V}$ is zero, so: $\vec{a}{}^{\prime }=\vec{a}.$ Both observers measure the same acceleration — the invariance the whole chapter rests on.

7. Prime marks and the hat

Prerequisite map

Read it bottom-up: origin, sign convention and vectors build the position arrow; add a clock and you get events; differentiate to get velocity, then acceleration; feed in the constant frame velocity $\vec{V}$ and you can build the Galilean transformation, whose payoff is that Newton's laws survive unchanged. See Relative velocity for the direct application ${\vec{u}}_{A/B}={\vec{u}}_A-{\vec{u}}_B$.

Equipment checklist

Test yourself — cover the right side and answer before revealing.