Position, Displacement, and Reference Frames

Every time your phone locks onto GPS it solves a position-vector problem far grander than any homework: the receiver measures its distance to four or more satellites and computes its position vector in a frame whose origin is the centre of the Earth — components of several thousand kilometres, updated every second, accurate to a few metres. Converting that fix to your map grid is exactly eq. (6) plus a rotation: a change of reference frame, performed billions of times a day, with Derivation C silently guaranteeing that no distance on the map gets distorted in the process. The distance–displacement split is now consumer-visible too. A running app reports the 8.0 km you jogged (distance, integrated step by step); a delivery app quotes "1.2 km away" as the crow flies (a displacement magnitude) and then routes you 2.1 km of actual streets. Robotics leans on the whole trio at once: a warehouse robot's wheel encoders accumulate distance (odometry), a running vector sum of little legs maintains its displacement from the dock — dead reckoning, eq. (3) evaluated live — and periodic fixes against beacons re-anchor its position in the warehouse frame, because odometry drifts and the vector sum inherits every small error. And when a drone's "return to home" button is pressed, it does not retrace its wandering outbound path; it flies the displacement vector, reversed — the straightest possible statement of everything this article just taught.

Stand at a fixed lamppost in a park and watch a drone hover. To pin down the drone completely you must say two things: how far it is from the lamppost, and in which direction. Draw that as an arrow — tail at the lamppost, head at the drone — and you have drawn the position vector $\vec r$. The lamppost is playing the role of the origin: the agreed reference point that all measurements in the problem share. Lay a coordinate grid over the park, axes crossing at the lamppost, and the arrow casts two "shadows": one along the $x$-axis of length $x$, one along the $y$-axis of length $y$. Those shadows are the components of the position vector, and they are just the drone's coordinates. The single arrow and the pair of numbers carry identical information — the arrow is the geometric costume, the components are the algebraic one, and the first skill of this topic is changing costumes quickly.

Now let a dog loose in the same park. It sprints in loops, doubles back, investigates three trees, and finally flops down 20 m northeast of where it started. Its GPS collar logs 300 m of running — that is distance, a scalar that only ever grows. Its displacement is the single straight arrow from the release point to the flop point: 20 m, northeast. A crow flying directly between those two points would trace exactly that arrow. Displacement is the crow's-eye summary of any journey, however chaotic — and if the dog trots back to its owner, the arrow shrinks to the zero vector while the odometer keeps its 300 m. The two measures answer different questions, and mechanics needs both: the kinematic machinery (velocity, the $\Delta x$ of every equation you will meet) runs on displacement, while path-dependent costs — fuel burned, work done against friction — run on distance.

One subtlety deserves respect from day one: the origin is your choice. Put it at the launch pad, the classroom door, the centre of the Earth — physics does not care, as long as you commit to one origin and stick with it for the whole problem. Two mission controllers tracking the same drone from different vans write down different position vectors, and both are right. What they always agree on is the arrow between any two things they both watch — because shifting the origin shifts every position vector by the same amount, and differences shrug that shift off. Measure the same trip from two different origins and every position vector changes, but the subtraction wipes the bookkeeping out: everyone, everywhere, gets the same displacement arrow.

The full package — an origin, a set of (usually orthogonal) axes, and a clock, all rigidly attached to some physical body — is a reference frame. The park frame, the train frame, the Earth-centred frame of GPS: each is a complete measuring apparatus, and the same physical event gets different numbers in each. That dependence is not a defect; it is the central organizing fact of kinematics. The art, developed below, is telling apart the quantities that are private to a frame (positions, and — once frames move relative to each other — displacements too) from the quantities every rigid frame agrees on (the distance between two particles at the same instant).

Everything in this topic is a conversation between an arrow's two costumes: the component triple $(x, y, z)$ and the magnitude-and-direction description. With axes chosen, the position of a point $P$ is written

where: $\vec r$ — position vector from origin $O$ to the particle; $x, y, z$ — signed coordinates (negative means the other side of the origin); $\hat{\imath}, \hat{\jmath}, \hat{k}$ — unit vectors, arrows of length exactly 1 along the axes, which do nothing but carry direction. Read (1) as walking instructions, not abstract algebra: go $x$ metres along the $x$-direction, then $y$ along $y$, then $z$ up, and the arrow points to wherever you end up. For motion in a plane, drop the last instruction. How long is the arrow? The components are the legs of a right triangle (in 3D, of two stacked right triangles) whose hypotenuse is the arrow itself — the drone 40 m east and 30 m north is not 70 m from the pad, because the straight line cuts the corner. Pythagoras does the cutting:

where: $|\vec r|$ — the particle's distance from the origin, always positive or zero. The squares kill the signs, which is why a magnitude can never be negative: a point at $(-40, -30)$ is exactly as far from the origin as one at $(40, 30)$. Notice also what (2) throws away — a bare distance is "5 km away", a whole circle of possibilities. The missing half of the answer lives in the angle: in the plane, $\theta = \arctan(y/x)$ measured anticlockwise from the $+x$-axis, plus $180°$ whenever $x < 0$, because $\arctan$ alone cannot tell a point from its diametric opposite — the worked examples spring this trap on a real radar fix. Now do the first genuine piece of vector algebra in mechanics. When a particle moves from $\vec r_i$ to $\vec r_f$, the arrow that gets you from the head of $\vec r_i$ to the head of $\vec r_f$ is the displacement. Arrows connect head-to-tail — $\vec r_i + \Delta\vec r = \vec r_f$ — so the definition writes itself:

where: $\Delta\vec r$ — displacement (the Greek $\Delta$ always reads "final minus initial"); $\vec r_i, \vec r_f$ — the endpoint positions measured from a common origin. Read the right-hand side as an instruction: locate the finish, locate the start, subtract. Nothing about the route in between ever enters, so path-independence is built into the definition at birth. Mind the order: final minus initial, always; swap the two and you have described the return trip, $-\Delta\vec r$. And each component settles its own account — a 2D displacement is two 1D displacements happening at once. Distance has no such shortcut — it must honour the route, because the route is exactly what it measures. Chop the path into instants: in each short interval $\mathrm{d}t$ the particle covers $|\mathrm{d}\vec r| = |\mathrm{d}\vec r/\mathrm{d}t|\,\mathrm{d}t$ of ground, and the odometer adds those unsigned lengths:

where: $s$ — path length (distance travelled), a scalar that can only grow while anything moves; $\mathrm{d}\vec r/\mathrm{d}t$ — the velocity, whose magnitude is the speed. The inequality is the headline: the straight arrow can never beat the odometer, and Derivation B below gives the one-line reason. Equality holds only for straight, never-reversing motion — the crow's route. Two objects at once bring the last pair of tools. The position of $B$ relative to $A$ — the arrow a sailor on ship $A$ would draw to ship $B$ — is another difference of positions,

where: $\vec r_A, \vec r_B$ — positions from a common origin; $\vec r_{AB}$ — the arrow from $A$ to $B$, origin-proof for exactly the same reason displacement is. And if a whole second frame $S'$ (a train, a drifting boat) carries its origin along a trajectory $\vec R(t)$ as seen from frame $S$, then one head-to-tail triangle relates the two descriptions of any particle:

where: $\vec r$ — position in frame $S$; $\vec r\,'$ — position of the same particle in frame $S'$; $\vec R(t)$ — where $S'$'s origin sits in $S$ at time $t$. When $\vec R$ changes with time the frames are in relative motion, and now even displacements disagree between frames — the passenger who walks 1 m in the train has walked 31 m over the ground. That is the doorway to Galilean transformations, opened properly in the next article; here it is enough to see that (6) is just (5) wearing a time-dependence. One organizing table before the derivations — each equation and the job it does:

A frame is more than an origin. Two observers may also tilt their axes relative to each other — the surveyor aligns $x$ with true north, the architect with the street grid — and, more drastically, one frame may move relative to the other. The most general relation between two rigid, non-moving frames is a rotation plus a translation: $\vec r\,' = R\,\vec r + \vec a$, where $R$ is a rotation matrix and $\vec a$ a constant offset. Positions change under both operations. Components change under rotation even though the arrow itself does not. So what, if anything, do all these bookkeepings agree on? The answer — the separation between two particles — is the deepest result of this article, and it deserves a real proof.

One more frame remains: the moving one, eq. (6) with $\vec R(t)$ genuinely changing. A passenger walking forward at $1\ \mathrm{m/s}$ in a train doing $30\ \mathrm{m/s}$ has position $\vec r_{\text{ground}} = \vec R_{\text{train}}(t) + \vec r_{\text{train}}$; in one second her displacement is 1 m in the train's books and 31 m in the ground's. Displacement is origin-proof but not motion-proof. Yet even here the invariant survives: two passengers sitting 10 m apart are 10 m apart in both frames at any instant, because the frames' relative motion shifts both passengers' positions identically at each moment. Choosing the frame that makes a problem trivial — the train frame for the passengers, the ground frame for the signal engineer — is the first genuinely professional skill in mechanics, and the systematic rules for converting velocities between such frames are the business of the Galilean-transformation article that follows.

1 · The drone on the pad (confidence builder)

A drone lifts off from its launch pad — take the pad as origin, east as $+x$, north as $+y$. It ends up hovering 40 m east and 30 m north of the pad. (a) Write its position vector. (b) How far is it from the pad? (c) In what direction?

Step 1. Components are read straight off: $x = 40$ m, $y = 30$ m, so $\vec r = 40\,\hat\imath + 30\,\hat\jmath$ m.

Step 2. Magnitude by (2): $|\vec r| = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = \mathbf{50\ m}$ — the 3-4-5 triangle scaled by ten.

Step 3. Direction: both components positive, first quadrant, no correction needed. $\theta = \arctan(30/40) = \arctan(0.75) = \mathbf{36.9^\circ}$ north of east.

Step 4. Sanity check by projecting back: $50\cos 36.9^\circ = 50(0.800) = 40$ m and $50\sin 36.9^\circ = 50(0.600) = 30$ m — the round trip returns the original components exactly.

2 · The ship on the radar (exam level)

A harbour radar takes the harbour master's tower as origin, with $+x$ east and $+y$ north. A ship shows up at coordinates $(-60\ \mathrm{m},\ 25\ \mathrm{m})$, and some minutes later at $(-36\ \mathrm{m},\ 18\ \mathrm{m})$. Find (a) the magnitude and direction of the ship's initial position vector, and (b) its displacement between the two fixes.

Step 1. Initial components: $x = -60$ m (west of the tower), $y = +25$ m (north of it) — second quadrant. Magnitude: $|\vec r| = \sqrt{(-60)^2 + 25^2} = \sqrt{3600 + 625} = \sqrt{4225} = \mathbf{65\ m}$. (A tidy 5–12–13 triangle scaled by 5.)

Step 2. Naive calculator angle: $\arctan(25/{-60}) = \arctan(-0.4167) = -22.6^\circ$. That points east-southeast — plainly wrong for a ship to the west.

Step 3. Quadrant correction ($x < 0$, so add $180°$): $\theta = -22.6^\circ + 180^\circ = \mathbf{157.4^\circ}$ anticlockwise from east — that is, 65 m away, $22.6^\circ$ north of west.

Step 4. Displacement by (3), final minus initial, component by component: $\Delta x = -36 - (-60) = +24$ m, $\Delta y = 18 - 25 = -7$ m. So $\Delta\vec r = 24\,\hat\imath - 7\,\hat\jmath$ m: the ship moved east and slightly south, toward the harbour mouth.

Step 5. Magnitude: $|\Delta\vec r| = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = \mathbf{25\ m}$ (a 7–24–25 triangle). Note that the displacement points nowhere near either position vector — net change and location are different animals, and only the subtraction connects them.

3 · One passenger, two frames (exam level)

A train moves due east at a steady $30\ \mathrm{m/s}$. A passenger walks toward the front of the train at $1.0\ \mathrm{m/s}$ (relative to the train) for $60\ \mathrm{s}$. Find her displacement in the train frame and in the ground frame. Then check that the separation between her and a friend seated 10 m behind her is the same in both frames at the end of the walk.

Step 1. Train frame: she walks a straight line at $1.0\ \mathrm{m/s}$ for 60 s, so $\Delta x' = 1.0 \times 60 = \mathbf{60\ m}$ toward the front. (Straight, no reversals — here distance equals displacement magnitude.)

Step 2. Ground frame: use eq. (6) in reverse, $\vec r = \vec r\,' + \vec R(t)$. The train's origin advances $\Delta R = 30 \times 60 = 1800$ m while she gains 60 m within it, so $\Delta x = 1800 + 60 = \mathbf{1860\ m} = 1.86$ km east.

Step 3. The two answers differ by exactly $\Delta R = 1800$ m — the accumulated shift of the moving origin. This is Derivation A's fine print made flesh: displacement is origin-proof under a fixed shift, but a moving origin shifts the two endpoints by different amounts, and the difference no longer cancels.

Step 4. The separation check. At the end of the walk the friend (who stayed seated) is 10 m behind her in the train frame. In the ground frame at that same instant, both positions have picked up the identical offset $\vec R(t_f)$, so the separation is $|(\vec r_{\text{her}} + \vec R) - (\vec r_{\text{friend}} + \vec R)| = \mathbf{10\ m}$ — the same in both books, exactly as Derivation C demands.

Step 5. Sanity check with speeds: in the ground frame she covers 1860 m in 60 s, an average of $31\ \mathrm{m/s}$ — the train's 30 plus her 1. Velocity addition, met here for the first time, is just the time derivative of eq. (6).

Texts. Morin, Introduction to Classical Mechanics, Ch. 1–2 · Taylor, Classical Mechanics, Ch. 1 · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 1 and Ch. 4 (orthogonal transformations) · Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 2–4 for the gentler on-ramp.

Provenance. Definitions of position, displacement, distance, and reference frame follow Morin Ch. 1–2, Taylor Ch. 1, Goldstein Ch. 1, and Landau-Lifshitz §1. A verifier should recompute: (i) the telescoping cancellation showing displacement is path/origin independent; (ii) the triangle-inequality bound $|\Delta\mathbf{r}|\leq s$; (iii) the invariance $|\mathbf{r}_B'-\mathbf{r}_A'|=|\mathbf{r}_B-\mathbf{r}_A|$ using $R^{\mathsf T}R=\mathbb{1}$; and (iv) the 3-4-5 hiker numbers ($\sqrt{9+16}=5$). No empirical constants are quoted beyond illustrative example speeds/distances.

Builds on: Vectors in Kinematics and Coordinate Systems. Leads to: Velocity and Acceleration · One-Dimensional Kinematics · Relative Motion and Galilean Transformations · Newton's First Law and Inertial Frames.

• Shift the origin by constant $\vec d$ — positions change; displacements, relative positions, separations all survive
• Rotate the axes — components change; lengths and separations survive ($R^{\mathsf T}R = \mathbb{1}$)
• Let the frame move — even displacements change; separations at one instant still survive
• Round trip — $\Delta\vec r = \vec 0$ always; distance keeps every metre

• Drone at $(40, 30)$ m: $\lvert\vec r\rvert = 50$ m at $36.9^\circ$ — the 3-4-5 triangle scaled by 10
• Ship at $(-60, 25)$ m: 65 m at $157.4^\circ$ — calculator says $-22.6^\circ$; $x<0$ forces $+180^\circ$
• Hike 5 km north + 3 km east: distance 8 km, displacement $\sqrt{34} = 5.83$ km
• Train walker: 60 m (train frame) vs 1860 m (ground frame) — origin-proof is not motion-proof

Step 1. Fix origin, axes, positive direction; keep them for the whole problem. Step 2. Positions: components → magnitude by Pythagoras, direction by quadrant-checked $\arctan$. Step 3. Displacement: subtract final minus initial, component by component; rebuild $\vec r_f = \vec r_i + \Delta\vec r$ as a check. Step 4. Sanity: $\lvert\Delta\vec r\rvert \le$ distance, and any frame-independent claim should survive shifting the origin.

• Distance is NOT the magnitude of displacement — a 400 m lap has $s = 400$ m, $\lvert\Delta\vec r\rvert = 0$
• $\arctan$ answers only between $-90^\circ$ and $+90^\circ$: no quadrant check, no credit
• $\vec r_i - \vec r_f$ is the return trip — order matters
• Bare coordinates without a declared origin and frame are meaningless
• "Origin-proof" holds for constant shifts only — a moving frame changes displacements