Non-inertial reference frames — pseudo forces: Visual walkthrough

Step 1 — Two watchers, one ball

WHAT. Picture a smooth (frictionless) bus floor with a small ball resting on it. One watcher, G, stands on the ground. Another watcher, P, sits inside the bus. The bus is about to speed up.

WHY. Before any algebra we must be crystal-clear about who measures what. A "reference frame" is just a chosen watcher plus the ruler and stopwatch they carry. Different watchers, moving differently, can disagree about how a thing moves — and that disagreement is the entire seed of the pseudo force.

PICTURE. Ground-watcher G is fixed to the road. Bus-watcher P is fixed to the bus. The ball sits between them, doing nothing yet.

Figure — Non-inertial reference frames — pseudo forces

Step 2 — Name the positions with arrows

WHAT. Draw three arrows from the ground origin:

$R$ — from G's origin to P's origin (where the bus is).
$r^{'}$ — from P's origin to the ball (what P measures).
$r$ — from G's origin to the ball (what G measures).

WHY. To compare two watchers we need one honest bookkeeping equation linking their measurements. A position "arrow" (a vector: an arrow with length and direction) lets us add journeys tip-to-tail. Going G→bus→ball must land on the same ball as going G→ball directly.

PICTURE. The two short arrows $R$ and $r^{'}$ laid tip-to-tail equal the long arrow $r$ .

Step 3 — Ask about acceleration, and why twice

WHAT. Differentiate the arrow equation twice with respect to time $t$ : $r = R + r^{'} \frac{d ^{2}}{d t ^{2}} a = a_{0} + a^{'}$ where $a = \ddot{r}$ is the ball's acceleration seen by G, $a_{0} = \ddot{R}$ is the bus's acceleration, and $a^{'} = \ddot{r}^{'}$ is the ball's acceleration seen by P.

WHY this tool — the second derivative? Newton's law talks about acceleration, not position. The tool " $\frac{d}{d t}$ " answers the question "how fast is this changing right now?" Position tells us where; its first derivative gives velocity (change of position); its second derivative gives acceleration (change of velocity). We need acceleration, so we ask the change-question twice. We use derivatives and not, say, a simple difference, because the bus's speed-up can vary from instant to instant — we want the instantaneous rate.

WHY it's just a sum. Differentiation is linear: the change of a sum is the sum of the changes. So the three arrows keep the same tip-to-tail shape after two derivatives. The velocity term $\dot{r}$ and its constant pieces drop or carry through cleanly, leaving pure accelerations.

PICTURE. The same tip-to-tail triangle, now with each arrow relabelled as an acceleration.

Step 4 — Let only G use Newton (he's allowed to)

WHAT. G is inertial, so for G the honest law holds with real forces only: $F_{real} = m a .$ Here $F_{real}$ is the sum of forces from actual bodies touching or pulling the ball (gravity, floor, string...), and $m$ is the ball's mass — its resistance to being accelerated.

WHY. The whole point of "inertial" is that $F = m a$ is true there with no fudge. On a smooth floor there is no horizontal real force, so G predicts $a = 0$ horizontally: the ball stays put. This is our anchor of truth; P's world must be made consistent with it.

PICTURE. G's view: ball motionless, bus sliding forward underneath it. No horizontal arrow touches the ball.

Step 5 — Substitute, so P's acceleration appears

WHAT. Put Step 3's $a = a_{0} + a^{'}$ into Step 4's law: $F_{real} = m (a_{0} + a^{'}) = m a_{0} + m a^{'} .$

WHY. We want an equation written in terms of $a^{'}$ — the acceleration P actually sees — because P is the one confused. So we swap out the quantity G measures ( $a$ ) for the pieces that include P's quantity.

PICTURE. The real-force arrow is shown splitting into two contributions: one that "pays for" the bus's motion ( $m a_{0}$ ) and one left over for P's ball-motion ( $m a^{'}$ ).

Step 6 — Rearrange, and name the leftover

WHAT. Move $m a_{0}$ to the left: $m a^{'} = F_{real} - m a_{0} .$ P wants this to look like Newton's law: "(mass)×(the acceleration I see) = (total force I feel)". The right side already reads that way if we christen the extra piece: $F_{pseudo} = - m a_{0}$

WHY. P insists on using $F = m a^{'}$ inside the bus — it's habit and it's convenient. The only way to keep that habit honest is to admit an extra term. That term is not a discovery of new physics; it is the price of using an accelerating watcher. The minus sign is the heart of it: whatever way the bus accelerates ( $a_{0}$ ), this invented force points the opposite way.

PICTURE. In P's view the ball slides backward; we draw the backward pseudo-force arrow $- m a_{0}$ right on the ball, exactly opposite the forward bus-acceleration arrow.

Step 7 — Check the smooth-bus numbers

WHAT. Smooth floor ⇒ $F_{real} = 0$ horizontally. With $a_{0} = 2 m/s^{2}$ and $m = 0.5 kg$ : $F_{pseudo} = - m a_{0} = - (0.5) (2) = - 1 N, a^{'} = \frac{F _{pseudo}}{m} = \frac{- 1}{0.5} = - 2 m/s^{2} .$

WHY. P predicts the ball accelerates backward at $2 m/s^{2}$ — exactly the backward slide P watches. And it matches G, who says the ball is still while the bus surges forward at $2 m/s^{2}$ (so relative to the bus the ball recedes at $- 2 m/s^{2}$ ). Two watchers, one reality. ✓

PICTURE. Side-by-side: G's frame (ball fixed, bus arrow forward) versus P's frame (ball arrow backward), the two accelerations equal and opposite in magnitude.

Step 8 — Every case: sign of $a_{0}$ , and the degenerate frame

WHAT. Read $F_{pseudo} = - m a_{0}$ for each situation:

Bus does	$a_{0}$ direction	Pseudo force $- m a_{0}$
Speeds up forward	forward	backward (thrown back)
Brakes (slows)	backward	forward (thrown into belt)
Lift accelerates up	up	down (feel heavier)
Lift accelerates down	down	up (feel lighter)
Free fall ( $a_{0} = g$ down)	down, size $g$	up, size $m g$ — cancels weight ⇒ weightless
Constant velocity ( $a_{0} = 0$ )	none	zero — frame is inertial again

WHY. The formula is one rule that covers all of them because every case is just a different arrow $a_{0}$ fed into $- m a_{0}$ . Notice the degenerate case: if the bus stops accelerating ( $a_{0} = 0$ ), the invented force vanishes and P's repaired law collapses back to ordinary Newton. That is the sanity check that we invented nothing spurious.

PICTURE. A dial of $a_{0}$ arrows (forward, backward, up, down, zero) each with its opposite pseudo-force arrow, and the zero case shown empty.

The one-picture summary

WHAT. One figure holds the whole story: the tip-to-tail acceleration triangle ( $a = a_{0} + a^{'}$ ), G's honest $F_{real} = m a$ , and the flip that turns $- m a_{0}$ into the backward pseudo force on the ball in P's world.

Recall Feynman retelling of the whole walkthrough

Two people watch a ball on a slippery bus floor. The ground guy is calm: no push on the ball, so the ball just sits — but the bus scoots forward under it, so relative to the bus the ball drifts back. The bus guy insists nothing is moving him, yet the ball drifts backward, and he wants his trusty "force = mass × acceleration" to still work. The only way is to pretend a ghost-hand shoves the ball backward — exactly as strong as the bus's forward surge times the ball's mass, and always pointing the opposite way (that's the minus sign). Turn the bus's push off and the ghost vanishes; the guy becomes an ordinary observer again. The ghost was never real — it's the toll you pay for doing physics inside something that speeds up.

Active recall

What single geometric equation starts the whole derivation?

r = R + r^{'}

(tip-to-tail positions).

Why differentiate twice, not once?

Once gives velocity; Newton's law needs acceleration, the change of velocity — so we take the change-question twice.

Which watcher is allowed to use

F = m a

untouched, and why?

The ground (inertial) watcher G — zero acceleration, no correction needed.

Define the pseudo force from the rearranged law.

The leftover

- m a_{0}

m a^{'} = F_{real} - m a_{0}

Ball on smooth bus floor,

m = 0.5

kg,

a_{0} = 2

m/s²: pseudo force and ball's acceleration in bus frame?

- 1

N and

- 2

m/s² (both backward).

What happens to

- m a_{0}

when the frame moves at constant velocity?

It is zero; the frame is inertial and ordinary Newton returns.

Connections

Parent — pseudo forces — this page is its visual derivation.
Newton's Second Law — the law we insist on keeping, forcing the invention.
Newton's Third Law — the pseudo force breaks it (Step 6: no partner).
Apparent weight & normal force — the up/down rows of the Step 8 table.
Centrifugal force — same trick with $a_{0}$ pointing to the centre.
Coriolis force — the velocity-dependent cousin in rotating frames.
Galilean relativity — defines the constant-velocity family where $a_{0} = 0$ .