Reference frames — Galilean transformations: Question bank

Before we start, a one-line reminder of the symbols so nothing here is a mystery:

$S$ = ground frame, $S^{'}$ = frame moving at constant velocity $V$ relative to $S$ .
$r$ position, $u$ velocity, $a$ acceleration; primed = measured in $S^{'}$ .
The dictionary: $r^{'} = r - V t$ , $t^{'} = t$ , $u^{'} = u - V$ , $a^{'} = a$ .

True or false — justify

A ball dropped inside a smoothly cruising train lands at the thrower's feet, so the train must be at rest.

False. Landing at your feet happens in any inertial frame — the ball shares the train's horizontal velocity the whole time; no experiment inside tells you the constant speed.

Two observers in different inertial frames always disagree about an object's velocity.

Mostly true, one exception. They differ by

V

, so they agree only in the special case

V = 0

(same frame) or when the object's velocity happens to be measured only along a direction perpendicular to a purely... no — they differ by

V

in every component along

V

, so genuine disagreement whenever

V \neq = 0

All inertial observers agree on the acceleration of any object.

True. Since

V

is constant,

d V / d t = 0

, so

a^{'} = a

; this invariance is exactly why

F = m a

holds in every inertial frame.

In Galilean physics two events that are simultaneous in

S

are also simultaneous in

S^{'}

True. Because

t^{'} = t

(universal time), if

t_{1} = t_{2}

S

then

t_{1}^{'} = t_{2}^{'}

S^{'}

— simultaneity is absolute, unlike in special relativity.

The time interval between two events is the same for all Galilean observers.

True. From

t^{'} = t

, any difference

Δ t^{'} = Δ t

; there is no time dilation in the Galilean world — that only appears in Special relativity — Lorentz transformation.

The spatial distance between two simultaneous events is the same for all Galilean observers.

True.

r^{'} = r - V t

; for two events at the same

t

, the

V t

shift is identical and cancels in the difference, so

Δ r^{'} = Δ r

The distance a single moving particle travels between two different times is frame-independent.

False. Here the two events have different

t

, so the

V t

terms don't cancel; path length and displacement both differ between frames (the platform sees a curve, the train sees a straight drop).

A frame attached to a car turning at constant speed is inertial.

False. Constant speed with changing direction means

V

is not constant (

A \neq = 0

), so pseudo-forces appear — see Inertial and non-inertial frames.

Kinetic energy of an object is the same in all inertial frames.

False. KE depends on

∣ u ∣^{2}

, and

u

shifts by

V

; a ball can have zero KE in the train yet large KE from the platform.

Spot the error

"A boat crosses a river heading north at

4

m/s while the current runs east at

3

m/s, so its ground speed is

4 + 3 = 7

m/s."

Error: scalar-adding perpendicular vectors. Galilean addition is a vector equation; the speed is

3^{2} + 4^{2} = 5

m/s, not

7

— see Relative velocity.

"Since acceleration is invariant, the object's velocity must also be the same in every inertial frame."

Error: invariance survives only after the derivative. Velocity differs by

V

; it is the time-derivative of velocity, with constant

V

killing the extra term, that is invariant.

"Two cars each do

30

m/s toward each other, so in one car's frame the other approaches at

30

m/s."

Error: sign of the relative velocity. With east positive,

u_{A / B} = u_{A} - u_{B} = 30 - (- 30) = 60

m/s; opposite motions make speeds add.

"To go from

S^{'}

back to

S

use

r = r^{'} - V t^{'}

Error: wrong sign on the inverse. Inverting

r^{'} = r - V t

gives

r = r^{'} + V t^{'}

; the reverse transform uses

+ V

, not

- V

"On a moving train I toss a ball straight up; the platform sees it also go straight up."

Error: ignoring the shared horizontal velocity. The ball keeps the train's

V

, so the platform sees a parabola moving forward, while you see a vertical line.

"At everyday speeds we should still write

t^{'} = t - V x / c^{2}

to be safe."

Error: importing a Lorentz correction where it's negligible. The

V x / c^{2}

term is

\sim v / c^{2}

tiny; Galilean physics defines

t^{'} = t

and is the correct model when

v ≪ c

Why questions

Why does the position transform contain a

V t

term but the acceleration transform contains nothing?

Each time-derivative knocks one power of

t

off the shift:

V t \to V \to 0

, since

V

is constant. Position carries

V t

, velocity carries

V

, acceleration carries zero.

Why must

V

be constant for Newton's laws to look identical in

S^{'}

Only then is

d V / d t = 0

, giving

a^{'} = a

; an accelerating frame adds

- A

and hence pseudo-forces that Newton's laws don't naturally contain.

Why do we differentiate the position rule to get the velocity rule, rather than assuming a new law?

Velocity is the time-derivative of position by definition, so the velocity transform is forced by the position transform — no extra physics is invented, only calculus applied.

Why is "velocity relative to what?" a necessary question but "acceleration relative to what?" less so?

Velocity changes by

V

between inertial frames, so it needs a reference frame stated; acceleration is the same in all of them, so it is unambiguous among inertial observers.

Why does the everyday success of "just add the speeds" (Example 1) not generalise to the river-boat?

Scalar addition is the special case of vector addition when the velocities are parallel; the river-boat velocities are perpendicular, so only the full vector rule

u = u^{'} + V

works.

Why does the Galilean transform fail near the speed of light?

It bakes in

t^{'} = t

(absolute time), but experiment shows clocks and lengths change with speed; Special relativity — Lorentz transformation replaces it once

v \sim c

Edge cases

What does the Galilean transform give when

V = 0

It reduces to the identity:

r^{'} = r

t^{'} = t

u^{'} = u

— the two frames coincide and all measurements agree.

What is an object's velocity in

S^{'}

if it moves at exactly

V

S

u^{'} = u - V = V - V = 0

; it is at rest in

S^{'}

— this is precisely a passenger sitting still on the train.

t = 0

the two origins coincide; what does the position transform say then?

r^{'} = r - V \cdot 0 = r

; positions agree at the instant the frames overlap, and diverge linearly in time afterward.

If the "object" is the origin

O^{'}

of the moving frame itself, what does the velocity rule give?

S

it moves at

V

; in its own frame

u^{'} = V - V = 0

, consistent with

O^{'}

being fixed in

S^{'}

What happens to the closing-speed formula if the two cars move in the same direction at equal speed?

u_{A / B} = 30 - 30 = 0

; they are mutually at rest, matching the intuition that same-velocity travellers see each other frozen.

Recall One-sentence self-test

If you can explain why "position carries $V t$ , velocity carries $V$ , acceleration carries nothing" using only the phrase "each derivative removes one power of $t$ ," you have understood the whole page.