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Galilean Transformation Equation

Key idea: Learn Galilean transformations for position and velocity, identify what stays invariant between inertial frames, and practise exam-style frame-conversion questions.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • recall and apply the Galilean transformation equations to solve problems relating observations in different inertial frames of reference

Galilean transformations relate how positions, times, and velocities are described in two inertial frames moving at constant relative speed. This is the mathematical backbone of “no preferred inertial frame” in Newtonian mechanics.

Frame S prime moves right at velocity u; an event at coordinate x in S is at x prime equals x minus u t in S prime.
With coincident origins at t = 0, the displacement in S splits as x = ut + x′, so x′ = x − ut.

1. Definitions (Must Know)

  • Inertial frames: Frames moving at constant velocity relative to each other (no relative acceleration).
  • Standard setup: Frame S' moves at constant speed u relative to S along the + x direction.
  • Galilean transformation (1D):
    x' = x - ut; t' = t
  • 3D extension: y' = y, z' = z (if relative motion is along x only).
  • Validity: Classical regime u ≪ c and speeds in the problem ≪ c.
  • Symbols used in this lesson: u relative speed (m s⁻¹), v velocity (m s⁻¹), a acceleration (m s⁻²), c speed of light (≈ 3.00 × 10⁸ m s⁻¹).

Summary table (1D, S' moves at + u relative to S):

QuantityTransformationWhat to remember
Positionx' = x - utSame event, different origin
Timet' = t“Absolute time” (classical)
Velocityvₓ' = vₓ - uBe consistent with signs
Accelerationaₓ' = aₓKey reason Newton’s laws match

2. Key Ideas (What Earns Marks)

  • A Galilean transformation is just a shift of origin moving at constant speed: same time, different x coordinate.
  • Between inertial frames: acceleration is invariant (a' = a), so Newton’s second law keeps its form.
  • Velocities add/subtract linearly: vₓ = v'ₓ + u (watch the sign convention).

3. Detailed Explanations

A. Deriving the coordinate transformation

At t = 0, let the origins coincide. After time t, the origin of S' has moved a distance ut in S.

So if an event has coordinate x in S, then relative to the moving origin: x' = x - ut Galilean relativity assumes the same time coordinate: t' = t

B. Transforming velocity and acceleration (why Newton’s laws match)

Differentiate x' = x - ut with respect to t:

v'ₓ = dx'/dt = dx/dt - u = vₓ - u; a'ₓ = dv'ₓ/dt = dvₓ/dt = aₓ

So a' = a. If the net real force is the same physical interaction in both frames, then: ∑ F = ma ⇒ ∑ F = ma'

C. The (Galilean) velocity addition rule

Rearrange v'ₓ = vₓ - u: vₓ = v'ₓ + u Interpretation: “velocity in S = velocity relative to S' + velocity of S' relative to S”.

4. Common Mistakes

  • Mixing up which frame is moving: define clearly whether S' moves at + u relative to S or vice versa.
  • Sign errors: if S' moves in + x, then x' = x - ut (not x + ut).
  • Forgetting that Galilean transformations assume absolute time (t' = t).
  • Using Galilean velocity addition for light or other relativistic situations (use the H3 special relativity results instead).

5. Exam Tips

  • Start by writing your frame statement: “Let S' move at speed u along + x relative to S.”
  • Quote the transformation before doing any algebra:
    x' = x - ut, t' = t
  • For data/word problems, draw a quick axis and label + x to keep the sign consistent.

6. Worked Examples

Modelled example 1

Transforming an event coordinate

Core

Problem

Frame S' moves at u = 5.0 m s⁻¹ along + x relative to S. At t = 3.0 s, an event occurs at x = 20 m in S. Find x'.
Study the worked solution

  1. Fix the frame convention

    Method

    S' moves in the positive direction relative to S.

    Reason

    With coincident origins at t = 0, its origin is at x = ut in S.

    Working

    x' = x-ut

  2. Transform the coordinate

    Method

    x' = 5.0 m.

    Reason

    Subtract the displacement of the moving origin from the S coordinate.

    Working

    x' = 20-(5.0)(3.0) = 5.0 m

Guided practice 2

Velocity seen in a second inertial frame

About 4 min

Problem

A cart moves at v'ₓ = 2.0 m s⁻¹ in S'. Frame S' moves at u = 5.0 m s⁻¹ along + x relative to S. Find vₓ.

Try this before viewing the solution

Hints

Hint 1: interpret both motions
In S, the cart has its motion within S' plus the motion of S'.
Hint 2: choose the sign
Rearrange v'ₓ = vₓ-u.
View solution step by step

  1. Select the inverse transform

    Method

    vₓ = v'ₓ + u.

    Reason

    Both stated velocities point along + x.

    Working

    vₓ = v'ₓ + u

  2. Evaluate

    Method

    vₓ = 7.0 m s⁻¹.

    Reason

    Add the cart’s velocity relative to S' to the frame velocity.

    Working

    vₓ = 2.0 + 5.0 = 7.0 m s⁻¹

Common misconception 3

Acceleration is invariant (quick check)

Find and correct the mistake

Learner claim

In S, vₓ = 5.0-2.0t in SI units. Frame S' moves at constant u = 3.0 m s⁻¹ along + x. A learner claims that both velocity and acceleration are unchanged between the frames. Find v'ₓ(t) and a'ₓ, then correct the claim.

Try this before viewing the solution

Quantity invariant under this transform

View solution step by step

  1. Transform velocity

    Method

    v'ₓ = 2.0-2.0t m s⁻¹.

    Reason

    The moving-frame velocity is v'ₓ = vₓ-u.

    Working

    v'ₓ = (5.0-2.0t)-3.0 = 2.0-2.0t

  2. Differentiate

    Method

    a'ₓ = -2.0 m s⁻².

    Reason

    The derivative of the constant frame speed is zero.

    Working

    a'ₓ = dv'ₓ/dt = -2.0 m s⁻²

  3. Correct the claim

    Method

    Velocity is frame-dependent, while acceleration is invariant between these inertial frames.

    Reason

    A Galilean boost subtracts a constant velocity.

    Working

    v'ₓ ≠ vₓ, a'ₓ = aₓ

Examiner practice 4

Transforming an equation of motion

3 marks

Examination question

In S, a particle follows x = 2.0 + 4.0t-t² in SI units. Frame S' moves at u = 3.0 m s⁻¹ along + x relative to S, with origins coincident at t = 0. Find x'(t). [3 marks]

Try this before viewing the solution

View solution step by step

  1. State the transformation

    1 mark

    Method

    x' = x-ut.

    Reason

    The origins coincide at t = 0 and S' moves along + x.

    Working

    x' = x-ut

  2. Substitute the motion

    1 mark

    Method

    Insert both x(t) and u.

    Reason

    The transform applies at the same Newtonian time t' = t.

    Working

    x' = (2.0 + 4.0t-t²)-(3.0)t

  3. Simplify

    1 mark

    Method

    x' = 2.0 + 1.0t-t².

    Reason

    Only the coefficient of the linear term changes.

    Working

    x' = 2.0 + 1.0t-t²

Challenge 5

Choosing a frame where an object is at rest

Minimal support

Independent transfer

In S, a cart moves at constant vₓ = +10.0 m s⁻¹. Determine the speed and direction of a frame S' in which the cart is at rest.

Try this before viewing the solution

Hints

Hint 1: translate at rest
Set the cart’s transformed velocity v'ₓ equal to zero.
View solution step by step

  1. Set the target condition

    Method

    v'ₓ = 0.

    Reason

    The cart must have no velocity in its rest frame.

    Working

    0 = vₓ-u

  2. Solve for frame motion

    Method

    u = +10.0 m s⁻¹.

    Reason

    The frame must move alongside the cart with the same velocity.

    Working

    u = vₓ = +10.0 m s⁻¹

7. Mind Stretchers

Mind stretcher 1: Two different “straight-line” paths can both be correctExtension

In S', a ball is thrown straight up and lands back in the thrower’s hand. Explain (without new forces) why an observer in S can see a parabolic path and still agree with Newton’s laws.

One way to phrase it

The vertical motion is identical in both frames because the vertical forces are the same. The horizontal motion differs because the ball has an extra constant horizontal velocity u in S, so combining “constant horizontal velocity” with “constant vertical acceleration” produces a parabola.

Mind stretcher 2: Why kinetic energy is frame-dependent (but physics is consistent)Extension

Two inertial observers measure different speeds for the same object, so they compute different kinetic energies. Does this mean energy conservation “depends on the frame”?

What to say in an exam

Kinetic energy depends on the frame because it depends on speed, and speeds change under v' = v-u. However, energy conservation is still consistent when you include the full system and apply the work–energy idea correctly within each frame (and account for which parts are doing work in that frame). Different frames can assign different numerical kinetic energies while still agreeing on the underlying dynamics.

8. Optional/Enrichment: Why Galilean Transformations Fail for Light

Galilean velocity addition would predict that a light pulse has speed c ± u in different inertial frames, but the H3 syllabus requires that the measured speed in vacuum is c for all inertial observers. This motivates special relativity:

Next step

Return to the Frames of Reference hub, or continue in sequence to the Centre of Mass Frame, where the same velocity transformation selects the zero-momentum frame.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027