Lorentz Transformation Equations

Key idea: Use Lorentz transformation equations and the difference form to solve special relativity problems on simultaneity, time dilation, and length contraction.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • discuss the concept of simultaneity
  • show an understanding of the terms proper time and proper length
  • apply the Lorentz transformation equations to solve one-dimensional problems
  • Derive the time dilation formula and the length contraction formula, making use of the Lorentz factor.
  • apply the time dilation formula and the length contraction formula in related situations (e.g. the lifetime of fast-moving muons) or to solve problems
Parallel inertial frames S and S prime with S prime moving at velocity v along the shared positive x direction
Standard configuration: the origins coincide at time zero, and frame S-prime moves at positive velocity v relative to frame S. State this convention before transforming coordinates.

Lorentz transformations are the coordinate-change rules that replace Galilean transformations when the speed of light is invariant. They let you convert (x,t) in one inertial frame into (x',t') in another moving at constant speed v along the x-axis.

Treat this maintained page as the mathematical anchor for the whole cluster; revisit it whenever sign conventions or frame conditions get fuzzy.

1. Definitions (Must Know)

  • Standard configuration: frame S' moves at speed v along + x relative to frame S; axes are parallel; origins coincide at t = t' = 0.

  • Lorentz factor: γ = 1/(square root of (1-v²/c²))

  • Lorentz transformations (1D motion along x): x' = γ(x-vt)

    t' = γ(t-vx/c²)

    with y' = y, z' = z.

  • Inverse transformations: x = γ(x' + vt')

    t = γ(t' + vx'/c²)

  • Symbols used in this lesson: x,x' (m), t,t' (s), v (m s⁻¹), c (m s⁻¹), γ (dimensionless).

2. Key Ideas (What Earns Marks)

  • Space and time mix: t' depends on both t and x.
  • The transformations reduce to Galilean ones in the low-speed limit: v≪ c ⇒ γ ≈ 1 and t' ≈ t.
  • Use the difference form for two events: it’s often cleaner for time dilation, length contraction, and simultaneity.

3. Detailed Explanations

A. The forward and inverse transformations

Forward (from S to S'): x' = γ(x-vt)

t' = γ(t-vx/c²)

Inverse (from S' to S) comes from swapping primes and replacing v → -v: x = γ(x' + vt')

t = γ(t' + vx'/c²)

B. The “difference form” for two events

For two events, define Δ x = x₂-x₁ and similarly for time. Then: Δ x' = γ(Δ x-vΔ t)

Δ t' = γ(Δ t-(vΔ x)/c²)

and inversely: Δ x = γ(Δ x' + vΔ t')

Δ t = γ(Δ t' + (vΔ x')/c²)

These are the forms you’ll use most in H3 problem-solving.

C. Classical limit (sanity check)

If c → ∞, then γ → 1 and vx/c² → 0, so: x' → x-vt, t' → t which matches Galilean transformations.

4. Common Mistakes

  • Swapping the sign of v (define clearly whether S' moves at + v relative to S).
  • Forgetting the vx/c² term in t' (that term is the source of relativity of simultaneity).
  • Mixing coordinates from different frames in the same equation.
  • Using degrees for v/c (it’s a pure ratio; no angles involved).

5. Exam Tips

  • Start by stating the standard setup: “S' moves at speed v along + x relative to S.”
  • Decide whether you should use the direct form (single event) or the difference form (two events).
  • Quick plausibility checks:
    • if v = 0, then x' = x and t' = t.
    • if Δ x = 0 (events at same place in S), the time relation simplifies.

6. Worked Examples

Modelled example 1

Transforming one event

Core

Problem

In S, an event occurs at x = 600 m and t = 3.0 μs. Frame S' moves at v = 0.60c along + x. Find x' and t'.
Study the worked solution
  1. Fix the convention and factor

    Method

    Use the forward transformation because the given coordinates are in S and the required coordinates are in S'.

    Reason

    S' moves at + v relative to S.

    Working

    γ = 1/(square root of (1-0.60²)) = 1.25
  2. Transform position

    Method

    Substitute consistently in SI units.

    Reason

    c(3.0 × 10⁻⁶) = 900 m, so vt = 540 m.

    Working

    x' = 1.25(600-540) = 75 m
  3. Transform time

    Method

    Retain the space-time mixing term.

    Reason

    vx/c² = (0.60)(600)/c = 1.2 × 10⁻⁶ s.

    Working

    t' = 1.25(3.0-1.2) × 10⁻⁶ = 2.3 × 10⁻⁶ s

Guided practice 2

Relativity of simultaneity via the difference form

About 6 min

Problem

Two events are simultaneous in S (Δ t = 0) and separated by Δ x = 300 m along + x. If S' moves at v = 0.80c, find and interpret Δ t'.

Try this before viewing the solution

Hints

Hint 1: choose the difference form
Use Δ t' = γ(Δ t-vΔ x/c²).
Hint 2: preserve the sign
With v > 0, Δ x > 0 and Δ t = 0, the transformed interval must be negative.
View solution step by step
  1. Calculate the factor

    Method

    Find γ for 0.80c.

    Reason

    The transformation is relativistic.

    Working

    γ = 1/(square root of (1-0.80²)) = 1.67
  2. Transform the interval

    Method

    Apply the simultaneous-in-S condition.

    Reason

    Δ t = 0 removes the first term but not the spatial term.

    Working

    Δ t' = -1.67(0.80c)(300)/c² ≈ -1.3 × 10⁻⁶ s
  3. Interpret

    Method

    The events are not simultaneous in S'.

    Reason

    The negative sign means event 2 has the earlier S' time coordinate under the stated ordering.

    Working

    t'₂ < t'₁

Common misconception 3

Using the inverse transformations

Find and correct the mistake

Learner attempt

In S', an event has x' = 120 m and t' = 1.0 μs; S' moves at 0.60c along + x. A learner uses the forward minus signs to find x,t. Explain the sign error, then calculate the coordinates in S.

Try this before viewing the solution

Required signs

View solution step by step
  1. Choose the direction

    Method

    Use the inverse transformations.

    Reason

    The known coordinates are primed and the target coordinates are unprimed.

    Working

    x = γ(x' + vt'), t = γ(t' + vx'/c²)
  2. Find position

    Method

    Use γ = 1.25 and vt' = 180 m.

    Reason

    All terms inside the position bracket must have units of metres.

    Working

    x = 1.25(120 + 180) = 375 m
  3. Find time

    Method

    Use vx'/c² = 72/c = 2.40 × 10⁻⁷ s.

    Reason

    The time-mixing term must have seconds.

    Working

    t = 1.25(1.0 × 10⁻⁶ + 2.40 × 10⁻⁷) = 1.6 × 10⁻⁶ s

Examiner practice 4

Time dilation from the difference form

4 marks

Examination question

Two events occur at the same position in S' with Δ t' = 4.0 s. The relative speed is 0.80c. Use the Lorentz difference form to find Δ t in S. [4 marks]

Try this before viewing the solution

View solution step by step
  1. Identify the condition

    1 mark

    Method

    The events are co-located in S'.

    Reason

    This is the proper-time frame condition.

    Working

    Δ x' = 0
  2. Select and simplify

    1 mark

    Method

    Use the inverse time-difference transformation.

    Reason

    The known interval is primed.

    Working

    Δ t = γ(Δ t' + (vΔ x')/c²) = γΔ t'
  3. Calculate gamma

    1 mark

    Method

    Evaluate the Lorentz factor.

    Reason

    v/c = 0.80.

    Working

    γ = 1/(square root of (1-0.80²)) = 1.67
  4. Calculate the interval

    1 mark

    Method

    Multiply the proper time by γ.

    Reason

    The interval in S is longer.

    Working

    Δ t = (1.67)(4.0) = 6.7 s

Challenge 5

Length contraction from the difference form

Minimal support

Independent transfer

A rod is at rest in S' with L₀ = Δ x' = 2.0 m. In S, simultaneous endpoint positions give Δ t = 0. At v = 0.60c, derive and calculate L = Δ x.

Try this before viewing the solution

Hints

Hint 1: start from the spatial difference
Orient with Δ x' = γ(Δ x-vΔ t) and apply simultaneity in the measuring frame.
View solution step by step
  1. Apply the measurement condition

    Method

    Set Δ t = 0 because the endpoints are recorded simultaneously in S.

    Reason

    A moving length requires same-frame simultaneous endpoint positions.

    Working

    Δ x' = γΔ x
  2. Map the lengths

    Method

    Identify Δ x' = L₀ and Δ x = L.

    Reason

    The rod is at rest in S', so its primed length is proper.

    Working

    L = L₀/γ
  3. Calculate

    Method

    Use γ = 1.25 at 0.60c.

    Reason

    The moving length must be smaller than the proper length.

    Working

    L = 2.0/1.25 = 1.6 m

7. Mind Stretchers

Mind stretcher 1: Why t' depends on xExtension

If t' depended only on t, could c be the same in all inertial frames? Explain briefly.

Answer

No. If time were absolute (t' = t), then the speed of light would follow Galilean velocity addition between frames. The vx/c² mixing term is what allows different frames to relate measurements while keeping c invariant.

Mind stretcher 2: Why the “difference form” is so usefulExtension

Why are Δ x and Δ t often easier to use than x and t directly in exam questions?

Answer

Most questions describe two events and ask about time/space intervals between them. Using differences removes the need to track absolute origins, and you can apply special-case conditions cleanly (e.g. Δ x' = 0 for proper time, Δ t = 0 for length measurement in the lab frame).

8. Optional/Enrichment: Relativity Results as Corollaries

Time dilation, length contraction, and relativity of simultaneity can all be derived cleanly from the difference form of Lorentz transformations:

Forward difference form:

Δ x' = γ(Δ x-vΔ t)

Δ t' = γ(Δ t-(vΔ x)/c²)

For the inverse form, replace v with -v and interchange primed and unprimed quantities.

Next in the maintained sequence: Relativistic Velocity Addition.

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