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Comparing events between inertial frames

Key idea: Transform the same two events in the same order. Clock ticks share a position in their rest frame; a moving rod's endpoints must be recorded simultaneously in the measuring frame.

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  • GCE A-Level H3 Physics 2027
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When a question compares two events, transform their differences, keeping the same event order in both frames. The main decision is usually which measurement condition is zero: simultaneous times, or co-located positions.

Check your understanding 1 Predict first

Two lamps 300 m apart flash at the same time in frame S. Are the flashes simultaneous in a frame S' moving along the line between them?

In S′ the flashes are
Show answer

Not simultaneous. With Δ t = 0 and Δ x ≠ 0, Δ t' = -γ vΔ x/c², which is not zero. Example B below finds its value.

Label the same two events

Call them event 1 and event 2, and define every difference as “2 minus 1”:

Δ x = x₂-x₁, Δ t = t₂-t₁.

Use that order again for Δ x' and Δ t'. Subtracting the single-event Lorentz equations gives

Δ x' = γ(Δ x-vΔ t),
Δ t' = γ(Δ t-(vΔ x)/c²).

The inverse equations are

Δ x = γ(Δ x' + vΔ t'),
Δ t = γ(Δ t' + (vΔ x')/c²).

Here S' moves at + v relative to S, with parallel axes and coincident origins at time zero. Reversing event order reverses all differences; changing frames does not change which events we compare.

Check your understanding 2 Why the “difference form” is so useful

In one inertial frame, two events have (x₁,t₁) = (40 m,2.0 s) and (x₂,t₂) = (100 m,5.0 s). The observer relabels the same records using x tilde = x-30 m and t tilde = t + 7.0 s. Find the event differences before and after relabelling. Would their Lorentz-transformed differences change at the same relative frame velocity?

Unit: m
Unit: s
Transformed differences at the same v
Show answer

Originally Δ x = 100-40 = 60 m and Δ t = 5.0-2.0 = 3.0 s. The relabelled records are (10 m,9.0 s) and (70 m,12.0 s), so the differences remain 60 m and 3.0 s. Constant origin and clock offsets cancel. With the same relative velocity, the transformed differences are unchanged because the Lorentz difference equations use only these intervals.

Translate the measurement condition before substituting

SituationZero conditionConsequence
Two events simultaneous in SΔ t = 0Δ t' = -γ vΔ x/c², generally non-zero
Two ticks of one clock at rest in S'Δ x' = 0Δ t = γΔ t'; Δ t' is proper time
Endpoints of a moving rod recorded together in SΔ t = 0Δ x' = γΔ x; for a rod at rest in S', Δ x' = L₀

“Same position” and “same time” are different conditions. Identify the frame named by the question before setting either quantity to zero. Do not apply a time-dilation shortcut to an arbitrary spatially separated event pair.

Guided practice 1

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

Show the steps Optional working
Unit: μs
In S′, which event happens first?

Hints

Hint 1

only the mixing term survives

Which term remains in the time transformation when the events are simultaneous in S?
Show full solution

  1. Calculate the factor

    Method

    Find γ for 0.80c.

    Reason

    The transformation is relativistic.

    Working

    Answer

    γ = 1/(square root of (1-0.80²)) = 5/3

  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

    Answer

    Δ t' = -(5/3)(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

    Answer

    t'₂ < t'₁

Spot the mistake 2

Events simultaneous in S, not in S'

About 6 min

Learner claim

In S, two events are simultaneous and separated by 240 m along + x. Frame S' moves at 0.60c along + x. A learner concludes Δ t' = 0 because the events are simultaneous. Test the claim.

Try this before viewing the solution

Transformed time interval
Show the steps Optional working
Unit: μs
Show solution step by step

  1. Select the equation

    Method

    Use the Lorentz time-difference transformation.

    Reason

    The question compares two spatially separated events across frames.

    Working

    Answer

    Δ t' = γ(Δ t-(vΔ x)/c²)

  2. Apply the condition

    Method

    Set Δ t = 0 and calculate γ = 1.25.

    Reason

    Simultaneity is given only in S; the spatial term remains.

    Working

    Answer

    Δ t' = -γ(vΔ x)/c²

  3. Calculate and interpret

    Method

    The events are not simultaneous in S'; event 2 occurs earlier there.

    Reason

    The transformed interval is negative under the stated ordering.

    Working

    Answer

    Δ t' = -1.25(0.60c)(240)/c² = -6.0 × 10⁻⁷ s

Derive the two measurement results

For the clock ticks, write Δ t' = Δτ. Their co-location in the clock rest frame gives

Δ t = γΔτ.

For the rod endpoint records, simultaneity belongs to the measuring frame S. The forward spatial difference gives L₀ = γ L, so

L = L₀/γ.

The two endpoint records generally have different S' times. The rod’s endpoints are stationary in S', so their coordinate separation is still L₀. Setting Δ t' = 0 instead would compare a different event pair and would not measure the rod’s moving length in S.

Exam-style question 1

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

Show the steps Optional working
Unit: s
Show 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

    Answer

    Δ x' = 0

  2. Select and simplify

    1 mark

    Method

    Use the inverse time-difference transformation.

    Reason

    The known interval is primed.

    Working

    Answer

    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

    Answer

    γ = 1/(square root of (1-0.80²)) = 5/3

  4. Calculate the interval

    1 mark

    Method

    Multiply the proper time by γ.

    Reason

    The interval in S is longer.

    Working

    Answer

    Δ t = (5/3)(4.0) ≈ 6.7 s

Try it yourself 3

Length contraction from the difference form

Minimal support

Problem

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

Show the steps Optional working
Unit: m

Hints

Hint 1

start from the spatial difference

Orient with Δ x' = γ(Δ x-vΔ t) and apply simultaneity in the measuring frame.
Show 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

    Answer

    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

    Answer

    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

    Answer

    L = 2.0/1.25 = 1.6 m

Check your understanding 3

On paper, derive both results above from the Lorentz difference equations: the clock interval in S and the moving rod’s length. Show which difference is zero in each case.

Show answer

The clock ticks occur at one position in S', so Δ x' = 0. The inverse time equation gives Δ t = γ(Δ t' + vΔ x'/c²) = γΔ t'. At 0.80c, γ = 5/3 and Δ t = 6.7 s.

The moving rod’s endpoints are recorded simultaneously in S, so Δ t = 0. The forward spatial equation gives L₀ = Δ x' = γΔ x = γ L. At 0.60c, γ = 1.25 and L = 2.0/1.25 = 1.6 m.

Check the frame of each zero condition, the choice of forward or inverse equation, and the final units. The clock result multiplies by γ; the rod result divides by it.

Common mistakes

  • Defining Δ x as “2 minus 1” but Δ t as “1 minus 2”, or changing the event order between frames.
  • Assuming events simultaneous in one frame are simultaneous in another when they are separated in space.
  • Setting Δ t' = 0 for a length measurement in S; the endpoints must be simultaneous in the measuring frame.
  • Using the time-dilation shortcut for two events that are not at one place in either frame.

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