Year 12 Physics Module 7 ⏱ ~45 min 5 MC · 2 Short Answer Lesson 12 of 14

Simultaneity and Relativistic Paradoxes

The GPS satellite constellation, 24 satellites orbiting at 20,200 km, first launched by the US Department of Defense in 1978 and fully operational by 1994, must account for both special relativity ($-7.2\,\mu$s/day due to orbital speed) and general relativity ($+45.9\,\mu$s/day due to weaker gravity). The net correction is $+38.7\,\mu$s/day; without it, GPS position errors would accumulate at 10 km per day. Relativistic effects are not abstract, they are built into every navigation system on Earth.

Today's hook: The US Department of Defense GPS system (1978–1994) requires a time correction of $+38.7\,\mu$s per day due to special and general relativity. Without this correction, your phone's GPS would drift by 10 km every day. But if special relativity says the satellites' clocks run slow relative to ground, and the satellites' frame says the ground clocks run slow, how can both be true, and which clock needs the correction?
0/5TASKS
Your guided path0 of 8 done · keep going
1
You’re here

Orient and predict

Predict whether two distant explosions happen at the same time for every observer, then set your goals and vocabulary for simultaneity.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Before you read, predict

Two supernovae explode in distant galaxies. Alice, floating in space exactly midway between them, sees both flashes arrive at the same instant. Bob is moving rapidly toward one galaxy and away from the other.

  1. Did the two supernovae explode simultaneously? How would Alice decide?
  2. Would Bob agree that the explosions were simultaneous? If not, which one would he say happened first?
  3. Does this mean one of them is wrong about what happened?

Write your predictions before reading on, you will revisit them at the end.

Warm-up, what does the "relativity of simultaneity" mean?

Learning Intentions
goals

Know, Relativity of Simultaneity

  • Events simultaneous in one frame may not be in another
  • Depends on relative motion and spatial separation
  • $\Delta t' = -\gamma v\,\Delta x/c^2$

Understand, Paradox Resolution

  • Barn-ladder paradox: simultaneity of door events is frame-dependent
  • Twin paradox: different spacetime paths accumulate different proper times; the turnaround identifies the frame-change asymmetry
  • No absolute time or absolute space

Can Do, Analyse Paradoxes

  • Explain resolution using correct frames
  • Apply Lorentz transformations conceptually
  • Evaluate validity of apparent contradictions
Scan these before reading
vocab
Relativity of simultaneityTwo events simultaneous in one inertial frame are not necessarily simultaneous in another frame moving relative to the first.
Twin paradoxOne twin travels at relativistic speed and returns younger; resolved because the travelling twin changes inertial frames (accelerates).
Proper timeThe time interval measured by a clock present at both events, the shortest measured time, recorded in the object's own rest frame.
Spacetime interval$s^2 = (c\Delta t)^2 - (\Delta x)^2$; a Lorentz-invariant quantity that all observers agree on, unlike $\Delta t$ or $\Delta x$ alone.
Barn-ladder paradoxA ladder longer than a barn (in their rest frames) appears to fit inside the barn due to length contraction; resolved by the relativity of simultaneity of the door-closing events.
Cross-lesson links: L14 derived $E = mc^2$. L15 explores the conceptual underpinning that makes all of L11–L14 consistent: the relativity of simultaneity. The GPS satellite system (US DoD, 1978–1994, 24 satellites at 20,200 km) is the most practically important application, special-relativistic time dilation ($-7.2\,\mu$s/day) and gravitational time dilation ($+45.9\,\mu$s/day) must both be corrected, or GPS drifts 10 km/day. This lesson wraps up the special relativity block; L16 and L20 return to quantum phenomena.

Core Content

2

See why simultaneity is frame-dependent

Work through Einstein's train-and-lightning thought experiment, then read off the condition that makes $\Delta t' \neq 0$.

1
The Relativity of Simultaneity
+5 XP

"At the same time" is not universal

Einstein's 1905 paper contains a beautifully simple thought experiment. A train moves at velocity $v$ past a platform. Two lightning bolts strike the platform simultaneously (in the platform frame) at positions equidistant from a platform observer. The light from both bolts reaches the platform observer at the same time, confirming they were simultaneous in that frame.

But a passenger on the train is moving toward one bolt and away from the other. The light from the bolt ahead reaches her before the light from the bolt behind. Since the speed of light is the same in all frames, and she is equidistant from where the bolts struck in her frame, she concludes that the bolt ahead struck first. The events are not simultaneous in her frame.

Neither observer is wrong. Simultaneity is frame-dependent. This is not an optical illusion or a signal delay, it is a fundamental feature of spacetime geometry.

Time difference between simultaneous events

$$\Delta t' = -\frac{\gamma v\,\Delta x}{c^2}$$

For events with $\Delta t=0$ in the unprimed frame, non-zero $v$ and $\Delta x$ generally give $\Delta t'\neq0$. If those already-simultaneous events also have $\Delta x=0$, they are the same event and every frame assigns them the same spacetime location.

Platform frame Lightning strikes simultaneously at A and B A B Observer (midpoint) sees both flashes together Train frame Platform moves left; bolts strike at different times A B Train observer sees A first (moving toward it) Time difference: Δt′ = −γvΔx/c² If Δx ≠ 0 and v ≠ 0, then Δt′ ≠ 0

Figure 1, Relativity of simultaneity: events simultaneous in the platform frame (A and B strike together) are not simultaneous in the train frame (A strikes first). Neither observer is wrong, "at the same time" is frame-dependent.

Key consequence: If two events are simultaneous and occur at the same position ($\Delta x = 0$) in one frame, they are simultaneous in all frames. But if they are separated in space, simultaneity is frame-dependent. This means two clocks synchronised in the train frame will appear out of sync to platform observers, and vice versa.

Stop & Check

Two clocks on a fast-moving train are synchronised in the train frame. An observer on the platform sees the train moving to the right. Which clock (front or back) does the platform observer measure as ahead? Explain your reasoning using $\Delta t' = -\gamma v\,\Delta x/c^2$.

Simultaneity is frame-dependent: events simultaneous in one frame ($\Delta t = 0$) are generally not simultaneous in another moving frame when $\Delta x \neq 0$. Formula: $\Delta t' = -\gamma v\,\Delta x/c^2$. If $\Delta x = 0$, simultaneity is preserved in all frames. Neither observer is wrong, "at the same time" has no absolute meaning.

Write the simultaneity formula and note the condition under which it equals zero.

Two events occur simultaneously in frame S and are separated by $\Delta x = 400$ m. In frame S' moving at $0.6c$ along the x-axis, are the events simultaneous?

3

Resolve the twin paradox

Find the asymmetry that decides which twin is younger, and put numbers on it at $0.8c$.

2
The Twin Paradox
+5 XP

The most famous paradox in physics, and its resolution

We just saw that two spatially separated events simultaneous in one frame are not simultaneous in another. That raises a question: if each twin sees the other's clock running slow, how can they both be right, and who is actually younger at reunion? This card answers it → compare the proper time along each twin's complete spacetime path.

Alice stays on Earth while her twin Bob travels to a distant star at $0.8c$, turns around, and returns. Due to time dilation, Alice predicts Bob's clock runs slow, Bob should be younger when they reunite. But Bob could equally argue that Alice is moving relative to him, so her clock should run slow. Each predicts the other is younger. Who is right?

The resolution: The situations are not symmetric. Bob must accelerate to turn around, he changes inertial frames. Alice remains in a single inertial frame throughout. Time dilation applies symmetrically only between inertial frames; accelerating frames require careful treatment. Bob's worldline between the two reunion events is shorter in spacetime (proper time is less), so Bob is indeed younger.

A conceptual way to see it: during the outbound journey, both see the other's clock running slow. During the inbound journey, both again see the other's clock running slow. But during the turnaround, Bob sees Alice's clock "jump ahead" because of the relativity of simultaneity, the "now" slice on Bob's spacetime diagram shifts dramatically during acceleration. The net effect: Bob has aged less.

Twin paradox, key numbers

$\Delta t = \gamma\,\Delta\tau$  , Time dilation (moving clock runs slow)

$\Delta\tau_{\text{Bob}} = \Delta t_{\text{Alice}}/\gamma$  , Bob's proper time is less

At $v = 0.8c$: $\gamma = 1.667$. For a 10-year round trip (Alice), Bob ages 6 years.

Stop & Check

Bob travels to Alpha Centauri (4.37 ly away) at $0.9c$ and returns. (a) How much time passes on Earth? (b) How much on Bob's clock? (c) What $\gamma$ would be needed for Bob to age only 1 year while Alice ages 10?

Twin paradox: both see the other's clock run slow, but the situations are NOT symmetric. Bob accelerates to turn around (changes inertial frames); Alice stays in one inertial frame throughout. Result: Bob ages less ($\Delta\tau_{\text{Bob}} = \Delta t_{\text{Alice}}/\gamma$). At $v = 0.8c$: for a 10-year Earth trip, Bob ages only 6 years. The "jump" in Alice's clock during Bob's turnaround arises from shifting simultaneity planes.

Write the key asymmetry and the formula $\Delta\tau_{\text{Bob}} = \Delta t_{\text{Alice}}/\gamma$.

In the twin paradox, the travelling twin returns younger because:

4

Analyse the barn-ladder paradox

Work a full frame-by-frame analysis at $0.8c$, and show exactly where the apparent contradiction dissolves.

3
Worked Example, The Barn-Ladder Paradox
+5 XP

A complete spacetime analysis

We just saw how the twin paradox is resolved by asymmetric acceleration. That raises a question: how do you write out a complete frame-by-frame analysis of the barn-ladder paradox, showing exactly why there is no contradiction? This card answers it → a three-step spacetime analysis.

Problem

A barn has proper length 10 m. A ladder has proper length 15 m. The ladder moves at $0.8c$ through the barn. Analyse the situation in both frames, explaining the role of simultaneity.

Step 1, Barn frame

$\gamma = 1/\sqrt{1-0.64} = 1.667$

Contracted ladder length: $L = 15/1.667 = 9.0$ m. This fits inside the 10 m barn.

In the barn frame, the two door-closing events (front of ladder reaches back door; back of ladder clears front door) can be simultaneous: both doors close at the same instant with the contracted ladder fully inside.

Step 2, Ladder frame

Contracted barn length: $L = 10/1.667 = 6.0$ m. The 15 m ladder does not fit.

The two door-closing events are not simultaneous in the ladder frame. The front of the ladder reaches the back door first (back door closes), and later the back of the ladder clears the front door (front door closes). At no single instant is the ladder fully inside the contracted barn.

Step 3, No contradiction

Both frames correctly describe the same physical events. The barn frame says the ladder was momentarily fully enclosed (both doors closed together). The ladder frame says one door closed first, then the other, with the ladder never fully inside at any instant. The disagreement about whether the two door-closing events were simultaneous is exactly what the relativity of simultaneity predicts. There is no contradiction, just a frame-dependent description of when the doors closed.

Barn frame (10 m barn) Barn (10 m) Ladder 9.0 m (contracted) Both doors close simultaneously Ladder fits! (9.0 m < 10 m) γ = 1.667 Events A & B: simultaneous tA = tB (same time, different places) Ladder frame (15 m ladder) Barn 6.0 m Ladder 15 m (proper length) Back door closes FIRST Front door closes later Ladder never fully inside Events A & B: NOT simultaneous tA ≠ tB in this frame

Figure 2, Barn-ladder paradox resolved: the two door-closing events (A and B) are simultaneous in the barn frame but not in the ladder frame. Both descriptions are correct, the "paradox" is an artefact of assuming simultaneity is absolute.

Stop & Check

A 20 m train moves through a 15 m station at $0.6c$. (a) Calculate $\gamma$. (b) In the station frame, can the train fit entirely inside? (c) What about in the train frame? (d) Explain the resolution using the relativity of simultaneity.

Barn-ladder at $v = 0.8c$ ($\gamma = 1.667$): barn frame, contracted ladder $= 9$ m fits inside 10 m barn, both doors close simultaneously. Ladder frame, contracted barn $= 6$ m, ladder $= 15$ m, doors close sequentially. No contradiction: the two door-closing events are simultaneous in one frame and not in the other, exactly what the relativity of simultaneity predicts.

Write the two frame descriptions and the one-sentence resolution.

The resolution of the barn-ladder paradox relies on:

5

Use the invariant spacetime interval

Classify timelike, spacelike and lightlike intervals, and review the experiments that confirm special relativity.

Beyond the syllabus. A qualitative statement that simultaneity is frame-dependent supports the postulates and is worth having. The Lorentz time transformation, spacetime interval calculations, barn/ladder and twin-paradox resolutions and GPS relativity arithmetic are extension — excluded from core completion and not required in the exam.
4
The Spacetime Interval and Experimental Evidence
+5 XP

The invariant quantity, and the data that confirms it all

We just saw how the barn-ladder paradox is resolved by showing that door-closing events are simultaneous only in one frame. That raises a question: is there any quantity that all observers actually agree on in special relativity? This card answers it → the spacetime interval and the experimental evidence that confirms the whole framework.

While $\Delta t$ and $\Delta x$ vary between frames, the spacetime interval $s^2 = (c\,\Delta t)^2 - (\Delta x)^2$ is the same for all inertial observers. It is a Lorentz invariant. Three cases:

  • Timelike interval ($s^2 > 0$): a causal signal can connect the events at $v \leq c$. All observers agree on which event came first. Proper time $\tau = s/c$.
  • Spacelike interval ($s^2 < 0$): no causal signal can connect the events. Different frames disagree on the time ordering, consistent with no violation of causality.
  • Lightlike interval ($s^2 = 0$): a photon can travel between the events. Proper time along a light ray is zero.

Experimental evidence for special relativity:

  • Muon decay: Cosmic-ray muons created at 15 km altitude reach Earth's surface because their extended lifetime (time dilation) or the contracted atmosphere (length contraction) allows them to survive. Both descriptions are correct and experimentally verified.
  • Particle accelerators: Protons at the LHC travel at $0.99999999c$ with $\gamma \approx 7500$. Without relativistic corrections, the LHC would not work.
  • GPS satellites: Both special relativistic (time dilation from orbital speed) and general relativistic (gravitational time dilation) corrections are applied, otherwise GPS positions would drift by kilometres per day.
  • Hafele-Keating (1971): Caesium clocks flown around the world lost/gained time matching relativistic predictions to within 10%.
Δx cΔt Timelike s² > 0, causal Timelike (past) Spacelike s² < 0, no causal link Spacelike Light cone s² = 0 event O

Figure 3, Spacetime interval classification. Events inside the light cone are timelike-separated (causal order preserved in all frames). Events outside are spacelike-separated (time ordering is frame-dependent, no causal connection possible). On the light cone: lightlike ($s^2 = 0$, photon path).

HSC Exam Strategy, Resolving Paradoxes

When asked to resolve a paradox, always identify the asymmetry:

  1. Twin paradox: Bob accelerates to turn around, Alice does not. The non-inertial frame breaks the symmetry.
  2. Barn-ladder paradox: the two door-closing events are simultaneous in one frame but not the other. Identify which frame sees them as simultaneous.
  3. Never say "both are right in their own frame" without explaining why there is no true contradiction. Always mention the relativity of simultaneity explicitly.
Stop & Check

Classify each interval: (a) Two events: $c\Delta t = 5$ m, $\Delta x = 3$ m. (b) Two events: $c\Delta t = 3$ m, $\Delta x = 5$ m. (c) Two events: $c\Delta t = \Delta x = 4$ m. For each, state whether the time ordering is invariant and whether a causal signal can connect them.

Spacetime interval: $s^2 = (c\Delta t)^2 - (\Delta x)^2$, Lorentz invariant (same in all frames). Timelike ($s^2 > 0$): causal, time order is frame-invariant; spacelike ($s^2 < 0$): no causal link, order is frame-dependent; lightlike ($s^2 = 0$): photon path, proper time $= 0$. Evidence: muon decay, LHC, GPS corrections ($+38\,\mu$s/day), Hafele-Keating all confirm SR.

Write the spacetime interval formula and the three interval classifications.

6

Apply and consolidate

Run your own simultaneity calculations, then write full paradox analyses for the barn-ladder and the twins.

Activities

Activity 1, Simultaneity Calculations
ApplyBand 5

Apply $\Delta t' = -\gamma v\,\Delta x/c^2$ to calculate time differences between frames

  1. Take the primed frame to move in the $+x$ direction with $v/c = 0.6$, and set $\Delta t=0$, $\Delta x=+300$ m. Calculate $\gamma$, then find $\Delta t'$ in microseconds. Verify the sign using $\Delta t' = -\gamma v\Delta x/c^2$.
  2. Two clocks at the front and back of a 400 m train are synchronised in the train frame. At $v = 0.8c$, how much do they differ in the platform frame? Which clock is ahead?
  3. When $\Delta x = 0$, what is $\Delta t'$? Explain what this means physically about events at the same location.
  4. Alice and Bob are 1.0 ly apart (in their shared rest frame). A flash occurs midway. Is it possible for a frame moving at $0.5c$ to see Alice's reception of the flash before Bob's? Explain.
Activity 2, Paradox Analysis
AnalyseBand 6

Apply the relativity of simultaneity to resolve the barn-ladder paradox and the twin paradox

  1. A 12 m ladder moves at $0.6c$ through an 8 m barn. Calculate $\gamma$ and find the contracted length of the ladder in the barn frame. Does it fit? Describe what happens with the door-closing events in each frame.
  2. A rocket ($v = 0.995c$) leaves Earth and returns after 2.0 years (ship time). Calculate $\gamma$. How many years have passed on Earth? What is the age difference between the astronaut and an Earth twin?
  3. A student argues: "The twin paradox proves that time travel is possible, just travel fast enough and you skip into the future." Evaluate this claim. Is the physics correct? What is the key distinction from science fiction "time travel"?
  4. Classify each spacetime interval: (a) $\Delta t = 10$ ns, $\Delta x = 1$ m; (b) $\Delta t = 1$ ns, $\Delta x = 1$ m; (c) $\Delta t = 3.33$ ns, $\Delta x = 1$ m. ($c = 3.0\times10^8$ m/s)