F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special RelativityThe Relativity of Simultaneity
F1.1 F2.1 F3About 65 minutes
Learning goals- I can explain why two events that are simultaneous in one inertial frame need not be simultaneous in another.
- I can analyse Einstein's train-and-lightning thought experiment using the second postulate.
- I can calculate the time difference between two events in a frame where they are not simultaneous.
- I can explain why the order of cause and effect is the same for all observers.
Words to know
simultaneous events
Two events that happen at the same time in a given frame of reference.
relativity of simultaneity
Events that are simultaneous in one inertial frame are generally not simultaneous in another frame moving relative to the first, if the events are at different places along the direction of motion.
proper observer (for simultaneity)
An observer who corrects the time a signal reaches them for its travel time, so they record when an event really happened in their frame.
causality
The principle that a cause always comes before its effect. In relativity, all observers agree on the order of events that could be cause and effect.
light-travel time
The time a light signal takes to travel from an event to an observer: distance ÷ c.
'At the same time' is not as simple as it sounds
Two firecrackers go off at opposite ends of a soccer field. You say they burst 'at the same time'. In everyday life, that is a statement everyone agrees with. Einstein's second thought-experiment target was exactly this idea. He showed that 'at the same time' depends on the observer's frame. This is the key that unlocks time dilation and length contraction, so it is worth getting very clear about it.
First, a measuring rule. Seeing an event is not the same as when it happens. Light (or sound) takes time to reach you. A careful observer always corrects for the travel time. Two events are simultaneous in a frame if, after correcting for travel time, they happened at the same time on that frame's synchronized clocks. An observer exactly halfway between two events who receives both signals at once knows the events were simultaneous, because the signals travelled equal distances at the same speed.
Einstein's train
A long train moves at high speed v along a straight track. Lightning strikes both ends of the train, leaving scorch marks A (back) and B (front) on the track and on the train.
- Sam stands on the ground, exactly halfway between the scorch marks A and B. The two flashes reach him at the same instant. Both flashes travelled the same distance at speed c, so Sam concludes: the strikes were simultaneous (in the ground frame).
- Mia sits in the middle of the train. At the instant of the strikes (in Sam's frame), she is beside Sam. But she is moving toward B and away from A. So the flash from B reaches her first.
Mia is in an inertial frame. By postulate 2, in her frame both flashes travel at c. She is exactly halfway between the scorch marks on the train. Since the flash from B arrived first and both travelled equal distances at the same speed, she must conclude: the strike at B happened first.
Who is right? Both. Neither frame is special (postulate 1). Simultaneity is not absolute. It is relative to the frame.
The key rule
If two events are simultaneous in one frame and are separated along the direction of motion, then in a frame moving along that direction they are not simultaneous. The event toward the front (in the direction the second frame moves relative to the first) happens first in the second frame.
Mia's train moves toward B, and in her frame B happened first. Two more facts complete the picture:
- If two events happen at the same place and same time (for example, two balls colliding), every observer agrees they are simultaneous.
- Separations perpendicular to the motion do not cause disagreement; only separation along the motion does.
Why didn't anyone notice before 1905? Because at everyday speeds the time difference is unimaginably small. The disagreement is about Lv/c² for two events a distance L apart, and for a car on a highway it is far less than a trillionth of a second.
Worked exampleExample 1: A flash in a moving train car
A train car is 30 m long, as measured in the ground frame, and moves at 0.60c. A lamp at the exact centre of the car flashes. In the car's frame, the light reaches the front and back doors at the same time. Find when the light reaches each door in the ground frame.
- In the ground frame, the light leaves the centre and travels at c in both directions (postulate 2). Each door is 15 m from the centre at the moment of the flash.
- The back door moves toward the light: closing speed c + v = 1.60c. t_back = 15 m ÷ (1.60 × 3.00 × 10⁸ m/s) = 3.125 × 10⁻⁸ s = 31.25 ns.
- The front door moves away from the light: closing speed c − v = 0.40c. t_front = 15 m ÷ (0.40 × 3.00 × 10⁸ m/s) = 1.25 × 10⁻⁷ s = 125 ns.
- Difference: 125 − 31.25 = 93.75 ns.
In the ground frame the light reaches the back door after about 31 ns and the front door after 125 ns, about 94 ns later. In the car's frame, the two arrivals are simultaneous.
In the ground frame (drawn to scale), the car moves right at 0.60c while light spreads both ways from where the flash happened at speed c. The back door runs into the light first; the front door runs away from it. For the passengers, both doors are lit at the same instant.Worked exampleExample 2: Which happened first?
A spacecraft travels east past Earth at high speed. Two beacons, one east and one west, flash simultaneously in Earth's frame. According to the crew, which beacon flashed first?
- The flashes are simultaneous in Earth's frame and separated along the direction of motion (east–west).
- The crew moves east relative to Earth, so the event toward the east (the front, in their direction of motion) happens first in their frame.
- Check with the train: Mia moved toward B and found B first.
The east beacon flashed first, according to the crew.
Worked exampleExample 3: Correcting for travel time (everyday version)
Raj stands 680 m from a fireworks barge and 170 m from a second barge in the opposite direction. He hears the bang from the far barge 1.5 s after the bang from the near barge. Sound travels at 340 m/s. Were the explosions simultaneous in the ground frame?
- Travel time from near barge: 170/340 = 0.50 s. From far barge: 680/340 = 2.00 s.
- If the explosions were simultaneous, Raj would hear the far one 2.00 − 0.50 = 1.50 s after the near one.
- He did hear a 1.5 s gap, so after correcting for travel time, the bursts happened at the same instant.
Yes, simultaneous (in the ground frame). Correcting for signal travel time is always the first step; relativity of simultaneity is a disagreement that remains after this correction.
Does this break cause and effect?
If observers disagree about the order of events, could one observer see an effect before its cause? No. The order of two events can only differ between frames when the events are so far apart in space, compared with their separation in time, that not even light could travel from one to the other in time. Such events cannot influence each other. For events where a signal at or below c could connect them (a bat hitting a ball, then the ball breaking a window), every observer agrees on the order. Relativity keeps causality intact, and this is one reason nothing can carry information faster than light.
Simultaneity checklist
- Always correct for signal travel time first. The relativity of simultaneity is what is left after correcting.
- Simultaneous in one frame + separated along the motion → not simultaneous in the other frame.
- The event at the front (in the direction the observer moves) happens first for the moving observer.
- Events at the same place and time are simultaneous for everyone. Cause always precedes effect for everyone.
Just for fun'I'll be there at the same time as you!' is a promise that, strictly speaking, only works if you and your friend share a frame of reference. Try using that excuse when you are late.
Hands-on activityClap-timing: correcting for travel time
Sound is slow enough that you can feel the travel-time effect that every relativity observer must correct for. You need: two helpers, a long outdoor space such as a park or quiet schoolyard (at least 60 m long), a measuring tape or a phone app that measures distance, and a phone stopwatch.
- Place helper P at one end and helper Q at the other end, at least 60 m apart. Stand right next to P.
- Agree on a countdown by phone or hand signal so P and Q clap two pieces of wood together (or two pot lids) at the same instant on 'zero'. Q should raise an arm at the moment of clapping so you can see it.
- You will hear P's clap first, even though they clapped together. Do you hear Q's clap later? Try to estimate the delay, or film it with a phone to measure it.
- Calculate the expected delay: distance ÷ 343 m/s. For 68.6 m it is 0.200 s.
- Now stand exactly halfway between P and Q. What do you hear?
What to look for: next to P you hear Q's clap noticeably late (about 0.2 s for 69 m). Halfway, the claps arrive together. Write a short paragraph explaining why both positions agree that the claps were simultaneous after correcting for travel time, and how Einstein's train is different: Mia corrects for travel time too, and still disagrees with Sam.
Your turn
1Explain what it means for two events to be simultaneous in a frame of reference.
After correcting for the time signals take to reach the observer, the two events happened at the same time on that frame's synchronized clocks.
2In Einstein's train thought experiment, why does Mia receive the flash from B first?
She is moving toward B and away from A, so she meets the light from B before the light from A reaches her.
3Why does Mia conclude that strike B happened first, rather than that the light from B was faster?
By postulate 2, both flashes travel at c in her frame. She is halfway between the scorch marks on the train, so equal distances at equal speeds means the one that arrived first was emitted first.
4Two events happen at the same place and at the same time in one frame. Will they be simultaneous in all frames?
Yes. Disagreement about simultaneity only arises for events separated in space along the direction of motion.
5A 40 m train car (measured in the ground frame) moves at 0.50c. A lamp at the centre flashes. In the ground frame, when does light reach the back and front doors?
Back: 20 ÷ (1.50 × 3.00 × 10⁸) = 44.4 ns. Front: 20 ÷ (0.50 × 3.00 × 10⁸) = 133 ns.
6A rocket moves west past two buoys that blink simultaneously in the ocean's frame, one east and one west. Which blinks first for the rocket crew?
The west buoy (the one toward which the rocket is moving).
7Can an observer ever see a baseball break a window before the bat hits it? Explain.
No. The bat hitting the ball causes the window to break, and the ball travels slower than light. Events that can be causally connected keep the same order in every frame.
More practice: warm-up
1Fill in: simultaneity is ___ , not absolute.
relative (to the frame of reference).
2Who first analysed the relativity of simultaneity using a train and lightning?
3Why don't we notice the relativity of simultaneity in daily life?
Everyday speeds are tiny compared with c, so the time disagreements (about Lv/c²) are far too small to detect.
4Light travels from an event 600 m away. How long does it take to reach you?
t = 600 ÷ 3.00 × 10⁸ = 2.00 × 10⁻⁶ s = 2.00 µs.
5Does the relativity of simultaneity depend on separations perpendicular to the motion?
No. Only separations along the direction of relative motion matter.
More practice: on your own
1A 24 m train car (ground measurement) moves at 0.80c. A centre lamp flashes. Find the time between the light reaching the back and front doors in the ground frame.
Back: 12 ÷ (1.80c) = 12 ÷ 5.40 × 10⁸ = 22.2 ns. Front: 12 ÷ (0.20c) = 12 ÷ 6.00 × 10⁷ = 200 ns. Difference: about 178 ns.
2Sam (on the ground) sees two lightning strikes A and B simultaneously, with the train moving toward B. A second train moves at the same speed toward A. Which strike happens first for a passenger on the second train?
Strike A happens first, because that passenger moves toward A.
3A thunderstorm: you see a lightning flash, then hear thunder 3.0 s later. Sound travels at 343 m/s. How far away was the strike, and why can you ignore light's travel time?
d ≈ 343 × 3.0 = about 1.0 km. Light takes only 1030 ÷ 3.00 × 10⁸ ≈ 3.4 µs, negligible compared with 3.0 s.
4Explain why two events that cannot be connected even by light can have different orders in different frames without breaking causality.
If no signal (not even light) can travel from one event to the other, neither can cause the other. Changing their order changes no cause-and-effect relationship, so causality is safe.
5A friend says: 'Mia is wrong because she is the one moving.' Use postulate 1 to respond.
Postulate 1 says no inertial frame is special. From Mia's point of view she is at rest and the ground moves. Her conclusion is just as valid as Sam's.
Challenge
1Show that, in the ground frame, the time difference between light reaching the back and front doors of a car of ground-length L moving at v is Δt = Lv/(c² − v²). Check with Example 1.
t_front − t_back = (L/2)/(c − v) − (L/2)/(c + v) = (L/2)[(c + v) − (c − v)]/(c² − v²) = Lv/(c² − v²). Check: L = 30 m, v = 0.60c: 30(0.60c)/(0.64c²) = 28.1/c = 28.1 ÷ 3.00 × 10⁸ = 9.375 × 10⁻⁸ s = 93.75 ns ✓.
2Estimate the disagreement about simultaneity for a 30 m bus moving at 30 m/s (use Δt ≈ Lv/c²).
Δt ≈ (30)(30)/(9.00 × 10¹⁶) = 1.0 × 10⁻¹⁴ s, ten femtoseconds: far too small to notice.
3Two supernovas are seen by astronomers on Earth. After correcting for light-travel time, they exploded simultaneously in Earth's frame, 10 000 light-years apart. Could a fast-moving alien spacecraft observer find that they exploded in the opposite order from another fast ship? Could one explosion have caused the other?
Yes: they are simultaneous in Earth's frame and widely separated, so a ship moving toward one sees it first, and a ship moving the other way sees the other first. No: since they happened at the same time 10 000 light-years apart, no signal (even light) could travel between them, so neither could cause the other.
Quiz
1. Two events are simultaneous in frame S and happen at different places along the x-axis. Frame S′ moves along x relative to S. What does S′ find?
2. In Einstein's train, the train moves toward strike B. Which strike happened first in the train's frame?
3. Which pair of events will ALL inertial observers agree are simultaneous?
4. A centre lamp flashes in a car moving at 0.60c. In the ground frame, which door does the light reach first?
5. Why is causality not violated by the relativity of simultaneity?
6. Why did nobody notice the relativity of simultaneity before 1905?
Big idea (F1.1, F2.1): 'at the same time' depends on the frame of reference. Two events that are simultaneous for one observer can happen one after the other for another observer moving relative to the first. This is the most important idea for understanding the rest of the unit.
How to help: Have your teen draw Einstein's train and explain, step by step, why Mia concludes B came first even though she knows light travels at c. Ask: 'Why isn't Sam more correct than Mia?' The clapping activity is best done outdoors with the helpers at least 60 m apart; never clap near anyone's ears.
Common difficulties:
- Confusing seeing an event with when it happened. Always correct for the signal's travel time first.
- Thinking one observer must be 'really' right. Neither frame is special.
- Mixing up which event comes first. The event toward which the observer moves comes first for that observer.
- Thinking causality is broken. Only events that cannot affect each other can change order.