A.5 Galilean and special relativity
Reference frames and Galilean transformations, Einstein's two postulates, the Lorentz transformations, time dilation, length contraction, simultaneity, the spacetime interval, spacetime diagrams and the muon experiment.
mentorriyaz.in
In 30 seconds
- A.5 is HL only. A reference frame is a set of rulers and clocks. Newton's laws hold in every inertial (non-accelerating) frame.
- Galilean relativity adds velocities simply, u′ = u − v. It fails near the speed of light, because light has the same speed c in every inertial frame (Einstein's second postulate).
- The Lorentz transformations replace the Galilean ones. They lead to time dilation Δt = γΔt0, length contraction L = L0/γ and the relativity of simultaneity.
- Observers disagree about times and lengths, but they all agree on the spacetime interval (Δs)2 = (cΔt)2 − (Δx)2.
By the end you can
- Apply the Galilean transformations and explain why they fail at high speed. HL
- State the two postulates of special relativity. HL
- Use the Lorentz transformations, γ and relativistic velocity addition. HL
- Identify proper time and proper length, and calculate time dilation and length contraction. HL
- Use the invariant spacetime interval and draw and read spacetime diagrams. HL
- Explain the muon experiment from both frames. HL
1Reference frames and Galilean relativity HL
Inertial frame — a frame moving at constant velocity, where an object with no resultant force keeps a constant velocity.
A train moves at 30 m s−1. A passenger walks forwards at 2.0 m s−1 relative to the train. Find her speed relative to the ground and the speed of a light beam from the train's headlight, according to Galileo.
- u = u′ + v = 2.0 + 30 = 32 m s−1 ✓ A1
- light: c + 30 m s−1, which contradicts experiment and Maxwell's equations R1
2Einstein's postulates HL
- Postulate 1: the laws of physics are the same in all inertial frames.
- Postulate 2: the speed of light in a vacuum is the same, c, for all inertial observers, whatever the motion of the source.
3Lorentz transformations HL
In S, an event happens at x = 100 m, t = 0.50 μs. Find its coordinates in S′ moving at 0.60c.
- γ = 1/√(1 − 0.36) = 1.25 A1
- x′ = 1.25(100 − 0.60 × 3.0 × 108 × 0.50 × 10−6) = 1.25(100 − 90) = 12.5 m M1A1
- t′ = 1.25(0.50 × 10−6 − 0.60 × 100/(3.0 × 108)) = 0.375 μs A1
Two spaceships approach the Earth from opposite sides, each at 0.60c. Find the speed of one relative to the other.
- Earth frame S: ship B has u = −0.60c; ship A is frame S′, moving at v = +0.60c M1
- u′ = (u − v)/(1 − uv/c2) = (−0.60c − 0.60c)/(1 + 0.36) = −0.88c A1
- speed 0.88c, not 1.2c A1
4Time dilation and length contraction HL
Proper time Δt0 — the time between two events measured by a clock present at both, so the events happen at the same place in its frame. It is the shortest time any observer measures.
Proper length L0 — the length of an object measured in its own rest frame. It is the longest length any observer measures.
A spaceship travels at 0.80c to a star 12 light-years away (Earth frame). Find the journey time in the Earth frame and for the crew.
- Earth: t = 12 ly ÷ 0.80c = 15 years A1
- γ = 1/0.60 = 1.67
- crew clock is present at departure and arrival, so it shows proper time: 15/1.67 = 9.0 years M1A1
- for the crew the distance is contracted to 12/1.67 = 7.2 ly, and 7.2/0.80 = 9.0 years ✓ R1
The muon experiment
- Muons have a mean lifetime of about 2.2 μs in their own frame. At 0.995c (γ ≈ 10) they would travel only 0.66 km on average without relativity, yet many reach the ground from 10 km up.
- Earth frame: their lifetime is dilated to about 22 μs, enough to travel about 6.6 km on average, so a measurable fraction survive 10 km. Muon frame: the lifetime is normal, but the atmosphere is contracted to about 1 km.
5Simultaneity and the spacetime interval HL
- Relativity of simultaneity: events that are simultaneous in one frame are not simultaneous in a frame moving relative to it (unless they happen at the same place). This is the vx/c2 term in the Lorentz transformation for t′.
Two events are 600 m apart and 3.0 μs apart in the lab. Find the proper time between them.
- cΔt = 3.0 × 108 × 3.0 × 10−6 = 900 m M1
- (Δs)2 = 9002 − 6002 = 4.5 × 105 m2 M1
- in the frame where Δx = 0: cΔt0 = 671 m, so Δt0 = 2.2 μs A1
6Spacetime diagrams HL
- Reading coordinates in S′: draw lines from the event parallel to the S′ axes. Lines parallel to the x′ axis are lines of equal t′, which is how the diagram shows simultaneity.
- World lines of objects slower than light are steeper than 45°; light travels along 45° lines.
Test yourself — cover the answers first
Quick check
- State Einstein's two postulates.
- Calculate γ at 0.60c.
- Which observer measures proper time?
- A rod moves along its length at 0.80c. Its proper length is 3.0 m. What length is measured?
- Is the spacetime interval between two events the same for all inertial observers?
- On a spacetime diagram, what angle does a light ray make with the axes?
Exam-style
- A particle with proper lifetime 26 ns moves at 0.90c in the lab. Find its lifetime and the distance it travels in the lab.
- A rocket moving at 0.50c fires a probe forwards at 0.80c relative to the rocket. Find the probe's speed relative to Earth.
Show answers
- Physics is the same in all inertial frames; c is the same for all inertial observers.
- 1.25
- the one for whom both events happen at the same place
- 3.0/1.67 = 1.8 m
- yes, it is invariant
- 45°
- γ = 1/√(1 − 0.81) = 2.29
lab lifetime = 2.29 × 26 = 60 ns
distance = 0.90 × 3.0 × 108 × 60 × 10−9 = 16 m - u = (u′ + v)/(1 + u′v/c2)
= 1.30c/(1 + 0.40)
= 0.93c
Send it to me on WhatsApp. I read every message.
Want regular help? Send this to your parent →