Special Relativity
CollegeThe Lorentz transformation and spacetime
Chapter 17 introduced time dilation and length contraction as consequences of two postulates. Here we build the full formalism. Events are labelled by coordinates (t, x, y, z) in an inertial frame. A second frame S′ moving at constant velocity v along the x-axis relates to S by the Lorentz transformation:
t′ = γ(t − vx/c²)
y′ = y, z′ = z γ = 1/√(1 − v²/c²)
These replace the Galilean transformation x′ = x − vt, which fails at high speed because it silently assumes absolute time. The quantity every observer agrees on is the spacetime interval:
A particle's own clock measures proper time τ, defined by c²dτ² = c²dt² − dx² − dy² − dz², so dτ = dt/γ. Velocities do not add linearly; combining a velocity u (in S′) with frame velocity v gives the relativistic velocity-addition formula:
which correctly caps at c for any u, v < c. Energy and momentum combine into the invariant relation:
Spacetime, simultaneity, and the light cone
Special relativity is best understood geometrically: instead of separate space and time, treat spacetime as one four-dimensional continuum where the interval Δs² plays the role ordinary distance plays in geometry. Events with Δs² > 0 are timelike separated — a signal slower than light can connect them, and all observers agree on their time order. Events with Δs² < 0 are spacelike separated — no signal can connect them, and different observers can disagree on which happened first. This is the relativity of simultaneity: two events simultaneous in one frame are generally not simultaneous in another moving frame.
The set of all events reachable from "here and now" by a light signal forms the light cone, and it bounds cause and effect absolutely — no Lorentz transformation can reorder a cause after its effect, because that would require Δs² < 0 for what should be a timelike (causal) pair. This is why c is not just "the speed of light" but the universe's speed limit for any influence.
τ); the lab sees a longer Δt = γτ because the lab frame and the muon are in relative motion.Worked example — energy and momentum of a fast electron
An electron (m = 9.11×10⁻³¹ kg) moves at v = 0.80c. Find its momentum, total energy, kinetic energy, and confirm the energy–momentum invariant.
- Lorentz factor:
γ = 1/√(1 − 0.80²) = 1/√0.36 = 1/0.60 = 1.667. - Momentum:
p = γmv = 1.667 × 9.11×10⁻³¹ × 0.80×(3.0×10⁸) = 3.64×10⁻²² kg·m·s⁻¹. - Total energy:
E = γmc² = 1.667 × 9.11×10⁻³¹ × (3.0×10⁸)² = 1.367×10⁻¹³ J. - Rest energy:
mc² = 9.11×10⁻³¹×(3.0×10⁸)² = 8.20×10⁻¹⁴ J. Kinetic energy:KE = E − mc² = 1.367×10⁻¹³ − 0.820×10⁻¹³ = 0.547×10⁻¹³ J. - Check invariant:
(pc)² + (mc²)² = (3.64×10⁻²²×3.0×10⁸)² + (8.20×10⁻¹⁴)² = (1.092×10⁻¹³)² + (8.20×10⁻¹⁴)² = 1.367×10⁻¹³ J, matching E. ✓
Note the kinetic energy (0.547×10⁻¹³ J) is a substantial fraction of the rest energy — the non-relativistic formula ½mv² would badly underestimate it at 0.80c.
Derive the low-speed limit of E = γmc² and show it reduces to Newtonian kinetic energy.
For v ≪ c, expand γ ≈ 1 + v²/2c² (binomial approximation). Then E ≈ mc² + ½mv². Subtracting the rest energy mc² gives KE ≈ ½mv² — the familiar Newtonian formula emerges as the low-speed limit of the relativistic one.
Two spaceships each move at 0.9c toward each other as seen from Earth. What speed does one see the other approaching at?
Naive addition would give 1.8c, violating relativity. Using u′ = (u+v)/(1+uv/c²) with u = v = 0.9c: u′ = 1.8c/(1+0.81) = 1.8c/1.81 = 0.9945c — very fast, but still under c.
Why can two spacelike-separated events have disputed time order, but timelike-separated ones cannot?
Spacelike separation means no signal (even at light speed) could travel between them, so no causal link is possible — Lorentz transformations are free to reorder them for different observers without breaking cause and effect. Timelike-separated events could be causally connected, so preserving their order is required by Δs² > 0 being invariant.
A particle has E = 5.0×10⁻¹⁰ J and p = 1.6×10⁻¹⁸ kg·m·s⁻¹. Find its rest mass.
From (mc²)² = E² − (pc)²: pc = 1.6×10⁻¹⁸ × 3.0×10⁸ = 4.8×10⁻¹⁰ J. So (mc²)² = (5.0×10⁻¹⁰)² − (4.8×10⁻¹⁰)² = 2.5×10⁻¹⁹ − 2.304×10⁻¹⁹ = 1.96×10⁻²⁰, giving mc² = 1.4×10⁻¹⁰ J, so m = 1.56×10⁻²⁷ kg — close to a proton's mass.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1A rod is 2.0 m long at rest. How long is it moving at 0.80c?
- gamma = 1/sqrt(1 - 0.64) = 1/0.6 = 1.667
- L = L0/gamma = 2.0 / 1.667
Example 2An electron moves at 0.80c. Find its total energy in terms of rest energy E0. (E0 = m*c^2)
- gamma = 1/sqrt(1 - 0.64) = 1.667
- E = gamma*E0
Now you try
Work each one out first, then tap to reveal the worked answer.