Relativistic Time, Length, Momentum, Energy

The previous topic left two postulates: the laws are identical in every inertial frame, and light travels at c in all of them. Build a clock from those alone. Two mirrors a distance d apart, a pulse bouncing between them, one round trip taking Δτ = 2d/c in the clock's own frame. Now watch that clock ride past at speed v. The pulse still travels at c, but its path is two diagonals, because the mirrors move sideways while it crosses. Each diagonal is √(d² + (vΔt/2)²) long, so cΔt = 2√(d² + (vΔt/2)²). Square, substitute d = cΔτ/2, and collect: Δt²(c² − v²) = c²Δτ², so Δt = Δτ/√(1 − v²/c²). The whole of relativistic kinematics sits in that denominator, so write it once as γ = 1/√(1 − β²) with β = v/c. At β = 0, γ = 1 and nothing changes. At β = 0.10, γ = 1.005; at β = 0.99, γ = 7.09; at β = 0.999, γ = 22.4. γ is never below 1, which is the check that catches most sign errors later.

Δt = γΔτ is unusable until you can say which interval is Δτ. The proper time between two events is what a single clock present at both of them reads — one clock, no synchronisation, no second observer. A muon carries such a clock between its creation and its decay, and by it the mean lifetime is 2.20 μs. Any frame in which those two events happen at different places needs two separated clocks and reads longer. Take muons at 0.980c, so γ = 5.03: their mean lifetime in the lab is 5.03 × 2.20 μs = 11.1 μs, and at 0.980c = 2.938 × 10⁸ m s⁻¹ that covers 3.25 km. Non-relativistically they would manage 646 m. Crossing 3.00 km of atmosphere takes 10.2 μs of lab time, which is 2.03 μs of muon time — 0.92 of a mean lifetime, so about exp(−0.92) = 40% survive, against 1% on the classical count. Nothing has gone wrong inside the muon. Dilation is what an interval is worth in a frame that sees its endpoints apart in space.

Ride with the muon and its own clock reads nothing unusual: 2.20 μs, which at 0.980c buys 646 m. Yet it still reaches the ground. In its frame the atmosphere does the moving, and the 3.00 km column — a proper length, measured in the frame where it sits still — contracts to L = L₀/γ = 3000 m / 5.03 = 597 m. Two frames, two mechanisms, one arrival, and both make the crossing 0.92 of a mean lifetime, so both predict the same 40%. Two cautions. Contraction acts only along the relative motion: a rod 30.0 m long and 4.0 m wide flying lengthwise at 0.800c, where γ = 5/3, measures 18.0 m by 4.0 m in the lab. And measuring the length of something moving means marking both ends at the same time, which is precisely the operation the postulates topic showed to be frame-dependent. Length contraction is the relativity of simultaneity converted into a number.

Keep p = mv and momentum conservation stops holding in every frame: a collision that balances in the lab fails to balance in a frame moving relative to it, once the velocities are combined relativistically. The repair is to differentiate position with respect to the object's own proper time rather than coordinate time. Since dt/dτ = γ, that gives p = m dr/dτ = γmv. Mass stays what it always was, a fixed property of the object with the same value in every frame, and γ carries the entire speed dependence. Newton's second law survives in its momentum form, F = dp/dt, but not as F = ma. Differentiate p = γmv along one axis and d(γv)/dv = γ³, so F = γ³ma. At 0.10c that is a 1.5% correction; at 0.90c a given force delivers 12 times less acceleration than the Newtonian estimate; as v → c the acceleration falls to zero however hard you push. No speed is forbidden by decree — the impulse needed for the next m s⁻¹ simply diverges.

Integrate that force. W = ∫F dx = ∫(dp/dt) dx = ∫v dp, and with p = γmv, parts from rest gives W = γmv² − ∫₀ᵛ γmv dv = γmv² − mc²(1 − 1/γ) = (γ − 1)mc². So K = (γ − 1)mc², and E = γmc² reads as a total energy whose v = 0 value is not zero but mc² — a term Newtonian mechanics has no slot for, and the reservoir that decay and reaction energies come out of. Numbers: give an electron K = 1.00 MeV. With mc² = 0.511 MeV, γ = 1 + K/mc² = 2.957 and β = √(1 − 1/γ²) = 0.941. The classical route, v = c√(2K/mc²), returns 1.98c and so disqualifies itself. Momentum comes from the invariant: E = 1.511 MeV gives pc = √(E² − (mc²)²) = 1.42 MeV, against √(2mc²K) = 1.01 MeV classically. Keep E² = (pc)² + (mc²)² in that form. It holds the same value in every frame and moves you between K, p and β without touching v at all.

Expand: γ = 1 + ½β² + (3/8)β⁴ + …, so K → ½mv²(1 + ¾β² + …) and p → mv(1 + ½β² + …). The classical expressions are the leading terms of a series, not a separate theory. Sizes: a proton at 0.10c has K = 4.727 MeV against a classical 4.692 MeV, 0.76% high, matching ¾β² = 0.75%. An airliner at 300 m s⁻¹ gives ½β² = 5.0 × 10⁻¹³, so a ten-hour flight leaves its clock 18 ns behind one on the ground — caesium clocks resolve that, and GPS satellites carry a 7.2 μs per day correction from orbital speed alone, alongside a larger gravitational term of opposite sign that this topic does not cover. Then the sign checks, which cost more marks than the physics. γ ≥ 1 always, so lab time multiplies the proper time and lab length divides the proper length. γ = 1 + K/mc², not K/mc². The invariant takes a plus sign; only pc = √(E² − (mc²)²) takes a minus.