University Physics II · Introduction to Modern Physics · 15.3
Relativistic Time, Length, Momentum, Energy
Every relation in this topic carries the same γ. What separates a right answer from a wrong one is naming which frame owns the proper quantity, then deciding whether γ multiplies or divides.
Build the model
Connect the measurement to the mechanism.
Special relativity adds one dimensionless number to mechanics: γ = 1/√(1 − v²/c²), equal to 1 at rest and unbounded as v → c. Everything else here is bookkeeping about where it goes. A clock present at both of two events reads the proper time Δτ; a frame that sees those events at different places reads Δt = γΔτ, longer.
An object measured at rest has proper length L₀; a frame it moves through measures L = L₀/γ, shorter, and only along the motion. Momentum stops being mv and becomes p = γmv, so no finite impulse takes a massive object to c. Energy becomes E = γmc², leaving mc² behind at v = 0 and K = (γ − 1)mc² above it.
Because γ → 1 + ½β² at small β, each of these collapses to its Newtonian form, which is why the classical results held up for two centuries. The discipline is to name the proper quantity first, then decide whether the answer should come out larger or smaller before choosing to multiply or divide.
- Simple definition
- Relativistic mechanics rescales time, length, momentum and energy by the Lorentz factor γ = 1/√(1 − v²/c²): intervals between separated events stretch, lengths along the motion shrink, and p = γmv with E = γmc².
- Example
- At 0.980c, γ = 5.03. A muon that lives 2.20 μs by its own clock lasts 11.1 μs in the lab, while in its own frame the 3.00 km of air below it shrinks to 597 m. Two accounts, one arrival.
γ = 1.005 at 0.10c, 1.155 at 0.50c, 7.09 at 0.99c, 22.4 at 0.999c.
Dimensionless, and never less than 1 · c = 2.998 × 10⁸ m s⁻¹
The proper time is the shortest reading; every other frame records a longer interval.
Δτ: proper time in s, read by one clock present at both events
The proper length is the longest; a frame the object moves through measures less.
L₀: proper length in m · dimensions across the motion are unchanged
Along the motion F = γ³ma, so a fixed force buys steadily less acceleration as v climbs.
p in kg m s⁻¹ · m is the invariant mass, the same in every frame
K is total minus rest energy. It is not ½mv², and not ½γmv² either.
In J or eV · electron mc² = 0.511 MeV, proton mc² = 938.3 MeV
The quick route between K, p and β. Setting m = 0 gives E = pc, the photon case.
Same value in every inertial frame · plus sign, not minus
A clock made of light fixes the factor
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.
Proper time is the shortest reading
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.
Length contraction is the same statement, inverted
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.
Momentum must be γmv, or conservation breaks
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.
Work in, and the energy that was already there
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.
Low-speed limits, then the sign checks
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 4.691 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.
Change one variable at a time
Make the relationship visible.
Push β up to 0.95 and watch the top bar stretch past three while the length bar collapses below a third — one γ, multiplying on one row and dividing on the next.
LORENTZ FACTOR γ1.250
2.20 μs MUON IN LAB2.75 μs
3.00 km SEEN BY MUON2.400 km
KINETIC (γ−1)mc²0.250 mc²
Live interpretationLORENTZ FACTOR γ: 1.250. 2.20 μs MUON IN LAB: 2.75 μs. 3.00 km SEEN BY MUON: 2.400 km. KINETIC (γ−1)mc²: 0.250 mc²
Catch the common trap
Explain before calculating.
A muon has mc² = 105.7 MeV and a proper mean lifetime of 2.20 μs. It is produced with total energy 1.00 GeV. Taking c = 3.00 × 10⁸ m s⁻¹, which statement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 90.0 m spacecraft passes Earth at 0.600c. An experiment on board is timed at 5.00 s by a single clock in the laboratory bay. Find γ, the duration Earth records, and the ship's length as Earth measures it.
- β = v/c = 0.600, so β² = 0.360 and 1 − β² = 0.640.
- γ = 1/√0.640 = 1/0.800 = 1.25 — above 1, which every γ must be.
- One clock is present at both the start and the end of the experiment, so 5.00 s is the proper time Δτ. Earth sees those events at different places and reads Δt = γΔτ = 1.25 × 5.00 s = 6.25 s, longer.
- The 90.0 m is measured on the ship, where it sits still, so it is the proper length L₀. Earth measures L = L₀/γ = 90.0 m ÷ 1.25 = 72.0 m, shorter — multiply for the time, divide for the length.
Answerγ = 1.25; Δt = 6.25 s; L = 72.0 m.
MediumA proton (mc² = 938.3 MeV) is accelerated to 0.900c. Find γ, its kinetic energy and its momentum, and compare the kinetic energy with the Newtonian ½mv².
- β = 0.900, so 1 − β² = 1 − 0.810 = 0.190 and γ = 1/√0.190 = 1/0.4359 = 2.294.
- K = (γ − 1)mc² = 1.294 × 938.3 MeV = 1214 MeV, that is 1.21 GeV.
- Momentum from p = γmv, written so the mass appears as an energy: pc = γβ · mc² = 2.294 × 0.900 × 938.3 MeV = 1937 MeV, so p = 1937 MeV/c.
- Check against the invariant E² = (pc)² + (mc²)²: the total energy is E = γmc² = 2.294 × 938.3 MeV = 2153 MeV, and √(E² − (mc²)²) = 1937 MeV, the same pc.
- The Newtonian estimate is ½mv² = ½(mc²)β² = 0.5 × 938.3 MeV × 0.810 = 380 MeV — low by a factor of 3.2, because at this speed γ − 1 is nothing like ½β².
Answerγ = 2.294; K = 1214 MeV; p = 1937 MeV/c. The Newtonian ½mv² gives 380 MeV, low by a factor of 3.2.
HardA muon (mc² = 105.7 MeV, proper mean lifetime 2.20 μs) is created 15.0 km above the ground with total energy 2.00 GeV and travels straight down. Taking c = 3.00 × 10⁸ m s⁻¹, work the trip in the ground frame and again in the muon's frame, and show the two agree on how many mean lifetimes it lasts.
- Total energy fixes γ with no reference to speed: γ = E/mc² = 2000 MeV ÷ 105.7 MeV = 18.92.
- β = √(1 − 1/γ²) = √(1 − 1/358.0) = √0.99721 = 0.99860, so v = 0.99860 × 3.00 × 10⁸ m s⁻¹ = 2.996 × 10⁸ m s⁻¹.
- Ground frame: the mean lifetime is stretched, Δt = γΔτ = 18.92 × 2.20 μs = 41.6 μs, while the fall takes 15 000 m ÷ (2.996 × 10⁸ m s⁻¹) = 50.1 μs — that is 50.1/41.6 = 1.20 mean lifetimes.
- Muon frame: its own clock still gives 2.20 μs per lifetime, but the 15.0 km column is a proper length that contracts to L = L₀/γ = 15 000 m ÷ 18.92 = 793 m, crossed in 793 m ÷ (2.996 × 10⁸ m s⁻¹) = 2.65 μs — that is 2.65/2.20 = 1.20 mean lifetimes.
- The frames disagree about every quantity and agree on the count, so both predict the same survival fraction e(−1.20) = 0.30.
Answerγ = 18.9, β = 0.9986; 41.6 μs lab lifetime against a 50.1 μs fall, or 793 m in 2.65 μs on board — 1.20 mean lifetimes either way, so about 30% survive.