Playground

Gamma vs v/c curve with a moving dot and numerical readout.

Variables

SymbolNameSIDimensionRange
γγLorentz factoroutput
Time dilation / length contraction factor
dimensionless11 – 100
vvVelocity
Relative velocity between frames
m/sL·T⁻¹0 – 299792457
ccSpeed of light
Speed of light in vacuum
m/sL·T⁻¹299792458 – 299792458

Deep dive

Derivation
From Einstein's two postulates (relativity + invariant c), require that the spacetime interval s² = c²t² − x² is preserved between frames. Demanding linearity and isotropy yields the Lorentz transformations x' = γ(x − vt), t' = γ(t − vx/c²) with γ = 1/√(1 − v²/c²). The factor pops out from enforcing c² t'² − x'² = c²t² − x².
Experimental verification
Muon lifetime experiments (Rossi–Hall 1941; Frisch–Smith 1963) measured γ ≈ 8.4 for cosmic-ray muons reaching Earth's surface. Atomic-clock flights (Hafele–Keating 1971) confirmed time dilation to <1%. GPS satellites apply γ corrections of ~7 μs/day.
Common misconceptions
  • γ is not 'how much faster a moving clock runs' — moving clocks run *slower* (Δt = γΔt₀, where Δt₀ is proper time).
  • Length contraction acts only along the direction of motion, not perpendicular.
  • Both observers see the *other's* clock running slow — there is no contradiction because of relativity of simultaneity.
Real-world applications
  • GPS satellite timing — without γ corrections, GPS would drift ~10 km/day.
  • Particle accelerators — beam optics rely on γ for relativistic mass increase.
  • Cosmic-ray muon detection in atmospheric showers.
  • Synchrotron radiation in light sources scales as γ⁴.

Worked examples

Muon at 0.99c

Given:
v:
0.99
c:
1
Δt0:
0.0000022
Find: Δt
Solution

γ = 1/√(1 − 0.99²) = 1/√0.0199 ≈ 7.09; Δt = γΔt₀ ≈ 15.6 μs

Ship at 0.6c

Given:
v:
0.6
c:
1
Find: γ and length contraction
Solution

γ = 1/√(1 − 0.36) = 1/0.8 = 1.25; a 100 m ship appears 80 m long.

Scenarios

What if…
  • scenario:
    What if v = 0.5c?
    answer:
    γ = 1/√0.75 ≈ 1.155 — only a 15% effect, which is why everyday physics looks Newtonian until v gets close to c.
  • scenario:
    What if you traveled at 0.999999c to a star 10 ly away?
    answer:
    γ ≈ 707, so the proper time experienced is only ~5 days, while ~10 years pass on Earth — the basis of the twin paradox.
  • scenario:
    What if c were infinite?
    answer:
    γ would always equal 1, recovering Galilean relativity. There would be no time dilation, no length contraction, and no upper speed limit.
Limiting cases
  • condition:
    v = 0
    result:
    γ = 1
    explanation:
    Rest frame: no time dilation, no length contraction — Newtonian limit.
  • condition:
    v ≪ c
    result:
    γ ≈ 1 + ½(v/c)²
    explanation:
    Binomial expansion gives the leading relativistic correction; recovers Newton at low speed.
  • condition:
    v → c
    result:
    γ → ∞
    explanation:
    Infinite energy required to reach c — massive objects can never get there.

Context

Hendrik Lorentz · 1904

Lorentz introduced the factor in his transformations to explain the Michelson-Morley null result; Einstein gave it physical meaning a year later.

Hook

Why does a muon born in the upper atmosphere reach the ground when it 'shouldn't'?

A muon traveling at 0.99c has a proper lifetime of 2.2 μs. How long does it live in the lab frame?

Dimensions: γ is dimensionless: [v²/c²] = (L·T⁻¹)²/(L·T⁻¹)² = 1 ✓
Validity: Valid in all inertial frames in special relativity. Strictly v < c for massive objects. In curved spacetime (general relativity), γ applies locally between freely-falling frames.

Related formulas