04 // Interactive Formula Compendium // PHYSICSE = γ m c², where γ = 1 / √(1 − v² / c²)
Einstein Mass-Energy Equivalence, Lorentz Factor & Time Dilation Sheet
In special relativity, a particle of rest mass m moving at speed v has Lorentz factor γ = 1/√(1 − v²/c²), total energy E = γmc², rest energy E₀ = mc², and relativistic momentum p = γmv.
Formula Reference
Main Formula
E = γ m c², where γ = 1 / √(1 − v² / c²)
Rearranged Forms
E₀ = m c², K_rel = (γ − 1) m c² (Rest Energy & Relativistic Kinetic Energy)
E² = (p c)² + (m c²)² (Energy-Momentum Dispersion Relation)
Δt = γ · Δτ, L = L₀ / γ (Time Dilation & Length Contraction)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| gamma | Lorentz Factor γ = 1 / √(1 − β²) | dimensionless |
| beta | Velocity Fraction of Light (β = v / c < 1) | c |
Inputs & Parameters
[c]
Calculated Result (Lorentz Factor γ = 1 / √(1 − β²))
gamma =1.666667dimensionless
Formula Notes & Background
- From the constancy of the speed of light c in all inertial frames, a light-clock moving transversely at speed v travels hypotenuse c Δt while vertical path is c Δτ and horizontal path is v Δt.
- By the Pythagorean theorem, (c Δt)² = (v Δt)² + (c Δτ)² ⇒ Δt = Δτ / √(1 − v²/c²) = γ Δτ.
- Integrating relativistic work dK = v dp with p = γmv yields K = (γ − 1)mc², identifying E₀ = mc² as rest-mass energy.
Step-by-Step Examples
#1: Find the Lorentz factor γ, time dilation, and length contraction for a spacecraft traveling at v = 0.80c.
β = v / c = 0.80
γ = 1 / √(1 − 0.80²) = 1 / √(1 − 0.64) = 1 / √0.36 = 1 / 0.60
→ γ = 5/3 ≈ 1.6667 (moving clocks run 1.67× slower; length contracts to 60% of L₀)
#2: Calculate the energy released by converting Δm = 1.0 gram (0.001 kg) of matter completely into energy.
m = 0.001 kgc = 2.99792 × 10⁸ m/s
E₀ = 0.001 × (2.99792 × 10⁸)² = 8.98755 × 10¹³ J
→ 89.88 TJ (24.97 GWh of electricity, or ~21.5 kilotons of TNT)
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Frequently Asked Questions
How does relativistic kinetic energy K = (γ − 1)mc² reduce to ½mv² at low speeds?+
By the binomial series for v ≪ c, γ = (1 − v²/c²)^(−1/2) ≈ 1 + ½(v²/c²), so (γ − 1)mc² ≈ ½mv².
How can a massless photon (m = 0) carry momentum and energy?+
From the general dispersion relation E² = (pc)² + (mc²)², setting m = 0 gives E = pc, so a photon of energy E = hf carries momentum p = E/c = h/λ.
Do GPS satellites require relativistic clock corrections?+
Yes: special relativity slows orbital clocks by ~7 μs/day while general relativity (weaker gravity aloft) speeds them up by ~45 μs/day, requiring a net +38 μs/day correction.