Playground
Compares classical p = mv with relativistic p = gamma*m*v. Moving dot tracks current beta.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Momentumoutput Relativistic momentum magnitude | kg·m/s | MLT^-1 | 0 – 10000000000 | |
| Rest Mass Invariant mass of the particle | kg | M | 1e-30 – 1 | |
| Velocity Particle velocity | m/s | LT^-1 | 0 – 299792458 |
Deep dive
Derivation
Four-momentum P^μ = m·dX^μ/dtau where tau is proper time. Since dt/dtau = gamma, the spatial components become p = m·dx/dtau = gamma·m·v. Conservation of P^μ across all frames forces this definition.
Experimental verification
Every collider (LEP, Tevatron, LHC) reconstructs events using relativistic momentum conservation; deviations would be measured at the ppm level.
Common misconceptions
- 'Relativistic mass' gamma·m is deprecated — rest mass m is the invariant quantity
- Momentum is not bounded as v → c
- p is not simply mv scaled; it is the spatial part of the 4-momentum
Real-world applications
- Collider physics kinematics
- Particle identification via p and energy
- Electron and ion beam optics
- Cosmic-ray spectra
Worked examples
Electron at 0.9c
Given:
- m:
- 9.1093837015e-31
- v:
- 269813212
Find: p
Solution
gamma ≈ 2.294 → p ≈ 2.294 × 9.109e-31 × 2.698e8 ≈ 5.64e-22 kg·m/s.
LHC proton
Given:
- m:
- 1.67262192369e-27
- v:
- 299792455
Find: p
Solution
gamma ≈ 7460 → p ≈ gamma·m·c ≈ 3.74e-15 kg·m/s ≈ 7 TeV/c.
Scenarios
What if…
- scenario:
- What if v = 0.5c?
- answer:
- gamma ≈ 1.155 — only a 15% correction over Newton's p = mv.
- scenario:
- What if the particle is a photon?
- answer:
- m = 0 but p = E/c = h·f/c — massless momentum from the dispersion relation.
- scenario:
- What if we double v from 0.9c to 0.99c?
- answer:
- v only grows by 10%, but gamma jumps from 2.29 to 7.09, so momentum triples.
Limiting cases
- condition:
- v << c
- result:
- p → mv
- explanation:
- Recovers Newton's non-relativistic momentum.
- condition:
- v → c
- result:
- p → ∞
- explanation:
- Infinite momentum required — c is unreachable for massive bodies.
- condition:
- m = 0
- result:
- p = E/c
- explanation:
- Massless particles carry momentum through their energy.
Context
Albert Einstein · 1905
Einstein redefined momentum in his relativity paper so that conservation laws remain valid in all inertial frames.
Hook
How much momentum does a 1 kg mass carry at 0.9c compared to Newton's prediction?
Compute p = gamma·m·v with gamma = 1/sqrt(1 - 0.81) ≈ 2.294 and compare with the classical mv.
Dimensions:
- lhs:
- p → [MLT^-1]
- rhs:
- gamma·m·v → [1]·[M]·[LT^-1] = [MLT^-1]
- check:
- Both sides [MLT^-1]. ✓
Validity: Valid for any massive particle in an inertial frame at any v < c = 299792458 m/s.