Playground
A Pythagorean triangle in energy-momentum space: adjust p and m to see E change as the hypotenuse.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Total energyoutput Total relativistic energy of the particle | J | M·L²·T⁻² | 0 – 1e-9 | |
| Momentum Relativistic momentum magnitude | kg·m/s | M·L·T⁻¹ | 0 – 1e-17 | |
| Rest mass Invariant rest mass | kg | M | 0 – 1e-25 | |
| Speed of light Speed of light in vacuum | m/s | L·T⁻¹ | 299792458 – 299792458 |
Deep dive
Derivation
Starting from the Lorentz-covariant four-momentum p^μ = (E/c, p), the invariant magnitude is p^μ p_μ = (E/c)² − |p|² = (mc)². Multiplying both sides by c² yields E² = (pc)² + (mc²)². This is a direct consequence of the Minkowski metric and the definition of four-momentum.
Experimental verification
Verified in every particle physics experiment. Particle accelerators routinely measure momentum (via curvature in magnetic fields) and energy (via calorimeters) independently, confirming the relation to high precision. The discovery of the Higgs boson at the LHC relied on reconstructing invariant masses from E and p measurements.
Common misconceptions
- Relativistic mass is not needed — this relation uses invariant rest mass only.
- E = mc² is NOT the full equation; it only applies to particles at rest (p = 0).
- Massless particles (m = 0) still satisfy this equation: E = pc, not E = 0.
Real-world applications
- Particle identification in collider detectors
- Cosmic ray energy reconstruction
- Medical PET scan photon energy calculations
- Neutrino mass measurements from energy-momentum mismatch
Worked examples
Total energy of a proton with known momentum
Given:
- p:
- 500 MeV/c
- m:
- 938.3 MeV/c²
Find: E
Solution
E = √((pc)² + (mc²)²) = √((500)² + (938.3)²) = √(250000 + 880426.9) = √(1130426.9) ≈ 1063.2 MeV
Momentum of a photon with known energy
Given:
- E:
- 2.0 MeV
- m:
- 0
Find: p
Solution
E² = (pc)² + 0 → p = E/c = 2.0 MeV/c ≈ 1.07 × 10⁻²¹ kg·m/s
Scenarios
What if…
- scenario:
- What if the particle is massless (m = 0)?
- answer:
- The relation reduces to E = pc. Energy and momentum become directly proportional, as seen for photons and gluons.
- scenario:
- What if momentum doubles?
- answer:
- Energy increases but does not double unless the particle is ultra-relativistic (p ≫ mc). For a non-relativistic particle, the increase is smaller because the mc² term dominates.
- scenario:
- What if we used Newtonian mechanics instead?
- answer:
- Newtonian KE = p²/2m would underestimate total energy at relativistic speeds. At v = 0.9c, the Newtonian formula gives ~60% of the correct relativistic energy.
Limiting cases
- condition:
- m → 0
- result:
- E → pc
- explanation:
- For massless particles like photons, energy is purely momentum-driven.
- condition:
- p → 0
- result:
- E → mc²
- explanation:
- A particle at rest has energy equal to its rest mass energy.
- condition:
- p ≫ mc
- result:
- E ≈ pc
- explanation:
- At ultra-relativistic speeds, kinetic energy dominates and the particle behaves like a massless particle.
Context
Albert Einstein · 1905
Derived in Einstein's special relativity papers, this relation unifies momentum and energy for all particles, massive or massless.
Hook
How can a massless photon still carry momentum?
A particle has momentum 500 MeV/c and rest mass 938 MeV/c². Find its total energy.
Dimensions:
- lhs:
- E² → [M·L²·T⁻²]² = [M²·L⁴·T⁻⁴]
- rhs:
- (pc)² + (mc²)² → [M·L·T⁻¹]²·[L·T⁻¹]² + [M]²·[L·T⁻¹]⁴ = [M²·L⁴·T⁻⁴] + [M²·L⁴·T⁻⁴]
- check:
- Both sides are [M²·L⁴·T⁻⁴]. Dimensions match. ✓
Validity: Exact relation valid for all particles in special relativity. Applies to massive and massless particles alike. Does not account for gravitational effects (requires general relativity).