Playground

Right-triangle visualization of E^2 = (pc)^2 + (mc^2)^2. Adjust beta to see pc and E grow while mc^2 stays fixed.

Variables

SymbolNameSIDimensionRange
EETotal Energyoutput
Total relativistic energy of the particle
JML^2T^-20 – 1e-9
ppMomentum
Relativistic momentum magnitude
kg·m/sMLT^-10 – 1e-18
mmRest Mass
Invariant rest mass
kgM0 – 1e-25

Deep dive

Derivation
From four-momentum P^μ = (E/c, p_x, p_y, p_z), the invariant P^μ P_μ = (E/c)^2 - p^2 must equal (mc)^2 in every frame. Multiplying by c^2 gives E^2 - (pc)^2 = (mc^2)^2.
Experimental verification
Verified daily in colliders: invariant mass reconstruction at the LHC (Higgs discovery) depends on this relation.
Common misconceptions
  • Total energy is not simply (1/2)mv^2 + mc^2 — only a low-speed expansion
  • Even a particle at rest has energy mc^2
  • Massless particles have energy and momentum, but no rest frame
Real-world applications
  • Invariant mass reconstruction at colliders
  • Photon–matter pair production thresholds
  • Compton scattering kinematics
  • Nuclear mass defects

Worked examples

Electron at 1 MeV momentum

Given:
p:
5.344e-22
m:
9.1093837015e-31
Find: E
Solution

pc = 1 MeV, mc^2 = 0.511 MeV → E = sqrt(1 + 0.511^2) ≈ 1.123 MeV.

Photon at 500 nm

Given:
p:
1.325e-27
m:
0
Find: E
Solution

E = pc = 3.97e-19 J ≈ 2.48 eV.

Scenarios

What if…
  • scenario:
    What if p = mc?
    answer:
    E = sqrt(2)·mc^2 ≈ 1.414 mc^2; the particle's kinetic energy equals 0.414 mc^2.
  • scenario:
    What if m is complex (tachyons)?
    answer:
    E^2 - (pc)^2 < 0 — hypothetical tachyons with spacelike four-momenta; none observed.
  • scenario:
    What if we use natural units c = 1?
    answer:
    E^2 = p^2 + m^2 — the cleanest form, used throughout high-energy physics.
Limiting cases
  • condition:
    p → 0
    result:
    E → mc^2
    explanation:
    Pure rest energy — Einstein's famous equation.
  • condition:
    m → 0
    result:
    E = pc
    explanation:
    Photons and other massless particles.
  • condition:
    pc >> mc^2
    result:
    E ≈ pc
    explanation:
    Ultra-relativistic limit dominated by momentum.

Context

Albert Einstein / Paul Dirac · 1905

Einstein's 1905 relativity paper implied this relation; Dirac used it as the mass-shell constraint for relativistic quantum mechanics.

Hook

What is the total energy of a proton with 1 GeV/c of momentum?

Using E^2 = (pc)^2 + (mc^2)^2 with pc = 1 GeV and m_p c^2 ≈ 0.938 GeV, compute E ≈ 1.37 GeV.

Dimensions:
lhs:
E^2 → [M^2 L^4 T^-4]
rhs:
(pc)^2 + (mc^2)^2 → [M^2 L^4 T^-4]
check:
Both sides energy². ✓
Validity: Universally valid for any free particle in special relativity, including massless particles.

Related formulas

Relativistic MomentumLorentz FactorTime Dilation