Energy–Momentum Relation

Equivalent forms

E = \sqrt{(pc)^2 + (mc^2)^2}

E^2 - p^2 c^2 = m^2 c^4

A single Lorentz-invariant that encapsulates E = mc^2 and p = gamma·mv at once.

Symbol	Name	SI unit	Dimension	Range
$E$	Total Energyoutput Total relativistic energy of the particle	J	ML^2T^-2	0 – 1e-9
$p$	Momentum Relativistic momentum magnitude	kg·m/s	MLT^-1	0 – 1e-18
$m$	Rest Mass Invariant rest mass	kg	M	0 – 1e-25

Unit systems

Symbol	SI	CGS	Imperial
$E$	J	erg	ft·lbf
$p$	kg·m/s	g·cm/s	slug·ft/s
$m$	kg	g	slug

Where it holds

Universally valid for any free particle in special relativity, including massless particles.

Dimensional analysis

E^2 \to [M^2\,\mathrm{L}^4\,\mathrm{T}^-4]

(pc)^2 + (mc^2)^2 \to [M^2\,\mathrm{L}^4\,\mathrm{T}^-4]

Both sides energy

^{2}

.

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Discovery

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.

Try this

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.

Research status: stable

Real-world applications

Invariant mass reconstruction at colliders
Photon–matter pair production thresholds
Compton scattering kinematics
Nuclear mass defects

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

Experimental verification

Verified daily in colliders: invariant mass reconstruction at the LHC (Higgs discovery) depends on this relation.

Derivation

From four-momentum

P^\mu = (E/c

, p_x, p_y, p_z), the invariant

P^\mu P_\mu = (E/c)^2 - p^2

must equal (mc)^2 in every frame.

Multiplying by c^2 gives

E^2 - (pc)^2 = (mc^2)^2

.

Limiting cases

p \to 0

⟶

E \to mc^2

Pure rest energy — Einstein's famous equation.

m \to 0

⟶

E = pc

Photons and other massless particles.

pc >> mc^2⟶

E \approx pc

Ultra-relativistic limit dominated by momentum.

What if…

What if

p = mc

?

$E = \sqrt{2}\cdot mc^2 \approx 1.414\,\mathrm{mc}^2$ ; the particle's kinetic energy equals 0.414 mc^2.

What if m is complex (tachyons)?

E^2 - (pc)^2 < 0 — hypothetical tachyons with spacelike four-momenta; none observed.

What if we use natural units

c = 1

?

$E^2 = p^2 + m^2$ — the cleanest form, used throughout high-energy physics.

1

Electron at 1 MeV momentum

Given ·

p:: 5.344e-22
m:: 9.1093837015e-31

Find · E

Steps

$pc = 1\,\mathrm{MeV} = 1.602e-13\,\mathrm{J}$
$mc^2 = 0.511\,\mathrm{MeV} = 8.187e-14\,\mathrm{J}$
$E = \sqrt((pc)^2 + (mc^2)^2) \approx 1.123\,\mathrm{MeV}$

Result ·

pc = 1\,\mathrm{MeV}

,

mc^2 = 0.511\,\mathrm{MeV} \to E = \sqrt{1 + 0.511^2} \approx 1.123\,\mathrm{MeV}

.

2

Photon at 500 nm

Given ·

p:: 1.325e-27
m:: 0

Find · E

Steps

$m = 0 \to E = pc$
$p = h/\lambda = 6.626e-34/5e-7 = 1.325e-27 kg\cdot m/s$
$E = 1.325e-27 \times 299792458 \approx 3.97e-19\,\mathrm{J}$

Result ·

E = pc = 3.97e-19\,\mathrm{J} \approx 2.48\,\mathrm{eV}

.

Relativistic Momentum→Lorentz Factor→Time Dilation→