Relativistic Energy–Momentum Relation

Equivalent forms

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

A Pythagorean theorem in spacetime: energy and momentum as two legs of the invariant mass hypotenuse.

Symbol	Name	SI unit	Dimension	Range
$E$	Total energyoutput Total relativistic energy of the particle	J	M·L²·T⁻²	0 – 1e-9
$p$	Momentum Relativistic momentum magnitude	kg·m/s	M·L·T⁻¹	0 – 1e-17
$m$	Rest mass Invariant rest mass	kg	M	0 – 1e-25
$c$	Speed of light Speed of light in vacuum	m/s	L·T⁻¹	299792458 – 299792458

Unit systems

Symbol	SI	Natural	CGS
$E$	J	MeV	erg
$p$	kg·m/s	MeV/c	g·cm/s
$m$	kg	MeV/c²	g
$c$	m/s	1	cm/s

Where it holds

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).

Dimensional analysis

E^{2} \to [M\cdot L^{2}\cdot T^{-2}]^{2} = [M^{2}\cdot L^{4}\cdot T^{-4}]

(pc)^{2} + (mc^{2})^{2} \to [M\cdot L\cdot T^{-1}]^{2}\cdot [L\cdot T^{-1}]^{2} + [M]^{2}\cdot [L\cdot T^{-1}]^{4} = [M^{2}\cdot L^{4}\cdot T^{-4}] + [M^{2}\cdot L^{4}\cdot T^{-4}]

Both sides are

[M^{2}\cdot L^{4}\cdot T^{-4}]

. Dimensions match.

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Discovery

Albert Einstein · 1905

Derived in Einstein's special relativity papers, this relation unifies momentum and energy for all particles, massive or massless.

Try this

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.

Research status: stable

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

Common misconceptions

Relativistic mass is not needed — this relation uses invariant rest mass only.
$E = mc^{2} is$ NOT the full equation; it only applies to particles at rest $(p = 0)$ .
Massless particles $(m = 0)$ still satisfy this equation: $E = pc$ , not $E = 0$ .

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.

Derivation

Starting from the Lorentz-covariant four-momentum

p^\mu = (E/c

, p), the invariant magnitude is

p^\mu p_\mu = (E/c)^{2} - |p|^{2} = (mc)^{2}

.

Multiplying both sides by

c^{2}

yields

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

.

This is a direct consequence of the Minkowski metric and the definition of four-momentum.

Limiting cases

m \to 0

⟶

E \to pc

For massless particles like photons, energy is purely momentum-driven.

p \to 0

⟶

E \to mc^{2}

A particle at rest has energy equal to its rest mass energy.

p ≫ mc⟶

E \approx pc

At ultra-relativistic speeds, kinetic energy dominates and the particle behaves like a massless particle.

What if…

What if the particle is massless

(m = 0)

?

The relation reduces to $E = pc$ . Energy and momentum become directly proportional, as seen for photons and gluons.

What if momentum doubles?

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 $mc^{2}$ term dominates.

What if we used Newtonian mechanics instead?

Newtonian $KE = p^{2}/2m$ would underestimate total energy at relativistic speeds. At $v = 0.9c$ , the Newtonian formula gives $\sim 60$ % of the correct relativistic energy.

1

Total energy of a proton with known momentum

Given ·

p:: 500 MeV/c
m:: $938.3\,\mathrm{MeV/c}^{2}$

Find · E

Steps

Step 1: Identify values: $pc = 500\,\mathrm{MeV}$ , $mc^{2} = 938.3\,\mathrm{MeV}$ .
Step 2: Square both: $(pc)^{2} = 250000\,\mathrm{MeV}^{2}$ , $(mc^{2})^{2} = 880426.9\,\mathrm{MeV}^{2}$ .
Step 3: Sum: $250000 + 880426.9 = 1130426.9\,\mathrm{MeV}^{2}$ .
Step 4: Take square root: $E = \sqrt{1130426.9} \approx 1063.2\,\mathrm{MeV}$ .

Result ·

E = \surd ((pc)^{2} + (mc^{2})^{2}) = \surd ((500)^{2} + (938.3)^{2}) = \sqrt{250000 + 880426.9} = \sqrt{1130426.9} \approx 1063.2\,\mathrm{MeV}

2

Momentum of a photon with known energy

Given ·

E:: 2.0 MeV
m:: 0

Find · p

Steps

Step 1: For a massless photon, $m = 0$ , so $E^{2} = (pc)^{2}$ .
Step 2: Solve for p: $p = E/c = 2.0\,\mathrm{MeV} / c$ .
Step 3: Convert to SI: $p = (2.0 \times 1.602176634\times 10^{-13} J) / (299792458\,\mathrm{m/s}) \approx 1.07 \times 10^{-21} kg\cdot m/s$ .

Result ·

E^{2} = (pc)^{2} + 0 \to p = E/c = 2.0\,\mathrm{MeV/c} \approx 1.07 \times 10^{-21} kg\cdot m/s

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