Relativistic Momentum

Equivalent forms

p = \frac{mv}{\sqrt{1 - v^2/c^2}}

A single gamma factor converts Newton's momentum into its fully covariant form.

Symbol	Name	SI unit	Dimension	Range
$p$	Momentumoutput Relativistic momentum magnitude	kg·m/s	MLT^-1	0 – 10000000000
$m$	Rest Mass Invariant mass of the particle	kg	M	1e-30 – 1
$v$	Velocity Particle velocity	m/s	LT^-1	0 – 299792458

Unit systems

Symbol	SI	CGS	Imperial
$p$	kg·m/s	g·cm/s	slug·ft/s
$m$	kg	g	slug
$v$	m/s	cm/s	ft/s

Where it holds

Valid for any massive particle in an inertial frame at any v <

c = 299792458\,\mathrm{m/s}

.

Dimensional analysis

p \to [MLT^-1]

\gamma \cdot m\cdot v \to [1]\cdot [M]\cdot [LT^-1] = [MLT^-1]

Both sides [MLT^-1].

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Discovery

Albert Einstein · 1905

Einstein redefined momentum in his relativity paper so that conservation laws remain valid in all inertial frames.

Try this

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.

Research status: stable

Real-world applications

Collider physics kinematics
Particle identification via p and energy
Electron and ion beam optics
Cosmic-ray spectra

Common misconceptions

'Relativistic mass' $\gamma \cdot m$ is deprecated — rest mass m is the invariant quantity
Momentum is not bounded as $v \to c$
p is not simply mv scaled; it is the spatial part of the 4-momentum

Experimental verification

Every collider (LEP, Tevatron, LHC) reconstructs events using relativistic momentum conservation; deviations would be measured at the ppm level.

Derivation

Four-momentum

P^\mu = m\cdot dX^\mu

/dtau where tau is proper time.

Since dt/dtau

= \gamma

, the spatial components become

p = m\cdot dx

/dtau

= \gamma \cdot m\cdot v

.

Conservation of

P^\mu

across all frames forces this definition.

Limiting cases

v << c⟶

p \to mv

Recovers Newton's non-relativistic momentum.

v \to c

⟶

p \to \infty

Infinite momentum required — c is unreachable for massive bodies.

m = 0

⟶

p = E/c

Massless particles carry momentum through their energy.

What if…

What if

v = 0.5c

?

$\gamma \approx 1.155$ — only a 15% correction over Newton' $s p = mv$ .

What if the particle is a photon?

$m = 0$ but $p = E/c = h\cdot f/c$ — massless momentum from the dispersion relation.

What if we double v from 0.9c to 0.99c?

v only grows by 10%, but gamma jumps from 2.29 to 7.09, so momentum triples.

1

Electron at 0.9c

Given ·

m:: 9.1093837015e-31
v:: 269813212

Find · p

Steps

$\beta = 0.9$ , $\gamma = 1/\sqrt{0.19} \approx 2.294$
$p = \gamma \cdot m\cdot v$
$p \approx 5.64e-22 kg\cdot m/s$

Result ·

\gamma \approx 2.294 \to p \approx 2.294 \times 9.109e-31 \times 2.698e_{8} \approx 5.64e-22 kg\cdot m/s

.

2

LHC proton

Given ·

m:: 1.67262192369e-27
v:: 299792455

Find · p

Steps

$\gamma \approx E/(m_p c^2) \approx 7460$
$p \approx \gamma \cdot m\cdot c$ for ultra-relativistic
$p \approx 7460 \times 1.673e-27 \times 2.998e_{8}$

Result ·

\gamma \approx 7460 \to p \approx \gamma \cdot m\cdot c \approx 3.74e-15 kg\cdot m/s \approx 7\,\mathrm{TeV/c}

.

Lorentz Factor→Energy–Momentum Relation→Relativistic Velocity Addition→