Relativistic momentum p = γmv: Question bank — Physics deep dive

Before we start, three symbols we lean on everywhere:

$β = v / c$ — the speed written as a fraction of light speed. $β = 0$ means at rest, $β = 1$ means moving at light speed. It is a pure number with no units.
$γ = 1/ 1 - β^{2}$ — the Lorentz factor. It is always $\geq 1$ and blows up to infinity as $β \to 1$ . Think of it as a "stubbornness dial".
$m$ — the invariant rest mass, a fixed label of the particle that every observer agrees on. It never changes with speed.

True or false — justify

p = γ m v

can be larger than

m c

True.

p = γ β m c

, and

γ β

grows without bound as

β \to 1

, so momentum has no upper limit even though speed is capped at

c

v = 0

the relativistic momentum equals the Newtonian one.

True. At rest

β = 0

γ = 1

, giving

p = m v = 0

from both formulas — relativity and Newton must agree when nothing is moving.

Doubling the speed always doubles the momentum.

False. Only in the Newtonian limit. Because

γ

itself grows with speed, doubling

v

raises

p

by more than a factor of two at high speeds.

The extra factor

γ

means the particle's mass has increased.

False. Mass

m

is the invariant rest mass and stays fixed; the

γ

sits on the velocity/kinematics. Calling

γ m

"mass" causes contradictions (it would depend on direction of the force).

A photon has zero momentum because it has zero mass.

False. For

m = 0

you cannot use

γ m v

(it is

0 \times \infty

), so you use

E = p c

instead — a photon carries momentum

p = E / c

despite zero mass.

γ

can equal

0.5

for a very fast particle.

False.

γ \geq 1

for every real speed, since

1 - β^{2} \leq 1

makes its reciprocal

\geq 1

. A value below 1 means you flipped the formula.

Relativistic momentum is still a vector pointing along the velocity.

True.

p = γ m v

multiplies the velocity vector by a positive scalar

γ m

, so it keeps the same direction as

v

Momentum is conserved in every inertial frame with the

γ

correction.

True. That is the entire reason for the redefinition — dividing displacement by invariant proper time makes

p

transform cleanly, so conservation in one frame guarantees it in all.

Spot the error

γ = 1 - v^{2} / c^{2}

, so

γ < 1

for moving particles."

The error is a flipped formula.

γ

is the reciprocal:

γ = 1/ 1 - v^{2} / c^{2} \geq 1

. The square root written alone is the time-dilation shrink factor, not

γ

"For

v = 0.9 c

we can still use

p = m v

since it's close enough."

β = 0.9

γ \approx 2.29

, so Newton underestimates momentum by more than half. The

γ

correction matters strongly once

v ≳ 0.1 c

"Since

p \to \infty

v \to c

, momentum becomes undefined at

c

For a massive particle

v = c

is simply unreachable —

γ

diverges before you get there. Momentum stays finite for every allowed speed

v < c

; it never has to be evaluated at

c

p = γ m v

p = γ \times m \times v = 1 - β^{2} \cdot m v

The writer substituted

1 - β^{2}

for

γ

, which is upside down. Correct:

p = m v / 1 - β^{2}

, momentum grows, not shrinks.

"We differentiate position by lab time

d t

to build 4-momentum."

Wrong clock. We divide by proper time

d τ

(the clock riding with the particle) because

d τ

is Lorentz-invariant while

d t

is frame-dependent; that invariance is what makes momentum transform properly.

E = p c

holds for all particles."

Only for massless particles like photons. For massive particles the full relation is

E^{2} = (p c)^{2} + (m c^{2})^{2}

;

E = p c

is the special case where the

(m c^{2})^{2}

term vanishes.

"Because

γ \geq 1

, relativistic momentum is always at least

m c

No. At low speed

γ \approx 1

but

v

can be tiny, so

p = γ m v

is tiny too.

γ

being

\geq 1

does not put a floor on

p

; the velocity factor still shrinks to zero.

Why questions

Why do we use proper time

τ

instead of lab time

t

to define momentum?

Proper time is the same for all observers (invariant), so displacement over

d τ

transforms as a proper 4-vector — this is exactly what rescues momentum conservation across frames.

Why does momentum grow faster than linearly with speed?

Because

p = γ m v

carries the extra factor

γ

, which itself increases with

v

; so as speed rises, both

v

and

γ

push

p

up together.

Why can't a massive particle ever reach the speed of light?

Reaching

v = c

would require

γ \to \infty

, hence infinite momentum and infinite energy to supply — physically impossible, so

c

is an unattainable ceiling.

Why does the formula reduce to

p = m v

at everyday speeds?

For

v ≪ c

β \to 0

γ \to 1

, leaving

p \to m v

. Any correct relativistic law must collapse to Newton's at low speed — the correspondence principle.

Why do we keep mass invariant rather than let it grow with

γ

An invariant

m

is a clean, frame-independent label consistent across all of relativity; putting

γ

on velocity avoids direction-dependent "mass" contradictions in force laws.

Why do energy and momentum have to be treated together in relativity?

They are the time and space parts of one 4-momentum whose invariant length is

m c

; a boost mixes them, so you cannot transform one without the other — hence

E^{2} = (p c)^{2} + (m c^{2})^{2}

Why is

γ β

(proper velocity) often the cleanest quantity in momentum problems?

Because

p = γ β m c

, so setting

p

in units of

m c

turns the problem into solving for the single combination

γ β

, avoiding a messy nested square root.

Edge cases

Case

v = 0

: what are

γ

and

p

γ = 1

and

p = 0

— a stationary particle has no momentum, matching Newton exactly.

Case

m = 0

(photon): does

p = γ m v

apply?

No — it gives the indeterminate

0 \times \infty

. The correct route is the energy relation, yielding

p = E / c

for a massless particle moving at

c

Case

v \to c

for a massive particle: limiting behaviour of

p

γ \to \infty

p \to \infty

. Momentum diverges, which is precisely why light speed is a hard limit for anything with mass.

Case negative velocity (particle moving in

- x

): what happens to

p

γ

depends only on

v^{2}

so it is unchanged, but

p = γ m v

points in

- x

— the sign of momentum follows the sign of velocity while the magnitude is unaffected.

Case

β

just above 1 (faster than light): is

γ

real?

No.

1 - β^{2}

becomes negative, so

γ

turns imaginary — a mathematical signal that superluminal speed is not physically allowed for real particles.

Case two particles, same speed, one twice the mass: momentum ratio?

Exactly 2.

γ

and

v

are identical, so

p

scales linearly with

m

; the relativistic corrections cancel in the ratio.

Recall One-line summary of every trap

$γ \geq 1$ always, sits on velocity not mass; $p$ has no ceiling though $v$ does; use $E = p c$ (not $γ m v$ ) for photons; and everything reduces to Newton when $v ≪ c$ .

Connections

Lorentz factor and time dilation — why $γ \geq 1$ and where $d t = γ d τ$ comes from
Why nothing exceeds the speed of light — the $γ \to \infty$ edge case in depth
Photon momentum and radiation pressure — the $m = 0$ trap resolved
Mass-energy equivalence E = γmc² — energy partner behind $E^{2} = (p c)^{2} + (m c^{2})^{2}$
Energy-momentum four-vector — why $E$ and $p$ travel together
Newtonian momentum p = mv — the low-speed limit these traps keep returning to