Relativistic momentum p = γmv: Worked examples — Physics deep dive

Intuition What this page is for

> The parent note Relativistic momentum $p=\gamma m v$ gave you the formula and three examples. Here we hunt down every kind of situation the formula can throw at you — slow speeds, near-light speeds, the exact $v = 0$ and $v \to c$ edges, the massless photon, a real-world accelerator problem, and an exam-style twist. Each one is worked from the first symbol so you never meet a case you have not seen.

Before we start, let us re-earn the two symbols we lean on the whole page:

$β$ (Greek "beta") is just a nickname for speed measured in units of light-speed: $β = v / c$ . If you move at half light-speed, $β = 0.5$ . It has no units — it is a pure fraction between $0$ and $1$ .
$γ$ (Greek "gamma", the Lorentz factor) is the "stubbornness multiplier": $γ = \frac{1}{1 - β ^{2}}$ . At rest $γ = 1$ ; it only ever grows.

The scenario matrix

Every problem about $p = γ m v$ falls into one of these cells. The worked examples below are tagged with the cell they cover, and together they fill the whole table.

#	Case class	What is special	Example
A	Everyday / slow ( $β ≲ 0.1$ )	$γ \approx 1$ , Newton nearly exact	Ex 1
B	Zero speed ( $v = 0$ )	degenerate: $γ = 1$ , $p = 0$	Ex 2
C	Moderately relativistic ( $β \sim 0.5$ – $0.9$ )	$γ$ matters a lot	Ex 3
D	Limit $v \to c$	$γ \to \infty$ , $p \to \infty$	Ex 4
E	Massless ( $m = 0$ , $v = c$ )	formula $0 \cdot \infty$ — need $E = p c$	Ex 5
F	Inverse problem (given $p$ , find $v$ )	must invert $γ β$	Ex 6
G	Real-world word problem (accelerator, energy)	mix momentum + energy	Ex 7
H	Exam twist / sign & direction (2-D, vector, negative component)	$γ$ uses speed not signed velocity	Ex 8

Ex 1 — Cell A: everyday slow speed

Worked example A

$2 g$ pellet fired at $300 m/s$ Find its relativistic momentum. Mass $m = 0.002 kg$ , $v = 300 m/s$ , $c = 3 \times 1 0^{8} m/s$ .

Forecast: guess first — will $γ$ differ from $1$ in any way you could ever measure?

Step 1 — Find $β$ . Why this step? $γ$ only depends on speed through $β = v / c$ , so we compute the fraction of light-speed first. $β = \frac{300}{3 \times 1 0 ^{8}} = 1 \times 1 0^{- 6}$

Step 2 — Find $γ$ . Why this step? $γ$ is the correction factor; we test whether it is worth keeping. $γ = \frac{1}{1 - ( 1 0 ^{- 6} ) ^{2}} = \frac{1}{1 - 1 0 ^{- 12}} \approx 1 + \frac{1}{2} \times 1 0^{- 12}$ That is $1.0000000000005$ — indistinguishable from $1$ .

Step 3 — Momentum. Why this step? Plug into $p = γ m v$ ; since $γ \approx 1$ this is just Newton. $p = γ m v \approx (1) (0.002) (300) = 0.6 kg\cdotm/s$

Verify: the relativistic correction is $\sim 5 \times 1 0^{- 13}$ , so $p = 0.6 kg\cdotm/s$ to 12 decimal places. This is the correspondence principle at work — see Newtonian momentum p = mv. At everyday speeds relativity is invisible.

Ex 2 — Cell B: the degenerate case $v = 0$

Ex 3 — Cell C: moderately relativistic (with picture)

Worked example Electron at

$0.6 c$ $m = 9.11 \times 1 0^{- 31} kg$ , $v = 0.6 c$ . Find $p$ and compare with the Newtonian guess.

Forecast: by how many percent will the true momentum beat the Newtonian $m v$ ?

Step 1 — $β = 0.6$ . Why this step? directly given.

Step 2 — $γ$ . Why this step? this is where relativity enters. $γ = \frac{1}{1 - 0.36} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25$

Step 3 — Momentum. Why this step? multiply $γ \cdot m \cdot v$ ; use $v = 0.6 \times 3 \times 1 0^{8} = 1.8 \times 1 0^{8} m/s$ . $p = 1.25 \times 9.11 \times 1 0^{- 31} \times 1.8 \times 1 0^{8} = 2.05 \times 1 0^{- 22} kg\cdotm/s$

Step 4 — Newtonian comparison. Why this step? to see the size of the correction. $p_{Newton} = m v = 9.11 \times 1 0^{- 31} \times 1.8 \times 1 0^{8} = 1.64 \times 1 0^{- 22} kg\cdotm/s$ The ratio is exactly $γ = 1.25$ , i.e. 25% larger.

Verify: look at the figure — the burnt-orange curve $p = γ m v$ pulls above the teal Newton line $p = m v$ as $β$ grows. At $β = 0.6$ the vertical gap is the $25%$ boost. Units of $p$ are $kg\cdotm/s$ throughout. ✓

Ex 4 — Cell D: the limit $v \to c$

Worked example Pushing toward light-speed

Compute $γ$ at $β = 0.9$ , $0.99$ , $0.999$ . What happens to $p$ ?

Forecast: does $γ$ approach some finite ceiling, or run away to infinity?

Step 1 — Table of $γ$ . Why this step? to watch the trend as $β \to 1$ . $γ (0.9) = \frac{1}{1 - 0.81} = \frac{1}{0.19} = 2.294$ $γ (0.99) = \frac{1}{1 - 0.9801} = \frac{1}{0.0199} = 7.089$ $γ (0.999) = \frac{1}{1 - 0.998001} = \frac{1}{0.001999} = 22.37$

Step 2 — Read the pattern. Why this step? the denominator $1 - β^{2} \to 0$ as $β \to 1$ , so $γ \to \infty$ .

Step 3 — Momentum consequence. Why this step? $p = γ m v$ with $γ \to \infty$ means $p \to \infty$ . No finite push can supply infinite momentum, so a massive particle can never reach $c$ — see Why nothing exceeds the speed of light.

Verify: the plum curve in the figure shoots up vertically as it nears $β = 1$ (the dotted asymptote). Each tenfold squeeze of $(1 - β)$ roughly triples $γ$ — a genuine blow-up, not a plateau. ✓

Ex 5 — Cell E: the massless case ( $m = 0$ )

Worked example Momentum of a photon

A photon has rest mass $m = 0$ and always travels at $v = c$ . What is its momentum?

Forecast: the formula gives $p = γ m v = \infty \times 0 \times c$ . Is that $0$ , $\infty$ , or undefined?

Step 1 — Spot the trouble. Why this step? at $v = c$ , $γ = 1/ 1 - 1 = 1/0 = \infty$ ; with $m = 0$ we get the indeterminate product $\infty \cdot 0$ . The formula $p = γ m v$ cannot be used directly for light.

Step 2 — Switch tools to the energy–momentum relation. Why this step? the invariant $E^{2} = (p c)^{2} + (m c^{2})^{2}$ from the parent note stays valid even when $m = 0$ , because it never divides by $1 - β^{2}$ . $E^{2} = (p c)^{2} + (0)^{2} \Rightarrow E = p c \Rightarrow p = \frac{E}{c}$

Step 3 — Numbers. Why this step? a green photon with $E = 3.0 \times 1 0^{- 19} J$ : $p = \frac{3.0 \times 1 0 ^{- 19}}{3 \times 1 0 ^{8}} = 1.0 \times 1 0^{- 27} kg\cdotm/s$

Verify: light carries momentum despite zero mass — this drives Photon momentum and radiation pressure. Units: $J / (m/s) = kg\cdotm^{2} s^{- 2} / (m/s) = kg\cdotm/s$ ✓. The indeterminate form was resolved by choosing the tool that stays finite.

Ex 6 — Cell F: inverse problem (given $p$ , find $v$ )

Worked example Find the speed for a stated momentum

A particle has $p = 3 m c$ . Find $β$ and $γ$ .

Forecast: guess — is $β$ above or below $0.9$ ?

Step 1 — Introduce the clean unknown $γ β$ . Why this step? write $p = γ m v = γ β m c$ . Dividing by $m c$ isolates the pure-number combination $γ β$ , which is the easiest unknown to solve for. (Careful with names: the dimensionful quantity $γ v = γ β c$ is what physicists call proper velocity; our $γ β$ is simply that proper velocity measured in units of $c$ .) $γ β = \frac{p}{m c} = 3$

Step 2 — Square and expand $γ^{2}$ . Why this step? squaring removes the square root inside $γ$ ; use $γ^{2} = 1/ (1 - β^{2})$ . $γ^{2} β^{2} = 3 \Rightarrow \frac{β ^{2}}{1 - β ^{2}} = 3$

Step 3 — Solve for $β$ . Why this step? algebra. $β^{2} = 3 - 3 β^{2} \Rightarrow 4 β^{2} = 3 \Rightarrow β^{2} = \frac{3}{4}$ Taking the square root gives $β = \pm \frac{3}{2}$ . We discard the negative root: by definition $β = v / c$ is a speed fraction, so $β \geq 0$ . Hence $β = \frac{3}{2} \approx 0.866$

Step 4 — Back out $γ$ . Why this step? now that $β$ is known. $γ = \frac{1}{1 - 3/4} = \frac{1}{1/4} = 2$ Check: $γ β = 2 \times \frac{3}{2} = 3$ ✓.

Verify: $β \approx 0.866$ (below $0.9$ , as you may have guessed), $γ = 2$ . Plugging back: $p = γ β m c = 3 m c$ — matches the given. ✓

Ex 7 — Cell G: real-world accelerator problem

Worked example A proton in the LHC-style beam

A proton ( $m = 1.67 \times 1 0^{- 27} kg$ ) is accelerated until its total energy $E = 7 GeV = 1.12 \times 1 0^{- 9} J$ . Its rest energy is $m c^{2} = 1.50 \times 1 0^{- 10} J$ ( $= 0.938 GeV$ ). Find its momentum $p$ .

Forecast: since $E$ is about $7.5 \times$ the rest energy, will $p c$ be very close to $E$ or well below it?

Step 1 — Choose the energy–momentum relation. Why this step? we are given energy, not speed, so the tool $E^{2} = (p c)^{2} + (m c^{2})^{2}$ (see Mass-energy equivalence E = γmc²) reaches $p$ directly without finding $γ$ first. $(p c)^{2} = E^{2} - (m c^{2})^{2}$

Step 2 — Plug in. Why this step? arithmetic with the given numbers. $(p c)^{2} = (1.12 \times 1 0^{- 9})^{2} - (1.50 \times 1 0^{- 10})^{2} = 1.254 \times 1 0^{- 18} - 2.25 \times 1 0^{- 20}$ $(p c)^{2} = 1.232 \times 1 0^{- 18} J^{2} \Rightarrow p c = 1.110 \times 1 0^{- 9} J$

Step 3 — Divide by $c$ . Why this step? $p c$ has energy units; divide by $c$ to get momentum. $p = \frac{1.110 \times 1 0 ^{- 9}}{3 \times 1 0 ^{8}} = 3.70 \times 1 0^{- 18} kg\cdotm/s$

Verify: $p c = 1.110 \times 1 0^{- 9} J$ is only $\sim 0.9%$ below $E = 1.12 \times 1 0^{- 9} J$ — so the proton is ultra-relativistic ( $E \approx p c$ , almost photon-like), exactly what " $7 \times$ rest energy" predicted. Units of $p$ : $J / (m/s) = kg\cdotm/s$ ✓.

Ex 8 — Cell H: exam twist — direction & sign

Worked example Two-component (2-D) momentum

An electron moves with velocity components $v_{x} = + 0.6 c$ , $v_{y} = - 0.6 c$ . Find the magnitude of its relativistic momentum. Note $m = 9.11 \times 1 0^{- 31} kg$ .

Forecast: the trap — do you compute $γ$ using $v_{x}$ alone, or using the total speed?

Step 1 — Find the true speed (not a single component). Why this step? $γ$ depends on the actual speed $v = v_{x}^{2} + v_{y}^{2}$ , because $γ$ measures how fast the clock ticks and that cares only about total motion, not direction or sign. $v = (0.6 c)^{2} + (0.6 c)^{2} = 0.6 c 2 = 0.8485 c$ So $β = 0.8485$ — note the minus sign on $v_{y}$ squared away; direction never enters $γ$ .

Step 2 — Compute $γ$ from the true speed. Why this step? using $β = 0.6$ here would be the classic exam mistake. $γ = \frac{1}{1 - 0.848 5 ^{2}} = \frac{1}{1 - 0.72} = \frac{1}{0.28} = 1.890$

Step 3 — Build each momentum component. Why this step? the vector formula $p = γ m v$ multiplies the same $γ$ onto each signed velocity component. $p_{x} = γ m v_{x} = 1.890 \times 9.11 \times 1 0^{- 31} \times (+ 1.8 \times 1 0^{8}) = + 3.10 \times 1 0^{- 22}$ $p_{y} = γ m v_{y} = 1.890 \times 9.11 \times 1 0^{- 31} \times (- 1.8 \times 1 0^{8}) = - 3.10 \times 1 0^{- 22}$ (units $kg\cdotm/s$ ; $p_{y}$ is negative because $v_{y}$ is.)

Step 4 — Magnitude. Why this step? combine the perpendicular components. $∣ p ∣ = p_{x}^{2} + p_{y}^{2} = 3.10 \times 1 0^{- 22} \times 2 = 4.38 \times 1 0^{- 22} kg\cdotm/s$

Verify: shortcut check — $∣ p ∣ = γ m v = 1.890 \times 9.11 \times 1 0^{- 31} \times (0.8485 \times 3 \times 1 0^{8}) = 4.38 \times 1 0^{- 22}$ ✓, matching the component route. The single $γ$ is set by speed; the signs live only in the components. See Energy-momentum four-vector for why the whole $p$ transforms as one object.

Recall Which speed sets

$γ$ in a 2-D problem? The total speed $v = v_{x}^{2} + v_{y}^{2}$ ::: not any single component; $γ$ ignores direction and sign.

Recall Why can't

$p = γ m v$ be used for a photon? Because $m = 0$ and $v = c$ give $\infty \times 0 \times c$ (indeterminate) ::: use $E = p c$ instead.

Recall Given

$p = 3 m c$ , what are $β$ and $γ$ ? $β = 3 /2 \approx 0.866$ , $γ = 2$ ::: solve $γ β = 3$ .

Connections

Parent: Relativistic momentum $p=\gamma m v$
Lorentz factor and time dilation — where $γ$ comes from
Mass-energy equivalence E = γmc² — used in Ex 5 & 7
Energy-momentum four-vector — the vector view of Ex 8
Why nothing exceeds the speed of light — the $v \to c$ limit of Ex 4
Newtonian momentum p = mv — the slow-speed cell A
Photon momentum and radiation pressure — the massless cell E