Mass-energy equivalence E² = (pc)² + (mc²)²: Worked examples

This is the practice arena for the parent note Mass-energy equivalence. There we derived the master formula. Here we use it on every kind of input the world (and your exam) can throw at you.

Definition The four symbols — a 30-second recap

Before any example, pin down exactly what each letter means (the parent note builds these in full):

$E$ = total relativistic energy of a particle (rest energy + kinetic energy). Always a non-negative number, in joules (J) or, in particle physics, electron-volts.
$p$ = magnitude of the momentum vector $∣ p ∣$ . Because it is a length of a vector, $p$ is always a non-negative scalar ( $p \geq 0$ ). If a particle moves left instead of right, only the direction of $p$ flips — its magnitude $p$ (and hence $p c$ ) is unchanged. That is why only $p \geq 0$ ever enters the formula.
$m$ = rest mass (invariant mass): the mass measured in the particle's own rest frame, the same in every frame. Units: kg.
$c$ = the speed of light in vacuum, $c \approx 3 \times 1 0^{8}$ m/s. It is a fixed constant of nature; multiplying by $c$ or $c^{2}$ is what converts mass and momentum into the same units as energy.

Read the master triangle below (Figure s01): the coral segment along the bottom is the momentum-energy leg $p c$ ; the lavender segment on the right is the fixed rest-energy leg $m c^{2}$ ; the dark slate slanted segment is the hypotenuse $E$ . As you move through the examples, picture which of these two legs shrinks to zero or dominates — that is literally what each "case class" does to this one triangle.

The scenario matrix

Before solving anything, let us list every distinct situation this formula can produce. Each row is a "case class". The worked examples afterward each hit one or more cells, and together they cover the whole table.

#	Case class	What is special	Which leg dominates / degenerates	Example
C0	Fully degenerate ( $m = 0$ and $p = 0$ )	no mass, no motion	both legs $= 0$ → $E = 0$ (no particle at all)	Ex 0
C1	Pure rest ( $p = 0$ )	no motion	$p c$ leg $= 0$ → $E = m c^{2}$	Ex 1
C2	Massless ( $m = 0$ )	photon	$m c^{2}$ leg $= 0$ → $E = p c$	Ex 2
C3	General massive	both legs nonzero	triangle with two real legs	Ex 3
C4	Non-relativistic limit ( $p c ≪ m c^{2}$ )	slow particle	tiny $p c$ leg → recover $\frac{1}{2} m v^{2}$	Ex 4
C5	Ultra-relativistic limit ( $p c ≫ m c^{2}$ )	near light-speed	huge $p c$ leg → $E \approx p c$	Ex 5
C6	Solve backwards (given $E$ , $m$ → find $p$ )	invert the formula	rearrange for a leg	Ex 6
C7	Real-world word problem	mass defect / binding	$Δ m c^{2}$ released	Ex 7
C8	Exam twist (find $v$ from $E, m$ )	chain through $γ$	combine formula + $γ$	Ex 8

Recall Quick self-test: which cell?

A neutron at rest decays. Which case describes "energy of the neutron before decay"? ::: C1 — pure rest, $E = m c^{2}$ . A gamma ray hits a detector. Which case? ::: C2 — massless, $E = p c$ . Both legs zero — what is left? ::: C0 — $E = 0$ , i.e. no particle at all.

Example 0 — Cell C0: the fully degenerate corner ( $m = 0$ and $p = 0$ )

Worked example What if BOTH legs vanish?

Suppose someone hands you the impossible-looking case: a "thing" with zero rest mass and zero momentum. What total energy does the master formula assign?

Forecast: If the vertical leg $m c^{2} = 0$ and the horizontal leg $p c = 0$ , the triangle shrinks to a single point. A triangle with both legs zero has hypotenuse zero, so $E = 0$ .

Step 1 — Collapse both legs simultaneously. $E^{2} = (0)^{2} + (0)^{2} = 0 \Rightarrow E = 0$ Why this step? This is the only input for which even the positive root gives zero — both the momentum-energy and rest-energy contributions are absent.

Step 2 — Interpret physically. $E = 0$ means there is no particle at all: a real photon must move at $c$ and therefore always has $p > 0$ (so $p c \neq = 0$ ), and any particle with mass has $m c^{2} \neq = 0$ . So the $m = 0, p = 0$ corner is not a physical object — it is the empty limit that completes the list of every possible triangle collapse. Why this step? It closes the case table: C1 kills one leg, C2 kills the other, C0 kills both. There is nothing left to enumerate.

Verify: $0 + 0 = 0 \geq 0$ , consistent with $E \geq 0$ . No real photon or massive particle can occupy this cell. ✓

Example 1 — Cell C1: pure rest energy

Worked example Rest energy of a proton

A proton sits perfectly still ( $p = 0$ ). Its rest mass is $m = 1.673 \times 1 0^{- 27}$ kg. Find its energy in joules and in MeV. (Use $c = 3 \times 1 0^{8}$ m/s.)

Forecast: With no momentum the horizontal leg vanishes, so the triangle collapses to a vertical line — pure $E = m c^{2}$ . Since a proton is roughly 1836 electrons, expect around $938$ MeV.

Step 1 — Kill the momentum leg. Set $p = 0$ and take the positive root: $E^{2} = (0)^{2} + (m c^{2})^{2} \Rightarrow E = + m c^{2}$ Why this step? When $p = 0$ , the $(p c)^{2}$ term is exactly zero; only the rest-energy leg survives. We keep the positive root because energy is non-negative.

Step 2 — Plug numbers (SI). $E = (1.673 \times 1 0^{- 27}) (3 \times 1 0^{8})^{2} = 1.506 \times 1 0^{- 10} J$ Why this step? $c = 3 \times 1 0^{8}$ m/s is the speed of light, so $c^{2} = 9 \times 1 0^{16}$ m $^{2}$ /s $^{2}$ ; multiply by the mass.

Step 3 — Convert to MeV. Divide by $1.602 \times 1 0^{- 13}$ J/MeV: $E = \frac{1.506 \times 1 0 ^{- 10}}{1.602 \times 1 0 ^{- 13}} \approx 940 MeV$ Why this step? $1 MeV = 1.602 \times 1 0^{- 13}$ J; particle physicists always quote rest energies in MeV. (The tiny gap from the book value $938$ MeV comes from rounding $c$ to $3 \times 1 0^{8}$ .)

Verify: $\sim 938$ – $940$ MeV is the textbook proton rest energy. Units: (kg)(m/s) $^{2}$ = kg·m $^{2}$ /s $^{2}$ = J. ✓

Example 2 — Cell C2: massless photon

Worked example Momentum of a gamma-ray photon

A gamma photon has energy $E = 1.0$ MeV. It has no rest mass. Find its momentum $p$ . (Use $c = 3 \times 1 0^{8}$ m/s.)

Forecast: No rest leg → the hypotenuse equals the momentum leg, so $E = p c$ exactly. Expect $p = E / c$ .

Step 1 — Kill the rest leg. Set $m = 0$ and take the positive root: $E^{2} = (p c)^{2} + 0 \Rightarrow E = + p c \Rightarrow p = \frac{E}{c}$ Why this step? A massless particle has $m c^{2} = 0$ ; the triangle degenerates to a flat horizontal line. Since $p \geq 0$ and $E \geq 0$ , the positive root is the physical one.

Step 2 — Convert energy to joules. $E = 1.0 MeV = 1.6 \times 1 0^{- 13} J$ Why this step? Momentum in SI needs energy in joules.

Step 3 — Divide by $c$ . $p = \frac{1.6 \times 1 0 ^{- 13}}{3 \times 1 0 ^{8}} = 5.33 \times 1 0^{- 22} kg\cdotm/s$ Why this step? $p = E / c$ ; units J/(m/s) = kg·m/s. ✓ ( $p$ is the magnitude, a non-negative scalar.)

Verify: See Photon Momentum — a real photon does carry momentum even with zero mass. Newton's $p = m v$ would wrongly give zero here.

Example 3 — Cell C3: general massive particle (the full triangle)

Worked example Total energy and kinetic energy of a fast electron

An electron ( $m c^{2} = 0.511$ MeV) has momentum $p c = 1.0$ MeV. Find total energy $E$ and kinetic energy $K E$ .

Forecast: Both legs are comparable ( $1.0$ vs $0.511$ ), so this is a genuine two-legged triangle. Expect $E$ a bit above $1.1$ MeV.

Step 1 — Draw the triangle and use Pythagoras. $E = (p c)^{2} + (m c^{2})^{2} = 1. 0^{2} + 0.51 1^{2}$ Why this step? Both legs nonzero → use the full formula, nothing collapses. Positive root, as always for energy.

Step 2 — Compute. $E = 1.0 + 0.261 = 1.261 = 1.123 MeV$ Why this step? Square each leg, add, take the root — literally the hypotenuse.

Step 3 — Strip out the rest energy for $K E$ . $K E = E - m c^{2} = 1.123 - 0.511 = 0.612 MeV$ Why this step? Total energy = rest + kinetic; whatever exceeds $m c^{2}$ is the kinetic part.

Verify: Sanity check the triangle: hypotenuse $1.123 >$ each leg ( $1.0$ , $0.511$ ). ✓ And $K E > 0$ as expected for a moving particle.

Read Figure s02 alongside this example: it draws this exact triangle to scale — the coral base is the $p c = 1.0$ MeV momentum leg, the lavender upright is the $m c^{2} = 0.511$ MeV rest leg, and the slate hypotenuse is labelled with the computed $E = 1.123$ MeV. The mint annotation shows the leftover slice $K E = E - m c^{2} = 0.612$ MeV. Notice visually that the hypotenuse is only slightly longer than the base — that is why $E$ ( $1.123$ ) sits just above $p c$ ( $1.0$ ).

The tool we lean on twice: the binomial approximation

Example 4 — Cell C4: non-relativistic limit (slow particle)

Worked example Show the formula recovers Newton for a slow proton

A proton ( $m c^{2} = 938$ MeV) has a tiny momentum $p c = 9.38$ MeV (so $p c / m c^{2} = 0.01$ ). Find $K E$ and show it becomes the Newtonian $\frac{1}{2} m v^{2}$ .

Forecast: When the momentum leg is much smaller than the rest leg, the hypotenuse barely grows. Expect $K E \approx \frac{( p c ) ^{2}}{2 m c ^{2}}$ , which — once we insert the slow-speed relation $p = m v$ — should collapse into the familiar $\frac{1}{2} m v^{2}$ .

Step 1 — Expand the square root for a small leg. Write $x = p c / m c^{2} = 0.01$ , so $u = x^{2}$ is the small quantity in $1 + u$ : $E = m c^{2} 1 + x^{2} \approx m c^{2} (1 + \frac{1}{2} x^{2})$ Why this step? This is exactly the binomial tool stated above with $u = x^{2}$ ; the neglected term is of order $x^{4}$ , utterly negligible for $x = 0.01$ .

Step 2 — Read off the kinetic energy. $K E = E - m c^{2} \approx \frac{1}{2} m c^{2} x^{2} = \frac{( p c ) ^{2}}{2 m c ^{2}} = \frac{p ^{2}}{2 m}$ Why this step? Subtracting $m c^{2}$ leaves the leftover $\frac{1}{2} m c^{2} x^{2}$ ; substituting $x = p c / m c^{2}$ cancels the $c$ 's to give $\frac{p ^{2}}{2 m}$ .

Step 3 — Insert the slow-speed momentum $p = m v$ to reach Newton. At $v ≪ c$ we have $γ \to 1$ (recall $γ = 1/ 1 - v^{2} / c^{2}$ , defined at the top), so relativistic momentum $p = γ m v$ reduces to the Newtonian $p = m v$ . Substitute: $K E \approx \frac{p ^{2}}{2 m} = \frac{( m v ) ^{2}}{2 m} = \frac{1}{2} m v^{2}$ Why this step? This is the actual bridge to Newton: only after replacing $p$ by $m v$ does $\frac{p ^{2}}{2 m}$ literally become the schoolbook $\frac{1}{2} m v^{2}$ .

Step 4 — Number-check against the exact formula. $K E_{approx} = \frac{9.3 8 ^{2}}{2 \times 938} = 0.0469 MeV, K E_{exact} = 9.3 8^{2} + 93 8^{2} - 938 = 0.0469 MeV$ Why this step? Confirms the binomial approximation loses nothing at these speeds.

Verify: Approximate and exact $K E$ agree to 4 decimals, and the algebra reduces to $\frac{1}{2} m v^{2}$ . ✓ Newton is just Einstein's slow-speed corner.

Example 5 — Cell C5: ultra-relativistic limit (near light-speed)

Worked example A 50 GeV electron at a collider

An electron ( $m c^{2} = 0.511$ MeV $= 0.000511$ GeV) is accelerated so its momentum is $p c = 50$ GeV. Find $E$ , and by how much it differs from $p c$ .

Forecast: Now the momentum leg dwarfs the rest leg by ~100 000×. The triangle is almost a flat line: $E \approx p c$ . Expect $E$ barely above $50$ GeV.

Step 1 — Full formula (positive root). $E = + (50)^{2} + (0.000511)^{2} GeV$ Why this step? Even in the ultra-relativistic case we start exact, then see the small leg drop out.

Step 2 — Approximate by factoring out the large leg. Since $m c^{2} ≪ p c$ , set $u = (m c^{2} / p c)^{2}$ (small) and use the binomial tool again: $E = p c 1 + u \approx p c (1 + \frac{1}{2} u) \approx 50 + \frac{0.00051 1 ^{2}}{2 \times 50} \approx 50 + 2.6 \times 1 0^{- 9} GeV$ Why this step? Same binomial approximation, now with $u = (m c^{2} / p c)^{2}$ ; neglected error is order $u^{2} \sim 1 0^{- 18}$ .

Step 3 — Interpret. The rest-mass correction is $\sim 1 0^{- 9}$ GeV — utterly negligible. Why this step? At colliders, $E \approx p c$ is used constantly; mass is "forgotten" at high energy.

Verify: $E - p c = 2.6 \times 1 0^{- 9}$ GeV $> 0$ (hypotenuse always exceeds a leg), and vanishingly small. ✓

Example 6 — Cell C6: solve backwards (given $E$ and $m$ , find $p$ )

Worked example Invert the formula

A muon has rest energy $m c^{2} = 105.7$ MeV and total energy $E = 200$ MeV. Find its momentum $p c$ .

Forecast: We know the hypotenuse and one leg; solve for the other leg. Expect $p c = E^{2} - (m c^{2})^{2}$ , somewhere near $170$ MeV.

Step 1 — Rearrange the master formula for the $p c$ leg. $(p c)^{2} = E^{2} - (m c^{2})^{2} \Rightarrow p c = + E^{2} - (m c^{2})^{2}$ Why this step? Pythagoras also runs backwards: a leg = $hyp^{2} - other leg^{2}$ . We take the positive root because $p \geq 0$ (it is a magnitude).

Step 2 — Plug in. $p c = 20 0^{2} - 105. 7^{2} = 40000 - 11172.5 = 28827.5 = 169.8 MeV$ Why this step? Direct substitution; keep everything in MeV so units cancel cleanly.

Step 3 — Degeneracy check. Is $E \geq m c^{2}$ ? Yes ( $200 > 105.7$ ), so the square root is real. Why this step? If someone gave $E < m c^{2}$ the answer would be imaginary — physically impossible, total energy can never dip below rest energy.

Verify: Reassemble: $169. 8^{2} + 105. 7^{2} = 28832 + 11172 \approx 200$ MeV. ✓

Example 7 — Cell C7: real-world word problem (mass defect)

Worked example Energy released fusing hydrogen into helium

When four protons fuse into one helium nucleus, the products are lighter than the reactants by $Δ m = 4.8 \times 1 0^{- 29}$ kg (the "mass defect"). How much energy is released per fusion? (Use $c = 3 \times 1 0^{8}$ m/s.)

Forecast: The lost mass reappears as energy via the pure-rest form $E = (Δ m) c^{2}$ . Tiny mass, but $c^{2}$ is enormous — expect a few MeV.

Step 1 — Recognise the case. The missing mass converts entirely to energy at rest (photons + kinetic), so use $E = Δ m c^{2}$ (C1-type, applied to the difference). Why this step? Binding energy = mass defect × $c^{2}$ ; no momentum term needed for the released total.

Step 2 — Compute in joules. $E = (4.8 \times 1 0^{- 29}) (3 \times 1 0^{8})^{2} = 4.32 \times 1 0^{- 12} J$ Why this step? Same $m c^{2}$ machine as Example 1, now with $Δ m$ and $c$ the speed of light.

Step 3 — Convert to MeV. $E = \frac{4.32 \times 1 0 ^{- 12}}{1.6 \times 1 0 ^{- 13}} = 27 MeV$ Why this step? Nuclear energies are quoted in MeV; this ~27 MeV per helium is what powers the Sun.

Verify: Known solar fusion releases ≈26–27 MeV per helium-4. ✓ Units: kg·(m/s) $^{2}$ = J. ✓

Example 8 — Cell C8: exam twist (find the speed)

Worked example From energy to velocity

A particle has total energy $E = 2 m c^{2}$ (twice its rest energy). Find its speed $v$ as a fraction of $c$ .

Forecast: $E = γ m c^{2}$ , so $γ = 2$ . A large $γ$ means high speed; expect $v$ somewhere around $0.87 c$ .

Step 1 — Relate $E$ to the Lorentz factor. Since $E = γ m c^{2}$ and $E = 2 m c^{2}$ : $γ = 2$ Why this step? The Lorentz factor $γ$ (defined at the top of this page) links total energy to rest energy directly.

Step 2 — Invert $γ$ for $v$ . From $γ = \frac{1}{1 - v ^{2} / c ^{2}}$ : $1 - \frac{v ^{2}}{c ^{2}} = \frac{1}{γ ^{2}} = \frac{1}{4} \Rightarrow \frac{v ^{2}}{c ^{2}} = \frac{3}{4}$ Why this step? Solve the definition of $γ$ algebraically for $v / c$ .

Step 3 — Take the root. $\frac{v}{c} = \frac{3}{4} = 0.866 \Rightarrow v \approx 0.866 c$ Why this step? $v / c$ is between 0 and 1, as required for a massive particle — it never reaches $c$ (the speed of light).

Cross-check with momentum leg: $p c = E^{2} - (m c^{2})^{2} = 4 - 1 m c^{2} = 3 m c^{2}$ , and $p = γ m v$ gives $p c = 2 \cdot 0.866 m c^{2} = 1.732 m c^{2} = 3 m c^{2}$ . ✓

Verify: $γ = 2$ , $v = 0.866 c$ , and both routes to $p c$ agree at $3 m c^{2}$ . ✓

Wrapping up the matrix

Recall Did we cover every cell?

C0 fully degenerate? ::: Ex 0 (both legs zero → E = 0, no particle). C1 rest? ::: Ex 1 (proton rest energy). C2 massless? ::: Ex 2 (photon momentum). C3 general? ::: Ex 3 (fast electron). C4 non-relativistic limit? ::: Ex 4 (slow proton → Newton). C5 ultra-relativistic limit? ::: Ex 5 (50 GeV electron). C6 solve backwards? ::: Ex 6 (muon momentum). C7 real-world / mass defect? ::: Ex 7 (fusion energy). C8 exam twist (find v)? ::: Ex 8 (E = 2mc² → 0.866c).

Connections

Parent · Mass-energy equivalence
Photon Momentum — Example 2's $p = E / c$
Relativistic Kinetic Energy — Examples 3, 4
Lorentz Factor — Example 8's $γ$
Nuclear Binding Energy — Example 7's mass defect
Special Relativity — the framework behind all of it
Four-Momentum — the deeper 4-vector view