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

This page is your self-test. Read the problem, cover the solution, try it yourself, then open the collapsible to check. The problems climb five levels — from just recognising the pieces, up to combining everything into a single relativistic scenario.

Definition The symbols and constants you need on this page

$c$ = speed of light, the fixed conversion factor between space and time. Numerically $c = 3 \times 1 0^{8} m/s$ . It appears everywhere because it turns a mass into an energy ( $m c^{2}$ ) and a momentum into an energy ( $p c$ ).
$E$ = total relativistic energy (rest + kinetic).
$p c$ = the momentum $p$ multiplied by $c$ , so it has energy units.
$m c^{2}$ = rest energy, built from the invariant rest mass $m$ .
$K E$ = kinetic energy, the motion-only part of the energy — what is left over after you subtract the rest energy from the total: $K E = E - m c^{2} = (γ - 1) m c^{2}$ . At low speed it reduces to the Newtonian $\frac{1}{2} m v^{2}$ .
$γ$ = Lorentz factor, defined as $γ = \frac{1}{1 - v ^{2} / c ^{2}}$ . It measures how much relativity "stretches" energy and momentum: at rest $γ = 1$ ; as $v \to c$ , $γ \to \infty$ . It links the triangle to speed through $E = γ m c^{2}$ and $p = γ m v$ . See Lorentz Factor.

Figure — Mass-energy equivalence E² = (pc)² + (mc²)²

Figure s01 — The master triangle. Red hypotenuse $E$ (total energy); bottom black leg $p c$ (motion); right black leg $m c^{2}$ (rest energy, fixed); corner square = right angle. Every exercise below is a way of finding one side of this triangle from the other two.

Before we start, a units reminder that removes 90% of arithmetic pain:

Level 1 — Recognition

Recall Solution 1.1

What triangle? The horizontal leg $p c = 0$ , so the triangle collapses to a vertical line. The hypotenuse equals the vertical leg. $E = (0)^{2} + (938)^{2} = 938 MeV$ This is the famous $E = m c^{2}$ — but only because $p = 0$ . Answer: $E = 938 MeV$ .

Recall Solution 1.2

What triangle? The vertical leg $m c^{2} = 0$ , so the triangle collapses to a horizontal line. The hypotenuse equals the horizontal leg. $E = (2.5)^{2} + (0)^{2} = 2.5 MeV$ This is $E = p c$ — the massless limit. Answer: $E = 2.5 MeV$ . Every photon obeys this; see Photon Momentum.

Recall Solution 1.3

What is being asked? Which quantity is invariant (frame-independent). $E$ changes if you run alongside the particle (its energy looks different). $p c$ changes too (momentum depends on your motion). But $m c^{2}$ — the rest energy — is built from the rest mass $m$ , which by definition is measured in the particle's own rest frame and is the same for everyone. Answer: $m c^{2}$ (the rest energy) is invariant. Deeper reason in Four-Momentum.

Level 2 — Application

Recall Solution 2.1

Step 1 (WHAT): Both legs are known, so take the hypotenuse. $E = (500)^{2} + (938)^{2} = 250000 + 880344 = 1130344 \approx 1063.2 MeV$ Step 2 (WHY subtract): Total energy contains the rest energy plus the motion energy. Kinetic energy $K E$ (defined in the symbol list) is whatever is left after removing the rest part: $K E = E - m c^{2} = 1063.2 - 938 = 125.2 MeV$ Answer: $E \approx 1063 MeV$ , $K E \approx 125 MeV$ . See Relativistic Kinetic Energy.

Recall Solution 2.2

Which leg is missing? We know the hypotenuse $E$ and the vertical leg $m c^{2}$ ; we want the horizontal leg $p c$ . Rearrange Pythagoras: $p c = E^{2} - (m c^{2})^{2} = (1.022)^{2} - (0.511)^{2}$ $= 1.044484 - 0.261121 = 0.783363 \approx 0.885 MeV$ Answer: $p c \approx 0.885 MeV$ . (Neat check: here $E = 2 m c^{2}$ , so this electron's total energy is exactly twice its rest energy.)

Recall Solution 2.3

Step 1 (WHY $E = p c$ ): A photon is massless, so the vertical leg vanishes and $E = p c$ , hence $p = E / c$ . Step 2 (convert eV → joules): $E = 2.0 \times 1.6 \times 1 0^{- 19} = 3.2 \times 1 0^{- 19} J$ . Step 3 (divide by $c$ ): $p = \frac{E}{c} = \frac{3.2 \times 1 0 ^{- 19}}{3 \times 1 0 ^{8}} \approx 1.07 \times 1 0^{- 27} kg\cdotm/s$ Answer: $p \approx 1.07 \times 1 0^{- 27} kg\cdotm/s$ .

Level 3 — Analysis

Recall Solution 3.1

(a) Missing horizontal leg: $p c = E^{2} - (m c^{2})^{2} = 130 0^{2} - 93 8^{2} = 1690000 - 880344 = 809656 \approx 899.8 MeV$ (b) WHY this next step: We need speed, and the bridge between momentum-energy and speed is the pair $E = γ m c^{2}$ , $p c = γ m v c$ , where $γ = 1/ 1 - v^{2} / c^{2}$ is the Lorentz factor from the definitions box above. Divide them so $γ$ and $m$ cancel: $\frac{p c}{E} = \frac{γ m v c}{γ m c ^{2}} = \frac{v}{c}$ This ratio $p c / E$ directly equals $v / c$ — a beautifully clean result. So: $\frac{v}{c} = \frac{899.8}{1300} \approx 0.692$ Answer: $p c \approx 900 MeV$ , $v \approx 0.692 c$ . The Lorentz factor here is $\gamma$ $= E / (m c^{2}) = 1300/938 \approx 1.386$ .

Recall Solution 3.2

What we use: For $m = 0$ , the master relation gives $E = p c$ , so $p c / E = 1$ . Interpretation: From Exercise 3.1 we found $p c / E = v / c$ . Setting this equal to 1 gives $v / c = 1$ , i.e. $v = c$ . Conclusion: Massless particles must travel at exactly $c$ — no faster, no slower. Answer: $p c / E = 1 \Rightarrow v = c$ .

Recall Solution 3.3

Step 1 (total energy): $E = K E + m c^{2} = m c^{2} + m c^{2} = 2 m c^{2}$ . Step 2 (momentum): plug into Pythagoras and isolate the horizontal leg. $p c = E^{2} - (m c^{2})^{2} = (2 m c^{2})^{2} - (m c^{2})^{2} = 4 (m c^{2})^{2} - (m c^{2})^{2} = 3 m c^{2}$ Answer: $E = 2 m c^{2}$ , $p c = 3 m c^{2} \approx 1.732 m c^{2}$ . (Speed check: $v / c = p c / E = 3 /2 \approx 0.866$ .)

Level 4 — Synthesis

Recall Solution 4.1

WHAT the minimum means: At threshold, the two created particles are produced with the least possible energy — that is, at rest, each carrying only its rest energy. Energy bookkeeping: the photon's energy must at least supply both rest energies: $E_{γ}^{m i n} = m c_{e^{-}}^{2} + m c_{e^{+}}^{2} = 0.511 + 0.511 = 1.022 MeV$ Answer: $E_{γ}^{m i n} = 1.022 MeV$ . (In reality a stray nucleus must absorb momentum, nudging the true threshold slightly higher, but the energy floor is $1.022 MeV$ .) This is a mass–energy conversion in action.

Recall Solution 4.2

WHY energy appears: Rest mass is not conserved in nuclear reactions; the lost rest mass reappears as kinetic energy of the products. The released energy $Q$ is the drop in rest energy: $Q = M_{i} c^{2} - M_{f} c^{2} = 3752.0 - 3728.4 = 23.6 MeV$ Answer: $Q = 23.6 MeV$ released. The vertical (rest) legs of the "before" and "after" triangles shrank; that missing rest energy became motion energy (longer horizontal legs). See Nuclear Binding Energy.

Recall Solution 4.3

Sign setup: take $+ x$ (the photon's direction) as positive. The photon carries positive $p c$ ; the atom starts with $p c = 0$ . Conserve energy (WHAT): total energy after = rest energy of atom + photon energy. $E = M c^{2} + E_{γ} = 6000 + 10 = 6010 MeV$ Conserve momentum (WHY separate law): the atom was at rest ( $p c = 0$ ), the photon carried $p c = + E_{γ} = + 10 MeV$ (massless). Adding signed momenta, all of it transfers to the atom, which moves in $+ x$ : $(p c)_{atom} = + 10 MeV$ Check the master relation — the atom's new rest energy $m^{'} c^{2}$ (only the magnitude of $p c$ enters, since it is squared): $m^{'} c^{2} = E^{2} - (p c)^{2} = 601 0^{2} - 1 0^{2} = 36120100 - 100 = 36120000 \approx 6009.99 MeV$ Answer: $E = 6010 MeV$ , $p c = + 10 MeV$ , and the excited atom's rest energy $\approx 6010 MeV$ — the absorbed energy went almost entirely into internal (rest) energy, with a tiny sliver as recoil kinetic energy.

Level 5 — Mastery

Recall Solution 5.1

(a) Total energy — energies are always positive and simply add: $E_{tot} = 511 + 511 = 1022 keV = 1.022 MeV$ (b) Total momentum (WHERE THE SIGNS BITE) — with $+ x$ positive: the photon moving right has $p c = + 511 keV$ ; the photon moving left has $p c = - 511 keV$ . We add them with their signs before squaring (opposite directions subtract): $(p_{tot} c) = (+ 511) + (- 511) = 0 keV$ Note we did not add magnitudes (which would wrongly give $1022$ ); the leftward photon's negative sign cancels the rightward one. (c) Invariant mass — now square the totals: $M c^{2} = E_{tot}^{2} - (p_{tot} c)^{2} = (1022)^{2} - (0)^{2} = 1022 keV = 1.022 MeV$ which matches two electron rest energies, $2 \times 0.511 MeV = 1.022 MeV$ — as it must, since this is what annihilated. WHY non-zero from massless pieces: invariant mass is not the sum of individual masses. It is built from total $E$ and total $p$ . Because the photons' signed momenta cancelled to zero, the system as a whole has energy but no net momentum — exactly like a massive object at rest. The system therefore behaves as if it has mass $1.022 MeV / c^{2}$ . Answer: $E_{tot} = 1.022 MeV$ , $p_{tot} c = 0$ , $M c^{2} = 1.022 MeV$ . This is the heart of Four-Momentum.

Recall Solution 5.2

(a) Exact: $E = (50000)^{2} + (938)^{2} = 2.5 \times 1 0^{9} + 879844 \approx 50008.80 MeV$ (b) WHY this approximation: When $p c ≫ m c^{2}$ , the rest leg is tiny compared with the momentum leg. Factor $p c$ out of the square root: $E = p c 1 + (\frac{m c ^{2}}{p c})^{2} \approx p c (1 + \frac{1}{2} (\frac{m c ^{2}}{p c})^{2}) = p c + \frac{( m c ^{2} ) ^{2}}{2 p c}$ using the binomial expansion $1 + x \approx 1 + \frac{x}{2}$ for tiny $x$ . Numerically: $E \approx 50000 + \frac{93 8 ^{2}}{2 \times 50000} = 50000 + \frac{879844}{100000} = 50000 + 8.798 = 50008.80 MeV$ Fractional error: the exact and approximate values agree to better than $1$ part in $1 0^{9}$ . Answer: $E \approx 50008.80 MeV$ ; approximation error $\sim 1 0^{- 9}$ .

Recall Solution 5.3

(a) Rest energy (invariant leg): $m c^{2} = E^{2} - (p c)^{2} = 250 0^{2} - 240 0^{2} = 6250000 - 5760000 = 490000 = 700 MeV$ (b) Speed via the golden ratio $v / c = p c / E$ (only the magnitude of $p c$ is needed for the speed): $\frac{v}{c} = \frac{2400}{2500} = 0.96$ (c) Lorentz factor via $γ = E / (m c^{2})$ : $γ = \frac{2500}{700} \approx 3.571$ Cross-check using the definition $γ = 1/ 1 - v^{2} / c^{2}$ : $γ = \frac{1}{1 - 0.9 6 ^{2}} = \frac{1}{1 - 0.9216} = \frac{1}{0.0784} = \frac{1}{0.28} \approx 3.571 ✓$ Answer: $m c^{2} = 700 MeV$ , $v = 0.96 c$ , $γ \approx 3.571$ .

Connections

Parent topic
Special Relativity · Lorentz Factor · Photon Momentum
Relativistic Kinetic Energy · Four-Momentum · Nuclear Binding Energy