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Derivation

Mass-Energy Equivalence E = mc^2

D-099 Home PU-105 Threads energy · matter · symmetry Depends on Relativistic Momentum from Conservation
Statement

For a body of invariant (rest) mass m moving with speed v in an inertial frame, requiring that the relativistic momentum p = γmv be conserved in every inertial frame forces the existence of a conserved scalar E = γmc2, where γ = (1 − v2/c2)−1/2. In the rest frame this reduces to E0 = mc2, the rest energy, so that a body possesses energy purely by virtue of its mass.

Why it matters

Mass-energy equivalence dissolves the pre-relativistic separation between matter and energy. Rest mass is a form of stored energy that can be released (fission, fusion, annihilation) or created (pair production, particle collisions), and the same constant c2 that appears in the kinematics of space and time reappears as the exchange rate between mass and energy.

Operationally it is the accounting rule for every nuclear reaction, every particle-physics collision, and stellar energetics: the mass deficit between reactants and products, multiplied by c2, is the energy budget. It also fixes the correct low-speed kinetic energy, recovering the Newtonian ½mv2 as the leading correction to mc2.

Assumptions
The principle of relativity holds.If the laws of mechanics were not identical in all inertial frames, a conserved quantity in one frame need not be conserved in another, and the argument that pins down E as the time-component partner of momentum collapses. Momentum p = γmv is conserved in every inertial frame (the assumed prior result).Without frame-independent conservation of this specific momentum there is no Lorentz-covariant quantity whose conservation would demand a companion scalar; the whole derivation has no anchor. The invariant mass m is a Lorentz scalar characterising the body.If m itself transformed between frames, "rest energy" would not be a well-defined frame-independent property and E0 = mc2 would carry no invariant meaning.
Spacetime is flat (special relativity) and the body is free between interactions.In curved spacetime or under external potentials there is no single global inertial frame, and total energy must be redefined (e.g. including gravitational potential energy); the clean split into γmc2 holds only locally.
The energy-momentum four-vector transforms as a Lorentz four-vector.If (E/c, p) did not transform contravariantly under boosts, its "length" E2 − p2c2 would not be invariant and could not equal the frame-independent m2c4.
Derivation
1
p = γmu,  γ = (1 − u2/c2)−1/2
Take the relativistic momentum as the conserved vector supplied by the prior result; u is the body's velocity, u = |u|. A
2
F = dp/dt = d(γmv)/dt
Define force as the rate of change of relativistic momentum along the line of motion (one-dimensional case, u = v x̂), the natural covariant generalisation of Newton's second law. A
3
dW = F dx = v d(γmv)
The work done by the force over displacement dx is F dx; use dx = v dt to convert. Work-energy is the bridge from momentum to energy. A
4
v d(γmv) = mv(γ dv + v dγ)
Expand the differential by the product rule, holding the invariant m constant. A
5
dγ = γ3 (v/c2) dv
Differentiate γ = (1 − v2/c2)−1/2: dγ/dv = (v/c2)(1 − v2/c2)−3/2, and (1 − v2/c2)−3/2 = γ3. B
6
dW = mγv dv + mv2·γ3(v/c2)dv = mγv dv[1 + γ2v2/c2]
Substitute dγ from step 5 into step 4 and factor out mγv dv. B
7
1 + γ2v2/c2 = γ2(1 − v2/c2) + γ2v2/c2 = γ2
Use the identity 1 = γ2(1 − v2/c2) to combine the bracket into a single power of γ. This is the key simplification. C
8
dW = mγ3v dv
Insert the collapsed bracket from step 7. The right side is now an exact differential in γ. B
9
d(γc2) = c2 dγ = c2·γ3(v/c2)dv = γ3v dv
Recognise the right-hand side of step 8 (divided by m) as the exact differential of γc2, again using step 5. Hence dW = mc2 dγ. C
10
W = ∫ mc2 dγ = mc2(γ − 1)
Integrate from rest (v = 0, γ = 1) to speed v (Lorentz factor γ). The work done equals the kinetic energy K gained. A
11
K = γmc2 − mc2  ⇒  E ≡ K + mc2 = γmc2
Read K as the difference of a velocity-dependent term γmc2 and a constant mc2. Identifying total energy E as the quantity whose change equals work makes E = γmc2, with the constant of integration mc2 the rest energy. B
12
E2 − p2c2 = γ2m2c4 − γ2m2v2c2 = γ2m2c4(1 − v2/c2) = m2c4
Form the combination E2 − p2c2 with E = γmc2 and p = γmv; the γ2(1 − v2/c2) = 1 identity leaves a frame-independent scalar. This proves (E/c, p) is a four-vector of invariant length mc, confirming E is the true energy partner of p. C
Result
E = γmc2,   E0 = mc2,   E2 = (pc)2 + (mc2)2

Reading. The total energy of a free body is its Lorentz factor times mc2. Setting v = 0 (so γ = 1) leaves the rest energy mc2 — energy a body carries simply for existing with mass m. The kinetic energy is the excess K = (γ − 1)mc2. The energy-momentum relation ties them together and holds even for massless particles (m = 0 ⇒ E = pc).

Units check. [m][c]2 = kg · (m s−1)2 = kg m2 s−2 = J. The factor γ is dimensionless. In E2 = (pc)2 + (mc2)2, [pc] = (kg m s−1)(m s−1) = kg m2 s−2 = J, so every term is an energy squared. Consistent.

Limiting cases
  • Low speed (v « c): expand γ ≈ 1 + ½v2/c2, giving K ≈ ½mv2 — the Newtonian kinetic energy is recovered as the first correction to mc2.
  • Rest (v = 0): γ = 1, so E = mc2 and p = 0; the energy-momentum relation collapses to E = mc2.
  • Massless limit (m → 0): γm stays finite only if v → c; the relation gives E = pc, the photon dispersion.
  • Ultra-relativistic (v → c, m ≠ 0): E ≈ pc + m2c3/(2p); rest energy becomes a negligible fraction of the total.
Breaks when
  • Strong gravity / curved spacetime. With no global inertial frame, γmc2 is only the locally measured energy; conserved total energy must include gravitational contributions and is defined via a timelike Killing vector where one exists.
  • Bound / composite systems. The invariant mass of a bound system is not the sum of constituent masses: binding energy lowers it (mass defect). Applying E = γmc2 with the naive summed mass double-counts or misses the interaction energy stored in fields.
  • Non-conservation of the assumed momentum. If p = γmv is not conserved (e.g. an ill-defined force law, radiation reaction neglected), step 3 has no invariant basis and the derived E is not conserved either.
  • Quantum field regime with particle number change. When creation/annihilation occurs, single-particle E = γmc2 must be replaced by the total energy of the field configuration; only the summed four-momentum of all quanta is conserved.
Failure modes
  • "Relativistic mass" fallacy. Writing mrel = γm and then quoting E = mrelc2 as "the" famous equation. Here m is the invariant rest mass; the celebrated result is E0 = mc2, not γmc2.
  • Using ½mv2 for relativistic K. The Newtonian form undershoots badly as v → c; the correct kinetic energy is (γ − 1)mc2, which diverges.
  • Adding rest masses of reaction products. Treating Σm as conserved. Mass is not separately conserved; only total energy and total momentum are. The mass difference is the released energy.
  • Setting m = 0 in E = γmc2 directly. This gives the indeterminate 0 × ∞; one must use E2 = (pc)2 + (mc2)2 to reach E = pc.
  • Dropping the constant of integration. Concluding E = K = (γ−1)mc2 and missing that the rest energy mc2 is the physically real integration constant.
Discussion

The derivation shows that mc2 is not an arbitrary label for kinetic energy at zero speed but the value of a genuinely conserved quantity. Because momentum conservation must hold in every inertial frame, and boosts mix space and time components, the spatial momentum p cannot be conserved alone: it demands a scalar partner E/c completing a four-vector. Rest energy is the time-component of that four-vector in the body's own frame. This is the deeper reason the result is inescapable — it is a symmetry statement (Lorentz covariance), not a special dynamical accident.

The invariant E2 − p2c2 = m2c4 is arguably the more fundamental form. It defines mass as the Lorentz-invariant length of the energy-momentum four-vector, applies to massless particles, and is the object that appears directly in particle physics (invariant-mass reconstruction of resonances from decay products). Energy and momentum are frame-dependent shadows; the invariant mass is what all observers agree on.

Physically, the equivalence means mass is a reservoir. Nuclear binding stores negative energy, so a 4He nucleus weighs less than its two protons and two neutrons; the deficit powers the Sun. Conversely, a hot gas, a compressed spring, or a charged capacitor all weigh very slightly more than their cold, relaxed, uncharged counterparts, because internal energy contributes to invariant mass. Even a box of photons has mass, though each photon does not.

At the field-theoretic level, rest energy is where the derivation must be refined. For an interacting system the total four-momentum is that of the stress-energy tensor integrated over a spacelike slice, and "the mass of the body" is the invariant length of that total. The Higgs mechanism further shows that much of the mass of fundamental particles arises from coupling to a background field, while most of the mass of ordinary matter (protons, neutrons) is the confinement energy of nearly massless quarks and gluons — a striking illustration that mc2 is overwhelmingly stored interaction energy, not "stuff."

Common misconceptions. E = mc2 does not say mass "turns into" energy in the sense of matter vanishing; energy is always conserved, and mass is one of its forms. It is not a licence to compute rocket kinetic energy as γmc2 minus nothing — the useful kinetic part is (γ−1)mc2. And the "c2" is a units conversion factor, not a speed anything travels.

Worked examples

Example 1 — Rest energy of an electron.

1
E0 = mec2
Rest energy from the boxed result at v = 0. A
2
E0 = (9.109 × 10−31 kg)(2.998 × 108 m s−1)2
Insert me = 9.109 × 10−31 kg and c = 2.998 × 108 m s−1. A
3
E0 = 9.109 × 10−31 × 8.988 × 1016 = 8.187 × 10−14 J
Multiply; c2 = 8.988 × 1016 m2 s−2. A
4
E0 = 8.187 × 10−14 J ÷ (1.602 × 10−19 J eV−1) = 5.11 × 105 eV
Convert to electron-volts by dividing by the elementary charge. A
E0 = 8.19 × 10−14 J = 0.511 MeV

Reading. The standard electron rest energy. This is the threshold scale for electron-positron pair creation (2mec2 = 1.022 MeV).

Example 2 — Mass defect and energy release in deuterium-tritium fusion.

1
2H + 3H → 4He + n,  Δm = (mD + mT) − (mHe + mn)
Energy released equals the mass defect times c2, since total energy is conserved but rest mass is not. A
2
Δm = (2.013553 + 3.015500) − (4.001506 + 1.008665) u = 0.018882 u
Insert nuclear masses in atomic mass units (u). A
3
Δm = 0.018882 u × 1.66054 × 10−27 kg u−1 = 3.135 × 10−29 kg
Convert the mass defect to kilograms. A
4
Q = Δm c2 = 3.135 × 10−29 × 8.988 × 1016 = 2.818 × 10−12 J
Apply E = Δm c2. A
5
Q = 2.818 × 10−12 J ÷ 1.602 × 10−13 J MeV−1 = 17.6 MeV
Convert to MeV. A
Q = 2.82 × 10−12 J = 17.6 MeV per reaction

Reading. A mass loss of under 0.02 u releases 17.6 MeV — the energy per D-T fusion event, carried mostly as kinetic energy of the neutron (14.1 MeV) and the alpha (3.5 MeV). This is the reaction targeted by tokamak and inertial-confinement fusion.

Problems
  1. An electron is accelerated through a potential difference of 1.00 MV. Find its Lorentz factor, speed, and total energy.
    Solution

    K = eV = 1.00 MeV. Since K = (γ−1)mec2 and mec2 = 0.511 MeV: γ = 1 + K/(mec2) = 1 + 1.00/0.511 = 2.957. Then v/c = √(1 − 1/γ2) = √(1 − 1/8.744) = √0.8856 = 0.941. So v = 2.82 × 108 m s−1. Total energy E = γmec2 = 2.957 × 0.511 = 1.511 MeV (= rest 0.511 + kinetic 1.00 MeV).

  2. A body of rest mass 1.00 kg is heated so its internal energy rises by 4.18 × 105 J (1 kcal). By how much does its invariant mass increase?
    Solution

    Δm = ΔE/c2 = 4.18 × 105 J ÷ 8.988 × 1016 m2 s−2 = 4.65 × 10−12 kg. The fractional change is 4.65 × 10−12, utterly unmeasurable on a kitchen scale but real in principle: internal (thermal) energy contributes to invariant mass.

  3. A proton (rest energy 938.3 MeV) has total energy 1400 MeV. Find its momentum (in MeV/c) and speed.
    Solution

    From E2 = (pc)2 + (mc2)2: (pc)2 = 14002 − 938.32 = 1.960 × 106 − 8.804 × 105 = 1.080 × 106 MeV2. So pc = 1039 MeV, i.e. p = 1039 MeV/c. Speed: v/c = pc/E = 1039/1400 = 0.742, so v = 2.23 × 108 m s−1. (Check: γ = E/mc2 = 1.492, and γ√(1−0.7422) = 1.492 × 0.670 = 1.00. ✓)

  4. An electron and positron, each with kinetic energy 1.00 MeV, annihilate head-on. Find the total energy released and, assuming it goes into two identical photons, the energy of each.
    Solution

    Each particle: E = K + mec2 = 1.00 + 0.511 = 1.511 MeV. Total energy of the system = 2 × 1.511 = 3.022 MeV. By energy conservation this equals the total photon energy. In the centre-of-momentum frame (which this is, being head-on with equal energies) the two photons share it equally and move oppositely: each photon has Eγ = 3.022/2 = 1.511 MeV. (Momentum balances: the photons are back-to-back with equal Eγ/c.)

  5. The Sun radiates at L = 3.85 × 1026 W. At what rate (kg s−1) is it losing mass, and what fraction of its mass 2.0 × 1030 kg would it lose over 10 billion years (3.15 × 1017 s)?
    Solution

    Rate of mass loss to radiation: dm/dt = L/c2 = 3.85 × 1026 ÷ 8.988 × 1016 = 4.28 × 109 kg s−1. Over t = 3.15 × 1017 s: Δm = 4.28 × 109 × 3.15 × 1017 = 1.35 × 1027 kg. Fraction = 1.35 × 1027 / 2.0 × 1030 = 6.7 × 10−4, about 0.07% of the Sun's mass radiated away as light over its main-sequence lifetime. (This is distinct from the ~0.7% converted in hydrogen fusion, most of which stays bound as helium.)