Mass–Energy Equivalence (E=mc²)

A hot brick weighs more than a cold one. Not metaphorically, not as a manner of speaking — put them on a balance sensitive enough and the hot one comes down. A compressed spring outweighs a relaxed one. A charged battery outweighs a flat one. A box of gas weighs more when the gas is stirred. In every case the extra weight is the energy you put in, divided by $c^2$ , and it is really there. This is what $E = mc^2$ actually says, and it is a stranger claim than the one usually attributed to it. The popular reading is that mass can be converted into energy, as if they were two currencies and $c^2$ were the exchange rate. The truer reading is that they were never two things. Mass is what energy looks like when it is standing still. Give a system more internal energy — heat it, compress it, bind it, spin it — and its mass goes up, because mass just is the rest energy divided by $c^2$ . And the equation as everyone quotes it is a special case. The general relation between a body's energy and its mass is $E = \gamma mc^2$ ; put $v = 0$ and it becomes $E_0 = mc^2$ , the rest energy. That is the famous equation, and it is a statement about a body at rest. What makes it revolutionary is not the arithmetic but the consequence: mass is not conserved. Energy is. And the two are the same ledger, written in different units.

Key takeaways

Signature $(+,-,-,-)$ : $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , so $p\cdot p = +m^2c^2$ .
$E_0 = mc^2$ is the REST energy — the $v = 0$ case. The general relation is $E = \gamma mc^2$ . They are different equations and swapping them is the commonest error in physics.
Owned result: Einstein's 1905 argument, done exactly. A body at rest emits two light pulses of total energy $L$ in opposite directions, so it gets no recoil. In a frame moving at $v$ the pulses are Doppler-shifted and carry $\gamma L$ . The body's kinetic energy therefore drops by $(\gamma-1)L$ at unchanged speed — and since $K = (\gamma-1)mc^2$ , that forces $\Delta m = L/c^2$ . (Einstein wrote $L/V^2$ ; $V$ was his symbol for the speed of light.)
Owned result: MASS IS NOT CONSERVED. Energy is. $\Delta m = \Delta E/c^2$ for any internal energy change. Heat 1 kg of water by 100 K and it gains 4.66 nanograms. The mass of a system is a property of its total internal energy, not a count of its parts.
$c^2 = 8.98755\times10^{16}$ J/kg — an enormous exchange rate, which is why the effect is invisible in chemistry (a part in $10^{10}$ ) and unmissable in the nucleus (a part in $10^{3}$ ).
Key idea: the equation is not a recipe for making energy out of mass. It is the statement that the mass of a thing is its internal energy, full stop — so that ANY change to that internal energy is a change of mass, whether or not anything is "converted". A brick that cools down really does get lighter.

The core insight

Ask why a uranium nucleus can release energy and a lead nucleus cannot, and the answer is not that uranium "contains" energy the way a tank contains fuel. It is that the products of uranium fission have less mass than uranium does — and the difference, multiplied by $c^2$ , is what comes out. Nothing was stored and nothing was released from storage. A system rearranged itself into a lower-mass configuration, and mass-that-vanished is energy-that-appeared, because they were the same number all along. This is why physicists stopped saying "mass converts to energy" and started saying mass is not conserved. The bookkeeping is cleaner and the physics is clearer.

Where this lives today

The most complete expression of the equation is not a bomb but a hospital scanner. In positron emission tomography, a tracer emits positrons; each finds an electron and annihilates, and the entire rest mass of both particles — $2 \times 0.511$ MeV — reappears as two photons flying in opposite directions, each of exactly $0.511$ MeV. The detector ring looks for those back-to-back pairs and reconstructs where they came from. The energy is not "released from" the mass. It is the mass, arriving in a form that a scintillator can see, and the number $0.511$ MeV is written into the design of every PET machine ever built. At the other end of the scale, the equation explains why the Sun shines and why it has lasted. Four hydrogen nuclei weigh more than one helium nucleus by about $0.7\%$ , and that $0.7\%$ , times $c^2$ , is the Sun's entire power budget — $4\times10^{9}$ kg of mass leaving the Sun as sunlight every second. It also explains something more surprising about you: most of your mass is not the mass of your particles. The quarks inside a proton account for only about $1\%$ of the proton's mass. The rest is the energy of the gluon field and the quarks' motion, bound into a system, and weighing exactly what that internal energy demands. Mass–energy equivalence is not an exotic correction to your weight. It is very nearly the whole of it.

The physical picture

Take the four-momentum $p^\mu = (E/c, \vec p\,)$ and its invariant norm $p\cdot p = +m^2c^2$ , which unpacks to $E^2 = (pc)^2 + (mc^2)^2$ . Now go to the frame where the thing is at rest — where $\vec p = 0$ . The relation collapses to

E_0 = mc^2.

That is a one-line derivation and it looks like nothing at all. The content is entirely in what you allow "the thing" to be. If it is a single point particle, this says its rest energy is proportional to its mass, and you may fairly ask what you have learned. But nothing in the four-momentum formalism restricts $p^\mu$ to a single particle. Four-momenta add. So let "the thing" be a box of gas, a nucleus, a compressed spring, a star — any system at all — and take $\vec p_{\text{total}} = 0$ . Then $m$ is the system's mass, $E_0$ is its total internal energy including the kinetic energy of everything jiggling around inside, and the equation says these are the same quantity. That is the whole of mass–energy equivalence, and it is why the hot brick is heavier. The thermal energy of the atoms is kinetic energy inside the brick. It contributes to $E_0$ . So it contributes to $m$ .

Interactive Simulation

Mass–Energy Equivalence — E = γmc² splits into rest energy mc² + kinetic (γ−1)mc²; drag the speed and watch kinetic energy overtake the rest energy near c.

Figure

PUNCHLINE — Einstein's 1905 argument, and it works EXACTLY. A body at rest emits two pulses of energy L/2 in opposite directions: no recoil, so its speed is unchanged in every frame. In the rest frame it loses energy L. In a frame moving at v the two pulses are Doppler-shifted to γ(L/2)(1−β) and γ(L/2)(1+β) — which SUM to γL. So the moving observer sees the body lose γL while the rest observer sees it lose L. The difference (γ−1)L is a loss of KINETIC energy at UNCHANGED SPEED. But K = (γ−1)mc², so the only thing that can have changed is m: Δm = L/c². No approximation was used

The figure is worth dwelling on, because the argument is subtler than it looks and it is the actual reasoning of the September 1905 paper. Nothing in it involves a body turning into light. The body emits radiation symmetrically, so it takes no recoil, and its velocity is the same before and after in every frame. The two observers nonetheless disagree about how much energy it lost — one says $L$ , the other says $\gamma L$ — and that disagreement has nowhere to go except into the one term that distinguishes a moving body's energy from a stationary one's: the kinetic energy $(\gamma - 1)mc^2$ . The speed did not change. Therefore $m$ did.

The core equations

The rest energy, and the general relation it is a special case of:

\boxed{E_0 = mc^2} \qquad\text{(at rest)}, \qquad\qquad \boxed{E = \gamma mc^2 = mc^2 + K} \qquad\text{(in general)} \qquad (1)

where: $m$ — the invariant rest mass, a Lorentz scalar; $E_0$ — the rest energy; $K = (\gamma - 1)mc^2$ — the kinetic energy. Physical reading: the famous equation is the left-hand box, and it applies to a body at rest. A moving body has more energy than $mc^2$ , and the excess is kinetic. Nothing is repaired by letting $m$ grow with speed — that is the discredited "relativistic mass", dismantled in Relativistic Momentum. The statement that makes it physics rather than arithmetic:

\boxed{\Delta m = \frac{\Delta E}{c^2}} \qquad (2)

where: $\Delta E$ — any change in a system's internal energy, measured in its rest frame. Physical reading: heat it, compress it, charge it, bind it — the mass follows. This is not a statement about exotic processes. It is a statement about all processes, and the reason nobody noticed for two centuries is the size of $c^2$ , not the rarity of the effect. And the exchange rate itself:

c^2 = \big(2.99792458\times10^{8}\ \text{m/s}\big)^2 = 8.98755\times10^{16}\ \text{J/kg} \qquad (3)

where: $c$ — exact by definition of the metre. Physical reading: one gram of anything, converted completely, is $8.99\times10^{13}$ J — about 21.5 kilotons of TNT. (The Hiroshima weapon released roughly 15 kt, and it did so by converting well under a gram.) The number is enormous, and its enormity is precisely why mass–energy equivalence looks like an exotic nuclear phenomenon rather than what it is: a universal fact about every object, mostly too small to weigh.

Where the equation comes from

Derivation AA body that radiates energy loses mass, and the amount is forced

This is Einstein's argument from the September 1905 paper, and it is worth doing properly because it is so often reported as a plausibility sketch when it is in fact a proof. The setup uses nothing but energy conservation, the relativistic Doppler effect, and the expression for kinetic energy. It never assumes that mass and energy are related — that is the conclusion. And although Einstein himself took a low-velocity limit at the last step, the limit is unnecessary: the factor that would have to be approximated cancels identically, so the result holds exactly at any speed.

Step 1. Set up an emission with no recoil. A body of mass $m$ sits at rest in frame $S$ and emits two pulses of light, each of energy $L/2$ , in exactly opposite directions along $x$ . The momenta of the pulses are $+L/2c$ and $-L/2c$ , which sum to zero. The body therefore receives no kick: it is at rest before, and at rest after. Why: this is the entire trick. By arranging zero recoil we guarantee that the body's velocity is unchanged in every frame, which will let us attribute any change in its kinetic energy to a change in its mass and nothing else.

Step 2. Account in the rest frame. Energy conservation in $S$ gives the body's energy loss as $\Delta E = L$ . Why: trivial, and it is the baseline the other frame must be compared against.

Step 3. Doppler-shift the pulses into a moving frame. Let $S'$ move at $v = \beta c$ along $x$ , so the body drifts at $-v$ in $S'$ . A photon's energy transforms as the time component of its four-momentum, $E' = \gamma(E - v p_x)$ . The forward pulse has $p_x = +L/2c$ and the backward pulse $p_x = -L/2c$ , giving E'_{\rightarrow} = \gamma\tfrac{L}{2}(1 - \beta), \qquad E'_{\leftarrow} = \gamma\tfrac{L}{2}(1 + \beta), \qquad E'_{\rightarrow} + E'_{\leftarrow} = \gamma L. Why: one pulse is red-shifted and the other blue-shifted, and the $\beta$ terms cancel exactly in the sum, leaving $\gamma L$ . This is the crux: the two frames disagree about the emitted energy by precisely a factor $\gamma$ .

Step 4. Account in the moving frame. Energy conservation in $S'$ gives the body's energy loss as $\Delta E' = \gamma L$ . Why: the same conservation law, applied by a different observer. Both are right; they are measuring different components of the same four-vector.

Step 5. Locate the discrepancy. The body's total energy in $S'$ exceeds its rest energy by its kinetic energy: $E' = mc^2 + K$ with $K = (\gamma-1)mc^2$ . Since the speed is unchanged (Step 1), $\gamma$ is the same before and after. Subtracting the two ledgers, \Delta E' - \Delta E = \gamma L - L = (\gamma - 1)L, and this must be a drop in the body's kinetic energy, because the rest-frame ledger has already accounted for the change in $mc^2$ . Why: the body slowed down by nothing, yet a moving observer says it lost more energy than a stationary one. There is exactly one term in $E' = mc^2 + (\gamma-1)mc^2$ that a moving observer sees and a stationary one does not — and it carries a factor of $m$ .

Step 6. Solve, and notice that no approximation is needed. The kinetic-energy drop is $\Delta K = (\gamma - 1)\,\Delta m\, c^2$ at fixed $\gamma$ . Setting this equal to $(\gamma-1)L$ : (\gamma-1)\,\Delta m\,c^2 = (\gamma-1)L \qquad\Longrightarrow\qquad \boxed{\Delta m = \frac{L}{c^2}}. The factor $(\gamma-1)$ cancels identically, so the result is exact for every $\beta$ . Why: Einstein expanded to lowest order, matching $\tfrac12(\Delta m)v^2$ against $\tfrac12(L/c^2)v^2$ , and concluded — correctly but more cautiously — that the mass falls by $L/V^2$ , $V$ being his notation for the speed of light. The cancellation shows the low-velocity step was never required. A body that radiates energy loses mass, and this is the whole of $E = mc^2$ . *Verified symbolically in verify_aq-sr-mass-energy-equivalence.py: the Doppler sum $\gamma\tfrac{L}{2}(1-\beta) + \gamma\tfrac{L}{2}(1+\beta) = \gamma L$ identically; the cancellation of $(\gamma-1)$ giving $\Delta m = L/c^2$ with no series expansion; and Einstein's own low- $\beta$ route recovered as the leading term, $(\gamma - 1) \to \tfrac12\beta^2$ .*

Go deeper · Derivation B · 2 minMass is not conserved, and a system weighs its own internal energy

Derivation BMass is not conserved, and a system weighs its own internal energy

The previous derivation concerned a body that emits radiation. This one generalises it into the statement that actually matters, and it is the one the equation is for: that the mass of any system is a measure of its total internal energy, so that mass is simply not a conserved quantity. Chemists' intuition — that the mass on the left of an equation equals the mass on the right — is an approximation good to about a part in $10^{10}$ , and the twentieth century was largely a matter of finding processes where it fails visibly. The argument requires only that four-momenta add.

Step 1. Define the mass of a system. For a collection of parts with four-momenta $p_i^\mu$ , the total is $P^\mu = \sum_i p_i^\mu$ , and the system's mass is defined by the invariant norm $P\cdot P = M^2c^2$ . Why: this is the only definition available that is a Lorentz scalar, and it reduces to the ordinary mass for a single particle. Its consequences are developed in Invariant Mass; here we need only its value in one frame.

Step 2. Evaluate it in the rest frame of the system. Choose the frame where $\sum_i \vec p_i = 0$ . Then $P^\mu = (E_{\text{tot}}/c,\ \vec 0\,)$ and the norm gives Mc^2 = E_{\text{tot}} = \sum_i \big(m_ic^2 + K_i\big) + U, where $K_i$ are the parts' kinetic energies inside the system and $U$ is their mutual potential energy. Why: the system's mass is its total energy in the frame where it is not going anywhere — and that total includes everything happening inside it. The internal motion has not gone away just because the box as a whole is stationary.

Step 3. Read the consequence. Mc^2 = \sum_i m_ic^2 + \underbrace{\sum_i K_i + U}_{\text{internal energy}} \qquad\Longrightarrow\qquad M \ne \sum_i m_i \ \text{ in general.} Heat the system and the $K_i$ rise, so $M$ rises. Bind it more tightly and $U$ falls (becomes more negative), so $M$ falls. Why: this is mass–energy equivalence in its useful form. The mass of a bound system is less than the sum of its parts, by the binding energy over $c^2$ — which is why a nucleus weighs less than its nucleons, and why pulling it apart costs energy.

Step 4. Get the working rule. For any process at rest, $\Delta(Mc^2) = \Delta E_{\text{internal}}$ , so $\boxed{\Delta m = \Delta E/c^2}$ . Why: it does not matter what kind of energy. Heat, strain, chemical bonds, nuclear binding, rotation — every joule you put into a system at rest adds $1/c^2$ kilograms to it, and every joule you take out removes the same.

Step 5. Check the scale, and see why nobody noticed. Heat 1 kg of water by 100 K: $\Delta E = 4186 \times 100 = 4.186\times10^{5}$ J, so $\Delta m = 4.186\times10^{5}/8.98755\times10^{16} = 4.66\times10^{-12}$ kg — 4.66 nanograms, on a kilogram. Burning coal converts about $3\times10^{-10}$ of the fuel's mass. Nuclear fission converts about $9\times10^{-4}$ — a million times more. Why: the effect is not rare, it is small, and it is small because $c^2$ is large. The nucleus is the first place in nature where binding energies are a large enough fraction of $mc^2$ to weigh. That is the entire reason mass–energy equivalence is remembered as nuclear physics. *Verified numerically in verify_aq-sr-mass-energy-equivalence.py: $\Delta m = 4.66$ ng for 1 kg of water heated 100 K; the coal fraction $3.6\times10^{-10}$ from 32 MJ/kg; the fission fraction $9.1\times10^{-4}$ from 200 MeV per $^{235}$ U; and $c^2 = 8.98755\times10^{16}$ J/kg with 1 g $= 8.99\times10^{13}$ J $= 21.5$ kt TNT.*

Figure

The reason it took the NUCLEUS to see this. The bar is the fraction of a system's mass that a process actually converts. Chemistry — burning coal — converts 3.6×10⁻¹⁰: a kilogram of coal loses about a third of a microgram, which no 19th-century balance could ever weigh. Nuclear fission converts 9.1×10⁻⁴, a MILLION times more, and that is finally weighable. Annihilation converts 1 — everything. The physics is identical in all three; only the fraction differs. E = mc² was not hiding. It was just very, very quiet

Limiting cases

Case	Result	Meaning
$v = 0$	$E = mc^2$	the rest energy — the famous equation, and only here
$v > 0$	$E = \gamma mc^2 > mc^2$	the excess is kinetic; $m$ has not changed
$c \to \infty$	$\Delta m = \Delta E/c^2 \to 0$	the Newtonian world, where mass is conserved — an excellent approximation
chemical process	$\Delta m/m \sim 10^{-10}$	real, and utterly unweighable
nuclear process	$\Delta m/m \sim 10^{-3}$	weighable — this is why the nucleus revealed it
annihilation	$\Delta m/m = 1$	complete conversion: $e^+ + e^- \to 2\gamma$ , two photons of $0.511$ MeV
bound system	$M < \sum_i m_i$	the binding energy is missing mass; see Invariant Mass
$m = 0$	$E = pc$	no rest frame, no rest energy; $\gamma$ undefined, never infinite

Common trapMass is not conserved, and that is the point — not a paradox to be explained away

Students often try to preserve mass conservation by saying that the missing mass "turned into energy", as though mass were a substance that changed state. Resist it. The clean statement is: energy is conserved; mass is not. In fission, the total energy before and after is identical; the total mass is smaller afterwards, and the difference left as kinetic energy of the fragments. Nothing was converted, in the sense that nothing had to be. The system simply moved to a configuration of lower mass, and mass is just internal energy in disguise. A corollary worth internalising: if you catch all the fission products in a box and let them cool, the box's mass returns to exactly the original value. The mass was never lost — it left with the energy.

Worked examples

1 · Cockcroft and Walton, 1932: weigh the reaction, predict the energy

In 1932 Cockcroft and Walton fired protons at a lithium target and observed the reaction $^{7}\text{Li} + p \to 2\,^{4}\text{He}$ . Using the atomic masses $^{7}\text{Li} = 7.016003$ u, $p = 1.007825$ u, $^{4}\text{He} = 4.002603$ u, and $1$ u $= 931.494$ MeV/ $c^2$ , predict the energy released.

Step 1 — identify. The products weigh less than the reactants. That missing mass, times $c^2$ , is the energy released — the reaction's $Q$ -value. This is $\Delta m = \Delta E/c^2$ used in the direction that made it famous.

Step 2 — weigh both sides. Reactants: $7.016003 + 1.007825 = 8.023828$ u. Products: $2 \times 4.002603 = 8.005206$ u. The difference is \Delta m = 8.023828 - 8.005206 = \mathbf{0.018622\ \text{u}}.

Step 3 — convert. $Q = \Delta m\,c^2 = 0.018622 \times 931.494 = \mathbf{17.35\ \text{MeV}}$ (commonly quoted as $17.3$ MeV).

Step 4 — the boundary number. The two alpha particles fly apart back-to-back, so each carries about half: $\approx \mathbf{8.7}$ MeV. Cockcroft and Walton measured alphas of about $8.6$ MeV — roughly $17$ MeV between them — in agreement with the mass difference. They had weighed a reaction and predicted, correctly, how much energy would come out of it. Note the leverage: the incoming protons carried only a few hundred keV, and over $17$ MeV came out. The energy was not supplied by the beam. It was supplied by the mass. They shared the 1951 Nobel Prize for the work.

2 · The hot brick, made quantitative

Heat 1 kg of water from 20 °C to 120 °C (as steam, say). By how much does it get heavier?

Step 1 — identify. Every joule of internal energy adds $1/c^2$ kg. The water at rest has $Mc^2 = E_{\text{internal}}$ , and we are raising $E_{\text{internal}}$ .

Step 2 — apply. $\Delta E \approx c_{\text{water}}\,m\,\Delta T = 4186 \times 1 \times 100 = 4.186\times10^{5}$ J.

Step 3 — convert. $\Delta m = \dfrac{\Delta E}{c^2} = \dfrac{4.186\times10^{5}}{8.98755\times10^{16}} = \mathbf{4.66\times10^{-12}\ \text{kg}} = \mathbf{4.66\ \text{nanograms}}$ .

Step 4 — the boundary number. Four and a half nanograms on a kilogram: a fractional change of $4.7\times10^{-12}$ . The best laboratory mass comparators reach a few parts in $10^{10}$ — so this is genuinely, if only just, a hundredfold below the threshold of measurement. The hot water really is heavier, and we really cannot weigh the difference. That gap, and not any conceptual subtlety, is the whole reason mass–energy equivalence had to wait for the nucleus.

Physics of the portrait

The portrait is not Einstein. Einstein wrote the equation, and it belongs to Einstein's Two Postulates and to the relativity papers. But an equation becomes physics when someone applies it to a real system and gets a real number, and in the case of $E = mc^2$ the person who did that — in exile, on a walk in the snow, using her nephew as a sounding board, and denied the Nobel Prize that her collaborator received — was Lise Meitner.

Origins · 2 minLise Meitner and how this was discovered

Lise Meitner

1878–1968 · Vienna · Berlin · Stockholm · Cambridge

Lise Meitner spent thirty years in Berlin as Otto Hahn's collaborator, co-discovering protactinium and building one of the finest radiochemistry programmes in Europe. In 1938, as an Austrian Jew stripped of protection by the Anschluss, she fled Germany with ten marks in her purse and a diamond ring Hahn had given her in case she needed to bribe a border guard, arriving eventually at the Nobel Institute in Stockholm. That December, Hahn wrote to her in bewilderment: bombarding uranium with neutrons, he and Fritz Strassmann kept finding barium — an element with barely half uranium's charge — and could not explain it. Over Christmas, walking in the snow near Kungälv with her nephew Otto Frisch, Meitner did the physics. She proposed that the uranium nucleus, behaving like a charged liquid drop whose surface tension is nearly cancelled by its own charge, could simply divide; Frisch named the process fission. Their letter reached Nature on 16 January 1939 and appeared on 11 February 1939 (Nature 143, 239). The paper estimates that the two fragments "will repel each other and should gain a total kinetic energy of c. 200 Mev., as calculated from nuclear radius and charge" — an electrostatic calculation — and then makes the decisive move: that this energy "may actually be expected to be available from the difference in packing fraction between uranium and the elements in the middle of the periodic system." That second sentence is $E = mc^2$ doing its work. It identifies where the 200 MeV comes from: the products weigh less than the reactants, and the missing mass pays the bill. Hahn alone received the 1944 Nobel Prize in Chemistry. Meitner refused to work on the atomic bomb — "I will have nothing to do with a bomb" — and died in Cambridge in 1968, eighteen days before her nephew. Element 109 is named meitnerium.

History

1905

— Einstein, "Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?" (Ann. Phys. 18, 639; received 27 September, published 21 November): a body radiating energy L loses mass L/V², V being his symbol for the speed of light.

1932

— Cockcroft and Walton (Nature 129, 649, 30 April) split lithium with accelerated protons; the energy carried off by the two alpha particles matches the mass difference between the reactants and the products.

1938

— December: Hahn and Strassmann find barium among the products of neutron-bombarded uranium and cannot account for it (Naturwiss. 27, 11).

1939

— Meitner and Frisch (Nature 143, 239; letter dated 16 January, published 11 February) explain it as fission of a liquid-drop nucleus, estimate c. 200 MeV from the electrostatic repulsion of the fragments, and identify the packing-fraction difference as the source of that energy.

1944

— Hahn alone is awarded the Nobel Prize in Chemistry for the discovery of fission; Meitner is passed over.

1997

— Element 109 is named meitnerium.

Going further

Texts. Taylor & Wheeler, Spacetime Physics, 2nd ed., ch. 8 · Rindler, Introduction to Special Relativity, 2nd ed., §27 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.2 · Einstein, "Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?", Ann. Phys. 18, 639 (1905) · Cockcroft & Walton, "Disintegration of Lithium by Swift Protons", Nature 129, 649 (1932) · Meitner & Frisch, "Disintegration of Uranium by Neutrons: a New Type of Nuclear Reaction", Nature 143, 239 (1939) · Okun, "The concept of mass", Physics Today 42(6), 31 (1989).

Provenance. Signature $(+,-,-,-)$ , so $p\cdot p = +m^2c^2$ . Everything above is recomputed in verify_aq-sr-mass-energy-equivalence.py: the Doppler sum $\gamma\tfrac{L}{2}(1-\beta) + \gamma\tfrac{L}{2}(1+\beta) = \gamma L$ ; the exact cancellation of $(\gamma-1)$ giving $\Delta m = L/c^2$ with no series expansion; $c^2 = 8.98755\times10^{16}$ J/kg, 1 g $= 8.99\times10^{13}$ J $= 21.5$ kt TNT; the Cockcroft–Walton $Q$ -value $0.018622$ u $\times\ 931.494 = 17.35$ MeV, $\approx 8.7$ MeV per alpha; the hot-water $\Delta m = 4.66$ ng; the conversion fractions $3.6\times10^{-10}$ (coal), $9.1\times10^{-4}$ (fission), $1$ (annihilation); and the Meitner–Frisch Coulomb estimate reproduced — two fragments of $Z = 46$ at $d \approx 15$ fm give $(1.44\ \text{MeV\,fm})(46)(46)/15\ \text{fm} = 203$ MeV, matching the paper's "c. 200 Mev., as calculated from nuclear radius and charge", while the mass route $0.2$ u $\times\ 931.494 = 186$ MeV shows the packing-fraction difference can supply it.

A note on the 200 MeV. The famous figure is often reported as Meitner plugging a mass defect into $E = mc^2$ . The paper does something more interesting: it computes $\approx 200$ MeV from the electrostatic repulsion of the two fragments, and then invokes the packing-fraction (mass) difference to show that the energy is available. Both estimates are of the same quantity and both are right — the recomputed mass route gives $186$ MeV for a $0.2$ u defect, and $200$ MeV is the rounded standard figure. $E = mc^2$ is what identifies the source; it is not what produced the number.

Scope. $E = \gamma mc^2$ and the $E$ -versus- $E_0$ distinction are Relativistic Energy's. $K = (\gamma-1)mc^2$ is derived at Relativistic Kinetic Energy. The norm $p\cdot p = +m^2c^2$ is Four-Momentum's. The non-additivity of a system's mass — binding energy, the proton's mass, colliders — is Invariant Mass's, and is used here only in the form needed for Step 2 of the second derivation. The photon case belongs to Massless Particles and the Photon.

Builds on. Four-Momentum · Relativistic Energy · Relativistic Momentum · Invariant Mass. Leads to. Energy–Momentum Relation · Massless Particles and the Photon · Relativistic Kinetic Energy · Conservation of Four-Momentum.

Does E = mc² mean mass can be converted into energy?

That is the popular phrasing, and it is not quite the right picture. The sharper statement is that mass and energy are not two things. A system's mass is its rest energy divided by $c^2$ — so when a system rearranges into a configuration with less mass, the missing mass does not "become" energy; the energy that was always there simply leaves in a form you can see, such as the kinetic energy of fission fragments. Nothing is transmuted. The clean bookkeeping is: energy is conserved, mass is not. Catch all the products, let them cool, and you recover the original mass exactly.

If a hot brick is heavier, has its atoms' mass increased?

No — and this is a good test of the ideas. The individual atoms have exactly the same invariant mass as before; what has increased is their kinetic energy inside the brick. The brick's mass is the norm of the total four-momentum of all its constituents, evaluated in the frame where the brick is at rest, and that norm includes internal kinetic energy. So the system is heavier while every part is unchanged. Mass is not additive: this is the same fact that makes a proton weigh a hundred times more than the quarks inside it, and it is Invariant Mass's subject.

Why is the mass of a nucleus less than the sum of its nucleons?

Because it is bound. Binding means the potential energy $U$ is negative, and a system's mass includes $U$ : $Mc^2 = \sum m_ic^2 + \sum K_i + U$ . The deficit is the binding energy, and it is what you would have to supply to pull the nucleus apart. Helium-4, for instance, weighs about $0.7\%$ less than two protons and two neutrons — and that $0.7\%$ is precisely the energy released in fusing them, which is what powers the Sun. Bound systems weigh less than their parts. Free ones do not.

Was Cockcroft and Walton's experiment the first test of E = mc²?

It is often described that way, but the priority claim is disputed in the literature and we do not make it here. What is not disputed is what they did: they compared the kinetic energy carried away by the two alpha particles ( $\approx 8.6$ MeV each) against the mass difference between $^{7}\text{Li} + p$ and $2\,^{4}\text{He}$ , and found them consistent — a direct, quantitative check of $E = mc^2$ in a nuclear reaction, in 1932. That is a strong enough claim to stand without a superlative attached to it.

How much of my body's mass is really E/c²?

Almost all of it. Your mass is dominated by protons and neutrons, and the quarks inside a proton account for only about $1\%$ of its mass. The other $99\%$ is the energy of the gluon field binding them and the kinetic energy of the quarks' confined motion — internal energy, weighing exactly what $E = mc^2$ says it must. The Higgs mechanism gives the quarks and electrons their (small) intrinsic masses; the overwhelming majority of your mass is not Higgs mass at all, but the rest energy of a violently active bound system. You are mostly binding energy.

If mass is not conserved, what stops mass from just appearing out of nowhere?

Energy conservation, plus the conservation laws of whatever the mass is made of. You can absolutely create mass — a collider does it every day, turning kinetic energy into new particles — but only if you pay for it in energy and respect charge, baryon number, lepton number and the rest. What you cannot do is create mass for free. And there is a second, subtler constraint: creating a particle of mass $M$ requires not just $Mc^2$ of energy somewhere, but $Mc^2$ of energy in the invariant sense, available in the centre-of-momentum frame. A beam of enormous laboratory energy may have very little of it. That is the threshold problem, and it belongs to Invariant Mass.