Relativistic Inelastic Collisions

Take two lumps of clay, one kilogram each, and throw them at one another at ten metres per second. They meet, stop, and stick. A hundred joules of kinetic energy have gone somewhere, and the Newtonian answer is that they went into heat — the clay is now very slightly warmer, and that is the end of the story. It is not the end of the story. Put the merged lump on a scale accurate enough and it weighs more than two kilograms. The excess is $\Delta M = 100\ \mathrm{J}/c^2 = 1.11\times10^{-15}\ \mathrm{kg}$ — about the mass of a single bacterium, added to a lump of clay by the act of stopping it. Nothing was created. The kinetic energy did not turn into mass in some alchemical sense: it is mass, and always was, and it took a system's-eye view to notice. Scale it up and the effect stops being cute. Fire a proton at $7m_pc^2 = 6.568\ \mathrm{GeV}$ of total energy into a proton at rest and let them merge. The invariant mass of what comes out is not $2m_p$ . It is exactly $4m_p$ — the mass has doubled, and the extra $1.877\ \mathrm{GeV}/c^2$ is nothing but the beam's kinetic energy, re-labelled. That is this article's whole content, and it is the most useful idea in relativistic dynamics: a system's invariant mass is not the sum of the masses of its parts. It is larger, for any collection of free particles not all moving together — larger by exactly the kinetic energy they carry in their own centre-of-momentum frame. Mass is not stuff you count. It is a property of a four-momentum, and four-momenta add as vectors.

Key takeaways

Signature $(+,-,-,-)$ : $p\cdot p = +m^2c^2$ on shell. $s$ is in energy² units, so $\sqrt{s}$ is an energy (Centre-of-Momentum Frame locks this).
Inelastic means the rest masses are not conserved. Four-momentum still is — it always is. What fails is the additional condition that made total kinetic energy conserved in the elastic case.
Perfectly inelastic (the parts merge into one body): in the CM frame the product is at rest, so its energy is the whole CM energy, and $\boxed{Mc^2 = \sqrt{s}}$ . The single body's invariant mass is the system's invariant mass — which it always was, even before the collision.
Owned result — mass is not additive. $M = \sum_i m_i + T^{*}_{\mathrm{tot}}/c^2$ , with $T^{*}_{\mathrm{tot}}$ the kinetic energy in the CM frame. So $M > \sum_i m_i$ strictly, for any free system whose parts are not all moving together. The absorbed kinetic energy becomes the system's invariant mass. The general theorem needs no merger: $Mc^2 = \sqrt{s} \geq \sum_i m_ic^2$ , with equality only when every part shares one velocity — so two photons flying apart, each massless, form a system of mass $2E_\gamma/c^2$ .
This is not "relativistic mass". Every $m_i$ is an invariant rest mass and none of them changes with speed. What grows is the system's invariant mass — a different object, and a real one, with real weight.
Key idea: kinetic energy in the CM frame is not merely EQUIVALENT to mass. In a merger it IS the mass, and you can put the result on a balance. Over 99 % of a proton's own mass is this effect.

The core insight

Add two four-momenta and take the norm. That is the entire subject. Because $P^\mu = p_1^\mu + p_2^\mu$ is a vector sum, its length is not the sum of the lengths — exactly as for two arrows in a plane — and the "length" of a four-momentum is precisely $mc$ . So a system's mass is not the sum of its parts' masses, and the excess is a statement about directions in spacetime: about how much the constituent four-momenta disagree over which way "future" points. When they all point the same way, the masses add. When they do not, the system weighs more, by the CM kinetic energy over $c^2$ .

Where this lives today

The proton is the headline. Its three valence quarks have masses summing to roughly $9\ \mathrm{MeV}/c^2$ — under one percent of the proton's $938.272\ \mathrm{MeV}/c^2$ . More than 99 % of the mass of every atom in your body is therefore not the mass of anything: it is $\sqrt{s}$ of a confined, seething system. (Quark masses are scheme-dependent, $\overline{\mathrm{MS}}$ at $2\ \mathrm{GeV}$ ; the conclusion is not.) Non-additivity is also the working tool of experimental particle physics. To find a new particle you do not look for it — you take the four-momenta of a set of decay products, add them, and take the norm. If they came from a single parent, that norm is the parent's mass, in every event, whatever the parent's speed, and the histogram shows a peak. If they did not, the norm is meaningless and the histogram is smooth. Every discovery plot in the field, from the $J/\psi$ to the Higgs, is a picture of this fact. Particle Decay Kinematics explains why the peak is sharp. For bound systems the effect runs in reverse, and it is why the sun shines: a helium-4 nucleus weighs $28.3\ \mathrm{MeV}/c^2$ less than its two protons and two neutrons, and that deficit is what fusion releases. Same theorem, opposite sign, because the interaction energy is negative. Binding Energy and Mass Defect owns that half.

The physical picture

An elastic collision keeps every rest mass, and therefore keeps total kinetic energy (Relativistic Elastic Collisions owns that). An inelastic one does not. Two lumps of clay merge; two protons fuse into an excited state; a photon is absorbed. In every such case four-momentum is conserved — it always is, in every collision without exception — but rest masses are not, and the accounting has to be done on the four-momentum, not on the masses. Go to the CM frame. Before the collision the two bodies approach with $\vec p_1^{\;*} = -\vec p_2^{\;*}$ , carrying total energy $\sqrt{s} = m_1c^2 + m_2c^2 + T_{\mathrm{tot}}^{*}$ : their rest energies, plus whatever kinetic energy they have in this frame. If they merge, the product must be at rest — the total momentum was zero and stays zero — and a body at rest has energy $Mc^2$ . So

Mc^2 = \sqrt{s} = m_1c^2 + m_2c^2 + T_{\mathrm{tot}}^{*}

and the kinetic energy has not gone anywhere. It is sitting in the mass of the product, which is heavier than its ingredients by precisely $T_{\mathrm{tot}}^{*}/c^2$ . Microscopically it is heat, deformation, internal excitation — but whatever it is microscopically, it weighs, and the balance does not care about the mechanism.

Figure

PUNCHLINE — the mass ledger of a perfectly inelastic collision. A 6.568 GeV proton (E = 7·m_p·c²) merges with a proton at rest. The ingredients weigh 2·m_p = 1.877 GeV/c². The product weighs 4·m_p = 3.753 GeV/c² — the invariant mass has EXACTLY DOUBLED, and the extra 1.877 GeV/c² is the beam's kinetic energy, re-labelled. Exactly one third of the 5.630 GeV of lab kinetic energy is absorbed; the other two thirds must stay as motion, because the momentum has nowhere to go

The core equations

Four-momentum conservation is unchanged — it is not a property of any kind of collision — but the mass condition of the elastic case is now abandoned:

\sum_{\mathrm{in}} p_i^\mu = \sum_{\mathrm{out}} p_f^\mu \quad\text{(always)}, \qquad \sum_{\mathrm{in}} m_i \;\neq\; \sum_{\mathrm{out}} m_f \quad\text{(inelastic)} \qquad (1)

where: $m_i$ — the invariant rest mass of each particle. Physical reading: the masses are the things that changed, not the four-momentum. And since $s$ is built from four-momenta alone, $s$ is still conserved and still invariant — which is why the entire analysis can be done with it. For a perfectly inelastic collision — the parts merge into a single body of mass $M$ — the CM frame does all the work at once:

\boxed{\;Mc^2 \;=\; \sqrt{s} \;=\; \sqrt{\Big(\textstyle\sum_i E_i\Big)^2 - \Big|\textstyle\sum_i \vec p_i\Big|^2c^2}\;} \qquad (2)

where: $M$ — the system's invariant mass, not "the total rest mass". Physical reading: in the CM frame the product is at rest, so all of $\sqrt{s}$ is its rest energy. But notice what equation (2) does not say: it does not say the mass appeared. $\sqrt{s}$ was the system's invariant mass before the collision as well, because $s$ is conserved. The merger did not create mass; it localised it into a single body where you can weigh it. Split $\sqrt{s}$ into rest and kinetic parts in the CM frame and the owned result falls out:

\boxed{\;M \;=\; \sum_i m_i \;+\; \frac{T^{*}_{\mathrm{tot}}}{c^2}\;} \qquad\text{where}\qquad T^{*}_{\mathrm{tot}} = \sum_i \big(\gamma_i^{*} - 1\big)m_ic^2 \qquad (3)

where: the asterisk means "evaluated in the CM frame". Physical reading: the mass excess is the CM-frame kinetic energy, and nothing else. Not the lab kinetic energy — the lab has more, and most of it is unspendable because the products must keep carrying the system's momentum (Centre-of-Momentum Frame derives that; Threshold Energy makes it the punchline). Since every $\gamma_i^{*} \geq 1$ , we get $M \geq \sum_i m_i$ , with equality only if every part is at rest in the CM frame — that is, only if they were all moving together already. And the general theorem, which needs no merger at all:

Mc^2 = \sqrt{s} \;\geq\; \sum_i m_ic^2 \qquad\text{for ANY free system, with equality iff all parts share one four-velocity} \qquad (4)

where: $M$ — again the system's invariant mass. Physical reading: the two-photon case is the extreme. Two photons of energy $E_\gamma$ flying in opposite directions have every $m_i = 0$ , yet $s = (2E_\gamma)^2 = 4E_\gamma^2$ , so $M = 2E_\gamma/c^2$ . Two massless particles make a massive system. Point them the same way and $M$ collapses to zero — the equality case, where the four-velocities agree.

Deriving the mass gain

Derivation AA perfectly inelastic collision converts CM kinetic energy into invariant mass, exactly

When two bodies merge, the resulting body is heavier than the two that made it, by an amount predictable to the last digit. We derive it in the CM frame, where it is almost trivial, then read it back into the lab, where it is not obvious at all — and the gap between those two views is where every misconception about this subject lives.

Step 1. Set up in the CM frame. By definition $\sum_i\vec p_i^{\;*} = \vec 0$ , and the total energy there is $\sqrt{s}$ . Write each particle's CM energy as rest plus kinetic: $E_i^{*} = m_ic^2 + T_i^{*}$ . Summing, $\sqrt{s} = \sum_i m_ic^2 + T^{*}_{\mathrm{tot}}$ . Why: this is just a re-labelling of $\sqrt{s}$ — but it is the re-labelling that makes the result visible, because it separates the part of the energy that is already mass from the part that is not yet.

Step 2. Impose the merger. The two bodies stick together, so the final state is a single body. Its momentum must be the total momentum, which in this frame is $\vec 0$ — so the product is at rest in the CM frame. A single body at rest has four-momentum $(Mc, \vec 0\,)$ and energy $Mc^2$ . Why: this is the only place any physics enters. Everything else is bookkeeping. And note that momentum conservation, not energy conservation, is what pins the product to rest.

Step 3. Equate. Energy conservation in the CM frame gives $Mc^2 = \sqrt{s}$ , and substituting Step 1, $\boxed{M = \sum_i m_i + T^{*}_{\mathrm{tot}}/c^2}$ . Why: the kinetic energy did not vanish and it did not "become" mass by some process. It was always part of $\sqrt{s}$ , which is conserved — the collision merely moved it from the "kinetic" column of the ledger into the "rest" column, by bringing the parts to rest relative to one another.

Step 4. Sign the inequality. Every $T_i^{*} \geq 0$ , so $M \geq \sum_i m_i$ , with equality iff every $T_i^{*} = 0$ — iff every part is already at rest in the CM frame, which means they were all moving with the same velocity and there was no collision to speak of. For any real merger, $\boxed{M > \sum_i m_i}$ strictly. Why: for two equal masses this reads $M > 2m$ , which is the marquee statement. There is no configuration, at any energy, in which a merger fails to gain mass.

Step 5. Guard against the standard misreading. Not one $m_i$ has changed. Each is still the invariant rest mass of its particle, still the same number in every frame, still unaffected by speed. What increased is the system's invariant mass, which was never $\sum_i m_i$ in the first place. Why: this is the one legitimate sense in which "mass increases" in relativity, and it is nothing to do with the discredited notion of relativistic mass. It is the norm of a vector sum failing to equal the sum of the norms — a fact about geometry, not about the particles. *Verified in verify_aq-sr-relativistic-inelastic-collisions.py: $Mc^2 = \sqrt{s}$ and $M = \sum m_i + T^{*}/c^2$ are derived symbolically; $M > \sum m_i$ is asserted over a grid of masses and speeds; and the equality case is confirmed to occur exactly when the four-velocities coincide.*

Go deeper · Derivation B · 2 minThe lab formula, and what fraction of the beam energy can actually be absorbed

Derivation BThe lab formula, and what fraction of the beam energy can actually be absorbed

In the lab the collision looks nothing like the CM picture: the product does not stop, it careers forward, and most of the beam's kinetic energy is still kinetic afterwards. That is not a contradiction but momentum conservation — and saying exactly how much was absorbed and how much was not is the practical content of this article.

Step 1. Evaluate $s$ in the lab. Beam of mass $m_1$ and total energy $E_1$ , target of mass $m_2$ at rest. Centre-of-Momentum Frame derives $s = m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1$ ; recalling it here and using $Mc^2 = \sqrt{s}$ :

\boxed{\;Mc^2 = \sqrt{m_1^2c^4 + m_2^2c^4 + 2m_2c^2E_1}\;}

Why: $s$ is conserved, so we may compute it entirely from the initial state — before we know anything at all about how the collision proceeds. That is the whole power of working with an invariant.

Step 2. Specialise to equal masses. With $m_1 = m_2 = m$ and $E_1 = \gamma mc^2$ : $Mc^2 = mc^2\sqrt{2(1+\gamma)}$ , so $M = m\sqrt{2(1+\gamma)}$ . At $\gamma = 1$ this gives $M = 2m$ (no collision — they were already at rest together); for any $\gamma > 1$ , $M > 2m$ . Why: and at $\gamma = 7$ this is $M = m\sqrt{16} = 4m$ exactly — the article's worked example, and not a coincidence but the reason the antiproton threshold sits where it does.

Step 3. Find the absorbed energy. The mass gain is $\Delta M c^2 = Mc^2 - 2mc^2$ , and this is what got absorbed. The lab kinetic energy supplied was $T_{\mathrm{lab}} = E_1 - mc^2 = (\gamma-1)mc^2$ . Why: these are different numbers, and the ratio between them is the efficiency of the collision as a mass-making machine.

Step 4. Account for the rest. The product carries the whole system momentum, so it must move: its energy is $E_{\mathrm{tot}} = (\gamma+1)mc^2$ and its Lorentz factor is $\gamma_M = E_{\mathrm{tot}}/Mc^2 = (\gamma+1)/\sqrt{2(1+\gamma)} = \sqrt{(\gamma+1)/2}$ — which is exactly $\gamma_{\mathrm{cm}}$ , as it must be, since the product is at rest in the CM frame. Its kinetic energy is $T_M = (\gamma_M - 1)Mc^2$ , and the ledger closes: $T_{\mathrm{lab}} = \Delta Mc^2 + T_M$ . Why: nothing is missing and nothing is lost. The lab kinetic energy splits cleanly into the part that became mass and the part that had to stay kinetic because momentum cannot be destroyed.

Step 5. Take the Newtonian limit as a check. For $v \ll c$ , expand $\gamma_i^{*} - 1 \approx \tfrac12\beta_i^{*2}$ in equation (3): $\Delta Mc^2 \approx T^{*}_{\mathrm{tot}} = \tfrac12\mu v_{\mathrm{rel}}^2$ , with $\mu = m_1m_2/(m_1+m_2)$ the reduced mass. Why: this is exactly the kinetic energy a Newtonian says is "lost to heat" in a perfectly inelastic collision — the standard result from first-year mechanics. Relativity does not change the number. It tells you where the energy went, and that it weighs. *Verified in verify_aq-sr-relativistic-inelastic-collisions.py: $M = m\sqrt{2(1+\gamma)}$ is derived symbolically; the ledger $T_{\mathrm{lab}} = \Delta Mc^2 + T_M$ is proved to close identically in $\gamma$ ; $\gamma_M = \gamma_{\mathrm{cm}}$ is confirmed; and the $v \ll c$ expansion is shown to reproduce $\tfrac12\mu v_{\mathrm{rel}}^2$ to leading order.*

Go deeper · Derivation C · 2 minMass is not additive: the general theorem, and its equality case

Derivation CMass is not additive: the general theorem, and its equality case

The merger result is a special case of something far more general, and the general statement is the one worth carrying away. It holds for any collection of free particles, colliding or not, massive or massless: a system's invariant mass is at least the sum of its parts' masses, and strictly greater unless they all move together. No collision is needed — this is a statement about adding four-vectors.

Step 1. Write the invariant mass of a two-particle system. $M^2c^4 = s = (E_1+E_2)^2 - |\vec p_1 + \vec p_2|^2c^2$ . Expanding and using each mass shell $E_i^2 - |\vec p_i|^2c^2 = m_i^2c^4$ :

M^2c^4 = m_1^2c^4 + m_2^2c^4 + 2\big(E_1E_2 - \vec p_1\cdot\vec p_2\,c^2\big)

Why: everything now hinges on the cross term, which is $2c^2\,(p_1\cdot p_2)$ — a Lorentz invariant, so we may evaluate it in whichever frame is convenient.

Step 2. Identify the cross term. Evaluate $p_1\cdot p_2$ in the rest frame of particle 1, where $p_1 = (m_1c, \vec 0\,)$ and particle 2 moves at the relative velocity with Lorentz factor $\gamma_{\mathrm{rel}}$ . Then $p_1\cdot p_2 = m_1c\cdot(\gamma_{\mathrm{rel}}m_2c) = \gamma_{\mathrm{rel}}m_1m_2c^2$ , so $E_1E_2 - \vec p_1\cdot\vec p_2\,c^2 = \gamma_{\mathrm{rel}}\,m_1m_2c^4$ . Why: this is the classic invariant-evaluation trick — compute a scalar in the frame that makes it trivial, then trust it everywhere. And it hands us a $\gamma$ , which is always at least $1$ .

Step 3. Bound it. Since $\gamma_{\mathrm{rel}} \geq 1$ , with equality iff the relative velocity is zero:

M^2c^4 = m_1^2c^4 + m_2^2c^4 + 2\gamma_{\mathrm{rel}}m_1m_2c^4 \;\geq\; m_1^2c^4 + m_2^2c^4 + 2m_1m_2c^4 = (m_1+m_2)^2c^4

hence $\boxed{M \geq m_1 + m_2}$ , with equality iff the two particles have the same velocity. Why: the whole non-additivity of mass is the single factor $\gamma_{\mathrm{rel}}$ sitting in the cross term. Masses add only when the parts agree about which direction the future is — that is what "same four-velocity" means.

Step 4. Push it to the massless extreme. Take $m_1 = m_2 = 0$ : two photons of energy $E_\gamma$ each. Back-to-back, $\vec p_1 + \vec p_2 = \vec 0$ and $s = (2E_\gamma)^2$ , so $\boxed{M = 2E_\gamma/c^2 \neq 0}$ . Collinear, $|\vec p_1 + \vec p_2|c = 2E_\gamma$ and $s = 0$ , so $M = 0$ . Why: the same two particles, with the same energies, give a massive system or a massless one depending only on their relative direction. If any doubt remained that mass is a property of a four-momentum rather than a count of stuff, this kills it.

Step 5. Note the one exception, and hand it off. Everything above assumed the particles are free. A bound system can have $M < \sum_i m_i$ , because its constituents carry negative potential energy — a helium-4 nucleus weighs $28.3\ \mathrm{MeV}/c^2$ less than two protons plus two neutrons. Why: the theorem is not violated, because $\sqrt{s} = \sum_i E_i^{\mathrm{CM}}$ counts the interaction energy too, and that is negative. Binding Energy and Mass Defect owns the bound case; this article owns the free one, where the inequality runs the other way. *Verified in verify_aq-sr-relativistic-inelastic-collisions.py: the cross-term identity $E_1E_2 - \vec p_1\cdot\vec p_2c^2 = \gamma_{\mathrm{rel}}m_1m_2c^4$ is proved symbolically; $M \geq m_1+m_2$ is asserted over a grid of masses and velocities and the equality case confirmed at $\vec v_1 = \vec v_2$ ; and both photon configurations are checked to give $M = 2E_\gamma/c^2$ and $M = 0$ .*

Limiting cases

Case	Result	Meaning
all parts share one velocity	$M = \sum_i m_i$	the only case in which mass is additive
perfectly inelastic merger	$Mc^2 = \sqrt{s}$ , $\;M = \sum m_i + T^{*}/c^2$	all of the CM kinetic energy becomes mass
elastic collision	$\sum m_f = \sum m_i$	no mass change — the previous article
$v \ll c$	$\Delta Mc^2 \to \tfrac12\mu v_{\mathrm{rel}}^2$	Newton's "kinetic energy lost to heat" — the same number, now weighed
equal masses, $\gamma$ on a target at rest	$M = m\sqrt{2(1+\gamma)}$	$M = 4m$ at $\gamma = 7$ ; $M \to m\sqrt{2\gamma}$ at high energy
two photons, back-to-back	$M = 2E_\gamma/c^2$	massless parts, massive system
two photons, collinear	$M = 0$	same particles, same energies, different directions — no mass
bound system	$M < \sum_i m_i$	binding energy — the inequality reverses. See Binding Energy and Mass Defect

Common trapThis is not relativistic mass, and the distinction is the whole point

Nothing in this article says that a particle gets heavier when it moves. It does not. Every $m_i$ above is an invariant rest mass — a Lorentz scalar, the same number in every frame, unchanged by any boost. The deprecated notion of "relativistic mass" ( $\gamma m$ ) is not used here and is not needed anywhere. What does increase is the system's invariant mass $M$ , and it is a completely different object. $M$ is not a property of any particle; it is the norm of the total four-momentum, $Mc^2 = \sqrt{s}$ . It is larger than $\sum_i m_i$ for the same reason that the length of a sum of vectors is not the sum of their lengths. And it is entirely real: it gravitates, it resists acceleration, and after a merger you can put it on a balance and read it off. The tell that you have slipped into the old error is the phrase "the total rest mass". There is no such quantity. Say "the system's invariant mass", and the sentence "the system's invariant mass is larger than the sum of its constituents' masses" is both true and unremarkable — which is exactly what it should be.

Common trapThe absorbed energy is the CM kinetic energy, never the lab kinetic energy

The commonest arithmetic error here is to set $\Delta Mc^2$ equal to the lab kinetic energy of the beam. It is not, and the gap is large: in the worked example below the beam brings $5.630\ \mathrm{GeV}$ of lab kinetic energy and exactly $1.877\ \mathrm{GeV}$ of it becomes mass — one third. The other two thirds are not lost or radiated; they remain as kinetic energy of the merged body, which is obliged to keep moving because the system carries momentum and momentum is conserved. Only in the CM frame is the whole kinetic energy available, which is why $\sqrt{s}$ — and no lab energy — decides what a collision can make. Threshold Energy turns this into the chapter's punchline.

Worked examples

1 · Two protons, and a mass that doubles

A proton with total energy $E_1 = 6.568\ \mathrm{GeV}$ strikes a proton at rest and the two merge into a single body. Find its invariant mass, its speed, and how much of the beam's kinetic energy was absorbed. Take $m_pc^2 = 0.938272\ \mathrm{GeV}$ , and note that $6.568 = 7\,m_pc^2$ .

Step 1 — identify. The product's mass is fixed by $s$ alone: $Mc^2 = \sqrt{s}$ . The merger mechanism is irrelevant.

Step 2 — apply. $s = 2m_p^2c^4 + 2m_pc^2E_1 = 2m_p^2c^4 + 14m_p^2c^4 = 16\,m_p^2c^4$ , so $\sqrt{s} = 4m_pc^2$ and $M = 4m_p = \mathbf{3.753\ GeV/c^2}$ , exactly. The ingredients weighed $2m_p = 1.877\ \mathrm{GeV}/c^2$ : the invariant mass has doubled.

Step 3 — the speed of the blob. $E_{\mathrm{tot}} = 8m_pc^2 = 7.506\ \mathrm{GeV}$ , so $\gamma_M = E_{\mathrm{tot}}/Mc^2 = 8/4 = \mathbf{2}$ exactly, and $\beta_M = \sqrt{1 - 1/4} = \mathbf{0.866}$ .

Step 4 — the energy ledger. Absorbed: $\Delta Mc^2 = (4-2)m_pc^2 = \mathbf{1.877\ GeV}$ . Still kinetic: $T_M = (\gamma_M - 1)Mc^2 = \mathbf{3.753\ GeV}$ . Supplied: $T_{\mathrm{lab}} = 6m_pc^2 = \mathbf{5.630\ GeV}$ . And $1.877 + 3.753 = 5.630$ — the ledger closes exactly.

Step 5 — the boundary number. The absorbed fraction is $1.877/5.630 = \mathbf{1/3}$ , exactly. Two thirds of the beam's kinetic energy could never have become mass, whatever the collision did, because the momentum had nowhere to go. Not an inefficiency of the merger — a theorem about the frame.

Step 6 — the tie-in. $\sqrt{s} = 4m_pc^2$ is precisely the antiproton threshold, and not by coincidence: $E_1 = 7m_pc^2$ is exactly where the system's invariant mass reaches four proton masses, which is what $p + p \to p + p + p + \bar p$ requires. Threshold Energy owns the consequence.

2 · Two lumps of clay, and a bacterium's worth of mass

Two 1.00 kg lumps of clay approach head-on at $10.0\ \mathrm{m\,s^{-1}}$ each and stick together. By how much does the mass increase?

Step 1 — identify. The lab is the CM frame here, by symmetry. So the entire kinetic energy is CM kinetic energy, and all of it is absorbed: $\Delta Mc^2 = T^{*}_{\mathrm{tot}}$ .

Step 2 — apply. $T^{*}_{\mathrm{tot}} = 2\times\tfrac12 m v^2 = 1.00\times(10.0)^2 = \mathbf{100\ J}$ . (Cross-check with the reduced-mass formula: $\tfrac12\mu v_{\mathrm{rel}}^2 = \tfrac12(0.500)(20.0)^2 = 100\ \mathrm{J}$ . ✓)

Step 3 — the boundary number. $\Delta M = 100/c^2 = 100/(8.98755\times10^{16}) = \mathbf{1.11\times10^{-15}\ kg}$ — about 1.1 picograms, roughly the mass of a single bacterium. As a fraction of the 2 kg total it is $5.6\times10^{-16}$ : eleven orders of magnitude below the best laboratory balance. Newton was never going to catch this, which is precisely why "kinetic energy is lost to heat" survived for three centuries as a complete-sounding statement.

Figure

The general theorem, drawn. The invariant mass of a system is the LENGTH of the vector SUM of its parts' four-momenta — and the length of a sum is not the sum of the lengths. LEFT: two four-momenta pointing the same way; the lengths add, M = m₁ + m₂ exactly, the only additive case. RIGHT: the same two four-momenta at an angle in spacetime; the sum is LONGER than either, and M ≥ m₁ + m₂ strictly. The excess is γ_rel, the relative Lorentz factor, sitting in the cross term

Physics of the portrait

The inelastic collision was the embarrassment of seventeenth-century mechanics. Momentum survived it; "vis viva" did not; and for two hundred and fifty years the missing energy was written off to heat with no account of where the heat's inertia had gone.

Origins · 2 minInelastic collision and how this was discovered

Inelastic collision

1668–2008 · Wallis's imperfect bodies · Einstein's E = mc² · Cockcroft and Walton weigh it · the proton weighs itself

In 1668 the Royal Society asked its members for an account of the laws of impact, and received three: Wren and Huygens on the elastic case, and John Wallis on the imperfectly elastic one — bodies that collide and do not bounce. Wallis had the right conservation law (momentum) and no account at all of the energy: the quantity later called kinetic simply disappeared. For two and a half centuries the answer was "heat" — true, and quietly concealing the real puzzle, since heat has inertia, so where did the missing mass come from? Einstein answered in September 1905, in the three-page sequel to the relativity paper: a body giving off energy $E$ loses mass $E/c^2$ . This article is the converse — a body absorbing energy $E$ gains mass $E/c^2$ , and merging bodies gain exactly the kinetic energy they had in their common centre-of-momentum frame. The first direct weighing came on 14 April 1932, when John Cockcroft and Ernest Walton drove accelerated protons into lithium and split it into two helium nuclei: the products emerged with about $17\ \mathrm{MeV}$ more kinetic energy than went in, matching the mass deficit between ingredients and products to the accuracy of the day (Nature 129, 649). The strangest vindication is that the idea explains us: lattice-QCD calculations of the light hadron spectrum (Dürr et al., Science 322, 1224, 2008) confirmed from first principles that a proton's $938\ \mathrm{MeV}/c^2$ is overwhelmingly not the mass of its quarks — those sum to roughly $9\ \mathrm{MeV}/c^2$ — but the invariant mass of a confined, violently moving system.

History

1668

— John Wallis sends the Royal Society his rules for the collision of imperfectly elastic bodies: momentum survives, but the bodies' "force" does not, and the lost quantity has no home in the mechanics of the day.

1905

— Einstein's September paper argues that a body radiating energy E loses mass E/c², establishing the equivalence that makes the missing energy weighable.

1932

— On 14 April, Cockcroft and Walton disintegrate lithium with accelerated protons into two helium nuclei, releasing about 17 MeV; the energy released matches the mass deficit, the first mass-energy balance measured in a nuclear reaction (Nature 129, 649).

1934

— Einstein's own review of the mass-energy relation makes explicit that a system's inertia includes its internal kinetic and potential energy, not merely the masses of its parts.

1974

— Bubble-chamber and collider practice standardises invariant-mass reconstruction: a particle is identified as a peak in the invariant mass of its decay products, a quantity that exists only because mass is not additive.

2008

— Lattice-QCD calculations of the light hadron masses (Durr et al., Science 322, 1224) confirm from first principles that the proton's mass is overwhelmingly the invariant mass of its confined constituents, not the sum of its quark masses.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §31 · Griffiths, Introduction to Elementary Particles, 2nd ed., §3.3 · Taylor & Wheeler, Spacetime Physics, 2nd ed., ch. 8 · Okun, "The concept of mass", Physics Today 42(6), 31 (1989) · Cockcroft & Walton, Nature 129, 649 (1932) · Dürr et al., Science 322, 1224 (2008).

Provenance. Signature $(+,-,-,-)$ ; $p\cdot p = +m^2c^2$ ; $s$ in energy² units, $\sqrt{s}$ an energy. No relativistic mass is used anywhere — every $m_i$ is an invariant rest mass, and the quantity that grows is the system's invariant mass. verify_aq-sr-relativistic-inelastic-collisions.py parses this article and re-derives every number it prints: $E_1 = 6.568 = 7m_pc^2$ , $s = 16m_p^2c^4$ , $M = 4m_p = 3.753$ , $2m_p = 1.877$ , $\gamma_M = 2$ , $\beta_M = 0.866$ , $T_{\mathrm{lab}} = 5.630$ , $\Delta Mc^2 = 1.877$ , $T_M = 3.753$ , the ledger $1.877 + 3.753 = 5.630$ , the absorbed fraction $1/3$ , the clay numbers ( $100$ J, $1.11\times10^{-15}$ kg, $5.6\times10^{-16}$ ), and the quark sum $\approx 9$ MeV against $938.272$ MeV.

Scope. Bound systems, where $M < \sum m_i$ , are Binding Energy and Mass Defect's payload and are only cited here. Thresholds are Threshold Energy's. The base article ships a fixed-target threshold derivation and a collider-versus-fixed-target comparison; both are recalled by name here and derived there, not in this article.

Builds on. Centre-of-Momentum Frame · Conservation of Four-Momentum · Invariant Mass · Mass-Energy Equivalence · Relativistic Kinetic Energy. Leads to. Threshold Energy · Particle Decay Kinematics · Binding Energy and Mass Defect · Relativistic Elastic Collisions.

Is this the same thing as "relativistic mass"?

No, and conflating them is the error this article exists to prevent. "Relativistic mass" was the discredited idea that a single particle's mass grows with speed, as $\gamma m$ . It is not used here and is not needed anywhere: a particle's mass is an invariant, and $E = \gamma mc^2$ already says everything the idea was groping for. What grows here is the system's invariant mass $M$ , defined by $Mc^2 = \sqrt{s}$ — the norm of the total four-momentum, a property of the collection and not of any particle. And it grows for a purely geometric reason: the norm of a sum of vectors is not the sum of their norms.

Where does the mass physically live after the clay balls stick?

In the internal energy of the merged lump — thermal motion of its molecules, elastic strain, sound waves still ringing through it. Each is energy in the lump's own rest frame, and every joule of rest-frame energy is $1/c^2$ kilograms of mass. Let the lump cool by radiating $100\ \mathrm{J}$ away and it loses exactly $1.11\times10^{-15}\ \mathrm{kg}$ again, returning to $2.000\ \mathrm{kg}$ . The gain is neither permanent nor mysterious: a hot object weighs more than a cold one, by $\Delta E/c^2$ — the same sentence as $E = mc^2$ , read from the other end.

If s is conserved, how can the invariant mass "increase"?

It does not increase — and spotting that is the sharpest thing you can take from this article. $\sqrt{s}$ is conserved, so the system's invariant mass is the same before and after the collision. Before, it was $4m_p$ distributed over two protons that were flying apart; after, it is $4m_p$ localised in one body. What changed is not the system's mass but the sum of the constituents' masses, which went from $2m_p$ to $4m_p$ . The collision did not create mass. It converted kinetic energy into rest energy, which moved mass from the "invisible in a particle-by-particle census" column to the "sitting on the scale" column.

Why can only the CM kinetic energy be absorbed?

Because momentum is conserved and cannot be destroyed. If the system carries momentum $\vec P_{\mathrm{tot}}$ , then whatever it turns into must also carry $\vec P_{\mathrm{tot}}$ , so the products must keep moving, so they must retain enough kinetic energy to supply that momentum. That portion is unavailable — permanently. Only in the frame where $\vec P_{\mathrm{tot}} = \vec 0$ is all the kinetic energy free to be absorbed, and that frame is the CM frame. Hence $\Delta Mc^2 \leq T^{*}_{\mathrm{tot}}$ always, with equality for a perfectly inelastic merger. In the worked example the lab supplies $5.630\ \mathrm{GeV}$ and only $1.877\ \mathrm{GeV}$ — one third — is spendable.

Can a system of massless particles really have mass?

Yes, and it is the cleanest demonstration that mass is not "amount of stuff". Two photons of energy $E_\gamma$ flying in opposite directions have $\sum\vec p_i = \vec 0$ and $\sum E_i = 2E_\gamma$ , so $s = 4E_\gamma^2$ and the pair has invariant mass $2E_\gamma/c^2$ . Box them in a mirrored cavity and the box is genuinely heavier — it gravitates more and resists acceleration more. Point the same two photons the same way and $s = 0$ : same particles, same energies, zero mass. Nothing about the photons changed; only the relative direction of their four-momenta did, and that is the only thing mass ever depended on.

Does this mean the mass of the proton comes from kinetic energy?

Largely, yes — with one caveat. The proton is a bound system governed by QCD, so both confining field energy and quark kinetic energy contribute. The three valence quarks sum to roughly $9\ \mathrm{MeV}/c^2$ against the proton's $938.272$ , so over 99 % is not quark mass, and lattice-QCD calculations of the light hadron spectrum (Dürr et al., 2008) reproduce the observed masses from first principles. The concept is exactly the one derived above — a system's invariant mass is $\sqrt{s}$ of its contents, not a sum over them — but the dynamics is QCD's, not kinematics', so the derivation here does not extend to it.