The invariant that survives a boost

Two observers watching the same electron will not agree about its energy, and will not agree about its momentum. Each has a definite number; the numbers are different; neither is wrong. What is surprising is that a particular combination of the two comes out the same for both, to whatever precision either can measure, and that the combination is not an obscure by-product but the thing physics has been calling mass all along.

Everything below is where that combination comes from, what it makes possible, and where reading it carelessly produces a paradox that is not there.

Two quantities nobody agrees about, and one they do #

The energy of a moving body is $E = \gamma mc^2$ and its momentum is $\mathbf{p} = \gamma m\mathbf{v}$. Both contain $\gamma$, both therefore depend on who is measuring, and the first of them diverges at the speed of light.

Eliminating the speed between those two expressions takes one line. Squaring the energy gives $\gamma^2m^2c^4$, squaring the momentum and multiplying by $c^2$ gives $\gamma^2m^2v^2c^2$, and subtracting leaves $\gamma^2m^2c^4(1 - v^2/c^2)$, which is $m^2c^4$ because that is what $\gamma$ is for. So

$$E^2 - (pc)^2 = (mc^2)^2.$$

The left side is built from two frame-dependent numbers; the right side contains no speed. Two observers in relative motion write down different $E$ and different $p$ and get the same answer, in exactly the sense that they disagree about a time and a distance and agree about the interval — not an analogy but the same theorem on a different pair of quantities.

The object being described is the four-momentum: energy over $c$ in the time slot, the three components of momentum in the space slots, transforming under a boost precisely as time and position do. The relation above says its length is $mc$. Mass is not a quantity of stuff. It is the length of a vector whose components every observer disagrees about, and that is the whole of what this rung adds to the statement that mass is a form of energy.

The same factor appears in both components: 1.15 at half light speed, 2.29 at 0.9c, 7.09 at 0.99c — and 7,461 for a proton in the Large Hadron Collider. Energy and momentum each carry it, which is why neither is agreed on between frames, and why their particular combination is: the factor cancels out of $E^2 - p^2c^2$ exactly.

A worked case makes the arithmetic checkable. An electron accelerated through 100 kilovolts, the working voltage of a modest transmission electron microscope, has total energy $511 + 100 = 611$ keV against a rest energy of 511 keV, and the relation fixes its momentum at $pc = \sqrt{611^2 - 511^2} = 335$ keV. Those three numbers are the sides of a right triangle, and a reader with a calculator can confirm that $611^2 - 335^2 = 261{,}121 = 511^2$ to the digit. The speed follows as $\beta = pc/E = 0.548$ and γ as $E/mc^2 = 1.196$.

Two of those three sides depend on the frame and the third does not, so it is worth saying plainly what the third is not. “Relativistic mass” is $\gamma m$, and it is $E/c^2$ under another name. It carries no information the energy does not, it makes the invariant look like an accident rather than a definition, and it invites the false inference that a fast particle resists a sideways push as strongly as a forward one — resistance along and across the motion differ by $\gamma^2$, so no single number called mass describes both. The modern convention keeps $m$ for the invariant alone, and every statement in the rest of this essay becomes unwritable in the other language.

The construction that stretches a tick stretches energy and momentum in the same proportion, and it is worth seeing that they are not three separate results. At $\beta = 0.8$ light travels 1.67 times as far and $\gamma$ is 1.667; the clock, the energy and the momentum all inherit that one number from the same triangle.

A particle with no mass is not a contradiction #

Written as $E = \gamma mc^2$, the case of a photon is an embarrassment: $\gamma$ is infinite, $m$ is zero, and the product is undefined. Written as the invariant, setting $m = 0$ gives

$$E = pc,$$

which is finite, non-zero and correct. A photon of wavelength 71.1 picometres — the molybdenum line Compton used — has energy 17.44 keV and momentum 17.44 keV/$c$. Nothing in the relation objects, and nothing had to be added to accommodate it.

The usual framing has it backwards. A four-momentum with zero length is the unremarkable case, being what any vector on the light cone has. The special condition is mass, which requires $E > pc$ strictly and confines the particle inside the cone rather than on it. Massless particles are not an exception grudgingly admitted to a theory built for matter; they are the boundary it was built around, and matter is what happens strictly inside.

That third class has no counterpart among particles, and its absence is physics rather than convention: a four-momentum with $pc > E$ would move faster than light and carry negative energy in some frames, and nothing of the kind has been detected. Timelike, null and spacelike are the same three classes relabelled — matter inside the cone, light on it, and nothing at all outside.

The mass of a system is not the sum of its parts #

Here the bookkeeping produces something a reader has no reason to expect. Four-momenta add — that is what a conservation law means — but their lengths do not. The mass of a system is the length of the sum, and the sum of vectors is shorter or longer than the sum of their lengths depending on how they point.

The cleanest case has no mass in it anywhere. Take the two 511 keV photons produced when an electron meets a positron. Each has $E = pc$ and therefore zero mass. Flying back to back, the system’s total energy is 1.022 MeV and its total momentum is zero, so the invariant gives

$$M^2c^4 = 2E_1E_2(1 - \cos\theta),$$

which at $\theta = 180°$ is $4E^2$: a system mass of 1.022 MeV/$c^2$, exactly the mass of the electron and positron that were annihilated. Nothing was conserved by accident. Two objects of no mass form a system whose mass is 1.022 MeV/$c^2$, and if the same two photons travel parallel the same formula gives zero. At 90° it gives $511\sqrt{2} = 722.7$ keV; at 60°, exactly 511 keV.

So mass is a property of a configuration, not a stock held by the parts. The same two photons, of the same energies, are a thing of mass or a thing of no mass according to the angle between them alone.

The nucleus is the same statement with the sign reversed. A helium-4 nucleus weighs 3,727.379 MeV/$c^2$ while two protons and two neutrons weigh 3,755.674 separately — a deficit of 28.295 MeV, which is 0.75 per cent. The four particles’ four-momenta, added inside the bound system, give a total shorter than the sum of the lengths, because the potential energy holding them together is negative. In the two-photon case the vectors point apart and the sum is longer than the parts; in the nucleus they are bound and it is shorter. One relation covers both, and neither is a conversion event.

This is also the honest version of the claim that a hot object weighs more. A one-kilogram brick warmed by 100 K holds about 80 kilojoules more, and $80{,}000/c^2$ is 0.89 nanograms. Not one molecule in it has changed mass; the extra 0.89 nanograms is the kinetic energy of molecules going nowhere in particular, counted as mass in exactly the sense the invariant defines and in no other sense at all.

The threshold, and why every machine collides #

The consequence with the largest budget attached follows immediately. Whether a collision can make a new particle is decided by the invariant mass of the colliding system, $\sqrt{s}$, and not by the energy of the beam.

For a beam particle of total energy $E$ striking a stationary target of the same mass, the total energy is $E + mc^2$ and the total momentum is $p$, so

$$s = (E + mc^2)^2 - (pc)^2 = 2mc^2E + 2m^2c^4.$$

The square root is the punchline. The available energy grows as the square root of the beam energy, because the struck system keeps most of what the beam brought as momentum of the whole, and momentum of the whole cannot be spent on making anything. A 7 TeV proton — the Large Hadron Collider’s design energy, at γ = 7,461 — striking a stationary proton of rest energy 938.272 MeV yields $\sqrt{s} = 114.6$ GeV. Two 7 TeV protons meeting head-on have zero total momentum, so every joule is available and $\sqrt{s} = 14$ TeV. The same machine, the same beam, a factor of 122.

Run the arithmetic the other way and it is starker. Reaching 14 TeV of invariant mass against a stationary proton would need a beam of $s/2mc^2 = 104{,}000$ TeV, seven thousand times what the LHC accelerates. No such machine will be built. That single square root is why every high-energy facility since CERN’s Intersecting Storage Rings began running in 1971 collides two beams rather than striking a target.

Speed against rapidity is where the composition becomes addition: 0.75c composed with 0.75c gives rapidities of 0.973 and 0.973, which add to 1.946, whose hyperbolic tangent is 0.96. That is why a threshold calculation is done in invariant mass rather than in velocities — the invariant does not have to be composed at all.

The frame in which the total momentum vanishes is the centre-of-momentum frame, and $\sqrt{s}$ is the total energy measured there. Every threshold in particle physics is a statement in that frame — the relativistic version of the point that keeps moving as if nothing had happened, and the same privileged frame that makes a collision easier to analyse than the forces in it, now defined by a four-vector rather than a weighted average of positions.

What it costs #

An electron microscope’s resolution is set by $pc$, and the non-relativistic answer is wrong by several per cent. The wavelength of a massive particle is $h/p$, and taking $p$ from $\sqrt{2mK}$ rather than from the invariant gives 3.88 picometres at 100 kV where the correct value is 3.70 — a 4.6 per cent error. At 300 kV it is 1.97 picometres against 2.24, an error of 13.7 per cent. Instrument software does the relativistic calculation, and it has to.

A threshold sets a machine’s specification before it is designed. Making an antiproton requires $p + p \rightarrow p + p + p + \bar{p}$, so $\sqrt{s}$ must reach four proton masses, 3,753 MeV — against a stationary proton, a beam of total energy $7mc^2$, or 5.63 GeV of kinetic energy. The Bevatron at Berkeley was built to deliver 6.2 GeV, and in 1955 Chamberlain, Segrè, Wiegand and Ypsilantis found the antiproton with it. The specification was the threshold, computed years before the machine existed.

A discovery can be nothing but a histogram of an invariant. The Higgs boson’s cleanest channel is decay to two photons, and each event supplies two energies and an angle. Feeding them to $M^2c^4 = 2E_1E_2(1-\cos\theta)$ turns every pair into a mass, and the 2012 announcement was a bump at 125 GeV in that distribution. The photons are massless in every frame; what has the bump in it is the length of their summed four-momentum.

The invariant lets an unmeasured particle be eliminated. Compton’s scattering measurement works because the recoiling electron never has to be observed. Conservation says $p_\gamma + p_e = p_\gamma' + p_e'$; rearranging so the unobserved electron stands alone and taking the length of both sides sends its four-momentum into the known $m_ec$, leaving a relation between the photon’s two wavelengths and the angle.

What it costs is the habit of adding. Two photons each of zero mass have a total mass that is not zero, because the invariant of a sum is not the sum of the invariants — and that catches everybody once. The same arithmetic makes the mass of a proton mostly not the masses of its quarks: the binding and the internal motion contribute, and the invariant of the whole exceeds the sum of the parts by a factor of about fifty.

Where the model stops #

The relation holds for a free particle. For anything bound, the mass of the whole includes the binding energy and the parts’ masses stop being separately measurable from outside. A proton’s three valence quarks have rest energies summing to about one per cent of 938.272 MeV; the rest is gluon field energy and quark motion, and no experiment weighs a quark inside a proton. The relation still applies exactly to the proton as a unit, which is the only object the outside world is offered.

Four-momentum is defined relative to a global inertial frame, and general relativity does not supply one. Adding the four-momenta of two distant particles requires a way of comparing vectors at different places, and a curved spacetime has none — transporting a vector between two points depends on the path taken. The total mass of a gravitating system is therefore not the sum this essay describes but a subtler object defined by the metric far away, available only where spacetime is asymptotically flat and undefined for a closed universe. The invariant survives in curved spacetime only locally, in the frame of a freely falling observer.

The size of the error is worth putting a number to. Near the Earth’s surface the gravitational potential is $10^{-9}$ of $c^2$, so treating a laboratory as a global inertial frame misstates energies by parts in a billion — invisible at the LHC, and 36 microseconds a day for a satellite clock. The flat-space bookkeeping fails outright only where the potential approaches $c^2$, within a few Schwarzschild radii of a black hole.

Nothing here says what mass is. The relation says how mass combines, what it does not depend on, and how it constrains what a collision can make. The question of why a given particle has the mass it has is untouched, and answering it takes a mechanism the invariant has no opinion about.

The same construction somewhere else #

Because the four-momentum is a vector, anything true of vectors is available, and two consequences arrive with no new physics.

Boosting a photon’s four-momentum gives the Doppler effect. A photon has $E = pc$, so its components are $(E/c, E/c)$ along its direction of travel, and the transformation gives $E' = \gamma E(1 - \beta\cos\theta)$ directly. The longitudinal shifts fall out at $\theta = 0$ and $180°$; at $\theta = 90°$ the factor is γ alone, which is the shift that survives at right angles and which no classical treatment predicts. The invariant stays zero throughout — a boost changes a photon’s energy freely and cannot give it mass, because a rotation cannot change a length from zero to something.

Squaring one conservation statement gives Compton scattering, and the fact that the result contains only $\lambda_C = h/m_ec$ is the invariant’s fingerprint: the target’s binding, its element and the incident wavelength all drop out, because the only property of the electron the algebra keeps is the length of its four-momentum.

The third instance is where the arithmetic becomes least expected. A cosmic-ray proton flying through the microwave background meets photons of mean energy $6.3\times10^{-4}$ eV, and for a head-on encounter the invariant mass of the pair reaches the threshold for making a pion at a proton energy of about $1\times10^{20}$ eV. Above that the proton bleeds energy into the background and cannot travel far. The observed suppression begins somewhat lower, near $5\times10^{19}$ eV, because the Planck spectrum has a tail of more energetic photons. The fixed-target threshold formula that sized the Bevatron therefore also caps the energy of particles arriving from other galaxies, with the coldest thing in the universe as the target.

Minkowski, Planck, and the word that would not die #

The invariant relation is not in Einstein’s 1905 papers. Planck wrote the relativistic momentum $\gamma m\mathbf{v}$ in 1906, and Minkowski assembled energy and momentum into a single four-vector in 1908, in the same work that made the interval geometrical. The observation that its length is the rest mass is Minkowski’s, and its status was clear to him in a way it was not to the physics that followed: mass is not a component, it is the norm, and the components are its frame-dependent shadows.

What followed instead was the other convention. Lewis and Tolman in 1909 introduced a mass that grows with speed, which let $\mathbf{p} = m\mathbf{v}$ keep its Newtonian form, and for sixty years textbooks taught it. The cost was hidden and large. In that language the invariant reads as a curiosity about a quantity nobody had named, and the mass of a two-photon system is unstateable: each photon’s relativistic mass is $E/c^2$, the two add to $2E/c^2$ regardless of angle, and that is the wrong answer at every angle but 180°.

The convention turned in the 1960s and finally in 1989, when Lev Okun argued in Physics Today that the pedagogical mass had cost the subject its clearest idea. Most textbooks changed. The phrase survives in popular writing, where it reliably produces the two paradoxes it always produced — that a fast enough object should collapse into a black hole, and that mass should be additive — both of which dissolve the moment mass is the length of a vector, which it has been since 1908.

What the picture cannot show #

The hero figure is drawn in $t$ and $x$, and the essay’s claims live in $E$ and $p$. The transposition is exact — the same hyperbolae, the same three classes, the same invariance — but no figure on this site draws the momentum-space version directly. A reader who wants the mass shell has to relabel the axes and remember that the curve is now a set of possible states of one particle rather than a set of events.

A diagram with one space dimension and one time dimension is the limit of what a page carries, and what it hides is the transverse structure. A boost along one axis leaves the perpendicular components of momentum untouched, so the two-dimensional picture is exactly right about what it shows and silent about the two directions it has suppressed — which is where the transverse momentum of a collision product lives.

Every diagram here has one spatial dimension, and the angle between two photons needs two. The formula $2E_1E_2(1-\cos\theta)$ is entirely about that angle, and in one dimension only $\theta = 0$ and $\theta = 180°$ exist. The figures can show the two extreme answers, 1.022 MeV and zero, and cannot show the continuum between them where every real measurement sits.

A hyperbola shows a relation between states and not a process. A particle does not travel along the mass shell; a boost relocates it there, and a boost is a change of description rather than an event. Nothing in any of these figures distinguishes accelerating a particle from walking past it, because at the level of the invariant there is nothing to distinguish — which is the same silence that makes a contracted length an unfamiliar geometry rather than a squeeze.

Conservation is drawn as a closed triangle, which works only for three vectors. The Compton figure closes because two photons and one electron make three arrows in a plane. A collision at the LHC produces hundreds of outgoing four-momenta, and the invariant mass of a chosen subset is the entire method — an arithmetic operation on a list, with no picture available at all.

The ladder from here #

Later rungs on this anchor: pair production and the threshold for making matter from light, where the two-photon calculation runs forwards; binding energy across the periodic table and the iron peak, which is why fusion pays below iron and fission above it; the four-force, and Newton’s second law written when mass is a norm; invariant-mass reconstruction, including the missing transverse momentum that betrays a neutrino nobody detected; and energy as a source of gravity, where the flat-space bookkeeping of this page runs out.

The neighbouring ladders are the interval, which is this same theorem about time and position; the spacetime diagram, whose calibration curves are the mass shells relabelled; velocity composition, the boost parameter that leaves the invariant alone; and time dilation with the twin who returns younger, where the same γ that multiplies a tick multiplies both components this essay combines.