Four-Momentum

Newtonian mechanics has two conservation laws. Energy is conserved; momentum is conserved. They are stated separately, proved separately, applied separately — and Newton could not tell you why there were two of them, or why both should happen to be true. It is one of those facts so familiar that the strangeness has worn off it. Relativity says: there is one law.

p^\mu = m\,U^\mu = \left(\frac{E}{c},\ \vec{p}\,\right), \qquad \sum_{\text{in}} p^\mu = \sum_{\text{out}} p^\mu.

That single four-vector equation is energy conservation (in its $\mu = 0$ slot) and momentum conservation (in its $\mu = 1,2,3$ slots), simultaneously. And because $\Lambda$ is a linear map, if the equation holds in one inertial frame it holds in every one — automatically, with nothing further to check. Newton's two laws were not two laws. They were one law, seen in a frame, split along an axis that turns out to have no absolute meaning. That is the biggest idea in this article and quite possibly in the chapter. The four-momentum's length carries the rest of the physics. In this chapter's signature $(+,-,-,-)$ ,

p\cdot p = \frac{E^2}{c^2} - |\vec{p}\,|^2 = +\,m^2c^2 \qquad\Longleftrightarrow\qquad E^2 = (pc)^2 + (mc^2)^2,

and the invariant mass $m$ is simply the Minkowski length of a particle's four-momentum — the one thing about it that every observer agrees on. Set $m = 0$ and the relation collapses to $E = pc$ : massless particles are possible, they have momentum without mass, and they are forced to move at exactly $c$ . Set $v \ll c$ and it expands to $E \approx mc^2 + \tfrac12 mv^2$ , with a constant term Newton could not see because it never changed.

A warning that is not decoration: the AllQFT base for this article states the norm as $p\cdot p = -m^2c^2$ . That is correct in the other signature and wrong here, and the error is invisible — every numeric answer in the article would come out the same, while every sign in the algebra would be upside down. We derive the sign from $\eta = \mathrm{diag}(+1,-1,-1,-1)$ in front of you.

Key takeaways

Signature $(+,-,-,-)$ : $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , so $A\cdot B = A^0B^0 - \vec{A}\cdot\vec{B}$ . Recalled by name from `aq-sr-four-vectors`: $U\cdot U = +c^2$ .
$p^\mu \equiv m\,U^\mu = m\gamma(c,\vec{v}\,) = \big(E/c,\ \vec{p}\,\big)$ , with $\boxed{E = \gamma mc^2}$ and $\boxed{\vec{p} = \gamma m\vec{v}}$ . Here $m$ is the invariant rest mass, a Lorentz scalar — the only mass there is.
Owned result: $\boxed{p\cdot p = +m^2c^2}$ , hence $\boxed{E^2 = (pc)^2 + (mc^2)^2}$ . The sign is positive, derived from $\eta$ below. The base's $-m^2c^2$ belongs to $(-,+,+,+)$ and does not ship here.
Owned result — conservation of four-momentum is ONE equation meaning energy AND momentum conservation, IN EVERY FRAME. Since $\Lambda$ is linear, $\sum p^\mu_{\text{in}} = \sum p^\mu_{\text{out}}$ in one frame implies it in all. Newton needed two laws and could not say why both held. Relativity has one, and frame-independence is free.
Owned result: $\boxed{\vec{v} = \vec{p}\,c^2/E}$ — velocity is a ratio of four-momentum components, echoing the four-velocity's lesson. For $m = 0$ this forces $v = c$ exactly.
Owned result: the low-speed limit $E \approx mc^2 + \tfrac12mv^2$ , $\ \vec{p}\approx m\vec{v}$ . The rest energy $mc^2$ is the term Newton could not see, because a constant never shows up in a difference.
Massless: $m = 0 \Rightarrow p\cdot p = 0 \Rightarrow E = pc$ . $\gamma$ is UNDEFINED, not infinite, and $p^\mu = mU^\mu$ does not apply — a photon has no four-velocity. Define $p^\mu$ directly.
Key idea: $E = mc^2$ is the REST energy, the $v = 0$ case. The general relation is $E = \gamma mc^2$ . Conflating them is the most common error in all of physics pedagogy, and this chapter will not make it.

The core insight

The invariant mass is the length of the four-momentum vector; energy and momentum are its components. Components depend on who is looking; lengths do not. That is the whole reason relativistic dynamics comes out easier than Newtonian dynamics rather than harder: compute the frame-independent thing once, and every frame's answer is a projection of it.

Where this lives today

Four-momentum is not a pedagogical construct; it is the data format of experimental particle physics. Every reconstructed object in an LHC event — every track, every jet, every photon — is stored as a four-vector, and essentially every question asked of the event is a contraction of them. Is this pair of photons a Higgs? Compute $(p_1+p_2)\cdot(p_1+p_2)$ and see whether it clusters at $(125\ \mathrm{GeV})^2$ . Did anything invisible escape? Sum the four-momenta and look at what is missing from the transverse components — a neutrino, or something less familiar, announces itself as a failure of $\sum \vec{p}_\perp = 0$ . That inference works only because conservation of four-momentum is exact and frame-independent, which is the result derived above. The mass shell of the punchline figure has a life of its own in quantum field theory, where the condition $p\cdot p = m^2c^2$ is called being on shell. Real, observable particles are on shell; the internal lines of a Feynman diagram are not. They are "virtual", meaning precisely that their four-momentum fails $p\cdot p = m^2c^2$ — permitted only because they are never observed. Perturbative quantum field theory is, in large part, a bookkeeping of how far off the shell things may stray and for how long, and the shell in question is the curve in that figure.

Origins · 1 minFour-momentum and how this was discovered

Four-momentum

1906–1908 · Planck's relativistic equations of motion · Minkowski's four-dimensional momentum vector · the mass shell

The four-momentum was assembled, not discovered — by two people in quick succession. Max Planck took the first step in 1906, in "Das Prinzip der Relativität und die Grundgleichungen der Mechanik" (Verhandlungen der Deutschen Physikalischen Gesellschaft), where he put the relativistic equations of motion into essentially their modern form and identified $\vec{p} = \gamma m\vec{v}$ as the momentum that is actually conserved. Planck came to relativity early and championed it when almost no one else would; his advocacy did as much as anything to make the theory respectable. Hermann Minkowski supplied the geometry two years later: in the 1908 memoir "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern" he defined the four-dimensional momentum vector as the rest mass times the four-velocity, showed that its time component is the energy, and thereby made energy and momentum components of a single object — with the invariant length we now call the mass. The synthesis is his: Planck had the physics, but it was Minkowski who saw that the two conservation laws were one. He died on 12 January 1909, a few months after presenting "Raum und Zeit", and never saw the formalism become universal.

History

1906

— Planck's "Das Prinzip der Relativität und die Grundgleichungen der Mechanik" (Verh. Deutsch. Phys. Ges.) puts the relativistic equations of motion in modern form and identifies p = γmv as the conserved momentum.

1908

— Minkowski's "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern" defines the four-momentum as m times the four-velocity and identifies its time component with the energy.

1909

— Minkowski dies on 12 January; "Raum und Zeit", his Cologne lecture of September 1908, is published posthumously.

1928

— Dirac's equation is written with the four-momentum operator, and the mass-shell condition becomes the defining constraint of a relativistic wave equation.

1932

— Cockcroft and Walton split lithium with accelerated protons, and the energy release is checked against the mass difference: four-momentum conservation confirmed in a nuclear reaction.

2012

— ATLAS and CMS announce the Higgs boson as a peak in the invariant mass of reconstructed four-momenta near 125 GeV.

The physical picture

The four-velocity $U^\mu = \gamma(c,\vec{v}\,)$ is a four-vector, and $m$ — the invariant rest mass — is a Lorentz scalar. Multiplying a four-vector by a scalar gives a four-vector. So $p^\mu \equiv m\,U^\mu$ is a four-vector, with no further argument required. That is the entire construction, and it is worth noticing how little work it took: the difficult step (building $U^\mu$ by dividing by the invariant $d\tau$ ) was done at `aq-sr-four-velocity`, and everything below is bookkeeping on top of it. Write out the components: $p^\mu = m\gamma(c,\vec{v}\,) = (\gamma m c,\ \gamma m\vec{v}\,)$ . The spatial part is manifestly a momentum — it reduces to $m\vec{v}$ when $\gamma \to 1$ — so we name it $\vec{p} = \gamma m\vec{v}$ . The time part, times $c$ , has dimensions of energy and reduces (as we shall see) to $mc^2 + \tfrac12mv^2$ at low speed, so we name it $E/c$ , with $E = \gamma mc^2$ .

The names are the only physics input. Everything else is forced. And notice what has not happened: nowhere did we let $m$ depend on $v$ . The $\gamma$ sits in the components, where it belongs, because components are frame-dependent. The mass sits in the length, where it belongs, because the length is not.

Figure

PUNCHLINE — the mass shell. Every particle of mass m lives on the hyperbola E² − (pc)² = (mc²)², and a boost SLIDES it along that curve without ever moving it off. The length of the arrow is mc² in every frame — that is the invariant mass. A massless particle sits on the 45° asymptote itself, E = pc, which is why it is stuck at v = c and has no rest frame to be at rest in. Proton anchor: at K = 100 MeV, E = 1038.272 MeV and pc = 444.58 MeV, and 1038.272² − 444.58² = 938.272² — the proton's rest energy, recovered exactly

Read the picture once more, because it encodes the article's central discipline. $E = mc^2$ is a point on that curve — the intercept, where $p = 0$ . It is the rest energy. The curve itself, the thing that describes a particle at any speed, is $E = \gamma mc^2$ . Conflating a point with a curve is a small geometric sin and an enormous physical one, and it is committed constantly.

Figure

PUNCHLINE — Newton needed TWO conservation laws and could not say why both held. Relativity has ONE. The four-vector equation Σp_in = Σp_out has four slots: its TIME slot is conservation of energy, its SPACE slots are conservation of momentum. They were never independent laws — they are one geometric statement, chopped along an axis that has no absolute meaning. And because Λ is LINEAR, a linear map sends zero to zero: if the equation holds in one frame it holds in ALL of them, with nothing further to check. Newton had to assume frame-independence separately; here it is free

The core equations

p^{\mu} \equiv m\,U^{\mu} = m\gamma\,(c,\ \vec{v}\,) = \left(\frac{E}{c},\ \vec{p}\,\right), \qquad \boxed{\;E = \gamma mc^{2}\;}, \qquad \boxed{\;\vec{p} = \gamma m\vec{v}\;} \qquad (1)

where: $m$ — the invariant rest mass, a Lorentz scalar, the same number in every frame. Physical reading: there is exactly one mass in relativity and this is it. The factor $\gamma$ lives in the components, not in the mass. Writing " $\gamma m$ " and calling it a mass was a historical device, abandoned because it is not a scalar and because it tempts $F = ma$ with the wrong object — see `aq-sr-relativistic-momentum`, which owns that argument. Now the norm. Derive it, do not copy it. With $\eta = \mathrm{diag}(+1,-1,-1,-1)$ :

p\cdot p \;=\; \eta_{\mu\nu}p^{\mu}p^{\nu} \;=\; \left(\frac{E}{c}\right)^{2} - |\vec{p}\,|^{2} \;=\; m^{2}\,\big(U\cdot U\big) \;=\; m^{2}c^{2} \qquad\Longrightarrow\qquad \boxed{\;p\cdot p = +\,m^{2}c^{2}\;} \qquad (2)

where: the middle step uses $U\cdot U = +c^2$ , derived at `aq-sr-four-vectors` and recalled by name. Physical reading: the sign is positive. $p^\mu$ is timelike, as it must be — it is a scalar multiple of the timelike $U^\mu$ . In $(-,+,+,+)$ this reads $-m^2c^2$ ; that is the base's value, and it is wrong in this chapter. Multiplying (2) out by $c^2$ :

\boxed{\;E^{2} = (pc)^{2} + \big(mc^{2}\big)^{2}\;} \qquad (3)

where: $p = |\vec{p}\,|$ . Physical reading: this is the most-used equation in particle physics, and it is born here. Note it contains no $\gamma$ and no $v$ — it relates only invariant and measurable quantities, which is exactly why experimentalists live in it. Dividing the spatial part of (1) by the temporal part:

\boxed{\;\vec{v} = \frac{\vec{p}\,c^{2}}{E}\;} \qquad\Longleftrightarrow\qquad \vec{\beta} = \frac{\vec{p}c}{E} \qquad (4)

where: velocity is again a ratio of components, exactly as at `aq-sr-four-velocity`. Physical reading: for $m = 0$ , equation (3) gives $E = pc$ , and substituting into (4) gives $|\vec{v}| = c$ exactly. A massless particle does not merely happen to travel at $c$ ; it has no choice.

Where the four-momentum earns its keep

Derivation AThe norm, derived from eta — and why the sign is positive

This is the keystone of the chapter and the place where a silent error would do the most damage, so we derive it rather than quote it. The claim is that $p\cdot p = +m^2c^2$ , with a plus sign, and that the energy–momentum relation follows in one line. Everything hangs on the metric: the same physics in the opposite signature gives $-m^2c^2$ , every numerical answer in this article is unchanged, and every intermediate sign is inverted. There is no numeric check that catches that, which is why the derivation must be done in front of the reader and not inherited.

Step 1. Fix the metric and nothing else. $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , so for any four-vector $A\cdot A = (A^0)^2 - |\vec{A}|^2$ . Why: this single choice determines every sign below. Read the metric before you read any formula — a formula copied without its signature is a formula copied without its meaning.

Step 2. Contract $p^\mu = mU^\mu$ with itself. Because $m$ is a scalar it comes straight out: $p\cdot p = \eta_{\mu\nu}(mU^\mu)(mU^\nu) = m^2\,\eta_{\mu\nu}U^\mu U^\nu = m^2\,(U\cdot U)$ . Why: no components have been used yet. This step is pure algebra, and it means the answer is completely determined by the four-velocity's norm — which we already know.

Step 3. Insert the recalled norm. From `aq-sr-four-vectors`, $U\cdot U = +c^2$ for every massive particle at every speed. Hence $\boxed{p\cdot p = +m^2c^2}$ . Positive. Why: $p^\mu$ is a scalar multiple of a timelike vector, so it is timelike, and timelike means positive norm in this signature. The sign is not a convention we are choosing here — it is forced by the convention we chose at the top of the chapter.

Step 4. Read it in components, and out falls the famous relation. $p\cdot p = (E/c)^2 - |\vec{p}\,|^2 = m^2c^2$ ; multiply by $c^2$ : $\boxed{E^2 = (pc)^2 + (mc^2)^2}$ . Why: a Pythagorean relation, with $E$ as the hypotenuse — no coincidence, since the metric's minus sign has moved $|\vec p\,|^2$ across the equals sign. This is the equation on which every collider analysis rests.

Step 5. Verify it directly, and confirm no $\gamma$ survives. Substitute $E = \gamma mc^2$ , $p = \gamma mv$ : $(\gamma mc^2)^2 - (\gamma m v c)^2 = \gamma^2m^2c^2(c^2 - v^2) = \gamma^2m^2c^4(1 - \beta^2) = m^2c^4$ . ✓ Why: the $\gamma^2$ cancels against $(1-\beta^2)$ , exactly as it did in $U\cdot U$ . And the final relation (3) contains no $\gamma$ and no $v$ — which is why it survives the massless limit, where $\gamma$ does not even exist.

Step 6. Take $m = 0$ . Equation (2) gives $p\cdot p = 0$ : the four-momentum is null. Equation (3) gives $\boxed{E = pc}$ . Why: massless particles carry momentum without mass and energy without rest energy. Note carefully that $p^\mu = mU^\mu$ is useless here — it would give $0 \times$ (undefined) — so a photon's four-momentum is defined directly as $(E/c, \vec{p}\,)$ with $E = pc$ , and never via $\gamma$ . `aq-sr-massless-particles` develops this. *Verified symbolically in verify_aq-sr-four-momentum.py: $p\cdot p = +m^2c^2$ from $\eta = \mathrm{diag}(+1,-1,-1,-1)$ with $E = \gamma mc^2$ , $p = \gamma mv$ , for symbolic $m$ , $v$ ; the sign asserted positive (the base's $-m^2c^2$ explicitly rejected); $E^2 - (pc)^2 - (mc^2)^2 = 0$ identically; the massless limit $p\cdot p = 0 \Rightarrow E = pc$ ; and the numbers $E = 1038.272$ , $pc = 444.58 \Rightarrow \sqrt{E^2-(pc)^2} = 938.272$ MeV.*

Go deeper · Derivation B · 2 minOne equation, two conservation laws, every frame

Derivation BOne equation, two conservation laws, every frame

This is the article's biggest idea and it is almost entirely structural — there is barely any algebra in it, which is exactly why it is so often skipped. The claim: Newtonian mechanics needed two independent conservation laws, and had no account of why both should hold; relativity needs one, and gets frame-independence thrown in for nothing. The argument turns on a single property of the Lorentz transformation, and that property is linearity.

Step 1. State the law. For an isolated system, $\displaystyle\sum_{\text{in}} p^\mu_i = \sum_{\text{out}} p^\mu_f$ — one four-vector equation, i.e. four component equations. Why: nothing has been assumed about the interaction. It may be a collision, a decay, an annihilation; the particles need not even be the same ones before and after.

Step 2. Read its components, and recognise old friends. The $\mu = 0$ component is $\sum E_i/c = \sum E_f/c$ — conservation of energy. The $\mu = 1,2,3$ components are $\sum \vec{p}_i = \sum \vec{p}_f$ — conservation of momentum. Why: Newton's two laws are the time-slot and the space-slots of a single statement. They were never independent; they only looked independent because a frame had been chosen and the four-vector chopped along an axis that has no absolute meaning.

Step 3. Now prove it holds in every frame — and watch how little work it takes. Define $\Delta^\mu \equiv \sum_{\text{in}}p^\mu - \sum_{\text{out}}p^\mu$ . This is a sum and difference of four-vectors, hence itself a four-vector, so it transforms as $\Delta'^\mu = \Lambda^\mu{}_\nu\Delta^\nu$ . If $\Delta^\mu = 0$ in one frame, then $\Delta'^\mu = \Lambda^\mu{}_\nu\cdot 0 = 0$ in every frame. Why: this is the whole argument, and it is one line. It works because $\Lambda$ is linear: a linear map sends zero to zero. Had four-momentum not been a four-vector — had we tried to conserve, say, $(E, \vec p\,)$ with a stray $c$ , or $\gamma m$ as a "mass" — the sum would not transform with $\Lambda$ and this step would collapse.

Step 4. Appreciate what has been explained. In Newtonian physics the frame-independence of each law is a separate fact, and the two laws' coexistence is a brute fact with no explanation. Here both are consequences of one four-vector statement plus the linearity of $\Lambda$ . Why: this is what people mean when they say relativity unifies energy and momentum. Not that they are "the same thing" — they plainly are not; they have different units — but that they are components of one geometric object, and a geometric object is conserved or it is not, as a whole.

Step 5. Note the practical consequence, the one experimentalists use daily. Because the law is a four-vector equation, you may evaluate it in whichever frame makes the algebra easiest and trust the answer everywhere. Why: that freedom is worth more in practice than any individual formula here. `aq-sr-conservation-of-four-momentum` and `aq-sr-centre-of-momentum-frame` develop the machinery; what is owned here is the reason it works. *Verified in verify_aq-sr-four-momentum.py: $\Lambda^{\mathsf T}\eta\Lambda = \eta$ for the standard boost; $\Delta^\mu = 0 \Rightarrow \Lambda\Delta = 0$ shown symbolically; and an explicit two-body test — a randomly generated momentum-conserving configuration is boosted through $\beta = 0.6$ and shown to conserve $\sum E$ and $\sum \vec p$ in the new frame to machine precision, while neither $\sum E$ nor $\sum|\vec p\,|$ individually is the same number in the two frames.*

Go deeper · Derivation C · 2 minVelocity is a ratio of components — and the low-speed limit gives Newton back

Derivation CVelocity is a ratio of components — and the low-speed limit gives Newton back

Two loose ends, and they are the ones that connect this chapter to everything on either side of it. Backwards: does relativistic dynamics reduce to Newtonian dynamics at ordinary speeds, and if so, where did the rest energy go? Forwards: what fixes a massless particle's speed at exactly $c$ ? Both questions are answered by the same manoeuvre — take a ratio of four-momentum components — and the second answer is startlingly rigid.

Step 1. Form the ratio. From $p^\mu = (\gamma mc,\ \gamma m\vec{v}\,)$ , divide the spatial part by the temporal part: $\dfrac{\vec{p}}{E/c} = \dfrac{\gamma m\vec{v}}{\gamma mc} = \dfrac{\vec{v}}{c}$ , hence $\boxed{\vec{v} = \vec{p}\,c^2/E}$ . Why: the $\gamma m$ cancels top and bottom — the same cancellation that produced velocity addition at `aq-sr-four-velocity`. Velocity is not a fundamental quantity in this formalism; it is a quotient. That is a theme, not a coincidence.

Step 2. Apply it to a massless particle, and watch it be forced. For $m = 0$ , equation (3) gives $E = pc$ exactly. Substituting into Step 1: $v = \dfrac{pc^2}{E} = \dfrac{pc^2}{pc} = \boxed{c}$ . Why: there is no freedom here. A massless particle cannot be slow, cannot be fast, cannot be stopped. It travels at $c$ because its energy and momentum are locked in the ratio $E/p = c$ , and $v$ is that ratio. It also has no rest frame — which is another way of saying $\gamma$ is not merely large but undefined.

Step 3. Now go the other way: expand $E = \gamma mc^2$ for $v \ll c$ . With $\gamma = (1-\beta^2)^{-1/2} = 1 + \tfrac12\beta^2 + \tfrac38\beta^4 + \cdots$ :

E = mc^2\left(1 + \frac{v^2}{2c^2} + \frac{3v^4}{8c^4} + \cdots\right) = \boxed{mc^{2} + \tfrac12 mv^{2}} + \frac{3mv^4}{8c^2} + \cdots

Why: the second term is exactly Newton's kinetic energy. It is not an approximation to it or an analogue of it; it is it. Newtonian mechanics is the first two terms of a series, and the higher terms are what `aq-sr-relativistic-kinetic-energy` is about.

Step 4. Account for the first term, which is the interesting one. $mc^2$ is a constant — it does not depend on $v$ at all. Newtonian mechanics only ever measures energy differences (work done, kinetic energy gained), and a constant cancels out of every difference. So Newton could not have seen the rest energy even in principle, not because his experiments were imprecise but because his framework never asked a question whose answer contained it. Why: it took a process that changes $m$ — a nuclear reaction — to make the term visible. That is why $E_0 = mc^2$ had to wait for radioactivity, and it is the subject of `aq-sr-mass-energy-equivalence`.

Step 5. And the momentum, for completeness. $\vec{p} = \gamma m\vec{v} \to m\vec{v}$ as $\gamma \to 1$ . ✓ Why: both slots of the four-momentum reduce to their Newtonian counterparts, which is the minimum any successor theory must do. *Verified in verify_aq-sr-four-momentum.py: $\vec v = \vec p c^2/E$ symbolically; $m = 0 \Rightarrow v = c$ exactly; the series $\gamma = 1 + \tfrac12\beta^2 + \tfrac38\beta^4 + \cdots$ (coefficients $\tfrac12, \tfrac38$ sympy-confirmed) giving $E = mc^2 + \tfrac12mv^2 + \tfrac38 mv^4/c^2 + \cdots$ ; and the worked proton at $\beta = 0.4282$ , where $\tfrac12mv^2 = 86.0$ MeV against the true $K = 100$ MeV — Newton low by $14.0\%$ .*

Limiting cases

Case	Result	Meaning
$v = 0$	$p^\mu = (mc,\ \vec{0}\,)$ , $\ E = mc^2$	the rest energy — this, and only this, is what $E = mc^2$ says
$v \ll c$	$E \approx mc^2 + \tfrac12mv^2$ , $\ \vec{p} \approx m\vec{v}$	Newton recovered, plus a constant he could never have detected
$m = 0$	$p\cdot p = 0$ , $\ E = pc$ , $\ v = c$	null four-momentum. $\gamma$ is UNDEFINED, not infinite
$E \gg mc^2$	$E \approx pc\left(1 + \dfrac{m^2c^2}{2p^2}\right)$	ultrarelativistic: a massive particle looks massless to leading order
any frame	$p\cdot p = +m^2c^2$	the invariant. $E$ and $\vec p$ change under a boost; this does not
isolated system	$\sum p^\mu$ conserved	one law, and it holds in every frame because $\Lambda$ is linear

Common trapE = mc squared is the REST energy. The general relation is E = gamma m c squared

This is the most-conflated pair in all of physics pedagogy, and the distinction is not pedantry — it is the difference between a point and a curve. $E = mc^2$ is the $v = 0$ case: the energy a particle has when it is not moving, its rest energy, written $E_0 = mc^2$ . The relation that holds at any speed is $\boxed{E = \gamma mc^2}$ , reducing to $mc^2$ only when $\gamma = 1$ . Every popular account that writes " $E = mc^2$ " and then talks about fast-moving particles has silently swapped one for the other. The safe habit: $E_0 = mc^2$ for the rest energy, $E = \gamma mc^2$ for the total, and $E^2 = (pc)^2 + (mc^2)^2$ when you want to avoid $\gamma$ altogether — which, for a massless particle, you must, because $\gamma$ does not exist there.

Common trapThere is one mass, and it does not depend on speed

$m$ in every equation above is the invariant rest mass: a Lorentz scalar, the Minkowski length of $p^\mu$ , the same number in every frame. The old device of absorbing $\gamma$ into the mass — writing $\gamma m$ and calling it "the mass of a moving body" — is not used in this corpus, and it is worth knowing why: $\gamma m$ is not a scalar (it is a frame-dependent component of a four-vector), and treating it as a mass tempts you to write $\vec{F} = (\gamma m)\vec{a}$ , which is simply false — the acceleration is not even parallel to the force. `aq-sr-relativistic-momentum` owns that argument. The discipline here: $m$ is invariant, $\gamma$ lives in the components, and $E = \gamma mc^2$ already says everything the old device tried to say.

Worked examples

1 · A 100 MeV proton: energy, momentum, speed — and how badly Newton misses

A proton is accelerated to a kinetic energy of $K = 100$ MeV. Find its total energy, its momentum, and its speed. Then compare with what Newtonian mechanics would have predicted.

Step 1 — identify. $m_pc^2 = 938.272$ MeV. Total energy is $E = mc^2 + K$ ; momentum from $E^2 = (pc)^2 + (mc^2)^2$ ; speed from $\vec{v} = \vec{p}c^2/E$ , i.e. $\beta = pc/E$ .

Step 2 — total energy and $\gamma$ . $E = 938.272 + 100 = \mathbf{1038.272\ MeV}$ , so $\gamma = E/(mc^2) = \mathbf{1.1066}$ .

Step 3 — momentum, from the invariant. $(pc)^2 = E^2 - (mc^2)^2 = (E - mc^2)(E + mc^2) = 100 \times 1976.544 = 197654.4\ \mathrm{MeV^2}$ , so $pc = \mathbf{444.58\ MeV}$ . (Factorising the difference of squares is the numerically stable route, and worth the habit.)

Step 4 — speed. $\beta = \dfrac{pc}{E} = \dfrac{444.58}{1038.272} = \mathbf{0.4282}$ — the proton is at $42.8\%$ of light speed.

Step 5 — the boundary number. Newton would say $K = \tfrac12mv^2 = \tfrac12 \times 938.272 \times (0.4282)^2 = \mathbf{86.0\ MeV}$ — but the true kinetic energy is $100$ MeV. Newton is low by $14.0\%$ at only $43\%$ of light speed. And his momentum, $p = mv \Rightarrow pc = 938.272 \times 0.4282 = \mathbf{401.8\ MeV}$ , is low against the true $444.58$ MeV. Relativity is not a small correction to a fast-moving proton; it is a $14\%$ effect in a machine most people would call low-energy.

2 · The invariant is invariant: boost the same proton and check

Take the proton of Example 1 and view it from a frame moving at $V = 0.6c$ along its direction of motion. Show that $E$ and $p$ both change, while $E^2 - (pc)^2$ does not.

Step 1 — identify. Boost the four-momentum $p^\mu = (E/c,\ p)$ with $\Lambda$ , exactly as we would any four-vector: $E' = \gamma_V(E - Vp)$ , $\ p'c = \gamma_V(pc - \beta_V E)$ .

Step 2 — the boost factor. $\gamma_V = 1/\sqrt{1-0.36} = \mathbf{1.25}$ .

Step 3 — the new components. $E' = 1.25\,(1038.272 - 0.6 \times 444.58) = 1.25 \times 771.52 = \mathbf{964.40\ MeV}$ , and $p'c = 1.25\,(444.58 - 0.6 \times 1038.272) = 1.25 \times (-178.38) = \mathbf{-222.97\ MeV}$ . Both changed, and the momentum even reversed sign — in this frame the proton is moving backwards, because we overtook it.

Step 4 — the invariant. $E'^2 - (p'c)^2 = 964.40^2 - (-222.97)^2 = 930{,}072 - 49{,}718 = \mathbf{880{,}354}$ , and $\sqrt{880{,}354} = \mathbf{938.272\ MeV}$ — the rest energy, recovered exactly, and identical to the lab-frame value $1038.272^2 - 444.5834^2 = 880{,}354$ .

Step 5 — the boundary number. $m = 938.272$ MeV $/c^2$ in both frames, and in every other. The components of the four-momentum are opinions; its length is a fact. That is why particle physicists reconstruct invariant masses rather than energies — and why a bump in an invariant-mass histogram means a particle, while a bump in an energy spectrum means someone changed the beam. (`aq-sr-invariant-mass` extends this from one particle to a system.)

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §§ on relativistic mechanics · Taylor & Wheeler, Spacetime Physics, 2nd ed., ch. 7 · Morin, Special Relativity, ch. 5 · Griffiths, Introduction to Elementary Particles, 2nd ed., ch. 3 · Jackson, Classical Electrodynamics, 3rd ed., §11.5 · Planck, M., "Das Prinzip der Relativität und die Grundgleichungen der Mechanik", Verh. Deutsch. Phys. Ges. 8, 136 (1906) · Minkowski, H., "Die Grundgleichungen für die elektromagnetischen Vorgänge in bewegten Körpern", Nachr. Ges. Wiss. Göttingen (1908).

Provenance. Signature $(+,-,-,-)$ , $\eta = \mathrm{diag}(+1,-1,-1,-1)$ , so $U\cdot U = +c^2$ and $\boxed{p\cdot p = +m^2c^2}$ . The AllQFT base for this article gives $p_\mu p^\mu = -m^2c^2$ and $u_\mu u^\mu = -c^2$ — correct in $(-,+,+,+)$ , WRONG here. The norm is re-derived from $\eta$ in the first derivation card and the base's sign is not used. Everything is recomputed in verify_aq-sr-four-momentum.py, which also greps this article for each printed value: $p\cdot p = +m^2c^2$ and $E^2 = (pc)^2 + (mc^2)^2$ symbolically; $\Lambda^{\mathsf T}\eta\Lambda = \eta$ and the linearity argument for frame-independent conservation, with an explicit boosted two-body check; $\vec v = \vec p c^2/E$ and $m = 0 \Rightarrow v = c$ ; the series $E = mc^2 + \tfrac12mv^2 + \tfrac38mv^4/c^2 + \cdots$ ; and the proton at $K = 100$ MeV — $E = 1038.272$ MeV, $\gamma = 1.1066$ , $pc = 444.58$ MeV, $\beta = 0.4282$ , Newton's $\tfrac12mv^2 = 86.0$ MeV ( $14.0\%$ low) and $p = mv \Rightarrow 401.8$ MeV — boosted through $V = 0.6c$ to $E' = 964.40$ MeV, $p'c = -222.97$ MeV, invariant $938.272$ MeV recovered exactly. Canon constants: $m_pc^2 = 938.272$ MeV.

Scope. $U\cdot U = +c^2$ is derived at `aq-sr-four-vectors` and recalled by name. The extended energy–momentum treatment (ultrarelativistic expansions, dispersion relations) is `aq-sr-energy-momentum-relation`; massless particles are `aq-sr-massless-particles`; the four-force is `aq-sr-four-force`; the collision machinery and the centre-of-momentum frame are `aq-sr-conservation-of-four-momentum` and `aq-sr-centre-of-momentum-frame`. The system invariant mass — where $M \ne \sum m_i$ — is `aq-sr-invariant-mass`. The argument that killed " $\gamma m$ as a mass" is owned by `aq-sr-relativistic-momentum`.

Builds on. Four-Velocity · Relativistic Momentum · Relativistic Energy · Invariant Mass. Leads to. The Energy–Momentum Relation · Conservation of Four-Momentum · Four-Force · Massless Particles.

Why is the norm +m²c² and not −m²c²? I have seen both.

Both are correct — in different signatures. In $(+,-,-,-)$ , which this chapter uses throughout, timelike vectors have positive norm, so $p\cdot p = +m^2c^2$ . In $(-,+,+,+)$ , favoured by much of the general-relativity literature, timelike is negative and the same physics reads $-m^2c^2$ . What is not acceptable is mixing them, and the mixing is insidious because no numerical answer changes: $E$ , $p$ , $m$ and every threshold come out identical while every intermediate sign is inverted, so no unit test catches it. The base for this article carries the $(-,+,+,+)$ value; we re-derived from $\eta$ rather than copy. Read the metric before you read the formula.

Does a fast-moving object get heavier?

No. Its energy increases, as $E = \gamma mc^2$ , and its momentum increases, as $\vec{p} = \gamma m\vec{v}$ — but $m$ , the invariant rest mass, does not change at all. It cannot: it is the Minkowski length of the four-momentum, and boosts change components, not lengths. The older habit of calling $\gamma m$ "the relativistic mass" packaged the $\gamma$ into the mass so that $\vec{p} = m_{\text{rel}}\vec{v}$ would look Newtonian, and it was abandoned for good reasons: $\gamma m$ is not a Lorentz scalar, and it tempts you to write $\vec F = m_{\text{rel}}\vec a$ , which is false — the acceleration is not even parallel to the force. `aq-sr-relativistic-momentum` gives the proof. Modern practice: one mass, invariant, with $\gamma$ in the components where it belongs.

If a photon has no mass, how can it have momentum?

Because momentum is not $m\vec{v}$ — that is the low-speed limit of $\vec{p} = \gamma m\vec{v}$ , and neither expression is the definition. Momentum is the spatial part of the four-momentum, and the four-momentum of a massless particle is defined directly as $p^\mu = (E/c,\ \vec{p}\,)$ with $|\vec{p}\,| = E/c$ , so that $p\cdot p = 0$ . It carries momentum $p = E/c$ with $m = 0$ , and there is no contradiction anywhere, because at no point did anything require $p \propto m$ . This is not a special exception carved out for light: it is what $E^2 = (pc)^2 + (mc^2)^2$ says when you set $m = 0$ , and that equation contains no $\gamma$ and no $m$ in a denominator.

Where did the rest energy go in Newtonian mechanics?

Nowhere — it was always there and simply invisible, for a structural reason rather than an experimental one. Expanding $E = \gamma mc^2 = mc^2 + \tfrac12mv^2 + \cdots$ , the leading term is a constant: it does not depend on $v$ . Newtonian mechanics only ever measures energy differences — work done, kinetic energy gained — and a constant cancels identically out of every difference. So no Newtonian experiment, however precise, could have revealed it. It took a process that actually changes $m$ — a nuclear reaction, where the products weigh less than the reactants — to make the term observable. That is `aq-sr-mass-energy-equivalence`.

Why do particle physicists work with invariant mass rather than energy?

Because energy is a component and mass is a length. Two detectors in different frames — or one detector reconstructing a decay whose parent was moving — disagree about $E$ and $\vec{p}$ , but agree exactly about $\sqrt{p\cdot p}$ . So a peak in an invariant-mass histogram is a statement about a particle, while a peak in an energy spectrum is at best a statement about a particle and the beam. That is why the Higgs discovery plot is a histogram of $\sqrt{(p_1+p_2)\cdot(p_1+p_2)}$ . The extension from one particle to a system — where the invariant mass is emphatically not the sum of the parts' masses — is `aq-sr-invariant-mass`.

Is the four-momentum conserved, or is it the energy and momentum separately?

It is one statement, and asking which of the two it "really" is misses the point. $\sum p^\mu_{\text{in}} = \sum p^\mu_{\text{out}}$ has four components; its time component is energy conservation and its space components are momentum conservation. They are not two facts that happen to coincide — they are one geometric fact, read in a frame. The proof that this survives a change of frame is a single line ( $\Lambda$ is linear, so it maps zero to zero), whereas in Newtonian mechanics the frame-independence of each law is a separate argument and their coexistence has no explanation at all. That economy is the strongest argument for the whole four-vector apparatus.