Energy and Momentum from Symmetry: a Unified Derivation

This note derives the kinetic energy and momentum of a free particle in both Galilean and special relativity from a single common axiom set. The two theories are two specializations of one functional equation; the only difference is a small “package” — the boost (velocity vs rapidity) together with one companion axiom (mass conserved vs total energy conserved). Everything else is identical.

The detailed case-by-case derivations live in Kinetic Energy is Quadratic: a Galilean-Invariance Derivation; this note’s role is to exhibit their common root.

The common axioms¶

Both theories share, verbatim:

• (A1) Energy is a state function of a velocity variable $\xi$, extensive in mass, isotropic. A body of mass $m$ at “velocity-state” $\xi$ has energy $m\,\mathcal E(\xi)$, with $\mathcal E$ even: $\mathcal E(-\xi)=\mathcal E(\xi)$. (The variable $\xi$ is velocity $v$ for Galilean, rapidity $\phi$ for relativity — see the packages.)

• (A2) Relativity principle. The laws are the same in every inertial frame; frames are related by boosts.

• (A3) Parity. The symmetric collision — two equal masses approaching at $\pm\xi$ — leaves the blob at rest. (The setup is invariant under $\xi\mapsto-\xi$, so the blob cannot move either way.)

• (A4) Collision balance with frame-invariant bookkeeping. In an inelastic collision,

  $$\sum_i m_i\,\mathcal E(\xi_i)=M\,\mathcal E(0)+Q,$$

  (2)

  where $M$ is the blob’s mass and $Q\ge0$ the dissipated energy. Both $M$ and $Q$ are frame-invariant scalars (the same number in every inertial frame).

• (A5) Regularity. $\mathcal E$ is Lebesgue measurable.

Axiom (A4) is the heart: it says the collision’s two bookkeeping quantities — the mass $M$ and the dissipation $Q$ — are each frame-invariant. What distinguishes the two theories is which of them is independently fixed.

The package: the only difference¶

Each theory selects a boost and fixes one of the two frame-invariant scalars in (A4):

The two rows are not independent choices: they come locked as a pair (see Why the packages are locked). Picking the Galilean boost forces mass conservation for a sensible theory; picking the Lorentz boost forces total-energy conservation.

The single functional equation¶

Derive (1). Take the symmetric collision: two masses $m$ at $\pm\xi$, blob at rest (A3) of mass $M$. The lab-frame balance (A4) is

View it from a frame boosted by $\beta$ (A2). The particles move to $\xi-\beta$ and $-\xi-\beta$; the blob, still of mass $M$ (frame-invariant, A4), moves to $-\beta$. The balance there is

(The blob’s energy is $M\mathcal E(\beta)$ by evenness; the dissipation $Q$ is the same number, frame-invariant by A4.) Subtract (3) from (4) and use evenness:

This is the entire theory before the package is chosen. Now specialize.

Galilean specialization¶

$M=2m$ (mass conserved), $\xi=v$. Then $M/m=2$ and (5) becomes

Writing the kinetic energy $T(v):=\mathcal E(v)-\mathcal E(0)$ (so $T(0)=0$),

the parallelogram law. With regularity (A5) its solutions are $T(v)=c\,v^2$ (the Hamel-basis pathologies excluded by measurability). Hence

The constant $c$ is fixed by energy units (Newton’s convention $c=\tfrac12$, giving $\tfrac12mv^2$); the rest energy $\mathcal E(0)$ is a free additive constant — in Galilean mechanics energy is shiftable, so we conventionally set $\mathcal E(0)=0$.

Relativistic specialization¶

$Q=0$ (total energy conserved), so (3) fixes $M=2m\,\mathcal E(\xi)/\mathcal E(0)$, i.e. $M/m=2\,\mathcal E(\xi)/\mathcal E(0)$. With $\xi=\phi$, (5) becomes

d’Alembert’s equation. With regularity and the physical selection (energy $\ge$ rest, minimized at $\phi=0$), its solution is

The non-relativistic limit fixes the constant: $E_{\text{kin}}=m[\mathcal E(\phi)-\mathcal E(0)]=m\,\mathcal E(0)(\cosh\phi-1)\approx m\,\mathcal E(0)\,v^2/(2c^2)$, and matching $\tfrac12mv^2$ forces

Note the contrast with the Galilean case: here $\mathcal E(0)=c^2$ is forced, not a free constant — Lorentz covariance forbids shifting the rest energy. And the blob’s mass, $M=2m\cosh\phi=2\gamma m$, grows: the dissipated kinetic energy became rest mass, $Q_{\text{rel}}=\Delta M\,c^2$ — itself a derived theorem, not an assumption.

Momentum, unified¶

Momentum is derived from the same collision by the same move (write momentum balance in the boosted frame), giving a second functional equation that is the exact companion of (5).

Assume momentum is extensive in mass (a body of mass $m$ at $\xi$ carries momentum $m\,\pi(\xi)$, with $\pi$ odd) and conserved in every frame. The boosted-frame balance for the symmetric collision reads

Oddness on both sides gives the unified momentum equation

Galilean ($M/m=2$, $\xi=v$): $\pi(v+b)-\pi(v-b)=2\pi(b)$. Swapping $v\leftrightarrow b$ and using oddness yields $\pi(v+b)+\pi(v-b)=2\pi(v)$; adding gives $\pi(v+b)=\pi(v)+\pi(b)$, Cauchy’s equation, whose regular solution is $\pi(v)=v$ (the slope is a free unit convention, set to 1 so that $p=mv$). Thus

Relativistic ($M/m=2\cosh\phi$, $\xi=\phi$): $\pi(\phi+\beta)-\pi(\phi-\beta)=2\cosh\phi\,\pi(\beta)$. Swapping and adding gives $\pi(\phi+\beta)=\cosh\phi\,\pi(\beta)+\cosh\beta\,\pi(\phi)$, the $\sinh$ addition formula, whose regular solution is $\pi(\phi)=c\sinh\phi$ — here the prefactor is forced to the invariant speed $c$ by the non-relativistic limit $p\to mv$. Thus

So the same two equations, (5) and (13), produce both $\{E_{\text{kin}}=\tfrac12mv^2,\ p=mv\}$ and $\{E_{\text{kin}}=(\gamma-1)mc^2,\ p=\gamma mv\}$, distinguished only by the package. The energy equation fixes the “even” function ($\cosh$ or quadratic) and the momentum equation its “odd” companion ($\sinh$ or linear); together they assemble the Galilean pair $(p,E_{\text{kin}})=(mv,\tfrac12mv^2)$ or the relativistic four-momentum $P^\mu=(E/c,p)=(\gamma mc,\gamma mv)$.

Four-momentum and the conservation law¶

The momentum derived above is not an isolated three-vector. In both spacetimes it is the spatial part of a four-momentum $P^\mu=m\,u^\mu$ (mass times four-velocity). The relativistic time slot is $E/c$; the Galilean time slot is $m$. Thus the Galilean object is more precisely a mass-momentum four-vector, not an energy-momentum four-vector.

This section builds that object in full for both theories and proves a key structural fact: once the appropriate components are conserved, they assemble into a single frame-independent conservation law; conversely, frame-independence couples the components, so momentum conservation cannot hold without the time component (mass, or energy) being conserved too.

What “four-vector” means here¶

There are two levels of covariance to keep distinct.

First, in flat spacetime with inertial Cartesian/adapted frames, the admissible inertial frame changes are the ten continuous symmetries: rotations, translations, boosts, and time translations (plus optional discrete reflections). These are the Galilei transformations in Galilean spacetime and the Poincare transformations in Minkowski spacetime. Checking how $P^\mu$ transforms under these transformations proves that its components form the correct inertial-frame four-vector.

Second, in the differential-geometric sense, a tensor is a coordinate-independent object whose components transform tensorially under arbitrary smooth coordinate changes. That stronger statement is proved not by checking every coordinate system, but by giving a coordinate-free definition: $u^\mu$ is the tangent to the worldline, and $P^\mu=m u^\mu$ with $m$ a scalar. Then the tensor transformation law follows by the chain rule. The Galilei and Poincare formulas below are the inertial-coordinate special cases of this general tensorial statement.

Also, tensorial means covariant, not invariant. Components may change from one frame to another; what matters is that they change by the correct transformation law.

Definition versus conservation¶

Two claims must be kept separate.

First, $P^\mu=m u^\mu$ is the definition of four-momentum. Since $u^\mu$ is the worldline tangent and $m$ is a scalar, this also proves that $P^\mu$ is a vector. This is a kinematic/geometric statement: it determines how $P^\mu$ transforms between frames, but it does not by itself say that $P^\mu$ is conserved.

Second, conservation is an additional dynamical statement:

The earlier collision setup supplies or assumes the component conservation laws: in Galilean mechanics, mass conservation plus momentum conservation; in relativity, energy conservation plus momentum conservation. Once these components are conserved in one inertial frame, the vector character of $P^\mu$ packages them into one law and makes that law frame-independent. Thus $P^\mu=m u^\mu$ proves tensoriality, while the collision/symmetry argument supplies conservation.

The four-velocity¶

Parametrize a particle’s worldline by the time natural to each spacetime.

Relativistic — Minkowski spacetime. The invariant parameter is the proper time $\tau$, related to coordinate time by $d\tau=dt/\gamma$. The four-velocity is

It is a four-vector; under a Lorentz transformation $u'^\mu=\Lambda^\mu{}_\nu u^\nu$, with constant Minkowski norm $\eta_{\mu\nu}u^\mu u^\nu=\gamma^2(c^2-v^2)=c^2$.

Galilean — absolute time. Time is absolute, so in adapted coordinates the natural parameter is coordinate time $t$. Equivalently, the Newton-Cartan clock one-form $\tau_\mu$ normalizes the worldline tangent by $\tau_\mu u^\mu=1$. In such adapted coordinates the four-velocity is

Under a Galilean boost, including a possible rotation ($\mathbf x'=R\mathbf x-\mathbf b\,t+\mathbf a$, $t'=t+a^0$),

so for $u^0=1$ one has $\mathbf v'=R\mathbf v-\mathbf b$ — precisely the transformation law of a Galilean four-vector.

The four-momentum $P^\mu=m\,u^\mu$¶

Multiply by the rest mass $m$ (a frame-invariant scalar in both theories):

The same formula $P^\mu=m\,u^\mu$ holds in both. Since $m$ is a scalar and $u^\mu$ is the worldline tangent, $P^\mu$ is a genuine vector in the generally covariant formulation. The only difference is what occupies the time slot in adapted inertial coordinates — $E/c$ (relativistic) or $m$ (Galilean).

For a Galilean boost this gives

The term $-m\mathbf b$ has the elementary meaning $m(\mathbf v-\mathbf b)=m\mathbf v-m\mathbf b$: changing to a frame moving with velocity $\mathbf b$ subtracts the frame’s velocity from every particle velocity, and therefore subtracts $m\mathbf b$ from its momentum. This is why $\mathbf p$ alone is not a Galilean spacetime vector. It is a three-vector under rotations, but under boosts it is not closed by itself; it needs the mass component $P^0=m$.

The metric then reads off the “mass-shell”:

• Relativistic, non-degenerate $\eta_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$:

  $$\eta_{\mu\nu}P^\mu P^\nu=(E/c)^2-|\mathbf p|^2=m^2c^2 \quad\Longleftrightarrow\quad E^2=p^2c^2+m^2c^4.$$

  (21)

• Galilean, the degenerate pair $\tau_\mu=(1,0,0,0)$ (temporal) and $h^{\mu\nu}=\mathrm{diag}(0,1,1,1)$ (spatial), with $\tau_\mu h^{\mu\nu}=0$: the temporal one-form reads off the mass,

  $$\tau_\mu P^\mu=m,$$

  (22)

  and the spatial part is $\mathbf p$. There is no non-degenerate quadratic form in $P^\mu$ yielding an energy; instead $E=|\mathbf p|^2/(2m)=|\mathbf p|^2/\bigl(2\,\tau_\mu P^\mu\bigr)$, a ratio — the algebraic fingerprint of the degenerate metric.

Conservation, and a one-line proof of frame-independence¶

For a collision with incoming momenta $P_i^\mu$ and outgoing $P_j^\mu$, conservation of four-momentum is the single statement

Because $P^\mu$ is a four-vector it transforms linearly, $P'^\mu=\Lambda^\mu{}_\nu P^\nu$, so (23) holds in every inertial frame:

Linearity of $\Lambda$ is the whole proof. Reading off components gives the conservation laws:

• relativistic: $\sum_{\rm in}E=\sum_{\rm out}E$ (energy) and $\sum_{\rm in}\mathbf p=\sum_{\rm out}\mathbf p$ (momentum);

• Galilean: $\sum_{\rm in}m=\sum_{\rm out}m$ (mass) and $\sum_{\rm in}\mathbf p=\sum_{\rm out}\mathbf p$ (momentum).

The components are coupled (the heart of the matter)¶

The frame-independence proof conceals the key point: a boost mixes the time and space components, so conserving one component alone does not survive a change of frame. The components are locked together. In relativity the mixing is two-way: $E/c$ and $\mathbf p$ are not separate Lorentz tensors, only components of the four-vector $P^\mu$. In Galilean spacetime the mixing is triangular: the mass component is invariant, but the spatial momentum still receives a contribution from it. Thus $\mathbf p$ is a tensor only for the reduced rotation subgroup, not for the full Galilei group with boosts.

Galilean — momentum conservation forces mass conservation. Assume only that momentum is conserved in the lab, $\sum_{\rm in}\mathbf p=\sum_{\rm out}\mathbf p$, and ask whether it holds in a frame boosted by $\mathbf b$. Since $\mathbf p'=\mathbf p-m\mathbf b$,

Subtracting,

For this to vanish for every boost $\mathbf b$,

So frame-independent momentum conservation entails mass conservation. The two are inseparable: they are the time and space slots of one four-vector, and the boost couples them.

Relativistic — momentum conservation forces energy conservation. Take a Lorentz boost of rapidity $\beta$ along $x$,

and assume only $\sum_{\rm in}p_x=\sum_{\rm out}p_x$. Then

which vanishes for every $\beta$ only if $\sum_{\rm in}E=\sum_{\rm out}E$. So in relativity frame-independent momentum conservation entails energy conservation, and conversely. The invariant $E^2-p^2c^2=m^2c^4$ is the same coupling in static form.

What this explains, and the Galilean asymmetry¶

The coupling is exactly what the package demanded and what our derivations produced, now seen geometrically. In the Galilean package the companion axiom was mass conservation; the four-vector shows it is forced on us by frame-independent momentum conservation — the two come as one. In the relativistic package the companion axiom was total-energy conservation; the four-vector shows it is forced by frame-independent momentum conservation just the same.

The asymmetry, now sharp: the Galilean four-momentum $(m,\mathbf p)$ contains mass and momentum, not energy. Energy conservation in Galilean mechanics is therefore a separate law — outside the four-vector — and (as derived in Kinetic Energy is Quadratic: a Galilean-Invariance Derivation) it follows from the frame-invariance of heat, not from four-vector structure. Relativity needs no such extra law: its four-momentum already carries the energy in the time slot, so the single statement (23) covers everything. One conservation law (relativistic) versus one plus an extra (Galilean) — that is the precise cost of the degenerate metric.

Adding gravity: from inertial forces to curvature¶

The derivation so far was the flat-spacetime theory. If we write the same theory in arbitrary coordinates, including accelerating or rotating coordinates, the equations already contain connection terms. Those terms are fictitious forces: they come from a non-inertial coordinate choice, and the corresponding curvature is still zero. True gravity begins when the connection has nonzero curvature, so that no coordinate choice can remove the effect everywhere.

The bottom-up extension to gravity follows the same pattern in the Galilean and relativistic theories. The common steps are:

1. Universal free fall. All test bodies fall the same way, independently of their mass and composition.

2. Geometrize the universal force. Since the acceleration is universal, it should not be a force attached separately to each body. It is encoded in a common spacetime connection $\nabla$, and free particles follow geodesics.

3. Only tidal effects are invariant. A uniform gravitational field can be removed locally by a freely falling frame. Thus the connection coefficients themselves are not the invariant gravitational field; curvature is.

4. Choose the conserved source. Galilean gravity is sourced by mass density. Relativistic gravity is sourced by the full stress-energy tensor.

5. Locality and second-order simplicity. The field equation should be local, covariant, and second order in the gravitational variables.

6. Fix the normalization. The remaining constant is fixed by the observed Newtonian weak/static limit; symmetry fixes the form, not the numerical value of $G$.

Universal free fall gives the geodesic equation¶

The equivalence principle is the shared starting point:

So the equation of motion is literally the same tensor equation in both theories:

The difference lies in the background geometric structure: Newton-Cartan has absolute time and a degenerate spatial metric; GR has a non-degenerate Lorentzian metric.

In adapted Newton-Cartan coordinates, $\tau=dt$ and $u^\mu=(1,\dot{\mathbf x})$. For an ordinary Newtonian potential $\Phi$, the gravitational connection component is

so (30) becomes

In GR, the same geodesic equation in the weak, slowly moving, static limit with

also reduces to

Thus both theories agree on the motion law in the Newtonian limit.

Curvature is the gravitational field¶

A freely falling coordinate system can set the connection coefficients to zero at one event. Therefore $\Gamma^\mu_{\nu\rho}$ itself is not the invariant gravitational field. The invariant obstruction to removing gravity throughout a region is curvature.

In Newton-Cartan theory, with $\Gamma^i_{00}=\partial^i\Phi$, the tidal tensor is the second derivative of the potential:

A uniform field has $\partial_i\Phi=\text{constant}$ and hence zero curvature; it is locally removable by an accelerating frame. Tidal gravity is what remains.

In GR, the same statement is expressed by the Riemann tensor $R^\mu{}_{\nu\rho\sigma}$. In a local inertial frame the connection can be made to vanish at a point, but the curvature tensor generally cannot.

The field equation: curvature sourced by the conserved quantity¶

The final step is to connect curvature to matter.

Galilean / Newton-Cartan. The source selected by the flat theory is mass. Let $\rho$ be the mass density. Locality, spatial rotational invariance, and second-order simplicity select the scalar second-derivative equation

Writing $C=4\pi G$ gives Poisson’s equation,

In Newton-Cartan tensor form this is

In adapted coordinates, the only nontrivial component is $R_{00}=4\pi G\rho$, exactly (37).

For a point mass, $\rho=M\delta^{(3)}(\mathbf x)$, the rotationally symmetric solution is

Thus the inverse-square law follows from locality, three-dimensional rotational symmetry, and the second-order Poisson equation; $G$ is the empirical coupling constant.

Relativistic / GR. The source selected by the relativistic flat theory is the full stress-energy tensor $T_{\mu\nu}$, not just mass density. The left-hand side must be a symmetric, generally covariant, divergence-free tensor built locally from the metric and at most two derivatives. In four dimensions, the Lovelock uniqueness theorem gives

as the unique possibility of this type, where $G_{\mu\nu}:=R_{\mu\nu}-\tfrac12 Rg_{\mu\nu}$ and $\nabla^\mu G_{\mu\nu}=0$ by the Bianchi identity. Therefore the field equation must be

The coefficient is fixed by the weak/static limit: with $T_{00}\approx\rho c^2$ and $g_{00}=1+2\Phi/c^2$, the 00 component of (41) reduces to

The cosmological constant $\Lambda$ is allowed by the same covariance and second-order axioms; it is an additional empirical constant, set to zero if one demands exactly flat empty spacetime.

The parallel structure is therefore:

With gravity present, the flat-spacetime global conservation law is also replaced by a local covariant statement. In GR this is $\nabla_\mu T^{\mu\nu}=0$ for matter (with the gravitational field included geometrically rather than as a separate local stress-energy tensor). In Newton-Cartan theory, mass conservation remains the local continuity equation for the mass current, while momentum balance is expressed using the Newton-Cartan connection. Global conserved energies or momenta require extra spacetime symmetries, just as in GR.

The Galilean theory is the $c\to\infty$ limit¶

The two packages are not parallel alternatives; one is the singular limit of the other. In the relativistic solution let $c\to\infty$ at fixed $v$. Then rapidity $\phi=\operatorname{artanh}(v/c)\to v/c\to0$, $\cosh\phi\to 1+v^2/(2c^2)$, and

The rest energy $c^2$ diverges and drops out of the kinetic energy $E_{\text{kin}}=\mathcal E-c^2\to\tfrac12 v^2$; the blob’s mass $M\to2m$ (mass conservation re-emerges); the mass excess $\Delta M=2m(\cosh\phi-1)\to mv^2/c^2\to0$ vanishes, while the heat $Q=\Delta M\,c^2\to mv^2$ stays finite. In words: as $c\to\infty$ the relativistic bookkeeping “$Q$ becomes rest mass” degenerates into the Galilean bookkeeping “mass is conserved, $Q$ is heat” — the heat is the $c\to\infty$ shadow of the relativistic rest-mass gain. This is why the two packages are locked: the Galilean one is the singular limit of the relativistic one, not an independent option.

Why the packages are locked¶

One might hope to mix and match — e.g. Lorentz boosts with mass conservation. Equation (5) shows this fails: with $M=2m$ (mass conserved) but $\xi=\phi$ (rapidity), it becomes the parallelogram law in the rapidity variable, $T(\phi+\beta)+T(\phi-\beta)=2T(\phi)+2T(\beta)$, whose solution $T(\phi)\propto\phi^2$ grows only quadratically with rapidity and does not match the observed $\gamma mc^2$. Conversely, Galilean boosts with $Q=0$ (total energy conserved) give d’Alembert’s equation in the velocity variable, $T(v)\propto\cosh(v)$, exponential in speed. Both cross-combinations are mathematically consistent but describe worlds that are not ours.

So the boost and the bookkeeping are not independent dials: a physically correct theory requires them matched. Choosing the boost effectively chooses its companion axiom.

Summary¶

The common axioms (A1)–(A5) — energy extensive and isotropic, relativity, parity, the frame-invariant collision balance, regularity — produce one energy equation (5) and one momentum equation (13). The sole difference is the package: {velocity boost, mass conserved} versus {rapidity boost, energy conserved}. From this single fork, both $\tfrac12mv^2,\ mv$ and $(\gamma-1)mc^2,\ \gamma mv$ unfold.