Relativistic Momentum and Energy

The kinematics of special relativity forces a matching change in dynamics. Classical momentum $\vec{p}=m\vec{u}$ is not conserved in every inertial frame once velocities transform by the relativistic rule, and Newton's $\vec{F}=m\vec{a}$ cannot be right if it leads to conservation of $m\vec{u}$. Repairing conservation gives new definitions of momentum and energy, both reducing to the classical forms at low speed, and it ties mass to energy.¹

Relativistic momentum

Consider an elastic glancing collision of two identical balls, symmetric between frames $S$ and $S^{\prime }$ that move relative to each other at speed $v$. Requiring the $y$-momentum to balance in both frames, in the limit where each ball is nearly at rest in its home frame, forces the momentum of a ball moving at speed $u$ to be

Here $m$ is the rest mass, the mass measured in the ball's own frame, and ${\gamma }_u$ uses the particle's speed $u$ relative to the observer (distinct from the $\gamma$ of a frame's relative velocity). The definition meets the two requirements: momentum stays conserved across frames, and for $u\ll c$ the denominator is $\approx 1$ and $\vec{p}\to m\vec{u}$.

Because ${\gamma }_u\to \infty$ as $u\to c$, the momentum grows without bound as the speed approaches $c$, while the classical $mu$ would level off at $mc$. No finite momentum transfer brings a massive particle to the speed of light.

The relativistically correct force keeps Newton's second law in the form $\vec{F}=\mathbf{d}\vec{p}/\mathbf{d}t$, now with the relativistic $\vec{p}$:

Relativistic energy

Kinetic energy is the work a net force does in accelerating a particle from rest to speed $u$. In one dimension,

Using $\mathbf{d}({\gamma }_{u}mu)=m(1-u^2/c^2{)}^{-3/2}\mathbf{d}u$ and integrating gives

For $u\ll c$, the binomial expansion ${\gamma }_u\approx 1+\frac{1}{2}{u}^2/c^2$ recovers $E_k\approx \frac{1}{2}m{u}^2$. The exact expression diverges as $u\to c$: bringing a massive particle to the speed of light would take infinite energy, the dynamical statement of the same speed limit the momentum shows.

The kinetic energy splits into a speed-dependent piece ${\gamma }_{u}m{c}^2$ and a constant $m{c}^2$. The constant is the rest energy, and the total energy is their sum:

Total energy $E$, not $E_k$ and not $m{c}^2$ separately, is the conserved quantity for an isolated system in each frame. Conservation and invariance are different properties: a conserved quantity is unchanged before and after an interaction within one frame, while an invariant quantity has the same value in all frames. Total energy is conserved but not invariant; different frames measure different $E$.

The energy-momentum four-vector

Energy and momentum transform between frames together, exactly as time and space do. Writing out the transformation for a particle's $E$ and $\vec{p}$ between $S$ and $S^{\prime }$ moving at relative speed $v$,

The momentum $\vec{p}$ transforms like the position $\vec{r}$, and the energy $E$ transforms like the time $t$. Momentum and energy are the components of a four-vector, and like the invariant interval $(\Delta s{)}^2=(c\Delta t{)}^2-(\Delta x{)}^2$, this four-vector has an invariant length. Squaring the definitions of $E$ and $p$ and subtracting gives

The relation is a right triangle in $(E,pc,m{c}^2)$: the total energy is the hypotenuse, momentum times $c$ and rest energy the legs. At low speed $pc\ll m{c}^2$ and $E\approx m{c}^2+p^2/2m$; at high speed $pc\gg m{c}^2$ and $E\approx pc$.

Massless particles

The relation $E^2=(pc{)}^2+(m{c}^2{)}^2$ allows $m=0$, which has no classical analog. Setting $m{c}^2=0$,

and combined with $u/c=pc/E$ this gives $u=c$. A massless particle moves at the speed of light in every frame, carrying energy and momentum despite zero rest mass; the product ${\gamma }_{u}m$ stays finite as $m\to 0$ and ${\gamma }_u\to \infty$. The photon is the established example, with the gluon and the hypothetical graviton the others. For the photon the interval and the rest mass both vanish, and rest frame has no meaning, since light moves at $c$ relative to all frames.

The invariant mass of a system is not the sum of its parts' masses when the parts move relative to one another. Two photons of energies $5\mathrm{MeV}$ and $2\mathrm{MeV}$ approaching each other have total energy $7\mathrm{MeV}$ and net momentum $(5-2)/c=3\mathrm{MeV}/c$, so the system rest energy is $\sqrt{7^2-3^2}=6.3\mathrm{MeV}$, nonzero even though each photon is massless.

Binding energy

When particles are bound together, energy $E_b$ must be supplied to separate them, so the bound system's mass is less than the sum of its parts by $E_b/c^2$.

The deuteron makes the effect visible. It is a proton and neutron bound together. The rest energies of the free proton and neutron sum to $938.272+939.565=1877.837\mathrm{MeV}$, while the deuteron's rest energy is $1875.613\mathrm{MeV}$. The difference,

is the deuteron's binding energy: $2.224\mathrm{MeV}$ must be added to split it, and the same energy is released when a proton and neutron fuse into a deuteron.

The same accounting drives every energy release from the nucleus. Fission of a heavy nucleus and fusion of light ones both convert a mass defect into kinetic energy of the products, and the amounts are large because nuclear binding energies are millions of times the electron-volt binding energies of atoms. The nuclear physics module follows that thread; the mass-energy relation established here is its foundation. The last lesson leaves inertial frames behind and asks what happens under acceleration and gravity, in general relativity.