Relativistic kinematics and dynamics

Intuition [Beginner]

In Newtonian mechanics, momentum is $\mathbf{p}=m\mathbf{v}$ and kinetic energy is $\frac{1}{2}m{v}^2$. These formulae work at everyday speeds but fail as $v$ approaches the speed of light $c$. Special relativity replaces them with expressions that involve the Lorentz factor $\gamma =1/\sqrt{1-v^2/c^2}$.

The relativistic momentum is

$$\mathbf{p}=\gamma m\mathbf{v}.$$

At low speeds ($v\ll c$), $\gamma \approx 1$ and you recover $\mathbf{p}=m\mathbf{v}$. At high speeds, $\gamma >1$ and the momentum grows faster than the velocity. As $v\to c$, $\gamma \to \infty$: the momentum grows without bound.

The relativistic energy is

$$E=\gamma m{c}^2.$$

When the particle is at rest ($v=0$, $\gamma =1$), this gives $E=m{c}^2$. This is the rest energy — the energy a particle has just by existing. A particle of mass $m$ at rest contains the energy equivalent of $m{c}^2$.

The kinetic energy is the total energy minus the rest energy:

$$K=(\gamma -1)m{c}^2.$$

At low speeds this reduces to $\frac{1}{2}m{v}^2$. At high speeds it grows far beyond the Newtonian prediction.

Total energy and momentum satisfy the energy-momentum relation:

$$E^2=(pc{)}^2+(m{c}^2{)}^2.$$

This is the master equation. For a massive particle at rest ($p=0$), it gives $E=m{c}^2$. For a massless particle ($m=0$), it gives $E=pc$ — photons carry momentum $p=E/c=h\nu /c$ even though they have no mass.

As $v$ approaches $c$, $\gamma$ grows without bound. It takes infinite energy to accelerate a massive object to the speed of light. The speed $c$ is an unattainable barrier for anything with mass.

Worked example [Beginner]

A proton ($m=1.67\times 10^{-27}$ kg, rest energy $m{c}^2=938$ MeV) is accelerated to $v=0.99c$. Find $\gamma$, the kinetic energy, and the total energy.

Step 1. Compute $\gamma$. With $v/c=0.99$, the ratio $v^2/c^2=0.9801$:

$$\gamma =\frac{1}{\sqrt{1-0.9801}}=\frac{1}{\sqrt{0.0199}}\approx \mathrm{7.09.}$$

Step 2. Compute the kinetic energy:

$$K=(\gamma -1)m{c}^2=6.09\times 938\text{ MeV}=5716\text{ MeV}\approx 5.72\text{ GeV}.$$

The Newtonian formula $\frac{1}{2}m{v}^2$ would give $\frac{1}{2}(0.99{)}^2\times 938\approx 460$ MeV — wrong by more than a factor of 12. The relativistic correction is enormous at this speed.

Step 3. Compute the total energy:

$$E=\gamma m{c}^2=7.09\times 938=6650\text{ MeV}\approx 6.65\text{ GeV}.$$

Step 4. Verify the energy-momentum relation. The momentum is $p=\gamma mv=7.09\times 938/c\times 0.99c\approx 6584$ MeV/$c$. Check: $E^2=6650^2\approx 4.42\times 10^7$ MeV${}^2$. And $(pc{)}^2+(m{c}^2{)}^2=6584^2+938^2\approx 4.33\times 10^7+0.88\times 10^6=4.42\times 10^7$ MeV${}^2$. Consistent.

Check your understanding [Beginner]

Formal definition [Intermediate+]

A particle of rest mass $m$ moving along a worldline $x^{\mu }(\tau )$ has four-velocity

$$u^{\mu }:=\frac{d{x}^{\mu }}{d\tau }=\gamma \left(c,\mathbf{v}\right),$$

where $\tau$ is proper time and $\gamma =(1-v^2/c^2{)}^{-1/2}$. The four-velocity satisfies the normalisation ${\eta }_{\mu \nu }{u}^{\mu }{u}^{\nu }=c^2$ (timelike, future-directed).

The four-momentum is

$$\boxed{p^{\mu }:=m{u}^{\mu }=\left(\gamma mc,\gamma m\mathbf{v}\right)=\left(\frac{E}{c},\mathbf{p}\right).}$$

The time component is $p^0=E/c$ and the spatial components are the relativistic three-momentum $\mathbf{p}=\gamma m\mathbf{v}$. The Minkowski inner product gives

$${\eta }_{\mu \nu }{p}^{\mu }{p}^{\nu }=(p^0{)}^2-|\mathbf{p}{|}^2=\frac{E^2}{c^2}-|\mathbf{p}{|}^2=m^2{c}^2,$$

which rearranges to $E^2=|\mathbf{p}{|}^2{c}^2+m^2{c}^4$. This is the energy-momentum relation, and $m^2{c}^2$ is a Lorentz invariant — the same in every frame.

The relativistic kinetic energy is $K=E-m{c}^2=(\gamma -1)m{c}^2$. Expanding for $v\ll c$:

$$K=m{c}^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)=m{c}^2\left(1+\frac{v^2}{2c^2}+\frac{3v^4}{8c^4}+\cdots -1\right)=\frac{1}{2}m{v}^2+\frac{3}{8}\frac{m{v}^4}{c^2}+\cdots .$$

The leading term is the Newtonian kinetic energy. Relativistic corrections appear at order $v^4/c^2$.

Force in special relativity is defined as the rate of change of relativistic momentum:

$$\mathbf{F}=\frac{d\mathbf{p}}{dt}=\frac{d}{dt}(\gamma m\mathbf{v}).$$

This is not the same as $m\mathbf{a}$ because $\gamma$ depends on $v$. The acceleration is not in general parallel to the force. For a force perpendicular to $\mathbf{v}$, $\gamma$ is constant and $\mathbf{F}=\gamma m\mathbf{a}$. For a force parallel to $\mathbf{v}$, one finds $\mathbf{F}={\gamma }^{3}m\mathbf{a}$. The longitudinal mass ${\gamma }^{3}m$ and transverse mass $\gamma m$ differ — a reflection of the fact that Newton's $\mathbf{F}=m\mathbf{a}$ is not Lorentz-covariant.

Compton scattering

A photon of energy $E_{\gamma }$ scatters off a stationary electron of mass $m$. Conservation of four-momentum $p_{\gamma }^{\mu }+p_e^{\mu }=p_{\gamma }^{\prime \mu }+p_e^{\prime \mu }$ gives the Compton formula for the scattered photon energy at angle $\theta$:

$$\boxed{\frac{1}{E_{\gamma }^{\prime }}-\frac{1}{E_{\gamma }}=\frac{1}{m{c}^2}(1-\cos \theta ).}$$

The wavelength shift is $\Delta \lambda =\frac{h}{mc}(1-\cos \theta )$, where $h/(mc)=2.43\times 10^{-12}$ m is the Compton wavelength of the electron. This result, confirmed by Compton (1923), is direct experimental evidence that photons carry momentum $p=E/c$.

Particle production thresholds

To create a particle of mass $M$ in a collision, the centre-of-momentum (CM) energy must satisfy $\sqrt{s}\ge M{c}^2$ where $s=(p_1^{\mu }+p_2^{\mu })(p_{1\mu }+p_{2\mu })$ is the Mandelstam variable. For a fixed-target experiment with beam energy $E_{\mathrm{beam}}$ on a stationary target of mass $m_t$:

$$s=m_t^2{c}^4+m_b^2{c}^4+2m_t{c}^2{E}_{\mathrm{beam}}.$$

The threshold beam energy is $E_{\mathrm{thr}}=((M^2-m_b^2-m_t^2{)}^2-4m_b^2{m}_t^2)/(2m_t{c}^2)$, which can far exceed $M{c}^2$ because much of the beam energy goes into kinetic energy of the products rather than rest mass.

Counterexamples to common slips

• "Mass increases with speed" is a deprecated interpretation. The rest mass $m$ is a Lorentz scalar — the same in every frame. What changes with speed is $\gamma$, not $m$. The "relativistic mass" $\gamma m$ is a historical artefact; modern treatments use rest mass exclusively and let $\gamma$ carry the speed dependence.
• $\mathbf{F}=d\mathbf{p}/dt$ is frame-dependent — it is not a four-vector equation. The covariant generalisation is the four-force $f^{\mu }=d{p}^{\mu }/d\tau$, developed in the Master tier.
• $E=m{c}^2$ applies only to particles at rest. A moving particle has $E=\gamma m{c}^2>m{c}^2$. The correct general relation is $E^2=(pc{)}^2+(m{c}^2{)}^2$.
• The Newtonian formula $K=\frac{1}{2}m{v}^2$ is a low-speed approximation. It fails dramatically at relativistic speeds — by a factor of 12 for $v=0.99c$.

Key theorem with proof [Intermediate+]

Theorem (Conservation of four-momentum in relativistic collisions). If $N$ free particles with four-momenta $p_a^{\mu }$ ($a=1,\dots ,N$) interact and emerge as $N^{\prime }$ particles with four-momenta $p_b^{\prime \mu }$ ($b=1,\dots ,N^{\prime }$), and the interaction respects Lorentz invariance, then

$$\boxed{\sum _{a=1}^N{p}_a^{\mu }=\sum _{b=1}^{N^{\prime }}{p}_b^{\prime \mu }.}$$

Proof. Lorentz invariance of the interaction means the $S$-matrix (or the collision map on the space of asymptotic states) commutes with Lorentz transformations. Consider the total four-momentum operator $P^{\mu }$. In the asymptotic past, the system is a collection of free particles, each carrying four-momentum $p_a^{\mu }$. The operator $P^{\mu }$ is the generator of spacetime translations, and Lorentz invariance of the dynamics means $[S,P^{\mu }]=0$ — the $S$-matrix commutes with translations.

Equivalently, at the classical level: the interaction Lagrangian is a Lorentz scalar. Noether's theorem for spacetime translation symmetry yields a conserved energy-momentum four-vector $P^{\mu }$. In the asymptotic past ($t\to -\infty$), the particles are non-interacting and $P^{\mu }={\sum }_a{p}_a^{\mu }$. In the asymptotic future ($t\to +\infty$), the outgoing particles are non-interacting and $P^{\mu }={\sum }_b{p}_b^{\prime \mu }$. Conservation of $P^{\mu }$ — which holds at all times because it is a Noether charge of a translation-invariant theory — requires the two expressions to be equal. $\square$

The theorem is an operator equation in quantum field theory and a classical identity in particle mechanics. Its physical content is that energy and momentum are conserved together as components of a single four-vector — not as separate three-vector statements.

Corollary (Threshold for particle production). To produce a particle of mass $M$ at rest in the CM frame, the CM energy must satisfy $\sqrt{s}\ge M{c}^2$, with equality at threshold (all products at rest in the CM frame).

The Mandelstam variable $s={\eta }_{\mu \nu }(p_1^{\mu }+p_2^{\mu })(p_1^{\nu }+p_2^{\nu })$ is a Lorentz scalar, so the threshold condition is frame-independent.

Worked example at intermediate level: Compton scattering

A photon of energy 1.0 MeV scatters off a free electron at rest. The photon is detected at angle $\theta =90^{\circ }$. Find the scattered photon energy and the electron recoil kinetic energy.

Conservation of four-momentum: $p_{\gamma }^{\mu }+p_e^{\mu }=p_{\gamma }^{\prime \mu }+p_e^{\prime \mu }$. Square both sides (contract with ${\eta }_{\mu \nu }$) after rearranging to isolate one unknown. From the Compton formula:

$$\frac{1}{E_{\gamma }^{\prime }}=\frac{1}{E_{\gamma }}+\frac{1}{m{c}^2}(1-\cos 90^{\circ })=\frac{1}{1.0}+\frac{1}{0.511}(1-0)=1.0+1.957=2.957{\text{ MeV}}^{-1}.$$

So $E_{\gamma }^{\prime }=1/2.957\approx 0.338$ MeV. The electron kinetic energy is $K_e=E_{\gamma }-E_{\gamma }^{\prime }=1.0-0.338=0.662$ MeV.

Exercises [Intermediate+]

Covariant formulation of relativistic dynamics [Master]

The three-force $\mathbf{F}=d\mathbf{p}/dt$ is frame-dependent and does not transform as part of a four-vector. The covariant replacement is the four-force (Minkowski force):

$$f^{\mu }:=\frac{d{p}^{\mu }}{d\tau }=\gamma \frac{d{p}^{\mu }}{dt}=\gamma \left(\frac{1}{c}\frac{dE}{dt},\mathbf{F}\right).$$

Since ${\eta }_{\mu \nu }{p}^{\mu }{p}^{\nu }=m^2{c}^2$ is constant, differentiating gives $p_{\mu }{f}^{\mu }=0$: the four-force is always orthogonal (in the Minkowski sense) to the four-momentum. This constraint reduces the four independent components of $f^{\mu }$ to three.

Relativistic Lagrangian and Hamiltonian

The relativistic free-particle Lagrangian is

$$L=-m{c}^2\sqrt{1-v^2/c^2}=-m{c}^2/\gamma .$$

The conjugate momentum is $\mathbf{p}=\gamma m\mathbf{v}$ (the relativistic three-momentum) and the Hamiltonian, obtained by Legendre transform, is

$$H=c\sqrt{|\mathbf{p}{|}^2+m^2{c}^2}.$$

This is the relativistic energy $E=\sqrt{p^2{c}^2+m^2{c}^4}$. For a charged particle in an electromagnetic field with four-potential $A^{\mu }=(\varphi /c,\mathbf{A})$, the minimal-coupling Lagrangian is

$$L=-m{c}^2/\gamma +e\mathbf{A}\cdot \mathbf{v}-e\varphi ,$$

and the Hamiltonian becomes

$$H=c\sqrt{|\mathbf{p}-e\mathbf{A}{|}^2+m^2{c}^2}+e\varphi .$$

Covariant Lorentz force

The equation of motion for a charge $q$ in an electromagnetic field $F^{\mu \nu }$ is

$$\frac{d{p}^{\mu }}{d\tau }=q{F}^{\mu \nu }{u}_{\nu },$$

where $u_{\nu }={\eta }_{\nu \alpha }{u}^{\alpha }$ is the covariant four-velocity. The spatial components ($\mu =1,2,3$) recover the Lorentz force law $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ with relativistic momentum. The time component ($\mu =0$) gives the power equation $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. The magnetic field does no work because $\mathbf{v}\times \mathbf{B}$ is perpendicular to $\mathbf{v}$.

The stress-energy tensor

For a system of particles and fields, the symmetric stress-energy tensor $T^{\mu \nu }$ encodes the density and flux of energy and momentum. For a perfect fluid with energy density $\rho$, pressure $p$, and four-velocity $u^{\mu }$:

$$T^{\mu \nu }=(\rho +p/c^2)u^{\mu }{u}^{\nu }-p{\eta }^{\mu \nu }.$$

Conservation of four-momentum for a continuous system is ${\partial }_{\mu }{T}^{\mu \nu }=0$. The time component ($\nu =0$) is energy conservation; the spatial components ($\nu =1,2,3$) are momentum conservation (the relativistic generalisation of the Cauchy momentum equation). For the electromagnetic field alone, $T_{\mathrm{EM}}^{\mu \nu }=F^{\mu \alpha }{F}^{\nu }{}_{\alpha }/{\mu }_0-\frac{1}{4}{\eta }^{\mu \nu }{F}_{\alpha \beta }{F}^{\alpha \beta }/{\mu }_0$, and ${\partial }_{\mu }{T}_{\mathrm{EM}}^{\mu \nu }=-F^{\nu \sigma }{J}_{\sigma }$ is the rate at which the field transfers momentum to charges.

Centre-of-momentum frame [Master]

For a system of $N$ particles, the centre-of-momentum (CM) frame is defined by ${\mathbf{P}}_{\mathrm{tot}}={\sum }_a{\mathbf{p}}_a=0$. The total four-momentum in this frame is $P_{\mathrm{CM}}^{\mu }=(\sqrt{s}/c,0)$, so the CM energy $\sqrt{s}=c\sqrt{{\eta }_{\mu \nu }{P}^{\mu }{P}^{\nu }}$ is a Lorentz invariant — it has the same value in every frame.

The CM frame is the natural frame for analysing collisions and decays because:

1. All three-momentum components vanish, simplifying kinematics.
2. The invariant $\sqrt{s}$ directly determines what particle production is possible.
3. Angular distributions of final-state particles are defined unambiguously.

The velocity of the CM frame relative to the lab is ${\mathbf{v}}_{\mathrm{CM}}={\mathbf{P}}_{\mathrm{tot}}{c}^2/E_{\mathrm{tot}}$. For equal-mass particles with equal and opposite momenta, the lab frame is the CM frame. For a fixed-target experiment, ${\mathbf{v}}_{\mathrm{CM}}$ is in the beam direction and $\sqrt{s}$ grows only as $\sqrt{E_{\mathrm{beam}}}$ — far slower than the beam energy itself.

Connections [Master]

To covariant electromagnetism 10.06.01 pending. The four-momentum and four-force developed here are the building blocks of covariant electrodynamics. The Lorentz force becomes $d{p}^{\mu }/d\tau =q{F}^{\mu \nu }{u}_{\nu }$, and the field tensor $F^{\mu \nu }$ unifies the electric and magnetic fields into a single geometric object. The distinction between $\mathbf{E}$ and $\mathbf{B}$ is frame-dependent.

To relativistic quantum mechanics 12.11.01 pending. The Dirac equation is obtained by taking the "square root" of $E^2=p^2{c}^2+m^2{c}^4$ as an operator equation. The requirement that the square root produce a first-order differential equation (not a second-order one) forces the introduction of $4\times 4$ Dirac matrices and spinor wavefunctions. The entire framework of relativistic quantum mechanics rests on the energy-momentum relation derived here.

To radiation from accelerating charges 10.07.01 pending. The relativistic Larmor formula $P=(q^2/6\pi {ϵ}_{0}c){\gamma }^6[a^2-|\mathbf{v}\times \mathbf{a}{|}^2/c^2]$ requires the four-acceleration ${\alpha }^{\mu }=d{u}^{\mu }/d\tau$ and the invariant $-{\alpha }^{\mu }{\alpha }_{\mu }={\gamma }^4{a}^2+{\gamma }^6(\mathbf{v}\cdot \mathbf{a}{)}^2/c^2$. The factor ${\gamma }^6$ for parallel acceleration explains why synchrotron radiation losses grow catastrophically at high energies.

To general relativity 13.01.01 pending. The stress-energy tensor $T^{\mu \nu }$ is the source term in the Einstein field equations $G^{\mu \nu }=8\pi G{T}^{\mu \nu }/c^4$. The conservation law ${\nabla }_{\mu }{T}^{\mu \nu }=0$ (covariant derivative replacing the flat-space ${\partial }_{\mu }$) encodes the local conservation of energy and momentum in curved spacetime. The framework developed here — four-momentum, four-force, $T^{\mu \nu }$ — lifts directly to GR with minimal modification.

Historical and philosophical context [Master]

Einstein's 1905 paper "Does the Inertia of a Body Depend Upon Its Energy Content?" derived $E=m{c}^2$ from the Lorentz transformation and the conservation of momentum applied to a body emitting two equal and opposite light pulses. The argument is elegant: the emitted light carries energy and momentum; the body loses energy; by momentum conservation the body's velocity is unchanged; but by energy conservation its kinetic energy is reduced. The only resolution is that the body's mass has decreased by $\Delta m=\Delta E/c^2$.

Minkowski's 1908 reformulation recast energy and momentum as components of a single four-vector $p^{\mu }=(E/c,\mathbf{p})$, making $E^2-|\mathbf{p}{|}^2{c}^2=m^2{c}^4$ a statement about the Minkowski norm of $p^{\mu }$. This geometric insight — that energy and momentum are different projections of the same geometric object — unified the separate conservation laws of energy and momentum into the single statement "four-momentum is conserved."

Dirac's 1928 relativistic wave equation started from the energy-momentum relation and demanded a linear (first-order) form. The resulting equation predicted spin-1/2 particles and antimatter — an outcome no one could have anticipated from the classical kinematics alone. The energy-momentum relation is not merely a bookkeeping identity; it is the gateway to relativistic quantum field theory.

The philosophical import of $E=m{c}^2$ is that mass is not a conserved quantity. In Newtonian mechanics, mass is an additive, conserved property of matter. In special relativity, mass can be created and destroyed: kinetic energy can become mass (particle production), and mass can become energy (nuclear fission, annihilation). The conservation law that survives is conservation of four-momentum — energy and momentum together, not mass separately.

Bibliography [Master]

• Einstein, A. "Ist die Tragheit eines Korpers von seinem Energieinhalt abhangig?" Annalen der Physik 18, 639–641 (1905). The $E=m{c}^2$ paper.
• Minkowski, H. "Die Grundgleichungen fur die elektromagnetischen Vorgange in bewegten Korpern." Nachrichten der Gesellschaft der Wissenschaften zu Gottingen, 53–111 (1908). Four-vector formalism.
• Compton, A. H. "A Quantum Theory of the Scattering of X-rays by Light Elements." Physical Review 21, 483–502 (1923). Experimental proof that photons carry momentum.
• Dirac, P. A. M. "The Quantum Theory of the Electron." Proceedings of the Royal Society A 117, 610–624 (1928). The relativistic wave equation.
• Jackson, J. D. Classical Electrodynamics, 3rd ed. Wiley (1999). Ch. 11, §11.5–11.10.
• Landau, L. D. & Lifshitz, E. M. The Classical Theory of Fields, 4th ed. (Course of Theoretical Physics Vol. 2). Butterworth-Heinemann (1975). §8–9.
• Griffiths, D. J. Introduction to Electrodynamics, 4th ed. Cambridge University Press (2017). Ch. 12.2.
• Taylor, E. F. & Wheeler, J. A. Spacetime Physics, 2nd ed. Freeman (1992). Ch. 7.
• Susskind, L. & Friedman, A. Special Relativity and Classical Field Theory. Basic Books (2017). Lectures 8–9.
• Tong, D. "Lectures on Electromagnetism." §5. University of Cambridge (2015).