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For a massive test particle with electric charge $e$, choose the electromagnetic four-potential $A_t=-Q/r$ and proper time $\tau$. The cyclic coordinates $t,\varphi$ in the equatorial Lagrangian give the conserved quantities $E=mF\dot{t}+eQ/r$ and $l=r^2\dot{\varphi }$. Here $l$ is angular momentum per unit mass. Substituting these quantities into the four-velocity normalization $g(U,U)=-1$ givesThe energy includes the gauge-dependent electrostatic term; the displayed convention fixes that gauge by taking the potential to vanish at infinity.

For identical particles of rest mass $m$ and electric charge $q$, the four-current is $j^{\mu }=(q/m)\rho u^{\mu }$. The matter stress-energy tensor obeys ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, with the opposite force density in the electromagnetic stress-energy tensor. Antisymmetry of the electromagnetic field tensor makes its contraction with two four-velocities vanish, so mass continuity still holds. Removing that term leaves the displayed covariant Lorentz force law. The matter tensor alone is not divergence-free when an electromagnetic force acts.

Charge invariance means that the total electric charge of a specified collection of particles is unchanged by a Lorentz transformation. It differs from conservation of electric charge, which compares the same isolated system at different times. For a convective charge distribution with rest density ${\rho }_0$, the four-current is $J^{\mu }={\rho }_0{U}^{\mu }$, where $U^{\mu }$ is its four-velocity; its measured charge density increases under a boost while the simultaneous spatial volume contracts, leaving total charge unchanged.

For a constant electromagnetic field tensor, the relativistic Lorentz force is a constant-coefficient linear system on four-velocity, with solutionThe electric-type blocks of $F$ generate hyperbolic rotations, while magnetic-type blocks generate ordinary rotations.

In units $c=1$, a receiver on the hyperbolic motion worldline $t=a^{-1}\sinh (a\tau )$, $x=a^{-1}\cosh (a\tau )$ has retarded null coordinate $p=t-x=-a^{-1}{e}^{-a\tau }$. A right-moving light ray emitted at $(t_0,L)$ has $p=t_0-L$, so it meets the receiver only if $t_0<L$. The photon wave covector is proportional to $(1,-1)$, and contraction with the receiver's four-velocity gives its measured frequency as ${\nu }_0(\dot{t}-\dot{x})={\nu }_0{e}^{-a\tau }$. The intersection condition gives the second displayed equality. Emission at $t_0=L$ lies on the Rindler horizon and never reaches the receiver at finite proper time; $L$ is the supremum of receivable emission times, not an attained latest time.

A pressureless fluid with proper number density $n_0$, rest mass $m$ and four-velocity $u$ has tensor $m{n}_0{u}^{\mu }{u}^{\nu }$. In its local rest frame there is energy density but no spatial stress. Particle-number conservation and the Lorentz force make its divergence the electromagnetic force density.

A source on a timelike circular Schwarzschild geodesic at radius $r>3GM/c^2$ emits light to a static observer at infinity on the orbital symmetry axis. Such a ray has zero conserved axial angular momentum. The locally measured photon energy is the contraction of its four-momentum with the observer's four-velocity. Comparing the emitter with the observer at infinity therefore gives ${\nu }_{\mathrm{em}}=U^t{\nu }_{\infty }$, where $U^t=[1-3GM/(c^{2}r){]}^{-1/2}$, proving the formula.

Equivalently the gravitational redshift factor $\sqrt{1-2GM/(c^{2}r)}$ multiplies the transverse Doppler effect factor $1/\gamma$ measured by a local static observer. For $R_S=2GM/c^2\ll r$,The coefficient contains both gravitational and orbital time dilation. It assumes an observer at infinity on the symmetry axis; other viewing directions generally introduce a first-order longitudinal Doppler effect.

Transport defining a nonrotating frame along an accelerated timelike curve. With four-velocity normalization $u^{\mu }{u}_{\mu }=-c^2$ and four-acceleration $a^{\mu }=D{u}^{\mu }/d\tau$, the displayed equation preserves inner products and keeps a spatial spin orthogonal to $u$. Along a geodesic, $a^{\mu }=0$ and it reduces to parallel transport.

The four-acceleration is the covariant derivative of four-velocity along a massive observer's worldline: $A^{\mu }=d{U}^{\mu }/d\tau +{\Gamma }_{\nu \rho }^{\mu }{U}^{\nu }{U}^{\rho }$. In Cartesian Minkowski spacetime coordinates the Christoffel symbols vanish, leaving the ordinary proper time derivative. Differentiating the constant velocity norm gives $U\cdot A=0$; in the instantaneous rest frame, $A=(0,\mathbf{a})$, so $A\cdot A=-a^2$ with $+---$ metric signature. The magnitude $a$ is the proper acceleration.

A four-vector is a vector whose four vector components transform under a Lorentz transformation as $U^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{U}^{\nu }$. Its individual vector components generally change between inertial frames; the Minkowski metric combines them into invariant scalar products. Listing four quantities is not enough to make a four-vector: their transformation law matters.

The most direct example is the displacement between two spacetime events, written $(c\Delta t,\Delta x,\Delta y,\Delta z)$. Multiplying the time difference by the speed of light gives all four vector components the same length units. A position four-vector describes an event relative to a chosen origin, while a displacement does not depend on a common translation of the origin. Four-momentum and four-velocity use the same transformation rule for different physical quantities.

A massive observer's four-velocity is the tangent $U^{\mu }=d{X}^{\mu }/d\tau$ to its worldline, parametrized by proper time. Its invariant squared Minkowski norm is $c^2$ in the $+---$ metric signature, and $-c^2$ in the opposite convention. Unlike the coordinate values $X^{\mu }$, the tangent transforms as a vector under a general coordinate change.

Let $K={\partial }_v$ generate a Killing horizon with constant surface gravity $\kappa$, so ${\nabla }_{K}K=\kappa K$. In geometric optics, an affinely transported light wavevector on the same generator is $k=q(v)K$. The geodesic equation gives $q^{\prime }+\kappa q=0$. Observers related by the isometry generated by $K$ have the same $-K\cdot W$ at their crossings, where $W$ is their four-velocity, so their measured angular frequencies $\omega =-k\cdot W$ obey the displayed relation. A degenerate Killing horizon gives ratio one for equally prepared observers; changing their crossing velocities can produce a separate Doppler effect.

For metric signature $(+---)$, a stationary perfect fluid follows a timelike Killing vector field $K$ with four-velocity $U=K/N$. Its Killing-flow four-acceleration is $U^b{\nabla }_b{U}_a=-{\nabla }_a\log N$. The relativistic Euler equation and stationarity of the pressure give the displayed hydrostatic equilibrium equation. This uses stationarity along $K$ and does not require hypersurface orthogonality.

The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are

The covariant force on a charged particle is its charge times the electromagnetic field-strength tensor contracted with its four-velocity. Antisymmetry keeps the four-force orthogonal to that velocity and preserves the rest mass.

In Minkowski spacetime of metric signature $(-,+,+,+)$, let $u$ be a constant future-directed four-velocity with $\eta (u,u)=-c^2$. The time difference assigned to two spacetime events is $\Delta t_u=-\eta (u,Q-P)/c^2$. Thus the observer's simultaneous events form an affine hyperplane orthogonal to $u$. This is an invariant relation of the three arguments $u,P,Q$; the Minkowski metric alone does not single out one $u$. For fixed $u$, it defines an equivalence relation and the hyperplanes form a foliation. Changing $u$ changes this foliation.

The average of future-directed normalized four-velocities is generally timelike but is not normalized. For an isotropic ensemble with common Lorentz factor $\gamma$, the mean is $(\gamma c,0,0,0)$ and has squared norm $-{\gamma }^2{c}^2$. A normalized bulk velocity instead equals $c⟨v⟩/\sqrt{-g(⟨v⟩,⟨v⟩)}$. In a perfect fluid in general relativity its rest-frame energy eigenvector is the natural bulk choice; the rest-frame energy density includes random kinetic energy.

Compare two observers at one event using a common static orthonormal frame outside a Schwarzschild horizon. If one observer moves tangentially at speed $u$ and another radially at speed $w$, their four-velocity contraction is ${\gamma }_u{\gamma }_w{c}^2$. Their relative speed therefore has the displayed form. The radial component measured by the tangential observer is $w/{\gamma }_u$, so Euclidean subtraction of the two original three-velocities fails. For a circular timelike geodesic and radial fall from rest at infinity at radius $R$, the relative speed is $c\sqrt{3\mu /R}$, where $\mu =GM/c^2$ and $R>3\mu$.

The number current is $J^a=n{u}^a$, where $n$ is the rest-frame number density and $u^a$ the unit four-velocity. Its covariant continuity equation is ${\nabla }_a{J}^a=0$, equivalently $u^a{\nabla }_{a}n+n{\nabla }_a{u}^a=0$. Integration over a spacetime volume and the divergence theorem equate the particle numbers crossing its boundary. This conservation law is separate from stress-energy conservation.

Covered by the official notes: definitions of four-force and four-velocity and the vanishing antisymmetric contraction (PDF p. 15; handwritten p. 10).^[ref]

To make the necessity statement precise, mass preservation must hold for every allowed timelike four-velocity, with the same field tensor and nonzero charge. It is not necessary for one prescribed trajectory alone. Use metric signature $(-,+,+,+)$, so $U_a{F}^a=-c^{2}dm/d\tau$. Put $H^{ab}=({\mathcal{F}}^{ab}+{\mathcal{F}}^{ba})/2$. The mass-preservation condition requires the quadratic form $H^{ab}{U}_a{U}_b=0$ for every timelike $U$. Every timelike vector can be rescaled to a physical four-velocity; by homogeneity the quadratic form vanishes on the open timelike cone. A polynomial vanishing on an open set is identically zero, so $H^{ab}=0$. Thus the electromagnetic field tensor must be antisymmetric under this universal quantifier. For a single velocity, the scalar contraction can vanish even for a nonzero symmetric part, which is why the quantifier is essential.