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Take $\tau$ as particle proper time, dots as $d/d\tau$, and $f(r)=1-2\mu /r$ with $\mu =GM/c^2$. In the equatorial plane the geodesic Lagrangian isThe cyclic coordinates give $d(c^{2}f\dot{t})/d\tau =0$ and $d(-r^2\dot{\varphi })/d\tau =0$. Set the resulting constants to $c^{2}k$ and $-h$ respectively. The explicit second-order equations areThe polar equation is satisfied by $\theta =\pi /2$, $\dot{\theta }=0$, consistently with spherical symmetry. The proper-time normalization $d{s}^2=c^{2}d{\tau }^2$ supplies the third first integral:The normalization is $2\mathcal{L}=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.

Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=m{c}^{2}f\dot{t}=m{c}^{2}k$ and $P_{\varphi }=-m{r}^2\dot{\varphi }=-mh$. Thus $m{c}^{2}k$ is the conserved energy normalized at infinity and $mh$ is the signed axial angular momentum. This is Killing energy with positive time signature; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the geodesic equation makes these cyclic constants immediate, but its parametrization must be understood: for $p^{\mu }=d{x}^{\mu }/d\lambda$ it is exactly as written. For physical $P^{\mu }=md{x}^{\mu }/d\tau$, proper-time differentiation instead gives $d{P}_{\mu }/d\tau =({\partial }_{\mu }{g}_{\nu \sigma })P^{\nu }{P}^{\sigma }/(2m)$; equivalently, the displayed mass-free form uses $\lambda =\tau /m$. This affine momentum normalization does not change either conservation law.

Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial energy equationFor radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the speed at infinity vanishes. Choose the outward root:Hence the marginal escape geodesic in Schwarzschild spacetime hasThe constant $a$ has units of length to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper distance.

A Schwarzschild static observer at exterior radius $r$ uses local radial rulers $d\ell =dr/\sqrt{f}$ and a local clock $dT=\sqrt{f}dt$. Therefore the local radial speed of a Schwarzschild geodesic isFor the marginal trajectory $k=1$ this givesIts equality to $\dot{r}$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate velocity is instead $dr/dt=f\sqrt{2GM/r}$.

For nonzero asymptotic speed $0\le u<c$, the same energy constant is $k={\gamma }_{\infty }=(1-u^2/c^2{)}^{-1/2}$. Combining the radial equation with the local-ruler relation gives radial Schwarzschild escape with asymptotic speedThe positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has four-velocity $U=(1/\sqrt{f},0,0,0)$ in these coordinates, giving relative Lorentz factor $g(U,u_{\mathrm{particle}})/c^2=\sqrt{f}\dot{t}=k/\sqrt{f}$ and the same speed. As $r\to \infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow 2GM/c^2$, but that is a limit through exterior observers: no static timelike observer exists on or inside the horizon.

Nearby freely falling particles can accelerate relative to one another even though each follows a geodesic. A local inertial frame removes the common gravitational acceleration at one event; it cannot remove the variation across an extended laboratory. That variation is a tidal force, measured covariantly by geodesic deviation.

Let $u^{\mu }$ be the reference four-velocity and ${\xi }^{\mu }$ an infinitesimal separation to a neighboring geodesic, with matching affine parameters. With the convention $[{\nabla }_{\beta },{\nabla }_{\gamma }]V^{\mu }=R^{\mu }{}_{\nu \beta \gamma }{V}^{\nu }$,This relates measured relative acceleration to the Riemann curvature tensor. Reversing the curvature-tensor convention reverses the displayed curvature sign, without changing the physical relative motion.

In a small freely falling frame outside a spherical mass, the weak-field tidal tensor has radial and transverse components proportional to $2,-1,-1$. ThusAn initially spherical cloud stretches radially and compresses in both transverse directions. Two separated test masses, with their changing separation measured by light, form the basic idea of a gravity gradiometer; the common falling motion drops out. A uniform apparent acceleration in an accelerated laboratory can occur even in flat Minkowski spacetime, so it is the separation-dependent tidal response that diagnoses curvature.

Let $\mathbf{a}=d\mathbf{u}/dt$ denote coordinate acceleration and use $A^{\mu }$ for four-acceleration, to distinguish these two objects. Applying the chain rule to the Lorentz factor givesSince $dt/d\tau ={\gamma }_u$, differentiation of the four-velocity $U^{\mu }={\gamma }_u(c,\mathbf{u})$ with respect to proper time yieldsThis is also ${\gamma }_u^2[{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})/c,\mathbf{a}+{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})\mathbf{u}/c^2]$, as required.

The invariant Minkowski norm of the four-velocity is $U^{\mu }{U}_{\mu }=c^2$. Differentiating this identity gives $U^{\mu }{A}_{\mu }=0$. In the instantaneous rest frame, $U^{\mu }=(c,0)$, so $A^0=0$ and the spatial components equal the physically experienced proper acceleration $\mathbf{\alpha }$. Therefore ${\alpha }^2=-A^{\mu }{A}_{\mu }$ in the $+---$ metric signature.

For a direct calculation, put $q=\mathbf{u}\cdot \mathbf{a}$ and expand the temporal and spatial squares:The last step uses $1-u^2/c^2={\gamma }_u^{-2}$. This proper acceleration from coordinate velocity and acceleration also reads ${\alpha }^2={\gamma }_u^6{a}_{\parallel }^2+{\gamma }_u^4{a}_{\perp }^2$, providing the familiar one-dimensional limit $\alpha ={\gamma }_u^3|a_{\parallel }|$. Each expression has units ${\mathrm{m}}^2{\mathrm{s}}^{-4}$.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

For motion along one spatial axis, introduce the rapidity $\eta$ so that the four-velocity is $u^{\mu }=c(\cosh \eta ,\sinh \eta ,0,0)$. Its four-acceleration isA constant signed proper acceleration $g$ therefore gives $d\eta /d\tau =g/c$, hence $\eta ={\eta }_0+g\tau /c$. Differentiating the four-velocity twice yieldsThe zero transverse components also satisfy this equation. The constant acceleration measured on board is the proper acceleration, rather than a constant coordinate acceleration in the laboratory.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

Write $A(r)=1-2\mu /r$, with $\mu =GM/c^2$. On the equatorial plane, the geodesic Lagrangian isThe coordinates $t$ and $\varphi$ are cyclic, giving $\boxed{A\dot{t}=k,r^2\dot{\varphi }=h}$. The conserved physical energy per unit rest mass is $c^{2}k$, while $h$ is the signed angular momentum per unit rest mass. The radial Euler-Lagrange equation, with $A^{\prime }=2\mu /r^2$, isThese are the requested geodesic equations. The equatorial condition is consistent with the polar equation by the spherical symmetry of the Schwarzschild metric.

The normalization of the four-velocity is $c^2=c^{2}A{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2$. Insert the two conserved quantities and multiply by $A$ to obtainThis is a one-dimensional radial energy equation. Half of its left side is radial kinetic energy per unit mass plus an effective potential: the Newtonian centrifugal and gravitational terms acquire the extra attractive contribution $-\mu h^2/r^3$.

A timelike circular Schwarzschild geodesic requires $\dot{r}=\ddot{r}=0$. Equivalently, set the derivative of $V(r)=h^2/r^2-2\mu h^2/r^3-2\mu c^2/r$ to zero. At $r=R$ this gives $h^2(R-3\mu )=\mu c^2{R}^2$, and substitution in the radial energy equation yieldsSuch timelike circular orbits require $R>3\mu$. Their coordinate angular velocity satisfies ${\Omega }^2=\mu c^2/R^3$, and their proper clock rate is $d{\tau }_s/dt=A/k=\sqrt{1-3\mu /R_s}$.

For the static North-Pole clock, $d{\tau }_E/dt=\sqrt{1-2\mu /R_E}$. Consecutive overhead passages occur one coordinate orbital period apart, and the repeated Earth-satellite geometry gives equal signal travel delays. The comparison of circular-orbit and static Schwarzschild clocks therefore givesThis includes both gravitational and orbital-motion effects; a separate Newtonian time-dilation correction must not be added again.

For an estimate, Kepler's third law gives $R_s=42100\mathrm{km}(3/24{)}^{2/3}=10525\mathrm{km}$. The same approximate geostationary radius and period give $\mu =4{\pi }^2{R}_{\mathrm{geo}}^3/(c^2{T}_{\mathrm{geo}}^2)\simeq 4.4\mathrm{mm}$. Expanding the ratio to first order,The satellite clock consequently gains about $0.7\mu \mathrm{s}$ over a three-hour orbit in this approximation. The surface observer is taken to be static, and the spacetime model is the spherical Schwarzschild metric specified in the problem.

For rest mass $m$ and electric charge $q$, the generally covariant Lorentz force law isHere $F_{ab}={\partial }_a{A}_b-{\partial }_b{A}_a$ is the electromagnetic field tensor for the chosen electromagnetic four-potential normalization, and the covariant derivative includes the spacetime Levi-Civita connection. Antisymmetry of $F_{ab}$ ensures $u_{a}D{u}^a/d\tau =0$, preserving the four-velocity norm and the particle's rest mass. An uncharged particle follows a geodesic.

An affine geodesic has constant squared tangent norm, but that constant depends on the scale of its parameter. WriteUsing $W=\xi$ in the line element givesThe affine normalization in a Kaluza-Klein geodesic reduction matters: the specific formula requested in the paper uses $\mathcal{N}=1$, namely $\lambda$ equal to the positive five-dimensional arclength. With that normalization,A timelike four-dimensional projection has $d{u}^2=c^{2}d{\tau }^2>0$, requiring $|\xi |<1$ in this normalization. At $|\xi |=1$ the projection is null and has no proper-time four-velocity. For an arbitrary affine parameter the first boxed relation, with $\mathcal{N}$ retained, is the general result.

For coordinates $(t,x)$ the inverse metric is $g^{tt}=1/({\alpha }^2{x}^2)$, $g^{xx}=-1$. Only ${\partial }_x{g}_{tt}=2{\alpha }^{2}x$ is nonzero, giving the Christoffel symbolswith all remaining components zero.

On the specified constant-$x_0$ worldline, assume $x_0>0$. The four-velocity is $u^t=c/(\alpha x_0)$, $u^x=0$, and its norm is $c^2$, so $\tau$ is indeed proper time. The coordinate components are constant, but the covariant acceleration is nonzero:The proper acceleration of a static Rindler observer isThe acceleration has constant components in this accelerated chart and constant invariant magnitude. It is not a covariantly constant vector: its inertial-frame components change along the hyperbolic trajectory. For $x_0<0$, the magnitude is $c^2/|x_0|$, with the appropriate choice of future time orientation.

A fundamental observer follows the coarse-grained cosmological matter, with fixed comoving coordinates. The cosmological principle asserts spatial homogeneity and cosmological isotropy on large scales for this family of observers. A moving observer generally sees a preferred direction, for example a Doppler dipole in the cosmic microwave background, even when the underlying cosmology is isotropic.

In an FLRW metric, these worldlines are orthogonal to homogeneous spatial slices. Synchronize their clocks to define cosmic time $t$, which is their proper time. Along a fixed spatial label, $d{s}^2=c^{2}d{t}^2$, giving $g_{tt}=c^2$. Orthogonality of the spatial coordinate directions to the observers' four-velocity gives $g_{ti}=0$. Thus, in the $+---$ metric signature,where the spatial matrix $g_{ij}$ is negative definite. Spatial homogeneity and cosmological isotropy further restrict it to $-a(t{)}^2{\gamma }_{ij}(\mathbf{x})$, with $\gamma$ a constant-curvature spatial metric.

For radial motion, use the geodesic LagrangianThe Euler-Lagrange equation for $t$ gives $c^2\ddot{t}+a(da/dt){\dot{\chi }}^2=0$. Since $\chi$ is a cyclic coordinate, its conjugate momentum is constant. HenceFor a massive particle, take $\tau$ as proper time; the four-velocity normalization is $c^2{\dot{t}}^2-a^2{\dot{\chi }}^2=c^2$. For a photon, use an affine parameter rather than proper time, and the corresponding tangent is a null vector. ThereforeThe normalization of the null affine parameter remains arbitrary.

At any event on a massive observer's worldline, choose an orthonormal tetrad with timelike basis vector $e_{(0)}=U/c$. Its components obey $g(e_{(a)},e_{(b)})={\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, so locally $d{s}^2={\eta }_{ab}d{X}^{(a)}d{X}^{(b)}$: the Minkowski metric. A local inertial frame can also remove the Christoffel symbols at that event. Curvature cannot be removed on a finite region, and a frame following an accelerated observer need not remain inertial in its neighbourhood.

In general coordinates, $X^{\mu }(\tau )$ labels the trajectory, while the four-velocity and four-acceleration areCoordinate values $X^{\mu }$ are not themselves a vector under arbitrary nonlinear coordinate changes; their tangent $U$ is. For fixed rest mass $m$, the four-momentum is $P^{\mu }=m{U}^{\mu }$ and the four-force is $f^{\mu }=D{P}^{\mu }/d\tau =m{A}^{\mu }$. With the $+---$ metric signature,where $a$ is the proper acceleration. The middle identity follows by differentiating the constant velocity norm; in the instantaneous rest frame, $U=(c,0)$ and $A=(0,\mathbf{a})$, proving the last identity.

The proper time normalization of the astronaut's four-velocity gives ${\dot{t}}^2-{\dot{x}}^2=1$. In the light-cone coordinates this isThe four-acceleration has squared norm ${\ddot{t}}^2-{\ddot{x}}^2=-a^2$, because $a$ is the constant proper acceleration. The same coordinate change therefore givesEquivalently write $\dot{q}=e^{\eta }$, $\dot{p}=e^{-\eta }$. Their second derivatives give $\ddot{p}\ddot{q}=-{\dot{\eta }}^2$, so the rapidity form of one-dimensional proper acceleration makes $\dot{\eta }=\pm a$, with sign fixing the acceleration direction.

Differentiate each segment and express the four-velocity using a continuous rapidity:At $T$ both rapidities are $\omega T$; at $3T$ both are $-\omega T$. Hence the four-velocity is continuous at both junctions. This expression also gives $g(U,U)=c^2$ everywhere, so the parameter remains proper time on the later segments.

Away from the junctions,with ${\theta }^{\prime }$ successively $+\omega ,-\omega ,+\omega$. The rapidity is continuous but its slope reverses. Thus the four-acceleration has a finite jump to its negative at each junction, while its magnitude stays $c\omega$ on either side. There is no velocity jump; the idealized path is continuously differentiable but not twice differentiable at the switching events.

Take spacetime coordinates $X=(ct,x,y,z)$. Differentiating the initial segment gives the four-velocityThe identity ${\cosh }^{2}r-{\sinh }^{2}r=1$ gives $g(U,U)=c^2$, and $dt/ds=\cosh \omega s>0$. The proper time element therefore satisfies $d{\tau }^2=g(dX,dX)/c^2=d{s}^2$, with the future orientation selecting $d\tau =ds$. Thus $s$ is proper time along this path.

Use the metric $\eta =\mathrm{diag}(-1,1,1,1)$ and four-velocity $u^{\mu }=\gamma (c,\mathbf{v})$. Its derivative with respect to proper time is the four-acceleration. Sincea purely transverse laboratory acceleration has $d\gamma /dt=0$. With $d/d\tau =\gamma d/dt$,The invariant proper acceleration is the magnitude measured in the instantaneous rest frame, $a_0^2=A^{\mu }{A}_{\mu }$. Consequently,For longitudinal acceleration the factor is instead ${\gamma }^3$ multiplying the laboratory acceleration. The distinction matters for circular storage-ring motion, where the acceleration is centripetal and transverse even though its magnitude is nonzero.

Regard $f$ as a scalar field evaluated on a specified worldline $x^{\lambda }(\tau )$. The ordinary multivariable chain rule givesThis is the directional derivative along the four-velocity. With $x^0=ct$, it reads $df/d\tau =\gamma ({\partial }_{t}f+\mathbf{v}\cdot \nabla f)$. The spatial term is present because the worldline moves through the field; it would be lost by identifying a total derivative along the particle with a partial derivative at a fixed spatial position. For a scalar the contracted expression has the same value in all coordinate systems. For a tensor field the analogous covariant derivative also transports its indices.

Use $u^{\lambda }{u}_{\lambda }=-c^2$ and a constant Minkowski metric. Vary the velocity-dependent square root before imposing the proper-time normalization. Equivalently, vary a reparametrization-invariant worldline action first and then choose proper time. Its canonical covector momentum isThe coordinate derivative at fixed velocity is $\partial \mathcal{L}/\partial x^a=q{u}^{\lambda }{\partial }_a{A}_{\lambda }$. The Euler-Lagrange equation and the chain rule from a therefore giveDefine the antisymmetric field tensor byThe particle equation becomesAntisymmetry ensures $u^a{F}_{a\lambda }{u}^{\lambda }=0$, so the force is orthogonal to the four-velocity and preserves $u^2=-c^2$. The worldline variation before proper-time normalization is essential: substituting $\sqrt{-u^2}=c$ into the Lagrangian before taking its velocity derivative would turn the kinetic term into a constant and incorrectly remove the inertial momentum. The normalization is a property of the proper-time solution, not permission to discard the variational dependence.