Geodesic Equation from Extremal Proper Time

Derivation

Geodesic Equation from Extremal Proper Time

D-385 Home PU-403 Threads force · energy · symmetry Depends on euler-lagrange-from-stationary-action, Equivalence Principle Forces a Curved Metric

Statement

A free test particle in a spacetime with metric \( g_{\mu\nu}(x) \) follows the worldline that extremizes the proper time \( \tau = \int d\tau \) between two fixed events, with \( c^2\,d\tau^2 = -g_{\mu\nu}\,dx^\mu dx^\nu \) (signature \( -{+}{+}{+} \)). Extremization yields the geodesic equation \( \ddot x^\alpha + \Gamma^\alpha_{\mu\nu}\dot x^\mu \dot x^\nu = 0 \) with the Levi-Civita connection \( \Gamma^\alpha_{\mu\nu} = \tfrac12 g^{\alpha\beta}\!\left(\partial_\mu g_{\beta\nu} + \partial_\nu g_{\beta\mu} - \partial_\beta g_{\mu\nu}\right) \), where the overdot is \( d/d\tau \).

Why it matters

The geodesic equation is the law of inertia for curved spacetime. It states that gravitation is not a force but the shape of the paths that free particles are compelled to follow, and it packages the entire gravitational field into one geometric object, the Christoffel symbols. Everything from planetary orbits to light bending to the precession of Mercury's perihelion is a solution of this single equation for the appropriate metric.

Deriving it from extremal proper time — rather than postulating it — shows that general relativity inherits Hamilton's principle from classical mechanics essentially unchanged: the free particle still makes a functional stationary; only the functional is now the arc length of the metric. This is the cleanest route from the equivalence principle to dynamics, and it makes manifest that the connection is metric-compatible and torsion-free by construction.

Assumptions

Spacetime is a smooth pseudo-Riemannian manifold with a metric \( g_{\mu\nu}(x) \) of Lorentzian signature.Without a nondegenerate metric there is no proper time to extremize and no way to raise/lower indices; the inverse \( g^{\alpha\beta} \) in \( \Gamma \) would not exist.

The particle is a structureless test body: it carries negligible mass-energy and no spin, charge, or internal multipoles.A massive body back-reacts on the metric; a spinning body obeys the Mathisson–Papapetrou equations with a curvature–spin coupling term, not a pure geodesic.

The worldline is timelike, so \( g_{\mu\nu}\dot x^\mu \dot x^\nu < 0 \) everywhere and \( d\tau \) is real.For a null (light) worldline \( d\tau=0 \) identically, so proper time cannot be the action; one must instead extremize an affine-parametrized functional (see Breaks when).

The endpoints are held fixed and the extremal is a smooth non-self-intersecting curve on which an affine parameter can be chosen globally.If conjugate points occur between the endpoints the stationary curve is a saddle, not a maximum, of proper time, and "longest proper time" reasoning fails even though the geodesic equation still holds locally.

Derivation

\[ \tau[x] = \int_{\lambda_1}^{\lambda_2} L\,d\lambda,\qquad L \equiv \frac{1}{c}\sqrt{-\,g_{\mu\nu}(x)\,\dot x^\mu \dot x^\nu},\quad \dot x^\mu \equiv \frac{dx^\mu}{d\lambda} \]

Write proper time as a reparametrization-invariant functional of an arbitrary parameter \( \lambda \); the integrand is homogeneous of degree one in \( \dot x \), so \( \tau \) is independent of the choice of \( \lambda \). A

\[ \frac{d}{d\lambda}\!\left(\frac{\partial L}{\partial \dot x^\alpha}\right) - \frac{\partial L}{\partial x^\alpha} = 0 \]

Apply the Euler–Lagrange condition (prior result: euler-lagrange-from-stationary-action) to the stationary curves of \( \tau[x] \) with fixed endpoints. A

\[ \frac{\partial L}{\partial \dot x^\alpha} = \frac{1}{c}\,\frac{-\,g_{\alpha\nu}\dot x^\nu}{\sqrt{-g_{\mu\nu}\dot x^\mu \dot x^\nu}} = -\frac{g_{\alpha\nu}\dot x^\nu}{c^2}\,\frac{d\lambda}{d\tau} \]

Differentiate \( L \) using \( \partial(g_{\mu\nu}\dot x^\mu \dot x^\nu)/\partial \dot x^\alpha = 2 g_{\alpha\nu}\dot x^\nu \) (metric symmetry) and the chain rule \( \tfrac1c\sqrt{-g_{\mu\nu}\dot x^\mu\dot x^\nu}=d\tau/d\lambda \). B

\[ \frac{\partial L}{\partial x^\alpha} = -\frac{1}{2c^2}\,\big(\partial_\alpha g_{\mu\nu}\big)\dot x^\mu \dot x^\nu\,\frac{d\lambda}{d\tau} \]

Only the metric depends explicitly on position; differentiate under the square root and use the same \( d\tau/d\lambda \) factor. B

\[ \lambda = \tau \;\Rightarrow\; \frac{d\tau}{d\lambda}=1,\qquad L\big|_{\lambda=\tau}=1 \]

Fix the gauge by choosing proper time itself as the parameter. This is legitimate because \( \tau \) is monotonic along a timelike curve, and it removes the awkward \( d\lambda/d\tau \) factors precisely because \( L \) is then constant. C

\[ \frac{d}{d\tau}\!\big(g_{\alpha\nu}\dot x^\nu\big) - \frac{1}{2}\big(\partial_\alpha g_{\mu\nu}\big)\dot x^\mu \dot x^\nu = 0 \]

Insert Steps 3–4 into Step 2 at \( \lambda=\tau \) and cancel the common factor \( -1/c^2 \); now the overdot means \( d/d\tau \). B

\[ g_{\alpha\nu}\ddot x^\nu + \big(\partial_\mu g_{\alpha\nu}\big)\dot x^\mu \dot x^\nu - \frac{1}{2}\big(\partial_\alpha g_{\mu\nu}\big)\dot x^\mu \dot x^\nu = 0 \]

Expand the total derivative: \( \tfrac{d}{d\tau}(g_{\alpha\nu}\dot x^\nu) = (\partial_\mu g_{\alpha\nu})\dot x^\mu\dot x^\nu + g_{\alpha\nu}\ddot x^\nu \) by the chain rule. B

\[ \big(\partial_\mu g_{\alpha\nu}\big)\dot x^\mu \dot x^\nu = \frac{1}{2}\big(\partial_\mu g_{\alpha\nu} + \partial_\nu g_{\alpha\mu}\big)\dot x^\mu \dot x^\nu \]

Symmetrize on the dummy pair \( (\mu,\nu) \): since \( \dot x^\mu \dot x^\nu \) is symmetric, only the symmetric part of the coefficient survives. This is the key algebraic move that builds the connection. C

\[ g_{\alpha\nu}\ddot x^\nu + \frac{1}{2}\big(\partial_\mu g_{\alpha\nu} + \partial_\nu g_{\alpha\mu} - \partial_\alpha g_{\mu\nu}\big)\dot x^\mu \dot x^\nu = 0 \]

Substitute Step 8 into Step 7; the three metric-derivative terms combine into the fully symmetric bracket. B

\[ \ddot x^\alpha + \underbrace{\frac{1}{2}g^{\alpha\beta}\big(\partial_\mu g_{\beta\nu} + \partial_\nu g_{\beta\mu} - \partial_\beta g_{\mu\nu}\big)}_{\displaystyle \Gamma^\alpha_{\mu\nu}}\,\dot x^\mu \dot x^\nu = 0 \]

Contract Step 9 with the inverse metric \( g^{\alpha\beta} \), using \( g^{\alpha\beta}g_{\beta\nu}=\delta^\alpha_\nu \) to free the acceleration index; identify the bracket as the Christoffel symbol. B

Result

\[ \frac{d^2 x^\alpha}{d\tau^2} + \Gamma^\alpha_{\mu\nu}\,\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau} = 0,\qquad \Gamma^\alpha_{\mu\nu} = \frac{1}{2}g^{\alpha\beta}\!\left(\partial_\mu g_{\beta\nu} + \partial_\nu g_{\beta\mu} - \partial_\beta g_{\mu\nu}\right) \]

Reading. The four-acceleration measured in the coordinate basis is not zero in curved spacetime; instead it is exactly cancelled by the connection term \( \Gamma^\alpha_{\mu\nu}\dot x^\mu\dot x^\nu \). The invariant statement is that the covariant acceleration vanishes, \( \dot x^\nu \nabla_\nu \dot x^\alpha = 0 \): the four-velocity is parallel-transported along its own worldline. The particle "feels" nothing — it is in free fall — yet its coordinate path curves because the coordinate grid itself is curved. The Christoffel symbols encode how basis vectors twist from event to event; they are built from first derivatives of the metric alone, so the connection is fully determined by, and compatible with, the geometry.

Units check. Use dimensionless coordinates scaled so \( x^0 = ct \) and arc length \( s=c\tau \) (all \( x^\alpha \) in metres). Then \( d^2x^\alpha/ds^2 \) has units \( \mathrm{m^{-1}} \). The metric \( g_{\mu\nu} \) is dimensionless, so \( \Gamma^\alpha_{\mu\nu}\sim \partial g \) has units \( \mathrm{m^{-1}} \), and \( dx^\mu/ds\, dx^\nu/ds \) is dimensionless. Both terms are \( \mathrm{m^{-1}} \) — consistent.

Limiting cases

Flat spacetime, Cartesian coordinates: \( g_{\mu\nu}=\eta_{\mu\nu} \) is constant, so all \( \partial g=0 \), hence \( \Gamma^\alpha_{\mu\nu}=0 \) and \( \ddot x^\alpha=0 \) — Newton's first law, straight worldlines at constant four-velocity.
Weak, slow, static field: \( g_{00}=-(1+2\Phi/c^2) \) reduces the spatial equation to \( \ddot{\vec x}=-\nabla\Phi \), recovering Newtonian gravity with \( \Gamma^i_{00}=\partial_i\Phi/c^2 \).
Free particle in curvilinear flat coordinates: nonzero \( \Gamma \) (e.g. polar) but zero curvature; the connection terms are pure coordinate artifacts (fictitious forces such as centrifugal/Coriolis) that a coordinate change removes globally.
Photon limit: as the worldline approaches null, \( d\tau\to0 \); the geodesic equation survives if re-parametrized by an affine \( \lambda \), giving the null geodesics that describe light rays.

Breaks when

Null and spacelike curves. For light, \( d\tau\equiv0 \), so proper time is not a valid action and Step 5 (setting \( \lambda=\tau \)) is impossible. One must instead extremize the "energy" functional \( \int g_{\mu\nu}\dot x^\mu\dot x^\nu\,d\lambda \) with an affine parameter; the square-root derivation given here silently assumes a timelike path.
Conjugate points / caustics. Beyond a point conjugate to the start (e.g. antipodes on a sphere, or the focus of a gravitational lens), the geodesic ceases to be even a local maximum of proper time. The Euler–Lagrange curve still solves the equation, but "extremal" no longer means "longest," so variational uniqueness fails.
Non-test bodies. A finite-mass or spinning body sources or couples to curvature; back-reaction (self-force) and the Mathisson–Papapetrou spin–curvature term \( \propto R^\alpha{}_{\beta\mu\nu}S^{\mu\nu}\dot x^\beta \) add to the right-hand side, so the pure geodesic law is violated.
Non-gravitational forces or degenerate metric. A charged particle in an electromagnetic field obeys \( \ddot x^\alpha+\Gamma^\alpha_{\mu\nu}\dot x^\mu\dot x^\nu=(q/mc)F^\alpha{}_\nu\dot x^\nu\neq0 \); and where \( \det g_{\mu\nu}=0 \) (coordinate or physical singularity) the inverse \( g^{\alpha\beta} \) and hence \( \Gamma \) blow up.

Failure modes

Forgetting to symmetrize (Step 8). Dropping the symmetrization gives \( \Gamma^\alpha_{\mu\nu}=g^{\alpha\beta}(\partial_\mu g_{\beta\nu}-\tfrac12\partial_\beta g_{\mu\nu}) \), which is not symmetric in \( \mu\nu \) and gives wrong geodesics unless contracted with the symmetric \( \dot x^\mu\dot x^\nu \) — a trap when reusing \( \Gamma \) in the covariant derivative.
Treating \( \Gamma^\alpha_{\mu\nu} \) as a tensor. The connection transforms inhomogeneously; a student who "raises and lowers its indices" or concludes gravity is real because \( \Gamma\neq0 \) forgets that \( \Gamma \) can be set to zero at any single point (local inertial frame).
Setting \( \lambda=\tau \) before differentiating. Fixing the gauge inside \( L \) before taking \( \partial/\partial\dot x^\alpha \) loses the constraint \( g_{\mu\nu}\dot x^\mu\dot x^\nu=-c^2 \) and yields inconsistent equations; the parametrization must be fixed only after the Euler–Lagrange step.
Sign/signature slips. Using \( +{-}{-}{-} \) without flipping the sign under the square root gives an imaginary \( L \) for timelike curves; mismatched conventions produce a spurious overall sign in \( \Gamma^i_{00} \) and hence "antigravity."
Confusing coordinate acceleration with proper acceleration. Reading \( \ddot x^i\neq0 \) as "a force acts" ignores the \( \Gamma \) term; the physically felt acceleration is the covariant one, which is zero for a geodesic.

Discussion

The derivation makes precise the sense in which gravity "is" geometry. Once the equivalence principle promotes the flat metric \( \eta_{\mu\nu} \) to a general \( g_{\mu\nu}(x) \) (prior result: equivalence-principle-to-metric-gravity), the only Lorentz-invariant scalar one can build along a worldline is its proper time. Demanding that free particles extremize this scalar is the minimal dynamical postulate, and it automatically produces a torsion-free, metric-compatible connection — the Christoffel symbols are symmetric in their lower indices precisely because they arose from the symmetric second derivative of the path. No independent "force law" is needed.

A powerful structural feature is the link to symmetry and conservation. If the metric is independent of some coordinate \( x^k \) (a Killing direction), then \( \partial_k g_{\mu\nu}=0 \) and the Euler–Lagrange equation at Step 6 immediately gives \( \tfrac{d}{d\tau}(g_{k\nu}\dot x^\nu)=0 \): the covariant momentum component \( p_k=m g_{k\nu}\dot x^\nu \) is conserved. Time-translation symmetry gives conserved energy; axial symmetry gives conserved angular momentum. This is Noether's theorem living inside the geodesic equation, and it is what makes the Schwarzschild orbit problem integrable.

Geometrically, the equation says the four-velocity is parallel-transported along itself: a geodesic is "as straight as possible" given the curvature. The deviation of neighbouring geodesics is governed by the Riemann tensor through the geodesic deviation equation, which is the tidal-force content of gravity — the part of \( \Gamma \) that cannot be transformed away at a point. This cleanly separates real gravity (curvature, tidal effects) from coordinate effects (a nonzero \( \Gamma \) in an accelerating or curvilinear frame).

At the level of the action, extremizing \( \int d\tau \) is equivalent to extremizing the quadratic "energy" functional \( \tfrac12\int g_{\mu\nu}\dot x^\mu\dot x^\nu\,d\lambda \) for an affinely parametrized curve, and the two agree exactly on the constraint surface \( g_{\mu\nu}\dot x^\mu\dot x^\nu=\text{const} \). The quadratic form is analytic even where \( d\tau\to0 \), which is why it, not the square root, is the correct starting point for null geodesics and for the path integral of a relativistic particle. The square-root action's reparametrization invariance is a gauge redundancy whose "Hamiltonian" is the mass-shell constraint \( g^{\mu\nu}p_\mu p_\nu+m^2c^2=0 \) — a first-class constraint generating worldline diffeomorphisms.

Common misconceptions. (i) "Free particles take the path of longest proper time" is only locally true; globally a geodesic may be a saddle. (ii) A nonzero Christoffel symbol does not mean spacetime is curved — Cartesian-to-polar already produces nonzero \( \Gamma \) in flat space; curvature is nonzero Riemann, not nonzero connection. (iii) The particle is not "pulled" by gravity; it moves inertially, and only our coordinate description makes the path look accelerated.

Worked examples

Example 1 — Newtonian limit: recovering \( \vec g \) at Earth's surface.

\[ ds^2 = -\Big(1+\frac{2\Phi}{c^2}\Big)(dx^0)^2 + \delta_{ij}\,dx^i dx^j,\qquad x^0=ct \]

Weak static field: \( |\Phi|\ll c^2 \), metric time-independent, spatial part flat to first order. A

\[ \Gamma^i_{00} = -\tfrac12\,\delta^{ij}\,\partial_j g_{00} = -\tfrac12\,\partial_i\!\Big(\!-1-\tfrac{2\Phi}{c^2}\Big) = \frac{1}{c^2}\,\partial_i\Phi \]

Only \( \partial_j g_{00}\neq0 \) contributes for a static metric; symbols before numbers. B

\[ \frac{d^2x^i}{d\tau^2} + \Gamma^i_{00}\Big(\frac{dx^0}{d\tau}\Big)^2 = 0,\qquad \frac{dx^0}{d\tau}=c\frac{dt}{d\tau}\approx c,\;\; \frac{d}{d\tau}\approx\frac{d}{dt} \]

Slow motion: spatial velocities \( \ll c \), so \( d\tau\approx dt \) and only the \( 00 \) connection term matters. B

\[ \frac{d^2x^i}{dt^2} = -c^2\,\Gamma^i_{00} = -c^2\cdot\frac{1}{c^2}\partial_i\Phi = -\partial_i\Phi \quad\Rightarrow\quad \ddot{\vec x}=-\nabla\Phi \]

Substitute Step 2; the \( c^2 \) factors cancel, leaving Newton's gravity. A

\[ g = |\nabla\Phi| = \frac{GM_\oplus}{R_\oplus^2} = \frac{(6.674\times10^{-11})(5.972\times10^{24})}{(6.371\times10^{6})^2}\ \mathrm{m\,s^{-2}} \]

Insert Earth values \( M_\oplus=5.972\times10^{24}\,\mathrm{kg} \), \( R_\oplus=6.371\times10^{6}\,\mathrm{m} \). A

\[ g \approx 9.82\ \mathrm{m\,s^{-2}} \]

Reading. The geodesic equation for the weak-field metric reproduces surface gravity to three figures. The tiny metric perturbation \( 2\Phi/c^2 \approx -1.4\times10^{-9} \) at Earth's surface is enough to bend timelike geodesics into everyday falling trajectories.

Units check. \( [GM/R^2]=(\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{kg})/\mathrm{m^2}=\mathrm{m\,s^{-2}} \). Correct.

Example 2 — Geodesics on a 2-sphere (Riemannian analogue). The same equation extremizes arc length; here it selects great circles. Radius \( a=1\ \mathrm{m} \).

\[ ds^2 = a^2\big(d\theta^2 + \sin^2\!\theta\,d\phi^2\big),\qquad g_{\theta\theta}=a^2,\; g_{\phi\phi}=a^2\sin^2\!\theta \]

Round metric on the sphere; diagonal, so the inverse is \( g^{\theta\theta}=a^{-2},\,g^{\phi\phi}=a^{-2}\sin^{-2}\theta \). A

\[ \Gamma^\theta_{\phi\phi} = -\tfrac12 g^{\theta\theta}\partial_\theta g_{\phi\phi} = -\sin\theta\cos\theta,\qquad \Gamma^\phi_{\theta\phi}=\Gamma^\phi_{\phi\theta} = \tfrac12 g^{\phi\phi}\partial_\theta g_{\phi\phi} = \cot\theta \]

Only \( \partial_\theta g_{\phi\phi} = 2a^2\sin\theta\cos\theta \) is nonzero; all other Christoffels vanish. B

\[ \ddot\theta + \Gamma^\theta_{\phi\phi}\,\dot\phi^2 = 0,\qquad \ddot\phi + 2\Gamma^\phi_{\theta\phi}\,\dot\theta\dot\phi = 0 \]

Write out the two components of the geodesic equation (index \( \alpha=\theta,\phi \)). B

\[ \text{Trial: } \theta(s)=\tfrac{\pi}{2}\ (\text{equator}),\ \dot\theta=0:\quad \ddot\theta + \big(-\sin\tfrac\pi2\cos\tfrac\pi2\big)\dot\phi^2 = 0 + (-1\cdot0)\dot\phi^2 = 0 \;\checkmark \]

Test the equator: \( \cos(\pi/2)=0 \) kills the source term, so \( \theta=\pi/2 \) is consistent — a geodesic. A

\[ \text{Non-equator, } \theta_0=\tfrac{\pi}{3}:\quad \ddot\theta = \sin\tfrac\pi3\cos\tfrac\pi3\,\dot\phi^2 = \tfrac{\sqrt3}{2}\cdot\tfrac12\,\dot\phi^2 = \tfrac{\sqrt3}{4}\dot\phi^2 \neq 0 \]

A latitude circle at \( 60^\circ \) has nonzero \( \ddot\theta \): it cannot be traversed at constant \( \theta \), so it is not a geodesic. B

\[ \text{Great circles }(\theta=\tfrac\pi2)\text{ are geodesics; latitude circles }(\theta\neq\tfrac\pi2)\text{ are not.} \]

Reading. The connection term is exactly the "centripetal" push that would be needed to hold a particle on a non-great circle; only where it vanishes (the equator, and by symmetry any great circle) does a curve of constant \( \theta \) free-fall. This is why aircraft great-circle routes are the shortest paths on the globe.

Units check. \( \Gamma^\theta_{\phi\phi}=-\sin\theta\cos\theta \) is dimensionless (angles), and \( \ddot\theta,\dot\phi^2 \) both carry \( \mathrm{(arc)^{-2}} \) when \( s \) is arc length — consistent.

Problems

(A) From \( L=\tfrac1c\sqrt{-g_{\mu\nu}\dot x^\mu\dot x^\nu} \), compute the canonical momentum \( p_\alpha=\partial L/\partial\dot x^\alpha \) and show that, in proper-time gauge, \( p_\alpha = -g_{\alpha\nu}\dot x^\nu/c^2 \). Interpret its magnitude.
Solution
With \( L=\tfrac1c(-g_{\mu\nu}\dot x^\mu\dot x^\nu)^{1/2} \), differentiate: \( p_\alpha=\tfrac1c\cdot\tfrac{-2g_{\alpha\nu}\dot x^\nu}{2(-g\dot x\dot x)^{1/2}} = -\tfrac{1}{c}\tfrac{g_{\alpha\nu}\dot x^\nu}{(-g_{\mu\nu}\dot x^\mu\dot x^\nu)^{1/2}} \). In proper-time gauge \( (-g_{\mu\nu}\dot x^\mu\dot x^\nu)^{1/2}=c \), so \( p_\alpha=-g_{\alpha\nu}\dot x^\nu/c^2 \). Times \( mc^2 \) this is the four-momentum \( P_\alpha=m g_{\alpha\nu}\dot x^\nu \) (up to sign convention). Its norm is \( g^{\alpha\beta}P_\alpha P_\beta=m^2 g_{\mu\nu}\dot x^\mu\dot x^\nu=-m^2c^2 \), the mass shell.
(A) Show directly that \( \Gamma^\alpha_{\mu\nu}=\Gamma^\alpha_{\nu\mu} \), and count the independent components in 4 dimensions.
Solution
Swapping \( \mu\leftrightarrow\nu \) in \( \Gamma^\alpha_{\mu\nu}=\tfrac12 g^{\alpha\beta}(\partial_\mu g_{\beta\nu}+\partial_\nu g_{\beta\mu}-\partial_\beta g_{\mu\nu}) \) leaves the bracket unchanged: the first two terms exchange and \( \partial_\beta g_{\mu\nu}=\partial_\beta g_{\nu\mu} \) by metric symmetry. Hence symmetric (torsion-free). Independent components: \( \alpha \) takes 4 values, the symmetric pair \( (\mu\nu) \) takes \( \binom{4+1}{2}=10 \), giving \( 4\times10=40 \).
(B) For the flat plane in polar coordinates \( ds^2=dr^2+r^2d\phi^2 \), find all nonzero Christoffel symbols and write the geodesic equations. Verify that a straight line has zero covariant acceleration even though \( \ddot r\neq0 \).
Solution
\( g_{rr}=1,\ g_{\phi\phi}=r^2 \). Nonzero: \( \Gamma^r_{\phi\phi}=-\tfrac12 g^{rr}\partial_r g_{\phi\phi}=-r \); \( \Gamma^\phi_{r\phi}=\Gamma^\phi_{\phi r}=\tfrac12 g^{\phi\phi}\partial_r g_{\phi\phi}=1/r \). Equations: \( \ddot r - r\dot\phi^2=0 \), \( \ddot\phi + (2/r)\dot r\dot\phi=0 \). The first is exactly Newton's radial equation with the centrifugal term; a straight line \( r\cos(\phi-\phi_0)=d \) satisfies it. Physically \( \ddot r=r\dot\phi^2\neq0 \), but the connection term cancels it so the covariant acceleration is zero — the "acceleration" is a coordinate artifact of polar axes.
(B) The Rindler metric is \( ds^2=-\big(1+\tfrac{a x}{c^2}\big)^2 c^2 dt^2 + dx^2 \). Compute \( \Gamma^x_{tt} \) and find the proper acceleration required to hold an observer static at \( x=0 \) for \( a=9.8\ \mathrm{m\,s^{-2}} \).
Solution
Let \( f=1+ax/c^2 \), so \( g_{tt}=-f^2c^2 \), \( g_{xx}=1 \). Then \( \Gamma^x_{tt}=-\tfrac12 g^{xx}\partial_x g_{tt}=-\tfrac12\partial_x(-f^2c^2)=c^2 f\,\partial_x f = c^2 f\cdot\tfrac{a}{c^2}=a f \). A static observer has only \( \dot t\neq0 \); the proper acceleration magnitude is \( |a_{\text{proper}}|=\Gamma^x_{tt}(\dot t)^2\cdot(\ldots) \) which evaluates to \( a/f \). At \( x=0 \), \( f=1 \), so the required proper acceleration is exactly \( a=9.8\ \mathrm{m\,s^{-2}} \): Rindler coordinates describe a uniformly accelerating frame, and a geodesic (free-faller) has \( x \) decreasing — it is left behind by the accelerating grid, reproducing the equivalence principle.
(C) In a stationary axisymmetric metric with \( \partial_t g_{\mu\nu}=\partial_\phi g_{\mu\nu}=0 \), show that \( E=-g_{t\nu}\dot x^\nu \) and \( \ell=g_{\phi\nu}\dot x^\nu \) are conserved along geodesics. For Schwarzschild \( g_{tt}=-(1-2GM/rc^2) \), write the conserved energy per unit mass.
Solution
The proper-time Lagrangian \( \tfrac12 g_{\mu\nu}\dot x^\mu\dot x^\nu \) has no explicit \( t \) or \( \phi \) dependence, so \( \partial\mathcal L/\partial t=\partial\mathcal L/\partial\phi=0 \). Euler–Lagrange then gives \( \tfrac{d}{d\tau}(\partial\mathcal L/\partial\dot t)=0 \) and likewise for \( \phi \). Since \( \partial\mathcal L/\partial\dot t=g_{t\nu}\dot x^\nu \) and \( \partial\mathcal L/\partial\dot\phi=g_{\phi\nu}\dot x^\nu \), both are constants of motion; identify \( E=-g_{t\nu}\dot x^\nu \) (energy) and \( \ell=g_{\phi\nu}\dot x^\nu \) (angular momentum), the Noether charges of time-translation and rotation. For Schwarzschild, \( E=\big(1-\tfrac{2GM}{rc^2}\big)c^2\,\tfrac{dt}{d\tau} \) per unit mass — the redshift factor times coordinate-time rate, reducing to \( c^2+\tfrac12 v^2+\Phi \) in the weak-field slow limit.