Conserved Quantities Along Geodesics

Solving the geodesic equation head-on means integrating four coupled second-order nonlinear ODEs — a wall in any generic spacetime. Conserved quantities knock that wall down. Each constant of motion is a first integral: it lowers the order of the system by one, and when enough accumulate the trajectory collapses to a set of quadratures — ordinary integrals you evaluate one at a time. The systematic source of these constants is symmetry, and the exact statement "the metric is unchanged when you slide it along the direction $\xi$ " is that $\xi^\mu$ is a Killing vector field. The central theorem is almost suspiciously short. If $\xi^\mu$ satisfies Killing's equation $\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0$ and $u^\mu$ is tangent to an affinely parametrized geodesic, then $Q=\xi_\mu u^\mu$ is constant along the curve. The proof is two lines and turns on a single algebraic fact: a symmetric object contracted with an antisymmetric one is zero. The geodesic equation kills one term of $\mathrm{d}Q/\mathrm{d}\lambda$ ; Killing's equation kills the other. This is Noether's theorem wearing geometric clothing. The practical payoff shows up in coordinates. If the metric does not depend on some coordinate $x^\alpha$ — a cyclic coordinate — then $\partial_\alpha$ is Killing and the covariant momentum $p_\alpha$ is conserved: time translation gives energy, rotation gives angular momentum. In Schwarzschild those two charges plus the ever-present norm reduce orbital motion to a one-dimensional energy problem with an effective potential — the machinery behind Mercury's precession, light bending, and the innermost stable circular orbit. And Killing vectors are not the last word: Kerr hides a rank-2 Killing tensor whose quadratic charge, the Carter constant, makes even spinning-black-hole orbits fully integrable.

Key takeaways

A Killing vector field $\xi^\mu$ encodes a continuous symmetry of the metric; contract it with a geodesic's tangent $u^\mu$ and the scalar $Q=\xi_\mu u^\mu$ is constant along the curve.
The proof is two lines: $\mathrm{d}Q/\mathrm{d}\lambda=u^\nu u^\mu\nabla_\nu\xi_\mu+\xi_\mu u^\nu\nabla_\nu u^\mu=0$ , since the geodesic equation kills the second term and Killing's equation $\nabla_{(\mu}\xi_{\nu)}=0$ kills the first.
A coordinate the metric ignores is cyclic: $\partial_\alpha$ is Killing and $p_\alpha=g_{\alpha\mu}p^\mu$ is conserved — the curved-space Noether theorem for canonical momenta.
Stationarity ( $\partial_t$ ) gives conserved energy $E$ ; axisymmetry ( $\partial_\phi$ ) gives conserved angular momentum $L$ . In Schwarzschild these plus the norm reduce orbits to quadratures with $V_{\text{eff}}=\tfrac12(1-2GM/r)(1+L^2/r^2)$ .
The norm $g_{\mu\nu}u^\mu u^\nu$ is conserved on any geodesic by metric compatibility, fixing the curve as timelike, null, or spacelike for all $\lambda$ — even when there is no symmetry at all.
Killing tensors $K_{\mu\nu}$ give quadratic charges $K_{\mu\nu}u^\mu u^\nu$ ; the Carter constant in Kerr is the famous case, the fourth integral that makes spinning-hole geodesics integrable.
Key idea: a symmetry of the metric is a Killing vector $\xi^\mu$ ; contracting it with a geodesic's tangent $u^\mu$ produces $Q=\xi_\mu u^\mu$ , conserved because the geodesic equation and Killing's equation each annihilate one of the two terms in $\mathrm{d}Q/\mathrm{d}\lambda$ .

Where this lives today

Conserved quantities are the daily working tools of relativistic astrophysics. The innermost stable circular orbit sets the inner edge of black-hole accretion disks, and its spin dependence is how X-ray spectroscopists read black-hole rotation off the broadened iron $K\alpha$ line. The Event Horizon Telescope images of M87\ (2019) and Sgr A\ (2022) show a bright photon ring made by null geodesics circling near the photon sphere at $r=3GM$ — a locus fixed purely by the conserved impact parameter $b=L/E$ . In 2020 the GRAVITY collaboration measured the Schwarzschild precession of the star S2 around Sgr A\* — the same $6\pi GM/[c^2a(1-e^2)]$ that gives Mercury its $43''$ , now four million solar masses larger. And modelling extreme-mass-ratio inspirals for LISA is tractable only because the Carter constant renders the underlying geodesics fully integrable.

The physical picture

Picture a direction in which you can slide the entire spacetime and find it exactly as it was — like spinning a perfectly round bowl about its axis. That invariance is a Killing vector, and the rigorous version of "nothing changed" is $\nabla_{(\mu}\xi_{\nu)}=0$ . A particle sliding freely in the bowl conserves its angular momentum about the axis because no azimuthal direction is special; in general relativity that free particle is a geodesic and its conserved charge is $\xi_\mu u^\mu$ . The first figure makes the symmetry, the geodesic, and the two-term cancellation one picture.

Figure

A Killing vector ξ = ∂ϕ generates a symmetry — slide the whole spacetime along it and the metric is unchanged. Contract it with a geodesic tangent u and the charge Q = ξ·u is conserved, because the symmetric tensor uu contracted with the antisymmetric Killing gradient ∇ξ vanishes while the geodesic term u·∇u vanishes on its own

The engine is one algebraic fact: the symmetric tensor $u^\mu u^\nu$ contracted with the antisymmetric $\nabla_\mu\xi_\nu$ that Killing's equation forces is zero, always. And each independent charge peels one degree of freedom off the motion; stack time translation, rotation, and the norm and the Schwarzschild system collapses to the single first-order radial equation the second figure plots as an effective potential.

Figure

Schwarzschild effective potential V_eff = ½(1−2GM/r)(1+L²/r²) for L = 3.9 GM: a stable circular orbit sits at the well minimum (r ≈ 11 GM), an unstable one at the barrier peak (r ≈ 4 GM), and as L is lowered to L² = 12 (GM)² the two merge at the innermost stable circular orbit r = 6 GM where the energy is E = √(8/9) = 0.943

The core equations

Everything begins with the precise statement of a symmetry. Killing's equation says flowing along $\xi$ drags the metric onto itself — its Lie derivative vanishes — which, written with the Levi-Civita connection, is the vanishing of the symmetric part of $\nabla_\mu\xi_\nu$ .

\nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu=0 \qquad (1)

where: $\xi_\nu$ — covariant components of a Killing vector field, the generator of a one-parameter isometry; $\nabla_\mu$ — Levi-Civita covariant derivative. Reading: the symmetrized gradient vanishes, so $\nabla_\mu\xi_\nu$ is purely antisymmetric. The charge itself is the simplest scalar you can build from the symmetry and the motion: contract the Killing field with the geodesic tangent. The claim is that this number does not change along the curve.

Q=\xi_\mu u^\mu=\text{const along the geodesic} \qquad (2)

where: $u^\mu=\mathrm{d}x^\mu/\mathrm{d}\lambda$ — tangent of an affinely parametrized geodesic; $\lambda$ — affine parameter (proper time $\tau$ for matter, an affine $\lambda$ for light); $Q$ — the constant of motion tied to the symmetry $\xi$ . Reading: one Killing vector, one conserved scalar. Why $Q$ is constant is the whole content, and it fits on one line. Differentiate along the curve and apply the Leibniz rule: two terms appear, each annihilated by a different equation.

\frac{\mathrm{d}Q}{\mathrm{d}\lambda}=u^\nu\nabla_\nu(\xi_\mu u^\mu)=\underbrace{u^\nu u^\mu\nabla_\nu\xi_\mu}_{\text{symmetric}\,\times\,\text{antisymmetric}=0}+\underbrace{\xi_\mu\,u^\nu\nabla_\nu u^\mu}_{\text{geodesic}=0}=0 \qquad (3)

where: $u^\nu\nabla_\nu u^\mu$ — the geodesic acceleration, zero for an affine geodesic; $u^\nu u^\mu\nabla_\nu\xi_\mu$ — symmetric pair $u^\nu u^\mu$ contracted with $\nabla_{[\nu}\xi_{\mu]}$ , zero by Killing's equation. Reading: two independent facts, one per term. In practice you rarely solve Killing's equation directly; you read the symmetry off the metric. Any coordinate the components ignore is cyclic, and the corresponding basis vector is automatically Killing.

\partial_\alpha g_{\mu\nu}=0\;\Rightarrow\;\xi=\partial_\alpha\ \text{is Killing}\;\Rightarrow\;p_\alpha=g_{\alpha\mu}p^\mu=\text{const} \qquad (4)

where: $x^\alpha$ — a coordinate absent from the metric (cyclic); $p^\mu=m\,u^\mu$ — the four-momentum (use $p^\mu=u^\mu$ with affine $\lambda$ for massless particles); $p_\alpha$ — the conserved covariant, canonical momentum. Reading: no dependence on $x^\alpha$ means no force there, so $p_\alpha$ is constant. One conserved quantity needs no symmetry at all. Metric compatibility, $\nabla_\rho g_{\mu\nu}=0$ , guarantees the tangent's norm is dragged unchanged along any geodesic.

\frac{\mathrm{d}}{\mathrm{d}\lambda}\big(g_{\mu\nu}u^\mu u^\nu\big)=0 \qquad (5)

where: $g_{\mu\nu}u^\mu u^\nu$ — equals $-1$ (timelike, $\lambda=\tau$ ), $0$ (null), or $+1$ (spacelike) in signature $-+++$ . Reading: a geodesic keeps its character — a timelike curve stays timelike forever. Some spacetimes hide symmetries that are not directions at all but symmetric tensors. A Killing tensor $K_{\mu\nu}$ generates a quadratic charge, irreducible to any combination of Killing-vector charges.

K_{\mu\nu}u^\mu u^\nu=\text{const},\qquad \nabla_{(\rho}K_{\mu\nu)}=0 \qquad (6)

where: $K_{\mu\nu}$ — a symmetric tensor obeying the totally-symmetrized derivative condition; $\nabla_{(\rho}K_{\mu\nu)}$ — symmetrization over all three indices. Reading: the same symmetric-times-symmetrized-derivative argument as before, one rank higher. Assemble the two Killing charges $E,L$ with the conserved norm in Schwarzschild and the four equations of motion boil down to one first-order radial equation with an effective potential.

\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2=E^2-\left(1-\frac{2GM}{r}\right)\left(1+\frac{L^2}{r^2}\right) \qquad (7)

where: $E=(1-2GM/r)\,\mathrm{d}t/\mathrm{d}\tau$ — conserved energy per unit mass; $L=r^2\,\mathrm{d}\phi/\mathrm{d}\tau$ — conserved angular momentum per unit mass; $GM$ — geometrized mass ( $c=1$ ); $r$ — areal radial coordinate. Reading: $\tfrac12\dot r^2+V_{\text{eff}}=\tfrac12E^2$ with $V_{\text{eff}}=\tfrac12(1-2GM/r)(1+L^2/r^2)$ — a particle in one dimension.

Derivation AConservation of the Killing charge along a geodesic

The flagship theorem states that the scalar built by contracting any Killing vector with the tangent of an affinely parametrized geodesic is a constant of the motion. The goal here is to prove it by splitting a single derivative into two pieces and annihilating each with a separate structural equation — the geodesic equation for one, Killing's equation for the other. Let $u^\mu$ be tangent to an affine geodesic, $u^\nu\nabla_\nu u^\mu=0$ , let $\xi_\mu$ be Killing, and compute how $Q=\xi_\mu u^\mu$ changes along the curve.

Step 1. The rate of change of any scalar along the geodesic is the directional derivative $\mathrm{d}/\mathrm{d}\lambda=u^\nu\nabla_\nu$ , so $\mathrm{d}Q/\mathrm{d}\lambda=u^\nu\nabla_\nu(\xi_\mu u^\mu)$ . Why: for a scalar the covariant and ordinary directional derivatives agree, so this is an exact rewriting, not an approximation.

Step 2. Apply the Leibniz rule to the product $\xi_\mu u^\mu$ to obtain $\mathrm{d}Q/\mathrm{d}\lambda=u^\nu u^\mu\nabla_\nu\xi_\mu+\xi_\mu\,u^\nu\nabla_\nu u^\mu$ . Why: the covariant derivative is a derivation, so it distributes across the contraction and produces exactly two terms — one where $\nabla$ hits $\xi$ , one where it hits $u$ .

Step 3. The second term contains $u^\nu\nabla_\nu u^\mu$ , the geodesic acceleration, which vanishes identically for an affine geodesic, so $\xi_\mu\,u^\nu\nabla_\nu u^\mu=0$ . Why: an affine geodesic parallel-transports its own tangent — the coordinate-free "straightest curve" — which is why affine parametrization is required.

Step 4. In the first term $u^\nu u^\mu$ is symmetric under $\nu\leftrightarrow\mu$ , so only the symmetric part of $\nabla_\nu\xi_\mu$ survives: $u^\nu u^\mu\nabla_\nu\xi_\mu=u^\nu u^\mu\nabla_{(\nu}\xi_{\mu)}$ . Killing's equation states exactly $\nabla_{(\nu}\xi_{\mu)}=0$ , so this term vanishes too. Why: this is the single algebraic engine of the subject — symmetric contracted with antisymmetric is always zero, and Killing's equation is what makes $\nabla\xi$ antisymmetric.

Step 5. Both terms vanish, so $\mathrm{d}Q/\mathrm{d}\lambda=0$ and $Q=\xi_\mu u^\mu$ is constant along the geodesic. Why: nothing about the specific spacetime entered — only the geodesic and Killing equations — so the theorem holds for every isometry of every metric.

Go deeper · Derivation B · 1 minCyclic coordinates and conserved momenta

Derivation BCyclic coordinates and conserved momenta

The abstract theorem becomes a calculational tool the moment you can spot Killing vectors by inspection. This derivation shows that whenever the metric is independent of some coordinate $x^\alpha$ , the basis vector $\partial_\alpha$ is automatically Killing and its charge is just the covariant momentum $p_\alpha$ — the exact curved-space analogue of the mechanics rule that a coordinate missing from the Lagrangian yields a conserved canonical momentum. Work in a chart where $\partial_\alpha g_{\mu\nu}=0$ for one fixed index $\alpha$ and take $\xi^\mu=\delta^\mu{}_\alpha$ .

Step 1. Write the Lie derivative of the metric along $\xi^\rho=\delta^\rho{}_\alpha$ : $(\mathcal L_{\partial_\alpha}g)_{\mu\nu}=\xi^\rho\partial_\rho g_{\mu\nu}+g_{\rho\nu}\partial_\mu\xi^\rho+g_{\mu\rho}\partial_\nu\xi^\rho$ . Why: the Lie derivative, not the covariant one, is the coordinate-friendly test for an isometry, and it needs only partial derivatives of the components and of $\xi$ .

Step 2. The components $\xi^\rho=\delta^\rho{}_\alpha$ are constants, so their derivatives vanish and only the first term remains: $(\mathcal L_{\partial_\alpha}g)_{\mu\nu}=\partial_\alpha g_{\mu\nu}=0$ by assumption. Hence $\partial_\alpha$ is Killing. Why: coordinate-independence of the metric is literally the vanishing of one Lie derivative, so a cyclic coordinate is a genuine symmetry, not a coordinate artifact.

Step 3. Form the charge from the general theorem: $Q=\xi_\mu u^\mu=g_{\mu\nu}\delta^\nu{}_\alpha u^\mu=g_{\alpha\mu}u^\mu=u_\alpha$ . Why: contracting with a basis vector selects and lowers one component, so the abstract $\xi_\mu u^\mu$ becomes the concrete covariant $u_\alpha$ .

Step 4. Multiply by the mass to identify the conserved momentum, $p_\alpha=m\,u_\alpha=\text{const}$ . For $\alpha=t$ this is (minus) the energy; for $\alpha=\phi$ the angular momentum. Why: this is the Hamiltonian statement $\dot p_\alpha=-\partial H/\partial x^\alpha=0$ for a cyclic coordinate — the relativistic and canonical pictures are one result in two languages.

Go deeper · Derivation C · 1 minSchwarzschild orbits reduced to a one-dimensional energy problem

Derivation CSchwarzschild orbits reduced to a one-dimensional energy problem

This derivation demonstrates the power of conserved quantities by collapsing the four-component geodesic system of Schwarzschild spacetime into a single first-order radial equation, using the two Killing charges from stationarity and axisymmetry together with the always-conserved norm — the reduction underlying perihelion precession, light bending, and the innermost stable circular orbit. Work in the equatorial plane $\theta=\pi/2$ with metric $\mathrm{d}s^2=-(1-2GM/r)\mathrm{d}t^2+(1-2GM/r)^{-1}\mathrm{d}r^2+r^2\mathrm{d}\phi^2$ for a timelike geodesic in proper time $\tau$ , overdots meaning $\mathrm{d}/\mathrm{d}\tau$ .

Step 1. The metric ignores $t$ and $\phi$ , so $\partial_t$ and $\partial_\phi$ are Killing and give two charges: $E=-\xi_{(t)\mu}u^\mu=(1-2GM/r)\dot t$ and $L=\xi_{(\phi)\mu}u^\mu=r^2\dot\phi$ . Why: these are the cyclic-coordinate momenta from the previous derivation, and they let us eliminate $\dot t$ and $\dot\phi$ in favor of the two constants.

Step 2. The conserved norm of a timelike geodesic is $-1$ : $-(1-2GM/r)\dot t^2+(1-2GM/r)^{-1}\dot r^2+r^2\dot\phi^2=-1$ . Why: metric compatibility guarantees this holds at every point with the same value, so it functions as a third first integral alongside $E$ and $L$ .

Step 3. Substitute $\dot t=E/(1-2GM/r)$ and $\dot\phi=L/r^2$ into the norm to get $-E^2/(1-2GM/r)+\dot r^2/(1-2GM/r)+L^2/r^2=-1$ . Why: the two Killing charges convert the norm from a relation among four unknown velocities into a relation in $r$ and $\dot r$ alone.

Step 4. Multiply through by $(1-2GM/r)$ and isolate the radial term: $\dot r^2=E^2-(1-2GM/r)(1+L^2/r^2)$ . Why: the four-dimensional second-order system is now a single first-order equation $\tfrac12\dot r^2+V_{\text{eff}}(r)=\tfrac12E^2$ — a particle of energy $\tfrac12E^2$ in one dimension.

Step 5. Read off $V_{\text{eff}}=\tfrac12(1-2GM/r)(1+L^2/r^2)$ and, with $\mathrm{d}\phi/\mathrm{d}r=L/(r^2\dot r)$ , write the orbit as the quadrature $\phi=\int L\,\mathrm{d}r/(r^2\sqrt{E^2-(1-2GM/r)(1+L^2/r^2)})$ . Why: expanding $V_{\text{eff}}=\tfrac12-GM/r+L^2/2r^2-GML^2/r^3$ exposes the Newtonian potential plus the purely relativistic $-GML^2/r^3$ term that precesses Mercury and bends starlight.

Limiting cases

How many independent constants a geodesic carries is dictated entirely by the isometries — and the hidden symmetries — of the spacetime it lives in.

Case	Result	Meaning
Minkowski	$\dim(\text{isometry group})=10$	Ten Killing vectors (4 translations, 3 rotations, 3 boosts) — the maximal set; free particles move straight with conserved 4-momentum and angular momentum.
Schwarzschild	$E=-\xi_{(t)\mu}u^\mu,\ L=\xi_{(\phi)\mu}u^\mu$	Stationary and spherically symmetric: $\partial_t$ and $\partial_\phi$ give energy and angular momentum; with the norm, equatorial orbits reduce to quadratures.
Kerr	$E,\ L,\ \mu^2=-p\cdot p,\ \mathcal Q$	Only $\partial_t,\partial_\phi$ are Killing, but an irreducible rank-2 Killing tensor supplies the Carter constant $\mathcal Q$ — four integrals make geodesics integrable.
Generic spacetime	$\nabla_{(\mu}\xi_{\nu)}=0$ has no solution	No continuous symmetry: only the norm $g_{\mu\nu}u^\mu u^\nu$ survives; geodesics need numerical integration.

Common trapThe conserved energy is not everyone's energy

Conservation of $\xi_\mu u^\mu$ needs two things at once: an affinely parametrized geodesic and a genuine Killing vector. Reparametrize non-affinely and $u^\nu\nabla_\nu u^\mu=f(\lambda)u^\mu\neq0$ , leaving a residual $fQ$ that spoils constancy. Use a basis vector $\partial_\alpha$ when the metric actually depends on $x^\alpha$ and there is no conservation, because $\partial_\alpha$ is then not Killing. The subtler trap is meaning: the conserved energy $E=-\xi_{(t)\mu}p^\mu$ is a global label tied to the timelike Killing field, existing only in stationary spacetimes. It is not the energy a freely falling observer measures, which is the frame-dependent projection $E_{\text{loc}}=-p\cdot u_{\text{obs}}$ and is generally not conserved. Confusing the two is the commonest error in orbit problems.

Worked examples

1 · The innermost stable circular orbit and accretion efficiency (fully worked)

From the Schwarzschild effective potential $V_{\text{eff}}=\tfrac12(1-2GM/r)(1+L^2/r^2)$ , locate the innermost stable circular orbit, its angular momentum and energy, and the rest-energy fraction an infalling particle can radiate before crossing it. Then quote $r_{\text{ISCO}}$ for Sgr A\ ( $M=4.30\times10^6\,M_\odot$ ).*

Step 1. Circular orbits require $\mathrm{d}V_{\text{eff}}/\mathrm{d}r=0$ . Differentiating $V_{\text{eff}}=\tfrac12-GM/r+L^2/2r^2-GML^2/r^3$ gives $GM r^2-L^2 r+3GML^2=0$ , with roots $r_\pm=\dfrac{L^2\pm L\sqrt{L^2-12G^2M^2}}{2GM}$ — $r_+$ the stable orbit (well minimum), $r_-$ the unstable one (barrier peak). Two circular orbits exist only when $L^2>12G^2M^2$ .

Step 2. The two orbits merge — and stability is marginally lost — exactly when the square root vanishes, $L^2=12G^2M^2$ , i.e. $L_{\text{ISCO}}=2\sqrt3\,GM=3.464\,GM$ . Substituting back, $r_{\text{ISCO}}=L^2/2GM=12G^2M^2/2GM=\mathbf{6GM}$ . This is the innermost stable circular orbit; inside it, no stable circular motion is possible.

Step 3. The energy on a circular orbit is $E^2=\dfrac{(1-2GM/r)^2}{1-3GM/r}$ . At $r=6GM$ , $E^2=\dfrac{(2/3)^2}{1/2}=\dfrac{8}{9}$ , so $E_{\text{ISCO}}=\sqrt{8/9}=\mathbf{0.9428}$ per unit rest mass.

Step 4. A particle spiralling adiabatically from rest at infinity ( $E=1$ ) to the ISCO must shed the difference: the fractional binding energy is $1-E_{\text{ISCO}}=1-0.9428=\mathbf{0.0572}$ . About 5.7% of the rest energy is radiated — eight times the 0.7% of hydrogen fusion, which is why black-hole accretion is the most efficient steady power source known (rising to ~42% for a maximal Kerr hole).

Step 5. Scale to Sgr A\*. The gravitational radius is $r_g=GM/c^2=(1.477\text{ km})\times(4.30\times10^6)=6.35\times10^6\text{ km}$ , so $r_{\text{ISCO}}=6r_g=\mathbf{3.8\times10^7\ km}\approx0.25\ \mathrm{AU}$ — a whole cascade of physics that exists only because $E$ and $L$ are conserved.

2 · Mercury's perihelion precession (fully worked)

The relativistic $-GML^2/r^3$ term in $V_{\text{eff}}$ makes bound orbits precess by $\Delta\phi=6\pi GM/[c^2a(1-e^2)]$ per orbit. Evaluate it for Mercury ( $a=5.79\times10^7$ km, $e=0.2056$ , period $87.97$ d).

Step 1. A nearly circular bound orbit oscillates radially with frequency set by $V_{\text{eff}}''(r_0)$ and revolves with $\dot\phi=L/r_0^2$ . To first order in $GM/r$ the ratio of the two frequencies departs from unity by $3GM/[c^2a(1-e^2)]$ per radian, accumulating to $\Delta\phi=6\pi GM/[c^2a(1-e^2)]$ per orbit. The excess over the Newtonian closed ellipse comes entirely from the $1/r^3$ term.

Step 2. Insert numbers. With $GM_\odot/c^2=1.476$ km and $a(1-e^2)=5.79\times10^7\times(1-0.2056^2)=5.79\times10^7\times0.9577=5.545\times10^7$ km,

\Delta\phi=\frac{6\pi\,(1.476)}{5.545\times10^{7}}=5.02\times10^{-7}\ \text{rad}=\mathbf{0.1035''\ per\ orbit}.

Step 3. Mercury completes $36525/87.97=415.2$ orbits per century, so the accumulated advance is $0.1035''\times415.2=\mathbf{43.0''\ per\ century}$ . This matched the residual left unexplained by Newtonian perturbations — the first quantitative triumph of general relativity, and a direct consequence of the two Killing charges reducing the orbit to a quadrature.

Physics of the portrait

The subject sits at a three-way crossroads: the geometer who wrote down what a symmetry is, the mathematician who proved symmetries and conservation laws are the same thing, and the physicist who found the hidden symmetry no one expected. The portrait belongs to the first.

Origins · 1 minWilhelm Killing and how this was discovered

Wilhelm Killing

1847–1923 · the geometry of symmetry: isometries and the equation that bears his name

Wilhelm Killing was a German mathematician who spent most of his career at the Lyceum Hosianum in Braunsberg, an isolated seminary in East Prussia far from the great research centres — and who nonetheless built two of the load-bearing structures of modern geometry and physics. Working on the classification of "space forms", the geometries of constant curvature, he was led around 1892 to the condition an infinitesimal transformation must satisfy to be a symmetry of a metric: the vanishing of the symmetrized covariant derivative of its generating vector field. That is Killing's equation, and its solutions — Killing vector fields — are the exact objects that turn continuous symmetries into conserved quantities along geodesics. Killing arrived at them years before general relativity existed, purely as geometry; Einstein's theory later made the metric physical and the Killing charges into energy and angular momentum. The through-line from Killing's equation to this article runs through Emmy Noether, whose 1918 theorem proved every continuous symmetry yields a conservation law, and reaches its most surprising modern form in Brandon Carter's 1968 discovery that Kerr hides a Killing tensor — a symmetry that is not a direction at all — whose charge, the Carter constant, is the fourth integral that makes spinning-black-hole orbits exactly solvable.

History

1892

— Wilhelm Killing formulates the equation for the infinitesimal isometries of a metric

1915

— Einstein's field equations make the spacetime metric dynamical

1916

— Schwarzschild solves the equations for a spherical mass, exposing conserved E and L

1918

— Emmy Noether proves every continuous symmetry yields a conserved quantity

1968

— Brandon Carter finds the fourth constant that makes Kerr geodesics integrable

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §25.2–25.5 · Wald, General Relativity (1984), §C.3 and Ch. 6 · Carroll, Spacetime and Geometry (2004), §3.8, §5.4 · Hartle, Gravity (2003), Ch. 8–9 · Schutz, A First Course in General Relativity (2009), Ch. 7 · Carter, Phys. Rev. 174, 1559 (1968) — the Carter constant. The key numerical benchmarks are: $r_{\text{ISCO}}=6GM$ , $E_{\text{ISCO}}=\sqrt{8/9}=0.943$ , ISCO binding energy $5.7\%$ , Mercury's perihelion advance $43''$ /century (more precisely $42.98''$ /century), light bending $1.75''$ , and photon sphere at $r=3GM$ .

Builds on. The Geodesic Equation · Affine Parameters · Covariant Derivative · Metric Compatibility · The Levi-Civita Connection Leads to. Tidal Forces in General Relativity · Geodesic Deviation Equation · Geodesics as Extremal Paths.

Ask the tutor: Want to see how the same $E$ and $L$ that fix Mercury's $43''$ also carve out the ISCO and the photon ring — or how Kerr's hidden Carter constant differs from an ordinary symmetry?

Why is a symmetry the same thing as a conserved quantity?

This is Noether's theorem in clean geometric form. A continuous symmetry of the metric is a direction $\xi$ you can flow along without changing anything — a Killing vector. Contract it with a freely falling particle's momentum and the result cannot change, because the two ways it could are killed off: the geodesic equation removes the piece from the particle accelerating, Killing's equation the piece from the geometry varying along $\xi$ . So "the geometry looks the same this way" and "this momentum is conserved" are one fact — time translation becomes energy, rotation becomes angular momentum.

What is the difference between a Killing vector and a Killing tensor?

A Killing vector $\xi_\mu$ is a symmetry you can picture as a direction: slide the spacetime along it and the metric is unchanged. It gives a charge linear in the momentum, $\xi_\mu p^\mu$ . A Killing tensor $K_{\mu\nu}$ is a symmetric tensor obeying $\nabla_{(\rho}K_{\mu\nu)}=0$ ; it gives a quadratic charge $K_{\mu\nu}p^\mu p^\nu$ . The metric itself is a trivial Killing tensor, giving the norm. What makes Kerr remarkable is that it admits an irreducible one — not built from the metric and Killing vectors — so its charge, the Carter constant, is genuinely new. Such "hidden symmetries" have no visualizable flow, which is why they eluded physicists until 1968.

Is the conserved energy E the same as the energy I would measure locally?

No, and conflating them causes real errors. The conserved $E=-\xi_{(t)\mu}p^\mu$ is a global bookkeeping quantity set by the timelike Killing field of a stationary spacetime — constant along the whole geodesic, the natural "energy at infinity". A specific observer with four-velocity $u_{\text{obs}}$ instead measures $E_{\text{loc}}=-p\cdot u_{\text{obs}}$ , the projection onto their own time. For a static observer at radius $r$ the two are related by the redshift factor $E_{\text{loc}}=E/\sqrt{1-2GM/r}$ , so the locally measured photon energy grows without bound near the horizon even though $E$ stays fixed. Gravitational redshift is exactly this: one conserved $E$ , different local energies at different depths.

Why does the parameter have to be affine?

The whole proof rests on $u^\nu\nabla_\nu u^\mu=0$ , and that clean form of the geodesic equation holds only for affine parameters — proper time for a massive particle, or a parameter linearly related to it for light. With a non-affine parameter the equation picks up a right-hand side, $u^\nu\nabla_\nu u^\mu=f(\lambda)\,u^\mu$ , describing the same path traversed at a varying rate. That term feeds a residual $f(\lambda)\,Q$ into $\mathrm{d}Q/\mathrm{d}\lambda$ , so $Q$ drifts even though the underlying curve is unchanged. The conservation law is about the geodesic and the clock you read it with — re-expressing against proper time restores constancy.

How does the norm being conserved actually help solve for an orbit?

It supplies a free first integral that costs no symmetry. On any timelike geodesic the norm equals $-1$ for all proper time — one algebraic relation among the velocity components. In Schwarzschild, after the two Killing charges eliminate $\dot t$ and $\dot\phi$ , the norm is precisely the equation that becomes the radial energy relation $\dot r^2=E^2-(1-2GM/r)(1+L^2/r^2)$ : it is the third constant that closes a system of two charges and four unknowns. Where there is no symmetry at all the norm is the only constant you have — the last line of defence for numerical integration.

Do these conserved quantities still hold for a charged particle or a rocket?

The Killing charges do not, because the particle is no longer on a geodesic — only the norm survives unconditionally. A charged particle obeys $u^\nu\nabla_\nu u^\mu=(q/m)F^\mu{}_\nu u^\nu$ , so the second term in the two-term proof no longer vanishes and $\xi_\mu u^\mu$ drifts. What is conserved instead is a modified charge absorbing the field: for a symmetry $\xi$ shared by the metric and the potential $A_\mu$ , the canonical momentum $\xi_\mu(p^\mu+qA^\mu)$ is constant. A rocket firing its engine breaks the geodesic condition outright. The conservation law is about free fall in a symmetric geometry; add a force and you must add its potential to the charge.