The Carter Constant

Energy and axial angular momentum come from Kerr’s two Killing vectors, but they do not completely integrate four-dimensional geodesic motion. The missing first integral is quadratic, rather than linear, in the momentum. It is usually called the Carter constant, and its existence is the cleanest operational sign that Kerr has a hidden symmetry.

Required background. Killing Fields and Isometry Groups supplies ordinary Noether integrals; Affine and Variational Geodesics supplies geodesic momentum and Hamiltonian flow.

Helpful background. Einstein’s Equation fixes the vacuum background; The Riemann Curvature Tensor supplies index conventions; Exact Solutions of Einstein’s Equations describes the Kerr solution; Frames, Coframes, and Anholonomy prepares the principal-frame construction.

From separation constant to tensor invariant

Let $u^a$ be an affinely parametrized geodesic tangent and $p_a=\mu u_a$ for a massive particle. A symmetric tensor $K_{ab}$ is a rank-two Killing tensor when

$$\nabla_{(a}K_{bc)}=0.$$

Then

$$\frac{d}{d\lambda}(K_{ab}u^au^b) =u^c\nabla_cK_{ab}\,u^au^b +2K_{ab}u^a u^c\nabla_cu^b=0.$$

The geodesic equation removes the second term, while total symmetry of $u^au^bu^c$ selects the vanishing symmetrized derivative in the first. Thus

$$K\equiv K^{ab}p_ap_b$$

is coordinate-independent and conserved. In Kerr it agrees, with the normalization used on the preceding page, with

$$K=Q+(L_z-aE)^2.$$

The quantity $Q$ itself is often convenient because $Q=0$ for equatorial geodesics. It is not, however, invariantly meaningful until the Killing tensor, its normalization, and the manifest integrals $E=-p_t$ and $L_z=p_\phi$ have been specified. Frolov, Krtouš, and Kubizňák 2017, §§2.2, 3.5, and 6.1, pp. 17–22, 61–65, and 117–121, Open PDF give the tensor and separated descriptions together.

Why no Killing vector can supply it

A Killing vector $\xi^a$ produces the linear integral $\xi^ap_a$. Any fixed linear combination of Kerr’s stationary and axial generators therefore yields $-c_tE+c_\phi L_z$. The Carter integral contains $p_\theta^2$ and, away from special orbit families, cannot equal such a linear expression.

One might try to obtain $K_{ab}$ by symmetrized products of the metric and known Killing vectors. Such a tensor is called reducible. In Schwarzschild, restored rotational symmetry makes the total-angular-momentum tensor reducible in terms of the three rotation generators. For $a\neq0$, Kerr has only the two manifest continuous generators, and its Carter tensor is irreducible: it contains information not generated by their products. This distinction answers why the fourth integral is a hidden, rather than ordinary, symmetry.

The normalization is not unique. Adding $c\,g_{ab}$ shifts $K$ by the already conserved mass-shell quantity $-c\mu^2$, and adding symmetrized products of Killing vectors adds combinations of $E^2$, $EL_z$, and $L_z^2$. Different-looking Carter constants can therefore describe the same independent integral. A comparison must first reduce both expressions modulo these trivial conserved pieces.

Principal form and independent check

Kerr possesses a Killing–Yano two-form $f_{ab}$ satisfying

$$\nabla_{(a}f_{b)c}=0.$$

Its square,

$$K_{ab}=f_{ac}f_b{}^c,$$

is a Killing tensor. The proof follows by differentiating the square and using antisymmetry of $f_{ab}$ together with the Killing–Yano equation. This factorization is stronger than merely discovering a quadratic integral: it also supports symmetry operators for field equations. The next page constructs the closely related principal closed conformal Killing–Yano form.

Two limits test the interpretation. For $a\to0$, $K$ can be shifted by manifest constants to the total $L^2$, reflecting the enlarged $SO(3)$ symmetry. For motion confined to $\theta=\pi/2$, $Q=0$ but $K=(L_z-aE)^2$ generally remains nonzero; saying “the Carter constant vanishes on the equator” is therefore convention-dependent. Carter’s original derivation is Carter 1968, pp. 1559–1571, and the quadratic Killing-tensor interpretation was developed by Walker and Penrose 1970, pp. 265–274.

Exercise

Prove that the metric itself is a Killing tensor and explain why its quadratic integral is not new.

Solution

Metric compatibility gives $\nabla_{(a}g_{bc)}=0$, so $g_{ab}$ obeys the rank-two Killing-tensor equation. Its contraction is $g^{ab}p_ap_b=-\mu^2$, the mass-shell condition. This is already fixed by the geodesic Hamiltonian and supplies no independent integral. It also explains the freedom to replace $K_{ab}$ by $K_{ab}+c g_{ab}$.

References

• Carter, Brandon. “Global Structure of the Kerr Family of Gravitational Fields.” Physical Review 174 (1968): 1559–1571. DOI.
• Frolov, Valeri P., Pavel Krtouš, and David Kubizňák. “Black Holes, Hidden Symmetries, and Complete Integrability.” Living Reviews in Relativity 20 (2017): 6. DOI; Open PDF.
• Walker, Martin, and Roger Penrose. “On Quadratic First Integrals of the Geodesic Equations for Type {22} Spacetimes.” Communications in Mathematical Physics 18 (1970): 265–274. DOI.