Null Geodesics and Affine Parameters

A null curve accumulates zero proper time, so its own clock cannot label its points. The replacement is an affine parameter: a label for which the tangent parallel-transports itself. Affine parameters are not unique, but their freedom is only linear; an arbitrary parameter instead introduces a tangential nonaffinity term.

Required background. Affine and Variational Geodesics supplies affine geodesics, Jacobi fields, and their variational characterization; What the Equivalence Principles Do and Do Not Imply supplies local-inertial reasoning and the scope of universal free fall.

Helpful background. Events, Intervals, and Minkowski Geometry supplies causal types and timelike and null worldlines; Tensor Fields and Natural Operations supplies coordinate-independent tensor operations; Presymplectic Potentials and Currents supplies disciplined endpoint and boundary variations.

Null curves and null geodesics

For a smooth curve $x^\mu(\lambda)$ with tangent $k^\mu=dx^\mu/d\lambda$, nullness is the algebraic condition

$$k^\mu k_\mu=0.$$

It does not by itself imply geodesic motion. A null curve is a pregeodesic precisely when

$$k^\nu\nabla_\nu k^\mu=f(\lambda)k^\mu$$

for some scalar $f$. The right-hand side is purely tangent, so it changes the rate at which the curve is traversed without bending its image. A transverse component of acceleration would instead describe a genuinely nongeodesic null curve.

For the Levi-Civita connection, metric compatibility gives

$$\frac{d}{d\lambda}(k^2)=2f k^2=0,$$

so the null constraint is preserved. This check is necessary but not sufficient: $k^2=0$ can remain true even when transverse acceleration is present. For example, in Cartesian Minkowski coordinates the curve

$$x^\mu(\lambda) =\bigl(\lambda,b\cos(\lambda/b),b\sin(\lambda/b),0\bigr), \qquad b\gt0,$$

has tangent $k^\mu=(1,-\sin(\lambda/b),\cos(\lambda/b),0)$ and hence $k^2=0$ everywhere. Its acceleration is a nonzero inward spatial vector with squared norm $1/b^2$; it cannot be proportional to the null tangent. This is a valid null curve, but it is not a freely propagating vacuum light ray. Nullness specifies the allowed instantaneous direction, while the geodesic equation specifies how that direction evolves.

Constructing an affine parameter

Let $s$ be a new, orientation-preserving parameter and define

$$\widetilde k^\mu=\frac{dx^\mu}{ds} =\alpha k^\mu, \qquad \alpha=\frac{d\lambda}{ds}.$$

A direct covariant differentiation gives

$$\widetilde k^\nu\nabla_\nu\widetilde k^\mu =\alpha^2\left(f+\frac{d\ln\alpha}{d\lambda}\right)k^\mu.$$

Therefore $s$ is affine if

$$\frac{d\ln\alpha}{d\lambda}=-f.$$

This first-order equation always has a local solution where the original parametrization is regular. Once an affine parameter is found, every other affine parameter has the form $s'=as+b$ with $a\ne0$; $a>0$ preserves time orientation. Reall states this affine freedom and its special significance for null curves in Reall 2013, §4.3, pp. 45–46, PDF.

Proper time fixes the scale for a timelike geodesic through $u^2=-1$. Nullness supplies no analogous normalization because multiplying $k^\mu$ by a constant leaves $k^2=0$. Initial null direction and initial tangent are therefore different data: the latter also fixes the affine scale.

Null geodesics also expose what conformal geometry does and does not preserve. Under a metric rescaling $\widetilde g_{\mu\nu}=\Omega^2g_{\mu\nu}$, an affinely parametrized null tangent for $g_{\mu\nu}$ obeys

$$k^\nu\widetilde\nabla_\nu k^\mu =2\frac{d\ln\Omega}{d\lambda}k^\mu.$$

The same image is therefore a null pregeodesic of the rescaled metric, but $\lambda$ is generally no longer affine. Light cones and unparametrized null geodesics are conformally invariant; affine intervals and null completeness are not. The tangential form of the change provides a direct check of the pregeodesic criterion.

Massless action and conserved quantities

The massless einbein action is

$$S_0[x,e]=\frac12\int d\lambda\, e^{-1}g_{\mu\nu}\dot x^\mu\dot x^\nu.$$

Variation of $e$ imposes $\dot x^2=0$; variation of $x^\mu$ gives a pregeodesic with $f=\dot e/e$. Constant-e gauge makes the parameter affine. This avoids the degenerate expression $-m\int d\tau$, which vanishes when $m=0$ and $d\tau=0$.

If $\xi^\mu$ is a Killing field and $k^\mu$ is affinely parametrized, then

$$\frac{d}{ds}(\xi_\mu k^\mu) =k^\nu k^\mu\nabla_{(\nu}\xi_{\mu)}=0.$$

The numerical constant rescales under $s\mapsto as+b$. A physical photon frequency instead acquires its normalization from a wave phase and an observer through $\omega=-u_\mu k^\mu$. Geometry alone does not choose that scale.

The arbitrary scale cancels from some observables. If one affinely parametrized ray connects emitter and receiver, a constant rescaling of $k^\mu$ multiplies both measured frequencies by the same factor, so their ratio is unchanged. By contrast, constants assigned to different rays cannot be compared until a common wave-phase or endpoint normalization is specified. An affine parameter is therefore neither an intrinsic null distance nor an observer’s clock reading.

When null geodesics represent light

In general relativity, a null geodesic is a geometric curve. A light ray follows one only in the geometric-optics limit of Maxwell theory, when wavelength is short compared with background variation scales and the beam is considered away from diffraction and caustic breakdown. A quantum photon is not literally a classical point on a curve. The derivation from parallel transport, length variation, and affine parametrization is given in Carroll 1997, Ch. 3, pp. 68–73, PDF; the field-theory limit belongs to Light Rays as Null Geodesics.

Affine extension is also a global question. A locally affine null geodesic may have only a finite maximal parameter interval; the local construction does not establish null geodesic completeness.

Exercise

Start with an affinely parametrized geodesic $x^\mu(s)$ for $s>0$ and relabel it by $\lambda=\ln s$. Find its nonaffinity $f$.

Solution

Here $ds/d\lambda=s=e^\lambda$, so $k^\mu=dx^\mu/d\lambda=s\,dx^\mu/ds$. Since the $s$ tangent is parallel transported,

$$k^\nu\nabla_\nu k^\mu =\frac{d\ln(ds/d\lambda)}{d\lambda}k^\mu =k^\mu.$$

Thus $f=1$. Solving $d\ln\alpha/d\lambda=-1$ gives $\alpha\propto e^{-\lambda}$ and recovers an affine parameter proportional to $e^\lambda=s$.

References

• Carroll, Sean M. “Lecture Notes on General Relativity.” NSF-ITP/97-147, arXiv:gr-qc/9712019v1 (1997), Ch. 3, printed pp. 54–97. arXiv.
• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes (2013), § 4.3, printed pp. 45–46. PDF.