The Test-Particle Action

A point-particle action isolates the universal motion of a structureless probe from the details of the clock used to label its worldline. For a massive probe, extremizing elapsed proper time gives the geodesic equation. An auxiliary one-dimensional metric—the einbein—extends the same construction to massless probes and makes the distinction between arbitrary and affine parameters explicit.

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.

Proper-time action for a massive probe

Let $x^\mu(\lambda)$ be a future-directed timelike curve and write $\dot x^\mu=dx^\mu/d\lambda$. Its proper time is

$$\tau[x]=\int d\lambda\, \sqrt{-g_{\mu\nu}(x)\dot x^\mu\dot x^\nu},$$

so the massive worldline action is

$$S_{\rm length}[x]=-m\tau[x].$$

Under an orientation-preserving change $\lambda\mapsto\lambda'(\lambda)$, the square root and $d\lambda$ acquire inverse Jacobian factors. The action therefore depends on the oriented curve, not its labeling. Varying $x^\mu$ with fixed endpoints and only afterward choosing $\lambda=\tau$ gives

$$u^\nu\nabla_\nu u^\mu=0, \qquad u^\mu=\frac{dx^\mu}{d\tau}, \qquad u^\mu u_\mu=-1.$$

To see what is varied, set $w=\sqrt{-g_{\mu\nu}\dot x^\mu\dot x^\nu}$. At fixed parameter values,

$$\delta w=-\frac{1}{2w} \left( \partial_\rho g_{\mu\nu}\dot x^\mu\dot x^\nu\,\delta x^\rho +2g_{\mu\nu}\dot x^\mu\delta\dot x^\nu \right).$$

Integrate the $\delta\dot x$ term by parts, and then parametrize the reference curve by its proper time. The result is

$$\delta S_{\rm length} =m[u_\mu\delta x^\mu]_{\tau_1}^{\tau_2} -m\int_{\tau_1}^{\tau_2}a_\mu\delta x^\mu\,d\tau, \qquad a^\mu=\nabla_u u^\mu.$$

Fixed endpoints remove the first term. Arbitrary interior variations then require $a_\mu=0$. The mass multiplies the entire variation and cancels from the equation, giving universality within the test-body approximation. Substituting $u^2=-1$ into the action before varying would erase the path dependence of $d\tau$ and lose this equation. The detailed variation and its local-extremum qualification appear in Reall 2013, §3.3, pp. 36–38, PDF.

This action cannot describe a null curve directly: its integrand then vanishes identically. Treating that zero functional as though it had an ordinary Euler–Lagrange equation loses the null constraint.

Einbein action and the null limit

Introduce a positive worldline field $e(\lambda)$ and take

$$S[x,e]=\frac12\int d\lambda \left(e^{-1}g_{\mu\nu}\dot x^\mu\dot x^\nu-e m^2\right).$$

Reparametrization invariance is restored if $e$ transforms so that $e'(\lambda')d\lambda'=e(\lambda)d\lambda$. Variation with respect to $e$ gives the mass-shell constraint

$$g_{\mu\nu}\dot x^\mu\dot x^\nu+e^2m^2=0.$$

For $m>0$, eliminating $e=\sqrt{-\dot x^2}/m$ returns $S_{\rm length}$. Variation with respect to $x^\mu$ gives

$$\frac{D\dot x^\mu}{d\lambda} =\frac{\dot e}{e}\dot x^\mu.$$

Thus an arbitrary gauge for $e$ generally produces a nonaffine parametrization: the acceleration is tangent to the curve and changes its labeling, not its image in spacetime. A gauge with constant $e$ makes $\lambda$ affine. When $m=0$, the $e$ equation instead imposes $\dot x^2=0$, while the $x$ equation still supplies a null geodesic. This is why the einbein form, unlike the length form, has a regular massless limit.

The order of operations matters. For $m>0$, $e$ is algebraic and nonzero, so solving its equation and substituting back is equivalent to the square-root formulation. For $m=0$, the constraint cannot be solved by dividing by $m$; it must remain as an independent Euler–Lagrange equation. Likewise, fixing $e$ before variation is legitimate only if the lost constraint is imposed separately. Otherwise the quadratic equation finds affine geodesics of every causal type without selecting the desired mass shell.

Quadratic action and affine gauge

After fixing $e$ to a constant, one often uses the shorter quadratic functional

$$S_{\rm quad}[x]=\frac12\int d\lambda\, g_{\mu\nu}\dot x^\mu\dot x^\nu.$$

Its variation is particularly transparent:

$$\delta S_{\rm quad} =[\dot x_\mu\delta x^\mu]_{\lambda_1}^{\lambda_2} -\int d\lambda\, g_{\mu\nu}\frac{D\dot x^\nu}{d\lambda}\delta x^\mu.$$

The fixed-endpoint equation is therefore $D\dot x^\mu/d\lambda=0$, with no tangential nonaffinity term. But $S_{\rm quad}$ is not invariant under an arbitrary reparametrization; it is a gauge-fixed action. For a timelike solution, imposing $\dot x^2=-1$ identifies $\lambda$ with proper time up to an additive constant. For a null solution, no normalization removes the remaining constant rescaling of the affine parameter. Carroll derives the length variation and separates affine from nonaffine parameters in Carroll 1997, Ch. 3, pp. 68–72, PDF.

Checks and domain of the idealization

Three checks distinguish the actions. First, substituting the $e$ constraint must recover $-m\int d\tau$ for $m>0$. Second, metric compatibility gives $d(\dot x^2)/d\lambda=0$ in affine gauge, preserving timelike, null, or spacelike character. Third, in Cartesian Minkowski coordinates the equations reduce to $d^2x^\mu/d\lambda^2=0$.

The model assumes a body small compared with the curvature scale, negligible spin and higher multipoles, negligible charge or other applied force, and negligible disturbance of the background by its own field. The action predicts the leading trajectory only inside that regime; it is not an action for self-force, radiation reaction, internal dynamics, or a finite body’s choice of center.

Exercise

Show directly that eliminating $e$ gives the massive length action, while setting $m=0$ before varying does not make the dynamics trivial.

Solution

The $e$ equation gives $e=\sqrt{-\dot x^2}/m$ for a future-directed timelike curve. Substitution makes each of the two terms in the integrand equal to $-m\sqrt{-\dot x^2}/2$, hence their sum is $-m\sqrt{-\dot x^2}$. If $m=0$, variation of $e$ still gives $\dot x^2=0$ and variation of $x^\mu$ gives $D(e^{-1}\dot x^\mu)/d\lambda=0$. Choosing constant $e$ therefore yields a nontrivial affinely parametrized null geodesic.

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), §§ 3.3 and 4.3, printed pp. 36–46. PDF.