Timelike Congruences and Spatial Projection

A timelike congruence is a smooth family of nonintersecting timelike curves filling an open region. Its unit tangent field $u^\mu$ selects, at each event, a local rest space and a derivative tensor that measures the instantaneous relative motion of neighboring curves. This page constructs those objects for an arbitrary, possibly accelerated congruence in an $n$-dimensional Lorentzian spacetime.

The construction is meaningful only on a region where a single smooth tangent is defined. Neighboring members may converge to a caustic, beyond which their labels no longer define a one-to-one coordinate system. That failure does not prevent an individual curve from continuing. It marks the boundary of this particular congruence description and is one reason to keep the tangent field, its projected derivative, and the connecting fields conceptually distinct.

Required background. The Riemann Curvature Tensor fixes the curvature convention used when the congruence equations are evolved later in the chapter; Affine and Variational Geodesics supplies affine geodesics and the curve-variation setup; Jacobi Fields and Geodesic Deviation explains how connecting fields arise from a two-parameter family and when the geodesic specialization satisfies the Jacobi equation. Helpful background. Calculus and Linear Algebra for Relativity supplies projectors, traces, and bilinear forms; Frames, Coframes, and Anholonomy supplies orthonormal components in the local rest space.

The local rest projector

Choose the future-directed normalization

$$u^\mu u_\mu=-1.$$

The tensors

$$h_{\mu\nu}=g_{\mu\nu}+u_\mu u_\nu, \qquad h^\mu{}_\nu=\delta^\mu{}_\nu+u^\mu u_\nu$$

annihilate $u^\mu$. Direct multiplication gives

$$h^\mu{}_\alpha h^\alpha{}_\nu=h^\mu{}_\nu, \qquad h^\mu{}_\mu=n-1.$$

Thus $h^\mu{}_\nu$ projects vectors onto the $(n-1)$-dimensional subspace orthogonal to $u^\mu$. The metric restricted to that subspace is positive definite. Although “space” here is operationally meaningful at one event, a general congruence need not assemble these local rest spaces into hypersurfaces.

A tensor is spatial relative to $u^\mu$ when contraction with $u^\mu$ in every index vanishes. Projection with one copy of $h^\mu{}_\nu$ per index produces its spatial part. This is invariant information once the congruence has been chosen; it is not the same as discarding the time component in an arbitrary coordinate chart.

Acceleration and the deformation tensor

The four-acceleration of the congruence is

$$a^\mu=u^\nu\nabla_\nu u^\mu.$$

Differentiating $u^\mu u_\mu=-1$ shows $u_\mu a^\mu=0$. Define the doubly projected derivative

$$B_{\mu\nu} =h_\mu{}^\alpha h_\nu{}^\beta\nabla_\beta u_\alpha.$$

Both indices of $B_{\mu\nu}$ are spatial. The full derivative is recovered exactly as

$$\nabla_\nu u_\mu=B_{\mu\nu}-u_\nu a_\mu.$$

Contracting this expression with $u^\nu$ returns $a_\mu$, which fixes the minus sign. The distinction matters for accelerated observers: projecting too early and then pretending that $B_{\mu\nu}$ is the entire derivative loses the term that later appears as $\nabla_\mu a^\mu$ in the Raychaudhuri identity.

The trace and irreducible parts of $B_{\mu\nu}$ are developed on Expansion, Shear, and Vorticity. For now, $B_{\mu\nu}$ is best understood as a linear map on the local rest space.

Neighbor separation

Let $\xi^\mu$ connect neighboring curves, chosen so that it is Lie transported by the flow:

$$[u,\xi]^\mu=0.$$

For a torsion-free connection this implies $\nabla_u\xi=\nabla_\xi u$. A Lie-transported connecting field need not remain spatial when the congruence accelerates: differentiation gives $d(u\cdot\xi)/d\tau=a\cdot\xi$. Introduce its spatial representative

$$s^\mu=h^\mu{}_\nu\xi^\nu =\xi^\mu+(u\cdot\xi)u^\mu.$$

The physically relevant relative-motion equation is

$$\frac{D_\perp s^\mu}{d\tau} \equiv h^\mu{}_\alpha u^\nu\nabla_\nu s^\alpha =B^\mu{}_\nu s^\nu.$$

To see why projecting the vector itself matters, write $c=u\cdot\xi$. The derivative of $\xi$ contributes $B^\mu{}_\nu s^\nu-c a^\mu$ after projection, while the derivative of $cu$ contributes $+c a^\mu$. The acceleration terms cancel. Projecting only $\nabla_u\xi$ would omit that cancellation and would agree with the displayed equation only at events where $c=0$ (or for a geodesic congruence).

For the squared separation $L^2=g_{\mu\nu}s^\mu s^\nu=h_{\mu\nu}\xi^\mu\xi^\nu$, the symmetric part controls its rate:

$$\frac12\frac{dL^2}{d\tau} =B_{(\mu\nu)}s^\mu s^\nu.$$

The antisymmetric part rotates the pattern of separations relative to a nonrotating rest frame but makes no direct contribution to $L^2$. The projected calculation above extends the geodesic decomposition in Kar and SenGupta 2007, §II.A, pp. 6–7, PDF to accelerated timelike curves; the source discusses acceleration separately in §II.E, p. 14.

Flat-spacetime checks

For inertial observers in Cartesian Minkowski coordinates, $u=\partial_t$, so $a^\mu=0$ and $B_{\mu\nu}=0$. By contrast, Milne observers inside the future cone are also geodesic, but neighboring inertial rays separate: $B_{ij}=h_{ij}/\tau$. Flat curvature therefore does not imply vanishing congruence deformation.

Rindler observers give the opposite pattern. In

$$ds^2=-\rho^2d\eta^2+d\rho^2+dy^2+dz^2,$$

take $\rho\gt0$ and $u=\rho^{-1}\partial_\eta$. Their acceleration is $a^\rho=1/\rho$, while their rest-space deformation vanishes. The observers accelerate while the infinitesimal rest-space distances between their worldlines remain constant. This is Born rigidity; it does not mean that different accelerated observers assign the same finite radar distance. The example exposes the error in identifying acceleration with expansion.

A uniformly rotating Born-rigid congruence likewise has vanishing symmetric deformation but nonzero antisymmetric deformation and acceleration. It cannot be orthogonal to a family of simultaneity hypersurfaces throughout the rotating region.

Common pitfalls

A congruence is local. Curves cannot remain a congruence through an intersection, because the tangent field would become multivalued. Caustics therefore mark failure of this description even when every individual curve continues smoothly.

Rest spaces need not form slices. The distribution orthogonal to $u^\mu$ is integrable only when its vorticity vanishes. A projector at each event alone does not establish global simultaneity.

Coordinates do not define the projection. A coordinate time vector may fail to be unit, orthogonal to coordinate slices, or even timelike. Normalize the physical tangent and construct $h^\mu{}_\nu$ before taking spatial components.

Exercises

Projector algebra

Show directly that $h^\mu{}_\nu$ is a projector and that $a^\mu$ is spatial. Then decompose an arbitrary vector $V^\mu$ into temporal and rest-space parts.

Solution

Using $u^\alpha u_\alpha=-1$,

$$(\delta^\mu{}_\alpha+u^\mu u_\alpha) (\delta^\alpha{}_\nu+u^\alpha u_\nu) =\delta^\mu{}_\nu+u^\mu u_\nu.$$

Differentiating the normalization along $u^\mu$ gives

$$0=\frac12u^\nu\nabla_\nu(u^\mu u_\mu)=u_\mu a^\mu.$$

Finally,

$$V^\mu=(-u_\nu V^\nu)u^\mu+h^\mu{}_\nu V^\nu.$$

Contracting with $u_\mu$ verifies the temporal coefficient, and applying $h^\mu{}_\nu$ verifies the spatial part.

An accelerated connecting field

In the Rindler chart above, suppress the two spectator coordinates and take

$$u=\rho^{-1}\partial_\eta, \qquad \xi=\partial_\rho-\frac\eta\rho\partial_\eta.$$

Verify $[u,\xi]=0$, find $s=h\xi$, and compare $h\nabla_u\xi$ with $h\nabla_u s$. Explain why the distinction matters despite $B=0$.

Solution

The two terms in the $\eta$ component of the Lie bracket are $-1/\rho^2$ and $+1/\rho^2$, so they cancel. Since $u\cdot\xi=\eta$, spatial projection gives $s=\partial_\rho$. The relevant connection coefficients are $\Gamma^\eta{}_{\eta\rho}=1/\rho$ and $\Gamma^\rho{}_{\eta\eta}=\rho$. They give

$$h\nabla_u\xi=-\frac\eta\rho\partial_\rho, \qquad h\nabla_u s=0.$$

The first expression contains $-(u\cdot\xi)a$ and therefore is not $Bs$. The second is the correct spatial relative-motion equation and agrees with $B=0$: these neighboring accelerated observers keep a fixed infinitesimal rest-space separation.

References

• Kar, Sayan, and Soumitra SenGupta. “The Raychaudhuri Equations: A Brief Review.” Pramana 69 (2007), 49–76. arXiv:gr-qc/0611123.