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research32 min studyLesson 841 in Physics

Three-plus-one spacetime decomposition

Separate spacetime geometry into spatial data and coordinate evolution with declared normal and curvature signs.

Before you begin
On this page

The question for today

What you’ll understand

  • Recover spacetime metric components from lapse, shift, and spatial metric.
  • Distinguish normal proper time from coordinate-stationary proper time.
  • Compute extrinsic curvature with a fixed sign convention.

Numerical relativity evolves a family of spatial geometries rather than storing an entire spacetime at once. A three-plus-one decomposition organizes that family into a spatial metric, the way each slice bends in spacetime, and coordinate choices connecting successive slices. The distinction between geometry and coordinates is essential: a moving grid point need not follow a physical observer, and a change of lapse can alter coordinate evolution without changing the underlying solution.

Choose a normal and a metric signature

Throughout this module use spacetime signature minus, plus, plus, plus and geometric units with gravitational constant and light speed one. Time and spatial coordinates have length units, masses have length units, and metric components are dimensionless. Each spacelike slice has a future-directed unit normal n whose squared norm is minus one. Its induced spatial metric gamma is positive definite and supplies distances between nearby points within that slice.

The coordinate time vector decomposes into lapse times the normal plus a tangential shift. Positive lapse measures normal proper-time separation per coordinate-time interval. Shift describes the tangential displacement of coordinate labels. Neither function is an extra propagating gravitational polarization. This local description presumes a regular spacelike foliation; a coordinate system can become unusable even when spacetime curvature remains perfectly finite.

ds2=−α2dt2+γij(dxi+βidt)(dxj+βjdt),∂t=αn+βds^2=-\alpha^2dt^2+\gamma_{ij}(dx^i+\beta^i dt)(dx^j+\beta^jdt),\qquad \partial_t=\alpha n+\beta
Alpha is positive lapse, beta is tangential shift, and gamma is the spatial metric. Coordinate speeds and lapse are dimensionless in the declared units.↔ Scroll wide equations horizontally.

Distinguish coordinate lines from normal observers

A normal observer moves between coordinate labels with velocity minus the shift. Substituting that trajectory into the line element leaves proper time equal to lapse times coordinate time. In contrast, a coordinate-stationary curve has all spatial coordinate increments zero and retains a shift contribution in its norm. Such a curve is timelike only where the lapse squared exceeds the spatial squared norm of the shift.

The inverse metric and normal components provide algebraic checks on an implementation. Lowering the normal gives a covector with only a time component in adapted coordinates, even when its contravariant spatial components are nonzero. Confusing these two sets of components often produces incorrect normalization or a wrong advection sign. A spacetime diagram may help organize local frames, but a Lorentz boost of one event does not implement an arbitrary curved foliation.

nμ=α−1(1,−βi),nμ=(−α,0,0,0),dτnormal=α dtn^\mu=\alpha^{-1}(1,-\beta^i),\qquad n_\mu=(-\alpha,0,0,0),\qquad d\tau_{\rm normal}=\alpha\,dt
The normal is unit timelike. The proper-time identity applies to a normal trajectory, not automatically to a line of fixed spatial coordinates.↔ Scroll wide equations horizontally.

Fix the extrinsic-curvature sign before evolving

Define extrinsic curvature as minus one half the Lie derivative of the spatial metric along the normal. This is a common numerical-relativity convention, but the opposite sign also occurs in the literature. In adapted coordinates the metric evolution contains a negative lapse times curvature term plus a spatial Lie derivative along the shift. Every later constraint and gauge formula will use this same choice.

For zero shift, an expanding spatial direction has increasing metric component and therefore negative principal extrinsic curvature under this convention. Its coordinate component must be divided by the corresponding spatial metric component before interpreting the mixed tensor or its trace. Extrinsic curvature is not the intrinsic curvature of the slice: a flat spatial metric can still have nonzero extrinsic curvature because its embedding evolves. Keeping these geometric roles separate prevents a flat-looking grid from being mistaken for a static spacetime.

Kij=−12Lnγij,∂tγij=−2αKij+Lβγij,K=γijKijK_{ij}=-\tfrac12\mathcal L_n\gamma_{ij},\qquad \partial_t\gamma_{ij}=-2\alpha K_{ij}+\mathcal L_\beta\gamma_{ij},\qquad K=\gamma^{ij}K_{ij}
K has inverse-length units. The Lie derivative includes shift-gradient terms for a tensor, so it is not generally just beta times a component derivative.↔ Scroll wide equations horizontally.

Work it through

A normal step on a shifted grid

At one point let α=0.8\alpha=0.8, βx=0.3\beta^x=0.3, and γij=diag⁡(4,1,1)\gamma_{ij}=\operatorname{diag}(4,1,1). Find the t-x metric block, the normal components, and a normal observer displacement and proper time over dt=0.5Ldt=0.5L, treating coefficients as constant during this local step.

  1. The metric components are gtt=−0.82+4(0.3)2=−0.28g_{tt}=-0.8^2+4(0.3)^2=-0.28, gtx=4(0.3)=1.2g_{tx}=4(0.3)=1.2, and gxx=4g_{xx}=4. The coordinate-time vector is timelike here but does not equal the normal.

  2. The normal is (nt,nx)=(1.25,−0.375)(n^t,n^x)=(1.25,-0.375). Contracting with the displayed metric gives minus one; lowering it gives nt=−0.8,nx=0n_t=-0.8,n_x=0, checking both signs.

  3. A normal curve has dx/dt=−0.3dx/dt=-0.3, hence dx=−0.15Ldx=-0.15L during the stated step. Its proper time is 0.8(0.5L)=0.4L0.8(0.5L)=0.4L. This is a local constant-coefficient calculation, not a global spacetime integration.

The result

The metric block is negative 0.28, off-diagonal 1.2, and spatial 4; the normal step moves minus 0.15 L and accumulates 0.4 L of proper time.

The living notebook

Explore the model

Interactive lab

Interactive lab · spacetime

A local frame comparison

Reset and set relative velocity zero and event time one; predict transformed time one and position 0.6. Compare a fresh Reset run at velocity 0.6 and event time one; predict transformed time 0.8 and position zero. This is a flat Lorentz transformation of the fixed event, not lapse-shift evolution or a curved-spacetime slicing solver.

A local frame comparisonA fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Current readings: Lorentz factor γ: 1.12; Event x′: 0.168; Event ct′: 0.817. All parameters are adjustable with the labeled controls below.−4.69−2.3502.354.69−2.2−1.101.12.2x or x′ · c = 1ct or ct′EE′ct² − x² = 0.64
Original eventTransformed eventLight cone
Lorentz factor γ
1.12
Event x′
0.168
Event ct′
0.817
t = 0 s

A fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forLocal observer and coordinate descriptions must be distinguished; the lab does not construct a three-plus-one geometry.

Pause & reason

Check your understanding

Which trajectory accumulates proper time alpha times dt in the declared adapted coordinates?

Your turn

Try it yourself

Take lapse one and zero shift. At an instant let γxx=a2=4\gamma_{xx}=a^2=4, other diagonal entries one, and a˙=0.2/L\dot a=0.2/L. Find K-xx and the trace K, and explain their signs.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate a squared before using the curvature convention, and raise an index before tracing.

Compare with the solution

Here a=2a=2, so ∂tγxx=2aa˙=0.8/L\partial_t\gamma_{xx}=2a\dot a=0.8/L. Thus Kxx=−0.4/LK_{xx}=-0.4/L. The inverse spatial metric has xx component one quarter, giving K=−0.1/LK=-0.1/L because the other directions are static. The direction is expanding despite the negative curvature sign; that sign follows the declared definition. No intrinsic spatial curvature was required for this local embedding change.

Keep the idea

Take it with you

  • Lapse and shift specify how coordinates connect slices.
  • Normal proper time differs from fixed-coordinate proper time when shift is present.
  • The extrinsic-curvature convention fixes the sign of spatial metric evolution.

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