A Lorentzian metric distinguishes clocks, light cones, and volume
Declare the −+++ convention, compute proper time, and construct the coordinate-invariant integration measure.
On this page
The question for today
What you’ll understand
- Classify tangent intervals with a declared Lorentzian signature.
- Compute proper time along a specified timelike curve.
- Use the determinant factor in a volume or area element.
A smooth manifold becomes a spacetime geometry after a nondegenerate Lorentzian metric is specified. The metric measures tangent inner products and separates timelike, null, and spacelike directions. Its determinant also supplies the correct integration density. We will use signature −+++ throughout this module and write x⁰=ct when the time coordinate has length units. Keeping those conventions explicit prevents an interval sign or an angular component’s units from being mistaken for a physical disagreement.
The metric gives a local cone rather than a positive distance norm
At each event, a Lorentzian metric has one negative and three positive directions in an orthonormal frame. For our convention a nonzero timelike tangent has negative squared norm, a null tangent zero norm, and a spacelike tangent positive norm. The signature cannot change under a smooth invertible coordinate transformation. A metric becoming genuinely degenerate inside a valid chart is therefore not simply a switch between coordinate conventions. Metric components can still carry different units when some coordinates are lengths and others dimensionless angles.
Along a future-directed timelike worldline, the proper-time element is the positive square root of minus the line element divided by c. Integrating it requires the actual path, not only its endpoints. In flat spacetime a straight inertial segment has the familiar endpoint interval, while accelerated paths between the same events can accumulate different proper times. Timelike geodesics locally maximize proper time within suitable neighborhoods, not necessarily globally after conjugate points or across arbitrary spacetime topology. Null curves accumulate zero proper time and require another parameter.
Coordinate density and invariant volume are different objects
The bare coordinate measure changes by a Jacobian when coordinates change. The metric determinant supplies the inverse factor needed for a coordinate-invariant density. In four-dimensional Lorentzian signature the determinant is negative in a regular chart, so the density is the square root of minus that determinant. An oriented volume form additionally requires a choice of orientation; the positive density used for ordinary integration does not by itself choose one. A boundary integration also needs an induced measure and a consistent directed surface element.
Flat space in spherical coordinates illustrates the units. With x⁰ and r in metres and θ,φ dimensionless, the line element has angular coefficients r² and r²sin²θ. The four-volume density is r²sinθ on a chart with 0<θ<π and r>0. Its zeros at the polar-coordinate exclusions do not mean Minkowski geometry degenerates there. A regular Cartesian chart restores a nonsingular component matrix. In a purely spatial polar plane the corresponding area element is r dr dφ rather than dr dφ or r² dr dφ.
Read the lab’s sign convention before comparing intervals
Light cones are local structures: they classify tangents and, with a time orientation, future causal directions. They do not by themselves determine whether a distant point is reachable or whether closed timelike curves exist. Those questions depend on the metric and topology over a region. Likewise the proper-time expression is invariant, but a coordinate time read by an arbitrary chart need not be a physical clock. Static or accelerated observer families require their own normalization by the metric.
The spacetime lab uses c=1, a fixed event position x=0.6, and a readout ct²−x², which is the opposite sign to our ds². Reset and set both controls to β=0 and ct=1, predicting transformed coordinates (x′,ct′)=(0.6,1). Reset and set β=0.6 and ct=1, predicting (0,0.8). In both runs its invariant readout is +0.64, corresponding to ds²=−0.64 in our convention. The lab is a single flat-space Lorentz transformation, not a curved metric or arbitrary worldline integrator.
Work it through
Convert a timelike interval into a clock reading
Two events in flat spacetime have Δx⁰=cΔt=5 m and Δx=3 m, with other spatial differences zero. Find their squared interval, classify it, and compute the proper time of the straight inertial segment. Use c=3×10⁸ m/s.
With signature −+++, the squared interval is −5²+3²=−16 m². Its negative sign identifies timelike separation.
For the straight inertial segment, cΔτ=√16=4 m. This endpoint formula is valid because the metric is constant and the path is inertial.
Divide by c: Δτ=4/(3×10⁸) s≈1.33×10⁻⁸ s=13.3 ns. A different timelike path between the events requires a path integral.
The interval is −16 m² and timelike; the straight inertial clock accumulates about 13.3 ns.
The living notebook
Explore the model
Interactive lab · spacetime
Compare interval conventions under a boost
Reset and set BOTH controls to β=0,ct=1: predict (x′,ct′)=(0.6,1). Reset and set β=0.6,ct=1: predict (0,0.8). The fixed x is 0.6. Its ct²−x² readout is +0.64, opposite in sign to our ds²=−0.64. This is flat-space kinematics only.
- Lorentz factor γ
- 1.12
- Event x′
- 0.168
- Event ct′
- 0.817
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 forInvariant intervals agree physically once their signature and units are declared.
Pause & reason
Check your understanding
Your turn
Try it yourself
Find the area of a flat half-annulus with 1 m<r<2 m and 0<φ<π. Use the polar metric and state whether the excluded coordinate endpoints change the area.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Integrate the density r over radius and angle. Boundary curves have zero two-dimensional area.
Compare with the solution
The area is ∫₀^πdφ∫₁²rdr=π(4−1)/2=3π/2 m²≈4.71 m². The open versus closed boundary curves do not change this area. The nonconstant coordinate density reflects polar cell geometry, not intrinsic curvature of the plane.
Keep the idea
Take it with you
- The module uses −+++ and x⁰=ct when time is length valued.
- Proper time is a path-dependent invariant along timelike curves.
- The metric determinant converts coordinate measure into invariant volume density.
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