Minkowski Spacetime and Standard Charts

Minkowski spacetime is globally regular and flat, yet polar, null, Rindler, and Milne coordinates all have boundaries or degeneracies. The decisive test is the coordinate map: state the region it covers, check its Jacobian, and verify that the Riemann tensor remains zero.

Helpful background. Lorentzian Metrics and Causal Cones supplies causal classification; Exact Solutions of Einstein’s Equations supplies the chart-domain method.

Flat metric and polar charts

In inertial Cartesian coordinates,

$$ds^2=-dT^2+dX^2+dY^2+dZ^2, \qquad (T,X,Y,Z)\in\mathbb R^4.$$

The metric determinant is $-1$ and $R^a{}_{bcd}=0$. This is the control chart for every apparent boundary below. Reall 2013, §13.1, printed pp. 149–151 places Minkowski spacetime in the constant-curvature family with $K=0$.

For spherical polar coordinates,

$$X=r\sin\theta\cos\phi, \quad Y=r\sin\theta\sin\phi, \quad Z=r\cos\theta,$$ $$ds^2=-dT^2+dr^2+r^2d\theta^2 +r^2\sin^2\theta\,d\phi^2,$$

with $r\ge0$, $0\le\theta\le\pi$, and $\phi\sim\phi+2\pi$. The Jacobian $r^2\sin\theta$ vanishes at the origin and polar axis because the angular labels become redundant there. Those points remain regular in the Cartesian chart.

Cylindrical coordinates $X=R\cos\phi$, $Y=R\sin\phi$ give

$$ds^2=-dT^2+dR^2+R^2d\phi^2+dZ^2, \qquad R\ge0.$$

Again $R=0$ is an axis of the chart, not a curvature singularity.

Radial null coordinates

With $u=T-r$ and $v=T+r$,

$$ds^2=-du\,dv+\frac{(v-u)^2}{4}d\Omega_2^2, \qquad v\ge u.$$

The cross-term convention is $g_{uv}=g_{vu}=-1/2$. Surfaces of constant $u$ are outgoing radial null hypersurfaces and surfaces of constant $v$ are ingoing. The equality $u=v$ is the regular spherical origin, where all angular labels collapse. Thus $(u,v,\theta,\phi)$ is useful on radial orbit space but is not one smooth global four-dimensional chart.

Rindler wedge

On the right wedge $X\gt\lvert T\rvert$, define

$$T=\rho\sinh\eta, \qquad X=\rho\cosh\eta, \qquad \rho\gt0,$$

while retaining $Y$ and $Z$. The metric becomes

$$ds^2=-\rho^2d\eta^2+d\rho^2+dY^2+dZ^2.$$

A curve at fixed $(\rho,Y,Z)$ has proper acceleration $1/\rho$. The boost generator $\partial_\eta=X\partial_T+T\partial_X$ is timelike in the wedge, null on $X=\pm T$, and spacelike in the future and past cones. Those null planes are Rindler horizons for this accelerated observer family, not boundaries of Minkowski spacetime. The explicit wedge maps and boost flow are given in Crispino, Higuchi, and Matsas 2008, §II.B, arXiv PDF pp. 6–7.

The map’s Jacobian is $\rho$, so it ceases to be invertible at the limiting value $\rho=0$. Moreover, taking $\rho\to0$ at fixed $\eta$ reaches only the bifurcation two-plane. Generic horizon points require a correlated limit $\eta\to\pm\infty$. One cannot repair the chart by simply adjoining one $\rho=0$ surface; use inertial or null coordinates to cross it. The left wedge requires a separate sign choice.

Milne chart

Inside the future cone $T\gt0$ and $T^2-r^2\gt0$, set

$$T=\tau\cosh\chi, \qquad r=\tau\sinh\chi, \qquad \tau\gt0, \quad \chi\ge0.$$

Then

$$ds^2=-d\tau^2 +\tau^2\left(d\chi^2+\sinh^2\chi\,d\Omega_2^2\right).$$

This resembles an open FLRW geometry with scale factor $a(\tau)=\tau$, but the four-dimensional curvature is exactly zero. Constant-$\tau$ slices have negative intrinsic curvature $-1/\tau^2$; their extrinsic curvature cancels it in the spacetime Riemann tensor. Ellis and van Elst 2008, §4.3.1, printed p. 23, PDF identifies this Milne member as flat spacetime in expanding coordinates. At fixed finite $\chi$, the limit $\tau\to0$ reaches the tip of the future cone. Generic future-null-cone points instead require $\tau\to0$, $\chi\to\infty$ with $\tau e^\chi$ held finite. Both are boundaries of this chart, not big-bang curvature singularities.

Bičák 2000, §1.3, printed pp. 8–10 emphasizes that background identification requires both local invariants and global domain data. The complete Minkowski coordinate atlas adds rotating and conformally compactified charts; accelerated-observer interpretation belongs to Uniform Acceleration and Rindler Wedges.

Independent checks

Check	Result
Pull back the Cartesian metric	Produces each displayed metric, including the null cross-term and the hyperbolic spatial metric in Milne coordinates.
Compute the coordinate Jacobian	Identifies the polar axis, Rindler horizon limit, and the Milne boundary—its tip at fixed $\chi$ and its null cone under the correlated limit—as failures of the chosen map.
Contract the Riemann tensor	Gives $R^a{}_{bcd}=0$ in every regular chart region.
Follow an inertial geodesic	Crosses a Rindler horizon or Milne cone boundary smoothly in Cartesian coordinates and finite affine parameter.

Common pitfalls

Flat spacetime need not have flat spatial slices. Milne slices are hyperbolic even though the spacetime Riemann tensor vanishes.

A chart horizon is observer-dependent. Rindler horizons encode the causal limits of one accelerated congruence. Inertial observers encounter no geometric surface there.

Coordinate range does not prove global coverage. Both $\eta$ and $\chi$ run to infinity, but their charts cover only a wedge or cone selected by explicit Cartesian inequalities.

References

• Bičák, Jiří. “The Role of Exact Solutions of Einstein’s Equations in the Developments of General Relativity and Astrophysics: Selected Themes.” In Einstein’s Field Equations and Their Physical Implications, Lecture Notes in Physics 540, 1–126. Springer, 2000. arXiv:gr-qc/0004016.
• Crispino, Luís C. B., Atsushi Higuchi, and George E. A. Matsas. “The Unruh Effect and Its Applications.” Reviews of Modern Physics 80 (2008): 787–838. DOI. Open PDF.
• Ellis, George F. R., and Henk van Elst. “Cosmological Models.” In Theoretical and Observational Cosmology, 1–116. Springer, 1999. DOI. Open PDF.
• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes, 2013. Open PDF.