Conformal Rescaling of Geometric Fields

Conformal compactification replaces a physical spacetime whose radiative end is infinitely far away by a regular metric with a finite boundary. The replacement is not an isometry: connections, Ricci curvature, wave operators, unit normals, and extrinsic curvature all acquire definite powers and derivatives of the conformal factor. This page derives those changes and separates genuine singular behavior from expressions that only look singular before the field equations and boundary gauge are used.

Required background. The Riemann Curvature Tensor supplies the connection-difference method and the Ricci contractions used below.

Helpful background. We use The Einstein Equation, Hypersurface Normals and Tangential Projectors, Formulations of General Relativity, and Causality and Global Structure.

Physical and unphysical metrics

Tildes denote physical fields. On the physical region $\Omega\gt 0$, define the unphysical metric by

$$g_{\mu\nu}=\Omega^2\widetilde g_{\mu\nu}, \qquad \widetilde g_{\mu\nu}=\Omega^{-2}g_{\mu\nu}.$$

A regular conformal boundary is a hypersurface $\mathscr I=\{\Omega=0\}$ on which $d\Omega\ne0$ and $g_{\mu\nu}$ extends smoothly and nondegenerately. We take $\Omega\gt 0$ toward the physical interior, so $d\Omega$ points toward increasing $\Omega$; reversing that orientation reverses normal derivatives but not the powers of $\Omega$. A further conformal-gauge change

$$g_{\mu\nu}\mapsto g'_{\mu\nu}=\omega^2g_{\mu\nu}, \qquad \Omega\mapsto\Omega'=\omega\Omega, \qquad \omega\gt 0,$$

leaves $\widetilde g$ unchanged. Thus smoothness of a particular representative and conformally invariant boundary statements must not be conflated.

Let $\nabla$ and $\widetilde\nabla$ be the Levi-Civita connections of $g$ and $\widetilde g$. Directly differentiating $\widetilde g=\Omega^{-2}g$ gives the difference tensor

$$\widetilde\Gamma^{\rho}{}_{\mu\nu}-\Gamma^{\rho}{}_{\mu\nu} =-\Omega^{-1} \left( \delta^\rho{}_\mu\nabla_\nu\Omega +\delta^\rho{}_\nu\nabla_\mu\Omega -g_{\mu\nu}\nabla^\rho\Omega \right).$$

The minus sign is fixed by which metric carries the tilde. For a covector $v_\nu$ this implies

$$\widetilde\nabla_\mu v_\nu =\nabla_\mu v_\nu +\Omega^{-1} \left( v_\mu\nabla_\nu\Omega+v_\nu\nabla_\mu\Omega -g_{\mu\nu}v^\rho\nabla_\rho\Omega \right).$$

Ricci curvature and the scalar wave operator

Applying the connection-difference identity to curvature and contracting in four spacetime dimensions yields

$$\widetilde R_{\mu\nu} =R_{\mu\nu} +2\Omega^{-1}\nabla_\mu\nabla_\nu\Omega +g_{\mu\nu}\Omega^{-1}\Box\Omega -3g_{\mu\nu}\Omega^{-2}(\nabla\Omega)^2,$$

where $\Box=g^{\mu\nu}\nabla_\mu\nabla_\nu$ and $(\nabla\Omega)^2=g^{\mu\nu}\nabla_\mu\Omega\nabla_\nu\Omega$. Since $\widetilde g^{\mu\nu}=\Omega^2g^{\mu\nu}$, the scalar curvature is

$$\widetilde R =\Omega^2R+6\Omega\Box\Omega-12(\nabla\Omega)^2.$$

These are the same transformations used to obtain the conformal Einstein source tensor in Zenginoğlu 2008, equations (1)–(2). Negative powers are visible here, but they do not by themselves prove curvature blow-up: in a smooth vacuum completion, the equations force relations among $(\nabla\Omega)^2$, $\Box\Omega$, and the Hessian of $\Omega$ at $\mathscr I$.

For an ordinary scalar $\phi$ of conformal weight zero,

$$\widetilde\Box\phi =\Omega^2\left( \Box\phi-2\Omega^{-1}\nabla^\mu\Omega\nabla_\mu\phi \right).$$

This is not the conformally covariant wave operator. In four dimensions, put $\widetilde\phi=\Omega\phi$. Substitution of the scalar-curvature formula gives the exact identity

$$\left(\widetilde\Box-\frac16\widetilde R\right)(\Omega\phi) =\Omega^3\left(\Box-\frac16R\right)\phi.$$

The derivatives of $\Omega$ cancel only with this field rescaling. A mass term or a scalar of weight zero does not satisfy the same formula. In particular, a physical radiative scalar with $\widetilde\phi\sim r^{-1}$ can have a finite rescaled field $\phi=\Omega^{-1}\widetilde\phi$ when $\Omega\sim r^{-1}$; finiteness of the field does not by itself prove regularity of every derivative.

Normals, induced metrics, and extrinsic curvature

Let a spacelike hypersurface have future-directed unit normals $n^\mu$ for $g$ and $\widetilde n^\mu$ for $\widetilde g$. Their normalization gives

$$\widetilde n^\mu=\Omega n^\mu, \qquad \widetilde n_\mu=\Omega^{-1}n_\mu, \qquad \widetilde h_{\mu\nu}=\Omega^{-2}h_{\mu\nu}.$$

Using the site convention $K_{\mu\nu}=-h_\mu{}^\rho h_\nu{}^\sigma\nabla_\rho n_\sigma$, projection of the covector formula gives

$$\widetilde K_{\mu\nu} =\Omega^{-1}K_{\mu\nu} +\Omega^{-2}h_{\mu\nu}n(\Omega), \qquad \widetilde K=\Omega K+3n(\Omega).$$

The projectors with one index up agree, but the induced metrics with both indices down do not. If both physical and unphysical extrinsic curvatures are defined with the opposite sign, $K^+=-K$, the translated relation is $\widetilde K^+_{\mu\nu}=\Omega^{-1}K^+_{\mu\nu}-\Omega^{-2}h_{\mu\nu}n(\Omega)$. The coefficient of the newly defined $K^+$ stays positive, while the normal-derivative term changes sign. Andersson records the negative-sign scaling in Andersson 2002, § 2, equations (2a)–(2b).

Compactified Minkowski check

In outgoing coordinates, physical Minkowski spacetime is

$$d\widetilde s^2=-du^2-2\,du\,dr+r^2d\omega_2^2.$$

Set $\Omega=r^{-1}$, so $dr=-\Omega^{-2}d\Omega$. Then

$$ds^2=\Omega^2d\widetilde s^2 =-\Omega^2du^2+2\,du\,d\Omega+d\omega_2^2.$$

Every coefficient is regular at $\Omega=0$. The inverse of the $(u,\Omega)$ block gives $g^{\Omega\Omega}=\Omega^2$, hence

$$(\nabla\Omega)^2=\Omega^2\longrightarrow0.$$

Thus the boundary is null even though $d\Omega\ne0$. This check simultaneously fixes the cross-term sign, the factor $\Omega^2$ multiplying $du^2$, and the orientation $\Omega\gt 0$ for finite positive $r$.

Common pitfalls

Reversing the conformal direction. Writing $\widetilde g=\Omega^2g$ exchanges every power and makes a regular $g$ degenerate at null infinity. State which metric is physical before using any transformation formula.

Calling every pole a singularity. Terms such as $\Omega^{-2}(\nabla\Omega)^2$ are formally singular expressions. Their limits require the vacuum equations, a smooth completion, and an appropriate conformal gauge; algebraic cancellation must be shown, not assumed.

Exercises

For the compactified Minkowski metric above, invert the $(u,\Omega)$ block and compute the unphysical gradient vector $\nabla^\mu\Omega$. Is it tangent to $\mathscr I^+$?

Solution

The block and its inverse are

$$g_{ab}= \begin{pmatrix}-\Omega^2&1\\1&0\end{pmatrix}, \qquad g^{ab}= \begin{pmatrix}0&1\\1&\Omega^2\end{pmatrix},$$

in the order $(u,\Omega)$. Therefore

$$\nabla^\mu\Omega=g^{\mu\Omega} =(\partial_u)^\mu+\Omega^2(\partial_\Omega)^\mu.$$

At $\Omega=0$ this becomes $\partial_u$, which is tangent to the level surface $\Omega=0$ and is null there. A null hypersurface is special precisely because its normal vector is also tangent to it.

References

• Andersson, Lars. “Construction of Hyperboloidal Initial Data.” In The Conformal Structure of Spacetimes: Geometry, Analysis, Numerics, edited by Jörg Frauendiener and Helmut Friedrich, Lecture Notes in Physics 604, 183–194. Springer, 2002. arXiv:gr-qc/0205083.
• Zenginoğlu, Anıl. “Hyperboloidal Evolution with the Einstein Equations.” Classical and Quantum Gravity 25 (2008): 195025. DOI. Open PDF.