Trace, Trace Reversal, and Vacuum Equations

Tracing the Einstein equation separates its scalar and trace-free content and exposes when “vacuum” really means Ricci flatness. In dimension $D>2$, vacuum implies $R_{\mu\nu}=0$ only when both the non-cosmological stress tensor and $\Lambda$ vanish. A trace-free source is not vacuum, an Einstein metric need not have constant curvature, and two dimensions are algebraically exceptional.

Required background. We use the Riemann curvature tensor for Ricci and scalar curvature, Hilbert stress–energy for $T_{\mu\nu}$ and its trace, and the scope of the equivalence principles to keep a field equation distinct from local inertial physics.

Helpful background. Minkowski geometry supplies the flat benchmark, natural tensor operations clarify contractions, and presymplectic potentials prepare variational applications.

The trace in arbitrary dimension

Let $D\ge2$ and write the coupling as $\kappa_D=8\pi G_D$:

$$R_{\mu\nu}-\frac12Rg_{\mu\nu} +\Lambda g_{\mu\nu}=\kappa_D T_{\mu\nu}.$$

Contracting with $g^{\mu\nu}$ uses $g^{\mu\nu}g_{\mu\nu}=D$ and gives

$$\left(1-\frac D2\right)R+D\Lambda=\kappa_D T, \qquad T=g^{\mu\nu}T_{\mu\nu}.$$

For $D\ne2$,

$$R=\frac{2D\Lambda-2\kappa_D T}{D-2}.$$

Substituting this scalar back into the original equation yields the trace-reversed form

$$R_{\mu\nu} =\kappa_D\left(T_{\mu\nu}-\frac{1}{D-2}Tg_{\mu\nu}\right) +\frac{2\Lambda}{D-2}g_{\mu\nu}.$$

This derivation is purely algebraic and loses no information for $D>2$. In four dimensions it reduces to $R_{\mu\nu}=8\pi G(T_{\mu\nu}-\tfrac12Tg_{\mu\nu})+\Lambda g_{\mu\nu}$. Carroll performs the four-dimensional trace reversal in Carroll 1997, § 4, printed p. 111, Eqs. (4.43)–(4.45), PDF.

Trace reversal is not the same operation as removing the trace. In four dimensions, $\overline T_{\mu\nu}=T_{\mu\nu}-\tfrac12Tg_{\mu\nu}$ has trace $\overline T=-T$; applying this same operation twice returns $T_{\mu\nu}$. The trace-reversed source therefore retains all ten components, including the trace information. In other dimensions the inverse of $A_{\mu\nu}\mapsto A_{\mu\nu}-\tfrac12Ag_{\mu\nu}$ uses the coefficient $1/(D-2)$ derived above and is not generally the same map.

It is useful to separate the trace-free source explicitly. Define

$$S_{\mu\nu}=T_{\mu\nu}-\frac1D Tg_{\mu\nu}, \qquad g^{\mu\nu}S_{\mu\nu}=0.$$

Taking the trace-free part of Einstein’s equation gives

$$R_{\mu\nu}-\frac1D Rg_{\mu\nu}=\kappa_D S_{\mu\nu}.$$

The cosmological term drops out of this equation because it is pure trace. It controls the scalar Ricci curvature, while anisotropic or trace-free matter controls the trace-free Ricci part. This split is algebraic; it should not be confused with the Ricci–Weyl decomposition of the full Riemann tensor.

Vacuum and trace-free matter

If $T_{\mu\nu}=0$ and $D>2$, then

$$R_{\mu\nu}=\frac{2\Lambda}{D-2}g_{\mu\nu}.$$

Therefore:

• vacuum with $\Lambda=0$ is Ricci flat;
• vacuum with $\Lambda\ne0$ is an Einstein metric with nonzero Ricci curvature; and
• $T=0$ alone does not imply vacuum, because the trace-free tensor $T_{\mu\nu}$ can remain nonzero.

Reall derives $G_{\mu\nu}=0\Longleftrightarrow R_{\mu\nu}=0$ for four-dimensional zero-$\Lambda$ vacuum in Reall 2013, § 6.7, printed p. 64, Eq. (6.33), PDF. The stated assumptions are essential. For trace-free matter and $\Lambda=0$, the scalar curvature vanishes, yet the equation remains $R_{\mu\nu}=\kappa_D T_{\mu\nu}$; electromagnetic radiation is a standard four-dimensional example.

An equation $R_{\mu\nu}=\lambda g_{\mu\nu}$ defines an Einstein metric. The contracted Bianchi identity makes $\lambda$ constant on a connected manifold when $D>2$. It does not force the Weyl tensor to vanish. Constant curvature is the stronger condition

$$R_{\mu\nu\rho\sigma} =K(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho}),$$

which implies $R_{\mu\nu}=(D-1)Kg_{\mu\nu}$. Thus Ricci flatness also need not mean flat spacetime: Weyl curvature can carry nonzero tidal information.

Several quick tests expose mistaken vacuum reasoning. A four-dimensional electromagnetic field has $T=0$ but generally $T_{\mu\nu}\ne0$, so $R=4\Lambda$ while the trace-free Ricci tensor remains sourced. A zero-$\Lambda$ gravitational vacuum has $R_{\mu\nu}=0$ but can possess nonzero Riemann and Weyl curvature. A nonzero-$\Lambda$ vacuum is not Ricci flat even though there is no non-cosmological matter. Each example changes a different condition, so the word “vacuum” should always be accompanied by the treatment of $\Lambda$.

Why two dimensions are exceptional

In $D=2$, the Riemann tensor has only its scalar-curvature component and

$$R_{\mu\nu}=\frac12Rg_{\mu\nu}, \qquad G_{\mu\nu}\equiv0.$$

The Einstein equation collapses to

$$\Lambda g_{\mu\nu}=\kappa_2T_{\mu\nu}.$$

It does not determine $R$, so division by $D-2$ in the trace-reversed formula is illegal. In zero-$\Lambda$ vacuum the metric equation is identically satisfied; additional fields or a different gravitational action are needed for local dynamics. This exceptional algebra is also why dimension must be stated in uniqueness claims.

Consistency checks and limitations

Tracing the trace-reversed equation must return the scalar equation. Contracting its right-hand side produces

$$\kappa_D\left(T-\frac{D}{D-2}T\right) +\frac{2D\Lambda}{D-2},$$

which simplifies to the expression for $R$ above. This catches both the sign of the matter trace and the factor multiplying $\Lambda$. Substitution into the original equation is a second check and confirms equivalence only when $D\ne2$.

Trace reversal does not solve the differential equations or determine Weyl curvature. It is often convenient because $R_{00}$ has a direct weak-field interpretation and because an Einstein metric is easy to recognize. Yet boundary data, topology, regularity, and matter dynamics remain necessary for a spacetime solution. Named solution families and their global properties lie outside this local algebraic treatment.

The operation is also convention-sensitive only through the definitions already fixed. Once $G_{\mu\nu}=R_{\mu\nu}-\tfrac12Rg_{\mu\nu}$ and the sign of the cosmological term are chosen, the factors above follow from dimension alone. Changing the Riemann-sign convention changes both Ricci quantities consistently; importing a trace-reversed formula without translating those definitions can therefore conceal a sign error. Retracing and substituting back are the reliable checks.

Exercises

In $D>2$ with $\Lambda=0$, suppose $T=0$ but $T_{\mu\nu}\ne0$. Determine $R$ and $R_{\mu\nu}$, and explain why “scalar-flat” does not mean “Ricci-flat.”

Solution

The traced equation gives $R=-2\kappa_DT/(D-2)=0$. The trace-reversed equation gives

$$R_{\mu\nu}=\kappa_DT_{\mu\nu},$$

which is nonzero wherever the source is nonzero. Scalar curvature is only one contraction of the Ricci tensor; its vanishing removes the trace but not the trace-free Ricci components.

References

• Carroll, Sean M. “Lecture Notes on General Relativity.” NSF-ITP/97-147, arXiv:gr-qc/9712019v1 (1997). arXiv record.
• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes (2013), § 6.7, printed pp. 64–65. PDF.