The Cosmological Constant

The cosmological constant is the unique spacetime-independent coefficient multiplying the metric in the Einstein equation. Moving that term from geometry to an effective stress tensor changes no solutions when every parameter and field is held fixed. It becomes physically consequential only when the rewrite is mistaken for a change in what is varied, which fields exist, or which parameter the theory assigns to vacuum stress.

Required background. We use the Riemann curvature tensor to interpret vacuum curvature, Hilbert stress–energy to define metric response, and the scope of the equivalence principles to avoid treating local metric physics as a derivation of $\Lambda$.

Helpful background. Minkowski geometry supplies observer conventions, natural tensor operations clarify why the metric term is covariant, and presymplectic potentials prepare its action interpretation.

Two sides of one equation

Start with

$$G_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi G\,T_{\mu\nu}.$$

Algebraically, this is equivalent to

$$G_{\mu\nu}=8\pi G\left(T_{\mu\nu}+T^{(\Lambda)}_{\mu\nu}\right), \qquad T^{(\Lambda)}_{\mu\nu}=-\frac{\Lambda}{8\pi G}g_{\mu\nu}.$$

No observable changes under this rearrangement. Reall gives the geometric and effective-fluid forms together in Reall 2013, § 6.7, printed p. 65, Eq. (6.35), PDF. The term is automatically conserved because $\Lambda$ is constant and $\nabla_\rho g_{\mu\nu}=0$.

For any unit timelike observer $u^\mu$, let $h_{\mu\nu}=g_{\mu\nu}+u_\mu u_\nu$. In four dimensions the measured energy density and isotropic pressure are

$$\rho_\Lambda=T^{(\Lambda)}_{\mu\nu}u^\mu u^\nu =\frac{\Lambda}{8\pi G}, \qquad p_\Lambda=\frac13h^{\mu\nu}T^{(\Lambda)}_{\mu\nu} =-\frac{\Lambda}{8\pi G}.$$

Thus $p_\Lambda=-\rho_\Lambda$ and $w=p/\rho=-1$ when $\Lambda\ne0$. Positive $\Lambda$ has positive effective energy density in the site’s $(-,+,+,+)$ convention.

The observer decomposition is independent of which timelike observer is chosen. Because the tensor is proportional to $g_{\mu\nu}$, it has no momentum density or anisotropic stress in any orthonormal frame. Ordinary perfect fluids select a preferred four-velocity unless $p=-\rho$; vacuum-form stress does not. This invariant property is more precise than saying that it is merely “uniform.”

Vacuum curvature

In dimension $D>2$, tracing the vacuum equation and substituting back gives

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

In four dimensions this becomes $R_{\mu\nu}=\Lambda g_{\mu\nu}$ and $R=4\Lambda$. Such a metric is an Einstein metric, but need not have constant sectional curvature: the Weyl tensor can remain nonzero. If the stronger constant-curvature form holds,

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

then $K=2\Lambda/[(D-1)(D-2)]$. The exceptional $D=2$ case cannot be obtained by dividing by $D-2$ because its Einstein tensor vanishes identically.

The sign can be checked through geodesic deviation in the maximally symmetric case. For positive $\Lambda$ in four dimensions, initially comoving timelike geodesics tend to accelerate apart; for negative $\Lambda$, the corresponding curvature contribution tends to refocus them. This statement concerns the local curvature term and does not, by itself, establish a global cosmological history or a particular coordinate expansion.

Rewriting versus changing the theory

A constant contribution to a matter Lagrangian is not invisible to gravity. If

$$S_{\rm vac}=-\int d^4x\,\sqrt{-g}\,\rho_{\rm vac},$$

metric variation yields $T^{\rm vac}_{\mu\nu}=-\rho_{\rm vac}g_{\mu\nu}$. The field equation then depends on the combination

$$\Lambda_{\rm eff}=\Lambda_{\rm bare}+8\pi G\rho_{\rm vac}.$$

Calling the combined quantity geometric or material is bookkeeping; changing $\rho_{\rm vac}$ or $\Lambda_{\rm bare}$ while holding the other fixed changes the theory’s solutions. Carroll derives the constant action term and its metric field equation in Carroll 1997, § 4, printed pp. 118–119, Eqs. (4.73)–(4.74), PDF.

Nor may one replace $\Lambda$ by an arbitrary function $\Lambda(x)$ without adding dynamics. Taking the divergence gives

$$\partial_\nu\Lambda=8\pi G\,\nabla^\mu T_{\mu\nu}.$$

If ordinary matter is conserved, $\Lambda$ must be constant on each connected component. A varying scalar with potential energy is an additional field, with its own kinetic terms and equation of motion, rather than a cosmological constant.

Newtonian-scale check

Trace reversing the four-dimensional equation and taking weak, stationary, slowly moving matter gives

$$R_{00}\simeq4\pi G\rho-\Lambda.$$

Since $R_{00}\simeq\nabla^2\Phi$, the potential satisfies

$$\nabla^2\Phi\simeq4\pi G\rho-\Lambda.$$

In a source-free spherical region, one local solution is $\Phi=-GM/r-\Lambda r^2/6$ up to an additive constant. Its radial acceleration is $-GM/r^2+\Lambda r/3$. This is a sign and dimension check, not permission to use the weak-field expression near a horizon or on curvature scales comparable with $|\Lambda|^{-1/2}$. The expansion requires both $GM/r\ll1$ and $|\Lambda|r^2\ll1$.

The Newtonian expression also shows why moving the term across the equation does not remove its effect: the same $-\Lambda$ appears whether described as geometry or as the active gravitational contribution of a fluid with $\rho+3p=-2\rho$. Only the assignment of the term changes, not the resulting metric or acceleration.

Scope and limitations

This page establishes the local field-equation meaning of $\Lambda$. It does not infer its measured value, identify a microscopic vacuum state, or replace a cosmological model. Energy-condition language applied to $T^{(\Lambda)}_{\mu\nu}$ is an effective-source interpretation; moving the term back to the left does not remove its focusing effect from the curvature equation.

Nor does the algebra settle the vacuum-energy problem. Renormalized matter contributions, a bare gravitational parameter, and the measured effective curvature belong to different levels of description. Relating them requires a quantum-field-theory prescription and empirical cosmology, neither of which follows from the classical rewrite above.

Exercises

Assume a $D$-dimensional vacuum spacetime is maximally symmetric. Derive its curvature scale $K$ in terms of $\Lambda$.

Solution

Maximal symmetry gives $R_{\mu\nu}=(D-1)Kg_{\mu\nu}$. The vacuum Einstein equation for $D>2$ gives $R_{\mu\nu}=2\Lambda g_{\mu\nu}/(D-2)$. Equating coefficients yields

$$K=\frac{2\Lambda}{(D-1)(D-2)}.$$

For $D=4$, $K=\Lambda/3$. The derivation does not apply in $D=2$, where $G_{\mu\nu}$ is identically zero.

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 p. 65. PDF.