Interpreting a constraint on a simplified static spherically symmetric metric

Question

Writing the following simplified metric

$$ ds^2 = - dt^2 + dr^2 + A(r)^2 d\Omega^2 $$

Under the most general static spherically symmetric matter:

$$ T^{\mu}_{\nu} = - \rho(r) \delta^{\mu}_0 \delta^0_{\nu} + p_R(r) \delta^{\mu}_1 \delta^1_{\nu} + p_T(r) ( \delta^{\mu}_2 \delta^2_{\nu} + \delta^{\mu}_3 \delta^3_{\nu} ) $$

If I add $(0,0)$ and $(1,1)$ components of the EFE, I reach this equation:

$$ 2 A''(r) + \kappa A(r) \Big[ \rho(r) + p_R(r) \Big] = 0 $$

While $(2,2)$ and $(3,3)$ are essentially the same equation:

$$ A''(r) - \kappa A(r) p_T(r) = 0 $$

Substituting this in the first, results in the following identity:

$$ \rho(r) + p_R(r) + 2 p_T(r) = 0 $$

Now, I understand that by fixing the time and radial distance I'm overdetermining the metric, and this overdetermination reflects that not all possible spherically symmetric static matter distributions can be consistently described with the above metric.

Nonetheless, I'm curious why the constraint is this particular form, or if this quantity has some meaningful physical interpretation. I initially thought this was constraining the stress-energy tensor to be traceless, but this quantity is actually not the trace, since $\rho(r)$ has the opposite sign it should have in the trace

Answer 1

OP's metric is an example of ultrastatic spacetime, i.e. it is a direct product of a Riemannian manifold (“space”) and $\mathbb R$ (“time”) and thus $-g_{tt}$ can be set to unity everywhere.

This means that a test body initially at rest anywhere in such a spacetime will remain at rest: the spacetime does not have gravitational attraction or repulsion (at least for static bodies). Another way to look at this is that “gravitational potential” of this spacetime is constant everywhere and all curvature is purely spatial.

The constraint $\rho+p_R+2p_T\equiv 0$ has a simple physical interpretation: active quasilocal gravitational mass of an arbitrary (spatial) volume in this spacetime is always exactly zero.

To clarify this, let us consider a general static (not just ultrastatic and spherically symmetric) metric: $$ ds^2=-\Phi^2(x^i)+g_{ij}(x^i)dx^i dx^j.$$ The equation for the $g_{tt}$ metric component is the following: $$ \Delta_\text{3d} \Phi=4\pi \,\Phi\cdot(\rho + p^i{}_i).\tag{*}$$ We use units with $G=c=1$, the function $\Phi\equiv \sqrt{|g_{tt}|}$ can be called the relativistic gravitational potential, $ p^i{}_i$ is the trace of pressure tensor (it would be $3p$ for an ideal fluid and $p_R+2p_T$ in spherically symmetric case of OP) and $ \Delta_\text{3d}$ is the spatial Laplace–Beltrami operator: $$ \Delta_\text{3d} f \equiv \frac{1}{\sqrt{\hat g}}\left(\sqrt{\hat g} g^{ik} f_{,i}\right)_{,k}.$$ Here $\hat g\equiv \det( g_{ij})$ and commas denote partial derivatives.

The equation ($*$) is a straightforward relativistic generalization of a Poisson equation for Newtonian gravitational potential. The most obvious difference is the presence of pressure: positive pressures (or rather quantity $p/c^2$) contribute to gravitational potential in the same as positive densities. So if the constraint $\rho + p^i{}_i=0$ is satisfied within some spatial volume then this volume does not serve as a source of gravitational potential and if this condition is satisfied everywhere and the boundary conditions allow this, then the spacetime would be ultrastatic.

Couple of examples of well known solutions where this condition is satisfied:

• Einstein static universe. If we combine constant density of dust matter with a fine-tuned cosmological constant, then the negative pressure of cosmological constant would balance joint density of dust and cosmological constant resulting in ultrastatic FLRW spacetime. Note, that it is also a spherically symmetric spacetime so OP's ansatz would work for it.

• Cosmic string For a thin cosmic string its linear density equals to its tension (the one negative component of pressure tensor), so condition is satisfied for a $\delta$-like source. If we consider a large number of static strings oriented randomly in all the directions then such system would after averaging correspond to an isotropic fluid with $p= -\rho/3$ equation of state (a string gas).

A good discussion on the role pressure plays for relativistic gravitational potential can be found here:

• Ehlers, J., Ozsváth, I., & Schucking, E. L. (2006). Active mass under pressure. American journal of physics, 74(7), 607-613, doi:10.1119/1.2198881, arXiv:gr-qc/0505040.