In general relativity, on the one hand, asymptotical flatness is assumed to derive a solution to the EFE which is a good approximation in the solar system (Schwarzschild, Kerr...) On the other hand, the scale of the whole universe, the cosmological constant and the expansion leads to the definition of the boundary conditions.

Regarding the scale in beween these two extrema, the galaxies: We measure the velocity of the stars in the galaxies and they do not fit the boundary condition of the flat space.

Of course, the additional curvature measured can be a result of additional matter which we do not see and know yet.

Apart from that, could it be the result of unknown boundary conditions for galaxies which we do not account for yet, because we do not know them?

  • $\begingroup$ The issue of boundary conditions exists in Newtonian gravity as well; we solve $\nabla^2 \Phi = - 4 \pi G \rho$ subject to the boundary conditions that $\Phi \to 0$ as $r \to \infty$. In this context, the impetus to modify Newtonian dynamics is that you don't have $\nabla^2 \Phi = - 4 \pi G \rho$ in the interior, a problem which is independent of the boundary conditions you choose. $\endgroup$ Nov 30, 2021 at 17:56

1 Answer 1


Actually, there are two exterior Schwarzschild solutions (the name vacuum solution is traditionally reserved for $T_{\mu \nu}$ identically zero). If one assumes $p=0$, then from Einstein equations it follows that $\varepsilon=0$, too. On the other hand, if one assumes $\varepsilon=0$, it does not imply necessarily that $p=0$. The second solution is unlike the first, not asymptotically flat. Thus, it is a cosmological solution. It describes spacetime without energy density, but with pressure. The question whether such exterior metric would explain galaxy rotational curves would need a closer investigation. I speculate here: a spacetime without matter but with a non-vanishing mean hydrostatic stress (=pressure) could be an interesting new candidate for the mysterious dark matter.


The line element in static spherically-symmetric space-time can be written as \begin{equation} \label{metric} {\rm d}s^2={\rm e}^{2\nu}c^2{\rm d}t^2-{\rm e}^{2\lambda}{\rm d}r^2-r^2{\rm d}\Omega^2~, \end{equation} with curvature radius $r$ and the infinitesimal surface element ${\rm d}\Omega$ of a 2-sphere. The metric components satisfy Einstein field equations (EFE) \begin{equation}\label{einstein} {\rm{R}}_{\mu}^{~\nu}-\frac{1}{2}{\rm{R}}_{\lambda}^{~\lambda}~\delta_{\mu}^{\nu}=\kappa {\rm{T}}_{\mu}^{~\nu}, \end{equation} where, in case of a perfect fluid, the stress energy tensor has diagonal form ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$. Multiplying both sides of EFE equation with square of curvature radius of fluid sphere $R^{2}$ makes this equation dimensionless. The dimensionless pressure and energy density in subsequent considerations are $\varepsilon\equiv {\varepsilon}~{\kappa}~c^2R^2$ and $p\equiv {p}~{\kappa}~R^2$, with $\kappa$, the Einstein's gravitational constant $8\pi G/c^4$. The metric components functions $\nu$ and $\lambda$ depend only on the dimensionless variable $u\equiv r^2/ R^2$ and compactness parameter $\alpha\equiv r_S/R$.

The requirement for isotropic pressure reduces Einstein's field equations to these three differential equations \begin{equation}\label{compactform} {\rm e}^{-\lambda}\frac{\rm d}{{\rm d} u} \left( {\rm e}^{-\lambda}\frac{\rm d}{{\rm d}u}{\rm e}^{\nu}\right) = \frac{1}{4}\frac{\rm d}{{\rm d}u}\left(\frac{1-{\rm e}^{-2\lambda}}{u}\right)~{\rm e}^{\nu}~,\tag{1} \end{equation} \begin{equation}\label{pressure} p=\frac{4}{{\rm e}^{\nu}}\frac{{\rm d}{\rm e}^{\nu}}{{\rm d}u}{\rm e}^{-2\lambda}-\frac{1-{\rm e}^{-2\lambda}}{u}~,\tag{2} \end{equation} \begin{equation} \label{density} \varepsilon=\frac{1-{\rm e}^{-2\lambda}}{u}-2~\frac{{\rm d}{\rm e}^{-2\lambda}}{{\rm d}u}~.\tag{3} \end{equation} The first equation can be read as a first-order linear non-homogeneous equation for ${\rm e}^{-2\lambda}$, or as a second-order homogeneous linear differential equation for ${\rm e}^{\nu}$. Therefore, in total there are three boundary conditions which can be set on the star surface ($u=1$). They express continuity of metric and requirement for total mass $M$ being enclosed by spherical surface of area $4\pi R^{2}$. The boundary condition for ${\rm e}^{-2\lambda}$ is \begin{equation}\label{BC1} {\rm e}^{-2\lambda}(1,\alpha)=1-\alpha~,\tag{4} \end{equation} and the two boundary conditions for ${\rm e}^{\nu}$ are \begin{equation}\label{BC2&3} {\rm e}^{\nu}(1,\alpha)=\sqrt{1-\alpha},\tag{5} \end{equation} \begin{equation} \frac{{\rm d}{\rm e}^{\nu}}{{\rm d}u}(1,\alpha) = \frac{1}{4}\frac{p_{1}+\alpha}{\sqrt{1-\alpha}}~,~~~p_{1}\equiv{p(1,\alpha)}.\tag{6} \end{equation} The boundary condition (6) in not independent. It results from equation (2) by stetting (4) and (5) into it.

Let's solve EFE for static spacetime with a sphere of constant energy density \begin{equation} \label{Schwarzschildsolution} \varepsilon =\left\{ \begin{array} {rcl} 3~\alpha & \mbox{for} & 0 \leq u \leq 1 \\\\ 0 & \mbox{for} & 1 > u \leq \infty~. \end{array}\right.\tag{7} %\right\parskip \end{equation} The solution of differential equation (3) for metric function $\rm{e}^{-2\lambda}$ reads \begin{equation} \label{e-2lambda} {\rm e}^{-2\lambda} =\left\{ \begin{array}{rcl} 1-\alpha~u~ & \mbox{for} & 0\leq u \leq 1 \\ \\1-\alpha/\sqrt{u}~ & \mbox{for} & 1 > u \leq \infty~. \end{array}\right.\tag{8} \end{equation} The solution of equation (1) yields \begin{equation} \label{enulambdazero} {\rm e}^{\nu} =\left\{\begin{array}{rcl} (3/2+p_{1}/\alpha)~\sqrt{1-\alpha }-(1/2+p_{1}/\alpha)~\sqrt{1-\alpha u } & \mbox{for} & 0\leq u \leq 1 \\ \\ \Bigl(1-p_{1}/8~f(1,\alpha)\Bigr)~\sqrt{1-\alpha/\sqrt {u}}+p_{1}/8~\sqrt{1-\alpha}~f(u,\alpha)& \mbox{for} & 1 < u \leq \infty~. \end{array}\right.\tag{9} \end{equation} where auxiliary function $f$ is defined as \begin{equation} f(u,\alpha)\equiv{15~\alpha^{2}~\sqrt{1-\alpha/\sqrt{u}}~\tanh^{-1}{\sqrt{1-\alpha/\sqrt{u}}}-15~\alpha^{2}+5\alpha\sqrt{u}+2u}.\tag{10} \end{equation} Because for large $u$, $f(u)$ behaves like $\sim 2u$, an asymptotically flat solution is possible only if $p_{1}$ is zero. In that case the metric reduces to the well-known interior and exterior Schwarzschild metric solution.

The special case - Universe without matter but with pressure

Setting $\alpha$ to zero ($\varepsilon=0$) one obtains metric \begin{equation} \label{metricwithoutmatter} {\rm d}s^2=(1-p_{1}/4+p_{1}/4~u)^2~c^2{\rm d}t^2-{\rm d}r^2-r^2{\rm d}\Omega^2~,\tag{11} \end{equation} where $p_{1}\equiv p(R)$ and $u\equiv (r/R)^{2}$.

The pressure and energy density are given as \begin{equation} \label{metricwithoutmatterpressure} p=p_{1}~{\rm e}^{-\nu}\equiv p_{1}/(1-p_{1}/4+p_{1}/4~u),~~~~~~\varepsilon=0.\tag{12} \end{equation} For $p_{1}=4$, the metric (11) reduces to \begin{equation} \label{metricwithoutmatterpressurespecial} {\rm d}s^2=(r/R)^4~c^2{\rm d}t^2-{\rm d}r^2-r^2{\rm d}\Omega^2~,\tag{13} \end{equation} and the pressure and energy density are \begin{equation} \label{metricwithoutmatterpressure2} \varepsilon=0,~~~~~~p=2~c^{4}/G\cdot 1/(4\pi r^{2}).\tag{14} \end{equation} This solution could be possibly interpreted as a black hole without mass.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Buzz
    Nov 28, 2021 at 21:56

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