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Are there solutions of Einstein's equation

$$ R_{i,k}=0 $$

which imply there are 'curvature' without matter or energy?

For example, space is curved but there are no matter or any energy there so an spaceship could detect that there is gravity but no star planet or something out there

Qmechanic
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Jose Javier Garcia
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1 Answers1

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Symmetries

Symmetries of the Riemann curvature tensor:

$$\begin{align}{R_{\left(ab\right)c}}^d&=0\tag{1}\\ R_{ab\left(cd\right)}&=0\tag{2}\\{R_{abc}}^d+{R_{bca}}^d+{R_{cab}}^d&=0\tag{3}\\ R_{abcd}&=R_{cdab}\tag{4}\end{align}$$

where parenthesis enclosing indices, $\left(\ldots\right)$, denotes the symmetric part of the tensor in these indices: $T_{\left(ab\right)}\equiv\frac{1}{2}\left(T_{ab}+T_{ba}\right)$.

It can be shown from these symmetries that in an $N$ dimensional spacetime, there are $\frac{1}{12}N^2\left(N^2-1\right)$ independent and generally non-zero components of the Riemann curvature tensor.

Constraint

$$\begin{align}R_{ab}&\equiv {R_{cab}}^c\\\therefore R_{ab}=0\implies0&={R_{cab}}^c\tag{5}\end{align}$$

As switching $a$ and $b$ will give an identical equation by symmetry (4) followed by both symmetries (1) and (2) then there are $\frac{1}{2}N\left(N+1\right)$ linearly independent equations. Each linearly independent equation reduces the number of independent generally non-zero components by one.

When $N=1$, constraint (5) is equivalent to all four symmetries and so the number of non-zero components is still zero. When $N=2$, constraint 5 gives the following 3 equations:

$$\cases{{R_{000}}^0+{R_{100}}^1=0&(6a)\\{R_{001}}^0+{R_{101}}^1=0&(6b)\\{R_{011}}^0+{R_{111}}^1=0&(6c)}$$

Both terms in (6b) are zero by symmetries (1) and (2), respectively. While the first term of (6a) is zero by symmetry (1) or (2) and likewise the second term of (6c) is zero. Finally, ${R_{100}}^1\equiv{R_{011}}^0$ by symmetry (4) followed by both (1) and (2) so these three equations reduce to a single linearly independent equation and so once again, the number of independent generally non-zero components is zero.

However, for $N\ge3$, I believe constraint (5) is linearly independent of the four symmetries and so the number of independent generally non-zero components $C$ will be:

$$C=\begin{cases}0,&N\in\left\{1,2\right\}\\\displaystyle\frac{1}{2}\left(N+1\right)\left(\frac{1}{6}N^2\left(N-1\right)-N\right),&N\ge3\\\end{cases}$$

However, if I have missed any uses of the above four symmetries for $N\ge3$ (please let me know in the comments) there will be less linearly independent equations and so $C$ will increase, but must still be bounded by $C\le\frac{1}{12}N^2\left(N^2-1\right)$. Therefore, the looser bound on $C$ is:

$$\frac{1}{2}\left(N+1\right)\left(\frac{1}{6}N^2\left(N-1\right)-N\right)\le C\le\frac{1}{12}N^2\left(N^2-1\right)\quad\text{for }N\ge3$$

These bounds are plotted below:

enter image description here

Thus, in 3D spacetime, $C=0$ and interestingly 4D spacetime (in which we live) has the first non-zero $C=10$.

Curved or Flat

Finally, spacetime is only flat if ${R_{abc}}^d$ vanishes, and so spacetime is flat in 1D, 2D and 3D when $R_{ab}=0$ but is in general still curved in 4D and any higher dimensions. Thus, 4D spacetime can be curved at a point where the energy-momentum tensor vanishes, but whether it is curved or flat will be dictated by boundary conditions. For example, the spacetime around a massive object is curved due to the object's mass even though the space around the object is empty.

Finally, this means in an empty universe space could indeed be curved! Without going into much detail we apply the principles of homogeneity and isotropy to get $R_{abcd}=K\left(\gamma_{ac}\gamma_{bd}-\gamma_{ad}\gamma_{bc}\right)$ where $g_{ij}=-a^2\left(t\right)\gamma_{ij}$ and $i,j\ne0$ are the spatial components of the metric - this is the Friedmann–Lemaître–Robertson–Walker metric. This gives $R_{00}=3\frac{\ddot a}{a}$ and $R_{ij}=-\frac{1}{c^2}\left(\ddot aa+2\dot a^2+2Kc^2\right)\gamma_{ij}$ and so with the earlier constraint, this gives $\ddot a=0$ and so $K=-\frac{\dot a^2}{c^2}$. Thus, in an empty universe $R_{abcd}=-\frac{\dot a^2}{c^2}\left(\gamma_{ac}\gamma_{bd}-\gamma_{ad}\gamma_{bc}\right)$ where $\dot a$ is some real constant and $a,b,c,d\ne0$.

Chris Long
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    This is not a vacuum solution, $R_{ac}=-3(\dot{a}/c)^2\gamma_{ac}$, unless $\dot{a}=0$, i.e. Minkowski metric – KP99 Jul 17 '21 at 14:15
  • My bad i made a typo in my algebra it should have been $R_{00}=3\frac{\ddot a}{a}$. Thats where the $\ddot a=0$ came from so unless I made another error I think it is a vacuum solution? – Chris Long Jul 17 '21 at 15:41
  • The question asks for vacuum solution $R_{ab}=0$. Maximally symmetric spacetimes like FLRW metric (K=1,-1) can account for Lambda vacuum equation $R_{ab}=4\Lambda g_{ab}$, but not the true vacuum solution. If we impose conformal flatness, then vacuum solution becomes trivial (which follows from decomposition of Riemann tensor). So we have to consider space-times with non-zero Weyl curvature. Standard examples- Schwarzschild, Kerr , Rindler geometry etc etc – KP99 Jul 17 '21 at 15:58
  • @KP99 so where have I gone wrong in stating that $R_{00}=3\frac{\ddot a}{a}$ and $R_{ij}=-\frac{1}{c^2}\left(\ddot aa+2\dot a^2+2Kc^2\right)\gamma_{ij}$ for the FLRW metric? – Chris Long Jul 17 '21 at 17:24
  • In the final expression for $R_{abcd}$, if you contract both sides with metric, you will get $R_{ij}=-3(\dot{a}/c)^2\gamma_{ij}\neq 0$ unless a(t) is a constant, which will be the trivial case. The issue here is that, you have additionally assumed Maximally symmetric space time. If you ignore the maximal symmetry, there won't be any problem – KP99 Jul 17 '21 at 17:27
  • @KP99 but why have you neglected the $K$ term? See equation 8.13 from these notes https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll8.html – Chris Long Jul 17 '21 at 17:58
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    I am referring to the last line: "Thus in an empty universe, $R_{abcd}=-(\dot{a}/c)^2(\gamma_{ac}\gamma_{bd}-\gamma_{ad}\gamma_{bc})$...''. If you contract with $\gamma^{bd}$, you won't get vanishing $R_{00}$ and $R_{ij}$ which you have claimed in previous lines. I am not neglecting K, I'm simply back calculating from your final remarks – KP99 Jul 17 '21 at 18:38
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    @KP99 ah thank you, I now understand your point. I should have been explicit in stating $R_{abcd}=-\frac{\dot a^2}{c^2}\left(\gamma_{ac}\gamma_{bd}-\gamma_{ad}\gamma_{bc}\right)$ only holds for $a,b,c,d\ne0$ the time-like components will be defined differently and so when you contract the time-like components will compensate making the Ricci tensor zero. This is because $\gamma^{ij}$ is the spatial metric as defined in my answer. – Chris Long Jul 17 '21 at 19:09