According to this paper, the only vacuum solution to Einstein field equation which is static and plane-symmetric is the Taub metric, given by:

$$ ds^2 = z^{-2/3} dt^2 - dz^2 - z^{4/3}(dx^2 + dy^2)$$

with a curvature singularity at $z=0$ (where $0 < z < +\infty$). According to the paper, this metric has the peculiar property that objects are repelled from the singularity, and therefore they write that

... the solution must be interpreted as representing the exterior gravitational field due to a negative mass distribution

On the other hand I found this web page that discusses the gravitational field of an infinite "wall", and by requiring planar symmetry they arrive at the following metric (where I change the coordinates labeling to match with the paper) :

$$ ds^2 = z^{4/3} dt^2 - z^{-2} dz^2 - z^{2/3}(dx^2 + dy^2)$$

[Edit : Apparently this metric does not satisfy the vacuum equations (see comments) , not sure if it is just a typo in the web page or some other mistake]

This is said to satisfy the vacuum equations and gives the same proper acceleration from rest for all positions, as in the Newtonian case (but it's not clear to me if in this coordinates the singularity is at $z=0$ or $z=\infty$)

So now if the Taub metric is the only possible planar-symmetric vacuum metric, and it represents a non-physical (negative mass/pressure) matter distribution, what would happen if we actually built very large wall, made of ordinary matter ? surely if we stand close enough to the wall there should be approximate planar symmetry. How can that be reconciled with the non-physicality of the Taub metric ?

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    $\begingroup$ I am not sure of what the metric looks like, but an infinite plane domain wall has reulsive gravity. Such a domain wall has a constant positive energy density $T$ but it also has tension $T$ (negative pressure) in two direction and the effective gravity source is $E+P_1+P_2+P_3$ = +T-2T=-T<0$. $\endgroup$
    – mike stone
    Feb 6, 2022 at 13:41
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    $\begingroup$ this metric $$ ds^2 = z^{4/3} dt^2 - z^{-2} dz^2 - z^{2/3}(dx^2 + dy^2)$$ doesn't stratify the Einstein field equation $~G_{\mu\nu}=0~$ $\endgroup$
    – Eli
    Feb 6, 2022 at 14:49
  • $\begingroup$ @Eli Can you elaborate on how do you see it ? $\endgroup$
    – J. Delaney
    Feb 6, 2022 at 15:34
  • $\begingroup$ I use Maple symbolic program. Your first metric is o.k $\endgroup$
    – Eli
    Feb 6, 2022 at 15:43
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    $\begingroup$ @Felicia I believe he meant to say "satisfy" $\endgroup$
    – J. Delaney
    Feb 8, 2022 at 13:35

1 Answer 1


I follow the web documentation

I) Metric Ansatz

$$\mathbf G= \left[ \begin {array}{cccc} -{{\rm e}^{2\,u \left( x \right) }}&0&0&0 \\ 0&{{\rm e}^{2\,v \left( x \right) }}&0&0 \\ 0&0&{{\rm e}^{2\,w \left( x \right) }}&0 \\ 0&0&0&{{\rm e}^{2\,w \left( x \right) }} \end {array} \right] $$

II) Ricci Tensor $~\mathbf{RC}~$

$$RC_{1,1}={\frac {d^{2}}{d{x}^{2}}}u \left( x \right) + \left( {\frac {d}{dx}}u \left( x \right) \right) ^{2}- \left( {\frac {d}{dx}}u \left( x \right) \right) {\frac {d}{dx}}v \left( x \right) +2\, \left( { \frac {d}{dx}}u \left( x \right) \right) {\frac {d}{dx}}w \left( x \right) =0\tag 1$$

$$ RC_{2,2}=\left( {\frac {d}{dx}}u \left( x \right) \right) {\frac {d}{dx}}w \left( x \right) +{\frac {d^{2}}{d{x}^{2}}}w \left( x \right) +2\, \left( {\frac {d}{dx}}w \left( x \right) \right) ^{2}- \left( { \frac {d}{dx}}v \left( x \right) \right) {\frac {d}{dx}}w \left( x \right) =0\tag 2$$

$$RC_{3,3}={\frac {d^{2}}{d{x}^{2}}}u \left( x \right) + \left( {\frac {d}{dx}}u \left( x \right) \right) ^{2}- \left( {\frac {d}{dx}}u \left( x \right) \right) {\frac {d}{dx}}v \left( x \right) +2\,{\frac {d^{2}} {d{x}^{2}}}w \left( x \right) +2\, \left( {\frac {d}{dx}}w \left( x \right) \right) ^{2}-2\, \left( {\frac {d}{dx}}v \left( x \right) \right) {\frac {d}{dx}}w \left( x \right) =0\tag 3$$

now if you substitute the function that the author obtains

$$u(x)=\frac 23\,\ln(x)~,v(x)=-\ln(x)~,w(x)=\frac 13\,\ln(x)$$

you obtain that the Ricci tensor unequal zero!, so those solutions are wrong

$$\mathbf{RC}= \left[ \begin {array}{cccc} -{\frac {8}{9}}\,{x}^{4/3}&0&0&0 \\ 0&2/3\,{x}^{-2}&0&0\\ 0&0&4/9\, {x}^{2/3}&0\\ 0&0&0&4/9\,{x}^{2/3}\end {array} \right] \ne \mathbf 0$$

III the Solution

solving the equations (1),(2) and (3) you obtain

$$u(x)=\text{arbitrary}\\ v(x)=-3\,u(x)+ln(u'(x))\\ w(x)=-2\,u(x)$$


$$u(x)=\frac 13\ln(-3x)\\ v(x)=\ln(-x)+\ln(-x^{-1})\\ w(x)=-2\,u(x)$$

the new metric is now $$ds^2=-(3x)^{-2/3}\,dt^2+dx^2+(3x)^{4/3}(dy^2+dz^2)$$

you obtain a metric that has the same structure as the Taub metric!!

$$ ds^2 = -x^{-2/3} dt^2 + dx^2 +x^{4/3}(dy^2 + dz^2)$$

  • $\begingroup$ Thanks for your answer, it looks to me like the metric you derived is equivalent to the Taub metric, with the coordinate transformation $X=x^3$ (and appropriate scaling of the other coordinates). Is that correct ? If so this supports the claim that Taub is the only plane-symmetric solution. Still the question is why can't we have this type of solution with just ordinary matter $\endgroup$
    – J. Delaney
    Feb 11, 2022 at 11:43
  • $\begingroup$ I will check it for you $\endgroup$
    – Eli
    Feb 11, 2022 at 12:03
  • $\begingroup$ you can obtain u(x) from the Taub metric the new metric is almost the Taub metric , but the coefficient of $~dx^2~$ is $~{{\rm e}^{-2\,\ln \left( x \right) +2\,\ln \left( 1/3\,{x}^{-1} \right) }} ~$ it must be equal one $\endgroup$
    – Eli
    Feb 11, 2022 at 12:36
  • $\begingroup$ the metric is \begin{align*} G= \left[ \begin {array}{cccc} -{x}^{2/3}&0&0&0\\ 0&{ {\rm e}^{-2\,\ln \left( x \right) +2\,\ln \left( 1/3\,{x}^{-1} \right) }}&0&0\\ 0&0&{x}^{-4/3}&0 \\ 0&0&0&{x}^{-4/3}\end {array} \right] \end{align*} you make the Ansatz $~G_{2,2}=X^a~$ and solve it for $~x=f(X)~\Rightarrow~dx=f'(x)\,dX~$ $\endgroup$
    – Eli
    Feb 11, 2022 at 12:38
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    $\begingroup$ Good. maybe you could edit your answer to make this point clear. also note that $v(x) = -2\ln x + \ln(2/3x) = -3\ln(x) + const.$ $\endgroup$
    – J. Delaney
    Feb 11, 2022 at 18:54

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