The Taub metric and the gravitational field of an infinite wall

Question

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 ?

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

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

with

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

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!!

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