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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
$$v(x)=0\Rightarrow\\ u(x)=\frac 13\ln(-3x)\\ w(x)=-2\,u(x)$$

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!!

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

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!!

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

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
$$v(x)=0\Rightarrow\\ u(x)=\frac 13 +\ln\left(-\frac{1}{3\,x} \right)\\ w(x)=-2\,u(x)$$

the new metric now

$$\mathbf G=\left[ \begin {array}{cccc} - \left( -1/3\,{x}^{-1} \right) ^{2/3}&0&0 &0\\ 0&1&0&0\\ 0&0& \left( -1/3\,{ x}^{-1} \right) ^{-4/3}&0\\ 0&0&0& \left( -1/3\,{x}^ {-1} \right) ^{-4/3}\end {array} \right] $$

with $~3x=X~$ you obtain the Taub metric!!

this is the Taub metric

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

which also satisfy the Einstein field equation

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
$$v(x)=0\Rightarrow\\ u(x)=\frac 13\ln(-3x)\\ w(x)=-2\,u(x)$$

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!!

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

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
$$u(x)=\frac 23\ln(x)\quad\Rightarrow\\ v(x)=-2\ln(x)+\ln\left(\frac{2}{3x}\right)\\ w(x)=-\frac 43\ln(x)$$

the new metric is now

$$\mathbf G=\left[ \begin {array}{cccc} -{x}^{4/3}&0&0&0\\ 0&{ {\rm e}^{-4\,\ln \left( x \right) +2\,\ln \left( 2/3\,{x}^{-1} \right) }}&0&0\\ 0&0&{x}^{-8/3}&0 \\ 0&0&0&{x}^{-8/3}\end {array} \right] $$

with $$X=~{{\rm e}^{-4\,\ln \left( x \right) +2\,\ln \left( 2/3\,{x}^{-1} \right) }}\quad\Rightarrow\\ x=\frac 13\,\sqrt [3]{2}{3}^{2/3}\sqrt [6]{{X}^{-1}}\\ dx=-\frac{1}{18}\,{\frac {\sqrt [3]{2}{3}^{2/3}{\it dX}}{ \left( {X}^{-1} \right) ^{5/6}{X}^{2}}}\quad\Rightarrow $$

\begin{align*} &G_{11}=-\frac{1}{3}\,\frac{2^{4/9}\,3^{5/9}}{X\,\left(\frac{1}{X}\right)^{7/9}}\\ &G_{22}=\frac{1}{108}\,\frac{2^{2/3}\,3^{1/2}}{X^2\,\left(\frac{1}{X}\right)^{2/3}}\\ &G_{33}=\frac{1}{2}\,\frac{3^{8/9}\,2^{1/9}}{\left(\frac{1}{X}\right)^{4/9}}\\ &G_{44}=G_{33} \end{align*} this metric satisfy the Einstein field equation $~\mathcal G_{\mu\nu}=0$

this is the Taub metric

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

which also satisfy the Einstein field equation

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
$$v(x)=0\Rightarrow\\ u(x)=\frac 13 +\ln\left(-\frac{1}{3\,x} \right)\\ w(x)=-2\,u(x)$$

the new metric now

$$\mathbf G=\left[ \begin {array}{cccc} - \left( -1/3\,{x}^{-1} \right) ^{2/3}&0&0 &0\\ 0&1&0&0\\ 0&0& \left( -1/3\,{ x}^{-1} \right) ^{-4/3}&0\\ 0&0&0& \left( -1/3\,{x}^ {-1} \right) ^{-4/3}\end {array} \right] $$

with $~3x=X~$ you obtain the Taub metric!!

this is the Taub metric

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

which also satisfy the Einstein field equation