% Bernd Berg Oct 27 2018.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2018 
Solutions} \medskip

\centerline{\bf Set 9} 

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\item {\bf 22. Rindler 9.8: Transformation to a static metric.}
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Completion of the square gives
$$ d\vec{s}\,^2 = x^2\,(dt-3x^2y\,dx-x^3 dy)^2 
   - 9\,x^6y^2dx^2 - x^8dy^2 - dx^2\,.$$
Now we substitute $t'=t+f(x,y)$ and want
$$ dt' = dt+df = dt+f_xdx+f_ydy~~{\rm with}~~
   f_x=-3\,x^2y~~{\rm and} ~~f_y=-x^2\,.$$
Integrating $f_x$ gives
$$ \int f_xdx = - \int 3\,x^2y\,dx = -x^3y + g(y)\,, $$
which yields $f_y=-x^3+g'$ so that $g'=0$ and $g(y)$ is
a constant, and we are free to choose $g(y)=0$. Hence,
$$ dt' = dt-3x^2y\,dx-x^3 dy $$
holds and our final result is the static metric
$$ d\vec{s}\,^2 = x^2\,dt'\,^2-9\,x^6y^2dx^2-x^8dy^2-dx^2\,.$$

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