GR metric gauge transformation, deduce 'generating' vector

[binbagsss]

1. Problem

$g_{uv}'=g_{uv}+\nabla_v C_u+\nabla_u C_v$

If $g_{uv}'$ is given by $ds^2=dx^2+2\epsilon f'(y) dx dy + dy^2$
And $g_{uv}$ is given by $ds^2=dx^2+dy^2$, Show that $C_u=2\epsilon(f(y),0)$?

Homework Equations

Since we are in flatspace we have $g_{uv}'=g_{uv}+\partial_v C_u+\partial_u C_v$

The Attempt at a Solution

[/B]
Since $g_{xx}=g'_{xx}$ and $g_{yy}=g'_{yy}$

I get $\partial_x C_x=\partial_y C_y=0$

$\implies C_x=c(y)$, $c(y)$ the constant of $x$ from integration, some function of $y$
$C_y=k(x)$ $k(x)$ some function of $x$

From the cross term $g_{xy}'=g_{yx}'=\epsilon f'(y)$ I get:

$\epsilon f'(y)=\partial_x C_y + \partial_y C_x$

So $\epsilon f'(y)= k'(x) + c'(y)$

I have no idea what to do now..

Many thanks in advance

 

[binbagsss]

bump please. thank you very much.