- #1
binbagsss
- 1,254
- 11
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)##?
Since we are in flatspace we have ##g_{uv}'=g_{uv}+\partial_v C_u+\partial_u C_v##
[/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
##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