# General relativity- Coordinate/metric transformations

jgarrel

## Homework Statement

Consider the metric ds2=(u2-v2)(du2 -dv2). I have to find a coordinate system (t,x), such that ds2=dt2-dx2. The same for the metric: ds2=dv2-v2du2.

## The Attempt at a Solution

I started with a general transformation xa→x'a so the new metric is g'μν=gab(dxa/dx'μ)(dxb/dx'ν). The components of g (old metric) and g'(new metric) are known and the unknowns are the derivatives of the old coordinates with respect to the new ones. That's where I'm stuck.

I started with a general transformation xa→x'a so the new metric is g'μν=gab(dxa/dx'μ)(dxb/dx'ν).
Those derivatives should be partial derivatives. Write it in latex like this: $$g'_{\mu\nu} ~=~ g_{ab}\, \frac{\partial x^a}{\partial x'^\mu} \, \frac{\partial x^b}{\partial x'^\nu} ~.$$ (Look under the Info->Help/How-To menu to find a Latex primer. You'll need to master at least basic latex on this forum.)

Both metrics are diagonal, which makes it reasonably easy to write out the above transformation equations more explicitly.

I.e., write it out explicitly (where ##\mu,\nu## are ##t,x## and ##a,b## are ##u,v##). You should get 2 partial-differential equations where each right hand side has 2 terms. I'll wait to see if you can get that far before giving more hints.

jgarrel
I already had ended up with these two differential equations, but I thought there was another way, because they seem difficult to solve. I put them here:

##1=(u^2-v^2) \frac {\partial^2 u} {\partial t^2} -(u^2-v^2) \frac {\partial^2 v} {\partial t^2}##

##-1=(u^2-v^2) \frac {\partial^2 u} {\partial x^2} -(u^2-v^2) \frac {\partial^2 v} {\partial x^2}##

Btw, if those partial derivatives become too tedious to keep writing out fully in latex, you can always use the briefer "comma" notation, e.g., $$\frac{\partial u}{\partial t} ~\equiv~ u,_t ~~;~~~~ \mbox{and}~~~~~~ \frac{\partial^2 u}{\partial t^2} ~\equiv~ u,_{tt} ~~;~~~~ \mbox{(etc)}.$$
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