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.

Homework Equations

General coordinate transformation, ds²=g_abdx^adx^b

The Attempt at a Solution

I started with a general transformation x^a→x'^a so the new metric is g'_μν=g_ab(dx^a/dx'^μ)(dx^b/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.

 

[strangerep]

I started with a general transformation x^a→x'^a so the new metric is g'_μν=g_ab(dx^a/dx'^μ)(dx^b/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]

Thanks for the reply!
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:

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

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

 

[strangerep]

Try writing out the corresponding PDEs for the inverse transformation first. I.e., treating $t,x$ as functions of $u,v$. They turn out a bit easier.

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., $$ \frac{\partial u}{\partial t} ~\equiv~ u,_t ~~;~~~~ \mbox{and}~~~~~~ \frac{\partial^2 u}{\partial t^2} ~\equiv~ u,_{tt} ~~;~~~~ \mbox{(etc)}.$$