# Transformation from de Sitter to flat spacetime coordinates

Markus Kahn
Homework Statement:
A metric in a two-dimensional spacetime with coordinates ##(u,v)## is
$$d s^{2}=d u^{2}-u^{2} d v^{2}.$$
Transform the line element ##ds^2## from ##(u, v)## to new coordinates ##(x, t)## defined by
$$x=u \cosh v, \quad t=u \sinh v.$$
Relevant Equations:
All given above.
Let me begin by stating that I'm aware of the fact that this is a metric of de Sitter spacetime, aka I know the solution, my problem is getting there. My idea/approach so far: in the coordinates ##(u,v)## the metric is given by
$$g_{\mu\nu}= \begin{pmatrix}1 & 0\\ 0 & -u^2\end{pmatrix}.$$
The general transformation law for the metric is given by
$$\bar{g}_{\alpha\beta} = g_{\mu\nu}\frac{\partial x^\mu}{\partial\bar x ^\alpha}\frac{\partial x^\nu}{\partial\bar x ^\beta}.$$ So we first need ##u(x,t)## and ##t(u,v)##, which are given by
$$u = \sqrt{x^2-t^2}\quad \text{and}\quad v=\mathrm{arctanh}(t/x).$$
We now can calculate
\begin{align*}\frac{\partial u}{\partial x} &= \frac{x}{\sqrt{x^2-t^2}}, &\frac{\partial u}{\partial t} = - \frac{t}{\sqrt{x^2-t^2}} \\ \frac{\partial v}{\partial x} &= -\frac{t}{x^2}\frac{1}{1-\frac{t^2}{x^2}}, &\frac{\partial v}{\partial t} = \frac{1}{x}\frac{1}{1-\frac{t^2}{x^2}}\end{align*}.
With this we find
\begin{align*} \bar g _ {xx} &= g_{uu} \frac{\partial u}{\partial x }\frac{\partial u}{\partial x} + g_{vv} \frac{\partial v}{\partial x }\frac{\partial v}{\partial x}\\ &=\frac{x^2}{x^2-t^2}+ (x^2-t^2) \frac{t^2}{x^4}\frac{1}{(1-\frac{t^2}{x^2})^2} \neq \pm 1, \end{align*}
which (I think) it should be. So, am I doing something fundamentally wrong? If not, where exactly do I make my mistake?

$$\bar g _ {xx} = g_{uu} \frac{\partial u}{\partial x }\frac{\partial u}{\partial x} + g_{vv} \frac{\partial v}{\partial x }\frac{\partial v}{\partial x}\\ =\frac{x^2}{x^2-t^2}+ (x^2-t^2) \frac{t^2}{x^4}\frac{1}{(1-\frac{t^2}{x^2})^2} \neq \pm 1$$