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Transformation from de Sitter to flat spacetime coordinates

  • #1

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
$$x=u \cosh v, \quad t=u \sinh v.$$

Homework 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?
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,237
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With this we find
$$
\bar g _ {xx} =
g_{uu} \frac{\partial u}{\partial x }\frac{\partial u}{\partial x} + g_{vv} \frac{\partial v}{\partial x }\frac{\partial v}{\partial x}\\
=\frac{x^2}{x^2-t^2}+ (x^2-t^2) \frac{t^2}{x^4}\frac{1}{(1-\frac{t^2}{x^2})^2} \neq \pm 1
$$
Note that ##g_{vv} = -u^2##. Did you include the negative sign when you substituted for ##g_{vv}##?
 
  • #3
You were right, there is a minus sing missing and with it everything works out... Thanks for the help!
 

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