6
$\begingroup$

From what I understand of Lorentz surfaces (spacetimes of dimension 2), it seems that, according to Kulkarni's theorem, two reasonable enough Lorentz surfaces (I am only interested in surfaces with topology $\Bbb R^2$) are conformally equivalent, that is, $g_1 = \Omega^2 g_2$. This includes Minkowski space, meaning that they must all be conformally flat.

To find the equivalent conformally flat metric, I assumed that since they are conformal, the metric's eigenvalues should be $-\Omega^2$ and $\Omega^2$. This would then mean that, given a real symmetric $2\times 2$ matrix with negative determinant, the eigenvalues should always be inverses of each other.

From some calculations, this seems not to be the case. Did I misunderstand Kulkani's theorem or is the method I tried incorrect for such a task?

$\endgroup$
1
  • $\begingroup$ $\uparrow$ Which calculations? Which method did you try? $\endgroup$
    – Qmechanic
    Jul 2, 2016 at 13:53

1 Answer 1

1
$\begingroup$

The mistake here is to forget that the only coordinate-independent property of the eigenvalues of the metric is the sign. Consider the Minkowski metric in usual Cartesian coordinates $$d s^2 = -dt^2 + dx^2$$ Or in matrix form $$\left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right)$$ I.e., the eigenvalues are always in a ratio $\lambda_1/\lambda_2=-1$.

Now let us use a new time coordinate coordinate $\tau$ such that $$t=\frac{2}{3} \tau^{3/2}$$ which is restricted to $t \in (0,\infty) \to \tau \in (0,\infty)$. Now the metric transforms to $$ ds^2 = -\tau d\tau^2 + dx^2$$ or in matrix form $$\left( \begin{array}{cc} -\tau & 0 \\ 0 & 1 \end{array} \right)$$ Now we see that the eigenvalue ratio is $\lambda_1/\lambda_2=-\tau$ and as we go through $\tau \in (0,\infty)$, the ratio goes through all the possible values admissible for a non-degenerate matrix of signature $(-+)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.