Physics Asked on January 2, 2022

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 $2times 2$ 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?

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 $$ ds^2 = -tau dtau^2 + dx^2$$ 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 $(-+)$.

Answered by Void on January 2, 2022

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