# Conformal relation for 2-dim Lorentz space-times

I have two 2-dimensional space-times ($\mathbb{S}^1\times\mathbb{R}$) with signature $(-,+)$. One of them is flat the other one has non-vanishing curvature (Riemann tensor), both have vanishing Ricci tensor. But they seem to have a similar global and causal structure. Of course, because of the 2-dimensional case they are local conformally flat.

I am looking for a global relation between them that could explain the similar causal and global structure and I think that a (global) conformal transformation would be a possible approach.

********************Edit in response to comments*******************

The two metrics in question are $$ds^{2}=Td\psi^{2}+2dTd\psi \quad\text{ defined on }\quad S^{1}\times\mathbb{R}$$ and $$ds^{2}=-(\frac{2m}{r}-1)d\nu^{2}+2d\nu dr \quad\text{ defined on }\quad S^{1}\times(0,\infty).$$

Note that $\psi$ and $\nu$ are the according periodic variables.

I already could calculate the local conformal relation. But what about the global relation?

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I don't really know any way to answer this, but what if you look at possible obstructions to construct a global map form a cover that gives you a local conformal equivalence to the Lorentz space at each chart? Was there any attempt to do this? – Yuri Vyatkin Oct 25 '12 at 5:29

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A friend of mine gave to me two examples of metrics on $S^1 \times \Bbb R$, one is $d\theta^2-dx^2$, and the other is $-d\theta^2 + dx^2$. They are not globally conformally equivalent because the former does not admit closed timelike curves (it is causal), while the former does. Therefore, to show the existence you need to specify the metrics that you are using. – Yuri Vyatkin Nov 28 '12 at 2:48

## 1 Answer

The two metrics in question are $$ds^{2}=Td\psi^{2}+2dTd\psi \text{ defined on } S^{1}\times\mathbb{R}$$ and $$ds^{2}=-(\frac{2m}{r}-1)d\nu^{2}+2d\nu dr \text{ defined on } S^{1}\times(0,\infty).$$ Note that $\psi$ and $\nu$ are the according periodic variables.

I already could calculate the local conformal relation. But what about the global relation?

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This should have been an edit to the question, but apparently you have multiple unregistered accounts. – user53153 Feb 8 at 6:06
– Willie Wong♦ Feb 8 at 8:47
Now, while I copied the text verbatim into your question statement, there seems to be some typos on your part on the definition of your metrics: a metric is a nondegenerate quadratic form on the space of tangent vectors. Yet neither of your "metrics" are quadratic forms. The first one contains $T\psi^2$ which is a pure scalar term, and the latter one contains a linear term $(1-2m/r) d\nu$. – Willie Wong♦ Feb 8 at 8:53
Also, you should probably state which of the coordinates is the one in the $\mathbb{S}^1$ direction, and which is in the $\mathbb{R}_{(+)}$ direction. – Willie Wong♦ Feb 8 at 8:57
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