Conformal Transformation: Minkowski sheet to cylinder

Question

What conformal transformation can I make to 2d Minkowski with metric $ds^2=-dt^2+dx^2$ to show that it is conformal to a cylinder?

Answer 1

I'd say none. If you are talking about a Riemannian cylinder, its metric (say, induced from its embedding in $\Bbb{R}^{3}$) is

$$ g=dz^{2}+d\phi^{2} $$

If the two were conformally equivalent, a conformal transformation $(z,\phi)=(z(t,x),\phi(t,x))$ would pull back the metric as $$ \begin{aligned} g'&=\bigg(\frac{\partial z}{\partial t}dt+\frac{\partial z}{\partial x}dx\bigg)^{2}+\bigg(\frac{\partial \phi}{\partial t}dt+\frac{\partial \phi}{\partial x}dx\bigg)^{2} \\ &= \bigg[\bigg(\frac{\partial z}{\partial t}\bigg)^{2}+\bigg(\frac{\partial \phi}{\partial t}\bigg)^{2}\bigg]dt^{2}+\bigg[\bigg(\frac{\partial z}{\partial x}\bigg)^{2}+\bigg(\frac{\partial \phi}{\partial x}\bigg)^{2}\bigg]\ dx^{2}+2\bigg[\frac{\partial z}{\partial t}\frac{\partial z}{\partial x}+\frac{\partial \phi}{\partial t}\frac{\partial \phi}{\partial x}\bigg]\ dtdx \end{aligned} $$ You may choose the transformation in order to give $\frac{\partial z}{\partial t}\frac{\partial z}{\partial x}+\frac{\partial \phi}{\partial t}\frac{\partial \phi}{\partial x}=0$ or

$$ g'=\bigg[\bigg(\frac{\partial z}{\partial t}\bigg)^{2}+\bigg(\frac{\partial \phi}{\partial t}\bigg)^{2}\bigg]dt^{2}+\bigg[\bigg(\frac{\partial z}{\partial x}\bigg)^{2}+\bigg(\frac{\partial \phi}{\partial x}\bigg)^{2}\bigg]\ dx^{2} $$ but, as you can see, you won't find one that gives you the correct Lorentzian signature. This is because conformal equivalence is an equivalence relation amongst manifolds that carry the same metric signature. A Riemannian cylinder has signature $(+,+)$ while a Lorentzian 2D manifold has signature $(-,+)$.