On the Gaussian Curvature of time-like surfaces

[deRoy]

Firstly, I am asking for your patience and understanding because my maths formalism is not going to be rigorous.

In another thread here in this forum, I set an example for which now I am asking further instructions.

I am going to ask about time-like surfaces immersed in Minkowskian space-time and how to find the Gaussian curvature at a point.

So, in my example I am starting with line: $t = \sqrt {x^2 - \frac {1} {x^2}}$ and embed this in Minkowski space-time. I am going to rotate this about the t-axis and get a time-like surface resembling a hyperboloid. Time-like in the sense that the normal vector to the surface is always space-like.

Ok, now parametrize a bit with φ for latitude and θ for longitude coordinates to find position vector $r$.
I am getting: $(\sqrt{Coshφ}Cosθ ,\sqrt{Coshφ}Sinθ , i \sqrt{SinhφTanhφ})$ and I am using the imaginary unit $i$ for the t-axis coordinate in order to get the right metric.

Proceeding with the usual dot product after differentiating to find $dr$, I am getting the required metric: $ds^2 =-Sechφ(3+Sech^2φ)dφ^2/4+(Coshφ)dθ^2$. Calculations were done with Mathematica.
Finally, I am using the usual Gauss formula to find the curvature.

It turns out that at φ=0 curvature is 1/2 and at infinity curvature is infinite, a result which I am very happy with.

My question is, am I always allowed to use this trick with the imaginary unit to evaluate the metric and Gaussian curvature of a time-like surface? I have to admit that my knowledge in this subject is very limited.
No-one ever told me, never read it in a book, it just works!

 

[vanhees71]

Well, besides I strongly discourage everybody of the use of the oldfashioned $\mathrm{i} c t$ convention, as far as I can see your metric looks right. You can do this entirely within real linear algebra using the Lorentz bilinear form. Sine you are using the east-coast convention ("mostly +") the signature in your case is (3,1). For a Galilean frame the pseudo-metric components are thus given by $(\eta_{\mu \nu})=\mathrm{diag}(1,1,1,-1)$. For an elementary introduction to SR (however written in the west-coast convention since I'm used to it, working in high-energy hadron physics (heavy-ion collisions), where this convention is more frequently used than the east-coast one), see

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[deRoy]

Well, besides I strongly discourage everybody of the use of the oldfashioned ictict\mathrm{i} c t convention, as far as I can see your metric looks right. You can do this entirely within real linear algebra using the Lorentz bilinear form. Sine you are using the east-coast convention ("mostly +") the signature in your case is (3,1). For a Galilean frame the pseudo-metric components are thus given by (ημν)=diag(1,1,1,−1)(ημν)=diag(1,1,1,−1)(\eta_{\mu \nu})=\mathrm{diag}(1,1,1,-1).

Thank you for your answer. I like this metric because it represents a spatial ring-like section moving from singularity to singularity in finite co-moving time ( proper time for co-moving observers riding on the ring. )

But I must have evaluated the Gaussian Curvature wrong. It should be negative throughout I believe. What do you think?

I believe the standard formula given by Gauss still applies here for 2 dimensions ( I've checked derivation and it's the usual dot product to find the infinitesimal triangle area with Cosh of the angle to replace Cosines if required, I am not sure if this method still applies in higher dimensions though. )