- #1
deRoy
- 37
- 5
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!
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!