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I am trying to obtain the embedding diagram of Minkowski space. I took the metric:

$$ d s^{2} = - d t^{2} + d r^{2} + r^{2} \left( d \theta^{2} + \sin^{2} \theta d \phi^{2}\right) $$

for $t =$ cte and $\theta = \pi /2$,

$$ d s^{2} = d r^{2} + r^{2} d \phi^{2} $$

I know this should give me a disk, like the one below:

enter image description here

But I can't understand how to graph or if I should do any parameterization before. I know it's something basic but I would really appreciate if someone could guide me.

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  • $\begingroup$ You need to define suitable co-ordinates for it. In the case of Minkowski, I think it will be the hyperbolic ones. $\endgroup$ Commented Mar 16, 2024 at 8:11
  • $\begingroup$ @codebpr In hyperbolic coordinates, the Minkowski line element is expressed as $$ d s^{2} = d \rho^{2} + \sinh^{2} \rho d \theta ^{2} $$ but I can't figure out how to graph it as a 3D surface with Mathematica $\endgroup$ Commented Mar 16, 2024 at 12:00
  • $\begingroup$ What assumptions did you use to transform in hyperbolic coordinates? $\endgroup$ Commented Mar 16, 2024 at 13:21
  • $\begingroup$ @codebpr I considered minkowski at 2+1, $$ d s^{2} = -\mathrm{d} X_0^2+\mathrm{d} X_1^2+\mathrm{d} X_2^2 $$ and the transformation: $$ \begin{aligned} & X_0=\cosh \rho \\ & X_1=\sinh \rho \cos \theta \\ & X_2=\sinh \rho \sin \theta \end{aligned} $$ $\endgroup$ Commented Mar 16, 2024 at 13:43

1 Answer 1

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Let us follow your assumptions and say you had Minkowski spacetime in rectangular coordinates:

$$ds^2=-dX_0^2+dX_1^2+dX_2^2$$

Now, using your transformation:

$$X_0=\cosh\rho,$$ $$X_1=\sinh\rho\cos\theta,$$ $$X_2=\sinh\rho\sin\theta.$$

Then, after solving, we get the following, for a unit hyperbolic radius:

$$ds^2=d\rho^2+\sinh^2\rho d\theta^2$$

You have everything; all you need is just to plug your transformation rules in Mathematica:

ParametricPlot3D[{Sinh[ρ] Cos[θ], Sinh[ρ] Sin[θ], Cosh[ρ]}, {ρ, 0, 0.01}, {θ, 0, 2 π}, PlotRange -> Automatic] 

which gives you the required embedding diagram,

embedding

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  • $\begingroup$ amazing, thank you very much! $\endgroup$ Commented Mar 16, 2024 at 15:07

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