Principal and Gaussian curvature of the FRW metric

  • #1
shinobi20
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TL;DR Summary
I'd like to confirm my calculations of the principal and Gaussian curvature of the spatial part of the Friedmann-Robertson-Walker (FRW) metric. Specifically, if the spatial part has negative curvature.
I would like to calculate the principal and Gaussian curvature of the spatial part of the Friedmann-Robertson-Walker (FRW) metric; specifically, the negative Gaussian curvature . The FRW metric is,



For a certain time slice , the spatial part is,



If we set , then we get the metric (also called the first fundamental form) for the 3d hyperboloid in angular coordinates,





The 3d hyperboloid can be embedded in 4d Minkowski space. Let be the Minkowski metric and be the 3d hyperboloid metric in angular coordinates. The 3d hyperboloid in coordinates is described by the equation,



The parameterization is,



Given the point on the 3d hyperboloid, the tangent vectors are,



The normal vector on the 3d hyperboloid can be calculated as,



The normalized normal vector is,



We also need the derivative of the tangent vectors . The second fundamental form can be calculated as,



The principal curvatures for are the eigenvalues of the matrix ,



The Gaussian curvature is .

Questions:
(1) Are my calculations correct?
(2) Are the answers for the principal curvatures correct?
(3) I'm wondering about . As I know the Gaussian curvature should be for , but only for . Can anyone help clarify and explain more about the principal curvatures of the 3d hyperboloid?
(4) Another thing I'm unsure of is the sign of the normal vector, if I remove the minus sign then becomes negative and becomes positive while maintaining the same magnitude.
 
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  • #2
The second fundamental form looks a bit suspect (i.e. the cosh in the to left component). The 3d hyperboloid (H3) is maximally symmetric and should have constant intrinsic curvature everywhere. You should then find that , specifically . And you should get constant principal curvatures.
 
  • #3
ergospherical said:
The second fundamental form looks a bit suspect (i.e. the cosh in the to left component). The 3d hyperboloid (H3) is maximally symmetric and should have constant intrinsic curvature everywhere. You should then find that , specifically . And you should get constant principal curvatures.
Indeed, if I remove then I get . However, I'm not yet sure where I went wrong. The necessary quantities to calculate are listed above, although I did not explicitly type out the vectors , but that is just the derivative of the tangent vectors; I've used Mathematica to calculate those quantities. Is my normal vector correct?

The issue is with the component . The vector is given by,



Note the minus sign in the metric is for the coordinate . is calculated as,



So, the issue is the term .
 
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  • #4

Reinserting the parameterizations of gives , which is indeed .
 
  • #5
ergospherical said:

Reinserting the parameterizations of gives , which is indeed .
Your last term should be due to the metric right? The minus sign in is what makes it minus. That's the reason the issue comes out.
 
  • #6
I already account for the signs. Albeit maybe I had chosen the other signature to you. With your signature convention, gives and gives . Then take , and that Einstein sum has a positive sign for the first three terms and a negative sign for the last term.
 
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  • #7
ergospherical said:
I already account for the signs. Albeit maybe I had chosen the other signature to you. With your signature convention, gives and gives . Then take , and that Einstein sum has a positive sign for the first three terms and a negative sign for the last term.
Actually, I made a mistake interpreting the vectors as lower index, i.e., . Actually, they are upper index vectors . So when computing , it is as you did, i.e., .

My normal vector is,



and



When you take the inner product it indeed gives . However, if this is the case then all have the same sign. If my normal vector has a minus sign, then . If I remove the minus sign, then . Shouldn't one of the principal curvatures have a different sign?

I believe the 3d hyperboloid is hyperbolic in some direction and spherical in another direction. As an example, for 2d hyperboloid the principal curvatures should be .

The only way for this to happen is if not .
 
  • #8
The spatial isotropy forces sectional curvature to be equal in any direction at a point (and the homogeneity means they are also the same for any point).
 
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  • #9
ergospherical said:
The spatial isotropy forces sectional curvature to be equal in direction at a point (and the homogeneity means they are also the same for any point).
Oh yeah! So the correct principal curvatures are ?

If this is the case then the Gaussian curvature will be which has the correct sign. If then which is the assumption from the start.
 

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