Given a family of 2D curves, find a 3D manifold whose geodesics project to the plane curves
Clash Royale CLAN TAG#URR8PPP
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Below is an image of the family of 2D curves for reference. The image is arbitrary. Still trying to formulate a concise question.
I know that the geodesics for flat Euclidean Space are straight lines. But I want to take these curved geodesics and tried to work backwards to determine the geometry of the manifold that results from these geodesic paths going across it. I can't really visualise the manifold because there's a lot going on. So how would I build up a representation of the manifold that matches up with these geodesics running across it? What would the steps be to calculate the metric of this space? Is there enough information given by the geodesics to provide a good representation of the manifold? Any other interesting questions you could ask about this graph? What topological information could I learn from these geodesics?
general-topology differential-geometry manifolds general-relativity manifolds-with-boundary
edited May 30 at 1:25
Narasimham
20.3k51957
asked May 27 at 22:20
George Thomas
95417
 |Â
show 4 more comments
up vote
7
down vote
favorite
Below is an image of the family of 2D curves for reference. The image is arbitrary. Still trying to formulate a concise question.
I know that the geodesics for flat Euclidean Space are straight lines. But I want to take these curved geodesics and tried to work backwards to determine the geometry of the manifold that results from these geodesic paths going across it. I can't really visualise the manifold because there's a lot going on. So how would I build up a representation of the manifold that matches up with these geodesics running across it? What would the steps be to calculate the metric of this space? Is there enough information given by the geodesics to provide a good representation of the manifold? Any other interesting questions you could ask about this graph? What topological information could I learn from these geodesics?
general-topology differential-geometry manifolds general-relativity manifolds-with-boundary
edited May 30 at 1:25
Narasimham
20.3k51957
asked May 27 at 22:20
George Thomas
95417
"Project" in what sense? From $(x,y,z)$ to $(x,y)$?
– mr_e_man
May 27 at 22:28
Yeah let's go with that
– George Thomas
May 27 at 22:33
1
Geodesics' curvature/acceleration vector is always perpendicular to the surface. So you can try to visualize the surface in 3D from this 2D picture by having its normal vector point in the same general direction as the geodesics' curvature. I don't think that's enough information, though.
– mr_e_man
May 27 at 22:39
How did you get these curves? Is the pictured plane supposed to be the $(u,v)$ plane, for some coordinate system on the surface? That's different from orthogonal projection.
– mr_e_man
May 27 at 22:42
1
Only a hunch this.. geodesics on a minimal surface film formed between two pipes cutting orthogonally may produce such geodesics. The geodesics around minimum radii should be tangential to a common circle of the pipe.
– Narasimham
May 30 at 9:42
 |Â
show 4 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Below is an image of the family of 2D curves for reference. The image is arbitrary. Still trying to formulate a concise question.
I know that the geodesics for flat Euclidean Space are straight lines. But I want to take these curved geodesics and tried to work backwards to determine the geometry of the manifold that results from these geodesic paths going across it. I can't really visualise the manifold because there's a lot going on. So how would I build up a representation of the manifold that matches up with these geodesics running across it? What would the steps be to calculate the metric of this space? Is there enough information given by the geodesics to provide a good representation of the manifold? Any other interesting questions you could ask about this graph? What topological information could I learn from these geodesics?
general-topology differential-geometry manifolds general-relativity manifolds-with-boundary
edited May 30 at 1:25
Narasimham
20.3k51957
asked May 27 at 22:20
George Thomas
95417
Below is an image of the family of 2D curves for reference. The image is arbitrary. Still trying to formulate a concise question.
I know that the geodesics for flat Euclidean Space are straight lines. But I want to take these curved geodesics and tried to work backwards to determine the geometry of the manifold that results from these geodesic paths going across it. I can't really visualise the manifold because there's a lot going on. So how would I build up a representation of the manifold that matches up with these geodesics running across it? What would the steps be to calculate the metric of this space? Is there enough information given by the geodesics to provide a good representation of the manifold? Any other interesting questions you could ask about this graph? What topological information could I learn from these geodesics?
general-topology differential-geometry manifolds general-relativity manifolds-with-boundary
edited May 30 at 1:25
Narasimham
20.3k51957
asked May 27 at 22:20
George Thomas
95417
edited May 30 at 1:25
Narasimham
20.3k51957
edited May 30 at 1:25
Narasimham
20.3k51957
edited May 30 at 1:25
Narasimham
20.3k51957
20.3k51957
asked May 27 at 22:20
George Thomas
95417
asked May 27 at 22:20
George Thomas
95417
asked May 27 at 22:20
George Thomas
95417
95417
"Project" in what sense? From $(x,y,z)$ to $(x,y)$?
– mr_e_man
May 27 at 22:28
Yeah let's go with that
– George Thomas
May 27 at 22:33
1
Geodesics' curvature/acceleration vector is always perpendicular to the surface. So you can try to visualize the surface in 3D from this 2D picture by having its normal vector point in the same general direction as the geodesics' curvature. I don't think that's enough information, though.
– mr_e_man
May 27 at 22:39
How did you get these curves? Is the pictured plane supposed to be the $(u,v)$ plane, for some coordinate system on the surface? That's different from orthogonal projection.
– mr_e_man
May 27 at 22:42
1
Only a hunch this.. geodesics on a minimal surface film formed between two pipes cutting orthogonally may produce such geodesics. The geodesics around minimum radii should be tangential to a common circle of the pipe.
– Narasimham
May 30 at 9:42
 |Â
show 4 more comments
"Project" in what sense? From $(x,y,z)$ to $(x,y)$?
– mr_e_man
May 27 at 22:28
Yeah let's go with that
– George Thomas
May 27 at 22:33
1
Geodesics' curvature/acceleration vector is always perpendicular to the surface. So you can try to visualize the surface in 3D from this 2D picture by having its normal vector point in the same general direction as the geodesics' curvature. I don't think that's enough information, though.
– mr_e_man
May 27 at 22:39
How did you get these curves? Is the pictured plane supposed to be the $(u,v)$ plane, for some coordinate system on the surface? That's different from orthogonal projection.
– mr_e_man
May 27 at 22:42
1
Only a hunch this.. geodesics on a minimal surface film formed between two pipes cutting orthogonally may produce such geodesics. The geodesics around minimum radii should be tangential to a common circle of the pipe.
– Narasimham
May 30 at 9:42
"Project" in what sense? From $(x,y,z)$ to $(x,y)$?
– mr_e_man
May 27 at 22:28
"Project" in what sense? From $(x,y,z)$ to $(x,y)$?
– mr_e_man
May 27 at 22:28
Yeah let's go with that
– George Thomas
May 27 at 22:33
Yeah let's go with that
– George Thomas
May 27 at 22:33
1
1
Geodesics' curvature/acceleration vector is always perpendicular to the surface. So you can try to visualize the surface in 3D from this 2D picture by having its normal vector point in the same general direction as the geodesics' curvature. I don't think that's enough information, though.
– mr_e_man
May 27 at 22:39
Geodesics' curvature/acceleration vector is always perpendicular to the surface. So you can try to visualize the surface in 3D from this 2D picture by having its normal vector point in the same general direction as the geodesics' curvature. I don't think that's enough information, though.
– mr_e_man
May 27 at 22:39
How did you get these curves? Is the pictured plane supposed to be the $(u,v)$ plane, for some coordinate system on the surface? That's different from orthogonal projection.
– mr_e_man
May 27 at 22:42
How did you get these curves? Is the pictured plane supposed to be the $(u,v)$ plane, for some coordinate system on the surface? That's different from orthogonal projection.
– mr_e_man
May 27 at 22:42
1
1
Only a hunch this.. geodesics on a minimal surface film formed between two pipes cutting orthogonally may produce such geodesics. The geodesics around minimum radii should be tangential to a common circle of the pipe.
– Narasimham
May 30 at 9:42
Only a hunch this.. geodesics on a minimal surface film formed between two pipes cutting orthogonally may produce such geodesics. The geodesics around minimum radii should be tangential to a common circle of the pipe.
– Narasimham
May 30 at 9:42
 |Â
show 4 more comments
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"Project" in what sense? From $(x,y,z)$ to $(x,y)$?
– mr_e_man
May 27 at 22:28
Yeah let's go with that
– George Thomas
May 27 at 22:33
1
Geodesics' curvature/acceleration vector is always perpendicular to the surface. So you can try to visualize the surface in 3D from this 2D picture by having its normal vector point in the same general direction as the geodesics' curvature. I don't think that's enough information, though.
– mr_e_man
May 27 at 22:39
How did you get these curves? Is the pictured plane supposed to be the $(u,v)$ plane, for some coordinate system on the surface? That's different from orthogonal projection.
– mr_e_man
May 27 at 22:42
1
Only a hunch this.. geodesics on a minimal surface film formed between two pipes cutting orthogonally may produce such geodesics. The geodesics around minimum radii should be tangential to a common circle of the pipe.
– Narasimham
May 30 at 9:42