Mixing
“Arclength†parameterization of a surface
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Given an arbitrary parameterization of a differentiable curve $r: Rrightarrow R^n$ in terms of some variable $t$, we can re-parameterize (although perhaps not in closed-form) $r$ in terms of its arclength $s$. This gives a handy representation where as $s$ increases by 1, we move 1 unit along $r$ in $R^n$.
Is is 1) possible and 2) comparably straightforward to do this for surfaces? I.e. if $r: R^m rightarrow R^n$ is differentiable can we reparameterize $r$ so that moving 1 unit in $R^m$ moves us 1 unit along the surface $r$ in $R^n$?
manifolds parametric
asked 1 hour ago
Scott
1519
add a comment |Â
up vote
4
down vote
favorite
Given an arbitrary parameterization of a differentiable curve $r: Rrightarrow R^n$ in terms of some variable $t$, we can re-parameterize (although perhaps not in closed-form) $r$ in terms of its arclength $s$. This gives a handy representation where as $s$ increases by 1, we move 1 unit along $r$ in $R^n$.
Is is 1) possible and 2) comparably straightforward to do this for surfaces? I.e. if $r: R^m rightarrow R^n$ is differentiable can we reparameterize $r$ so that moving 1 unit in $R^m$ moves us 1 unit along the surface $r$ in $R^n$?
manifolds parametric
asked 1 hour ago
Scott
1519
I like to think of the Gauss map of an analogue of "arclength reparametrization" for a surface.
– Sobi
1 hour ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given an arbitrary parameterization of a differentiable curve $r: Rrightarrow R^n$ in terms of some variable $t$, we can re-parameterize (although perhaps not in closed-form) $r$ in terms of its arclength $s$. This gives a handy representation where as $s$ increases by 1, we move 1 unit along $r$ in $R^n$.
Is is 1) possible and 2) comparably straightforward to do this for surfaces? I.e. if $r: R^m rightarrow R^n$ is differentiable can we reparameterize $r$ so that moving 1 unit in $R^m$ moves us 1 unit along the surface $r$ in $R^n$?
manifolds parametric
asked 1 hour ago
Scott
1519
Given an arbitrary parameterization of a differentiable curve $r: Rrightarrow R^n$ in terms of some variable $t$, we can re-parameterize (although perhaps not in closed-form) $r$ in terms of its arclength $s$. This gives a handy representation where as $s$ increases by 1, we move 1 unit along $r$ in $R^n$.
Is is 1) possible and 2) comparably straightforward to do this for surfaces? I.e. if $r: R^m rightarrow R^n$ is differentiable can we reparameterize $r$ so that moving 1 unit in $R^m$ moves us 1 unit along the surface $r$ in $R^n$?
manifolds parametric
manifolds parametric
asked 1 hour ago
Scott
1519
asked 1 hour ago
Scott
1519
edited 40 mins ago
asked 1 hour ago
Scott
1519
asked 1 hour ago
Scott
1519
asked 1 hour ago
Scott
1519
1519
I like to think of the Gauss map of an analogue of "arclength reparametrization" for a surface.
– Sobi
1 hour ago
add a comment |Â
I like to think of the Gauss map of an analogue of "arclength reparametrization" for a surface.
– Sobi
1 hour ago
I like to think of the Gauss map of an analogue of "arclength reparametrization" for a surface.
– Sobi
1 hour ago
I like to think of the Gauss map of an analogue of "arclength reparametrization" for a surface.
– Sobi
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
No, this latter would mean that any manifold is locally isometric to an Euclidean space with its flat metric. This is only true in dimension $1$ by the mean of arclength parametrization as you pointed out, the first obstruction coming to my mind being the curvature.
answered 1 hour ago
Balloon
4,070521
The quantity of things that don't generalize neatly from 1-D to n-D causes me no end of amazement, delight and frustration.
– Scott
36 mins ago
I agree that the delight part is strong here: the article images.math.cnrs.fr/Un-theoreme-et-une-part-de-pizza (in french) explains curvature and how the fundamental theorem dealing with it helps us to properly eat pizzas.
– Balloon
28 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
No, this latter would mean that any manifold is locally isometric to an Euclidean space with its flat metric. This is only true in dimension $1$ by the mean of arclength parametrization as you pointed out, the first obstruction coming to my mind being the curvature.
answered 1 hour ago
Balloon
4,070521
The quantity of things that don't generalize neatly from 1-D to n-D causes me no end of amazement, delight and frustration.
– Scott
36 mins ago
I agree that the delight part is strong here: the article images.math.cnrs.fr/Un-theoreme-et-une-part-de-pizza (in french) explains curvature and how the fundamental theorem dealing with it helps us to properly eat pizzas.
– Balloon
28 mins ago
add a comment |Â
up vote
6
down vote
No, this latter would mean that any manifold is locally isometric to an Euclidean space with its flat metric. This is only true in dimension $1$ by the mean of arclength parametrization as you pointed out, the first obstruction coming to my mind being the curvature.
answered 1 hour ago
Balloon
4,070521
The quantity of things that don't generalize neatly from 1-D to n-D causes me no end of amazement, delight and frustration.
– Scott
36 mins ago
I agree that the delight part is strong here: the article images.math.cnrs.fr/Un-theoreme-et-une-part-de-pizza (in french) explains curvature and how the fundamental theorem dealing with it helps us to properly eat pizzas.
– Balloon
28 mins ago
add a comment |Â
up vote
6
down vote
up vote
6
down vote
No, this latter would mean that any manifold is locally isometric to an Euclidean space with its flat metric. This is only true in dimension $1$ by the mean of arclength parametrization as you pointed out, the first obstruction coming to my mind being the curvature.
answered 1 hour ago
Balloon
4,070521
No, this latter would mean that any manifold is locally isometric to an Euclidean space with its flat metric. This is only true in dimension $1$ by the mean of arclength parametrization as you pointed out, the first obstruction coming to my mind being the curvature.
answered 1 hour ago
Balloon
4,070521
answered 1 hour ago
Balloon
4,070521
answered 1 hour ago
Balloon
4,070521
answered 1 hour ago
Balloon
4,070521
4,070521
The quantity of things that don't generalize neatly from 1-D to n-D causes me no end of amazement, delight and frustration.
– Scott
36 mins ago
I agree that the delight part is strong here: the article images.math.cnrs.fr/Un-theoreme-et-une-part-de-pizza (in french) explains curvature and how the fundamental theorem dealing with it helps us to properly eat pizzas.
– Balloon
28 mins ago
add a comment |Â
The quantity of things that don't generalize neatly from 1-D to n-D causes me no end of amazement, delight and frustration.
– Scott
36 mins ago
I agree that the delight part is strong here: the article images.math.cnrs.fr/Un-theoreme-et-une-part-de-pizza (in french) explains curvature and how the fundamental theorem dealing with it helps us to properly eat pizzas.
– Balloon
28 mins ago
The quantity of things that don't generalize neatly from 1-D to n-D causes me no end of amazement, delight and frustration.
– Scott
36 mins ago
The quantity of things that don't generalize neatly from 1-D to n-D causes me no end of amazement, delight and frustration.
– Scott
36 mins ago
I agree that the delight part is strong here: the article images.math.cnrs.fr/Un-theoreme-et-une-part-de-pizza (in french) explains curvature and how the fundamental theorem dealing with it helps us to properly eat pizzas.
– Balloon
28 mins ago
I agree that the delight part is strong here: the article images.math.cnrs.fr/Un-theoreme-et-une-part-de-pizza (in french) explains curvature and how the fundamental theorem dealing with it helps us to properly eat pizzas.
– Balloon
28 mins ago
add a comment |Â
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I like to think of the Gauss map of an analogue of "arclength reparametrization" for a surface.
– Sobi
1 hour ago