Geodesic on plane without disc around $0$
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Given $M := mathbb R^2 setminus B_r(0)$ I want to determine the function
$$rho(x,y) = inf { int_a^b |gamma'(t)|,dt : gamma : [a,b] to M text{ is a smooth curve connecting } x text{ with }y }$$
on $M$.
I understand that for $overline{xy} := {rx + (1-r)y : rin [0,1]}$ not intersecting $B_r(0)$ we have $rho(x,y) = |x-y|$.
Otherwise I don't know how to procede, not even intuitively. When I think of connecting $x$ and $y$ then, I'm not sure whether to do this with two line segments or whether I should use something "curved".
Even if I could somehow prove that among "piecewise polynomial curves" only certain curves are "minimal", I wouldn't know why "piecewise polynomial curves" are the best ones to begin with.
What can I do?
multivariable-calculus differential-geometry
asked Nov 17 at 13:50
Jinx
154
add a comment |
up vote
0
down vote
favorite
Given $M := mathbb R^2 setminus B_r(0)$ I want to determine the function
$$rho(x,y) = inf { int_a^b |gamma'(t)|,dt : gamma : [a,b] to M text{ is a smooth curve connecting } x text{ with }y }$$
on $M$.
I understand that for $overline{xy} := {rx + (1-r)y : rin [0,1]}$ not intersecting $B_r(0)$ we have $rho(x,y) = |x-y|$.
Otherwise I don't know how to procede, not even intuitively. When I think of connecting $x$ and $y$ then, I'm not sure whether to do this with two line segments or whether I should use something "curved".
Even if I could somehow prove that among "piecewise polynomial curves" only certain curves are "minimal", I wouldn't know why "piecewise polynomial curves" are the best ones to begin with.
What can I do?
multivariable-calculus differential-geometry
asked Nov 17 at 13:50
Jinx
154
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $M := mathbb R^2 setminus B_r(0)$ I want to determine the function
$$rho(x,y) = inf { int_a^b |gamma'(t)|,dt : gamma : [a,b] to M text{ is a smooth curve connecting } x text{ with }y }$$
on $M$.
I understand that for $overline{xy} := {rx + (1-r)y : rin [0,1]}$ not intersecting $B_r(0)$ we have $rho(x,y) = |x-y|$.
Otherwise I don't know how to procede, not even intuitively. When I think of connecting $x$ and $y$ then, I'm not sure whether to do this with two line segments or whether I should use something "curved".
Even if I could somehow prove that among "piecewise polynomial curves" only certain curves are "minimal", I wouldn't know why "piecewise polynomial curves" are the best ones to begin with.
What can I do?
multivariable-calculus differential-geometry
asked Nov 17 at 13:50
Jinx
154
Given $M := mathbb R^2 setminus B_r(0)$ I want to determine the function
$$rho(x,y) = inf { int_a^b |gamma'(t)|,dt : gamma : [a,b] to M text{ is a smooth curve connecting } x text{ with }y }$$
on $M$.
I understand that for $overline{xy} := {rx + (1-r)y : rin [0,1]}$ not intersecting $B_r(0)$ we have $rho(x,y) = |x-y|$.
Otherwise I don't know how to procede, not even intuitively. When I think of connecting $x$ and $y$ then, I'm not sure whether to do this with two line segments or whether I should use something "curved".
Even if I could somehow prove that among "piecewise polynomial curves" only certain curves are "minimal", I wouldn't know why "piecewise polynomial curves" are the best ones to begin with.
What can I do?
multivariable-calculus differential-geometry
multivariable-calculus differential-geometry
asked Nov 17 at 13:50
Jinx
154
asked Nov 17 at 13:50
Jinx
154
asked Nov 17 at 13:50
Jinx
154
asked Nov 17 at 13:50
Jinx
154
asked Nov 17 at 13:50
Jinx
154
154
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
In the case of intersecting; the shorter path is constructed with the tangents to the disk from each point and the arc between the points of tangency.
Consider moving $F$ on the boundary towards $C$ (but still on the boundary), clearly the new path passes by $F$ too but deviates from the previous straight line connecting $C$ and $F$, the shortest path. Now, move $F$ in the opposite sense that of course cannot be on the border. All the figure is displaced outwards making the path larger. From here the total distance can be calculated.
To prove it analytically doesn't seem difficult.
answered Nov 17 at 16:02
Rafa Budría
5,4001825
Thanks for answering. - Unfortunately I probably have to ask new question because I asked the wrong question: I wanted $B_r$ to refer to the closed ball...
– Jinx
Nov 17 at 19:48
I think in this case there is no shortest path.
– Rafa Budría
Nov 17 at 19:54
So, would $rho(x,y)$ be the length of the path in your picture, nonetheless?
– Jinx
Nov 17 at 19:55
The length of the path in the drawing is the greater lower bound of the length of all possible paths.
– Rafa Budría
Nov 17 at 20:12
I'll take this as a 'yes' ;)
– Jinx
Nov 17 at 21:07
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In the case of intersecting; the shorter path is constructed with the tangents to the disk from each point and the arc between the points of tangency.
Consider moving $F$ on the boundary towards $C$ (but still on the boundary), clearly the new path passes by $F$ too but deviates from the previous straight line connecting $C$ and $F$, the shortest path. Now, move $F$ in the opposite sense that of course cannot be on the border. All the figure is displaced outwards making the path larger. From here the total distance can be calculated.
To prove it analytically doesn't seem difficult.
answered Nov 17 at 16:02
Rafa Budría
5,4001825
Thanks for answering. - Unfortunately I probably have to ask new question because I asked the wrong question: I wanted $B_r$ to refer to the closed ball...
– Jinx
Nov 17 at 19:48
I think in this case there is no shortest path.
– Rafa Budría
Nov 17 at 19:54
So, would $rho(x,y)$ be the length of the path in your picture, nonetheless?
– Jinx
Nov 17 at 19:55
The length of the path in the drawing is the greater lower bound of the length of all possible paths.
– Rafa Budría
Nov 17 at 20:12
I'll take this as a 'yes' ;)
– Jinx
Nov 17 at 21:07
|
show 3 more comments
up vote
2
down vote
accepted
In the case of intersecting; the shorter path is constructed with the tangents to the disk from each point and the arc between the points of tangency.
Consider moving $F$ on the boundary towards $C$ (but still on the boundary), clearly the new path passes by $F$ too but deviates from the previous straight line connecting $C$ and $F$, the shortest path. Now, move $F$ in the opposite sense that of course cannot be on the border. All the figure is displaced outwards making the path larger. From here the total distance can be calculated.
To prove it analytically doesn't seem difficult.
answered Nov 17 at 16:02
Rafa Budría
5,4001825
Thanks for answering. - Unfortunately I probably have to ask new question because I asked the wrong question: I wanted $B_r$ to refer to the closed ball...
– Jinx
Nov 17 at 19:48
I think in this case there is no shortest path.
– Rafa Budría
Nov 17 at 19:54
So, would $rho(x,y)$ be the length of the path in your picture, nonetheless?
– Jinx
Nov 17 at 19:55
The length of the path in the drawing is the greater lower bound of the length of all possible paths.
– Rafa Budría
Nov 17 at 20:12
I'll take this as a 'yes' ;)
– Jinx
Nov 17 at 21:07
|
show 3 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In the case of intersecting; the shorter path is constructed with the tangents to the disk from each point and the arc between the points of tangency.
Consider moving $F$ on the boundary towards $C$ (but still on the boundary), clearly the new path passes by $F$ too but deviates from the previous straight line connecting $C$ and $F$, the shortest path. Now, move $F$ in the opposite sense that of course cannot be on the border. All the figure is displaced outwards making the path larger. From here the total distance can be calculated.
To prove it analytically doesn't seem difficult.
answered Nov 17 at 16:02
Rafa Budría
5,4001825
In the case of intersecting; the shorter path is constructed with the tangents to the disk from each point and the arc between the points of tangency.
Consider moving $F$ on the boundary towards $C$ (but still on the boundary), clearly the new path passes by $F$ too but deviates from the previous straight line connecting $C$ and $F$, the shortest path. Now, move $F$ in the opposite sense that of course cannot be on the border. All the figure is displaced outwards making the path larger. From here the total distance can be calculated.
To prove it analytically doesn't seem difficult.
answered Nov 17 at 16:02
Rafa Budría
5,4001825
edited Nov 17 at 16:21
answered Nov 17 at 16:02
Rafa Budría
5,4001825
answered Nov 17 at 16:02
Rafa Budría
5,4001825
answered Nov 17 at 16:02
Rafa Budría
5,4001825
5,4001825
Thanks for answering. - Unfortunately I probably have to ask new question because I asked the wrong question: I wanted $B_r$ to refer to the closed ball...
– Jinx
Nov 17 at 19:48
I think in this case there is no shortest path.
– Rafa Budría
Nov 17 at 19:54
So, would $rho(x,y)$ be the length of the path in your picture, nonetheless?
– Jinx
Nov 17 at 19:55
The length of the path in the drawing is the greater lower bound of the length of all possible paths.
– Rafa Budría
Nov 17 at 20:12
I'll take this as a 'yes' ;)
– Jinx
Nov 17 at 21:07
|
show 3 more comments
Thanks for answering. - Unfortunately I probably have to ask new question because I asked the wrong question: I wanted $B_r$ to refer to the closed ball...
– Jinx
Nov 17 at 19:48
I think in this case there is no shortest path.
– Rafa Budría
Nov 17 at 19:54
So, would $rho(x,y)$ be the length of the path in your picture, nonetheless?
– Jinx
Nov 17 at 19:55
The length of the path in the drawing is the greater lower bound of the length of all possible paths.
– Rafa Budría
Nov 17 at 20:12
I'll take this as a 'yes' ;)
– Jinx
Nov 17 at 21:07
Thanks for answering. - Unfortunately I probably have to ask new question because I asked the wrong question: I wanted $B_r$ to refer to the closed ball...
– Jinx
Nov 17 at 19:48
Thanks for answering. - Unfortunately I probably have to ask new question because I asked the wrong question: I wanted $B_r$ to refer to the closed ball...
– Jinx
Nov 17 at 19:48
I think in this case there is no shortest path.
– Rafa Budría
Nov 17 at 19:54
I think in this case there is no shortest path.
– Rafa Budría
Nov 17 at 19:54
So, would $rho(x,y)$ be the length of the path in your picture, nonetheless?
– Jinx
Nov 17 at 19:55
So, would $rho(x,y)$ be the length of the path in your picture, nonetheless?
– Jinx
Nov 17 at 19:55
The length of the path in the drawing is the greater lower bound of the length of all possible paths.
– Rafa Budría
Nov 17 at 20:12
The length of the path in the drawing is the greater lower bound of the length of all possible paths.
– Rafa Budría
Nov 17 at 20:12
I'll take this as a 'yes' ;)
– Jinx
Nov 17 at 21:07
I'll take this as a 'yes' ;)
– Jinx
Nov 17 at 21:07
|
show 3 more comments
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