Relationship between Gaussian, Normal and Geodesic Curvatures
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How do I show that the square of the gaussian curvature is the sum of the squares of the normal and geodesic curvatures other than the one shown in page 38 of http://www.maths.lancs.ac.uk/~belton/www/notes/geom_notes.pdf? That is,
$$kappa^2=kappa_n^2 + kappa_g^2$$
Is there another way?
differential-geometry
edited Mar 28 '13 at 10:40
Martin Sleziak
44.7k9117272
asked Mar 28 '13 at 9:17
anon102938anon102938
242
$endgroup$
add a comment |
$begingroup$
How do I show that the square of the gaussian curvature is the sum of the squares of the normal and geodesic curvatures other than the one shown in page 38 of http://www.maths.lancs.ac.uk/~belton/www/notes/geom_notes.pdf? That is,
$$kappa^2=kappa_n^2 + kappa_g^2$$
Is there another way?
differential-geometry
edited Mar 28 '13 at 10:40
Martin Sleziak
44.7k9117272
asked Mar 28 '13 at 9:17
anon102938anon102938
242
$endgroup$
$begingroup$
^May I ask how you proved it? Thanks.
$endgroup$
– user70735
Apr 3 '13 at 6:43
add a comment |
$begingroup$
How do I show that the square of the gaussian curvature is the sum of the squares of the normal and geodesic curvatures other than the one shown in page 38 of http://www.maths.lancs.ac.uk/~belton/www/notes/geom_notes.pdf? That is,
$$kappa^2=kappa_n^2 + kappa_g^2$$
Is there another way?
differential-geometry
edited Mar 28 '13 at 10:40
Martin Sleziak
44.7k9117272
asked Mar 28 '13 at 9:17
anon102938anon102938
242
$endgroup$
How do I show that the square of the gaussian curvature is the sum of the squares of the normal and geodesic curvatures other than the one shown in page 38 of http://www.maths.lancs.ac.uk/~belton/www/notes/geom_notes.pdf? That is,
$$kappa^2=kappa_n^2 + kappa_g^2$$
Is there another way?
differential-geometry
differential-geometry
edited Mar 28 '13 at 10:40
Martin Sleziak
44.7k9117272
asked Mar 28 '13 at 9:17
anon102938anon102938
242
edited Mar 28 '13 at 10:40
Martin Sleziak
44.7k9117272
asked Mar 28 '13 at 9:17
anon102938anon102938
242
edited Mar 28 '13 at 10:40
Martin Sleziak
44.7k9117272
edited Mar 28 '13 at 10:40
Martin Sleziak
44.7k9117272
edited Mar 28 '13 at 10:40
Martin Sleziak
44.7k9117272
44.7k9117272
asked Mar 28 '13 at 9:17
anon102938anon102938
242
asked Mar 28 '13 at 9:17
anon102938anon102938
242
asked Mar 28 '13 at 9:17
anon102938anon102938
242
242
$begingroup$
^May I ask how you proved it? Thanks.
$endgroup$
– user70735
Apr 3 '13 at 6:43
add a comment |
$begingroup$
^May I ask how you proved it? Thanks.
$endgroup$
– user70735
Apr 3 '13 at 6:43
$begingroup$
^May I ask how you proved it? Thanks.
$endgroup$
– user70735
Apr 3 '13 at 6:43
$begingroup$
^May I ask how you proved it? Thanks.
$endgroup$
– user70735
Apr 3 '13 at 6:43
add a comment |
1 Answer
1
active
oldest
votes
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The shape of a surface influences the curvature of curves on the surface.Let $S subset mathbb{R}^n$ a regular surface and $alpha:(a, b) rightarrow S$ a unit speed curve. If, for instance, $alpha$ lies on a plane or a cylinder,
it can have zero curvature everywhere. This is not possible for curves on a sphere since no segment of a straight line can lie on a sphere. Thus the natural way to investigate how much a $S$ curves, consists in looking at the curvature of curves on the surface. In order to make everything crystal clar, please consider the picture below.
The geodesic and normal curvatures of $alpha$ are related to the Gaussian curvature by the formula you shown above. Now by remembering that $alpha'$ is a unit vector and it is tangent to $S$ we can conclude that $alpha'$ is perpendiculat to the norlal $N$ of $S$.
Thus we have that $alpha'$, $N$, and $N times alpha^{'}$ are mutually perpendicular unit vectors. Since $alpha'$ is the unit speed, $alpha''$ is perpendicular to $alpha'$ and then it can be written as a linear combination of $N$ and $N times alpha^{'}$.
In other words we have:
begin{equation}
alpha^{''}(s)=kappa_n N + k_g N times alpha^{'}(s)
end{equation}
Where
begin{equation}
kappa_n =alpha^{''} cdot N
end{equation}
and
begin{equation}
kappa_g =alpha^{''} cdot (N times alpha^{'})
end{equation}
Now:
begin{equation}
kappa_alpha(s)^2=||alpha^{''}(s)||^2=||alpha^{''}(s)^{parallel}||^2+||alpha^{''}(s)^{perp}||^2
end{equation}
i.e. the parallel and perpedicular components of the acceleration of $alpha(s)$. By definition $alpha^{''}(s)^{parallel}$ is the component tangent to $S$ and $alpha^{''}(s)^{perp}$ s the component parallel to a normal vector to the surface.
answered Mar 28 '16 at 12:42
UpaxUpax
1,507613
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The shape of a surface influences the curvature of curves on the surface.Let $S subset mathbb{R}^n$ a regular surface and $alpha:(a, b) rightarrow S$ a unit speed curve. If, for instance, $alpha$ lies on a plane or a cylinder,
it can have zero curvature everywhere. This is not possible for curves on a sphere since no segment of a straight line can lie on a sphere. Thus the natural way to investigate how much a $S$ curves, consists in looking at the curvature of curves on the surface. In order to make everything crystal clar, please consider the picture below.
The geodesic and normal curvatures of $alpha$ are related to the Gaussian curvature by the formula you shown above. Now by remembering that $alpha'$ is a unit vector and it is tangent to $S$ we can conclude that $alpha'$ is perpendiculat to the norlal $N$ of $S$.
Thus we have that $alpha'$, $N$, and $N times alpha^{'}$ are mutually perpendicular unit vectors. Since $alpha'$ is the unit speed, $alpha''$ is perpendicular to $alpha'$ and then it can be written as a linear combination of $N$ and $N times alpha^{'}$.
In other words we have:
begin{equation}
alpha^{''}(s)=kappa_n N + k_g N times alpha^{'}(s)
end{equation}
Where
begin{equation}
kappa_n =alpha^{''} cdot N
end{equation}
and
begin{equation}
kappa_g =alpha^{''} cdot (N times alpha^{'})
end{equation}
Now:
begin{equation}
kappa_alpha(s)^2=||alpha^{''}(s)||^2=||alpha^{''}(s)^{parallel}||^2+||alpha^{''}(s)^{perp}||^2
end{equation}
i.e. the parallel and perpedicular components of the acceleration of $alpha(s)$. By definition $alpha^{''}(s)^{parallel}$ is the component tangent to $S$ and $alpha^{''}(s)^{perp}$ s the component parallel to a normal vector to the surface.
answered Mar 28 '16 at 12:42
UpaxUpax
1,507613
$endgroup$
add a comment |
$begingroup$
The shape of a surface influences the curvature of curves on the surface.Let $S subset mathbb{R}^n$ a regular surface and $alpha:(a, b) rightarrow S$ a unit speed curve. If, for instance, $alpha$ lies on a plane or a cylinder,
it can have zero curvature everywhere. This is not possible for curves on a sphere since no segment of a straight line can lie on a sphere. Thus the natural way to investigate how much a $S$ curves, consists in looking at the curvature of curves on the surface. In order to make everything crystal clar, please consider the picture below.
The geodesic and normal curvatures of $alpha$ are related to the Gaussian curvature by the formula you shown above. Now by remembering that $alpha'$ is a unit vector and it is tangent to $S$ we can conclude that $alpha'$ is perpendiculat to the norlal $N$ of $S$.
Thus we have that $alpha'$, $N$, and $N times alpha^{'}$ are mutually perpendicular unit vectors. Since $alpha'$ is the unit speed, $alpha''$ is perpendicular to $alpha'$ and then it can be written as a linear combination of $N$ and $N times alpha^{'}$.
In other words we have:
begin{equation}
alpha^{''}(s)=kappa_n N + k_g N times alpha^{'}(s)
end{equation}
Where
begin{equation}
kappa_n =alpha^{''} cdot N
end{equation}
and
begin{equation}
kappa_g =alpha^{''} cdot (N times alpha^{'})
end{equation}
Now:
begin{equation}
kappa_alpha(s)^2=||alpha^{''}(s)||^2=||alpha^{''}(s)^{parallel}||^2+||alpha^{''}(s)^{perp}||^2
end{equation}
i.e. the parallel and perpedicular components of the acceleration of $alpha(s)$. By definition $alpha^{''}(s)^{parallel}$ is the component tangent to $S$ and $alpha^{''}(s)^{perp}$ s the component parallel to a normal vector to the surface.
answered Mar 28 '16 at 12:42
UpaxUpax
1,507613
$endgroup$
add a comment |
$begingroup$
The shape of a surface influences the curvature of curves on the surface.Let $S subset mathbb{R}^n$ a regular surface and $alpha:(a, b) rightarrow S$ a unit speed curve. If, for instance, $alpha$ lies on a plane or a cylinder,
it can have zero curvature everywhere. This is not possible for curves on a sphere since no segment of a straight line can lie on a sphere. Thus the natural way to investigate how much a $S$ curves, consists in looking at the curvature of curves on the surface. In order to make everything crystal clar, please consider the picture below.
The geodesic and normal curvatures of $alpha$ are related to the Gaussian curvature by the formula you shown above. Now by remembering that $alpha'$ is a unit vector and it is tangent to $S$ we can conclude that $alpha'$ is perpendiculat to the norlal $N$ of $S$.
Thus we have that $alpha'$, $N$, and $N times alpha^{'}$ are mutually perpendicular unit vectors. Since $alpha'$ is the unit speed, $alpha''$ is perpendicular to $alpha'$ and then it can be written as a linear combination of $N$ and $N times alpha^{'}$.
In other words we have:
begin{equation}
alpha^{''}(s)=kappa_n N + k_g N times alpha^{'}(s)
end{equation}
Where
begin{equation}
kappa_n =alpha^{''} cdot N
end{equation}
and
begin{equation}
kappa_g =alpha^{''} cdot (N times alpha^{'})
end{equation}
Now:
begin{equation}
kappa_alpha(s)^2=||alpha^{''}(s)||^2=||alpha^{''}(s)^{parallel}||^2+||alpha^{''}(s)^{perp}||^2
end{equation}
i.e. the parallel and perpedicular components of the acceleration of $alpha(s)$. By definition $alpha^{''}(s)^{parallel}$ is the component tangent to $S$ and $alpha^{''}(s)^{perp}$ s the component parallel to a normal vector to the surface.
answered Mar 28 '16 at 12:42
UpaxUpax
1,507613
$endgroup$
The shape of a surface influences the curvature of curves on the surface.Let $S subset mathbb{R}^n$ a regular surface and $alpha:(a, b) rightarrow S$ a unit speed curve. If, for instance, $alpha$ lies on a plane or a cylinder,
it can have zero curvature everywhere. This is not possible for curves on a sphere since no segment of a straight line can lie on a sphere. Thus the natural way to investigate how much a $S$ curves, consists in looking at the curvature of curves on the surface. In order to make everything crystal clar, please consider the picture below.
The geodesic and normal curvatures of $alpha$ are related to the Gaussian curvature by the formula you shown above. Now by remembering that $alpha'$ is a unit vector and it is tangent to $S$ we can conclude that $alpha'$ is perpendiculat to the norlal $N$ of $S$.
Thus we have that $alpha'$, $N$, and $N times alpha^{'}$ are mutually perpendicular unit vectors. Since $alpha'$ is the unit speed, $alpha''$ is perpendicular to $alpha'$ and then it can be written as a linear combination of $N$ and $N times alpha^{'}$.
In other words we have:
begin{equation}
alpha^{''}(s)=kappa_n N + k_g N times alpha^{'}(s)
end{equation}
Where
begin{equation}
kappa_n =alpha^{''} cdot N
end{equation}
and
begin{equation}
kappa_g =alpha^{''} cdot (N times alpha^{'})
end{equation}
Now:
begin{equation}
kappa_alpha(s)^2=||alpha^{''}(s)||^2=||alpha^{''}(s)^{parallel}||^2+||alpha^{''}(s)^{perp}||^2
end{equation}
i.e. the parallel and perpedicular components of the acceleration of $alpha(s)$. By definition $alpha^{''}(s)^{parallel}$ is the component tangent to $S$ and $alpha^{''}(s)^{perp}$ s the component parallel to a normal vector to the surface.
answered Mar 28 '16 at 12:42
UpaxUpax
1,507613
edited Apr 10 '16 at 17:45
answered Mar 28 '16 at 12:42
UpaxUpax
1,507613
answered Mar 28 '16 at 12:42
UpaxUpax
1,507613
answered Mar 28 '16 at 12:42
UpaxUpax
1,507613
1,507613
add a comment |
add a comment |
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$begingroup$
^May I ask how you proved it? Thanks.
$endgroup$
– user70735
Apr 3 '13 at 6:43