Orthogonality of principal curvature directions
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I was trying to find a counterexample to the theorem that states that principal curvature directions are orthogonal.
Obviously such an example doesn't exist but I'm having hard time understanding what's wrong with the following example.
I construct the surface as follows.
I take a planar curve, namely a parabola $a (r^2)$ and revolve it through the $Z$ axis while adjusting the parameter $a$ that controls the curvature of the parabola. The parametrization is given by:
$$ left [X = rcos(theta), quad quad Y = rsin(theta), quad quad Z = ar^2 right],$$
$$a(theta)=0.5(1 + sin(4theta)),$$
where the parametric domain is
$$ theta in [0,pi], quad quad r in [-0.5,0.5].$$
The "problematic" point is the origin $(0,0,0)$.
See a Matlab rendering of the surface in $3D$:
For $theta in {3pi/8,7pi/8}$ the $sin(4theta)$ in the expression for $a(theta)$ is at minimum ($-1$) and we get $a=0$.
The restriction of $theta$ to these two values corresponds to two orthogonal
straight lines on the surface (green in the figure).
Hence $kappa_2 = 0$.
For $theta in {pi/8,5pi/8}$ the $sin(4theta)$ in the expression for $a(theta)$ is at maximum ($1$) and we get $a=1$.
The restriction of $theta$ to these two values corresponds to two orthogonal parabolas on the surface (red in the figure) with $kappa_1=2$.
Here is a top view of the $XY$ plane:
The "contradiction" seems to be that the curves with minimal and maximal curvatures at the origin are attained at $45$ degrees rather than $90$.
differential-geometry
asked Nov 13 at 18:20
Wazowski
8718
add a comment |
up vote
1
down vote
favorite
I was trying to find a counterexample to the theorem that states that principal curvature directions are orthogonal.
Obviously such an example doesn't exist but I'm having hard time understanding what's wrong with the following example.
I construct the surface as follows.
I take a planar curve, namely a parabola $a (r^2)$ and revolve it through the $Z$ axis while adjusting the parameter $a$ that controls the curvature of the parabola. The parametrization is given by:
$$ left [X = rcos(theta), quad quad Y = rsin(theta), quad quad Z = ar^2 right],$$
$$a(theta)=0.5(1 + sin(4theta)),$$
where the parametric domain is
$$ theta in [0,pi], quad quad r in [-0.5,0.5].$$
The "problematic" point is the origin $(0,0,0)$.
See a Matlab rendering of the surface in $3D$:
For $theta in {3pi/8,7pi/8}$ the $sin(4theta)$ in the expression for $a(theta)$ is at minimum ($-1$) and we get $a=0$.
The restriction of $theta$ to these two values corresponds to two orthogonal
straight lines on the surface (green in the figure).
Hence $kappa_2 = 0$.
For $theta in {pi/8,5pi/8}$ the $sin(4theta)$ in the expression for $a(theta)$ is at maximum ($1$) and we get $a=1$.
The restriction of $theta$ to these two values corresponds to two orthogonal parabolas on the surface (red in the figure) with $kappa_1=2$.
Here is a top view of the $XY$ plane:
The "contradiction" seems to be that the curves with minimal and maximal curvatures at the origin are attained at $45$ degrees rather than $90$.
differential-geometry
asked Nov 13 at 18:20
Wazowski
8718
First of all, $theta=-pi/8$ and $theta=15pi/8$ are the same (mod $2pi$), hence the same direction in the plane. Your description is also inaccurate, when you suggest you're looking at a surface of revolution. You're certainly not. Can you please delete your duplicate $theta$ values and specify exactly what point we're looking at? The lines in the surface will only given principal directions when you have a point with $K=0$. I don't see what the problem actually is.
– Ted Shifrin
Nov 13 at 18:41
Haven't checked the details, but I think what's going on is that your function $Z$ (and thus your surface) is not twice-differentiable at the origin. Thus the directional second derivatives of $Z$ at the origin are not described by a Hessian matrix and the geodesic curvatures are not described by a second fundamental form; so you can't apply orthonormal diagonalization.
– Anthony Carapetis
2 days ago
The Hessian of $Z$ is $ left[ begin {array}{cc} 1+sin(4theta) & 4cos( 4theta) r\ 4cos(4theta) r & -8sin(4theta)r^2 end {array} right]$ so $Z$ is twice differentiable.
– Wazowski
2 days ago
I meant $Z$ as a function of $X,Y$, not $r,theta,$ but more importantly: existence of the Hessian/second partial derivatives (which is true for $Z(X,Y)$) does not imply twice-differentiability in the strong sense.
– Anthony Carapetis
2 days ago
I edited the question based on Ted Shifrin suggestions. Tried to clarify and simplify things and added some illustrations.
– Wazowski
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was trying to find a counterexample to the theorem that states that principal curvature directions are orthogonal.
Obviously such an example doesn't exist but I'm having hard time understanding what's wrong with the following example.
I construct the surface as follows.
I take a planar curve, namely a parabola $a (r^2)$ and revolve it through the $Z$ axis while adjusting the parameter $a$ that controls the curvature of the parabola. The parametrization is given by:
$$ left [X = rcos(theta), quad quad Y = rsin(theta), quad quad Z = ar^2 right],$$
$$a(theta)=0.5(1 + sin(4theta)),$$
where the parametric domain is
$$ theta in [0,pi], quad quad r in [-0.5,0.5].$$
The "problematic" point is the origin $(0,0,0)$.
See a Matlab rendering of the surface in $3D$:
For $theta in {3pi/8,7pi/8}$ the $sin(4theta)$ in the expression for $a(theta)$ is at minimum ($-1$) and we get $a=0$.
The restriction of $theta$ to these two values corresponds to two orthogonal
straight lines on the surface (green in the figure).
Hence $kappa_2 = 0$.
For $theta in {pi/8,5pi/8}$ the $sin(4theta)$ in the expression for $a(theta)$ is at maximum ($1$) and we get $a=1$.
The restriction of $theta$ to these two values corresponds to two orthogonal parabolas on the surface (red in the figure) with $kappa_1=2$.
Here is a top view of the $XY$ plane:
The "contradiction" seems to be that the curves with minimal and maximal curvatures at the origin are attained at $45$ degrees rather than $90$.
differential-geometry
asked Nov 13 at 18:20
Wazowski
8718
I was trying to find a counterexample to the theorem that states that principal curvature directions are orthogonal.
Obviously such an example doesn't exist but I'm having hard time understanding what's wrong with the following example.
I construct the surface as follows.
I take a planar curve, namely a parabola $a (r^2)$ and revolve it through the $Z$ axis while adjusting the parameter $a$ that controls the curvature of the parabola. The parametrization is given by:
$$ left [X = rcos(theta), quad quad Y = rsin(theta), quad quad Z = ar^2 right],$$
$$a(theta)=0.5(1 + sin(4theta)),$$
where the parametric domain is
$$ theta in [0,pi], quad quad r in [-0.5,0.5].$$
The "problematic" point is the origin $(0,0,0)$.
See a Matlab rendering of the surface in $3D$:
For $theta in {3pi/8,7pi/8}$ the $sin(4theta)$ in the expression for $a(theta)$ is at minimum ($-1$) and we get $a=0$.
The restriction of $theta$ to these two values corresponds to two orthogonal
straight lines on the surface (green in the figure).
Hence $kappa_2 = 0$.
For $theta in {pi/8,5pi/8}$ the $sin(4theta)$ in the expression for $a(theta)$ is at maximum ($1$) and we get $a=1$.
The restriction of $theta$ to these two values corresponds to two orthogonal parabolas on the surface (red in the figure) with $kappa_1=2$.
Here is a top view of the $XY$ plane:
The "contradiction" seems to be that the curves with minimal and maximal curvatures at the origin are attained at $45$ degrees rather than $90$.
differential-geometry
differential-geometry
asked Nov 13 at 18:20
Wazowski
8718
asked Nov 13 at 18:20
Wazowski
8718
edited yesterday
asked Nov 13 at 18:20
Wazowski
8718
asked Nov 13 at 18:20
Wazowski
8718
asked Nov 13 at 18:20
Wazowski
8718
8718
First of all, $theta=-pi/8$ and $theta=15pi/8$ are the same (mod $2pi$), hence the same direction in the plane. Your description is also inaccurate, when you suggest you're looking at a surface of revolution. You're certainly not. Can you please delete your duplicate $theta$ values and specify exactly what point we're looking at? The lines in the surface will only given principal directions when you have a point with $K=0$. I don't see what the problem actually is.
– Ted Shifrin
Nov 13 at 18:41
Haven't checked the details, but I think what's going on is that your function $Z$ (and thus your surface) is not twice-differentiable at the origin. Thus the directional second derivatives of $Z$ at the origin are not described by a Hessian matrix and the geodesic curvatures are not described by a second fundamental form; so you can't apply orthonormal diagonalization.
– Anthony Carapetis
2 days ago
The Hessian of $Z$ is $ left[ begin {array}{cc} 1+sin(4theta) & 4cos( 4theta) r\ 4cos(4theta) r & -8sin(4theta)r^2 end {array} right]$ so $Z$ is twice differentiable.
– Wazowski
2 days ago
I meant $Z$ as a function of $X,Y$, not $r,theta,$ but more importantly: existence of the Hessian/second partial derivatives (which is true for $Z(X,Y)$) does not imply twice-differentiability in the strong sense.
– Anthony Carapetis
2 days ago
I edited the question based on Ted Shifrin suggestions. Tried to clarify and simplify things and added some illustrations.
– Wazowski
yesterday
add a comment |
First of all, $theta=-pi/8$ and $theta=15pi/8$ are the same (mod $2pi$), hence the same direction in the plane. Your description is also inaccurate, when you suggest you're looking at a surface of revolution. You're certainly not. Can you please delete your duplicate $theta$ values and specify exactly what point we're looking at? The lines in the surface will only given principal directions when you have a point with $K=0$. I don't see what the problem actually is.
– Ted Shifrin
Nov 13 at 18:41
Haven't checked the details, but I think what's going on is that your function $Z$ (and thus your surface) is not twice-differentiable at the origin. Thus the directional second derivatives of $Z$ at the origin are not described by a Hessian matrix and the geodesic curvatures are not described by a second fundamental form; so you can't apply orthonormal diagonalization.
– Anthony Carapetis
2 days ago
The Hessian of $Z$ is $ left[ begin {array}{cc} 1+sin(4theta) & 4cos( 4theta) r\ 4cos(4theta) r & -8sin(4theta)r^2 end {array} right]$ so $Z$ is twice differentiable.
– Wazowski
2 days ago
I meant $Z$ as a function of $X,Y$, not $r,theta,$ but more importantly: existence of the Hessian/second partial derivatives (which is true for $Z(X,Y)$) does not imply twice-differentiability in the strong sense.
– Anthony Carapetis
2 days ago
I edited the question based on Ted Shifrin suggestions. Tried to clarify and simplify things and added some illustrations.
– Wazowski
yesterday
First of all, $theta=-pi/8$ and $theta=15pi/8$ are the same (mod $2pi$), hence the same direction in the plane. Your description is also inaccurate, when you suggest you're looking at a surface of revolution. You're certainly not. Can you please delete your duplicate $theta$ values and specify exactly what point we're looking at? The lines in the surface will only given principal directions when you have a point with $K=0$. I don't see what the problem actually is.
– Ted Shifrin
Nov 13 at 18:41
First of all, $theta=-pi/8$ and $theta=15pi/8$ are the same (mod $2pi$), hence the same direction in the plane. Your description is also inaccurate, when you suggest you're looking at a surface of revolution. You're certainly not. Can you please delete your duplicate $theta$ values and specify exactly what point we're looking at? The lines in the surface will only given principal directions when you have a point with $K=0$. I don't see what the problem actually is.
– Ted Shifrin
Nov 13 at 18:41
Haven't checked the details, but I think what's going on is that your function $Z$ (and thus your surface) is not twice-differentiable at the origin. Thus the directional second derivatives of $Z$ at the origin are not described by a Hessian matrix and the geodesic curvatures are not described by a second fundamental form; so you can't apply orthonormal diagonalization.
– Anthony Carapetis
2 days ago
Haven't checked the details, but I think what's going on is that your function $Z$ (and thus your surface) is not twice-differentiable at the origin. Thus the directional second derivatives of $Z$ at the origin are not described by a Hessian matrix and the geodesic curvatures are not described by a second fundamental form; so you can't apply orthonormal diagonalization.
– Anthony Carapetis
2 days ago
The Hessian of $Z$ is $ left[ begin {array}{cc} 1+sin(4theta) & 4cos( 4theta) r\ 4cos(4theta) r & -8sin(4theta)r^2 end {array} right]$ so $Z$ is twice differentiable.
– Wazowski
2 days ago
The Hessian of $Z$ is $ left[ begin {array}{cc} 1+sin(4theta) & 4cos( 4theta) r\ 4cos(4theta) r & -8sin(4theta)r^2 end {array} right]$ so $Z$ is twice differentiable.
– Wazowski
2 days ago
I meant $Z$ as a function of $X,Y$, not $r,theta,$ but more importantly: existence of the Hessian/second partial derivatives (which is true for $Z(X,Y)$) does not imply twice-differentiability in the strong sense.
– Anthony Carapetis
2 days ago
I meant $Z$ as a function of $X,Y$, not $r,theta,$ but more importantly: existence of the Hessian/second partial derivatives (which is true for $Z(X,Y)$) does not imply twice-differentiability in the strong sense.
– Anthony Carapetis
2 days ago
I edited the question based on Ted Shifrin suggestions. Tried to clarify and simplify things and added some illustrations.
– Wazowski
yesterday
I edited the question based on Ted Shifrin suggestions. Tried to clarify and simplify things and added some illustrations.
– Wazowski
yesterday
add a comment |
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First of all, $theta=-pi/8$ and $theta=15pi/8$ are the same (mod $2pi$), hence the same direction in the plane. Your description is also inaccurate, when you suggest you're looking at a surface of revolution. You're certainly not. Can you please delete your duplicate $theta$ values and specify exactly what point we're looking at? The lines in the surface will only given principal directions when you have a point with $K=0$. I don't see what the problem actually is.
– Ted Shifrin
Nov 13 at 18:41
Haven't checked the details, but I think what's going on is that your function $Z$ (and thus your surface) is not twice-differentiable at the origin. Thus the directional second derivatives of $Z$ at the origin are not described by a Hessian matrix and the geodesic curvatures are not described by a second fundamental form; so you can't apply orthonormal diagonalization.
– Anthony Carapetis
2 days ago
The Hessian of $Z$ is $ left[ begin {array}{cc} 1+sin(4theta) & 4cos( 4theta) r\ 4cos(4theta) r & -8sin(4theta)r^2 end {array} right]$ so $Z$ is twice differentiable.
– Wazowski
2 days ago
I meant $Z$ as a function of $X,Y$, not $r,theta,$ but more importantly: existence of the Hessian/second partial derivatives (which is true for $Z(X,Y)$) does not imply twice-differentiability in the strong sense.
– Anthony Carapetis
2 days ago
I edited the question based on Ted Shifrin suggestions. Tried to clarify and simplify things and added some illustrations.
– Wazowski
yesterday