Differential Geometry, why orthogonal eigenvectors have minimum and maximum curvature? why not they happen in...
Hello brothers and masters, I am undergrad studuent, learning introduction of differential geometry(DG).
I am vary curious that in the DG, the principal curvature appear in the eigenvectors which is orthogonal to each other(because linear algebra), given that a point of the regular surface.
But I cannot hard to understand with my heart..
Because it looks also possibile that the principal curvature may happen in the tangent vectors which are not orthogonal to each other.
For example,
countereample which I am suspecting
enter image description here
This surface has maximum curvature at $theta = 0$ and $theta = pi/2$, and has minimum curvature at $theta = pi/4$ and $theta = 3/4 * pi$
Is there a something wrong with this surface?
or... what can be a good explanation for this surface?
Please quench this curiosity!
Thank you in advance.
the wolfram alpha code of the surface
differential-geometry curvature
asked Nov 22 '18 at 2:25
SeonookCHUN
111
add a comment |
Hello brothers and masters, I am undergrad studuent, learning introduction of differential geometry(DG).
I am vary curious that in the DG, the principal curvature appear in the eigenvectors which is orthogonal to each other(because linear algebra), given that a point of the regular surface.
But I cannot hard to understand with my heart..
Because it looks also possibile that the principal curvature may happen in the tangent vectors which are not orthogonal to each other.
For example,
countereample which I am suspecting
enter image description here
This surface has maximum curvature at $theta = 0$ and $theta = pi/2$, and has minimum curvature at $theta = pi/4$ and $theta = 3/4 * pi$
Is there a something wrong with this surface?
or... what can be a good explanation for this surface?
Please quench this curiosity!
Thank you in advance.
the wolfram alpha code of the surface
differential-geometry curvature
asked Nov 22 '18 at 2:25
SeonookCHUN
111
1
Basically, it’s because curvature involves quadratic approximations to the surface, which brings the principal axis theorem into play. If I have time, I’ll expand that into a proper answer.
– amd
Nov 22 '18 at 19:29
add a comment |
Hello brothers and masters, I am undergrad studuent, learning introduction of differential geometry(DG).
I am vary curious that in the DG, the principal curvature appear in the eigenvectors which is orthogonal to each other(because linear algebra), given that a point of the regular surface.
But I cannot hard to understand with my heart..
Because it looks also possibile that the principal curvature may happen in the tangent vectors which are not orthogonal to each other.
For example,
countereample which I am suspecting
enter image description here
This surface has maximum curvature at $theta = 0$ and $theta = pi/2$, and has minimum curvature at $theta = pi/4$ and $theta = 3/4 * pi$
Is there a something wrong with this surface?
or... what can be a good explanation for this surface?
Please quench this curiosity!
Thank you in advance.
the wolfram alpha code of the surface
differential-geometry curvature
asked Nov 22 '18 at 2:25
SeonookCHUN
111
Hello brothers and masters, I am undergrad studuent, learning introduction of differential geometry(DG).
I am vary curious that in the DG, the principal curvature appear in the eigenvectors which is orthogonal to each other(because linear algebra), given that a point of the regular surface.
But I cannot hard to understand with my heart..
Because it looks also possibile that the principal curvature may happen in the tangent vectors which are not orthogonal to each other.
For example,
countereample which I am suspecting
enter image description here
This surface has maximum curvature at $theta = 0$ and $theta = pi/2$, and has minimum curvature at $theta = pi/4$ and $theta = 3/4 * pi$
Is there a something wrong with this surface?
or... what can be a good explanation for this surface?
Please quench this curiosity!
Thank you in advance.
the wolfram alpha code of the surface
differential-geometry curvature
differential-geometry curvature
asked Nov 22 '18 at 2:25
SeonookCHUN
111
asked Nov 22 '18 at 2:25
SeonookCHUN
111
asked Nov 22 '18 at 2:25
SeonookCHUN
111
asked Nov 22 '18 at 2:25
SeonookCHUN
111
asked Nov 22 '18 at 2:25
SeonookCHUN
111
111
1
Basically, it’s because curvature involves quadratic approximations to the surface, which brings the principal axis theorem into play. If I have time, I’ll expand that into a proper answer.
– amd
Nov 22 '18 at 19:29
add a comment |
1
Basically, it’s because curvature involves quadratic approximations to the surface, which brings the principal axis theorem into play. If I have time, I’ll expand that into a proper answer.
– amd
Nov 22 '18 at 19:29
1
1
Basically, it’s because curvature involves quadratic approximations to the surface, which brings the principal axis theorem into play. If I have time, I’ll expand that into a proper answer.
– amd
Nov 22 '18 at 19:29
Basically, it’s because curvature involves quadratic approximations to the surface, which brings the principal axis theorem into play. If I have time, I’ll expand that into a proper answer.
– amd
Nov 22 '18 at 19:29
add a comment |
1 Answer
1
active
oldest
votes
Through the power of Wolfram Alpha, one can check that
$$
cos(4arctan(t)) = 3 - frac{8t^2}{(t^2+1)^2},
$$
so that
$$
cos(4arctan(v/u))+2 = 5 - frac{8u^2v^2}{(u^2+v^2)^2},
$$
and hence that
$$
(u^2+v^2)(cos(4arctan(v/u))+2) = 5(u^2+v)^2 - frac{8u^2v^2}{u^2+v^2}.
$$
Thus, your surface is the graph of the function
$$
f(u,v) = 5(u^2+v^2) - frac{8u^2v^2}{u^2+v^2}.
$$
I claim that $f$ is not twice differentiable at $(0,0)$, which is what you need to have a well-defined shape operator at $(0,0)$, and hence well-defined principal curvatures at $(0,0)$; in fact, I claim that $f$ isn't even continuously differentiable at $(0,0)$.
Now, for $(u,v) neq (0,0)$, one can compute
$$
partial_1f(u,v) = 10u - frac{16uv^4}{(u^2+v^2)^2}, quad partial_2f(u,v) = 10v - frac{16u^4v}{(u^2+v^2)^2}.
$$
Let $t neq 0$, and consider the line $v = tu$. On that line (away from the origin), we find that
$$
partial_2f(u,tu) = 10tu - frac{16t}{(1+t^2)^2},
$$
and hence that the limit of $partial_2f(u,v)$ as $(u,v) to (0,0)$ along the line $v = tu$ is
$$
lim_{u to 0} partial_2f(u,tu) = -frac{16t}{(1+t^2)^2},
$$
which clearly gives different limits for different values of the slope $t$. Thus, the partial derivative $partial_2 f$ is discontinuous at $(0,0)$, and hence $f$ can't possibly be twice differentiable at $(0,0)$.
answered Nov 22 '18 at 3:48
Branimir Ćaćić
9,91522046
Thank you for your answer!
– SeonookCHUN
Nov 23 '18 at 12:33
But, you also think that the property of orthogonality is make sense in your heart?
– SeonookCHUN
Nov 23 '18 at 12:33
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Through the power of Wolfram Alpha, one can check that
$$
cos(4arctan(t)) = 3 - frac{8t^2}{(t^2+1)^2},
$$
so that
$$
cos(4arctan(v/u))+2 = 5 - frac{8u^2v^2}{(u^2+v^2)^2},
$$
and hence that
$$
(u^2+v^2)(cos(4arctan(v/u))+2) = 5(u^2+v)^2 - frac{8u^2v^2}{u^2+v^2}.
$$
Thus, your surface is the graph of the function
$$
f(u,v) = 5(u^2+v^2) - frac{8u^2v^2}{u^2+v^2}.
$$
I claim that $f$ is not twice differentiable at $(0,0)$, which is what you need to have a well-defined shape operator at $(0,0)$, and hence well-defined principal curvatures at $(0,0)$; in fact, I claim that $f$ isn't even continuously differentiable at $(0,0)$.
Now, for $(u,v) neq (0,0)$, one can compute
$$
partial_1f(u,v) = 10u - frac{16uv^4}{(u^2+v^2)^2}, quad partial_2f(u,v) = 10v - frac{16u^4v}{(u^2+v^2)^2}.
$$
Let $t neq 0$, and consider the line $v = tu$. On that line (away from the origin), we find that
$$
partial_2f(u,tu) = 10tu - frac{16t}{(1+t^2)^2},
$$
and hence that the limit of $partial_2f(u,v)$ as $(u,v) to (0,0)$ along the line $v = tu$ is
$$
lim_{u to 0} partial_2f(u,tu) = -frac{16t}{(1+t^2)^2},
$$
which clearly gives different limits for different values of the slope $t$. Thus, the partial derivative $partial_2 f$ is discontinuous at $(0,0)$, and hence $f$ can't possibly be twice differentiable at $(0,0)$.
answered Nov 22 '18 at 3:48
Branimir Ćaćić
9,91522046
Thank you for your answer!
– SeonookCHUN
Nov 23 '18 at 12:33
But, you also think that the property of orthogonality is make sense in your heart?
– SeonookCHUN
Nov 23 '18 at 12:33
add a comment |
Through the power of Wolfram Alpha, one can check that
$$
cos(4arctan(t)) = 3 - frac{8t^2}{(t^2+1)^2},
$$
so that
$$
cos(4arctan(v/u))+2 = 5 - frac{8u^2v^2}{(u^2+v^2)^2},
$$
and hence that
$$
(u^2+v^2)(cos(4arctan(v/u))+2) = 5(u^2+v)^2 - frac{8u^2v^2}{u^2+v^2}.
$$
Thus, your surface is the graph of the function
$$
f(u,v) = 5(u^2+v^2) - frac{8u^2v^2}{u^2+v^2}.
$$
I claim that $f$ is not twice differentiable at $(0,0)$, which is what you need to have a well-defined shape operator at $(0,0)$, and hence well-defined principal curvatures at $(0,0)$; in fact, I claim that $f$ isn't even continuously differentiable at $(0,0)$.
Now, for $(u,v) neq (0,0)$, one can compute
$$
partial_1f(u,v) = 10u - frac{16uv^4}{(u^2+v^2)^2}, quad partial_2f(u,v) = 10v - frac{16u^4v}{(u^2+v^2)^2}.
$$
Let $t neq 0$, and consider the line $v = tu$. On that line (away from the origin), we find that
$$
partial_2f(u,tu) = 10tu - frac{16t}{(1+t^2)^2},
$$
and hence that the limit of $partial_2f(u,v)$ as $(u,v) to (0,0)$ along the line $v = tu$ is
$$
lim_{u to 0} partial_2f(u,tu) = -frac{16t}{(1+t^2)^2},
$$
which clearly gives different limits for different values of the slope $t$. Thus, the partial derivative $partial_2 f$ is discontinuous at $(0,0)$, and hence $f$ can't possibly be twice differentiable at $(0,0)$.
answered Nov 22 '18 at 3:48
Branimir Ćaćić
9,91522046
Thank you for your answer!
– SeonookCHUN
Nov 23 '18 at 12:33
But, you also think that the property of orthogonality is make sense in your heart?
– SeonookCHUN
Nov 23 '18 at 12:33
add a comment |
Through the power of Wolfram Alpha, one can check that
$$
cos(4arctan(t)) = 3 - frac{8t^2}{(t^2+1)^2},
$$
so that
$$
cos(4arctan(v/u))+2 = 5 - frac{8u^2v^2}{(u^2+v^2)^2},
$$
and hence that
$$
(u^2+v^2)(cos(4arctan(v/u))+2) = 5(u^2+v)^2 - frac{8u^2v^2}{u^2+v^2}.
$$
Thus, your surface is the graph of the function
$$
f(u,v) = 5(u^2+v^2) - frac{8u^2v^2}{u^2+v^2}.
$$
I claim that $f$ is not twice differentiable at $(0,0)$, which is what you need to have a well-defined shape operator at $(0,0)$, and hence well-defined principal curvatures at $(0,0)$; in fact, I claim that $f$ isn't even continuously differentiable at $(0,0)$.
Now, for $(u,v) neq (0,0)$, one can compute
$$
partial_1f(u,v) = 10u - frac{16uv^4}{(u^2+v^2)^2}, quad partial_2f(u,v) = 10v - frac{16u^4v}{(u^2+v^2)^2}.
$$
Let $t neq 0$, and consider the line $v = tu$. On that line (away from the origin), we find that
$$
partial_2f(u,tu) = 10tu - frac{16t}{(1+t^2)^2},
$$
and hence that the limit of $partial_2f(u,v)$ as $(u,v) to (0,0)$ along the line $v = tu$ is
$$
lim_{u to 0} partial_2f(u,tu) = -frac{16t}{(1+t^2)^2},
$$
which clearly gives different limits for different values of the slope $t$. Thus, the partial derivative $partial_2 f$ is discontinuous at $(0,0)$, and hence $f$ can't possibly be twice differentiable at $(0,0)$.
answered Nov 22 '18 at 3:48
Branimir Ćaćić
9,91522046
Through the power of Wolfram Alpha, one can check that
$$
cos(4arctan(t)) = 3 - frac{8t^2}{(t^2+1)^2},
$$
so that
$$
cos(4arctan(v/u))+2 = 5 - frac{8u^2v^2}{(u^2+v^2)^2},
$$
and hence that
$$
(u^2+v^2)(cos(4arctan(v/u))+2) = 5(u^2+v)^2 - frac{8u^2v^2}{u^2+v^2}.
$$
Thus, your surface is the graph of the function
$$
f(u,v) = 5(u^2+v^2) - frac{8u^2v^2}{u^2+v^2}.
$$
I claim that $f$ is not twice differentiable at $(0,0)$, which is what you need to have a well-defined shape operator at $(0,0)$, and hence well-defined principal curvatures at $(0,0)$; in fact, I claim that $f$ isn't even continuously differentiable at $(0,0)$.
Now, for $(u,v) neq (0,0)$, one can compute
$$
partial_1f(u,v) = 10u - frac{16uv^4}{(u^2+v^2)^2}, quad partial_2f(u,v) = 10v - frac{16u^4v}{(u^2+v^2)^2}.
$$
Let $t neq 0$, and consider the line $v = tu$. On that line (away from the origin), we find that
$$
partial_2f(u,tu) = 10tu - frac{16t}{(1+t^2)^2},
$$
and hence that the limit of $partial_2f(u,v)$ as $(u,v) to (0,0)$ along the line $v = tu$ is
$$
lim_{u to 0} partial_2f(u,tu) = -frac{16t}{(1+t^2)^2},
$$
which clearly gives different limits for different values of the slope $t$. Thus, the partial derivative $partial_2 f$ is discontinuous at $(0,0)$, and hence $f$ can't possibly be twice differentiable at $(0,0)$.
answered Nov 22 '18 at 3:48
Branimir Ćaćić
9,91522046
edited Nov 22 '18 at 3:53
answered Nov 22 '18 at 3:48
Branimir Ćaćić
9,91522046
answered Nov 22 '18 at 3:48
Branimir Ćaćić
9,91522046
answered Nov 22 '18 at 3:48
Branimir Ćaćić
9,91522046
9,91522046
Thank you for your answer!
– SeonookCHUN
Nov 23 '18 at 12:33
But, you also think that the property of orthogonality is make sense in your heart?
– SeonookCHUN
Nov 23 '18 at 12:33
add a comment |
Thank you for your answer!
– SeonookCHUN
Nov 23 '18 at 12:33
But, you also think that the property of orthogonality is make sense in your heart?
– SeonookCHUN
Nov 23 '18 at 12:33
Thank you for your answer!
– SeonookCHUN
Nov 23 '18 at 12:33
Thank you for your answer!
– SeonookCHUN
Nov 23 '18 at 12:33
But, you also think that the property of orthogonality is make sense in your heart?
– SeonookCHUN
Nov 23 '18 at 12:33
But, you also think that the property of orthogonality is make sense in your heart?
– SeonookCHUN
Nov 23 '18 at 12:33
add a comment |
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1
Basically, it’s because curvature involves quadratic approximations to the surface, which brings the principal axis theorem into play. If I have time, I’ll expand that into a proper answer.
– amd
Nov 22 '18 at 19:29