Proof of Euler's Theorem involving curvature.
$begingroup$
Theorem: Let $φ$ be the angle, in the tangent plane, measured counterclockwise from the direction of minimum curvature $kappa_1$ . Then the normal curvature $kappa_n(φ)$ in direction $φ$ is given by $$kappa_n(varphi)=kappa_1 cos^2varphi+kappa_2sin^2varphi.$$
How do I prove this?
differential-geometry surfaces curvature
edited May 14 '16 at 7:41
none
103117
asked May 12 '16 at 20:01
zermelovaczermelovac
504212
$endgroup$
add a comment |
$begingroup$
Theorem: Let $φ$ be the angle, in the tangent plane, measured counterclockwise from the direction of minimum curvature $kappa_1$ . Then the normal curvature $kappa_n(φ)$ in direction $φ$ is given by $$kappa_n(varphi)=kappa_1 cos^2varphi+kappa_2sin^2varphi.$$
How do I prove this?
differential-geometry surfaces curvature
edited May 14 '16 at 7:41
none
103117
asked May 12 '16 at 20:01
zermelovaczermelovac
504212
$endgroup$
add a comment |
$begingroup$
Theorem: Let $φ$ be the angle, in the tangent plane, measured counterclockwise from the direction of minimum curvature $kappa_1$ . Then the normal curvature $kappa_n(φ)$ in direction $φ$ is given by $$kappa_n(varphi)=kappa_1 cos^2varphi+kappa_2sin^2varphi.$$
How do I prove this?
differential-geometry surfaces curvature
edited May 14 '16 at 7:41
none
103117
asked May 12 '16 at 20:01
zermelovaczermelovac
504212
$endgroup$
Theorem: Let $φ$ be the angle, in the tangent plane, measured counterclockwise from the direction of minimum curvature $kappa_1$ . Then the normal curvature $kappa_n(φ)$ in direction $φ$ is given by $$kappa_n(varphi)=kappa_1 cos^2varphi+kappa_2sin^2varphi.$$
How do I prove this?
differential-geometry surfaces curvature
differential-geometry surfaces curvature
edited May 14 '16 at 7:41
none
103117
asked May 12 '16 at 20:01
zermelovaczermelovac
504212
edited May 14 '16 at 7:41
none
103117
asked May 12 '16 at 20:01
zermelovaczermelovac
504212
edited May 14 '16 at 7:41
none
103117
edited May 14 '16 at 7:41
none
103117
edited May 14 '16 at 7:41
none
103117
103117
asked May 12 '16 at 20:01
zermelovaczermelovac
504212
asked May 12 '16 at 20:01
zermelovaczermelovac
504212
asked May 12 '16 at 20:01
zermelovaczermelovac
504212
504212
add a comment |
add a comment |
1 Answer
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$begingroup$
The normal curvature can be defined in terms of the second fundamental form as follows: the normal curvature $k_n$ of a vector $v in T_{p}S$ is defined as $k_n = text{II}_{p}(v)=<-dN_{p}(v), v>$. Recall that the differential of the gauss map is a self-adjoint linear map and so there exists an orthonormal basis ${e_1, e_2}$ for $T_{p}S$ such that $-dN_{p}(e_1) = k_{1}e_{1}$ and $-dN_{p}(e_2) = k_{2}e_{2}$.
Thus for an arbitrary direction vector described by your $varphi$, call it $v$ can we written as a linear combination of some orthonormal basis ${e_1, e_2}$ with the above properties as $v = e_{1}cosvarphi + e_{2}sinvarphi$.
Now following the definition we get:
begin{align}
k_{n} & = text{II}_{p}(v) = -<dN_{p}(v), v> \
& =-<dN_{p}(e_{1}cosvarphi + e_{2}sinvarphi), e_{1}cosvarphi + e_{2}sinvarphi> \
& =<k_{1}e_{1}cosvarphi+k_{2}e_{2}sinvarphi, e_{2}sinvarphi+ e_{1}cosvarphi> \
& =k_{1}cos^2varphi + k_{2}sin^2varphi \
end{align}
answered May 13 '16 at 4:47
nonenone
103117
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1 Answer
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1 Answer
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$begingroup$
The normal curvature can be defined in terms of the second fundamental form as follows: the normal curvature $k_n$ of a vector $v in T_{p}S$ is defined as $k_n = text{II}_{p}(v)=<-dN_{p}(v), v>$. Recall that the differential of the gauss map is a self-adjoint linear map and so there exists an orthonormal basis ${e_1, e_2}$ for $T_{p}S$ such that $-dN_{p}(e_1) = k_{1}e_{1}$ and $-dN_{p}(e_2) = k_{2}e_{2}$.
Thus for an arbitrary direction vector described by your $varphi$, call it $v$ can we written as a linear combination of some orthonormal basis ${e_1, e_2}$ with the above properties as $v = e_{1}cosvarphi + e_{2}sinvarphi$.
Now following the definition we get:
begin{align}
k_{n} & = text{II}_{p}(v) = -<dN_{p}(v), v> \
& =-<dN_{p}(e_{1}cosvarphi + e_{2}sinvarphi), e_{1}cosvarphi + e_{2}sinvarphi> \
& =<k_{1}e_{1}cosvarphi+k_{2}e_{2}sinvarphi, e_{2}sinvarphi+ e_{1}cosvarphi> \
& =k_{1}cos^2varphi + k_{2}sin^2varphi \
end{align}
answered May 13 '16 at 4:47
nonenone
103117
$endgroup$
add a comment |
$begingroup$
The normal curvature can be defined in terms of the second fundamental form as follows: the normal curvature $k_n$ of a vector $v in T_{p}S$ is defined as $k_n = text{II}_{p}(v)=<-dN_{p}(v), v>$. Recall that the differential of the gauss map is a self-adjoint linear map and so there exists an orthonormal basis ${e_1, e_2}$ for $T_{p}S$ such that $-dN_{p}(e_1) = k_{1}e_{1}$ and $-dN_{p}(e_2) = k_{2}e_{2}$.
Thus for an arbitrary direction vector described by your $varphi$, call it $v$ can we written as a linear combination of some orthonormal basis ${e_1, e_2}$ with the above properties as $v = e_{1}cosvarphi + e_{2}sinvarphi$.
Now following the definition we get:
begin{align}
k_{n} & = text{II}_{p}(v) = -<dN_{p}(v), v> \
& =-<dN_{p}(e_{1}cosvarphi + e_{2}sinvarphi), e_{1}cosvarphi + e_{2}sinvarphi> \
& =<k_{1}e_{1}cosvarphi+k_{2}e_{2}sinvarphi, e_{2}sinvarphi+ e_{1}cosvarphi> \
& =k_{1}cos^2varphi + k_{2}sin^2varphi \
end{align}
answered May 13 '16 at 4:47
nonenone
103117
$endgroup$
add a comment |
$begingroup$
The normal curvature can be defined in terms of the second fundamental form as follows: the normal curvature $k_n$ of a vector $v in T_{p}S$ is defined as $k_n = text{II}_{p}(v)=<-dN_{p}(v), v>$. Recall that the differential of the gauss map is a self-adjoint linear map and so there exists an orthonormal basis ${e_1, e_2}$ for $T_{p}S$ such that $-dN_{p}(e_1) = k_{1}e_{1}$ and $-dN_{p}(e_2) = k_{2}e_{2}$.
Thus for an arbitrary direction vector described by your $varphi$, call it $v$ can we written as a linear combination of some orthonormal basis ${e_1, e_2}$ with the above properties as $v = e_{1}cosvarphi + e_{2}sinvarphi$.
Now following the definition we get:
begin{align}
k_{n} & = text{II}_{p}(v) = -<dN_{p}(v), v> \
& =-<dN_{p}(e_{1}cosvarphi + e_{2}sinvarphi), e_{1}cosvarphi + e_{2}sinvarphi> \
& =<k_{1}e_{1}cosvarphi+k_{2}e_{2}sinvarphi, e_{2}sinvarphi+ e_{1}cosvarphi> \
& =k_{1}cos^2varphi + k_{2}sin^2varphi \
end{align}
answered May 13 '16 at 4:47
nonenone
103117
$endgroup$
The normal curvature can be defined in terms of the second fundamental form as follows: the normal curvature $k_n$ of a vector $v in T_{p}S$ is defined as $k_n = text{II}_{p}(v)=<-dN_{p}(v), v>$. Recall that the differential of the gauss map is a self-adjoint linear map and so there exists an orthonormal basis ${e_1, e_2}$ for $T_{p}S$ such that $-dN_{p}(e_1) = k_{1}e_{1}$ and $-dN_{p}(e_2) = k_{2}e_{2}$.
Thus for an arbitrary direction vector described by your $varphi$, call it $v$ can we written as a linear combination of some orthonormal basis ${e_1, e_2}$ with the above properties as $v = e_{1}cosvarphi + e_{2}sinvarphi$.
Now following the definition we get:
begin{align}
k_{n} & = text{II}_{p}(v) = -<dN_{p}(v), v> \
& =-<dN_{p}(e_{1}cosvarphi + e_{2}sinvarphi), e_{1}cosvarphi + e_{2}sinvarphi> \
& =<k_{1}e_{1}cosvarphi+k_{2}e_{2}sinvarphi, e_{2}sinvarphi+ e_{1}cosvarphi> \
& =k_{1}cos^2varphi + k_{2}sin^2varphi \
end{align}
answered May 13 '16 at 4:47
nonenone
103117
answered May 13 '16 at 4:47
nonenone
103117
answered May 13 '16 at 4:47
nonenone
103117
answered May 13 '16 at 4:47
nonenone
103117
103117
add a comment |
add a comment |
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