Mixing
geodesic curvature about boundary of geodesic disc
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Suppose $M^2$ is a surface in $mathbb{R}^3$, choose a point $xin M^2$, consider the geodesic disc $D_r(x)$. Why do we have following equality?
$$frac{d}{dt}(int_{partial D_t}1 ds)=int_{partial D_t}k_g ds$$
where $k_g$ is the geodesic curvature of $partial D_t$.
differential-geometry riemannian-geometry
asked Jan 29 at 21:39
STUDENTSTUDENT
1036
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add a comment |
$begingroup$
Suppose $M^2$ is a surface in $mathbb{R}^3$, choose a point $xin M^2$, consider the geodesic disc $D_r(x)$. Why do we have following equality?
$$frac{d}{dt}(int_{partial D_t}1 ds)=int_{partial D_t}k_g ds$$
where $k_g$ is the geodesic curvature of $partial D_t$.
differential-geometry riemannian-geometry
asked Jan 29 at 21:39
STUDENTSTUDENT
1036
$endgroup$
3
$begingroup$
Have you seen a first-variation-of-arclength computation that shows that geodesics are locally length-minimizing? This is the same computation.
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– Ted Shifrin
Jan 29 at 23:05
$begingroup$
Thanks @ Ted Shifrin, I think I get it now.
$endgroup$
– STUDENT
Jan 30 at 5:50
add a comment |
$begingroup$
Suppose $M^2$ is a surface in $mathbb{R}^3$, choose a point $xin M^2$, consider the geodesic disc $D_r(x)$. Why do we have following equality?
$$frac{d}{dt}(int_{partial D_t}1 ds)=int_{partial D_t}k_g ds$$
where $k_g$ is the geodesic curvature of $partial D_t$.
differential-geometry riemannian-geometry
asked Jan 29 at 21:39
STUDENTSTUDENT
1036
$endgroup$
Suppose $M^2$ is a surface in $mathbb{R}^3$, choose a point $xin M^2$, consider the geodesic disc $D_r(x)$. Why do we have following equality?
$$frac{d}{dt}(int_{partial D_t}1 ds)=int_{partial D_t}k_g ds$$
where $k_g$ is the geodesic curvature of $partial D_t$.
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Jan 29 at 21:39
STUDENTSTUDENT
1036
asked Jan 29 at 21:39
STUDENTSTUDENT
1036
asked Jan 29 at 21:39
STUDENTSTUDENT
1036
asked Jan 29 at 21:39
STUDENTSTUDENT
1036
asked Jan 29 at 21:39
STUDENTSTUDENT
1036
1036
3
$begingroup$
Have you seen a first-variation-of-arclength computation that shows that geodesics are locally length-minimizing? This is the same computation.
$endgroup$
– Ted Shifrin
Jan 29 at 23:05
$begingroup$
Thanks @ Ted Shifrin, I think I get it now.
$endgroup$
– STUDENT
Jan 30 at 5:50
add a comment |
3
$begingroup$
Have you seen a first-variation-of-arclength computation that shows that geodesics are locally length-minimizing? This is the same computation.
$endgroup$
– Ted Shifrin
Jan 29 at 23:05
$begingroup$
Thanks @ Ted Shifrin, I think I get it now.
$endgroup$
– STUDENT
Jan 30 at 5:50
3
3
$begingroup$
Have you seen a first-variation-of-arclength computation that shows that geodesics are locally length-minimizing? This is the same computation.
$endgroup$
– Ted Shifrin
Jan 29 at 23:05
$begingroup$
Have you seen a first-variation-of-arclength computation that shows that geodesics are locally length-minimizing? This is the same computation.
$endgroup$
– Ted Shifrin
Jan 29 at 23:05
$begingroup$
Thanks @ Ted Shifrin, I think I get it now.
$endgroup$
– STUDENT
Jan 30 at 5:50
$begingroup$
Thanks @ Ted Shifrin, I think I get it now.
$endgroup$
– STUDENT
Jan 30 at 5:50
add a comment |
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3
$begingroup$
Have you seen a first-variation-of-arclength computation that shows that geodesics are locally length-minimizing? This is the same computation.
$endgroup$
– Ted Shifrin
Jan 29 at 23:05
$begingroup$
Thanks @ Ted Shifrin, I think I get it now.
$endgroup$
– STUDENT
Jan 30 at 5:50