Geodesic on the catenoid
$begingroup$
Consider the catenoid ${(x,y,z) in mathbb{R}^3: cosh z = sqrt{x^2+y^2}}$, given with the parameterization
$$
r(u,v) = (cosh u cos v, cosh u sin v, u) .
$$
I'm trying to show that if $gamma(t) = r(u(t),v(t))$ is a geodisic curve, such that $gamma$ is bounded (for every $tinmathbb{R}$), then the image $gamma$ is the circle ${x^2 + y^2 =1, z=0}$.
By Clairaut's relation, it seems that that $gamma$ is geodesic if and on if the following are true:
$$
cosh(u)^2 dot v equiv c\
cosh(u)^2 (dot u^2 + dot v^2) equiv 1 ,
$$
where $c$ is constant, and $u,v$ are understood as functions of $t$. The first equation comes from Clairaut's relation, and the second comes from the first fundamental form.
However, trying all sorts of algebric manipulations, I couldn't find a way to show that if $gamma$ is not the said circle, then it is unbounded. I'd be glad for any hint.
differential-geometry geodesic
asked Dec 21 '18 at 22:54
j3Mj3M
659516
$endgroup$
add a comment |
$begingroup$
Consider the catenoid ${(x,y,z) in mathbb{R}^3: cosh z = sqrt{x^2+y^2}}$, given with the parameterization
$$
r(u,v) = (cosh u cos v, cosh u sin v, u) .
$$
I'm trying to show that if $gamma(t) = r(u(t),v(t))$ is a geodisic curve, such that $gamma$ is bounded (for every $tinmathbb{R}$), then the image $gamma$ is the circle ${x^2 + y^2 =1, z=0}$.
By Clairaut's relation, it seems that that $gamma$ is geodesic if and on if the following are true:
$$
cosh(u)^2 dot v equiv c\
cosh(u)^2 (dot u^2 + dot v^2) equiv 1 ,
$$
where $c$ is constant, and $u,v$ are understood as functions of $t$. The first equation comes from Clairaut's relation, and the second comes from the first fundamental form.
However, trying all sorts of algebric manipulations, I couldn't find a way to show that if $gamma$ is not the said circle, then it is unbounded. I'd be glad for any hint.
differential-geometry geodesic
asked Dec 21 '18 at 22:54
j3Mj3M
659516
$endgroup$
add a comment |
$begingroup$
Consider the catenoid ${(x,y,z) in mathbb{R}^3: cosh z = sqrt{x^2+y^2}}$, given with the parameterization
$$
r(u,v) = (cosh u cos v, cosh u sin v, u) .
$$
I'm trying to show that if $gamma(t) = r(u(t),v(t))$ is a geodisic curve, such that $gamma$ is bounded (for every $tinmathbb{R}$), then the image $gamma$ is the circle ${x^2 + y^2 =1, z=0}$.
By Clairaut's relation, it seems that that $gamma$ is geodesic if and on if the following are true:
$$
cosh(u)^2 dot v equiv c\
cosh(u)^2 (dot u^2 + dot v^2) equiv 1 ,
$$
where $c$ is constant, and $u,v$ are understood as functions of $t$. The first equation comes from Clairaut's relation, and the second comes from the first fundamental form.
However, trying all sorts of algebric manipulations, I couldn't find a way to show that if $gamma$ is not the said circle, then it is unbounded. I'd be glad for any hint.
differential-geometry geodesic
asked Dec 21 '18 at 22:54
j3Mj3M
659516
$endgroup$
Consider the catenoid ${(x,y,z) in mathbb{R}^3: cosh z = sqrt{x^2+y^2}}$, given with the parameterization
$$
r(u,v) = (cosh u cos v, cosh u sin v, u) .
$$
I'm trying to show that if $gamma(t) = r(u(t),v(t))$ is a geodisic curve, such that $gamma$ is bounded (for every $tinmathbb{R}$), then the image $gamma$ is the circle ${x^2 + y^2 =1, z=0}$.
By Clairaut's relation, it seems that that $gamma$ is geodesic if and on if the following are true:
$$
cosh(u)^2 dot v equiv c\
cosh(u)^2 (dot u^2 + dot v^2) equiv 1 ,
$$
where $c$ is constant, and $u,v$ are understood as functions of $t$. The first equation comes from Clairaut's relation, and the second comes from the first fundamental form.
However, trying all sorts of algebric manipulations, I couldn't find a way to show that if $gamma$ is not the said circle, then it is unbounded. I'd be glad for any hint.
differential-geometry geodesic
differential-geometry geodesic
asked Dec 21 '18 at 22:54
j3Mj3M
659516
asked Dec 21 '18 at 22:54
j3Mj3M
659516
asked Dec 21 '18 at 22:54
j3Mj3M
659516
asked Dec 21 '18 at 22:54
j3Mj3M
659516
asked Dec 21 '18 at 22:54
j3Mj3M
659516
659516
add a comment |
add a comment |
1 Answer
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$begingroup$
Your second equation isn't right. If you assume an arclength parametrization by $t$, then the second equation should be $dot u^2 + cosh^2u,dot v^2 = 1$, right? Then we have
$$frac{dv}{du} = frac{dot v}{dot u} = frac c{cosh usqrt{cosh^2u-c^2}},$$
and so
$$v = c int frac{du}{cosh usqrt{cosh^2 u-c^2}} + c'$$
for some constants $c$ and $c'$. This formulation is good for understanding how the geodesic winds around.
First thing to understand: What value of $c$ gives the belt circle?
Next thing to understand: Is there any way such an integral will diverge as $u$ stays bounded? How can a geodesic meet a parallel circle tangentially? It really is better to think about Clairaut geometrically: The expression $cosh^2 u,dot v = c$ tells you that the radius of the parallel circle times the cosine of the angle between the geodesic and the parallel must stay constant.
answered Dec 21 '18 at 23:44
Ted ShifrinTed Shifrin
64.1k44692
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add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
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votes
$begingroup$
Your second equation isn't right. If you assume an arclength parametrization by $t$, then the second equation should be $dot u^2 + cosh^2u,dot v^2 = 1$, right? Then we have
$$frac{dv}{du} = frac{dot v}{dot u} = frac c{cosh usqrt{cosh^2u-c^2}},$$
and so
$$v = c int frac{du}{cosh usqrt{cosh^2 u-c^2}} + c'$$
for some constants $c$ and $c'$. This formulation is good for understanding how the geodesic winds around.
First thing to understand: What value of $c$ gives the belt circle?
Next thing to understand: Is there any way such an integral will diverge as $u$ stays bounded? How can a geodesic meet a parallel circle tangentially? It really is better to think about Clairaut geometrically: The expression $cosh^2 u,dot v = c$ tells you that the radius of the parallel circle times the cosine of the angle between the geodesic and the parallel must stay constant.
answered Dec 21 '18 at 23:44
Ted ShifrinTed Shifrin
64.1k44692
$endgroup$
add a comment |
$begingroup$
Your second equation isn't right. If you assume an arclength parametrization by $t$, then the second equation should be $dot u^2 + cosh^2u,dot v^2 = 1$, right? Then we have
$$frac{dv}{du} = frac{dot v}{dot u} = frac c{cosh usqrt{cosh^2u-c^2}},$$
and so
$$v = c int frac{du}{cosh usqrt{cosh^2 u-c^2}} + c'$$
for some constants $c$ and $c'$. This formulation is good for understanding how the geodesic winds around.
First thing to understand: What value of $c$ gives the belt circle?
Next thing to understand: Is there any way such an integral will diverge as $u$ stays bounded? How can a geodesic meet a parallel circle tangentially? It really is better to think about Clairaut geometrically: The expression $cosh^2 u,dot v = c$ tells you that the radius of the parallel circle times the cosine of the angle between the geodesic and the parallel must stay constant.
answered Dec 21 '18 at 23:44
Ted ShifrinTed Shifrin
64.1k44692
$endgroup$
add a comment |
$begingroup$
Your second equation isn't right. If you assume an arclength parametrization by $t$, then the second equation should be $dot u^2 + cosh^2u,dot v^2 = 1$, right? Then we have
$$frac{dv}{du} = frac{dot v}{dot u} = frac c{cosh usqrt{cosh^2u-c^2}},$$
and so
$$v = c int frac{du}{cosh usqrt{cosh^2 u-c^2}} + c'$$
for some constants $c$ and $c'$. This formulation is good for understanding how the geodesic winds around.
First thing to understand: What value of $c$ gives the belt circle?
Next thing to understand: Is there any way such an integral will diverge as $u$ stays bounded? How can a geodesic meet a parallel circle tangentially? It really is better to think about Clairaut geometrically: The expression $cosh^2 u,dot v = c$ tells you that the radius of the parallel circle times the cosine of the angle between the geodesic and the parallel must stay constant.
answered Dec 21 '18 at 23:44
Ted ShifrinTed Shifrin
64.1k44692
$endgroup$
Your second equation isn't right. If you assume an arclength parametrization by $t$, then the second equation should be $dot u^2 + cosh^2u,dot v^2 = 1$, right? Then we have
$$frac{dv}{du} = frac{dot v}{dot u} = frac c{cosh usqrt{cosh^2u-c^2}},$$
and so
$$v = c int frac{du}{cosh usqrt{cosh^2 u-c^2}} + c'$$
for some constants $c$ and $c'$. This formulation is good for understanding how the geodesic winds around.
First thing to understand: What value of $c$ gives the belt circle?
Next thing to understand: Is there any way such an integral will diverge as $u$ stays bounded? How can a geodesic meet a parallel circle tangentially? It really is better to think about Clairaut geometrically: The expression $cosh^2 u,dot v = c$ tells you that the radius of the parallel circle times the cosine of the angle between the geodesic and the parallel must stay constant.
answered Dec 21 '18 at 23:44
Ted ShifrinTed Shifrin
64.1k44692
answered Dec 21 '18 at 23:44
Ted ShifrinTed Shifrin
64.1k44692
answered Dec 21 '18 at 23:44
Ted ShifrinTed Shifrin
64.1k44692
answered Dec 21 '18 at 23:44
Ted ShifrinTed Shifrin
64.1k44692
64.1k44692
add a comment |
add a comment |
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