Mixing
are geodesic shortest path or quickest path?
$begingroup$
I'm a bit confused with geodesics : are they the shortest path (in distance) or the quickest path (in time). For example, Let take a triangle ABC. I'm using a car. I'm in $A$ and I have to go in $B$. The path $AB$ is 2km long, but I can go at 10 km/h only, where as the path that path through C has 4 km length, but it's a free way and I can go at 100 km/h.
Clearly, the path through C is quicker, but the path AB is shorter. What is going to be the Geodesic ? The path through $C$ or the path $AB$ ?
differential-geometry
asked Jan 27 at 14:30
DylanDylan
3118
$endgroup$
add a comment |
$begingroup$
I'm a bit confused with geodesics : are they the shortest path (in distance) or the quickest path (in time). For example, Let take a triangle ABC. I'm using a car. I'm in $A$ and I have to go in $B$. The path $AB$ is 2km long, but I can go at 10 km/h only, where as the path that path through C has 4 km length, but it's a free way and I can go at 100 km/h.
Clearly, the path through C is quicker, but the path AB is shorter. What is going to be the Geodesic ? The path through $C$ or the path $AB$ ?
differential-geometry
asked Jan 27 at 14:30
DylanDylan
3118
$endgroup$
$begingroup$
They are the shortest path as far as I know. The quickest also depends on velocity, right?
$endgroup$
– Exp ikx
Jan 27 at 14:32
1
$begingroup$
It depend on your metric ! If on your chart 1 km represent really 1 km, then AB is the geodesic. If 1 km represent the the distance you make in 1h, then the path through C will be the geodesic.
$endgroup$
– Surb
Jan 27 at 14:34
$begingroup$
This is a great question, and while the answers address it, you might also be interested in the brachistochrone problem (en.wikipedia.org/wiki/Brachistochrone_curve), which uses the same methods we use for studying geodesics, to find "fastest path" curves (in a somewhat different context).
$endgroup$
– John Hughes
Jan 27 at 16:32
$begingroup$
I believe that general relativity uses geodesics which are the fastest path between two points in spacetime (and hence the path followed by light)—but that's the result of using a metric where distance can be interpreted as time.
$endgroup$
– timtfj
Jan 27 at 16:41
add a comment |
$begingroup$
I'm a bit confused with geodesics : are they the shortest path (in distance) or the quickest path (in time). For example, Let take a triangle ABC. I'm using a car. I'm in $A$ and I have to go in $B$. The path $AB$ is 2km long, but I can go at 10 km/h only, where as the path that path through C has 4 km length, but it's a free way and I can go at 100 km/h.
Clearly, the path through C is quicker, but the path AB is shorter. What is going to be the Geodesic ? The path through $C$ or the path $AB$ ?
differential-geometry
asked Jan 27 at 14:30
DylanDylan
3118
$endgroup$
I'm a bit confused with geodesics : are they the shortest path (in distance) or the quickest path (in time). For example, Let take a triangle ABC. I'm using a car. I'm in $A$ and I have to go in $B$. The path $AB$ is 2km long, but I can go at 10 km/h only, where as the path that path through C has 4 km length, but it's a free way and I can go at 100 km/h.
Clearly, the path through C is quicker, but the path AB is shorter. What is going to be the Geodesic ? The path through $C$ or the path $AB$ ?
differential-geometry
differential-geometry
asked Jan 27 at 14:30
DylanDylan
3118
asked Jan 27 at 14:30
DylanDylan
3118
asked Jan 27 at 14:30
DylanDylan
3118
asked Jan 27 at 14:30
DylanDylan
3118
asked Jan 27 at 14:30
DylanDylan
3118
3118
$begingroup$
They are the shortest path as far as I know. The quickest also depends on velocity, right?
$endgroup$
– Exp ikx
Jan 27 at 14:32
1
$begingroup$
It depend on your metric ! If on your chart 1 km represent really 1 km, then AB is the geodesic. If 1 km represent the the distance you make in 1h, then the path through C will be the geodesic.
$endgroup$
– Surb
Jan 27 at 14:34
$begingroup$
This is a great question, and while the answers address it, you might also be interested in the brachistochrone problem (en.wikipedia.org/wiki/Brachistochrone_curve), which uses the same methods we use for studying geodesics, to find "fastest path" curves (in a somewhat different context).
$endgroup$
– John Hughes
Jan 27 at 16:32
$begingroup$
I believe that general relativity uses geodesics which are the fastest path between two points in spacetime (and hence the path followed by light)—but that's the result of using a metric where distance can be interpreted as time.
$endgroup$
– timtfj
Jan 27 at 16:41
add a comment |
$begingroup$
They are the shortest path as far as I know. The quickest also depends on velocity, right?
$endgroup$
– Exp ikx
Jan 27 at 14:32
1
$begingroup$
It depend on your metric ! If on your chart 1 km represent really 1 km, then AB is the geodesic. If 1 km represent the the distance you make in 1h, then the path through C will be the geodesic.
$endgroup$
– Surb
Jan 27 at 14:34
$begingroup$
This is a great question, and while the answers address it, you might also be interested in the brachistochrone problem (en.wikipedia.org/wiki/Brachistochrone_curve), which uses the same methods we use for studying geodesics, to find "fastest path" curves (in a somewhat different context).
$endgroup$
– John Hughes
Jan 27 at 16:32
$begingroup$
I believe that general relativity uses geodesics which are the fastest path between two points in spacetime (and hence the path followed by light)—but that's the result of using a metric where distance can be interpreted as time.
$endgroup$
– timtfj
Jan 27 at 16:41
$begingroup$
They are the shortest path as far as I know. The quickest also depends on velocity, right?
$endgroup$
– Exp ikx
Jan 27 at 14:32
$begingroup$
They are the shortest path as far as I know. The quickest also depends on velocity, right?
$endgroup$
– Exp ikx
Jan 27 at 14:32
1
1
$begingroup$
It depend on your metric ! If on your chart 1 km represent really 1 km, then AB is the geodesic. If 1 km represent the the distance you make in 1h, then the path through C will be the geodesic.
$endgroup$
– Surb
Jan 27 at 14:34
$begingroup$
It depend on your metric ! If on your chart 1 km represent really 1 km, then AB is the geodesic. If 1 km represent the the distance you make in 1h, then the path through C will be the geodesic.
$endgroup$
– Surb
Jan 27 at 14:34
$begingroup$
This is a great question, and while the answers address it, you might also be interested in the brachistochrone problem (en.wikipedia.org/wiki/Brachistochrone_curve), which uses the same methods we use for studying geodesics, to find "fastest path" curves (in a somewhat different context).
$endgroup$
– John Hughes
Jan 27 at 16:32
$begingroup$
This is a great question, and while the answers address it, you might also be interested in the brachistochrone problem (en.wikipedia.org/wiki/Brachistochrone_curve), which uses the same methods we use for studying geodesics, to find "fastest path" curves (in a somewhat different context).
$endgroup$
– John Hughes
Jan 27 at 16:32
$begingroup$
I believe that general relativity uses geodesics which are the fastest path between two points in spacetime (and hence the path followed by light)—but that's the result of using a metric where distance can be interpreted as time.
$endgroup$
– timtfj
Jan 27 at 16:41
$begingroup$
I believe that general relativity uses geodesics which are the fastest path between two points in spacetime (and hence the path followed by light)—but that's the result of using a metric where distance can be interpreted as time.
$endgroup$
– timtfj
Jan 27 at 16:41
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First note that a Geodesic does not have to be the quickest path between two points: For example, there are two geodesics on the sphere from say the north pole to say London. The shortest path is going down the Greenwich meridian from the north pole to London. But going from the north pole to the south pole along that same meridian and then going from the south pole to London along the shortest path (again along the Greenwich meridian) is also a geodesic from the North pole to London.
However, the shortest path between two points (if such a shortest path exists!) always is a geodesic.
Now to your actual question: what is meant by distance? In my example above "distance" meant the usual distance on a sphere. However you can choose what you mean by "distance" by specifying a metric. Roughly speaking a metric lets you measure lengths (and angles, too) in space. Your metric can just measure distance between points but you could modify your metric to take into account that something is slowing down movement.
answered Jan 27 at 14:56
0x5390x539
1,445518
$endgroup$
add a comment |
$begingroup$
Is the shortest, but only locally.
Think about $mathbb{S}^{2}$ with the metric induced by euclidean topology, then great circles are geodesic.
Take two points $A=(0,0,1)$, $B=(0,frac{sqrt{3}}{2},frac{1}{2})$ then both $alpha(t)=(0,sin(t),cos(t))$, $tin[0,frac{pi}{3}]$ and $beta(t)=(0,sin(t),cos(t))$, $tin[0,-frac{5}{3}pi]$, are geodesic, but $alpha$ is shorter than $beta$.
Moreover, it's easy to show that exist paths shorter than $beta$ which aren't geodesic.
answered Jan 27 at 14:58
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
A geodesic (resp. a curve) $gamma:[0,T]to M,tmapstogamma(t)$ can be reparametrized by $tilde{gamma}:[0,frac{T}{c}],tmapstogamma(ct)$, the resulting curve still being a geodesic (resp. a curve) but $c$ times quicker than $gamma$. So, since talking about "quickest curves" can be confusing since it depends of the parametrizations you choose, you can :
choose a reparametrization with unit speed (that is $|gamma'(t)|equiv 1$, it exists and for geodesics it is even one of the reparametrizations of the form $tmapstogamma(ct)$ since a geodesic always has constant speed: $$D_t|gamma'|^2=D_t(gamma',gamma')=2(gamma',D_tgamma')=2(gamma',0)=0,$$
and so $|gamma'|$ is constant),read the $tilde T$ obtained in the new interval $tildegamma:[0,tilde T]to M$ (which now corresponds to a traveled distance!),
check that geodesics are (locally) the shortest paths, that is if there is a unit speed curve $gamma$ joining $a$ and $b$ such that $T_gammaleq T_beta$ for all other unit speed curves $beta$ joining $a$ and $b$, then $gamma$ is a geodesic.
answered Jan 28 at 9:57
BalloonBalloon
4,645822
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
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oldest
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active
oldest
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oldest
votes
$begingroup$
First note that a Geodesic does not have to be the quickest path between two points: For example, there are two geodesics on the sphere from say the north pole to say London. The shortest path is going down the Greenwich meridian from the north pole to London. But going from the north pole to the south pole along that same meridian and then going from the south pole to London along the shortest path (again along the Greenwich meridian) is also a geodesic from the North pole to London.
However, the shortest path between two points (if such a shortest path exists!) always is a geodesic.
Now to your actual question: what is meant by distance? In my example above "distance" meant the usual distance on a sphere. However you can choose what you mean by "distance" by specifying a metric. Roughly speaking a metric lets you measure lengths (and angles, too) in space. Your metric can just measure distance between points but you could modify your metric to take into account that something is slowing down movement.
answered Jan 27 at 14:56
0x5390x539
1,445518
$endgroup$
add a comment |
$begingroup$
First note that a Geodesic does not have to be the quickest path between two points: For example, there are two geodesics on the sphere from say the north pole to say London. The shortest path is going down the Greenwich meridian from the north pole to London. But going from the north pole to the south pole along that same meridian and then going from the south pole to London along the shortest path (again along the Greenwich meridian) is also a geodesic from the North pole to London.
However, the shortest path between two points (if such a shortest path exists!) always is a geodesic.
Now to your actual question: what is meant by distance? In my example above "distance" meant the usual distance on a sphere. However you can choose what you mean by "distance" by specifying a metric. Roughly speaking a metric lets you measure lengths (and angles, too) in space. Your metric can just measure distance between points but you could modify your metric to take into account that something is slowing down movement.
answered Jan 27 at 14:56
0x5390x539
1,445518
$endgroup$
add a comment |
$begingroup$
First note that a Geodesic does not have to be the quickest path between two points: For example, there are two geodesics on the sphere from say the north pole to say London. The shortest path is going down the Greenwich meridian from the north pole to London. But going from the north pole to the south pole along that same meridian and then going from the south pole to London along the shortest path (again along the Greenwich meridian) is also a geodesic from the North pole to London.
However, the shortest path between two points (if such a shortest path exists!) always is a geodesic.
Now to your actual question: what is meant by distance? In my example above "distance" meant the usual distance on a sphere. However you can choose what you mean by "distance" by specifying a metric. Roughly speaking a metric lets you measure lengths (and angles, too) in space. Your metric can just measure distance between points but you could modify your metric to take into account that something is slowing down movement.
answered Jan 27 at 14:56
0x5390x539
1,445518
$endgroup$
First note that a Geodesic does not have to be the quickest path between two points: For example, there are two geodesics on the sphere from say the north pole to say London. The shortest path is going down the Greenwich meridian from the north pole to London. But going from the north pole to the south pole along that same meridian and then going from the south pole to London along the shortest path (again along the Greenwich meridian) is also a geodesic from the North pole to London.
However, the shortest path between two points (if such a shortest path exists!) always is a geodesic.
Now to your actual question: what is meant by distance? In my example above "distance" meant the usual distance on a sphere. However you can choose what you mean by "distance" by specifying a metric. Roughly speaking a metric lets you measure lengths (and angles, too) in space. Your metric can just measure distance between points but you could modify your metric to take into account that something is slowing down movement.
answered Jan 27 at 14:56
0x5390x539
1,445518
answered Jan 27 at 14:56
0x5390x539
1,445518
answered Jan 27 at 14:56
0x5390x539
1,445518
answered Jan 27 at 14:56
0x5390x539
1,445518
1,445518
add a comment |
add a comment |
$begingroup$
Is the shortest, but only locally.
Think about $mathbb{S}^{2}$ with the metric induced by euclidean topology, then great circles are geodesic.
Take two points $A=(0,0,1)$, $B=(0,frac{sqrt{3}}{2},frac{1}{2})$ then both $alpha(t)=(0,sin(t),cos(t))$, $tin[0,frac{pi}{3}]$ and $beta(t)=(0,sin(t),cos(t))$, $tin[0,-frac{5}{3}pi]$, are geodesic, but $alpha$ is shorter than $beta$.
Moreover, it's easy to show that exist paths shorter than $beta$ which aren't geodesic.
answered Jan 27 at 14:58
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
Is the shortest, but only locally.
Think about $mathbb{S}^{2}$ with the metric induced by euclidean topology, then great circles are geodesic.
Take two points $A=(0,0,1)$, $B=(0,frac{sqrt{3}}{2},frac{1}{2})$ then both $alpha(t)=(0,sin(t),cos(t))$, $tin[0,frac{pi}{3}]$ and $beta(t)=(0,sin(t),cos(t))$, $tin[0,-frac{5}{3}pi]$, are geodesic, but $alpha$ is shorter than $beta$.
Moreover, it's easy to show that exist paths shorter than $beta$ which aren't geodesic.
answered Jan 27 at 14:58
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
Is the shortest, but only locally.
Think about $mathbb{S}^{2}$ with the metric induced by euclidean topology, then great circles are geodesic.
Take two points $A=(0,0,1)$, $B=(0,frac{sqrt{3}}{2},frac{1}{2})$ then both $alpha(t)=(0,sin(t),cos(t))$, $tin[0,frac{pi}{3}]$ and $beta(t)=(0,sin(t),cos(t))$, $tin[0,-frac{5}{3}pi]$, are geodesic, but $alpha$ is shorter than $beta$.
Moreover, it's easy to show that exist paths shorter than $beta$ which aren't geodesic.
answered Jan 27 at 14:58
ecrinecrin
3477
$endgroup$
Is the shortest, but only locally.
Think about $mathbb{S}^{2}$ with the metric induced by euclidean topology, then great circles are geodesic.
Take two points $A=(0,0,1)$, $B=(0,frac{sqrt{3}}{2},frac{1}{2})$ then both $alpha(t)=(0,sin(t),cos(t))$, $tin[0,frac{pi}{3}]$ and $beta(t)=(0,sin(t),cos(t))$, $tin[0,-frac{5}{3}pi]$, are geodesic, but $alpha$ is shorter than $beta$.
Moreover, it's easy to show that exist paths shorter than $beta$ which aren't geodesic.
answered Jan 27 at 14:58
ecrinecrin
3477
edited Jan 27 at 16:10
answered Jan 27 at 14:58
ecrinecrin
3477
answered Jan 27 at 14:58
ecrinecrin
3477
answered Jan 27 at 14:58
ecrinecrin
3477
3477
add a comment |
add a comment |
$begingroup$
A geodesic (resp. a curve) $gamma:[0,T]to M,tmapstogamma(t)$ can be reparametrized by $tilde{gamma}:[0,frac{T}{c}],tmapstogamma(ct)$, the resulting curve still being a geodesic (resp. a curve) but $c$ times quicker than $gamma$. So, since talking about "quickest curves" can be confusing since it depends of the parametrizations you choose, you can :
choose a reparametrization with unit speed (that is $|gamma'(t)|equiv 1$, it exists and for geodesics it is even one of the reparametrizations of the form $tmapstogamma(ct)$ since a geodesic always has constant speed: $$D_t|gamma'|^2=D_t(gamma',gamma')=2(gamma',D_tgamma')=2(gamma',0)=0,$$
and so $|gamma'|$ is constant),read the $tilde T$ obtained in the new interval $tildegamma:[0,tilde T]to M$ (which now corresponds to a traveled distance!),
check that geodesics are (locally) the shortest paths, that is if there is a unit speed curve $gamma$ joining $a$ and $b$ such that $T_gammaleq T_beta$ for all other unit speed curves $beta$ joining $a$ and $b$, then $gamma$ is a geodesic.
answered Jan 28 at 9:57
BalloonBalloon
4,645822
$endgroup$
add a comment |
$begingroup$
A geodesic (resp. a curve) $gamma:[0,T]to M,tmapstogamma(t)$ can be reparametrized by $tilde{gamma}:[0,frac{T}{c}],tmapstogamma(ct)$, the resulting curve still being a geodesic (resp. a curve) but $c$ times quicker than $gamma$. So, since talking about "quickest curves" can be confusing since it depends of the parametrizations you choose, you can :
choose a reparametrization with unit speed (that is $|gamma'(t)|equiv 1$, it exists and for geodesics it is even one of the reparametrizations of the form $tmapstogamma(ct)$ since a geodesic always has constant speed: $$D_t|gamma'|^2=D_t(gamma',gamma')=2(gamma',D_tgamma')=2(gamma',0)=0,$$
and so $|gamma'|$ is constant),read the $tilde T$ obtained in the new interval $tildegamma:[0,tilde T]to M$ (which now corresponds to a traveled distance!),
check that geodesics are (locally) the shortest paths, that is if there is a unit speed curve $gamma$ joining $a$ and $b$ such that $T_gammaleq T_beta$ for all other unit speed curves $beta$ joining $a$ and $b$, then $gamma$ is a geodesic.
answered Jan 28 at 9:57
BalloonBalloon
4,645822
$endgroup$
add a comment |
$begingroup$
A geodesic (resp. a curve) $gamma:[0,T]to M,tmapstogamma(t)$ can be reparametrized by $tilde{gamma}:[0,frac{T}{c}],tmapstogamma(ct)$, the resulting curve still being a geodesic (resp. a curve) but $c$ times quicker than $gamma$. So, since talking about "quickest curves" can be confusing since it depends of the parametrizations you choose, you can :
choose a reparametrization with unit speed (that is $|gamma'(t)|equiv 1$, it exists and for geodesics it is even one of the reparametrizations of the form $tmapstogamma(ct)$ since a geodesic always has constant speed: $$D_t|gamma'|^2=D_t(gamma',gamma')=2(gamma',D_tgamma')=2(gamma',0)=0,$$
and so $|gamma'|$ is constant),read the $tilde T$ obtained in the new interval $tildegamma:[0,tilde T]to M$ (which now corresponds to a traveled distance!),
check that geodesics are (locally) the shortest paths, that is if there is a unit speed curve $gamma$ joining $a$ and $b$ such that $T_gammaleq T_beta$ for all other unit speed curves $beta$ joining $a$ and $b$, then $gamma$ is a geodesic.
answered Jan 28 at 9:57
BalloonBalloon
4,645822
$endgroup$
A geodesic (resp. a curve) $gamma:[0,T]to M,tmapstogamma(t)$ can be reparametrized by $tilde{gamma}:[0,frac{T}{c}],tmapstogamma(ct)$, the resulting curve still being a geodesic (resp. a curve) but $c$ times quicker than $gamma$. So, since talking about "quickest curves" can be confusing since it depends of the parametrizations you choose, you can :
choose a reparametrization with unit speed (that is $|gamma'(t)|equiv 1$, it exists and for geodesics it is even one of the reparametrizations of the form $tmapstogamma(ct)$ since a geodesic always has constant speed: $$D_t|gamma'|^2=D_t(gamma',gamma')=2(gamma',D_tgamma')=2(gamma',0)=0,$$
and so $|gamma'|$ is constant),read the $tilde T$ obtained in the new interval $tildegamma:[0,tilde T]to M$ (which now corresponds to a traveled distance!),
check that geodesics are (locally) the shortest paths, that is if there is a unit speed curve $gamma$ joining $a$ and $b$ such that $T_gammaleq T_beta$ for all other unit speed curves $beta$ joining $a$ and $b$, then $gamma$ is a geodesic.
answered Jan 28 at 9:57
BalloonBalloon
4,645822
answered Jan 28 at 9:57
BalloonBalloon
4,645822
answered Jan 28 at 9:57
BalloonBalloon
4,645822
answered Jan 28 at 9:57
BalloonBalloon
4,645822
4,645822
add a comment |
add a comment |
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$begingroup$
They are the shortest path as far as I know. The quickest also depends on velocity, right?
$endgroup$
– Exp ikx
Jan 27 at 14:32
1
$begingroup$
It depend on your metric ! If on your chart 1 km represent really 1 km, then AB is the geodesic. If 1 km represent the the distance you make in 1h, then the path through C will be the geodesic.
$endgroup$
– Surb
Jan 27 at 14:34
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This is a great question, and while the answers address it, you might also be interested in the brachistochrone problem (en.wikipedia.org/wiki/Brachistochrone_curve), which uses the same methods we use for studying geodesics, to find "fastest path" curves (in a somewhat different context).
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– John Hughes
Jan 27 at 16:32
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I believe that general relativity uses geodesics which are the fastest path between two points in spacetime (and hence the path followed by light)—but that's the result of using a metric where distance can be interpreted as time.
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– timtfj
Jan 27 at 16:41