How can you determine if a point is inside a parametric 2D manifold?
$begingroup$
Asume I have an arbitrary, parametric, closed, orientable, surface; a sphere, ellipsoid, closed cylinder, weird general cone....
If you only have access to the parametrization, how can you determine whether a 3D point is inside or outside of the manifold?
calculus geometry manifolds parametric geometric-topology
edited Nov 29 '18 at 22:38
Bernard
119k639112
asked Nov 29 '18 at 21:42
MakoganMakogan
756217
$endgroup$
add a comment |
$begingroup$
Asume I have an arbitrary, parametric, closed, orientable, surface; a sphere, ellipsoid, closed cylinder, weird general cone....
If you only have access to the parametrization, how can you determine whether a 3D point is inside or outside of the manifold?
calculus geometry manifolds parametric geometric-topology
edited Nov 29 '18 at 22:38
Bernard
119k639112
asked Nov 29 '18 at 21:42
MakoganMakogan
756217
$endgroup$
add a comment |
$begingroup$
Asume I have an arbitrary, parametric, closed, orientable, surface; a sphere, ellipsoid, closed cylinder, weird general cone....
If you only have access to the parametrization, how can you determine whether a 3D point is inside or outside of the manifold?
calculus geometry manifolds parametric geometric-topology
edited Nov 29 '18 at 22:38
Bernard
119k639112
asked Nov 29 '18 at 21:42
MakoganMakogan
756217
$endgroup$
Asume I have an arbitrary, parametric, closed, orientable, surface; a sphere, ellipsoid, closed cylinder, weird general cone....
If you only have access to the parametrization, how can you determine whether a 3D point is inside or outside of the manifold?
calculus geometry manifolds parametric geometric-topology
calculus geometry manifolds parametric geometric-topology
edited Nov 29 '18 at 22:38
Bernard
119k639112
asked Nov 29 '18 at 21:42
MakoganMakogan
756217
edited Nov 29 '18 at 22:38
Bernard
119k639112
asked Nov 29 '18 at 21:42
MakoganMakogan
756217
edited Nov 29 '18 at 22:38
Bernard
119k639112
edited Nov 29 '18 at 22:38
Bernard
119k639112
edited Nov 29 '18 at 22:38
Bernard
119k639112
119k639112
asked Nov 29 '18 at 21:42
MakoganMakogan
756217
asked Nov 29 '18 at 21:42
MakoganMakogan
756217
asked Nov 29 '18 at 21:42
MakoganMakogan
756217
756217
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I presume you are trying to determine is if a point $p$ is inside or outside the region $Omega$ of space enclosed by the manifold, which can be seen as the boundary $partial Omega$ of $Omega$. A solution is to use use Stokes–Cartan theorem:
$$
int_{partial Omega} omega = int_{Omega} domega
$$
with a suitable choice of the differential form $omega$. In our case the following will work, compute:
$$
I = int_{partial Omega} frac{(x-p)}{|x-p|^3} dS
$$
over the manifold, where $x$ is the (3-dim) variable of integration, and $dS$ is the 2-form representing the area element of the surface. If $p$ is outside $Omega$ then $I=0$. If $p$ is inside then $|I|$ yields the area of a unit sphere. Replace the exponent 3 in the denominator with the number of dimensions for a more general result, e.g., for $n=2$ (a curve in the 2-dim space) what you get is the winding-number of the curve around $p$ multiplied by $2pi$.
answered Nov 29 '18 at 22:56
mlerma54mlerma54
1,152148
$endgroup$
1
$begingroup$
I am going to stare really hard at this until i understand at least half
$endgroup$
– Makogan
Nov 29 '18 at 23:00
$begingroup$
Informally, think of the integrand as a small solid angle determined by rays traced from $p$ to a small portion of your surface around a point $x$ selected on the surface. E.g. see how it looks if your surface is a sphere of radius $r$ and $p$ is the origin of coordinates, then compute the integral in polar coord. On the other hand, if $p$ is outside the region enclosed by the manifold then by the Stokes theorem the integral has to be zero because our $omega = frac{(x-p)dS}{|x-p|^3}$ is a closed form and $domega = 0$. I believe I saw a similar example in Spivak's book on diff.geom.
$endgroup$
– mlerma54
Nov 30 '18 at 3:34
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
I presume you are trying to determine is if a point $p$ is inside or outside the region $Omega$ of space enclosed by the manifold, which can be seen as the boundary $partial Omega$ of $Omega$. A solution is to use use Stokes–Cartan theorem:
$$
int_{partial Omega} omega = int_{Omega} domega
$$
with a suitable choice of the differential form $omega$. In our case the following will work, compute:
$$
I = int_{partial Omega} frac{(x-p)}{|x-p|^3} dS
$$
over the manifold, where $x$ is the (3-dim) variable of integration, and $dS$ is the 2-form representing the area element of the surface. If $p$ is outside $Omega$ then $I=0$. If $p$ is inside then $|I|$ yields the area of a unit sphere. Replace the exponent 3 in the denominator with the number of dimensions for a more general result, e.g., for $n=2$ (a curve in the 2-dim space) what you get is the winding-number of the curve around $p$ multiplied by $2pi$.
answered Nov 29 '18 at 22:56
mlerma54mlerma54
1,152148
$endgroup$
1
$begingroup$
I am going to stare really hard at this until i understand at least half
$endgroup$
– Makogan
Nov 29 '18 at 23:00
$begingroup$
Informally, think of the integrand as a small solid angle determined by rays traced from $p$ to a small portion of your surface around a point $x$ selected on the surface. E.g. see how it looks if your surface is a sphere of radius $r$ and $p$ is the origin of coordinates, then compute the integral in polar coord. On the other hand, if $p$ is outside the region enclosed by the manifold then by the Stokes theorem the integral has to be zero because our $omega = frac{(x-p)dS}{|x-p|^3}$ is a closed form and $domega = 0$. I believe I saw a similar example in Spivak's book on diff.geom.
$endgroup$
– mlerma54
Nov 30 '18 at 3:34
add a comment |
$begingroup$
I presume you are trying to determine is if a point $p$ is inside or outside the region $Omega$ of space enclosed by the manifold, which can be seen as the boundary $partial Omega$ of $Omega$. A solution is to use use Stokes–Cartan theorem:
$$
int_{partial Omega} omega = int_{Omega} domega
$$
with a suitable choice of the differential form $omega$. In our case the following will work, compute:
$$
I = int_{partial Omega} frac{(x-p)}{|x-p|^3} dS
$$
over the manifold, where $x$ is the (3-dim) variable of integration, and $dS$ is the 2-form representing the area element of the surface. If $p$ is outside $Omega$ then $I=0$. If $p$ is inside then $|I|$ yields the area of a unit sphere. Replace the exponent 3 in the denominator with the number of dimensions for a more general result, e.g., for $n=2$ (a curve in the 2-dim space) what you get is the winding-number of the curve around $p$ multiplied by $2pi$.
answered Nov 29 '18 at 22:56
mlerma54mlerma54
1,152148
$endgroup$
1
$begingroup$
I am going to stare really hard at this until i understand at least half
$endgroup$
– Makogan
Nov 29 '18 at 23:00
$begingroup$
Informally, think of the integrand as a small solid angle determined by rays traced from $p$ to a small portion of your surface around a point $x$ selected on the surface. E.g. see how it looks if your surface is a sphere of radius $r$ and $p$ is the origin of coordinates, then compute the integral in polar coord. On the other hand, if $p$ is outside the region enclosed by the manifold then by the Stokes theorem the integral has to be zero because our $omega = frac{(x-p)dS}{|x-p|^3}$ is a closed form and $domega = 0$. I believe I saw a similar example in Spivak's book on diff.geom.
$endgroup$
– mlerma54
Nov 30 '18 at 3:34
add a comment |
$begingroup$
I presume you are trying to determine is if a point $p$ is inside or outside the region $Omega$ of space enclosed by the manifold, which can be seen as the boundary $partial Omega$ of $Omega$. A solution is to use use Stokes–Cartan theorem:
$$
int_{partial Omega} omega = int_{Omega} domega
$$
with a suitable choice of the differential form $omega$. In our case the following will work, compute:
$$
I = int_{partial Omega} frac{(x-p)}{|x-p|^3} dS
$$
over the manifold, where $x$ is the (3-dim) variable of integration, and $dS$ is the 2-form representing the area element of the surface. If $p$ is outside $Omega$ then $I=0$. If $p$ is inside then $|I|$ yields the area of a unit sphere. Replace the exponent 3 in the denominator with the number of dimensions for a more general result, e.g., for $n=2$ (a curve in the 2-dim space) what you get is the winding-number of the curve around $p$ multiplied by $2pi$.
answered Nov 29 '18 at 22:56
mlerma54mlerma54
1,152148
$endgroup$
I presume you are trying to determine is if a point $p$ is inside or outside the region $Omega$ of space enclosed by the manifold, which can be seen as the boundary $partial Omega$ of $Omega$. A solution is to use use Stokes–Cartan theorem:
$$
int_{partial Omega} omega = int_{Omega} domega
$$
with a suitable choice of the differential form $omega$. In our case the following will work, compute:
$$
I = int_{partial Omega} frac{(x-p)}{|x-p|^3} dS
$$
over the manifold, where $x$ is the (3-dim) variable of integration, and $dS$ is the 2-form representing the area element of the surface. If $p$ is outside $Omega$ then $I=0$. If $p$ is inside then $|I|$ yields the area of a unit sphere. Replace the exponent 3 in the denominator with the number of dimensions for a more general result, e.g., for $n=2$ (a curve in the 2-dim space) what you get is the winding-number of the curve around $p$ multiplied by $2pi$.
answered Nov 29 '18 at 22:56
mlerma54mlerma54
1,152148
answered Nov 29 '18 at 22:56
mlerma54mlerma54
1,152148
answered Nov 29 '18 at 22:56
mlerma54mlerma54
1,152148
answered Nov 29 '18 at 22:56
mlerma54mlerma54
1,152148
1,152148
1
$begingroup$
I am going to stare really hard at this until i understand at least half
$endgroup$
– Makogan
Nov 29 '18 at 23:00
$begingroup$
Informally, think of the integrand as a small solid angle determined by rays traced from $p$ to a small portion of your surface around a point $x$ selected on the surface. E.g. see how it looks if your surface is a sphere of radius $r$ and $p$ is the origin of coordinates, then compute the integral in polar coord. On the other hand, if $p$ is outside the region enclosed by the manifold then by the Stokes theorem the integral has to be zero because our $omega = frac{(x-p)dS}{|x-p|^3}$ is a closed form and $domega = 0$. I believe I saw a similar example in Spivak's book on diff.geom.
$endgroup$
– mlerma54
Nov 30 '18 at 3:34
add a comment |
1
$begingroup$
I am going to stare really hard at this until i understand at least half
$endgroup$
– Makogan
Nov 29 '18 at 23:00
$begingroup$
Informally, think of the integrand as a small solid angle determined by rays traced from $p$ to a small portion of your surface around a point $x$ selected on the surface. E.g. see how it looks if your surface is a sphere of radius $r$ and $p$ is the origin of coordinates, then compute the integral in polar coord. On the other hand, if $p$ is outside the region enclosed by the manifold then by the Stokes theorem the integral has to be zero because our $omega = frac{(x-p)dS}{|x-p|^3}$ is a closed form and $domega = 0$. I believe I saw a similar example in Spivak's book on diff.geom.
$endgroup$
– mlerma54
Nov 30 '18 at 3:34
1
1
$begingroup$
I am going to stare really hard at this until i understand at least half
$endgroup$
– Makogan
Nov 29 '18 at 23:00
$begingroup$
I am going to stare really hard at this until i understand at least half
$endgroup$
– Makogan
Nov 29 '18 at 23:00
$begingroup$
Informally, think of the integrand as a small solid angle determined by rays traced from $p$ to a small portion of your surface around a point $x$ selected on the surface. E.g. see how it looks if your surface is a sphere of radius $r$ and $p$ is the origin of coordinates, then compute the integral in polar coord. On the other hand, if $p$ is outside the region enclosed by the manifold then by the Stokes theorem the integral has to be zero because our $omega = frac{(x-p)dS}{|x-p|^3}$ is a closed form and $domega = 0$. I believe I saw a similar example in Spivak's book on diff.geom.
$endgroup$
– mlerma54
Nov 30 '18 at 3:34
$begingroup$
Informally, think of the integrand as a small solid angle determined by rays traced from $p$ to a small portion of your surface around a point $x$ selected on the surface. E.g. see how it looks if your surface is a sphere of radius $r$ and $p$ is the origin of coordinates, then compute the integral in polar coord. On the other hand, if $p$ is outside the region enclosed by the manifold then by the Stokes theorem the integral has to be zero because our $omega = frac{(x-p)dS}{|x-p|^3}$ is a closed form and $domega = 0$. I believe I saw a similar example in Spivak's book on diff.geom.
$endgroup$
– mlerma54
Nov 30 '18 at 3:34
add a comment |
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