Is there such a thing as a “most parallel” vector on a curved manifold?
My question is in response to Dirac's assertion:
Suppose we have a vector $left{ A^{mu}right}$ at a point $mathscr{P}$. If the space is curved, we cannot give meaning to a parallel vector at a distant point $mathscr{Q}$, as one can easily see if one thinks of the example of a curved two-dimensional space in a three-dimensional Euclidean space.
The statement is from Chapter 6 of General Theory of Relativity by P.A.M Dirac.
Let us restrict consideration to the surface of a globe such as an idealized Earth surface. We shall further restrict consideration to an open "nice" region which is small relative to the radius of the globe. To be specific, let us describe a boundary $partial{mathcal{D}_{o}}$ which is at a constant geodesic distance of 1000 kilometers from some designated location $mathscr{P}_{o}$. We are assuming our globe is measured with units proportional in scale to the size of Earth. We call $mathcal{D}_{o}$ the open region bounded by $partial{mathcal{D}_{o}}$ with the smaller area.
Suppose now that we have a disk $deltamathcal{D}_{o}$ of geodesic radius 1 meter, centered on $mathscr{P}_{o}$, upon which we draw two perpendicular two-space diameters (which intersect at $mathscr{P}_{o}$). Since $deltamathcal{D}_{o}$ is practically indistinguishable form a disk living in the tangent Euclidean plane, using naive concepts of positively directed segments, we designate one of the (half-diameter) radial segments $mathfrak{hat{e}}_{x}$, and another $mathfrak{hat{e}}_{y}$. These constitute an orthonormal spanning basis of the tangent plane at $mathscr{P}_{o}$.
It seems intuitively obvious to me that this disk $,deltamathcal{D}_{o}$ can be parallel transported along geodesics intersecting at $mathscr{P}_{o}$ to every point $mathscr{Q}$ in $mathcal{D}_{o}$ so that, at every point $mathscr{Q}$ in $mathcal{D}_{o}$ there is uniquely determined image of $deltamathcal{D}_{o}$.
My question is this: in my restricted example, have I not provided a definition of "parallel" such that at every point $mathscr{Q}inmathcal{D}_{o}$ there is a one-to-one mapping between the tangent plane $mathcal{T}_{mathscr{Q}}$ and $mathcal{D}_{o}$?
I am not pretending to have refuted Dirac's assertion. I am merely attempting to determine its exact meaning.
differential-geometry riemannian-geometry mathematical-physics
edited Nov 27 '18 at 18:44
Jean Marie
28.8k41949
asked Nov 27 '18 at 17:40
Steven Hatton
721315
add a comment |
My question is in response to Dirac's assertion:
Suppose we have a vector $left{ A^{mu}right}$ at a point $mathscr{P}$. If the space is curved, we cannot give meaning to a parallel vector at a distant point $mathscr{Q}$, as one can easily see if one thinks of the example of a curved two-dimensional space in a three-dimensional Euclidean space.
The statement is from Chapter 6 of General Theory of Relativity by P.A.M Dirac.
Let us restrict consideration to the surface of a globe such as an idealized Earth surface. We shall further restrict consideration to an open "nice" region which is small relative to the radius of the globe. To be specific, let us describe a boundary $partial{mathcal{D}_{o}}$ which is at a constant geodesic distance of 1000 kilometers from some designated location $mathscr{P}_{o}$. We are assuming our globe is measured with units proportional in scale to the size of Earth. We call $mathcal{D}_{o}$ the open region bounded by $partial{mathcal{D}_{o}}$ with the smaller area.
Suppose now that we have a disk $deltamathcal{D}_{o}$ of geodesic radius 1 meter, centered on $mathscr{P}_{o}$, upon which we draw two perpendicular two-space diameters (which intersect at $mathscr{P}_{o}$). Since $deltamathcal{D}_{o}$ is practically indistinguishable form a disk living in the tangent Euclidean plane, using naive concepts of positively directed segments, we designate one of the (half-diameter) radial segments $mathfrak{hat{e}}_{x}$, and another $mathfrak{hat{e}}_{y}$. These constitute an orthonormal spanning basis of the tangent plane at $mathscr{P}_{o}$.
It seems intuitively obvious to me that this disk $,deltamathcal{D}_{o}$ can be parallel transported along geodesics intersecting at $mathscr{P}_{o}$ to every point $mathscr{Q}$ in $mathcal{D}_{o}$ so that, at every point $mathscr{Q}$ in $mathcal{D}_{o}$ there is uniquely determined image of $deltamathcal{D}_{o}$.
My question is this: in my restricted example, have I not provided a definition of "parallel" such that at every point $mathscr{Q}inmathcal{D}_{o}$ there is a one-to-one mapping between the tangent plane $mathcal{T}_{mathscr{Q}}$ and $mathcal{D}_{o}$?
I am not pretending to have refuted Dirac's assertion. I am merely attempting to determine its exact meaning.
differential-geometry riemannian-geometry mathematical-physics
edited Nov 27 '18 at 18:44
Jean Marie
28.8k41949
asked Nov 27 '18 at 17:40
Steven Hatton
721315
1
Of course you can do that. As soon you have a metric you have a notion of parallel transport along a curve, and in particular along geodesics. The question is whether the result is, in any way, natural, i.e. whether it depends on a choice you made. Critical is the choice of the curve -- and in general, in curved manifolds, the result depends on the curve chosen. See here: en.wikipedia.org/wiki/Holonomy. -- the image on the top right should be especially instructive.
– Thomas
Nov 27 '18 at 17:51
This is an useful construction -- in fact, I am using it in my work, both in spherical and in hyperbolic geometry -- but it does not have all the properties of "parallel" we would want (as mentioned by Thomas, if you transport v from A to B, and then from B to C, then the result is (generally) not the same as when you transport v directly from A to C)... It is less confusing if we do not insist on calling this parallel.
– Zeno Rogue
Nov 27 '18 at 23:05
add a comment |
My question is in response to Dirac's assertion:
Suppose we have a vector $left{ A^{mu}right}$ at a point $mathscr{P}$. If the space is curved, we cannot give meaning to a parallel vector at a distant point $mathscr{Q}$, as one can easily see if one thinks of the example of a curved two-dimensional space in a three-dimensional Euclidean space.
The statement is from Chapter 6 of General Theory of Relativity by P.A.M Dirac.
Let us restrict consideration to the surface of a globe such as an idealized Earth surface. We shall further restrict consideration to an open "nice" region which is small relative to the radius of the globe. To be specific, let us describe a boundary $partial{mathcal{D}_{o}}$ which is at a constant geodesic distance of 1000 kilometers from some designated location $mathscr{P}_{o}$. We are assuming our globe is measured with units proportional in scale to the size of Earth. We call $mathcal{D}_{o}$ the open region bounded by $partial{mathcal{D}_{o}}$ with the smaller area.
Suppose now that we have a disk $deltamathcal{D}_{o}$ of geodesic radius 1 meter, centered on $mathscr{P}_{o}$, upon which we draw two perpendicular two-space diameters (which intersect at $mathscr{P}_{o}$). Since $deltamathcal{D}_{o}$ is practically indistinguishable form a disk living in the tangent Euclidean plane, using naive concepts of positively directed segments, we designate one of the (half-diameter) radial segments $mathfrak{hat{e}}_{x}$, and another $mathfrak{hat{e}}_{y}$. These constitute an orthonormal spanning basis of the tangent plane at $mathscr{P}_{o}$.
It seems intuitively obvious to me that this disk $,deltamathcal{D}_{o}$ can be parallel transported along geodesics intersecting at $mathscr{P}_{o}$ to every point $mathscr{Q}$ in $mathcal{D}_{o}$ so that, at every point $mathscr{Q}$ in $mathcal{D}_{o}$ there is uniquely determined image of $deltamathcal{D}_{o}$.
My question is this: in my restricted example, have I not provided a definition of "parallel" such that at every point $mathscr{Q}inmathcal{D}_{o}$ there is a one-to-one mapping between the tangent plane $mathcal{T}_{mathscr{Q}}$ and $mathcal{D}_{o}$?
I am not pretending to have refuted Dirac's assertion. I am merely attempting to determine its exact meaning.
differential-geometry riemannian-geometry mathematical-physics
edited Nov 27 '18 at 18:44
Jean Marie
28.8k41949
asked Nov 27 '18 at 17:40
Steven Hatton
721315
My question is in response to Dirac's assertion:
Suppose we have a vector $left{ A^{mu}right}$ at a point $mathscr{P}$. If the space is curved, we cannot give meaning to a parallel vector at a distant point $mathscr{Q}$, as one can easily see if one thinks of the example of a curved two-dimensional space in a three-dimensional Euclidean space.
The statement is from Chapter 6 of General Theory of Relativity by P.A.M Dirac.
Let us restrict consideration to the surface of a globe such as an idealized Earth surface. We shall further restrict consideration to an open "nice" region which is small relative to the radius of the globe. To be specific, let us describe a boundary $partial{mathcal{D}_{o}}$ which is at a constant geodesic distance of 1000 kilometers from some designated location $mathscr{P}_{o}$. We are assuming our globe is measured with units proportional in scale to the size of Earth. We call $mathcal{D}_{o}$ the open region bounded by $partial{mathcal{D}_{o}}$ with the smaller area.
Suppose now that we have a disk $deltamathcal{D}_{o}$ of geodesic radius 1 meter, centered on $mathscr{P}_{o}$, upon which we draw two perpendicular two-space diameters (which intersect at $mathscr{P}_{o}$). Since $deltamathcal{D}_{o}$ is practically indistinguishable form a disk living in the tangent Euclidean plane, using naive concepts of positively directed segments, we designate one of the (half-diameter) radial segments $mathfrak{hat{e}}_{x}$, and another $mathfrak{hat{e}}_{y}$. These constitute an orthonormal spanning basis of the tangent plane at $mathscr{P}_{o}$.
It seems intuitively obvious to me that this disk $,deltamathcal{D}_{o}$ can be parallel transported along geodesics intersecting at $mathscr{P}_{o}$ to every point $mathscr{Q}$ in $mathcal{D}_{o}$ so that, at every point $mathscr{Q}$ in $mathcal{D}_{o}$ there is uniquely determined image of $deltamathcal{D}_{o}$.
My question is this: in my restricted example, have I not provided a definition of "parallel" such that at every point $mathscr{Q}inmathcal{D}_{o}$ there is a one-to-one mapping between the tangent plane $mathcal{T}_{mathscr{Q}}$ and $mathcal{D}_{o}$?
I am not pretending to have refuted Dirac's assertion. I am merely attempting to determine its exact meaning.
differential-geometry riemannian-geometry mathematical-physics
differential-geometry riemannian-geometry mathematical-physics
edited Nov 27 '18 at 18:44
Jean Marie
28.8k41949
asked Nov 27 '18 at 17:40
Steven Hatton
721315
edited Nov 27 '18 at 18:44
Jean Marie
28.8k41949
asked Nov 27 '18 at 17:40
Steven Hatton
721315
edited Nov 27 '18 at 18:44
Jean Marie
28.8k41949
edited Nov 27 '18 at 18:44
Jean Marie
28.8k41949
edited Nov 27 '18 at 18:44
Jean Marie
28.8k41949
28.8k41949
asked Nov 27 '18 at 17:40
Steven Hatton
721315
asked Nov 27 '18 at 17:40
Steven Hatton
721315
asked Nov 27 '18 at 17:40
Steven Hatton
721315
721315
1
Of course you can do that. As soon you have a metric you have a notion of parallel transport along a curve, and in particular along geodesics. The question is whether the result is, in any way, natural, i.e. whether it depends on a choice you made. Critical is the choice of the curve -- and in general, in curved manifolds, the result depends on the curve chosen. See here: en.wikipedia.org/wiki/Holonomy. -- the image on the top right should be especially instructive.
– Thomas
Nov 27 '18 at 17:51
This is an useful construction -- in fact, I am using it in my work, both in spherical and in hyperbolic geometry -- but it does not have all the properties of "parallel" we would want (as mentioned by Thomas, if you transport v from A to B, and then from B to C, then the result is (generally) not the same as when you transport v directly from A to C)... It is less confusing if we do not insist on calling this parallel.
– Zeno Rogue
Nov 27 '18 at 23:05
add a comment |
1
Of course you can do that. As soon you have a metric you have a notion of parallel transport along a curve, and in particular along geodesics. The question is whether the result is, in any way, natural, i.e. whether it depends on a choice you made. Critical is the choice of the curve -- and in general, in curved manifolds, the result depends on the curve chosen. See here: en.wikipedia.org/wiki/Holonomy. -- the image on the top right should be especially instructive.
– Thomas
Nov 27 '18 at 17:51
This is an useful construction -- in fact, I am using it in my work, both in spherical and in hyperbolic geometry -- but it does not have all the properties of "parallel" we would want (as mentioned by Thomas, if you transport v from A to B, and then from B to C, then the result is (generally) not the same as when you transport v directly from A to C)... It is less confusing if we do not insist on calling this parallel.
– Zeno Rogue
Nov 27 '18 at 23:05
1
1
Of course you can do that. As soon you have a metric you have a notion of parallel transport along a curve, and in particular along geodesics. The question is whether the result is, in any way, natural, i.e. whether it depends on a choice you made. Critical is the choice of the curve -- and in general, in curved manifolds, the result depends on the curve chosen. See here: en.wikipedia.org/wiki/Holonomy. -- the image on the top right should be especially instructive.
– Thomas
Nov 27 '18 at 17:51
Of course you can do that. As soon you have a metric you have a notion of parallel transport along a curve, and in particular along geodesics. The question is whether the result is, in any way, natural, i.e. whether it depends on a choice you made. Critical is the choice of the curve -- and in general, in curved manifolds, the result depends on the curve chosen. See here: en.wikipedia.org/wiki/Holonomy. -- the image on the top right should be especially instructive.
– Thomas
Nov 27 '18 at 17:51
This is an useful construction -- in fact, I am using it in my work, both in spherical and in hyperbolic geometry -- but it does not have all the properties of "parallel" we would want (as mentioned by Thomas, if you transport v from A to B, and then from B to C, then the result is (generally) not the same as when you transport v directly from A to C)... It is less confusing if we do not insist on calling this parallel.
– Zeno Rogue
Nov 27 '18 at 23:05
This is an useful construction -- in fact, I am using it in my work, both in spherical and in hyperbolic geometry -- but it does not have all the properties of "parallel" we would want (as mentioned by Thomas, if you transport v from A to B, and then from B to C, then the result is (generally) not the same as when you transport v directly from A to C)... It is less confusing if we do not insist on calling this parallel.
– Zeno Rogue
Nov 27 '18 at 23:05
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Of course you can do that. As soon you have a metric you have a notion of parallel transport along a curve, and in particular along geodesics. The question is whether the result is, in any way, natural, i.e. whether it depends on a choice you made. Critical is the choice of the curve -- and in general, in curved manifolds, the result depends on the curve chosen. See here: en.wikipedia.org/wiki/Holonomy. -- the image on the top right should be especially instructive.
– Thomas
Nov 27 '18 at 17:51
This is an useful construction -- in fact, I am using it in my work, both in spherical and in hyperbolic geometry -- but it does not have all the properties of "parallel" we would want (as mentioned by Thomas, if you transport v from A to B, and then from B to C, then the result is (generally) not the same as when you transport v directly from A to C)... It is less confusing if we do not insist on calling this parallel.
– Zeno Rogue
Nov 27 '18 at 23:05