A geometric argument for geodesics in manifolds also being geodesics in hypersurafces?
$begingroup$
I've read that when a hypersurface within a manifold contains a curve, if the curve is a geodesic in the manifold, it is also a geodesic in the hypersurface.
This is quite abstract for me, I've only recently started learning GR, could someone provide a geometric (or if necessary, algebraic) argument for why this is true? Is the converse ever/always/never true?
(Based on my limited intuition, I'd guess sometimes but not always, but I can't think of anything a professor wouldn't call 'handwavey').
Cheers
general-relativity
asked Nov 6 '18 at 12:29
T. SmithT. Smith
111
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migrated from physics.stackexchange.com Nov 7 '18 at 23:23
This question came from our site for active researchers, academics and students of physics.
add a comment |
$begingroup$
I've read that when a hypersurface within a manifold contains a curve, if the curve is a geodesic in the manifold, it is also a geodesic in the hypersurface.
This is quite abstract for me, I've only recently started learning GR, could someone provide a geometric (or if necessary, algebraic) argument for why this is true? Is the converse ever/always/never true?
(Based on my limited intuition, I'd guess sometimes but not always, but I can't think of anything a professor wouldn't call 'handwavey').
Cheers
general-relativity
asked Nov 6 '18 at 12:29
T. SmithT. Smith
111
$endgroup$
migrated from physics.stackexchange.com Nov 7 '18 at 23:23
This question came from our site for active researchers, academics and students of physics.
2
$begingroup$
This is a perfectly good question, but I think it belongs on the math site.
$endgroup$
– Javier
Nov 6 '18 at 12:44
$begingroup$
Well, the intuition is just that geodesics have zero acceleration. If a particle moving in the $xy$ plane has zero acceleration, it still has zero acceleration if you "forget" about the $z$ axis and restrict to the $xy$ plane alone.
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
However, this may be too simple. Are you looking for a more complicated proof?
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
knzhou, I like the point you made, restricting it to a 2d case I understand well has helped, I think I can take it from here, unless you have any smart insights you can add I likely wouldn't have reached myself. Javier, ill repost to the math site to see what they have to say, thanks.
$endgroup$
– T. Smith
Nov 6 '18 at 14:15
1
$begingroup$
The converse is clearly false. Just think of a 2-sphere imbedded in ordinary 3-space.
$endgroup$
– Bert Barrois
Nov 6 '18 at 14:21
add a comment |
$begingroup$
I've read that when a hypersurface within a manifold contains a curve, if the curve is a geodesic in the manifold, it is also a geodesic in the hypersurface.
This is quite abstract for me, I've only recently started learning GR, could someone provide a geometric (or if necessary, algebraic) argument for why this is true? Is the converse ever/always/never true?
(Based on my limited intuition, I'd guess sometimes but not always, but I can't think of anything a professor wouldn't call 'handwavey').
Cheers
general-relativity
asked Nov 6 '18 at 12:29
T. SmithT. Smith
111
$endgroup$
I've read that when a hypersurface within a manifold contains a curve, if the curve is a geodesic in the manifold, it is also a geodesic in the hypersurface.
This is quite abstract for me, I've only recently started learning GR, could someone provide a geometric (or if necessary, algebraic) argument for why this is true? Is the converse ever/always/never true?
(Based on my limited intuition, I'd guess sometimes but not always, but I can't think of anything a professor wouldn't call 'handwavey').
Cheers
general-relativity
general-relativity
asked Nov 6 '18 at 12:29
T. SmithT. Smith
111
asked Nov 6 '18 at 12:29
T. SmithT. Smith
111
asked Nov 6 '18 at 12:29
T. SmithT. Smith
111
asked Nov 6 '18 at 12:29
T. SmithT. Smith
111
asked Nov 6 '18 at 12:29
T. SmithT. Smith
111
111
migrated from physics.stackexchange.com Nov 7 '18 at 23:23
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Nov 7 '18 at 23:23
This question came from our site for active researchers, academics and students of physics.
2
$begingroup$
This is a perfectly good question, but I think it belongs on the math site.
$endgroup$
– Javier
Nov 6 '18 at 12:44
$begingroup$
Well, the intuition is just that geodesics have zero acceleration. If a particle moving in the $xy$ plane has zero acceleration, it still has zero acceleration if you "forget" about the $z$ axis and restrict to the $xy$ plane alone.
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
However, this may be too simple. Are you looking for a more complicated proof?
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
knzhou, I like the point you made, restricting it to a 2d case I understand well has helped, I think I can take it from here, unless you have any smart insights you can add I likely wouldn't have reached myself. Javier, ill repost to the math site to see what they have to say, thanks.
$endgroup$
– T. Smith
Nov 6 '18 at 14:15
1
$begingroup$
The converse is clearly false. Just think of a 2-sphere imbedded in ordinary 3-space.
$endgroup$
– Bert Barrois
Nov 6 '18 at 14:21
add a comment |
2
$begingroup$
This is a perfectly good question, but I think it belongs on the math site.
$endgroup$
– Javier
Nov 6 '18 at 12:44
$begingroup$
Well, the intuition is just that geodesics have zero acceleration. If a particle moving in the $xy$ plane has zero acceleration, it still has zero acceleration if you "forget" about the $z$ axis and restrict to the $xy$ plane alone.
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
However, this may be too simple. Are you looking for a more complicated proof?
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
knzhou, I like the point you made, restricting it to a 2d case I understand well has helped, I think I can take it from here, unless you have any smart insights you can add I likely wouldn't have reached myself. Javier, ill repost to the math site to see what they have to say, thanks.
$endgroup$
– T. Smith
Nov 6 '18 at 14:15
1
$begingroup$
The converse is clearly false. Just think of a 2-sphere imbedded in ordinary 3-space.
$endgroup$
– Bert Barrois
Nov 6 '18 at 14:21
2
2
$begingroup$
This is a perfectly good question, but I think it belongs on the math site.
$endgroup$
– Javier
Nov 6 '18 at 12:44
$begingroup$
This is a perfectly good question, but I think it belongs on the math site.
$endgroup$
– Javier
Nov 6 '18 at 12:44
$begingroup$
Well, the intuition is just that geodesics have zero acceleration. If a particle moving in the $xy$ plane has zero acceleration, it still has zero acceleration if you "forget" about the $z$ axis and restrict to the $xy$ plane alone.
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
Well, the intuition is just that geodesics have zero acceleration. If a particle moving in the $xy$ plane has zero acceleration, it still has zero acceleration if you "forget" about the $z$ axis and restrict to the $xy$ plane alone.
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
However, this may be too simple. Are you looking for a more complicated proof?
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
However, this may be too simple. Are you looking for a more complicated proof?
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
knzhou, I like the point you made, restricting it to a 2d case I understand well has helped, I think I can take it from here, unless you have any smart insights you can add I likely wouldn't have reached myself. Javier, ill repost to the math site to see what they have to say, thanks.
$endgroup$
– T. Smith
Nov 6 '18 at 14:15
$begingroup$
knzhou, I like the point you made, restricting it to a 2d case I understand well has helped, I think I can take it from here, unless you have any smart insights you can add I likely wouldn't have reached myself. Javier, ill repost to the math site to see what they have to say, thanks.
$endgroup$
– T. Smith
Nov 6 '18 at 14:15
1
1
$begingroup$
The converse is clearly false. Just think of a 2-sphere imbedded in ordinary 3-space.
$endgroup$
– Bert Barrois
Nov 6 '18 at 14:21
$begingroup$
The converse is clearly false. Just think of a 2-sphere imbedded in ordinary 3-space.
$endgroup$
– Bert Barrois
Nov 6 '18 at 14:21
add a comment |
1 Answer
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Ehlers (1961) in the classic paper "Contributions to the relativistic mechanics of continuous media" states:
A vortex-free flow is rigid precisely when its orthogonal hypersurfaces are totally geodesic.
(1993 translation). As for "flow", this need not be an actual fluid, but simply a 4-velocity field. Background fact: hypersurfaces orthogonal to the 4-velocities exist if and only if the motion is vorticity-free (this follows from Frobenius' theorem). Totally geodesic means every geodesic in the hypersurface is also a geodesic on the full manifold.
For example, think of static observers in Schwarzschild spacetime (the ones with fixed $r>2M$, $theta$, and $phi$ coordinates; with 4-velocities proportional to the Killing vector field which is timelike at infinity). This velocity field is rigid, and the orthogonal hypersurfaces have $t=textrm{const}$, and these must be totally geodesic. As another example, consider observers which "fall from rest at infinity". This is not a rigid velocity field, because tidal forces lead to shear and expansion, and it turns out the radial direction within the hypersurfaces is not a geodesic within the manifold. A final example, consider a homogeneous and isotropic universe, with the velocity field "comoving" with the average local matter flow. This is not rigid in an expanding universe, hence the hypersurfaces of constant cosmic time are not totally geodesic! This surprised me when I first realised it.
This response probably does not address everything in your question, but I hope it helps.
answered Jan 6 at 8:32
Colin MacLaurinColin MacLaurin
1065
$endgroup$
add a comment |
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$begingroup$
Ehlers (1961) in the classic paper "Contributions to the relativistic mechanics of continuous media" states:
A vortex-free flow is rigid precisely when its orthogonal hypersurfaces are totally geodesic.
(1993 translation). As for "flow", this need not be an actual fluid, but simply a 4-velocity field. Background fact: hypersurfaces orthogonal to the 4-velocities exist if and only if the motion is vorticity-free (this follows from Frobenius' theorem). Totally geodesic means every geodesic in the hypersurface is also a geodesic on the full manifold.
For example, think of static observers in Schwarzschild spacetime (the ones with fixed $r>2M$, $theta$, and $phi$ coordinates; with 4-velocities proportional to the Killing vector field which is timelike at infinity). This velocity field is rigid, and the orthogonal hypersurfaces have $t=textrm{const}$, and these must be totally geodesic. As another example, consider observers which "fall from rest at infinity". This is not a rigid velocity field, because tidal forces lead to shear and expansion, and it turns out the radial direction within the hypersurfaces is not a geodesic within the manifold. A final example, consider a homogeneous and isotropic universe, with the velocity field "comoving" with the average local matter flow. This is not rigid in an expanding universe, hence the hypersurfaces of constant cosmic time are not totally geodesic! This surprised me when I first realised it.
This response probably does not address everything in your question, but I hope it helps.
answered Jan 6 at 8:32
Colin MacLaurinColin MacLaurin
1065
$endgroup$
add a comment |
$begingroup$
Ehlers (1961) in the classic paper "Contributions to the relativistic mechanics of continuous media" states:
A vortex-free flow is rigid precisely when its orthogonal hypersurfaces are totally geodesic.
(1993 translation). As for "flow", this need not be an actual fluid, but simply a 4-velocity field. Background fact: hypersurfaces orthogonal to the 4-velocities exist if and only if the motion is vorticity-free (this follows from Frobenius' theorem). Totally geodesic means every geodesic in the hypersurface is also a geodesic on the full manifold.
For example, think of static observers in Schwarzschild spacetime (the ones with fixed $r>2M$, $theta$, and $phi$ coordinates; with 4-velocities proportional to the Killing vector field which is timelike at infinity). This velocity field is rigid, and the orthogonal hypersurfaces have $t=textrm{const}$, and these must be totally geodesic. As another example, consider observers which "fall from rest at infinity". This is not a rigid velocity field, because tidal forces lead to shear and expansion, and it turns out the radial direction within the hypersurfaces is not a geodesic within the manifold. A final example, consider a homogeneous and isotropic universe, with the velocity field "comoving" with the average local matter flow. This is not rigid in an expanding universe, hence the hypersurfaces of constant cosmic time are not totally geodesic! This surprised me when I first realised it.
This response probably does not address everything in your question, but I hope it helps.
answered Jan 6 at 8:32
Colin MacLaurinColin MacLaurin
1065
$endgroup$
add a comment |
$begingroup$
Ehlers (1961) in the classic paper "Contributions to the relativistic mechanics of continuous media" states:
A vortex-free flow is rigid precisely when its orthogonal hypersurfaces are totally geodesic.
(1993 translation). As for "flow", this need not be an actual fluid, but simply a 4-velocity field. Background fact: hypersurfaces orthogonal to the 4-velocities exist if and only if the motion is vorticity-free (this follows from Frobenius' theorem). Totally geodesic means every geodesic in the hypersurface is also a geodesic on the full manifold.
For example, think of static observers in Schwarzschild spacetime (the ones with fixed $r>2M$, $theta$, and $phi$ coordinates; with 4-velocities proportional to the Killing vector field which is timelike at infinity). This velocity field is rigid, and the orthogonal hypersurfaces have $t=textrm{const}$, and these must be totally geodesic. As another example, consider observers which "fall from rest at infinity". This is not a rigid velocity field, because tidal forces lead to shear and expansion, and it turns out the radial direction within the hypersurfaces is not a geodesic within the manifold. A final example, consider a homogeneous and isotropic universe, with the velocity field "comoving" with the average local matter flow. This is not rigid in an expanding universe, hence the hypersurfaces of constant cosmic time are not totally geodesic! This surprised me when I first realised it.
This response probably does not address everything in your question, but I hope it helps.
answered Jan 6 at 8:32
Colin MacLaurinColin MacLaurin
1065
$endgroup$
Ehlers (1961) in the classic paper "Contributions to the relativistic mechanics of continuous media" states:
A vortex-free flow is rigid precisely when its orthogonal hypersurfaces are totally geodesic.
(1993 translation). As for "flow", this need not be an actual fluid, but simply a 4-velocity field. Background fact: hypersurfaces orthogonal to the 4-velocities exist if and only if the motion is vorticity-free (this follows from Frobenius' theorem). Totally geodesic means every geodesic in the hypersurface is also a geodesic on the full manifold.
For example, think of static observers in Schwarzschild spacetime (the ones with fixed $r>2M$, $theta$, and $phi$ coordinates; with 4-velocities proportional to the Killing vector field which is timelike at infinity). This velocity field is rigid, and the orthogonal hypersurfaces have $t=textrm{const}$, and these must be totally geodesic. As another example, consider observers which "fall from rest at infinity". This is not a rigid velocity field, because tidal forces lead to shear and expansion, and it turns out the radial direction within the hypersurfaces is not a geodesic within the manifold. A final example, consider a homogeneous and isotropic universe, with the velocity field "comoving" with the average local matter flow. This is not rigid in an expanding universe, hence the hypersurfaces of constant cosmic time are not totally geodesic! This surprised me when I first realised it.
This response probably does not address everything in your question, but I hope it helps.
answered Jan 6 at 8:32
Colin MacLaurinColin MacLaurin
1065
answered Jan 6 at 8:32
Colin MacLaurinColin MacLaurin
1065
answered Jan 6 at 8:32
Colin MacLaurinColin MacLaurin
1065
answered Jan 6 at 8:32
Colin MacLaurinColin MacLaurin
1065
1065
add a comment |
add a comment |
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2
$begingroup$
This is a perfectly good question, but I think it belongs on the math site.
$endgroup$
– Javier
Nov 6 '18 at 12:44
$begingroup$
Well, the intuition is just that geodesics have zero acceleration. If a particle moving in the $xy$ plane has zero acceleration, it still has zero acceleration if you "forget" about the $z$ axis and restrict to the $xy$ plane alone.
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
However, this may be too simple. Are you looking for a more complicated proof?
$endgroup$
– knzhou
Nov 6 '18 at 13:26
$begingroup$
knzhou, I like the point you made, restricting it to a 2d case I understand well has helped, I think I can take it from here, unless you have any smart insights you can add I likely wouldn't have reached myself. Javier, ill repost to the math site to see what they have to say, thanks.
$endgroup$
– T. Smith
Nov 6 '18 at 14:15
1
$begingroup$
The converse is clearly false. Just think of a 2-sphere imbedded in ordinary 3-space.
$endgroup$
– Bert Barrois
Nov 6 '18 at 14:21