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How does GR handle metric transition for a spherical mass shell?

by Q-reeus
Tags: handle, mass, metric, shell, spherical, transition
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Q-reeus
#163
Oct29-11, 06:12 AM
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Quote Quote by PeterDonis View Post
Originally Posted by Q-reeus:
"This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before."

I think this is right, provided that the tetrahedron is large enough that spatial curvature is measurable over its size, and that we deform the tetrahedron to conform to the curvature. Consider the analogous case on a 2-D surface, where we make a triangle out of rigid tubes joined by free-hinging joints, and then measure the joint angles after carefully wrapping the triangle onto a sphere so it conforms to the sphere's curvature. This case is exactly like that of the circular rubber disk or annulus; to see any change in shape, we need to exert external force on the rubber to make it conform to the shape of the sphere, and we have to decide how it is going to conform.
Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent). It is understood here measurements are taken in the curved environment - not some distant coordinate reference. Hence the specification of free-hinging pinned joints, and rigid tubes that are not 'floppy'. Otherwise, what is implied is surely an intrinsic, locally measurable curvature of just one straight rod, in going from flat spacetime to curved. But in 3-curvature, how will the 'straight' rod/tube 'know' which way to bend?

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.
We also have to make the shape we are trying to wrap around the sphere large enough that it "sees" the curvature; if it's too small we will be unable to use it to detect any difference from a flat plane (like the tiny pieces of paper we cut out of the paper disk or annulus in my previous example).
Yes, and is it not just this size related differential that allows flat-landers to detect curvature induced angular changes by means of a small, 'standard' protractor that minimally 'feels' curvature. Thoughts?
PAllen
#164
Oct29-11, 08:24 AM
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Quote Quote by Q-reeus View Post
Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent). It is understood here measurements are taken in the curved environment - not some distant coordinate reference. Hence the specification of free-hinging pinned joints, and rigid tubes that are not 'floppy'. Otherwise, what is implied is surely an intrinsic, locally measurable curvature of just one straight rod, in going from flat spacetime to curved. But in 3-curvature, how will the 'straight' rod/tube 'know' which way to bend?

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

Yes, and is it not just this size related differential that allows flat-landers to detect curvature induced angular changes by means of a small, 'standard' protractor that minimally 'feels' curvature. Thoughts?
One part you don't get is the issue of embedding. Please think about how a curved spherical surface is embedded in flat 3-space without telling you anything about the curvature of the 3-space. Similarly, in curved spacetime you can embed flat planes, and any procedure looking for straight lines will pick out this embedded flat plane. Thus no procedure limited to a plane can detect curvature of a 4-manifold.

I have explained this multiple times and you have continued to ignore it.
Q-reeus
#165
Oct29-11, 09:17 AM
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Quote Quote by PAllen View Post
I have explained this multiple times and you have continued to ignore it.
Ignore is not perhaps the right word, as all I could pick up were assertions of how it is - may well be true, but to me they were just assertions.
One part you don't get is the issue of embedding. Please think about how a curved spherical surface is embedded in flat 3-space without telling you anything about the curvature of the 3-space. Similarly, in curved spacetime you can embed flat planes, and any procedure looking for straight lines will pick out this embedded flat plane. Thus no procedure limited to a plane can detect curvature of a 4-manifold.

I have quoted a proof by J.L. Synge that five vertices are the minimum to detect curvature of spacetime (that is, even a tetrahedron can always be constructed to conform to Euclidean expectations).
Allright, given that is so, what sense does one make of the 'popular' statement that the internal angles of a triangle do not generally add to 180 degrees in curved spacetime? Having not studied the subject, I took my que from such as:
"In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form." at http://en.wikipedia.org/wiki/Pythago...idean_geometry
[Another grab, from: http://en.wikipedia.org/wiki/Curved_...flat.2C_closed
"Triangles which lie on the surface of an open space will have a sum of angles which is less than 180. Triangles which lie on the surface of a closed space will have a sum of angles which is greater than 180. The volume, however, is not (4 / 3)πr3" In context this appears to me to be a generalized statement applicable to higher than 2D curvature. As I say , haven't studied this subject at all.]
PAllen
#166
Oct29-11, 10:02 AM
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Quote Quote by Q-reeus View Post
Ignore is not perhaps the right word, as all I could pick up were assertions of how it is - may well be true, but to me they were just assertions.

Allright, given that is so, what sense does one make of the 'popular' statement that the internal angles of a triangle do not generally add to 180 degrees in curved spacetime? Having not studied the subject, I took my que from such as:
"In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form." at http://en.wikipedia.org/wiki/Pythago...idean_geometry
If you look at that link, the context is geometry on a 2-surface. Popular statements don't get into the issue of embedding - they are over-simplifed. I am not just making assertions, I am asking you to think, as I'll do again. Apply your idea to 3-space "if a triangle is non-pythagorean, the *space* is non-euclidean' to flat 3-space. On a 2-sphere in 3-space, you conclude, correctly, that the 2-sphere is curved (triangle is non-pythagorean). What does that tell you about the 3-space: nothing. The 3-space is still flat.

One specific argument that flat planes are embeddable in a 4-manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any non-Euclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold.
Q-reeus
#167
Oct29-11, 10:40 AM
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Quote Quote by PAllen View Post
If you look at that link, the context is geometry on a 2-surface.
More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.
One specific argument that flat planes are embeddable in a 4-manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any non-Euclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold.
Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not?
PAllen
#168
Oct29-11, 10:55 AM
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Quote Quote by Q-reeus View Post
More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not?
The whole passage refers to geometry of a 2-surface. Euclid's Parrallel postulate is a postulate about plane geometry. Spherical geometry is the geometry of a 2-sphere - the *surface* of a sphere. And I ask you again to think about my simple embedding example. The surface of a globe is non-euclidean. The globe is sitting in a flat euclidean 3-space. Contradiction? No. The embedded space is curved, the space embedded in happens to be flat.
PAllen
#169
Oct29-11, 11:09 AM
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Quote Quote by Q-reeus View Post
More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not?
I am saying if you try to make Euclidean triangles in a general 4-d semi-riemannian manifold, you will succeed. For 3-space, I'm not sure whether you can *always* do it. There are fewer degrees of freedom, and the simple counting argument I used for 4-manifold does not work. It is certainly true that not all 2-manifolds can be embedded in flat 3-space, so it is plausible that some 3-manifolds will not embed a section of flat 2-surface.

Also recall my discussion with Peter: the SC geometry allows embedding of completely flat 3-space regions. The K factor is actually a feature of a particular class of observers (static observers, which are non-inertial observers), not something intrinsic to the geometry. It is analogous to distortions seen in an accelerating rocket in flat spacetime - a feature of the observer, not the intrinsic spacetime geometry. GP observers in the same SC geometry, experience absolutely flat 3-space.
JDoolin
#170
Oct29-11, 11:52 AM
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Quote Quote by PeterDonis View Post
Lorentz-contrated and time-dilated relative to what?
If you look carefully, there is one particle in the diagram that doesn't move. All motion in this diagram, then, is relative to that stationary particle.

This deserves further animations; I'd rather show you, if I can, rather than tell you. But suffice it to say, the animation above could be Lorentz transformed so as to place any particle in the center of the circle. However, because there are only a finite number of particles in the picture, there would be a gravitational asymmetry for every particle, except for one.

It would depend on what the shell was made of and how you specified the initial conditions of the expansion.

There are a couple different ways I might specify the initial conditions; (1) starting with an equipartition of rapidity (2) starting with an equipartition of momentum.

In this animation I assumed equal masses, and used equipartition of momentum.

Let

[tex]q\equiv \left \|\frac{\vec p}{m c} \right \| = \left \|\frac{(\vec v/c)}{\sqrt{1-v^2/c^2}} \right \|<3[/tex]


[tex]\frac{v}{c}=\sqrt{\frac{q^2}{q^2+1}}[/tex]

Once I calculated the velocities, I animated, finding position, by simply multiplying the velocities by t.



The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force.

There's no limit to what q might be. If I set q=10^100, then all the particles that are in this animation would be so close to the center that they would experience essentially no net force.

For a shell made of "normal" matter, i.e., with pressure no greater than 1/3 its energy density, the gravity of the shell would cause the expansion to decelerate, similar to a matter-dominated expanding FRW model. Depending on the initial conditions (velocity of expansion vs. shell energy density and pressure), the shell might stop expanding altogether and re-contract, or it might go on expanding forever, continually slowing down but never quite stopping.

If the shell's density is uniform (i.e., uniform throughout the shell at any particular "time" in the shell's comoving frame--the density could still change with time, as long as it remained uniform spatially within the shell), then I think you are right to guess that there would be no observable effect from the "potential" within the shell itself. There might still be an effect relative to the potential in the spacetime exterior to the shell. (The potential interior to the shell would be the same as the potential on the shell's inner surface, just as for a static shell.)
I have a plan in mind to present the animation from the reference frame of a particle on the edge of the mass by Lorentz Transforming all of the velocities. Definitely, as long as the total mass of the distribution is finite, then there would be particles on the edge that experience extreme accelerations toward the center. If the mass were infinite (and thus no edge), you could invoke symmetry, and there would be no net force in any direction for any particle.

But then you'd have the problem, being inside a shell of infinite mass, that at any given point inside, you are at an infinite negative gravitational potential

On the other hand, with a finite mass, the particles at the edge could experience, at least for a time, acceleration equivalent to a black hole. I'm not at all sure what theoretical ramifications that would have.

Anyway, I'll work on the other animations, and hope that makes my meaning clearer.
PeterDonis
#171
Oct29-11, 12:17 PM
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Quote Quote by Q-reeus View Post
Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent).
If the triangle is *assembled* in the curved space, in the right way, this will be true. But you specified that the triangle (or tetrahedron) was assembled in flat space, and then *moved* to curved space. That means the "natural" configuration of the sides is the flat configuration, Euclidean straight lines, and they must be deformed to get into the curved configuration, geodesics on a curved surface.

If, instead, you assembled the triangle carefully in a curved region, so that the "natural" configuration of its sides was as geodesics on the curved surface, and the "natural" angles at each vertex summed to more than 180 degrees, then the triangle would deform if you tried to make it conform to a flat Euclidean plane. Similarly, if you assembled a tetrahedron in a region of curved space, hovering over a gravitating body, so that it was unstressed in that configuration, and then took it far away from gravitating bodies where space was flat, it would undergo stress and deformation in the course of conforming to the flat space.

Quote Quote by Q-reeus View Post
I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.
It depends on what you want to make an analogy to. If we are trying to construct an analogy to what happens when we take an object constructed far away from gravitating bodies and move it into a gravity well, the above is *not* a good analogy for that, because deflating the balloon corresponds to contracting the spacetime as a whole. That would do as an analogy for a collapsing universe, but *not* for moving into a gravity well. A better analogy for that would be to consider a surface like the Flamm paraboloid...

http://en.wikipedia.org/wiki/File:Fl...paraboloid.svg

...and think of moving a triangle from the flat region to the curved region.
PeterDonis
#172
Oct29-11, 12:21 PM
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Quote Quote by JDoolin View Post
If you look carefully, there is one particle in the diagram that doesn't move. All motion in this diagram, then, is relative to that stationary particle.
Then I'm not sure I would call this a "shell". "Shell" implies a thin region of matter with complete vacuum inside and outside it, at least in the context of this thread.

Quote Quote by JDoolin View Post
The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force.
How are you calculating the force?
JDoolin
#173
Oct29-11, 01:28 PM
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Quote Quote by PeterDonis View Post
Then I'm not sure I would call this a "shell". "Shell" implies a thin region of matter with complete vacuum inside and outside it, at least in the context of this thread.



How are you calculating the force?
This is a good question, or more specifically, how should I (or how should we) calculate the force? So far, essentially, I'm not calculating the force. Invoking symmetry, there is no net force. There's a gravitational potential, but no net force on a particle at or near the center of the distribution.

But let's look at the other extreme now, from the very edge of the distribution.


Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right.

Let me give you a couple of premises of how I would go about calculating the force, and see if you agree with these premises, or if I am hopelessly naive.

#1 The net gravitational force on a particle is [tex]F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i[/tex]
#2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force.
#3 The mass of each particle affecting gravitation is the rest-mass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction.
#4 The location of the particle is not the "simultaneous" location, but rather the speed-of-light delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the past-light-cone of our particle of interest with the world-lines of the particles involved.

Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what common-sense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away.
Attached Thumbnails
edgeview2.gif  
PeterDonis
#174
Oct29-11, 02:51 PM
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Quote Quote by JDoolin View Post
Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right.
Well, the problem as a whole has spherical symmetry--or at least, you can impose that condition as a reasonable idealization to make the problem tractable. If the problem has spherical symmetry, then the "force" on any particle (a) must point in the radial direction, and (b) must be a function only of its radial coordinate. If the matter is all ordinary matter, as I described in my last post, then (a) can be further refined: the force on any particle must point radially *inward*, i.e., the expansion of the shell must be decelerating. As I note below, viewing this deceleration as caused by a "force" may not be the simplest way to view this problem.

Quote Quote by JDoolin View Post
#1 The net gravitational force on a particle is [tex]F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i[/tex]
This requires the problem to be non-relativistic. That is not consistent with your #4. In fact, the inconsistency between #1 and #4 was one of the stumbling blocks on the road to General Relativity, back in the early 1900's. See further note below.

Quote Quote by JDoolin View Post
#2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force.
A better way to state this would be: the 4-force on a particle is a covariant geometric object, it is the particle's rest mass times the particle's 4-acceleration, which is the covariant derivative of its 4-velocity with respect to its proper time.

Quote Quote by JDoolin View Post
#3 The mass of each particle affecting gravitation is the rest-mass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction.
A better way to state this would be: the "gravitational field" is determined by the stress-energy tensor of the matter, which is a covariant geometric object. The SET determines the field via the Einstein Field Equation. It also correctly accounts for the fact that "rest mass" is what affects gravity, not momentum due purely to kinematics.

Quote Quote by JDoolin View Post
#4 The location of the particle is not the "simultaneous" location, but rather the speed-of-light delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the past-light-cone of our particle of interest with the world-lines of the particles involved.
If you are assuming this, then you can't use #1 as your force equation. Consider a simple example: the force on the Earth at a given instant does *not* point towards where the Sun was 8 minutes ago by the Earth's clock. It points towards where the Sun is "now" by the Earth's clock. (Technically, there are some small correction factors, but they can be ignored for this discussion.) So if you use your #1, you have to plug in the position of the Earth relative to the Sun "now", not the "retarded" position, or you'll get the wrong answer. (Steve Carlip wrote an excellent paper some time ago that explains all this: it's at http://arxiv.org/abs/gr-qc/9909087.)

As I noted above, using #1 requires the problem to be non-relativistic, and you don't seem to be imposing that limitation. For the relativistic case, you can't really use a "force" equation for this problem, or at least it does not seem to be the easiest way to approach it. A better model would be an expanding matter-dominated FRW-type model, as I mentioned in a previous post, especially since it doesn't seem like your particles are a "shell", since there's no interior vacuum region, as far as I can see. This type of model does not view gravity as a "force"; it just solves for the dynamics of a curved spacetime using the EFE and an expression for the stress-energy tensor of the matter. You can still view individual particles as being subject to a "force", but that force can be more easily calculated *after* you have constructed the model from the EFE.

If the spatial extent of the expanding "shell" is limited, then at the surface of the shell, the FRW-type solution would be matched to an exterior Schwarzschild vacuum solution; this would basically be the time reverse of the Oppenheimer-Snyder solution for the gravitational collapse of a star.

Quote Quote by JDoolin View Post
Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what common-sense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away.
Huh? I don't understand what you're getting at here at all. How can you accelerate toward a past event? For that matter, how can you accelerate toward a future event? You accelerate toward a point in space, not an event in spacetime. Also, what kind of "reference frames" are you talking about? If you're talking about ordinary Lorentz frames, they are only valid locally; you can't use them to correctly evaluate distances to faraway objects.

It looks like this discussion might be better moved to a new thread, since it appears to be getting further from the topic of this one.
JDoolin
#175
Oct29-11, 06:13 PM
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Quote Quote by PeterDonis View Post
This requires the problem to be non-relativistic. That is not consistent with your #4. In fact, the inconsistency between #1 and #4 was one of the stumbling blocks on the road to General Relativity, back in the early 1900's. See further note below.
Okay, I see where I had one mistake.

http://en.wikipedia.org/wiki/Gauss's...oulomb.27s_law

Strictly speaking, Coulomb's law cannot be derived from Gauss's law alone, since Gauss's law does not give any information regarding the curl of E (see Helmholtz decomposition and Faraday's law). However, Coulomb's law can be proven from Gauss's law if it is assumed, in addition, that the electric field from a point charge is spherically-symmetric (this assumption, like Coulomb's law itself, is exactly true if the charge is stationary, and approximately true if the charge is in motion).
I was thinking that coulomb's law and gauss law were equivalent. But they are only equivalent when the charge is not moving. I'll try to read more of the article to see what I've missed. May take me a bit of time to catch up so I don't think I'm ready to start another thread at this time.

If indeed, the gravitational field from a receding object has some predictable variation based on its relative velocity, then we should, of course, use that modification. But if you want to claim that there is no predictable variation; no predictable "meaningful" position we can use for distant objects, I beg to differ.

On another topic, if possible, can you support your argument that "ordinary Lorentz frames...are valid only locally." I've heard this time and time again, but no one has ever explained what it means. Are you saying that when I point at a distant galaxy, that that direction that I am pointing only exists locally? Are you saying that the very concept of direction is only a local phenomenon?
PeterDonis
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Oct29-11, 06:59 PM
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Quote Quote by JDoolin View Post
If indeed, the gravitational field from a receding object has some predictable variation based on its relative velocity, then we should, of course, use that modification. But if you want to claim that there is no predictable variation; no predictable "meaningful" position we can use for distant objects, I beg to differ.
The answer to this will be easier to understand if I answer your other question first. See below.

Quote Quote by JDoolin View Post
On another topic, if possible, can you support your argument that "ordinary Lorentz frames...are valid only locally." I've heard this time and time again, but no one has ever explained what it means.
It means that you can set up a Lorentz frame centered around any event you like in spacetime, but the frame will only obey the laws of special relativity in a restricted local region of spacetime around that event. Strictly speaking, it will only obey the laws of special relativity exactly *at* that event; but in practice, there will be a finite region around the event where the deviations from the laws of special relativity due to the curvature of spacetime are too small to detect. The size of that region depends on how accurate your measurements are and how curved the spacetime is.

Here's a simple example: take an object that is freely falling towards the Earth from far away (say halfway between the Earth and the Moon), and set up a Lorentz frame using its worldline as the t axis. Since the object is in free fall, i.e., inertial motion, its worldline can be used this way according to the laws of SR. Now take a second object which is also in free fall towards the Earth, but which is slightly lower than the first object. Suppose there is an instant of time at which the second object is at rest relative to the first; we take this instant of time to define t = 0 in our Lorentz frame, and the position of the first object at this instant to define the spatial origin, so the frame is centered on that event on the first object's worldline.

It should be evident that, for a given accuracy in measuring the relative velocity of the two objects, there will be some time t > 0 at which it becomes apparent that the second object is no longer at rest relative to the first. (This is because it is slightly closer to the Earth and therefore sees a slightly higher acceleration due to gravity.) But SR says that' can't happen: both objects are moving inertially, both were at rest relative to each other at one instant, and SR says that they therefore should remain at rest relative to each other forever. They don't. So we can only set up a Lorentz frame around the first object's worldline for a short enough period of time that the effects of tidal gravity (which is what causes the difference in acceleration of the two objects) can't be measured. Similarly, if the second object were further away from the first, its relative acceleration would become evident sooner; so there is a limit to how far we can extend a Lorentz frame around the first object in space as well, before the effects of spacetime curvature become evident.

Quote Quote by JDoolin View Post
Are you saying that when I point at a distant galaxy, that that direction that I am pointing only exists locally? Are you saying that the very concept of direction is only a local phenomenon?
The direction you are pointing is the direction from which light rays emitted by the distant galaxy are entering your eyes. It is certainly reasonable to call that "the direction of the galaxy", but only if you are aware of the limitations of that way of thinking. Light paths are bent by gravity, so the direction you are seeing the light come from may not be "the" direction the galaxy is "actually" in. This is another manifestation of spacetime curvature, and it means you can only set up a Lorentz frame locally with regard to directions and positions as well as relative motion of objects.

This also applies to what you were saying about assigning a "distance" to distant objects. There is no unique definition of "distance" in a curved spacetime; there are at least three different ones that are used in cosmology. You can uniquely define distance in a local Lorentz frame, but that only works locally; once you are out of the local region where the laws of SR can be applied to the desired accuracy, your unique definition of distance no longer works.
pervect
#177
Oct29-11, 07:54 PM
Emeritus
Sci Advisor
P: 7,620
There's a well known solution for an expanding (or collapsing) sphere of pressureless dust. Which isn't quite what was asked for, but you might be able to graft onto it to get the solution. The expanding pressureles dust sphere solution is just the FRW solution of cosmology, by the way.

As far as forces go, they're pretty much not used in GR. You can calculate them as an afterthought when you have the metric by evaluating [itex]u^a \nabla_a u^a[/itex], where u^a is your velocity vector.

[add]Conceptually, what the above expression does is calculate what an accelerometer following the worldline would read.

The reason forces aren't used much in GR that they transform in a complex manner - i.e. they don't transform as a tensor.

In SR, you can still deal with 4-forces as a tensor under Lorentz boosts. In GR, with accelerated coordinate systems, forces obviously cannot transform as a tensor. For instance, if you have an unaccelerated system with zero force, an accelerated system will have a nonzero force, but a tensor that is zero in one coordinate system must be zero in all.
Q-reeus
#178
Oct30-11, 09:00 AM
P: 1,115
Quote Quote by PAllen View Post
And I ask you again to think about my simple embedding example. The surface of a globe is non-euclidean. The globe is sitting in a flat euclidean 3-space. Contradiction? No. The embedded space is curved, the space embedded in happens to be flat.
No problem in agreeing with that situation.
Also recall my discussion with Peter: the SC geometry allows embedding of completely flat 3-space regions. The K factor is actually a feature of a particular class of observers (static observers, which are non-inertial observers), not something intrinsic to the geometry. It is analogous to distortions seen in an accelerating rocket in flat spacetime - a feature of the observer, not the intrinsic spacetime geometry. GP observers in the same SC geometry, experience absolutely flat 3-space.
In that more generaized context I see the point, no problems.
Q-reeus
#179
Oct30-11, 09:03 AM
P: 1,115
Quote Quote by PeterDonis View Post
A better analogy for that would be to consider a surface like the Flamm paraboloid...
http://en.wikipedia.org/wiki/File:Fl...paraboloid.svg
Handy reminder; while I think DrGreg had posted on it much earlier, upon finding the original article at http://en.wikipedia.org/wiki/Flamm%2...27s_paraboloid , studied that piece with some more attention. Had never bothered to understand the significance of the w ordinate, but now appreciate how it gives a useful handle on visualising spatial part of SM. 'Inverting' w ordinate gives the sense of radial contraction as I had envisaged. Matching that to shell transition is the interesting exercise. Brings it back to on topic, and having come to appreciate the general view of the difficulties of applying a simple reading of SC's, conclude the original posting was an inappropriate vehicle for finding problems with SM. Thanks for all the inputs.
JDoolin
#180
Oct30-11, 04:02 PM
PF Gold
P: 706
Quote Quote by PeterDonis View Post
The answer to this will be easier to understand if I answer your other question first. See below.



It means that you can set up a Lorentz frame centered around any event you like in spacetime, but the frame will only obey the laws of special relativity in a restricted local region of spacetime around that event. Strictly speaking, it will only obey the laws of special relativity exactly *at* that event; but in practice, there will be a finite region around the event where the deviations from the laws of special relativity due to the curvature of spacetime are too small to detect. The size of that region depends on how accurate your measurements are and how curved the spacetime is.

Here's a simple example: take an object that is freely falling towards the Earth from far away (say halfway between the Earth and the Moon), and set up a Lorentz frame using its worldline as the t axis. Since the object is in free fall, i.e., inertial motion, its worldline can be used this way according to the laws of SR. Now take a second object which is also in free fall towards the Earth, but which is slightly lower than the first object. Suppose there is an instant of time at which the second object is at rest relative to the first; we take this instant of time to define t = 0 in our Lorentz frame, and the position of the first object at this instant to define the spatial origin, so the frame is centered on that event on the first object's worldline.

It should be evident that, for a given accuracy in measuring the relative velocity of the two objects, there will be some time t > 0 at which it becomes apparent that the second object is no longer at rest relative to the first. (This is because it is slightly closer to the Earth and therefore sees a slightly higher acceleration due to gravity.) But SR says that' can't happen: both objects are moving inertially, both were at rest relative to each other at one instant, and SR says that they therefore should remain at rest relative to each other forever. They don't. So we can only set up a Lorentz frame around the first object's worldline for a short enough period of time that the effects of tidal gravity (which is what causes the difference in acceleration of the two objects) can't be measured. Similarly, if the second object were further away from the first, its relative acceleration would become evident sooner; so there is a limit to how far we can extend a Lorentz frame around the first object in space as well, before the effects of spacetime curvature become evident.



The direction you are pointing is the direction from which light rays emitted by the distant galaxy are entering your eyes. It is certainly reasonable to call that "the direction of the galaxy", but only if you are aware of the limitations of that way of thinking. Light paths are bent by gravity, so the direction you are seeing the light come from may not be "the" direction the galaxy is "actually" in. This is another manifestation of spacetime curvature, and it means you can only set up a Lorentz frame locally with regard to directions and positions as well as relative motion of objects.

This also applies to what you were saying about assigning a "distance" to distant objects. There is no unique definition of "distance" in a curved spacetime; there are at least three different ones that are used in cosmology. You can uniquely define distance in a local Lorentz frame, but that only works locally; once you are out of the local region where the laws of SR can be applied to the desired accuracy, your unique definition of distance no longer works.
I think my main objection to this is that if two objects are in free-fall, you cannot claim that they are both moving inertially.

Although, depending on what you mean by "set up a Lorentz frame using its worldline as the t axis," I may have an even greater objection:

Are you suggesting that geodesic paths in a gravitational field can represent straight lines in global inertial Lorentz Frames (in which case, I cannot agree; curved paths are not straight lines) or are you saying that a momentarily comoving reference frame with time axis tangent to the geodesic worldline at one particular event represents an inertial Lorentz Frame?

In any case, this no longer has anything to do with shells, so I can invite you to read my comments here: http://www.physicsforums.com/showthread.php?t=545002 or if you prefer, you or I could start a new thread regarding whether Lorentz Transformations are purely a local phenomenon.


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