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

[PeterDonis]

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.

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.

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]

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 $u^a \nabla_a u^a$, 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]

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]

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's_paraboloid#Flamm.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]

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: https://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.

 

[PeterDonis]

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.

Um, excuse me? "Moving inertially" and "in free fall" are the same thing.

Are you suggesting that geodesic paths in a gravitational field can represent straight lines in global inertial Lorentz Frames

Obviously not; the whole point is that geodesic paths in a gravitational field are not straight lines in a global Lorentz frame. That's why SR cannot be applied on a large scale in a curved spacetime.

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?

Yes, that's a more precise way of stating what I meant by "use the worldline of one freely falling body as the t axis of the Lorentz frame". So it seems that you understand that I can only set up a momentarily comoving reference frame tangent to a worldline in a curved spacetime at one particular event. That's what "Lorentz frames are only valid locally" means. So I'm having trouble understanding why you're not clear about the meaning of "Lorentz frames are only valid locally".

I'll look at what you posted in the other thread and make further comments on that if need be.

 

[PeterDonis]

I'll look at what you posted in the other thread and make further comments on that if need be.

Quick answers to the questions you posed in the other thread; I won't comment in that thread unless it seems warranted.

(1) my answer is (b)

(2) my answer could be (a), but only in the vacuous sense; in the presence of gravity there are no flat regions of spacetime, there is some curvature everywhere. (b) is the strictly correct answer, but as I said in an earlier post, for a given accuracy of measurement there will be some finite region where the deviations from flatness are not observable, and within that region a Lorentz transformation on a local coordinate patch of "flat" Minkowski coordinates will work

(3) my answer is (b), but "similar" has to be interpreted carefully; Fredrik's comments in an earlier post in that thread on the properties a coordinate chart has to have to admit these transformations are worth reading

(4) my answer is "it depends"; (a) is correct *only* if the coordinate chart meets the requirements for having a rotation transformation in the first place, per Fredrik's post; otherwise the answer is undefined because there is no well-defined assignment of coordinates to objects beyond a small local coordinate patch; (b) is never correct, if the transformation is valid at all when extended to distant objects it will change their coordinate positions

 

[Passionflower]

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.

As Peter already wrote, inertial movement and free fall refers to the same thing.

One can have many objects in free fall all with lower and higher local velocities wrt a free falling object at escape velocity.

 

[JDoolin]

As Peter already wrote, inertial movement and free fall refers to the same thing.

One can have many objects in free fall all with lower and higher local velocities wrt a free falling object at escape velocity.

Newton's first law is often referred to as the law of inertia. The velocity of a body remains constant unless the body is acted upon by an external force.

An object in free fall is changing its velocity because it is acted upon by an external force. The force of gravity.

Inertial movement is movement in a straight line.

Free-fall is not inertial movement. It's accelerated movement.

This is a basic reality-check. You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?

 

[PeterDonis]

Newton's first law is often referred to as the law of inertia. The velocity of a body remains constant unless the body is acted upon by an external force.

An object in free fall is changing its velocity because it is acted upon by an external force. The force of gravity.

Ah, ok. We have a terminology problem. See below.

Inertial movement is movement in a straight line.

Not in GR. In SR, yes, but that's only because SR assumes spacetime is flat. In GR, where spacetime can be curved, inertial motion has to be defined physically. An object is moving inertially if it feels no force, i.e., is weightless, i.e., is in free fall.

Free-fall is not inertial movement. It's accelerated movement.

In Newtonian terms, yes. But that's because Newtonian physics defines "force" and "acceleration" differently than GR does. In GR, an object in free fall, that is weightless, *feels* no force, and therefore there is no force on it in any coordinate-independent, invariant sense, and it is *not* accelerated. More precisely, it is not accelerated in the coordinate-independent sense of "acceleration": the covariant derivative of its 4-velocity with respect to its proper time is zero. This is sometimes called proper acceleration or 4-acceleration. And since "force" in GR is defined as the object's rest mass times its proper acceleration, a freely falling object is experiencing zero force in GR.

The object is "accelerated" with respect to the Earth, but that is coordinate acceleration, not proper acceleration; similarly, the "force" of gravity in Newtonian terms is not a "force" in GR, because it does not cause any 4-acceleration of the object. That's how I was using the terms, and how they are standardly used in GR.

You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?

In GR, yes. More precisely, the object's *worldline* is a geodesic of the curved spacetime around the Earth, so it's as "straight" as any worldline can be in that spacetime. I was not saying that the object's path in *space* was straight; obviously it's not. But that's irrelevant to this discussion.

 

[Passionflower]

This is a basic reality-check. You mean to tell me that if an object is in orbit, you actually consider that to be a straight line?

It is not really relevant to the question what is free falling and inertial, every test observer that has no proper acceleration is free falling and inertial.

From one perspective one certainly could consider it a straight line as a test object in orbit differs only from a radially free falling test object by having a non-zero angular momentum. For instance do you feel yourself turned often when the seasons pass? The same for an astronaut who travels in orbit, as far as he is concerned he travels in a straight line as no forces act on him.

 

[Passionflower]

I was not saying that the object's path in *space* was straight; obviously it's not. But that's irrelevant to this discussion.

I disagree with that.

Whether a free falling object's path in space is straight depends solely on the chosen coordinates. In simple terms, the Sun just as much goes around the Earth as the Earth goes around the Sun it simply depends on the point of view.

 

[JDoolin]

Um, excuse me? "Moving inertially" and "in free fall" are the same thing.



Obviously not; the whole point is that geodesic paths in a gravitational field are not straight lines in a global Lorentz frame. That's why SR cannot be applied on a large scale in a curved spacetime.

Okay, basic reality check passed. RIGHT, geodesics are not straight lines in a global Lorentz Frame. Of course they aren't. But why would that be a failure of Special Relativity?

In fact, I would say that points to the sanity of Special Relativity. In fact, if Special Relativity somehow claimed that geodesic paths in spacetime were straight, I would find that to be much more alarming, and troubling.

Yes, that's a more precise way of stating what I meant by "use the worldline of one freely falling body as the t axis of the Lorentz frame".

You can't make the worldline of a freely falling body as the t-axis of an inertial frame. Because a freely falling body is changing velocity, and if there is a change in velocity, it's not the same inertial reference frame.

So it seems that you understand that I can only set up a momentarily comoving reference frame tangent to a worldline in a curved spacetime at one particular event. That's what "Lorentz frames are only valid locally" means.

Even if the object is only momentarily comoving with the reference frame at a single event, it's a GLOBAL LORENTZ frame. The Lorentz Frame extends for infinity into the past, the future, and in all directions.

In the very next instant, the body will be comoving with an entirely different GLOBAL Lorentz Frame.

So I'm having trouble understanding why you're not clear about the meaning of "Lorentz frames are only valid locally".

Because the Lorentz Transformation, just like the rotation transformation, takes as its input, the location of every event in space and time, and outputs the locations of every event in space and time.

It's like this. If I have a function whose domain is the real numbers, and its range is the real numbers, then I would say that function is valid globally. If I have a function whose domain is 0 to 1 and range is 0 to 1, then I would say the function is valid only locally.

The Lorentz Transformations take as input and output global inertial reference frames, representing every event that ever has and ever will happen in the entire universe.

 

[JDoolin]

It is not really relevant to the question what is free falling and inertial, every test observer that has no proper acceleration is free falling and inertial.

From one perspective one certainly could consider it a straight line as a test object in orbit differs only from a radially free falling test object by having a non-zero angular momentum. For instance do you feel yourself turned often when the seasons pass? The same for an astronaut who travels in orbit, as far as he is concerned he travels in a straight line as no forces act on him.

There are tidal forces in effect, if you have more than a point-observer. Even locally, free-fall is different from straight-line motion. The speed of light has a different geodesic from the falling box. Objects moving at different velocities have different geodesics, and with a careful bit of study, you might be able to figure out what you're orbiting around and how far away it is, even if you can't look outside your box.

The local phenomena within the area of a geodesic would be small, perhaps too small for your most sensitive equipment to detect, but traveling along a geodesic is different from traveling in a straight line.

 

[PeterDonis]

Okay, basic reality check passed. RIGHT, geodesics are not straight lines in a global Lorentz Frame. Of course they aren't. But why would that be a failure of Special Relativity?

Because SR requires initially parallel geodesics (freely falling worldlines) to remain parallel. Otherwise the whole mechanism for setting up Lorentz frames does not work.

In fact, I would say that points to the sanity of Special Relativity. In fact, if Special Relativity somehow claimed that geodesic paths in spacetime were straight, I would find that to be much more alarming, and troubling.

Then you should definitely be alarmed and troubled, because that's exactly what SR does claim. If you disagree, please provide an explicit counterexample: a geodesic path in a spacetime in which the laws of SR apply, that is not straight.

Of course it's easy to find geodesic paths in a curved spacetime that are not "straight" in the sense you're using the term, but the laws of SR don't apply in those spacetimes, precisely because they are curved. I've already given a specific example of such a law: SR requires initially parallel geodesics to remain parallel. (Or, in more ordinary language, SR requires that two objects, both in free fall and weightless, feeling no force, which are at rest relative to each other at one instant of time, must remain at rest relative to each other at all times.) In a curved spacetime, this law is violated, as my example of bodies falling towards Earth made clear.

You can't make the worldline of a freely falling body as the t-axis of an inertial frame. Because a freely falling body is changing velocity, and if there is a change in velocity, it's not the same inertial reference frame.

Exactly. And that violates the laws of SR. Therefore, SR only applies "locally" in a curved spacetime--in a small enough region that the changes in "velocity" you speak of are not observable within the accuracy of measurement being used.

Btw, it's also worth noting that you speak of "changing velocity" without defining what that means. The 4-velocity of a freely falling body does *not* change in the coordinate-independent sense I gave in my last post: its covariant derivative with respect to the body's proper time is zero. So if you think its velocity is changing, what is it changing relative to? Any such definition of velocity "changing" will be a coordinate-dependent definition. The GR definition is not; it's a genuine physical observable (whether or not the body feels acceleration).

Even if the object is only momentarily comoving with the reference frame at a single event, it's a GLOBAL LORENTZ frame. The Lorentz Frame extends for infinity into the past, the future, and in all directions.

And how are the coordinates in this supposed global Lorentz frame to be defined? Can you specify a way to do it that is both consistent with all the laws of SR, *and* works in a curved spacetime, where gravity is present? If so, please elucidate.

Because the Lorentz Transformation, just like the rotation transformation, takes as its input, the location of every event in space and time, and outputs the locations of every event in space and time.

If by "locations" you mean "coordinates", then yes, mathematically, this is what the LT does. That's the easy part. You have completely glossed over the hard part, *assigning* those coordinates in the first place in a way that is consistent with all the laws of SR. If you think you can do that in a curved spacetime, again, please elucidate.

 

[pervect]

There are tidal forces in effect, if you have more than a point-observer. Even locally, free-fall is different from straight-line motion. The speed of light has a different geodesic from the falling box. Objects moving at different velocities have different geodesics, and with a careful bit of study, you might be able to figure out what you're orbiting around and how far away it is, even if you can't look outside your box.

The local phenomena within the area of a geodesic would be small, perhaps too small for your most sensitive equipment to detect, but traveling along a geodesic is different from traveling in a straight line.

Well, there's never (or at least not usually) a straight line in our actual universe, because space-time is in general not flat. So geodesics are as close as we come.

Geodesics are of course described by the geodesic equation.

$$\frac{d^2x^\lambda }{dt^2} + \Gamma^{\lambda}{}_{\mu \nu }\frac{dx^\mu }{dt}\frac{dx^\nu }{dt} = 0$$

Futhermore, if we multiply through by m, and replace the right hand side by a force, this is about as close as GR comes to Newton's equations of motion.

Specifically, if we have a test particle moving under the action of an external non-gravitational force, (for instance, an electric field), we can write in GR

$$m\,\frac{d^2x^\lambda }{ds^2} + m\,\Gamma^{\lambda}{}_{\mu \nu }\frac{dx^\mu }{ds}\frac{dx^\nu }{ds} = F$$

So, it shouldn't be too surprising that the Christoffel symbols act pretty much like forces. In particular, we can identify some of them as being equal to what we used called the "force" of gravity in Newtonian theory.

And in GR, we can replace solving F=ma with solving the geodesic equations (well, there are occasions where this doesn't work, and we have to worry about the Paperpatrou equations, but this is rare)

BUT

As I remarked earlier, the components of the Christoffel symbols transform in a complex way. From wiki http://en.wikipedia.org/w/index.php?title=Christoffel_symbols&oldid=455650120 they transform like:

$$\overline{\Gamma^k_{ij}} = \frac{\partial x^p}{\partial y^i}\, \frac{\partial x^q}{\partial y^j}\, \Gamma^r_{pq}\, \frac{\partial y^k}{\partial x^r} + \frac{\partial y^k}{\partial x^m}\, \frac{\partial^2 x^m}{\partial y^i \partial y^j}$$

which is not the tensor transformation law. So it's convenient to think of Christoffel symbols as "forces" in anyone particular coordinate system that you want to work in, but it's a mistake to think they'll transform in the same manner as the forces in flat space-time that you may be thinking of them as being analogous to.

 

[JDoolin]

A couple of notes that I wanted to throw out before I go to work:

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

(1) I wanted to point out that this distribution is NOT an equipartition of rapidity, but an equipartition of v/(c sqrt(1-1/v/c^2)

I think if I redo it with equipartition of rapidity, we'll have distribution that looks the same (density-wise) in the center after Lorentz Transformation.

(2) I want to make an analogy. Let's say I take a cone, lay it out flat, and draw a straight line on it, and then wrap it back up in the cone shape again. We live in a cartesian coordinate system. Has that Cartesian Coordinate system failed us because that line that we have defined as straight is now curved? No. The cartesian coordinate system is alive and well, but it has a cone in it, and it has curved lines in it.

I'm not trying to criticize the application of geometry to take a cone and wrap it, and say, YES, you can draw "straight" lines on a curved object. I have no problem with that. But when you go on to say that the overlying cartesian geometry has somehow been invalidated because you can shape a paper into a cone, that's where I have a disagreement.

(3) I'm not saying that straight lines are always easy to detect. Of course if I look at the moon on the horizon, it looks distorted, because the light has NOT moved in a straight line (mostly because of atmospheric rather than gravitational effects). But the fact that the light doesn't move in a straight line does NOT mean that straight lines don't exist. But on the whole, how often does that happen? The great majority of things in the universe are not significantly affected by this sort of phenomenon. You point at them, and that's the direction they (not are) WERE, when the light left them. You can't see where the object IS, but you can see where the object WAS when the light left it.

 

[PeterDonis]

(2) I want to make an analogy. Let's say I take a cone, lay it out flat, and draw a straight line on it, and then wrap it back up in the cone shape again. We live in a cartesian coordinate system. Has that Cartesian Coordinate system failed us because that line that we have defined as straight is now curved? No. The cartesian coordinate system is alive and well, but it has a cone in it, and it has curved lines in it.

I'm not trying to criticize the application of geometry to take a cone and wrap it, and say, YES, you can draw "straight" lines on a curved object. I have no problem with that. But when you go on to say that the overlying cartesian geometry has somehow been invalidated because you can shape a paper into a cone, that's where I have a disagreement.

The cone's surface (barring the singularity at the apex) is still "flat" in the sense that it has zero intrinsic curvature. It has nonzero *extrinsic* curvature--the way in which it is embedded in the surrounding 3-D space is curved--but that's not what we're talking about in this discussion. If you calculated the connection coefficients in the formulas that pervect was writing, working with the Cartesian coordinates you assigned to the cone when it was laid out flat and you drew a straight line on it, they were zero when the cone was laid out flat and they are still zero when the cone is wrapped back up into a cone. So the cone is still flat in the intrinsic sense. (Technically, you could construct coordinates on the cone where the connection coefficients were nonzero, but the point is that you don't *have* to--there will always be *some* coordinate chart where they are all zero, whether the cone is laid out flat or wrapped up. The same is *not* true for a sphere--see below.)

Now try the same thought experiment with a sphere. You can't do it. There is no way to "lay a sphere out flat" and draw straight lines on the flat version, and then wrap it all back up into a sphere again. It's impossible--any such operation will distort the surface and invalidate its geometric invariants. That's one way of expressing the fact that a sphere has nonzero *intrinsic* curvature. The connection coefficients in pervect's formulas are nonzero on a sphere (i.e., there is *no* coordinate chart on the sphere that makes them all zero). And that's the kind of curvature we're talking about when we say that gravity is curvature of spacetime.

(3) I'm not saying that straight lines are always easy to detect. Of course if I look at the moon on the horizon, it looks distorted, because the light has NOT moved in a straight line (mostly because of atmospheric rather than gravitational effects). But the fact that the light doesn't move in a straight line does NOT mean that straight lines don't exist. But on the whole, how often does that happen? The great majority of things in the universe are not significantly affected by this sort of phenomenon. You point at them, and that's the direction they (not are) WERE, when the light left them. You can't see where the object IS, but you can see where the object WAS when the light left it.

No, you are not even seeing that. Light is bent by gravity. This has been measured for light passing close by the Sun: take two stars that are a little further apart in the sky than the apparent diameter of the Sun, when the Sun is in a different part of the sky--say the are separated by an angle a. When the Sun is in between them in the sky, they are separated by a *larger* angle, a + da, because the Sun bends their light towards itself. And what's true for the Sun is true for any large gravitating body.

The spacetime we are living in has intrinsic curvature because of gravity; it is a spacetime analogue of something like a sphere (actually more like a saddle, but the same argument I gave for a sphere would apply to a saddle too). It is *not* the spacetime analogue of something like a cone or a cylinder that can be laid out flat and wrapped up again without changing its intrinsic geometry. And in the presence of intrinsic curvature, all your intuitions about how things work in flat Euclidean space, or flat Minkowskian spacetime, are simply wrong on any large scale; they only approximately work over very small patches. You can only treat the Earth's surface as flat over a small area, and you can only treat spacetime as flat over a small region.

 

[JDoolin]

I have made several further posts in the other thread.

https://www.physicsforums.com/showthread.php?t=545002

I think I have explained myself better there.

I'm trying to find the most concise description of our disagreement: You seem to take the attitude that infinite straight lines do not exist. And I take the attitude that infinite straight lines do exist, or at least are definable. In my thinking, regardless of the curvature of paths caused by gravity, it is always possible to imagine what would happen if matter wasn't there. In your thinking, regardless of how we try to imagine what would happen if matter wasn't there, there would always be more matter, screwing things up. Now I won't argue with that, but the question is not whether we can really map the straight lines with perfect accuracy. The question is whether those straight lines exist at all. I think they do. You think they don't.

I explain why I think they do in the other thread.

 

[PeterDonis]

I'll comment in the other thread.

 

[Dale]

I will do that. It will take a few days.

This came up in another thread. I worked on this for a few days, got stuck, and did not complete it. The matter distribution in question had some discontinuities which made it difficult for me to handle, also, the matter distribution didn't have a nice convenient form like a fluid.

The bottom result is that I was not able to conclusively prove or disprove either position. I still think that I am probably correct, but not so strongly as before.

 

[George Jones]

This came up in another thread. I worked on this for a few days, got stuck, and did not complete it. The matter distribution in question had some discontinuities which made it difficult for me to handle, also, the matter distribution didn't have a nice convenient form like a fluid.

The bottom result is that I was not able to conclusively prove or disprove either position. I still think that I am probably correct, but not so strongly as before.

I'm having trouble tracing this back. What, precisely (and succinctly), is the matter distribution in question.

 

[Q-reeus]

I'm having trouble tracing this back. What, precisely (and succinctly), is the matter distribution in question.

Try #17 where the specs were set out.

 

[George Jones]

I am looking for a new post, written in succinct scientific style (without a lot commentary), of the form

"For a matter distribution given by ..., calculate ..."

 

[JDoolin]

Here's an idea for a non-stationary matter distribution:

Assume you have 5000 particles, originating from an event (t=0,x=0) and each is assigned a random x and y rapidity between -3 and 3, traveling away from each other in the xy-plane. Assume that these particles move with constant velocity until time t=1.

At time t=1, the particles spontaneously develop mass. Given one of the particles, calculate the force on this particle resulting from the other 4999 particles.

(The proper-time/coordinate-time of the spontaneous development of mass, and the delay before that mass is detected, would need to be more fully described to answer the question.)

 

[Dale]

I'm having trouble tracing this back. What, precisely (and succinctly), is the matter distribution in question.

The details are a little fuzzy, but if I recall correctly Q-reeus was claiming that GR was not self consistent because of the transition from the Schwarzschild metric outside a solid shell to the flat metric inside the solid shell.

Somehow the discussion became focused on the space-space components of the (Ricci) curvature tensor and the space-space components of the stress-energy tensor. Q-reeus though you could neglect the space-space components of the stress-energy tensor simply because the time-time component was larger (which somehow led to a contradiction, though I don't remember how).

So I was going to calculate the metric for a finite-thickness solid shell and show that the space-space components of the stress-energy tensor could not be neglected and that accounting for them resolved the supposed contradictions.

 

[pervect]

At the risk of hijacking the thread (which seems hopelessly confused anyway), I do have a specific problem along the lines of boundary conditions - one I think I solved correctly, that I presented earlier. I think there was some questions raised about it, but I didn't follow the questions.

If we consider one of the simplest possible forms for the interior metric of a photon star,

from http://arxiv.org/abs/gr-qc/9903044

eq (1)

$$\frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2$$

we might ask how do we go about enclosing said interior metric in a thin, massless shell, so we get the exterior Schwarzschild metric. I.e. how do we match up the exterior and interior Schwarzschild soultions at the boundary so that we have a solution for light in a spherical "box" by matching the interior solution given by (1) to some exterioor Schwarzschild solution.

I started with the line element from Wald for the spherically symmetric metric:

eq(2)

$$-f(r)\,dt^2 + h(r)\,dr^2 + r^2 \left(d \theta ^2 + sin \, \theta \: d\phi^2 \right)$$

Einsteins' equations give via equations 6.2.3 and 6.2.4 from Wald, General Relativity

$$8 \, \pi \rho = \frac{ \left( dh/dr \right) }{r \, h^2} + \frac{1}{r^2} \left( 1 - \frac{1}{h} \right) \; = \; \frac{1}{r^2} \frac{d}{dr} \left[r \, \left(1 - \frac{1}{h} \right) \right]$$

$$8 \, \pi \, P = \frac{ \left( df/dr \right) } {r \, f \, h} - \frac{1}{r^2} \left( 1 - \frac{1}{h} \right)$$

Here $\rho$ and P are the density and pressure in the spherical shell.

Setting $\rho$ to zero and using 6.2.3 immediately tells us that r (1 - 1/h) is constant through the shell. For a thin shell, this means that h is the same inside the shell and outside the shell, because r is the same at the interior of the shell and the exterior of the shell, so h-, h inside the shell, equals h+, h outside the shell.

We can add 6.2.3 and 6.2.4 together to get

$$8 \pi \left(\rho + P \right) \; = \; \frac{ \left(dh/dr \right) } {r h^2 }+ \frac{ \left( df/dr \right) } {r \, f \, h} \; = \; \left( \frac{1}{ r \, f \, h^2 } \right) \, \frac{d}{dr} \left[ f \, h \right]$$

So we can see that the product (f * h) can't be constant through the shell. So, we known that the right boundary conditions are that h is constant and f varies. In a shell of finite thickness, f will increase continuously throughout the shell. As we shrink it to zero width, f jumps discontinuously.

Simply put, for a _massless_ shell, we can say that the spatial curvature coefficient, h, is the same inside the shell and outside. This is a consequence of Einstein's equations. While h is constant, f, the time dilation metric coefficient, is NOT constant. This also follows from Einstein's equations.

We can do some more computation and find the exterior metric if we assume that the boundary of the shell is located at r=1. (It turns out we can place it wherever we like).

Then, the metric previously given in (1) is used for r<1, and for r> 1, we use

$$\frac{dr^2}{1-\frac{3}{7r}}+ r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \left( 1-\frac{3}{7\,r} \right) dt^2$$

We can do some more interesting stuff along the lines of comparing the Komar mass to the Schwarzschild mass parameter, but I think it suffices to say that the two agree for the total mass M as judged by the observer in asymptotically flat space-time, but are distrubuted differently in the interior.

 

[Q-reeus]

At the risk of hijacking the thread (which seems hopelessly confused anyway),...

Feel free to 'hijack' this confused thread, which evidently has kicked back into life, and hopefully end the confusion.
Won't comment on the specifics of your photon gas inside a containing shell model, other than to say that there shell self-gravitation as contribution to it's own stress seems to be, understandably, an entirely absent factor. In my scenario, it is the only contribution. If you go back to #1 hopefully my problem statement is made clear enough, and basically what DaleSpam said in #201 sums it up. Again, the specific model settled on was in #17, but that can be obviously generalized.
I walked away from this thread owing to a general failure to get agreement on being able to apply differential length, radial (dr) vs azimuthal (rdΩ), in coordinate measure, as suggested by the standard Schwarzschild coordinates. As expressed in earlier entries, I accept boundary matching from exterior to interior regions is always possible mathematically. Whether that math is properly based on a physical principle (and I was genuinely shocked when it was claimed shell stresses for an entirely self-gravitating shell would do that trick) is another matter!