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

by Q-reeus
Tags: handle, mass, metric, shell, spherical, transition
Physics
Sci Advisor
PF Gold
P: 5,507
 Quote by JDoolin 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.

 Quote by JDoolin 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.

 Quote by JDoolin 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.
Physics
Sci Advisor
PF Gold
P: 5,507
 Quote by 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
P: 1,555
 Quote by JDoolin 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.
PF Gold
P: 706
 Quote by Passionflower 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?
Physics
Sci Advisor
PF Gold
P: 5,507
 Quote by JDoolin 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.

 Quote by JDoolin 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.

 Quote by JDoolin 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.

 Quote by JDoolin 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.
P: 1,555
 Quote by JDoolin 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.
P: 1,555
 Quote by PeterDonis 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.
PF Gold
P: 706
 Quote by PeterDonis 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.
PF Gold
P: 706
 Quote by Passionflower 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.
Physics
Sci Advisor
PF Gold
P: 5,507
 Quote by JDoolin 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.

 Quote by JDoolin 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.

 Quote by JDoolin 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).

 Quote by JDoolin 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.

 Quote by JDoolin 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.
Emeritus
Sci Advisor
P: 7,436
 Quote by JDoolin 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?...ldid=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 any one 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.
PF Gold
P: 706
A couple of notes that I wanted to throw out before I go to work:

 Quote by JDoolin
 Quote by JDoolin 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.
Physics
Sci Advisor
PF Gold
P: 5,507
 Quote by JDoolin (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.

 Quote by JDoolin (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.
 PF Gold P: 706 I have made several further posts in the other thread. http://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.
 Physics Sci Advisor PF Gold P: 5,507 I'll comment in the other thread.
Mentor
P: 16,476
 Quote by DaleSpam 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.
Mentor
P: 6,038
 Quote by DaleSpam 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.
P: 1,115
 Quote by George Jones 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.

 Related Discussions Special & General Relativity 15 Quantum Physics 15 Advanced Physics Homework 3 Advanced Physics Homework 6 Quantum Physics 1