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

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?

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.

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.

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.

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

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

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

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.

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).

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.

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

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.

[tex]
\frac{d^2x^\lambda }{dt^2} + \Gamma^{\lambda}{}_{\mu \nu }\frac{dx^\mu }{dt}\frac{dx^\nu }{dt} = 0
[/tex]

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

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

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:

[tex]
\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}
[/tex]

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.

JDoolin

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

(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

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.

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.

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.

PeterDonis

I'll comment in the other thread.

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.

George Jones

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

Q-reeus

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.)

DaleSpam

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)

[tex]\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[/tex]

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)

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

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

[tex]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]
[/tex]

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

Here [itex]\rho[/itex] and P are the density and pressure in the spherical shell.

Setting [itex]\rho[/itex] 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

[tex]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] [/tex]

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

[tex]
\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
[/tex]

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

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!