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I am the first in the world

Haha, touché. However, it does open up a potential new avenue for fun, in simple ridicule of those who claim to be physicists but can't do basic physics. I think some difficult EM or Classical Mechanics problems would suffice, since most crackpots believe in "tangible" physics, as they call it.

Also, I believe I should have posted this here: http://www.internationalskeptics.com/forums/showthread.php?t=194562

Oh well. :boxedin:

It's their misplaced hubris that amuses me. They always seem to think that they are smarter than Einstein,Newton,Hawking and Feynman rolled into one,but can't do basic maths. Back to the topic,
 
Constant with respect to radius, I assume.

Correct. Inside of a given r (as radius) it has the stress-energy tensor as shown, outside of that r the stress-energy tensor is null.

For spherical symmetry, the exterior must be Schwarzschild because of Birkhoff's theorem. A hypothetical star of uniform density has the interior of a closed FRW universe; it is well-known that this can be smoothly joined to a Schwarzschild exterior.

I agree about the Birkhoff part (and my own solution I found is 'Schwarzschild' outside of the radius where the stress-energy tensor is zero), but I question the FRW universe claim. At the very least I will have to give that part some thought, so let me get back to you about that.

For this to be static, it is necessary for the scale factor a to not change, in particular ä = 0, so that
[latex]$\frac{\ddot{a}}{a} = -\frac{4\pi}{3}(\rho + 3p) = 0$[/latex]
from the Friedmann equations, so that we must have p = -ρ/3 rather than p = -ρ as you have (here, ρ is density rather than your radius). And even then, that's not sufficient for it to be static and it will be still be unstable.

Well, you can say what you want, but I DID find a solution with the stress-energy tensor I gave (the solution obeys the Einstein Field Equations). As to whether the solution is physical or not is another matter. Another problem is you would have to judge for yourself if my solution really is a solution (in the strict math sense) and that takes a lot of time on both our parts.

Does the FRW solution have any singularities or event horizons inside of the appropriate sphere (whether in the static case or not)? Just looked, it definitely has event horizons at least.

Also, this is the first time I have heard that the FRW solution should apply inside of some region. Thanks for pointing that out. I am self-taught on many matters concerning GR.

Well, sure, but that doesn't pertain to the the non-existence of black holes much. The interior can still collapse; the real question is can and must it collapse? There are singularity theorems (Penrose, Hawking, et. al) that address this in a very general context, without even assuming nice geometrical properties like spherical symmetry.

Yep, I am aware of the Hawking-Penrose Theorems. I am not questioning whether the math people have been doing in relation to blackholes is true or not, but whether the physical interpretations are correct.

That's virtually how the Scwarzschild coordinate is defined in the first place, so I don't understand the significance. By definition, the surface of Schwarzschild radius r = R has an area of 4πR² in any slice of constant Schwarzschild time. Since this is the same as that of an ordinary Euclidean 2-sphere of radius R, it's very natural to expect that curvature to be the same. And indeed, it is such a 2-sphere, as is obvious from Schwarzschild metric.

OK, depends on how you define radius. It seems to me the most natural definition of radius is as the length from the center to some point. Big problem if one does that for a totally Schwarzschild Universe, because that will give you infinity. In the solution I found there are no singularities or event horizons so an unambiguous radius and not a reduced circumference can be defined.

Ughghg, too much to cover. Let's forget about the radius not being the reduced radius claim for a moment. Whichever solution you use, the Schwarzschild only, Schwarzschild + FLRW or Schwarzschild + "Metric I found", the r in all such cases, as they are usually written, are not coordinates but geordinates. In a non-spherical geometry the r will not allow for unique designation of points in all cases. Coordinates must always allow for unique designation of points or else they are not coordinates, no matter what geometry is used.

P.S. Who is Crothers?

You do not want to know. He is this quackpot guy that didn't get a PhD and is saying a number of things about why GR does not make sense. The only thing I found for sure that he was correct about is that the radius in the Schwarzschild Solution (Oh, I also think he has a correct historical point to make that the Schwarzschild Solution is actually Hilbert's solution) is not the radius. Covered above some about the radius questions.

I don't think you found such a thing. A collapsing star of uniform density has the interior of a closed FRW universe, and that has a singularity in its future. Before the singularity theorems of Penrose, Hawking, et al., it was hoped that formation of singularities was an artifict of exact geometrical conditions (spherical symmetry) that could otherwise be dismissed as nonphysical. Those theorems proved that wrong.

Actually, I have a feeling you are faithfully representing the current viewpoint very well. I also think you could be correct that the inner region Stress-energy tensor should have the form of

[latex]$T_{\mu\nu} = \mbox{diag}(\rho, -p, -p, -p)$[/latex]

instead of

[latex]$T_{\mu\nu} = \mbox{diag}(p, -p, -p, -p)$[/latex]

for the inner region of the sphere where the mass-energy resides. I will try and solve the equation again with this change and see if I can.

And in general, I don't understand your basic logic either--how does the exhibition of a single solution, no matter how well-behaved it actually is, support the non-existence of a whole class of very general phenomena (black holes)?

It doesn't. I wanted to call into question the general logic of such claims. Here is something I have heard for an argument for spherically symmetric blackholes. If a mass gets squeezed further then the Schwarzschild radius, then since the mass is below the event horizon, nothing will stop it from further collapse eventually down into a singularity.

The problem is this statement is not taking into consideration the two-part nature of the mass in question. There is a solution outside of the sphere and inside of the sphere. The metric solution outside of the sphere will at the very least always have a singularity in the origin no matter how collapsed or uncollapsed the matter is (that is the nature of the Schw. sol.). This does not matter though since another metric takes over on the inside. In the solution I found the event horizon was always on the inside of the sphere where this is also not a problem since the inside metric does not have an event horizon there (the outside metric's event horizon and singularity reside inside the sphere of mass-energy).

Looking at the FRW solution I can probably guess it has a singularity at the origen. Let me know if I am wrong.

Lets step back a moment. Imagine you have a sphere of constant charge density. You want to know what the electric field should be. There will be a two part solution. The inside part of the field is proportional to the radius and the outside part is proportional to the inverse radius squared. This is good because there are no 'singularities' on the inside or outside (and certainly no event horizons).

Here is my take, if you choose a metric that blows up on the inside it is like choosing the wrong solution. It is like saying that the solution on the inside is proportional to the inverse of the radius to some power like in the simple electric field example (as opposed to proportional to the radius to some power).

I could go on, but for now what I am saying is that the physicists and mathematicians might definitely have messed up somewhere. Infinities in theories imply a limit to a theory (such as happened with Blackbody Radiation), a problem with a theory (nothing comes to mind off-hand), or a misinterpretation of a theory (...). I am opting for misinterpretation.

Oh well, get back to me. I love having gristle for thought.
 
I don't think you found such a thing. A collapsing star of uniform density has the interior of a closed FRW universe, and that has a singularity in its future. Before the singularity theorems of Penrose, Hawking, et al., it was hoped that formation of singularities was an artifict of exact geometrical conditions (spherical symmetry) that could otherwise be dismissed as nonphysical. Those theorems proved that wrong.

Hmm. Does this mean that the collapse to a black hole is like a "little big crunch" in some way (insofar as the interior of the collapsing object goes)?
 
It's their misplaced hubris that amuses me. They always seem to think that they are smarter than Einstein,Newton,Hawking and Feynman rolled into one,but can't do basic maths. Back to the topic,

So then there's such a thing as "well-placed" hubris? What's that? It's good to look down on and scathe your fellow humans?!
 
Haha, touché. However, it does open up a potential new avenue for fun, in simple ridicule of those who claim to be physicists but can't do basic physics. I think some difficult EM or Classical Mechanics problems would suffice, since most crackpots believe in "tangible" physics, as they call it.

Ths idea of having "fun" by hurting other people is not something I agree with. What's the problem with saying "you're wrong", perhaps with some rational argument as to why, and just leaving it there?
 
Hmm. Does this mean that the collapse to a black hole is like a "little big crunch" in some way (insofar as the interior of the collapsing object goes)?

Yes. When you're inside the event horizon, the singularity is not separated from you by distance, it's separated from you by time. The singularity becomes your future. It because every possible future for you. No matter how you try to move, you can't escape moving into the future.
 
I agree about the Birkhoff part (and my own solution I found is 'Schwarzschild' outside of the radius where the stress-energy tensor is zero), but I question the FRW universe claim. At the very least I will have to give that part some thought, so let me get back to you about that.
You're right--I just double-checked things, and the smoothing gluing of portions of FRW with portions of Schwarzschild exterior is for the case of dust (pressureless) FRW universes. See, e.g., MTW Box 32.1. For non-negligible pressure, the situation is qualitatively similar, but can't be obtained so simply.

Well, you can say what you want, but I DID find a solution with the stress-energy tensor I gave (the solution obeys the Einstein Field Equations). As to whether the solution is physical or not is another matter. Another problem is you would have to judge for yourself if my solution really is a solution (in the strict math sense) and that takes a lot of time on both our parts.
OK, but I can judge that you probably did it correctly already. For a spherically symmetric static star, assuming a uniform density ρ, the Tolman-Oppenheimer-Volkov equation says:
[latex]$\frac{dp}{dr} = -\frac{GM(r)}{r^2}{(1+3\frac{p}{\rho})}{(1+\frac{p}{\rho})}{(1-\frac{2GM(r)}{r})^{-1}}$[/latex]
(For a nonuniform star, a certain density averaging is required.) This means that constant-pressure solutions exist for p = -ρ/3, which is what I expected from the flawed FRW analogy, but also p = -ρ, which should be what you found by explicit calculation. However, they are both unstable, and in yours the fluid is light-like.

OK, depends on how you define radius. It seems to me the most natural definition of radius is as the length from the center to some point. Big problem if one does that for a totally Schwarzschild Universe, because that will give you infinity.
There is no such definition either in general relativity nor differential geometry. In general, distance depends on the precise way it is measured; different observers, or even the same observer varying the procedure, will give different results. That the Schwarzschild radial coordinate poorly represents radial distance is obvious from the dr² term in the metric, but that was never its purpose in the first place.

If you want a chart of the Schwazschild spacetime whose radial coordinate represents proper distance within its constant-time hypersurfaces, take a look at the Painlevé-Gullstrand coordinates, which are adapted to a family of observers freefalling from rest at infinity. The constant-time slices are exactly Euclidean, even.

Whichever solution you use, the Schwarzschild only, Schwarzschild + FLRW or Schwarzschild + "Metric I found", the r in all such cases, as they are usually written, are not coordinates but geordinates. In a non-spherical geometry the r will not allow for unique designation of points in all cases. Coordinates must always allow for unique designation of points or else they are not coordinates, no matter what geometry is used.
How do you define "geordinate"?

---

Yep, I am aware of the Hawking-Penrose Theorems. I am not questioning whether the math people have been doing in relation to blackholes is true or not, but whether the physical interpretations are correct.
I don't understand your argument, then. You have a single unstable solution of a luminal perfect fluid. Therefore... ?

It doesn't. I wanted to call into question the general logic of such claims. Here is something I have heard for an argument for spherically symmetric blackholes. If a mass gets squeezed further then the Schwarzschild radius, then since the mass is below the event horizon, nothing will stop it from further collapse eventually down into a singularity.
For the spherically symmetric, uniform-density perfect fluid that you're considering, that statement is even stronger that it needs to be: the central pressure tends to infinity when R/2M≤9/8 and with R/2M<4/3, the fluid at the center must be superluminal for the star to remain static. This is either derived or left as an exercise in several books. Note that your solution is right at the edge of violating this condition already.

The problem is this statement is not taking into consideration the two-part nature of the mass in question. There is a solution outside of the sphere and inside of the sphere. The metric solution outside of the sphere will at the very least always have a singularity in the origin no matter how collapsed or uncollapsed the matter is (that is the nature of the Schw. sol.).
That doesn't make sense. The metric outside the sphere has no singularity at the origin because it's not defined near the origin. You seem to be thinking that the claim that Schwarzschild has a singularity at the origin or horizon at the Schwarzchild radius follows from it them being singular there. That's not correct. Heck, polar coordinates on a standard Euclidean plane are singular at the origin, but the plane itself is well-behaved there.

Instead, calculate some curvature invariant in the Schwarzschild geometry, for example the Kretschmann or Ricci scalar, and you'll find divergence as r→0. But in your geometry, the metric is Schwarzschild only far away from the origin, so it makes no makes sense to say the Schwarzschild exterior has a singularity. The metric is just a rank-2 tensor field defined at every point on the manifold having certain properties. At every point, there is just one particular tensor, and it makes no sense to apply it to a point other at which it's defined.

Lets step back a moment. Imagine you have a sphere of constant charge density. You want to know what the electric field should be. There will be a two part solution. The inside part of the field is proportional to the radius and the outside part is proportional to the inverse radius squared. This is good because there are no 'singularities' on the inside or outside (and certainly no event horizons).
Exactly! Therefore, it makes no sense to say that the (Coulomb-like) vector field solution outside of the sphere is singular at the origin. So I'm very confused as to why you made the analogous statement about the exterior since you seem to realize that it's wrong...

Here is my take, if you choose a metric that blows up on the inside it is like choosing the wrong solution.
I think you might be ascribing a mistake to mainstream physics that as far as I can see, no one has ever made. See the discussion on stellar evolution in MTW Ch. 23 or Weinberg Ch. 11.1. Even in the highly idealized case of spherical symmetry, no one is silly enough to claim the solution becomes Schwarzschild just because the exterior is.
 
Vorpal, I am just glad to be able to discuss GR with someone who is read on the subject. I did know about the Kretschmann scalar and all that (in a Schw. Universe K ~ 1/r^6 or something like that). By MTW and the box referrence I think you are probably talking about the book "Gravity" by Misner, Thorne and Wheeler, no? Just wanted to get that out for everyone else that might be following along.

Man, I love it that you cite equations too. Now it is time to look up that as well (have heard of it, I will let you know if something is unheard of by me, but it is nice all the same to see an equation). Oh, I have not heard of Panleve-Gulstrand coordinates I think, so that is new... There are so many types of coordinates out there on the subject of blackholes.

Let me go over geordinates because it is hopefully the least contentious aspect of the post I made and can be understood in a purely mathematical way if one desires. If you think about it, a particular coordinate system (modulo frame considerations) is like an algorithm, use a ruler here, a compass there, a clock or two, do some calculations and it should give a unique set of numbers for any one point inside of a given region.

Coordinates are normally thought of as being independent of geometry. It should not matter after all whether your geometry is spherically symmetric or of any other form for one to be able to say that some point is located at (1,1,2) or some such.

Imagine though for the moment that we have (and I am just using this because it is the nearest example to hand) a spherically symmetric 3-D geometry. Out to any given point from the origin one could find the Gaussian radius of curvature, denoted by say u. In flat space the radius (as defined as the distance from a straight line to the origin) would be u = r. In some other space that has a geometry that is spherically symmetric about the origin of the coordinate system, u would be a monotonically increasing function of r so that u = u(r).

In this case if we have for instance a "coordinate" system of (u, \theta, \phi) with the \theta and \phi having their conventional meaning to them. We might be tempted to say that this system of numbers was good for any type of geometry, but that would be incorrect, since in another type of geometry (imagine a geometry that is cylindrically symmetric, or warped in any number of ways) and the u would not work as a coordinate all of the time. An ordinate that depends on what type of geometry exists I am for the moment calling a geordinate (my realization could very well have been done before me after all).

It is the case that in a spherically symmetric geometry the (u...) ordinate system will work for all points, but I hope you can see that it will not work in all geometries. This actually makes the (u...) a mixed coordinate / geordinate system I guess.

Now, how does that play with anything we have been talking about? OK, look at the Schwarzschild solution. I have heard of the r in the solution referred to as the radius, reduced circumference or aerial something or other. It can not be a radius though. If we set t and the angles to constant in the Schw. metric and find the distance from some point off of the origin (strictly should go from the origin but then the integral would explode in the Schw. sol. case) to a point out to some given r, then the formula one gets is not the same as (r - constant) but some other formula (f(r) - const).

For that reason the "r" in the Schw. Sol. is certainly not a radius. It acts like a radius, increasing monotonically away from the origin, but is not the radius. Hopefully this does not surprise you. In fact, for points far from the origin (and doing the integral of the metric with t and so forth constant as described before), f(r)/r -> 1 as r -> \infty.

But if r is not the radius, then what is it? This is thanks to Crothers. It is the Gaussian radius of curvature. It really is, look up the formula for the Gaussian Curvature, compute it for the metric in Schw. Sol. and call it some variable x, then the Gaussian radius of curvature is R = 1/x^(1/2). If you do it you get R = r. But if r is the Gaussian radius of curvature (GRC from now on), then r does not act like a normal coordinate. In another geometry the GRC plus the other coordinates would not necessarily give a unique designation to a set of points. The r does act like a coordinate though (for spherically symmetric geometries), so I call it a geordinate (term of my own coining).

That r is not like a normal coordinate should not be all that surprising perhaps. In the derivation of Sch. Sol. some assumptions as to the form of the metric are made in line with the spherically symmetry. It may turn out that doing this for various kinds of symmetry will lead to geordinates.

I saw the complaint that there is no such definition in GR for radius. Correct, the concept of a radius is independent of GR. The problem I think is that when you get used to coordinate independent equations one starts to think everything should be coordinate independent, but this does not make sense. Coordinates are not just labels, (x, y, z), what have you. Each coordinate for a specific case (and the Schw. Sol. is a specific case) has a specific meaning associated with it. Yes, laws should be coordinate independent, but specific solutions that can be tested (such as Schw. and the other exact solutions in GR) have to have a meaning, an algorithm or set of steps by which one can tell an experimentalist how to measure things. I think some people working in the field have forgotten this.

So, the algorithm one uses to find r in the Schw. solution is not the same (and can not be the same) as the algorithm used to find radii in the conventional sense, so why don't we stop calling it the radius or similarly contrived concept? It is in fact the GRC. The reduced circumference and aerial whatever do give you r, but they make you think perhaps you are doing so in a geometry independent manner, which is simply not the case, since the GRC depends on the geometry and even the reduced circ. / aerial coord. would not equal the GRC in such other geometries.

Next post in a bit, trying to break things up.
 
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Since the discussion so far has been of a congenial nature I hate to have to bring the following up.

In general, distance depends on the precise way it is measured; different observers, or even the same observer varying the procedure, will give different results.

Different observers should not change the result of distance measurements. I think this was a slip on your part because you seem to me to be very knowledgeable of other aspects of GR and so on. I am really towing the company line as it were in saying this too. Part of the whole point of GR is that propre-times will be agreed upon by all observers.

The path is important to the length though I agree. The length of a given path is a coordinate independent quantity however.

Radii can be thought of as paths originating from a point called the origin to some other point without variation in angle or time to that origin (as per the type of coords used in the Schw. sol. expressed in the coordinates it is usually given in).

Enough about radii though. My next topics should be about the more contentious topics being brought up.
 
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If you think about it, a particular coordinate system (modulo frame considerations) is like an algorithm, use a ruler here, a compass there, a clock or two, do some calculations and it should give a unique set of numbers for any one point inside of a given region.

Coordinates are normally thought of as being independent of geometry. It should not matter after all whether your geometry is spherically symmetric or of any other form for one to be able to say that some point is located at (1,1,2) or some such.
Coordinates are independent of geometry. Your algorithm analogy is mistaken: "a ruler here or a compass there" requires a metric. Coordinates are blind to this. With some very minimal assumptions about the manifold, they do reflect the local topology. That's about all they can do beyond being labels for points.

You're treating coordinates as if they had intrinsic meaning. If I tell you that I have a two-dimensional manifold and a coordinate chart with x,y satisfying 0<x,y<1, you would have no idea whether the manifold is infinite like a plane, or finite like a square, or anything else. Talking about size or shape doesn't even make sense with that information alone. The only thing you know is my choice of labels.

Imagine though for the moment that we have (and I am just using this because it is the nearest example to hand) a spherically symmetric 3-D geometry. Out to any given point from the origin one could find the Gaussian radius of curvature, denoted by say u. In flat space the radius (as defined as the distance from a straight line to the origin) would be u = r. In some other space that has a geometry that is spherically symmetric about the origin of the coordinate system, u would be a monotonically increasing function of r so that u = u(r).
Conventionally, one would say that the metric has the form [latex]$ds^2 = A(r)dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$[/latex] in that coordinate chart. It's possible to translate that into intrinsic, coordinate-free language, but it would still depend on the metric.

It is the case that in a spherically symmetric geometry the (u...) ordinate system will work for all points, but I hope you can see that it will not work in all geometries. This actually makes the (u...) a mixed coordinate / geordinate system I guess.
A coordinate chart is just a way of mapping some points to some n-tuples of real numbers in a nice, bicontinuous way (jargon: a homeomorphism between an open set of the manifold an an open set of Rn). If you're going to a different manifold, then you are also going to have a different coordinate chart--by definition!

You misunderstand coordinates, though that doesn't necessarily mean that "geordinates" are doomed. In order to make sense of having "the same" coordinates for different manifolds, one needs some sort of device to identify either points between those manifolds in a suitable way, or their coordinate charts directly. (A trivial example of the former kind of device would be a diffeomorphism, but that's not very interesting here because diffeomorphic spacetimes are also physically equivalent.)

Now, how does that play with anything we have been talking about? OK, look at the Schwarzschild solution. I have heard of the r in the solution referred to as the radius, reduced circumference or aerial something or other. It can not be a radius though. If we set t and the angles to constant in the Schw. metric and find the distance from some point off of the origin (strictly should go from the origin but then the integral would explode in the Schw. sol. case) to a point out to some given r, then the formula one gets is not the same as (r - constant) but some other formula (f(r) - const).
That's correct, and obvious from the dr² term of the Schwarzschild metric. I don't really understand why consider it to be a problem, though. The Schwarzschild geometry is spherically symmetric, and the Schwarzschild coordinate r is just a parametrization of the nested spheres (in t=const) such that they have the geometry Euclidean spheres of radius r. There was no claim that it represents distance in the first place.

It's far from the only such definition, even within the same Schwarzschild geometry. For example, the isotropic radius is characterized by round light-cones. The way 'radius' is used is having nested spheres satisfying such-and-such criterion. Different criteria give different kinds of radius.

Actually, isotropic coordinates are good examples: if you object to using "radius" in "Schwarzschild radius", for consistency's sake, you should object to things like "angle" for the Schwarzschild φ and θ, because they do not faithfully represent angular measurements (but isotropic ones do). And what about "Schwarzschild time", since there are so many ways of measuring time?

But if r is not the radius, then what is it? This is thanks to Crothers. It is the Gaussian radius of curvature. It really is, look up the formula for the Gaussian Curvature, compute it for the metric in Schw. Sol. and call it some variable x, then the Gaussian radius of curvature is R = 1/x^(1/2).
That's not thanks to Crothers; that's thanks to Schwarzschild. Some books also define the surface of constant r (and t=const, again) as being a Euclidean sphere of area 4πr², which is so completely and obviously equivalent that Crothers didn't need to discover it.

I saw the complaint that there is no such definition in GR for radius. Correct, the concept of a radius is independent of GR. The problem I think is that when you get used to coordinate independent equations one starts to think everything should be coordinate independent, but this does not make sense. Coordinates are not just labels, (x, y, z), what have you. Each coordinate for a specific case (and the Schw. Sol. is a specific case) has a specific meaning associated with it.
They are just labels. If you want them to be more, you'll have to define your own concept.
 
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In general, distance depends on the precise way it is measured; different observers, or even the same observer varying the procedure, will give different results.
Different observers should not change the result of distance measurements. I think this was a slip on your part because you seem to me to be very knowledgeable of other aspects of GR and so on. I am really towing the company line as it were in saying this too. Part of the whole point of GR is that propre-times will be agreed upon by all observers.
Of course lengths of each path are agreed-upon. What's not agreed upon is which path one should take when measuring distance. Suppose you have two observers, A and B, stationary in Schwarzschild spacetime, radially one on top of another (i.e., same Schwarzschild angular coordinates). Then, knowing that the speed of light is constant, they endeavor to measure the distance between them.
1) A bounces off a light pulse off B, and multiplies half the amount of time it takes by the speed of light.
2) B does the reverse, bouncing a light pulse off A.
3) Some third person, C, slowly crawls along a rope between A and B with his trusty ruler.
4) Yet another person, D, does the same as C, except while in freefall instead of crawling along.
None of them will get the same distance. What is the proper procedure, then?

Radii can be thought of as paths originating from a point called the origin to some other point without variation in angle or time to that origin (as per the type of coords used in the Schw. sol. expressed in the coordinates it is usually given in).
There are many mutually contradictory notion of "radii" that satisfy the definition you use here.
 
On your second post, I was referring specifically to the "different observers" part. Unless the different observers use a different path, they must agree on given path lengths, something I think we both agree on. A comma was used between "different observes" and "or even the same observer varying the procedure" parts, so I do not think it was untoward for me to point out what I did.
 
On your second post, I was referring specifically to the "different observers" part. Unless the different observers use a different path, they must agree on given path lengths, something I think we both agree on. A comma was used between "different observes" and "or even the same observer varying the procedure" parts, so I do not think it was untoward for me to point out what I did.
You can point it out, but it misses the original point, since it was a response to:
OK, depends on how you define radius. It seems to me the most natural definition of radius is as the length from the center to some point. Big problem if one does that for a totally Schwarzschild Universe, because that will give you infinity.
Using that definition, there many different radii, because "the length from the center to some point" is not at all well-defined.
 
So, the algorithm one uses to find r in the Schw. solution is not the same (and can not be the same) as the algorithm used to find radii in the conventional sense

It is the same as one such algorithm. Consider a set of concentric spheres. For each sphere, define its radius to be the square root of its area divided by 4 pi. In flat spacetime that's a perfectly good definition of radius, and in the Schwarzchild metric it gives you the coordinate r. Because it relies on spherical symmetry, it turns out to be quite useful.

So there's nothing wrong with calling the Schwarzschild coordinate "r" the radius, although I don't think most people do when they're being careful - they call it "r" or the "areal radius".
 
I myself have solved the Einstein Field Equations for the case of spherically symmetric system and having a stress-energy tensor of

T = diag (p, -p, -p, -p) for radius inside of some sphere.
T = diag (0, 0, 0, 0) for radius outside of some sphere.

The solution was two part and showed some interesting things to me. The first was that the overall solution had no singularities or event horizons.

The first thing you need to do is check T carefully at the cross-over radius. Since the Einstein tensor involves two derivatives of the metric g and the functions in your g may have discontinuities in their first derivatives, you may find a singular spherical shell of some mass and pressure. If so, you need to modify your claim.

Regardless of the result, like Vorpal I do not understand why you think such solutions have any bearing on the existence of black holes or the arguments for them. Stars exist. Stars are spherically symmetric (approximately), and their exterior metric is Schwarzschild (approximately). Most importantly, they must be larger than their Schwarzschild radius, and if they ever collapse to a size below it, they become black holes more or less by definition (because all their matter is inside the horizon).

So to prove that black holes do not exist, you need to prove that stars can never collapse to a size smaller than their own Swchawzschild radius. How does writing down some solution with bizarre stress-energy that doesn't describe a star have any bearing on that question?
 
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context is sometimes important!

Coordinates are independent of geometry.

Agreed.

Your algorithm analogy is mistaken: "a ruler here or a compass there" requires a metric. Coordinates are blind to this. With some very minimal assumptions about the manifold, they do reflect the local topology. That's about all they can do beyond being labels for points.

Disagree.

A coordinate system is a way to get coordinates. If one locally measures with a ruler, the metric does not matter (and one must locally measure in GR). Coordinates are blind to what exactly? The way in which one measures them? That does not make a lot of sense. Something else perhaps?

Stop thinking in highlevel math for a minute and consider how one measures things like coordinates. Here is a metric:

ds^2 = dx^2 + x^2 dy^2.

I bet you thought I meant that x and y were Cartesian huh? Nope, I meant for the x to act completely like r and y to act completely like [latex]\theta[/latex] in polar coordinates. Coordinates do not have any intrinsic meaning, coordinates though in a coordinate system have operational meaning. There is a fixed way in which to determine what a coordinate is for a given point for a given coordinate system and it does follow something like an algorithm (just a spatially based algorith: put your ruler here, measure that time when the light occurs, etc). Even if you just have a look-up table for points, for example, you would still have to look them up, which is an algorithm (finite number of steps, has an end, steps are definite in nature, etc. etc.).

If you take the metric I wrote above and interpret the x and y as 2-D cartesian then you would say that the space was warped, but since you know that the (x,y) are really like [latex](r, \theta)[/latex] as they are conventionally understood, you know that the space is flat. That is kind of what I am getting at when it concerns Schw. Sol. My critiques are from an operational point of view. What is being measured and how is it being measured are the questions.

To clearly understand what a metric is telling you, you have to know both how the coordinates work (as per what the coordinate system is) as well as what the metric is in terms of those coordinates, when dealing with a specific case. (mentally though you could just "look" at a particular manifold to see topology and shape without caring about coordinates, but that is another story).

You're treating coordinates as if they had intrinsic meaning. If I tell you that I have a two-dimensional manifold and a coordinate chart with x,y satisfying 0<x,y<1, you would have no idea whether the manifold is infinite like a plane, or finite like a square, or anything else. Talking about size or shape doesn't even make sense with that information alone. The only thing you know is my choice of labels.

I thought you might bring up coordinate charts, I was thinking I might want to head you off at the pass in one of the last posts, but decided not to. I kind of tried to by bringing up the "within a given region" phrase (basically the same idea as coordinate charts), as well as the "modulo some frame" bit so that you would not be able to bring in relativistic effects either as a point of contention. Since I was referring to only one "region", I thought you would pick up on the meaning of why geordinates could cause problems in said region, but alas, such is life.

In the above you did not give a coordinate system, which is the general topic I have been addressing, so while I have to agree with your conclusions in the post above, it does not change anything, or even address any issues I brought up.

Conventionally, one would say that the metric has the form [latex]$ds^2 = A(r)dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2)$[/latex] in that coordinate chart. It's possible to translate that into intrinsic, coordinate-free language, but it would still depend on the metric.

The metric you listed is already specialized. I guess we could bring up spaces such as metric spaces ala topology that are even more generalized then what is used in GR, but I think we should just stick to Riemannian Geometry for now. In Riemannian Geometry of course the most general metric is (as I am sure you know already)

[latex]ds^2 = g_{\mu\nu} dx^{\mu} dx^{\nu}[/latex].

The above statement for ds is in coordinate-free language. What in the world are you referring to then? Oh yeah, in coordinate-free language the most general form of the metric is always used (as well as perhaps any tensors in coordinate-free or covariant form). Let's not get into Differential Forms either, that would just confuse things (although when I hear "coordinate-free", differential forms immediately comes to mind for some reason).

If you can make the metric you gave above be coordinate-free as you claim, do it, because I have no idea what you are referring to exactly. Maybe I would learn something, my guess though is that your concepts are confused and trying to put the metric you gave in coordinate-free language would show that. Sounds like fun to me either way.

A coordinate chart is just a way of mapping some points to some n-tuples of real numbers in a nice, bicontinuous way (jargon: a homeomorphism between an open set of the manifold an an open set of Rn). If you're going to a different manifold, then you are also going to have a different coordinate chart--by definition!

Yes, a coordinate chart is a region of some space with a coordinate system (or mapping, which can be thought of as a kind of coordinate system if one wants to) and some rules about how regions that share between the charts have some bijective properties in terms of various point-set operations (you are not the only one who knows some math jargon, or math for that matter). GR uses Riemannian Geometry, so the coordinate charts, beyond just being coordinate charts, should also be diffeomorphic (usually one assumes [latex]C^\infty[/latex]) or whatever. But this is getting off-topic.

You misunderstand coordinates, though that doesn't necessarily mean that "geordinates" are doomed. In order to make sense of having "the same" coordinates for different manifolds, one needs some sort of device to identify either points between those manifolds in a suitable way, or their coordinate charts directly. (A trivial example of the former kind of device would be a diffeomorphism, but that's not very interesting here because diffeomorphic spacetimes are also physically equivalent.)

The point is not to try to save geordinate systems (something that acts like a coordinate system but has one or more ordinates that depend on the geometry) as a concept since that needs no saving, but to point out that -

a. It is possible to have ordinates in the kind of charts I was relaying to you that depend on the geometry of the space and yet still give a unique set of designators for points in said chart (I only had one chart for the whole space in the example I gave, but the same principle applies).

b. That if one is not careful (uses charts, makes sure the geometry is OK for the geordinates in question, etc.), then the possibility exists that points may not be uniquely defined by said geordinates + coordinates for a given geometry.

Points a. and b. are the only ones I care to defend. I do not misunderstand coordinates, I just have an understanding I think you have probably not considered too much yet. It is worth considering because I think it will give you a better insight into GR and Differential Geometry, but, whatever. I can only show a horse the water and all that.

That's correct, and obvious from the dr² term of the Schwarzschild metric. I don't really understand why consider it to be a problem, though. The Schwarzschild geometry is spherically symmetric, and the Schwarzschild coordinate r is just a parametrization of the nested spheres (in t=const) such that they have the geometry Euclidean spheres of radius r. There was no claim that it represents distance in the first place.

I care because I love precision when it comes to thought. There was no claim by whom? When I first learned about them in a class in school the instructor literally said that r is the radius. In another class I took the claim was that r was the reduced circumference (the book "Spacetime Physics" was used). Really, it is best understood as being the GRC as far as I am concerned.

It's far from the only such definition, even within the same Schwarzschild geometry. For example, the isotropic radius is characterized by round light-cones. The way 'radius' is used is having nested spheres satisfying such-and-such criterion. Different criteria give different kinds of radius.

(just a stupid point, it is "different kinds of radii", I suck at English some times so do not be afraid in pointing out any syntax or spelling errors on my part either)

The metric in Schw. Sol. is

[latex]ds^2 = (1-2M/r) dt^2 - (1-2M/r)^{-1} dr^2 - r^2 d\Omega^2[/latex]

(I prefer (1, -1, -1, -1) for my signature)

with

[latex]d\Omega^2 = d\theta^2 + \sin^2 \theta d\phi^2[/latex]
(square of solid angle line element)

If you look at the derivation for the Schw. Sol. given in books on the subject you will see that they start off with [latex](t, r, \theta, \phi)[/latex] and they really mean it as the coordinate system of 3D spherical coordinates with time added (for whatever kind of criterion you want for r as long as it meets the normal monotonically increasing condition, but I get the impression it is as one would normally imagine r with equal rulings and so on). Then they say they want spherical symmetry and so forth for the solution that is sought, which specializes the metric to

[latex]ds^2 = A(r) dt^2 - B(r) dr^2 - C(r) d\Omega^2[/latex].

At this point r still operationally means the same as the r from 3D spherical coordinate systems. Then they go, oh well, just make

[latex]C(r) = \rho^2[/latex],

but we don't like writing [latex]\rho[/latex], so lets just write [latex]\rho[/latex] as r instead, it doesn't matter, we are GR people and don't care about coordinate systems after all. That is a logical error. It implicitly says that

[latex]r = \rho[/latex], which is not the case.

How do I know? First off, note that no coordinate changes were made to t or the angles, so the argument that I should also be concerned about those coordinates when it comes to their operational meaning is invalid, as far as I can tell.

Let me just write down the Schw. Sol. without the implicit error (you shouldn't have a problem with me doing this, I am just keeping the original lable for the coordinate transform after all, and it is "just" a lable!).

[latex]
ds^2 = (1-2M/\rho) dt^2 - (1-2M/\rho)^{-1} d\rho^2 - r^2 d\Omega^2
[/latex]

Now, how does [latex]\rho[/latex] behave as a function of r? First off, there are two problems, we know that in the above metric that if t and the angles are set to constant and [latex]\rho[/latex] is integrated from 0 to whatever then the integral will diverge from below. Also, is it even the case that when r = 0 that [latex]\rho = 0[/latex] as well? That would need to be shown, or assumed, or something for goodness sakes.

For now, the best one can do is just integrate from some constant
[latex]\rho[/latex] to say [latex]\rho = \rho[/latex] and see what happens.
Please don't make me do the integral, but I can assure you that the result is something like

[latex]r = f(\rho) - C[/latex]

where f is not the identity function and C a constant of integration. This is important because

[latex]
\int dr = \int^{\rho = \rho}_{const} d\rho (1-2M/\rho)^{-1/2}
[/latex]

is the most sensible way to define the relation between r and
[latex]\rho[/latex], given the geometry and everything else. But then, I get that by understanding good old fashioned Analytic Geometry, so maybe you will have to think about that one for a bit because your head is so high in the clouds of Differential Geometry. Differential Geometry is just another tool, don't forget about your other math tools.

Actually, isotropic coordinates are good examples: if you object to using "radius" in "Schwarzschild radius", for consistency's sake, you should object to things like "angle" for the Schwarzschild φ and θ, because they do not faithfully represent angular measurements (but isotropic ones do). And what about "Schwarzschild time", since there are so many ways of measuring time?

I have heard of isotropic coordinates for Schw. Sol. before. Not sure what to say beyond general doubts. Would have to look into specifics before being able to comment.

That's not thanks to Crothers; that's thanks to Schwarzschild. Some books also define the surface of constant r (and t=const, again) as being a Euclidean sphere of area 4πr², which is so completely and obviously equivalent that Crothers didn't need to discover it.

The r as conventionally used is really [latex]\rho[/latex] and operationally means the same as the GRC, or reduced circumference, or the sphere example you gave, or the aerial whatsits, or who knows what else. The difference is that by knowing that the r that is conventionally used (my rho) is the GRC, one gains a much greater appreciation for what the geometry actually is (and isn't that the point of doing GR?). Unique shells just does not tell you as much, even if they obey some sane set of rules.

Oh yeah, any metric with

[latex]ds^2 = ... + r^2 d\Omega^2[/latex]

will have a GRC of r (assuming we want the GRC of that spatial slice and r is not a function of the angles), and was certainly well known before Crothers. The point was that he noticed this about the Schw. Sol. and was the first I could tell of to be annoyingly vocal about it, unless you can show me someone else who did the same.

They are just labels. If you want them to be more, you'll have to define your own concept.

Ughghg, by now I hope you see that coordinates in coordinate systems are not just labels. Think of them as lables when you want to state things in a coordinate-free way, such as laws of nature, generalized formulas to find length, area etc. But if coordinates are just labels, go tell an experimentalist to find what p is -- oh, and it is just a label by the way. Is that momentum, pressure, context please!

Sorry if that sounds smug, but I told you people in the field forget that coordinates in a coordinate system have an operational meaning when a specific solution is rendered, and you went and proved as much by your reply.

Or is it some kind of common fallacy, or myth. Not everything in physics is coordinate-free! Sometimes you need to know how a coordinate is measured to understand a problem when a specific solution is found!

Plus, I would rather be getting into the whole blackhole thing (this is much more contentious and my thoughts on the matter are still evolving, plus, you have made some remarks I find interesting). This should have been an easy one for you to agree with, given some thought. So before you reply, please think more carefully about what I am saying. I am trying to make a point and so far you have not gotten the point. It is a basic point really, in the grand scheme of things.

OK, best of luck.
 
The first thing you need to do is check T carefully at the cross-over radius. Since the Einstein tensor involves two derivatives of the metric g and the functions in your g may have discontinuities in their first derivatives, you may find a singular spherical shell of some mass and pressure. If so, you need to modify your claim.

Regardless of the result, like Vorpal I do not understand why you think such solutions have any bearing on the existence of black holes or the arguments for them. Stars exist. Stars are spherically symmetric (approximately), and their exterior metric is Schwarzschild (approximately). Most importantly, they must be larger than their Schwarzschild radius, and if they ever collapse to a size below it, they become black holes more or less by definition (because all their matter is inside the horizon).

So to prove that black holes do not exist, you need to prove that stars can never collapse to a size smaller than their own Swchawzschild radius. How does writing down some solution with bizarre stress-energy that doesn't describe a star have any bearing on that question?

Interesting that you bring up the derivatives of the metric matching. The solution I found has the metric matching on both sides but not the derivatives (or higher derivatives I would suppose).

In some later post I will get into more thoughts on blackholes specifically, but on the derivative of the metric topic, I do not see a good reason why the Einstein Tensor has to be the same on either side of a boundary. In EM theory for instance discontinuities come up quite often when considering various problems. I did calculate the Kretschmann scalar on either side of the boundary of the solution I found and it was also discontinuous.

That is all for now.
 
Interesting that you bring up the derivatives of the metric matching. The solution I found has the metric matching on both sides but not the derivatives (or higher derivatives I would suppose).

Then you are wrong that it is a solution to Einstein's equations with the stress tensor you gave. Be more careful, and you will discover than there are is an infinitely thin spherical shell of infinite density and/or pressure at the radius where you match the two metrics.

You might find some old papers by Werner Israel helpful in untangling that. I suspect you will find that the density and pressure at the shell is impossible physically - that it violates all the energy conditions. If you're curious, the reason I say that is that it looks to me that light rays will defocus as they cross the shell, which is impossible if the energy conditions hold.

In some later post I will get into more thoughts on blackholes specifically, but on the derivative of the metric topic, I do not see a good reason why the Einstein Tensor has to be the same on either side of a boundary. In EM theory for instance discontinuities come up quite often when considering various problems. I did calculate the Kretschmann scalar on either side of the boundary of the solution I found and it was also discontinuous.

You're misunderstanding the issue. There is nothing wrong with components of the Einstein tensor being discontinuous. But what will happen at your shell is different - the Einstein tensor is infinite there. Even that is not necessarily a problem (there's nothing really wrong with delta functions in the Einstein tensor either), except that in this case I suspect the energy conditions are violated, meaning that no such shell can exist even as an approximation.
 
Another point: the Kretschmann scalar will almost certainly be divergent at the shell (again - this is different than the discontinuity in the finite part, which is obviously there too).

But if the Kretschmann scalar (or any other curvature invariant) is divergent, the "solution" should be regarded with extreme suspicion. General relativity is a classical approximation - it cannot be the true theory of gravity. Quantum corrections introduce new terms into Einstein's equations, and the Kretschmann scalar is the simplest and most important such term. If it diverges somewhere, it means the "solution" is singular there and has no clear physical meaning.
 

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