# New Home for Crother's thread

1. Jun 24, 2006

### pervect

Staff Emeritus
The Crother's thread obviously needs a new home, as the OP of this thread doesn't want it hijacked to discuss Crother's issues.

I'm not sure how much longer the discussion is gonig to go on, but as a courtesy to the OP, I'll just move further discussion here, to avoid hijacking his thread.

I really don't see the point in rehashing Schwarzschild's derivation. One can directly calculate whether or not a metric satisfies Einstein's field equations. If there's a problem with the derivation, it should show up as a problem with the solution.

The Schwarzschild metric satisfies the EFE's but it contains some coordinate singularites.

The significance of these singularities is not particularly clear, until they are removed by use of a different coordinate system - the Kruskal extension.

No convincing "flaw" of the Kruskal extension has been demonstrated, only promises to "reveal it later".

One can confirm via direct calculation that the metric resulting from the Kruskal extension satisfies Einstein's field equations - so that isn't the flaw.

An important issue that has perhaps not been fully addressed is the completness of the solution.

Imagine a 4-d space-time, with a standard x,y,z,t cartesian coordinate system. But imagine that one refuses to label any part of the space time with x<=0 with a coordiante, i.e. one insists that x > 0.

This will be an incomplete description of the geometry of the space-time, ecause portions of the geometry exist that do not have coordinates.

The symptoms of this incompleteness will be geodesics that suddenly "stop" for no particularly good reason.

The Kruskal extension not only solve's EFE's, but it is as complete as possible. This is an important property that the non-extended solutions do not have.

The Abrams paper, in particular, seems to advocate an incomplete solution of this kind, where geodesics suddenly "stop" at the event horizon. The geodesics suddenly "stop" because the allowable domain of a particular variable, r, has been arbitrarily limited.

It appears that these incomplete solutions are being touted as "new and improved" solutions (even as "different" solutions) - when they are really just truncated versions of the Kruskal solution.

2. Jun 25, 2006

### Crothers

Schwarzschild is regular on 0 < r < oo. To force a black hole from Schwarzschild's solution requires his r to go down to -alpha, contrary to the very definition of r leading to Schwarzschild's eq. 6, namely 0 < r < oo. Alternatively, to force a black hole from Schwarzschild's solution would require his auxiliary quantity R to go down to 0, in contradiction to the definition of R on 0 < r < oo leading to his eq. 14 and the very statement of the problem. Hilbert's line-element is nothing but Schwarzschild's auxiliary line-element where Schwarzschild's R is merely called r and the range 0 < r < oo applied. This is inconsistent with Schwarzschild's solution, which is already regular everywhere on 0 < r < oo. Thus, there are two metrics, Schwarzschild's and Hilbert's, yet both allegedly defined on 0 < r < oo. This is not possible. One of them has to be incorrect. It is Hilbert who is incorrect. The black hole is based upon Hilbert's solution. The black hole cannot be obtained from Schwarzschild's solution. Hilbert's metric is a corruption of Schwarzschild's solution and is invalid. Hilbert's is the metric used by the black holers. Why is Hilbert's solution wrong? Because it violates the intrinsic geometry of the form of the line-element. Schwarzshild solution is consistent with the intrinsic geometry of the form of the line-element. This is a matter of pure mathematics. A geometry is completely determined by the form of the line-element. Quantities in the space described by the line-element must be obtained from the intrinsic geometrical structure of the line-element and the consequent geometrical relations between the components of the metric tensor.

I have asked several times for a mathematically rigorous justification for the arbitrary alteration of Schwarzschild's true solution. The requests have been ignored.

I have asked several times for a mathematically rigorous demonstration that General Relativity necessarily requires that a singularity must occur only where the Riemann tensor scalar curvature invariant (Kretschmann scalar) is unbounded. No one has ever provided the required proof (and that does include Hawking, Wald, Penrose, Thorne, etc), but without it the black hole has no theoretical possibility whatsoever. I have already stated that it is easily proved that curvature-type singularities do not occur in Einstein's gravitational field, and I have offered the proof subsequent to attempts to proof the so far unproved orthodox assumptions.

As for my promise to "reveal later" I have already explained that I will not deal with the alleged K-S coordinates until these preliminary issues are first addressed by you, or others. The K-S extension relies upon the unproved validity of the arbitrary manipulation of Schwarzschild's solution and the so far unproved assumption about curvature invariants. I have already cited the error in Kruskal's paper, quoting him in his unproved assumption concerning curvature invariants. If his assumption is false his whole programme is false. His argument rests upon the validity of his assumption. Therefore a rigorous proof is required of Kruskal's assumption, as is a rigorous poof that the manipulation of Schwarzschild's variable is mathematically valid.

I do not see how my requests have been in any way unreasonable. The issues are of fundamental importance. Please attempt the proofs, or provide the proofs given others, or explain why you will not do so.

We cannot go forward in this discussion if you insist upon disregarding the issues I have raised. That amounts to justification by just saying that I am wrong because that is convenient. That is not scientific method.

I remark that I have so far put aside your previous insults on this very site, for the sake of rational argument in science. You will recall that you have previously called me a crank. I took that on the chin in the hope that reason will be exercised. I ask again for reason.

3. Jun 25, 2006

### Wallace

This is I believe where you have gone wrong. The line element does not have naively interpretable geometric properties. It is not a duck that quacks that way. The same space-time can be described with an infinate number of different line elements that all apparently imply different geometries according to your way of thinking. However they all give the same physical predictions, as an orthonormal basis unique for each line element must be used to extract frame dependant observables.

Take for example standard Minkowski space. It has no curvature. However the exact same space-time can be described by a deSitter metric, the metric of an expanding empty universe. In this case there is spatial curvature, but this is merely a matter of co-ordinates and all physical observables are indentical. By your reasoning they imply different metric spaces but this is not the case. They are both valid descriptions of the same space-time. I note that there is no mathematically rigourous justification for your heterodox views on the meaning of the line element in any of your material.

4. Jun 25, 2006

### pervect

Staff Emeritus
...
I've been a bit too busy to write a detailed response. The response I write is probably in any event not going to approach the problem according to your suggestions, but is going to focus on why the Kruskal solution is a complete solution to EFE's without any singularities.

What I intend to show is how the proposed Abarams solution can be shown to be a diffeomorphism of the subset of the manifold described by the Kruskal solution, and how the Schwarzschild solution is also a diffeomorphism of a subset of this Kruskal manifold.

This will not in and of itself demonstrate that the Kruskal solution is complete, but will demonstrate that it is

satisfies the Einstein field equations
free of singularities
more complete than proposed alternatives (which are a subset of the Kruskal solution, basically).

It will also establish that outside the event horizon, the Kruskal solution is just a diffeomorphism of the Schwarzschild solution - which means that they are essentially the same solution, only the coordinate labels have changed.

On the issue of personal insults, the statement that I believe got the thread closed last time was:

I felt very insulted by this remark, because, gosh darn it, I didn't get any money out of defending the Kruskal solution. Your remark seems to suggest that there is money to be made, and that I'm just too stupid to get my fair share of the dough that is being thrown at people for doing so.

I'm sure that if the discussion gets too heated, again, that the moderators will lock it down, again.

5. Jun 26, 2006

### pervect

Staff Emeritus
Phase 0: how diffeomorphisms can make a line element of a well-behaved geometry appear to be singular: a simple example.

Conisder the 2-d line element:

ds^2 = -dt^2/t^4 + dx^2

How would we describe the geometry associated with this metric? It appears that there is some sort of singularty at t=0, as the metric coefficient of dt^2 becomes infinite there.

But let us make the transformation

T = 1/t.

then dT = -dt/t^2

So we can re-write our metric as:

ds^2 = -dT^2 + dx^2.

We can see that this metric is well-behaved everywhere. Basically,we have at heart a well behaved Minkowski metric - and the apparent ill-behavior of the metric at t=0 is simply an artifact of applying the ill-behaved diffeomorphism T = 1/t to a well-behaved metric.

A diffeomorphism is just a "relabelling" of the geometry with a different set of coordinates. It does not affect the underlying geometry, just how the points are labelled.

In the above example, we have a basically well-behaved geometry, with an ill-behaved diffeomorphism (a labelling of coordinates of the geometry) which makes the geometry appear to be singular when it is not.

Exactly the same thing happens with the Schwarzschild metric. The Schwarzschild metric appears to be singular, but it is diffeomorphic to a well-behaved underlying geometry. The ill-behavior is in the diffeomorphism, not in the actual geometry.

Phase 1: The Kruskal metric - the gory details.

Piggybaking a bit on a previous post
https://www.physicsforums.com/showpost.php?p=1019121&postcount=67

we consider the line element [1]

$$ds^2 = \frac{32 M^3 e^{r/2M}}{r}(-dT^2 + dX^2) + r^2 d \Omega^2$$

where r = r(X,T) = 2M*LambertW((X^2-T^2)/e) + 2M

The LambertW(x) is defined for the domain x>=-1/e, and is a monotonically increasing function that satisfies

LambertW(x)*exp(LambertW(x)) = x

(Note that LambertW(-1/e) = -1)

also
d (LambertW(x))/dx = LambertW(x) /[x (1+LambertW(x))]

This defintion is equivalent to the perhaps conceptually simpler implicit definition of r(X,T) given by the relationship

(r/2M - 1)* exp(r/2M) = X^2 - T^2

The first defintion is convenient for computer algebra, though, which is a practical necessity for what follows. It *could* in principle be done by hand, by someone sufficiently dedicated, I suppose, but would be extremely time consuming.

We can confirm by direct (computer-aided) calculation that the Ricci tensor for the for line element[1] is zero.

Thus the Kruskal metric represents a vacuum solution for Einstein's field equations.

The range of coordinates for which we have a non-singular metric is just X^2 - T^2 > -1 (equality is not allowed). This is equivalent to r > 0.

Now consider the diffeomorphism

(X,T) -> (r,t)

where r = r(X,T) = 2M*LambertW((X^2-T^2)/e) + 2M
and t = 2M ln[(X+T)/(X-T)]

Applying this diffeomorphism to the well-behaved Kruskal metric above yields the apparently singular Schwarzschild metric, i.e line element [2]:

ds^2 = dr^2 / (1-2*M/r) + dt^2 (-1 + 2M/r) + $r^2 d\Omega^2$

This can be demonstrated by substituting

dr = (dr/dX)*dX + (dr/dT)*dT
dt = (dt/dX)*dX + (dt/dT)*dT

into the line element of the Schwarzschild metric[2], and finding that it reduces to

$$ds^2 = 16\,{\frac { \left( {{\it dX}}^{2}-{{\it dT}}^{2} \right) {M}^{2}{\it LambertW} \left( \left( {X}^{2}-{T}^{2} \right) {e^{-1}} \right) }{ \left( {\it LambertW} \left( \left( {X}^{2}-{T}^{2} \right) {e^{-1}} \right) +1 \right) \left( {X}^{2}-{T}^{2} \right) }}$$

which can be further simplified to the line element [1] by using the expression for X^2-T^2 previously given.

This diffeomorphism is non-singular only for (X^2 - T^2) > 0, i.e. X>T, which is equivalent to r > 2M

We can thus see that the Schwarzschild metric [2] has a non-singular diffeomorphism to the Kruskal metric [1] in the region (X^2 - T^2) > 0, which is equivalent to the region (r > 2M), a region which is a subset of the entire manifold spanned by the Kruskal metric.

It is the ill-behavior of t(X,T) that leads to apparent ill-behavior in the Schwarzschild metric at r=2M.

Thus, as promised, we have found a diffeomorphism from a subset of the manifold covered by Kruskal coordinates to the Schwarzschild metric.

We can map the interior part of the Schwarzschild solution (0<r<2M) into the Kruskal metric as well, by the same set of diffeomorphisms, as long as we omit the event horizon (r=2M) itself.

We can furthermore see that the ill-behavior of the Schwarschild metric was induced by the ill-behavior of the diffeomorphism that defined the 't' coordinate. The diffeomorphism that defines the 'r' coordinate remains well-behaved, which is why coordinates such as Eddington-Finklestein which use the r coordinate and an additional coordinate (usually T+X/2 or T-X/2) are possible.

It should be reasonably obvious that as Abrams proposal maps into a subset of the Schwarzschild metric by a diffeomrophism r->r', it also maps into a subset of the Kruskal metric.

Thus the Kruskal metric represents the "complete" geometry. Other solutions all map, via a diffeomorphism, into some subset of the manifold covered by the Kruskal coordinates.

Because the Schwarzschild exterior metric has the desired properties of:
2) asymptotic flatness
3) time-stationary

and because the Kruskal metric is diffeomorphic to the Schwarzschild exterior metric, it also has the desired properties in the exterior region.

The Kruskal metric is not "time-stationary" in the interior region. This is the nature of the geometry itself - nothing can "stand still" beyond the event horizon of a black hole.

Last edited: Jun 26, 2006
6. Jun 26, 2006

### Crothers

to Wallace

You have clearly not understood my analysis. I have never claimed that the geometry varies with different admissible analytic functions of r in place of r. Please point to any statement in my work which makes this alleged claim, either explicitly or implicitly, to substantiate your allegation. One can easily generate an infinite number of equivalent metrics. They all describe the same geometry. That is little more than a platitude. However, the infinite number of equivalent metrics are not arbitrary. The intrinsic geometry of the line-element cannot be violated (in which case there is no equivalence). One must adduce a means whereby an infinite number of equivalent metrics can be constructed.

Contrary to your claim, the intrinsic geometry of the line-element is not something naive. It is a matter or pure mathematics. The entire nature of a geometry is known to be determined by the form of its line element, and so the choice of Minkowski space has completely fixed the character of the geometry to be used in Einstein's relativity, both special and general.

I ask yet again that some one or more of you prove the so far unproved basic assumptions that lead to the black hole. Prove that the arbitrary manipulation of Schwarzschild's true solution (which is regular on 0 < r < oo), a manipulation that gives rise to the Hilbert solution, alleged also to be defined on 0 < r < oo, is mathematically valid. (However, I remark that it is a rather trivial procedure that demonstrates that Hilbert's solution is not equivalent to Schwarzschild's solution. Consequently, they do not describe the same manifold.) And secondly, prove that General Relativity necessarily requires that a singularity must only occur where the Riemann tensor scalar curvature invariant is unbounded. The Kruskal-Szekeres extension is founded upon this unproved assumption. I have cited Kruskal himself in my previous postings, wherein he states the assumption he made (although he did not realise his statment was that of his assumption). Without the required proof the Kruskal-Szekeres coordinates are ill-defined. With a proof of the invalidity of the assumption the whole Kruskal-Szekeres objective is eliminated, and with it the black hole.

The two requested proofs have never been given by the relativists. Indeed, it appears that the relativists do not even perceive the problem to begin with, rather astonishingly. Their failure to perceive does not mean that the required proofs can simply be ignored.

7. Jun 26, 2006

### Crothers

to pervect

I repeat here some of my comments in reply to Wallace, as they are relevant.

I ask yet again that some one or more of you prove the so far unproved basic assumptions that lead to the black hole. Prove that the arbitrary manipulation of Schwarzschild's true solution (which is regular on 0 < r < oo) that gives rise to the Hilbert metric, alleged also to be regular on 0 < r < oo, is mathematically valid. (However, I remark that it is a rather trivial procedure that demonstrates that Schwarzschild's solution is not equivalent to Hilbert's solution. Consequently, they do not describe the same manifold.) And also prove that General Relativity necessarily requires that a singularity must only occur where the Riemann tensor scalar curvature invariant is unbounded. The Kruskal-Szekeres extension is founded upon this unproved assumption. I have cited Kruskal himself in my previous postings, wherein he states the assumption he made (although he did not realise his statment was that of his assumption). Without the required proof the Kruskal-Szekeres coordinates are ill-defined. With a proof of the invalidity of the assumption the whole Kruskal-Szekeres objective is eliminated, and with it the black hole.

Thus, there is no point in regurgitating the derivation of the Kruskal-Szekeres line-element, as that proves nothing. Giving the derivation of the Kruskal-Szekeres line-element to validate the Kruskal-Szekeres line-element is circular. Anyone can read of it in any of the textbooks. I have done that too, many times. So why repeat it here?

The two requested proofs have never been given by the relativists. Indeed, it appears that the relativists do not even perceive the problem to begin with, rather astonishingly. Their failure to perceive does not mean that the required proofs can simply be ignored. That will stifle science.

Why do you resist an attempt at the required proofs, and instead adduce arguments that do not deal with the fundamental factors which completely determine the validity or invalidity of the complications you introduce in the conventional manner? What we have here under discussion will not be found in the textbooks. It requires independent clear thinking. That's what research is all about. Also, argument on authority is outside scientific method. You are intelligent, so there is no good reason why you should not investigate the proofs I have called for, and instead rely upon the say so of say Hawking, Penrose, Kruskal, etc. And the fact that you have studied physics seriously, indicates that you are not afraid of hard work. Do you think I ask for scientific discussion to annoy, for want of better things to do with my time? I too have put in considerable hard work on these matters, and in doing so I have uncovered serious anomalies in the standard analysis. All I hope to achieve is to bring these quite serious anomalies to the attention of others interested in the problems this theory deals with. Unfortunately, it is clear to me that these anomalies, when squarely addressed, will have profound consequences that are not all that welcome, so I have sadly discovered. But that is the way of science.

As an aside, it is interesting to note that the Hilbert solution (and therefore along with it the Kruskal-Szekeres extension) gives rise to the absurdity of an infinite acceleration of a test particle where, according to the orthodox claims, there is no matter, as you can readily verify by simply calculating the accelerations. Don't take my word for it, try it and see for yourself.

8. Jun 27, 2006

### pervect

Staff Emeritus
I think it's pretty clear that the argument, if nothing else, is going in circles.

I'm not going to attempt to justify Schwarzschild's solution because the solution actually does have coordinate singularities. This is something that everyone can agree on.

Given that there are coordinate singularities in Schwarzschild's solution, the issue becomes - how do we handle them? Here is where the disagreement arises.

I believe that Croether's approach to handling them (the coordinate singularities) is wrong, and the textbook approach is correct. I haven't seen anything that changes my mind in this rather long discussion.

9. Jun 28, 2006

### Crothers

to pervect

I am disappointed that none of you will address the straightforward proofs I have called for. Referring to the textbook arguments is futile. Why will you not attempt the proofs? It only requires schoolboy level matehmatics to demonstrate that Hilbert's solution (the black hole solution) is inconsistent with Schwarzschiuld's tru solution. They cannot both be right. Schwarzschild is the one who is right, not Hilbert and the black holers. One only need study Schwarzschild's paper, a link to which i have previously provided. Here it is again

www.geocities.com/theometria/schwarzschild.pdf

Failing to even address the issue of the relativist arbitrary manipulation of the Schwarzschild true solution and the Kruskal assumption (actually demonstrably invalid) that a singularity can only occur in Einstein's gravitational field where the Riemann tensor scalar curvature invariant is unbounded is a rejection of scientific method. It is clear that you are not interested in scientific truth.