More on the controversy about the Schwarzschild radius and black holes.
LEJ Brouwer
gauntlet in defence of Crothers' papers which appear to demonstrate
clearly the non-existence of an interior Schwarzschild solution (see
the earlier thread "On the controversy about the Schwarzschild radius
and black holes"). Of course he came against some strong opposition
from the noble defenders of the mainstream view - Steve Carlip, Daryl
McCullough, Jan Bielawski and Tom Roberts all arguing for the validity
of the interior solution, making reference to the usual list of non-
Schwarzschild coordinates which provide a singularity-free mapping
over the maximal extension containing both the interior and exterior
solutions.
I can understand Crother's reasoning as I have been through the
mathematics in detail, and I hope I can understand the reasoning of
Steve Carlip et alii, as I have also been through the mathematics of
various solutions in detail. From my perspective then, any errors in
Crother's argument (or in the standard view) are not in the
mathematics, but in the physical understanding and interpretation of
the problem being solved and in particular of the coordinate systems
being used.
I have been trying to settle in my mind where the problem lies, and I
have come to the conclusion that the difference ultimately lies in the
physical significance (or lack thereof) which is attached to the
coordinates.
In Carlip et al's view (and in the standard interpretation of general
relativity), the coordinates are just labels, and coordinate
transformations are of no physical relevance. In Crother et al's view,
coordinates can be given a physical significance a priori - e.g. if
the direction of t is defined to be timelike, and the direction of r
is defined to be spacelike when a problem is stated, then it must
retain these properties in the solution. (Note that the latter
position is not in conflict with the former).
For the problem at hand, in Carlip's view, the fact that the spacelike
and timelike coordinates exchange 'labels' on passing from the
exterior to the interior Scwarzschild solution is of no consequence.
In Crother's view, the change of t (resp. r) from timelike (resp.
spacelike) to spacelike (resp. timelike) across the horizon means that
the interior solution does not satisfy the prior physical constraints
on t and r in the original statement of the problem, and therefore the
interior solution is unphysical.
When trying to find the metric corresponding to a centrally symmetric
gravitational field, the general solution can be written in the form
(see e.g. Landau & Lifschitz, Vol 2, p299):
(1) ds^2 = h(r,t) dr^2 + k(r,t) (sin^2 theta d phi^2 + d theta^2) +
l(r,t) dt^2 + a(r,t) dr dt
where r labels 'spheres' and t labels 'time' in some way. Without loss
of generality, coordinate transformations can be applied to bring this
to the form:
(2) ds^2 = exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta d
phi^2) - exp(lambda(r,t)) dr^2
where the new r and new t still label spheres and time respectively.
The point to note at this point is that r and t are _not_ merely
labels - they have a definite physical attribute determined by their
respective spacelike and timelike natures. The solution to equation
(2) above is of course the usual Schwarzschild solution.
What Crothers' argument shows is that the interior Schwarzschild
solution is _not_ compatible with the physical meaning attached to r
and t, even though it naively (i.e. mathematically) provides a
solution to the equation (2) above.
On the other hand, according to Carlip et alii, r and t are just
labels for coordinates, and general relativity holds irrespective of
the labels assigned to coordinates or the coordinate system used, so
the physical meanings attached to them are irrelevant, and so the
interior solution _is_ valid.
However, while it is true that general relativity does not care about
the coordinate system used, in the formulation of the problem above, a
physical meaning has indeed been attached to the coordinates r and t a
priori, and given that this is the case, only the exterior
Schwarzschild solution appears in my view to be a correct solution to
the problem.
But as shown by Carlip et al, the Kruskal-Szekeres and Eddington-
Finkelstein metrics (and numerous others) also satisfy the vacuum
field equations, and provide consistent, smooth and singularity free
coordinate systems covering the maximal extension which also includes
the Schwarzschild interior solution. They argue that the exterior
Schwarzschild solution is incomplete on its own and this
incompleteness is due to a quirk in the original choice of coordinates
r and t and in particular the physical constraints placed upon them,
i.e. constraining them a priori to be spacelike and timelike
respectively.
So who is right?
Well, it appears to me that Crothers et al are correct and Carlip et
al are wrong.
Why? Because the original problem, which seeks metrics corresponding
to a spherically symmetric gravitational field, explicitly defines the
coordinates r and t to be spacelike and timelike respectively.
If that is the case, then where are Carlip et al going wrong?
Well, while it is true that Carlip et al's metric solve the vacuum
field equations, and while it is true that their exterior solution can
be analytically continued into the interior, this interior solution
must, at best, be the solution to a different physical problem (i.e
which is not spherically symmetric, but as it happens can be sewn
smoothly to the spherically symmetric exterior solution), or, at
worst, it is indeed spherically symmetric in a mathematical sense
(i.e. has an SO(3) symmetry) but does not correspond to any physical
solution.
The conclusion to be reached then is that analytic continuation of a
solution is only valid if it respects the physical constraints of the
original problem. If the original constraints are ignored, well, a
different problem is being solved, with the possibly that the
resulting solutions are unphysical. [If Carlip et al disagree with
this conclusion, then I would very much like to see a proof that the
interior solution does indeed solve the original problem and
corresponds to a physically realisable vacuum].
- Sabbir.
Tom Roberts
LEJ Brouwer wrote:
> In Carlip et al's view (and in the standard interpretation of general
> relativity), the coordinates are just labels, and coordinate
> transformations are of no physical relevance.
This is just basic mathematics and physics, not just "Carlip et al's view".
> In Crother et al's view,
> coordinates can be given a physical significance a priori
It QUITE CLEARLY can not be "a priori" -- NOTHING in physics is "a priori".
> For the problem at hand, in Carlip's view, the fact that the spacelike
> and timelike coordinates exchange 'labels' on passing from the
> exterior to the interior Scwarzschild solution is of no consequence.
You are confused -- it is not the _coordinates_ that "exchange", it is
merely the way people assign labels to them that "exchange". The names
we humans assign to concepts cannot possibly affect either the math or
the physics.
Make no mistake: the conventional labeling of Schw. coordinates in the
region r>2M is {r,theta,phi,t} and the conventional labeling in the
region r<2M is {r,theta,phi,t}. But there is NO RELATIONSHIP WHATSOEVER
between these coordinates in these two regions, except for the labels
assigned to them. How one labels something cannot possibly affect how it
behaves, or what it is (good thing, or the multiple human languages
would be a disaster...).
> In Crother's view, the change of t (resp. r) from timelike (resp.
> spacelike) to spacelike (resp. timelike) across the horizon
THIS DOES NOT HAPPEN. t is INVALID on the horizon, which invalidates the
entire coordinate system, and there is no "across". The two regions in
which these coordinates are valid are DISJOINT and coordinates on such
disjoint regions are UNRELATED (even though one might use the same set
of labels).
> means that
> the interior solution does not satisfy the prior physical constraints
> on t and r in the original statement of the problem, and therefore the
> interior solution is unphysical.
This is nonsense. One cannot possibly "constrain" t to be timelike and r
to be spacelike -- the mathematics will DETERMINE that.
As best I can tell from your writing, the error is in ASSUMING that
there is a manifold in GR that is: a) static, b) spherically symmetric),
and c) surrounds a point mass with the remainder of the manifold in
vacuum. It is now well known that in GR there simply is no such manifold.
If, however, one relaxes (a) and (c) to be: d) static in a region
outside some radius and extending to spatial infinity, then the
Schwarzschild manifold meets the requirements (b) and (d), with "some
radius" = 2M.
The problem is that you (and presumably Crothers) are looking for
something that does not exist, and keep INSISTING that your desires be
met. You cannot impose your personal wishes and desires onto either
mathematics or physics. <shrug>
> When trying to find the metric corresponding to a centrally symmetric
> gravitational field, the general solution can be written in the form
> (see e.g. Landau & Lifschitz, Vol 2, p299):
> (1) ds^2 = h(r,t) dr^2 + k(r,t) (sin^2 theta d phi^2 + d theta^2) +
> l(r,t) dt^2 + a(r,t) dr dt
> where r labels 'spheres' and t labels 'time' in some way. Without loss
> of generality, coordinate transformations can be applied to bring this
> to the form:
> (2) ds^2 = exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta d
> phi^2) - exp(lambda(r,t)) dr^2
Except you forgot to mention that in this transformation there was a
sqrt(), and you implicitly chose the sign for its result.
Realizing this mistake, which appears to have been instrumental
generating your entire confusion, the right way to write this is:
ds^2 = K exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 +
sin^2 theta dphi^2) - K exp(lambda(r,t)) dr^2
where K is a constant, either +1 or -1.
> The point to note at this point is that r and t are _not_ merely
> labels - they have a definite physical attribute determined by their
> respective spacelike and timelike natures.
No, they don't -- you CONFUSED yourself by forgetting that sqrt() has
sign ambiguity. And you confuse yourself even further by thinking that
you can impose your personal wishes and desires onto mathematics and
physics. <shrug>
For K=+1, t is timelike, r is spacelike and the solution is valid in the
region r>2M; for K=-1, t is spacelike, r is timelike, and the solution
is valid in the region r<2M. There is a solution for either value of K,
but they are valid in different intervals of the r coordinate. You
CANNOT impose on them your "a priori" "physical attributes", you MUST
follow what the mathematics yields.
Remember that for EVERY solution of such differential equations you MUST
keep track of the region of validity of the solution. You forgot to do
that -- had you done so, you would then be presented with a conundrum:
why is there a HOLE in the middle of the manifold? (remember you kept
only the K=+1 portion of the original Ansatz, and that has no solution
in the region r<=2M.)
Note, please, that this equation has no solution that is valid at r=2M.
But that is a region of measure zero, and is not really important --
selecting better coordinates solves this problem.
> However, while it is true that general relativity does not care about
> the coordinate system used, in the formulation of the problem above, a
> physical meaning has indeed been attached to the coordinates r and t a
> priori,
Repeating this falsehood does not make it correct. In the original
Ansatz, one solved for the unknown functions by applying the Ansatz to
the Einstein field equation. Your "a priori" "physical attributes" are
merely what you WISH THEM TO BE, but the mathematics says different. <shrug>
> Well, it appears to me that Crothers et al are correct and Carlip et
> al are wrong.
You are wrong. You (and presumably Crothers) based your entire argument
on an insufficiently detailed mathematical analysis -- sqrt() _DOES_
have a sign ambiguity which you ignored, disjoint regions of a manifold
cannot be compared as you attempt to do, and you MUST keep track of the
region of validity of any solution. <shrug>
> Why? Because the original problem, which seeks metrics corresponding
> to a spherically symmetric gravitational field, explicitly defines the
> coordinates r and t to be spacelike and timelike respectively.
Nonsense. You attempt to impose your personal wishes and desires on
mathematics -- it does not permit that. <shrug>
> Well, while it is true that Carlip et al's metric solve the vacuum
> field equations, and while it is true that their exterior solution can
> be analytically continued into the interior, this interior solution
> must, [...]
You are confused and there is no such "must". You have mesmerized
yourself into thinking that by just seeking a spherically symmetric and
static solution that there somehow "must" be one. There isn't. All there
is is the Schwarzschild solution, which is only static in the region
r>2M. Your wishes and dreams do not affect this.
Here is the KEY QUESTION you must face: If your analysis is correct,
what happens in the region r<=2M? -- where did that "hole" come from?
Tom Roberts
Ben Rudiak-Gould
> The point to note at this point is that r and t are _not_ merely
> labels - they have a definite physical attribute determined by their
> respective spacelike and timelike natures.
All that's happening here is a standard formal technique for finding
solutions to a system of differential equations: guess the solution in a
parameterized form, and solve for the parameters. One can certainly restrict
the parameters enough that one obtains only the r>2M portion of the
Schwarzschild metric, but that doesn't change the fact that the 0<r<2M
portion is a valid solution of the same differential equations. You also
won't obtain the exterior Kerr metric, but that doesn't change the fact that
it's a solution. They all satisfy the equations; that's what it means to be
a solution. It doesn't matter how they're derived, and it doesn't matter
what letters of the alphabet are used for the coordinates.
Would you be happier with the two separate statements that
ds^2 = (...) dt^2 - (...) dr^2 - r^2 (d theta^2 + sin^2 theta d phi^2)
(r > 2M, 0 < theta < pi, 0 <= phi < 2pi)
is a solution of the vacuum field equations and
ds^2 = (...) dT^2 - (...) dZ^2 - T^2 (d theta^2 + sin^2 theta d phi^2)
(0 < T < 2M, 0 < theta < pi, 0 <= phi < 2pi)
is also a solution of the vacuum field equations? I assume everyone else
would be equally happy with that version. It's just a longer way of saying
the same thing.
> What Crothers' argument shows is that the interior Schwarzschild
> solution is _not_ compatible with the physical meaning attached to r
> and t, even though it naively (i.e. mathematically) provides a
> solution to the equation (2) above.
The whole idea of background independence is that you don't know the
physical meaning of the coordinates until you know the metric. That's what
makes general relativity different from all other physical theories. If you
attach a physical meaning to r and t before solving for the metric, you're
starting with an a priori fixed background. Background independence is a
very esthetically appealing idea, and you're never going to convince anyone
to abandon it without a good reason. By a good reason I mean a formal
argument starting with background independence and reaching a conclusion
that's either mathematically inconsistent or inconsistent with experiment.
You obviously have a different esthetic, but it's one that your opponents
don't share, and you're just wasting your time by trying to appeal to it.
-- Ben
JanPB
> I was glad to see that Craig Feinstein was brave enough to take up the
> gauntlet in defence of Crothers' papers which appear to demonstrate
> clearly the non-existence of an interior Schwarzschild solution (see
> the earlier thread "On the controversy about the Schwarzschild radius
> and black holes"). Of course he came against some strong opposition
> from the noble defenders of the mainstream view
It's not a "view" any more than 1+1=2 is a "view".
> - Steve Carlip, Daryl
> McCullough, Jan Bielawski and Tom Roberts all arguing for the validity
> of the interior solution, making reference to the usual list of non-
> Schwarzschild coordinates which provide a singularity-free mapping
> over the maximal extension containing both the interior and exterior
> solutions.
>
> I can understand Crother's reasoning as I have been through the
> mathematics in detail, and I hope I can understand the reasoning of
> Steve Carlip et alii, as I have also been through the mathematics of
> various solutions in detail. From my perspective then, any errors in
> Crother's argument (or in the standard view) are not in the
> mathematics, but in the physical understanding and interpretation of
> the problem being solved and in particular of the coordinate systems
> being used.
Crothers does make mistakes, e.g. (pointed out by Daryl McC) he
assumes certain complex number is real from which he derives some
amazing conclusions against the "mainstream" "views".
> I have been trying to settle in my mind where the problem lies, and I
> have come to the conclusion that the difference ultimately lies in the
> physical significance (or lack thereof) which is attached to the
> coordinates.
That too but that's not the main problem. Of course his assumption
that changing coordinates changes the metric is wrong as well (why
would anyone make such an obvious blunder is hard for me to guess -
HOW can physical length and/or angle depend on man-made labels??!) I
suspect Crothers is confused by the distinction between choosing a
unit of physical length, versus choosing a coordinate system. I don't
know how else one can make physical lengths "depend" on "coordinate
systems" the way he envisions it.
> In Carlip et al's view (and in the standard interpretation of general
> relativity), the coordinates are just labels, and coordinate
> transformations are of no physical relevance. In Crother et al's view,
> coordinates can be given a physical significance a priori - e.g. if
> the direction of t is defined to be timelike, and the direction of r
> is defined to be spacelike when a problem is stated, then it must
> retain these properties in the solution.
Sure, any assumption made cannot be contradicted in the end or else
you have reductio ad absurdum. Point is, the spacelike/timelike
assumption is NOT made. Unfortunately most of the textbooks are quite
imprecise about this and they just glide over the derivation of
Schwarzschild with the slight violation of logic. Crothers - lazily
and inexcusably - took this as a sign of a real problem. Had he spent
some time studying the subject, he would have seen that these are all
mere textbook-editorial issues.
> (Note that the latter
> position is not in conflict with the former).
>
> For the problem at hand, in Carlip's view, the fact that the spacelike
> and timelike coordinates exchange 'labels' on passing from the
> exterior to the interior Scwarzschild solution is of no consequence.
Yes. Note that in the usual derivation when no staticity is assumed
(just spherical symmetry), no assumption is made on the exact meaning
of the r- and t-coordinates. The fact that they label events in
certain ways on disjoint domains is per se of no physical consequence.
> In Crother's view, the change of t (resp. r) from timelike (resp.
> spacelike) to spacelike (resp. timelike) across the horizon means that
> the interior solution does not satisfy the prior physical constraints
> on t and r in the original statement of the problem, and therefore the
> interior solution is unphysical.
There are no "prior physical constraints" of the type he envisions. He
got tangled up and trapped in the logical twists of the slightly
sloppy textbook presentations I mentioned above.
> When trying to find the metric corresponding to a centrally symmetric
> gravitational field, the general solution can be written in the form
> (see e.g. Landau & Lifschitz, Vol 2, p299):
>
> (1) ds^2 = h(r,t) dr^2 + k(r,t) (sin^2 theta d phi^2 + d theta^2) +
> l(r,t) dt^2 + a(r,t) dr dt
>
> where r labels 'spheres' and t labels 'time' in some way. Without loss
> of generality, coordinate transformations can be applied to bring this
> to the form:
>
> (2) ds^2 = exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta d
> phi^2) - exp(lambda(r,t)) dr^2
>
> where the new r and new t still label spheres and time respectively.
This is an example of such textbook sloppiness. If one assumes just
spherical symmetry, it is NOT true that (2) is the only form. What is
true is that (1) can be transformed into one of TWO forms:
(form 1) same as (2) in Landau-Lifschitz,
or
(form 2) ds^2 = -exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta
d phi^2) + exp(lambda(r,t)) dr^2
(note the sign changes).
As to why so many authors disregard the obvious second form is a bit
of a mystery for me. Clearly both forms satisfy the given constraints
of:
(i) spherical symmetry,
(ii) nondegenerate,
(iii) signature 2.
Then when one solves the EFE for both forms, one obtains the exterior
and the interior solutions - both equally valid and logical. The
supposed inconsistency of the interior solution with the "assumptions"
resulted just from stating the "assumptions" sloppily.
> The point to note at this point is that r and t are _not_ merely
> labels - they have a definite physical attribute determined by their
> respective spacelike and timelike natures. The solution to equation
> (2) above is of course the usual Schwarzschild solution.
>
> What Crothers' argument shows is that the interior Schwarzschild
> solution is _not_ compatible with the physical meaning attached to r
> and t, even though it naively (i.e. mathematically) provides a
> solution to the equation (2) above.
Well, it's just a (small) logical error that Landau & Lifschitz & many
others make, and Crothers doesn't have a good enough grasp of the
subject to see it for what it is, and how trivial the fix is.
> On the other hand, according to Carlip et alii, r and t are just
> labels for coordinates, and general relativity holds irrespective of
> the labels assigned to coordinates or the coordinate system used, so
> the physical meanings attached to them are irrelevant, and so the
> interior solution _is_ valid.
It's not just that - point is (see above) that the assumption of
spherical symmetry does allow two forms of solutions right from the
start - Crothers just missed it. And textbook authors know it's
correct anyway :-) so they don't bother to clean up their act :-)
> However, while it is true that general relativity does not care about
> the coordinate system used, in the formulation of the problem above, a
> physical meaning has indeed been attached to the coordinates r and t a
> priori, and given that this is the case, only the exterior
> Schwarzschild solution appears in my view to be a correct solution to
> the problem.
>
> But as shown by Carlip et al, the Kruskal-Szekeres and Eddington-
> Finkelstein metrics (and numerous others) also satisfy the vacuum
> field equations, and provide consistent, smooth and singularity free
> coordinate systems covering the maximal extension which also includes
> the Schwarzschild interior solution. They argue that the exterior
> Schwarzschild solution is incomplete on its own and this
> incompleteness is due to a quirk in the original choice of coordinates
> r and t and in particular the physical constraints placed upon them,
> i.e. constraining them a priori to be spacelike and timelike
> respectively.
>
> So who is right?
The math and physics is right - one only has to go through the entire
derivation carefully which is almost never done in textbooks for some
reason. Perhaps it's the physicists' slight disdain for mathematical
pedantry. In this case, however, based on the sheer numbers of
confused students, I'd suggest to be excessively pedantic for just
this one time, something like I did in my PDF note.
> Well, it appears to me that Crothers et al are correct and Carlip et
> al are wrong.
What if 2+2=5? I mean, really, what if? It's a question of transparent
mathematics and physics - if it's not transparent to Mr. Crothers it
doesn't make it false, you know.
> Why? Because the original problem, which seeks metrics corresponding
> to a spherically symmetric gravitational field, explicitly defines the
> coordinates r and t to be spacelike and timelike respectively.
It doesn't. What you wrote above is a false statement.
> If that is the case, then where are Carlip et al going wrong?
Nowhere :-)
[snippage]
--
Jan Bielawski
LEJ Brouwer
> The whole idea of background independence is that you don't know the
> physical meaning of the coordinates until you know the metric. That's what
> makes general relativity different from all other physical theories. If you
> attach a physical meaning to r and t before solving for the metric, you're
> starting with an a priori fixed background. Background independence is a
> very esthetically appealing idea, and you're never going to convince anyone
> to abandon it without a good reason. By a good reason I mean a formal
> argument starting with background independence and reaching a conclusion
> that's either mathematically inconsistent or inconsistent with experiment.
> You obviously have a different esthetic, but it's one that your opponents
> don't share, and you're just wasting your time by trying to appeal to it.
>
> -- Ben
Well, I think that sums up the point quite well - if one's aim is
merely to find solutions of the vacuum field equations, then certainly
the Schwarzschild interior metric does that - but how can one prove
that this mathematical solution corresponds to a physically realised/
realisable vacuum?
On the other hand, if one starts off with the assumption of a
spherically symmetric matter distribution, then one is explicitly
making a prior assumption about the background, and this is what is
usually done when stating this particular problem. The danger in this
case is of course is that the assumed background may not be a
physically realisable one.
It may simply be a matter of aesthetics so long as we are ignorant of
the actual physical situation, but in reality either an interior does
physically exist, or it does not. The mathematics certainly allows an
interior solution, but whether physically there is one, we cannot
easily know. So while the 'background independent approach' to the
above problem may be more aesthetically pleasing in some respects
(though one caveat here would be that one would likely find a high
correlation between what is considered more aesthetically pleasing and
what one was taught at graduate school), I don't see the need for
physicists to be quite so dogmatic about taking one stance over the
other.
- Sabbir.
LEJ Brouwer
> [I don't have Crothers' papers, and they don't appear on arxiv.org.]
This is the specific paper you should take a look at:
http://www.ptep-online.com/index_files/2005/PP-02-01.PDF
Please go throught it first, THEN complain.
There are links to Crothers' other papers here:
http://www.geocities.com/theometria/papers.html
- Sabbir.
JanPB
> On Apr 23, 9:57 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > [I don't have Crothers' papers, and they don't appear on arxiv.org.]
>
> This is the specific paper you should take a look at:
>
> http://www.ptep-online.com/index_files/2005/PP-02-01.PDF
If I have time I'll demolish this paper to smithereens.
(Unfortunately, I do have a life outside Usenet :-)
--
Jan Bielawski
LEJ Brouwer
> On Apr 23, 4:36 am, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > Crother's argument (or in the standard view) are not in the
> > mathematics, but in the physical understanding and interpretation of
> > the problem being solved and in particular of the coordinate systems
> > being used.
>
> Crothers does make mistakes, e.g. (pointed out by Daryl McC) he
This is not a mistake. This comes back to my original point about how
the original problem is set up. If you just look for vacuum solutions
with SO(3) symmetry, and make no prior assumptions about the nature of
r and t, then sure - it does not matter if integrating dr gives a
complex number, as Daryl pointed out. However, if you look for
solutions assuming a spherical matter distribution, and take a priori
the physical interpretations of r and t as spacelike and timelike
directions respectively (as is usually done in the textbooks), then
the integral of dr becoming complex does not make sense and is
physically incorrect, as Crothers states.
The point I am trying to make is that two different and inequivalent
problems are being solved here. The first approach gives spherically
symmetric mathematical solutions to the field equations which may not
be physical, and the second approach assumes a priori a spherically
symmetric background and draws conclusions therefrom.
> > I have been trying to settle in my mind where the problem lies, and I
> > have come to the conclusion that the difference ultimately lies in the
> > physical significance (or lack thereof) which is attached to the
> > coordinates.
>
> That too but that's not the main problem. Of course his assumption
> that changing coordinates changes the metric is wrong as well
If you read his papers you will see that he does not make that
assumption at all, though for some reason this accusation is levelled
against him repeatedly. Quite the opposite in fact, as he gives
multiple choices of coordinate to describe the same background.
> > In Carlip et al's view (and in the standard interpretation of general
> > relativity), the coordinates are just labels, and coordinate
> > transformations are of no physical relevance. In Crother et al's view,
> > coordinates can be given a physical significance a priori - e.g. if
> > the direction of t is defined to be timelike, and the direction of r
> > is defined to be spacelike when a problem is stated, then it must
> > retain these properties in the solution.
>
> Sure, any assumption made cannot be contradicted in the end or else
> you have reductio ad absurdum. Point is, the spacelike/timelike
> assumption is NOT made. Unfortunately most of the textbooks are quite
> imprecise about this and they just glide over the derivation of
> Schwarzschild with the slight violation of logic.
This imprecision you mention is at the heart of the matter - the
standard textbooks usually solve the problem of finding the metric
outside a spherical matter distribution using r and t with their usual
interpretations and get the exterior Schwarzschild solution and rather
sloppily note that the interior also solves the equation (which is
incorrect in that context). An exception is Chandrasekhar's "The
Mathematical Theory of Black Holes", in which he rigorously derives
the maximal extension in Kruskal coordinates using the method
originally due to Synge. But like Synge, he solved the _inequivalent_
problem of finding the set of vacuum solutions with SO(3) invariance.
>
> > (Note that the latter
> > position is not in conflict with the former).
>
> > For the problem at hand, in Carlip's view, the fact that the spacelike
> > and timelike coordinates exchange 'labels' on passing from the
> > exterior to the interior Scwarzschild solution is of no consequence.
>
> Yes. Note that in the usual derivation when no staticity is assumed
> (just spherical symmetry), no assumption is made on the exact meaning
> of the r- and t-coordinates. The fact that they label events in
> certain ways on disjoint domains is per se of no physical consequence.
>
> > In Crother's view, the change of t (resp. r) from timelike (resp.
> > spacelike) to spacelike (resp. timelike) across the horizon means that
> > the interior solution does not satisfy the prior physical constraints
> > on t and r in the original statement of the problem, and therefore the
> > interior solution is unphysical.
>
> There are no "prior physical constraints" of the type he envisions. He
> got tangled up and trapped in the logical twists of the slightly
> sloppy textbook presentations I mentioned above.
No he didn't. As I have hopefully made clear now, the matter is more
subtle than that. If anything Crothers has avoided the sloppiness of
the textbooks and given the correct answer to the problem as usually
stated.
> As to why so many authors disregard the obvious second form is a bit
> of a mystery for me. Clearly both forms satisfy the given constraints
> of:
>
> (i) spherical symmetry,
> (ii) nondegenerate,
> (iii) signature 2.
The second form does not satisfy the constraints:
(iv) r direction is spacelike
(v) t direction is timelike
You are arguing that this an invalid constraint, and I agree with you
if your aim is only to look for spherically symmetric solutions, but
not if you consider the prior physically constraints which _have_ been
placed on r and t in the usual statement of the problem.
> > What Crothers' argument shows is that the interior Schwarzschild
> > solution is _not_ compatible with the physical meaning attached to r
> > and t, even though it naively (i.e. mathematically) provides a
> > solution to the equation (2) above.
>
> Well, it's just a (small) logical error that Landau & Lifschitz & many
> others make, and Crothers doesn't have a good enough grasp of the
> subject to see it for what it is, and how trivial the fix is.
I think Crother's argument is mathematically and physically sound. The
logical error made in the textbooks is to not distinguish between the
two different problems being solved, and not the one you assume above.
> > On the other hand, according to Carlip et alii, r and t are just
> > labels for coordinates, and general relativity holds irrespective of
> > the labels assigned to coordinates or the coordinate system used, so
> > the physical meanings attached to them are irrelevant, and so the
> > interior solution _is_ valid.
>
> It's not just that - point is (see above) that the assumption of
> spherical symmetry does allow two forms of solutions right from the
> start - Crothers just missed it. And textbook authors know it's
> correct anyway :-) so they don't bother to clean up their act :-)
I think it's more likely that no-one has really bothered to think
about the matter hard enough. Pre-1960s, there was no issue, and the
interior solution was correctly discarded. Post-1960s, they found the
maximal extension as a solution but did not realise that they had
actually solved a slightly different problem. This is the real source
of the confusion.
> > So who is right?
>
> The math and physics is right - one only has to go through the entire
> derivation carefully which is almost never done in textbooks for some
> reason. Perhaps it's the physicists' slight disdain for mathematical
> pedantry. In this case, however, based on the sheer numbers of
> confused students, I'd suggest to be excessively pedantic for just
> this one time, something like I did in my PDF note.
Well, I agree with you and think the matter needs to be clarified out
once and for all. The way the standard texts add on the interior
solution is artificial and mathematically incorrect, whereas the
maximal extension does clearly satisfy the field equations. This
bothers me for various reasons, and this thread is my latest effort to
clarify the source of the confusion.
> --
> Jan Bielawski- Hide quoted text -
Anyway, good to hear from you again.
Best wishes,
Sabbir.
JanPB
> On Apr 24, 12:52 am, JanPB <film...@gmail.com> wrote:
> > That too but that's not the main problem. Of course his assumption
> > that changing coordinates changes the metric is wrong as well
>
> If you read his papers you will see that he does not make that
> assumption at all, though for some reason this accusation is levelled
> against him repeatedly.
I'll answer in more detail later. Here I'll just say this: read the
opening paragraphs of Section 2 of his paper. There is NO OTHER
conclusion one can make from his statements there except that he
thinks metric depends on coordinates. Specifically, he says: "The form
of C(r) cannot be pre-empted, and must in fact be rigorously
determined from the general solution to (2a)" - a patently false
statement - it can only mean that he holds metric to be a coordinate-
dependent entity (since C(r) obviously disappears upon the appropriate
coordinate change).
--
Jan Bielawski
Eric Gisse
[...]
> What Crothers' argument shows is that the interior Schwarzschild
> solution is _not_ compatible with the physical meaning attached to r
> and t, even though it naively (i.e. mathematically) provides a
> solution to the equation (2) above.
What, did Crothers notice the obvious? That r becomes timelike and t
becomes spacelike within the black hole? Then did he argue that
somehow there is no interior solution because of this?
[...]
> Why? Because the original problem, which seeks metrics corresponding
> to a spherically symmetric gravitational field, explicitly defines the
> coordinates r and t to be spacelike and timelike respectively.
Yup - sounds like it.
[...]
Stop whining about the Schwarzschild solution. Use one of the various
solutions that covers the entire 0 < r < oo, 0 < theta < 2pi, 0 < phi
< pi coordinate chart.
JanPB
>
> > On Apr 23, 4:36 am, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > > ...From my perspective then, any errors in
> > > Crother's argument (or in the standard view) are not in the
> > > mathematics, but in the physical understanding and interpretation of
> > > the problem being solved and in particular of the coordinate systems
> > > being used.
>
> > Crothers does make mistakes, e.g. (pointed out by Daryl McC) he
> > assumes certain complex number is real...
>
> This is not a mistake. This comes back to my original point about how
> the original problem is set up. If you just look for vacuum solutions
> with SO(3) symmetry, and make no prior assumptions about the nature of
> r and t, then sure - it does not matter if integrating dr gives a
> complex number, as Daryl pointed out. However, if you look for
> solutions assuming a spherical matter distribution, and take a priori
> the physical interpretations of r and t as spacelike and timelike
> directions respectively
Problem is no-one makes those "physical interpretations" except
Crothers. This is one of his strawmen.
I looked at his newer paper that you referred to. Although I said
earlier I might discuss this paper in detail I feel now it would be a
waste of time simply because the paper is so full of errors,
inaccuracies, non-sequiturs, and just plain gobbledy-gook that it
would be a waste of time to debate it seriously.
I'll quickly comment first on the opening paragraph because it's so
symptomatic. Crothers writes:
"The variable r has given rise to much confusion. In the conventional
analysis, based upon Hilbert metric which is almost invariably
incorrectly called the 'Schwarzschild' solution, r is taken both as a
coordinate and a radius in the spacetime manifold of the point-mass."
This is just the first sentence and it's already incorrect-squared.
Let's see:
"In the conventional analysis, based upon Hilbert metric..."
One can guess from the context that by "Hilbert metric" he means the
line element everyone else calls "Schwarzschild line element". He is
apparently unaware that in his very first communication to Einstein
Schwarzschild wrote down the very "Hilbert" formulation. Let's
continue:
"...which is almost invariably incorrectly called the 'Schwarzschild'
solution..."
Since Schwarzschild wrote it in December 1915 it's only fair to call
it Schwarzschild's solution.
"...r is taken both as a coordinate and a radius in the spacetime
manifold of the point-mass."
No, r is no such thing, never was. r is simply a _number_ assigned to
each sphere of symmetry. It doesn't stand for any "distance",
"radius", any a priori assumption of "timelike" or "spacelike" or
anything of the sort. It's just a number assigned to every such sphere
as follows:
1. take the sphere,
2. assign to it a number called "e" as follows:
(a) take the area of the sphere in question,
(b) divide this area by 4 pi,
(c) take the square root of the result. This number is a label
called "e" assigned to the sphere.
Now zillions of people got confused by the idiotic alphabetic fact
that this label is typically denoted by the Latin letter "r" rather
than "e", "q", or "DonaldDuck". It's simply astounding how such a
simple alphabetic convention could create so much confusion only
because people are so used to seeing "r" denote a "radius" of some
sort.
"r" is just a function on (a subset of) spacetime in question, a
function which happens to be constant on the spheres of symmetry and
equal to sqrt(area/4pi) there. That's all. It is, and always has been,
just that - IOW, a coordinate. It's not a priori any "radius", "time",
or whatever else. The fact that these coordinates approximate
Minkowskian coordinates far away has no a priori implication for
regions of significant curvature.
The rest of his "Introduction" contain pearls of wisdom like:
"ludicrous Kruskal-Szekeres formulation",
"one cannot treat the r-parameter as a radius" (as if anyone did).
[cut]
> > > In Crother's view, the change of t (resp. r) from timelike (resp.
> > > spacelike) to spacelike (resp. timelike) across the horizon means that
> > > the interior solution does not satisfy the prior physical constraints
> > > on t and r in the original statement of the problem, and therefore the
> > > interior solution is unphysical.
>
> > There are no "prior physical constraints" of the type he envisions. He
> > got tangled up and trapped in the logical twists of the slightly
> > sloppy textbook presentations I mentioned above.
>
> No he didn't. As I have hopefully made clear now, the matter is more
> subtle than that. If anything Crothers has avoided the sloppiness of
> the textbooks and given the correct answer to the problem as usually
> stated.
He got confused and made false assumptions about coordinates labelled
by the misleading letters "r" and "t". From these false assumptions he
then laboriously derives a contradiction.
> > As to why so many authors disregard the obvious second form is a bit
> > of a mystery for me. Clearly both forms satisfy the given constraints
> > of:
>
> > (i) spherical symmetry,
> > (ii) nondegenerate,
> > (iii) signature 2.
>
> The second form does not satisfy the constraints:
>
> (iv) r direction is spacelike
> (v) t direction is timelike
These constraints are made up by Crothers. They are invalid.
> You are arguing that this an invalid constraint, and I agree with you
> if your aim is only to look for spherically symmetric solutions, but
> not if you consider the prior physically constraints which _have_ been
> placed on r and t in the usual statement of the problem.
What physical constraints?
[cut]
> > > On the other hand, according to Carlip et alii, r and t are just
> > > labels for coordinates, and general relativity holds irrespective of
> > > the labels assigned to coordinates or the coordinate system used, so
> > > the physical meanings attached to them are irrelevant, and so the
> > > interior solution _is_ valid.
>
> > It's not just that - point is (see above) that the assumption of
> > spherical symmetry does allow two forms of solutions right from the
> > start - Crothers just missed it. And textbook authors know it's
> > correct anyway :-) so they don't bother to clean up their act :-)
>
> I think it's more likely that no-one has really bothered to think
> about the matter hard enough. Pre-1960s, there was no issue, and the
> interior solution was correctly discarded. Post-1960s, they found the
> maximal extension as a solution but did not realise that they had
> actually solved a slightly different problem. This is the real source
> of the confusion.
All I can say to this is "I disagree". (You provide no arguments
here.)
> > > So who is right?
>
> > The math and physics is right - one only has to go through the entire
> > derivation carefully which is almost never done in textbooks for some
> > reason. Perhaps it's the physicists' slight disdain for mathematical
> > pedantry. In this case, however, based on the sheer numbers of
> > confused students, I'd suggest to be excessively pedantic for just
> > this one time, something like I did in my PDF note.
>
> Well, I agree with you and think the matter needs to be clarified out
> once and for all. The way the standard texts add on the interior
> solution is artificial and mathematically incorrect,
Yes, yes, but it's a trivial omission and silly carelessness, it's
nothing important. And it's very easily fixed by simply observing that
the algebraic constraint right from the start yields two equally valid
candidate forms for the metric, not one. And after the derivation is
completed, it turns out that _both_ candidate forms are in fact
realised as solutions.
I agree that most textbooks do it in a logically confusing (that's an
euphemism for "wrong") manner by pretending only one algebraic
candidate exists and then suddenly pulling the interior solution of
out thin air. This is just an annoying habit, nothing substantial.
--
Jan Bielawski
Koobee Wublee
> LEJ Brouwer wrote:
> All that's happening here is a standard formal technique for finding
> solutions to a system of differential equations: guess the solution in a
> parameterized form, and solve for the parameters. One can certainly restrict
> the parameters enough that one obtains only the r>2M portion of the
> Schwarzschild metric, but that doesn't change the fact that the 0<r<2M
> portion is a valid solution of the same differential equations. You also
> won't obtain the exterior Kerr metric, but that doesn't change the fact that
> it's a solution. They all satisfy the equations; that's what it means to be
> a solution. It doesn't matter how they're derived, and it doesn't matter
> what letters of the alphabet are used for the coordinates.
Since the field equations yield an infinite number of solutions, the
field equations must be BS in nature. <shrug>
> Would you be happier with the two separate statements that
>
> ds^2 = (...) dt^2 - (...) dr^2 - r^2 (d theta^2 + sin^2 theta d phi^2)
> (r > 2M, 0 < theta < pi, 0 <= phi < 2pi)
>
> is a solution of the vacuum field equations and
>
> ds^2 = (...) dT^2 - (...) dZ^2 - T^2 (d theta^2 + sin^2 theta d phi^2)
> (0 < T < 2M, 0 < theta < pi, 0 <= phi < 2pi)
>
> is also a solution of the vacuum field equations? I assume everyone else
> would be equally happy with that version. It's just a longer way of saying
> the same thing.
You have just emphasized the infinite numbers of solutions to the
field equations. Mr. Bielawski should take notes here.
> > What Crothers' argument shows is that the interior Schwarzschild
> > solution is _not_ compatible with the physical meaning attached to r
> > and t, even though it naively (i.e. mathematically) provides a
> > solution to the equation (2) above.
>
> The whole idea of background independence is that you don't know the
> physical meaning of the coordinates until you know the metric.
No, no. The metric does not mean anything if you don't specify what
coordinate you are using. The curvature of space or spacetime can
only be described by a coordinate system and nothing else. It also
makes no sense to talk about the curvature of space or spacetime
without specifying a set of coordinate system to describe this
curvature. This should be an axiom in the annual of physics of curved
space or spacetime.
> That's what
> makes general relativity different from all other physical theories.
GR is as much BS as other crack-pot hypotheses. <shrug>
> If you
> attach a physical meaning to r and t before solving for the metric, you're
> starting with an a priori fixed background.
Again, the metric cannot be anything useful until you specifically
specify a set of coordinate system that this metric is a function of.
<shrug>
> Background independence is a
> very esthetically appealing idea, and you're never going to convince anyone
> to abandon it without a good reason.
Well, this should be very simple. If (A * B = 1), then (A = 1 / B).
In other words, for every B, there is a different A. If (metric *
coordinate = invariant geometry), then (metric = geometry /
coordinate). This math is ridiculous simple. Why do Mr. Bielawski,
Dr. Roberts, Professor Carlip, etc. have so much trouble understand
this very simple mathematical concept? This mathematical concept
should be in the grade school level. <shrug>
> By a good reason I mean a formal
> argument starting with background independence and reaching a conclusion
> that's either mathematically inconsistent or inconsistent with experiment.
What experiment?
> You obviously have a different esthetic, but it's one that your opponents
> don't share, and you're just wasting your time by trying to appeal to it.
You are correct. If the folks with higher education cannot understand
(A = 1 / B) if (A * B = 1), then nothing can convince these so called
highly educated individuals. <sigh>
Surfer
wrote:
I was wondering if "The river model of black holes" could throw any
light on this question.
http://arxiv.org/abs/gr-qc/0411060
Abstract
"This paper presents an under-appreciated way to conceptualize
stationary black holes, which we call the river model. The river model
is mathematically sound, yet simple enough that the basic
picture can be understood by non-experts. In the river model, space
itself flows like a river through a flat background, while objects
move through the river according to the rules of special relativity.
In a spherical black hole, the river of space falls into the black
hole at the Newtonian escape velocity, hitting the speed of light at
the horizon. Inside the horizon, the river flows inward faster than
light, carrying everything with it...."
This model suggests to me that objects falling into the hole could
never travel faster than light relative to the river, so objects would
continue to experience the direction r as spacelike, even after
crossing the horizion.
Is this reasonable?
-- Surfer
LEJ Brouwer
> On Apr 23, 6:13 pm, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > On Apr 24, 12:52 am, JanPB <film...@gmail.com> wrote:
> > > On Apr 23, 4:36 am, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > This is not a mistake. This comes back to my original point about how
> > the original problem is set up. If you just look for vacuum solutions
> > with SO(3) symmetry, and make no prior assumptions about the nature of
> > r and t, then sure - it does not matter if integrating dr gives a
> > complex number, as Daryl pointed out. However, if you look for
> > solutions assuming a spherical matter distribution, and take a priori
> > the physical interpretations of r and t as spacelike and timelike
> > directions respectively
>
> Problem is no-one makes those "physical interpretations" except
> Crothers. This is one of his strawmen.
Well, the textbook derivations usually start off with just that
assumption. As you say, they tend to 'correct' it in a rather ad hoc
way at the end by adding the interior solution. I don't consider it to
be a strawman, but an indication that there are indeed two different
problems being solved here.
In my reading of his paper, his main assumption is that the point mass
lies on one of the spheres labelled by r, which is taken to be a
spacelike coordinate. When one looks at the standard solution, one
sees that the point mass does not satisfy this condition, as r can
only describe the spheres in the exterior solution, which does not
contain the point mass.
This means that the initial assumption that the point mass is
contained within the solution manifold is incorrect. But this then
begs the question of whether it is even meaningful to talk about a
spherically symmetric solution about the point mass when the solution
does not even contain it.
If instead one looks for SO(3) invariant solutions of the vacuum field
equations, then we are solving a different problem, but this time one
which has a solution containing the supposed point mass - but even in
that case it is not possible to find a set of coordinates continuous
across the event horizon where the spherical symmetry about the point
mass is manifest globally. Both the interior and exterior have
spherical symmetry it is true, but to claim that the exterior solution
has a spherical symmetry with the point mass at its 'centre' does not
seem to be a valid conclusion given the lack of a global coordinate
system with manifest SO(3) symmetry throughout the manifold in which
this statement might make sense. The symmetry somehow breaks down at
the event horizon, and this is what makes the whole matter so
nontrivial.
> I looked at his newer paper that you referred to. Although I said
> earlier I might discuss this paper in detail I feel now it would be a
> waste of time simply because the paper is so full of errors,
> inaccuracies, non-sequiturs, and just plain gobbledy-gook that it
> would be a waste of time to debate it seriously.
I think you are perhaps misunderstanding his position. I read the
paper and I did not come to that conclusion at all. And the
mathematics in the paper appears to be quite sound if you can overcome
your initial negative reactions to the claims he makes.
> I'll quickly comment first on the opening paragraph because it's so
>
> symptomatic. Crothers writes:
>
> "The variable r has given rise to much confusion. In the conventional
> analysis, based upon Hilbert metric which is almost invariably
> incorrectly called the 'Schwarzschild' solution, r is taken both as a
> coordinate and a radius in the spacetime manifold of the point-mass."
>
> This is just the first sentence and it's already incorrect-squared.
>
> Let's see:
>
> "In the conventional analysis, based upon Hilbert metric..."
>
> One can guess from the context that by "Hilbert metric" he means the
> line element everyone else calls "Schwarzschild line element". He is
> apparently unaware that in his very first communication to Einstein
> Schwarzschild wrote down the very "Hilbert" formulation. Let's
> continue:
>
> "...which is almost invariably incorrectly called the 'Schwarzschild'
> solution..."
>
> Since Schwarzschild wrote it in December 1915 it's only fair to call
> it Schwarzschild's solution.
To be fair, there is a genuine confusion about what historically
constitutes the Schwarzschild solution and what constitutes the
Hilbert solution - the two are in fact distinct, and the matter has
been discussed in some detail by Antoci and others. So it's not really
fair to pre-judge the entire paper on that basis.
> "...r is taken both as a coordinate and a radius in the spacetime
> manifold of the point-mass."
>
> No, r is no such thing, never was. r is simply a _number_ assigned to
> each sphere of symmetry. It doesn't stand for any "distance",
> "radius", any a priori assumption of "timelike" or "spacelike" or
> anything of the sort.
That is one of the main assumptions made - that if there exists a
solution describing the spacetime outside a point mass, then it is
natural to assume that the point itself belongs to that spacetime.
With the benefit of hindsight, we can see that the exterior solution
does not contain the point mass - but this could merely be an
indication that the original problem was in fact ill-defined.
I don't think that these are 'false' assumptions - they are physically
reasonable assumptions which happen to lead to a contradiction.
I've snipped the rest because I think it mostly covers the same ground
as above.
Best wishes,
Sabbir.
Tom Roberts
> On Apr 23, 9:57 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> [I don't have Crothers' papers, and they don't appear on arxiv.org.]
> This is the specific paper you should take a look at:
> http://www.ptep-online.com/index_files/2005/PP-02-01.PDF
>
> Please go throught it first, THEN complain.
I made it quite clear I was pointing out YOUR mistakes, not Crothers'.
You seem to have no response to my elementary points about YOUR mistakes:
sqrt() DOES have a sign ambiguity, which you ignored
disjoint regions of a manifold cannot be compared as you
attempt to do
One MUST keep track of the region of validity of any
solution to a differential equation
Which of these do you disagree with?
Tom Roberts
Ben Rudiak-Gould
> If you just look for vacuum solutions
> with SO(3) symmetry, and make no prior assumptions about the nature of
> r and t, then sure - it does not matter if integrating dr gives a
> complex number, as Daryl pointed out. However, if you look for
> solutions assuming a spherical matter distribution, and take a priori
> the physical interpretations of r and t as spacelike and timelike
> directions respectively (as is usually done in the textbooks), then
> the integral of dr becoming complex does not make sense and is
> physically incorrect, as Crothers states.
The fact that you wrote this an hour after responding to my last post makes
me think you didn't understand what I was trying to say. You still seem to
think that one can conclude something about a solution based on the
assumptions that went into its derivation. That's crazy talk. That's
postmodernism. In the real, scientific world, a solution is a satisfying
assignment. Even if you made an outright mistake in your derivation --
getting a sign wrong or writing x^0 for the antiderivative of x^-1 or using
complex numbers where only reals make sense -- even then, if what you end up
with is a satisfying assignment, then it's valid. I know you understand this
at some level, but I think at another level you don't. Otherwise you
wouldn't continue to harp on the derivation of the Schwarzschild solution,
as though that could possibly have any relevance to anything.
I'm not saying your conclusion is wrong. I'm not even saying your intuition
is wrong. I'm just saying that your current attempt to formalize your
intuition makes no sense, and you're going to have to find a different one.
Perhaps what you're really concerned about is the *formation* of a black
hole; perhaps your intuition is that an initial spherically symmetric
configuration of matter can't dynamically evolve into something resembling
the r<2M portion of the Schwarzschild solution. That's a potentially
interesting line of argument, but it's not what you're arguing now. What
you're arguing now is silly.
-- Ben
Koobee Wublee
> On Apr 23, 6:13 pm, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> Crothers writes:
>
> "The variable r has given rise to much confusion. In the conventional
> analysis, based upon Hilbert metric which is almost invariably
> incorrectly called the 'Schwarzschild' solution, r is taken both as a
> coordinate and a radius in the spacetime manifold of the point-mass."
>
> This is just the first sentence and it's already incorrect-squared.
Squared?
> "In the conventional analysis, based upon Hilbert metric..."
>
> One can guess from the context that by "Hilbert metric" he means the
> line element everyone else calls "Schwarzschild line element". He is
> apparently unaware that in his very first communication to Einstein
> Schwarzschild wrote down the very "Hilbert" formulation. Let's
> continue:
You are totally mistaken. Given a segment of spacetime, we have
ds^2 = A dt^2 - B dr^2 - C dO^2
Hilbert's metric also known as the Schwarzschild metric gives the
following.
** A = c^2 (1 - 2 U)
** B = 1 / (1 - 2 U)
** C = r^2
Where
** U = G M / c^2 / r
The metric Schwarzschild communicated to Einstein in 1916 has the
following form.
** A = c^2 (1 - Q / R)
** B = r^4 / R^3 / (R - Q)
** C = R^2
Where
** R = (r^3 + Q^3)^(1/3)
** Q = 2 G M / c^2
These two metrics share the same coordinate system but are drastically
different. Hilber's metric manifests black holes; Schwarzschild's
original metric does not.
>
> "...which is almost invariably incorrectly called the 'Schwarzschild'
> solution..."
>
> Since Schwarzschild wrote it in December 1915 it's only fair to call
> it Schwarzschild's solution.
>
> "...r is taken both as a coordinate and a radius in the spacetime
> manifold of the point-mass."
You can call it whatever you want. Hilbert's metric also known as the
Schwarzschild metric is merely one of the infinite numbers of
solutions to a static and spherically symmetric set of field
equations. Static means the metric is independent of time, and
spherically symmetric means the metric represented by (A, B, C) is
independent of angular displacements (longitude and latitude). Each
solution describes a completely different universe.
> No, r is no such thing, never was. r is simply a _number_ assigned to
> each sphere of symmetry. It doesn't stand for any "distance",
> "radius", any a priori assumption of "timelike" or "spacelike" or
> anything of the sort. It's just a number assigned to every such sphere
> as follows:
r cannot be just a number. Using just a number, you can never
describe a geometry. To properly describe a geometry, you need to
specify a set of coordinate system of your choice first. You are so
confused.
> 1. take the sphere,
> 2. assign to it a number called "e" as follows:
>
> (a) take the area of the sphere in question,
> (b) divide this area by 4 pi,
> (c) take the square root of the result. This number is a label
> called "e" assigned to the sphere.
This is the radius of the sphere and not just a number. You are very
confused.
> Now zillions of people got confused by the idiotic alphabetic fact
> that this label is typically denoted by the Latin letter "r" rather
> than "e", "q", or "DonaldDuck". It's simply astounding how such a
> simple alphabetic convention could create so much confusion only
> because people are so used to seeing "r" denote a "radius" of some
> sort.
Whatever you want to describe the radius, e, r, or q, the fact remains
that this quantity specifies the radius of a sphere. You are utterly
confused.
> "r" is just a function on (a subset of) spacetime in question, a
> function which happens to be constant on the spheres of symmetry and
> equal to sqrt(area/4pi) there. That's all. It is, and always has been,
> just that - IOW, a coordinate. It's not a priori any "radius", "time",
> or whatever else. The fact that these coordinates approximate
> Minkowskian coordinates far away has no a priori implication for
> regions of significant curvature.
Any radius can be a function of something. You are hopelessly
confused. <shrug>
> All I can say to this is "I disagree". (You provide no arguments
> here.)
All I can say is that you do not understand the most basic of the
basic mathematics. <shrug>
> Yes, yes, but it's a trivial omission and silly carelessness, it's
> nothing important. And it's very easily fixed by simply observing that
> the algebraic constraint right from the start yields two equally valid
> candidate forms for the metric, not one. And after the derivation is
> completed, it turns out that _both_ candidate forms are in fact
> realised as solutions.
This is not what you have been saying. You have said any solutions to
the field equations are the same. You need to stop talking with a
forked tongue for a change, please.
> I agree that most textbooks do it in a logically confusing (that's an
> euphemism for "wrong") manner by pretending only one algebraic
> candidate exists and then suddenly pulling the interior solution of
> out thin air. This is just an annoying habit, nothing substantial.
Hilbert's metric also known as the Schwarzschild metric cannot
describe physics beyond (into) the event horizon. Schwarzschild's
original solution and others do. <shrug>
carlip...@physics.ucdavis.edu
[...]
> In Carlip et al's view (and in the standard interpretation of general
> relativity), the coordinates are just labels, and coordinate
> transformations are of no physical relevance.
Of course. To say otherwise is magical thinking: the belief that
an object's properties are affected by what you name it.
> In Crother et al's view,
> coordinates can be given a physical significance a priori - e.g. if
> the direction of t is defined to be timelike, and the direction of r
> is defined to be spacelike when a problem is stated, then it must
> retain these properties in the solution.
How do you "define [a coordinate] to be timelike"? What does that mean?
You can *say* "t is timelike," but why should that affect anything?
[...]
> When trying to find the metric corresponding to a centrally symmetric
> gravitational field, the general solution can be written in the form
> (see e.g. Landau & Lifschitz, Vol 2, p299):
> (1) ds^2 = h(r,t) dr^2 + k(r,t) (sin^2 theta d phi^2 + d theta^2) +
> l(r,t) dt^2 + a(r,t) dr dt
> where r labels 'spheres' and t labels 'time' in some way. Without loss
> of generality, coordinate transformations can be applied to bring this
> to the form:
> (2) ds^2 = exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta d
> phi^2) - exp(lambda(r,t)) dr^2
Not if you believe that "coordinates can be given a physical significance
a priori"! If you believe this, you don't have any business applying
coordinate transformations to bring a general metric to a more restricted
form, and you *certainly* don't have any business claiming that this is
done "without any loss of generality."
In fact, the transformation you have made eliminates Painleve-Gullstrand
coordinates, a nice set of coordinates which are well-behaved through
the horizon and in which t remains a time coordinate everywhere. By
parametrizing the metric the way you do, you've also changed h(r,t) and
l(r,t), whose signs are arbitrary in your original metric, to positive
functions. That's certainly *not* "without loss of generality," unless
you allow nu and lambda to be complex.
This is all fine if you allow me to switch back to better coordinates
after the field equations are solved. But you don't want to do that --
you want to allow coordinate transformations before you write down the
field equations, but not after.
> where the new r and new t still label spheres and time respectively.
Why? What tells you this, apart from the names you have chosen for them?
> The point to note at this point is that r and t are _not_ merely
> labels - they have a definite physical attribute determined by their
> respective spacelike and timelike natures.
What physical attributes are those? How do you determine them? Or is
this more magical thinking: you chant, "t shall be time" and that makes
it time?
[...]
> However, while it is true that general relativity does not care about
> the coordinate system used, in the formulation of the problem above, a
> physical meaning has indeed been attached to the coordinates r and t a
> priori,
How? What attaches this meaning, apart from "because I say so"?
[...]
> The conclusion to be reached then is that analytic continuation of a
> solution is only valid if it respects the physical constraints of the
> original problem. If the original constraints are ignored, well, a
> different problem is being solved, with the possibly that the
> resulting solutions are unphysical. [If Carlip et al disagree with
> this conclusion, then I would very much like to see a proof that the
> interior solution does indeed solve the original problem and
> corresponds to a physically realisable vacuum].
I've answered this in the past. If you want to ask whether the solution
is physically reasonable, set up physically reasonable initial data and
see how they evolve. In this case, for example, you can write down an
exact solution for a collapsing spherically symmetric shell or sphere of
matter, with a time coordinate that has the direct physical meaning as
the proper time as measured by an observer standing on the surface of the
collapsing matter. In the resulting exact solution the matter passes
through the horizon with nothing strange happening, and the solution
evolves into an ordinary Schwarzschild black hole -- both exterior and
interior -- typically described in Painleve-Gullstrand coordinates. You
can find two simple examples in Adler, Bjorken, Chen, and Liu, "Simple
analytical models of gravitational collapse," American Journal of Physics
73 (December 2005) 1148-1159.
Steve Carlip
JanPB
> On Apr 24, 7:37 am, JanPB <film...@gmail.com> wrote:
>
> > On Apr 23, 6:13 pm, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > > On Apr 24, 12:52 am, JanPB <film...@gmail.com> wrote:
> > > > On Apr 23, 4:36 am, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > > This is not a mistake. This comes back to my original point about how
> > > the original problem is set up. If you just look for vacuum solutions
> > > with SO(3) symmetry, and make no prior assumptions about the nature of
> > > r and t, then sure - it does not matter if integrating dr gives a
> > > complex number, as Daryl pointed out. However, if you look for
> > > solutions assuming a spherical matter distribution, and take a priori
> > > the physical interpretations of r and t as spacelike and timelike
> > > directions respectively
>
> > Problem is no-one makes those "physical interpretations" except
> > Crothers. This is one of his strawmen.
>
> Well, the textbook derivations usually start off with just that
> assumption. As you say, they tend to 'correct' it in a rather ad hoc
> way at the end by adding the interior solution. I don't consider it to
> be a strawman, but an indication that there are indeed two different
> problems being solved here.
It would be better if we could focus on a concrete issue. Your post
makes many claims following Crothers which are a bit imprecise so it's
hard to explain quickly why they are wrong. To explain in detail his
mistakes would take a paper about twice as long as his. It's not worth
it.
--
Jan Bielawski
Koobee Wublee
> LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > In Carlip et al's view (and in the standard interpretation of general
> > relativity), the coordinates are just labels, and coordinate
> > transformations are of no physical relevance.
>
> Of course. To say otherwise is magical thinking: the belief that
> an object's properties are affected by what you name it.
I do agree with you on this point. However, in order to describe a
set of curvature correction factors known as the metric, you need to
specify a set of coordinate to go along with it and stick with that
choice through out the duration of any derivation into something more
evolutionary in mathematics that may occur. All the solutions to the
field equations do just that. They all are based on a specific set of
coordinate system that is the common spherically symmetric polar
coordinate system. As a solution to the field equations, each metric
has an equal weight of being such a solution, and there are infinite
numbers of them. What you refer to as coordinate transformation is
not really a coordinate transformation but a simple and power trick or
short cut to find other solutions to the field equations if one
particular solution is first discovered. The beauty of this
mathematical trick but certainly no mathemagic is that you can start
with any solution to find any other solution. Again, to specify a
metric, you must have a choice of coordinate system. The field
equations are totally based on this metric with your choice of
coordinate system. You must carry out any mathematical operations
based on this choice of coordinate system and no others, or else your
field equations are not valid any more. The following two metrics
have equal weight of each being a solution out of an infinite numbers
of them that satisfy as a solution to the field equations in the
common spherically symmetric polar coordinate system.
ds^2 = c^2 (1 - 2 K / r) dt^2 - dr^2 / (1 - 2 K / r) - r^2 dO^2
And
ds^2 = c^2 dt^2 / (1 + 2 K / r) - (1 + 2 K / r) dr^2 - (r + K)^2 dO^2
Where
** K = G M / c^2
Both of them reduce to the Newtonian limit. To say one is a shadow
transformation of the other one is totally blasphemous against science
and mathematics without any basis. <shrug>
> [...]
The rest of your argument is like having a litigation against a
ghost. <shrug>
JanPB
> On Apr 24, 7:37 am, JanPB <film...@gmail.com> wrote:
[Something was wrong with Google Groups 2 days ago - long delays]
> > On Apr 23, 6:13 pm, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > > On Apr 24, 12:52 am, JanPB <film...@gmail.com> wrote:
> > > > On Apr 23, 4:36 am, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > > This is not a mistake. This comes back to my original point about how
> > > the original problem is set up. If you just look for vacuum solutions
> > > with SO(3) symmetry, and make no prior assumptions about the nature of
> > > r and t, then sure - it does not matter if integrating dr gives a
> > > complex number, as Daryl pointed out. However, if you look for
> > > solutions assuming a spherical matter distribution, and take a priori
> > > the physical interpretations of r and t as spacelike and timelike
> > > directions respectively
>
> > Problem is no-one makes those "physical interpretations" except
> > Crothers. This is one of his strawmen.
>
> Well, the textbook derivations usually start off with just that
> assumption. As you say, they tend to 'correct' it in a rather ad hoc
> way at the end by adding the interior solution. I don't consider it to
> be a strawman, but an indication that there are indeed two different
> problems being solved here.
1. I told you earlier there were two candidate forms for the metric,
not one. Hence no problem. Yes, the standard textbook presentation is
a small annoyance.
2. What is the second problem there? (You wrote "there are indeed two
different problems...")
> In my reading of his paper, his main assumption is that the point mass
> lies on one of the spheres labelled by r,
Well, it's just wrong.
> which is taken to be a spacelike coordinate.
Also wrong to assume this a priori.
> When one looks at the standard solution, one
> sees that the point mass does not satisfy this condition, as r can
> only describe the spheres in the exterior solution, which does not
> contain the point mass.
Why are you saying that "r can only describe the spheres in the
exterior solution"? r is defined as a certain labelling of the spheres
of symmetry. In other words, a real-valued function on (an open subset
of) spacetime. No restrictions otherwise besides the obvious
requirements like the combined four functions (t,r,theta,phi) forming
a homeomorphism compatible with the standard smooth structure on
spacetime plus for purely logical reasons we must leave open the
possibility of a restriction of the domain of the metric coordinate
representation imposed by the Einstein equations themselves or by any
of the derivation steps we decide to employ.
> This means that the initial assumption that the point mass is
> contained within the solution manifold is incorrect.
This assumption is not being made. What's assumed is a spherically
symmetric spacetime. (Even in Newtonian mechanics the gravitational
field due to a point mass has a singularity there.) The usual
derivation actually excludes the locus r=0 from the candidate
spacetime right from the start since (t,r,theta,phi) is not a
homeomorphism there (not 1-1 in fact). So in the standard approach to
the derivation one solves the Einstein equations using the usual two
candidate forms of the metric for the maximal potential domain r>0.
Having done that, one can go back to the special case r=0 and examine
it separately as it was excluded only by a coordinate choice - IOW, it
was just for human convenience. (Same thing must be done for the other
excluded locus that popped up during the derivation: r=2M.) And only
at this point one realises that the solution so obtained does not
extend smoothly (or even continuously) over the origin but it does
extend smoothly over r=2M.
> But this then
> begs the question of whether it is even meaningful to talk about a
> spherically symmetric solution about the point mass when the solution
> does not even contain it.
Why not? The only objections would be philosophical as in "where does
all this matter go" or "does it make philosophical sense to have an
edge of space and time inside a bounded region", etc.
> If instead one looks for SO(3) invariant solutions of the vacuum field
> equations, then we are solving a different problem,
What do you mean by "different problem"? Isn't the subject of this
thread the solution of the vacuum field equations in the spherically
symmetric case?
> but this time one
> which has a solution containing the supposed point mass - but even in
> that case it is not possible to find a set of coordinates continuous
> across the event horizon where the spherical symmetry about the point
> mass is manifest globally.
No, "it is not possible to find a set of coordinates continuous across
the event horizon" only if you insist on coordinates of a certain form
designed to facilitate the calculations. This is a "feature" of
certain labelling choice made for human convenience. It was merely not
obvious from the start that this labelling choice would cause the
coordinate lines (non-physical entities, lines of constant label
values) to bend away from a certain surface.
> Both the interior and exterior have
> spherical symmetry it is true, but to claim that the exterior solution
> has a spherical symmetry with the point mass at its 'centre' does not
> seem to be a valid conclusion given the lack of a global coordinate
> system with manifest SO(3) symmetry throughout the manifold in which
> this statement might make sense.
Explain - as written this is incorrect. Take Kruskal-Szekeres - what's
wrong with its symmetry properties?
> The symmetry somehow breaks down at
> the event horizon, and this is what makes the whole matter so
> nontrivial.
Again, in what way does "symmetry somehow breaks down at the event
horizon"?
> > I looked at his newer paper that you referred to. Although I said
> > earlier I might discuss this paper in detail I feel now it would be a
> > waste of time simply because the paper is so full of errors,
> > inaccuracies, non sequiturs, and just plain gobbledy-gook that it
> > would be a waste of time to debate it seriously.
>
> I think you are perhaps misunderstanding his position. I read the
> paper and I did not come to that conclusion at all. And the
> mathematics in the paper appears to be quite sound if you can overcome
> your initial negative reactions to the claims he makes.
It's not that - it's the way he writes: overcomplex (the standard
crank sign), false statements, non sequiturs. Here is an example - the
whole paper reads like this:
"The conventional analysis simply looks at the Hilbert metric and
makes the following unjustified assumptions, tacitly or otherwise:
(a) The variable r is a radius and/or coordinate of some kind in the
gravitational field.
(b) The regions 0 < r < 2m and 2m < r < infinity are both valid.
(c) A singularity in the gravitational field must occur only where the
Riemann tensor scalar curvature invariant (Kretschmann scalar)
f=R_abcd R^abcd is unbounded.
The orthodox analysis has never proved these assumptions, but
nonetheless takes them as given [...]"
Almost every sentence in the above quote is incorrect or misleading:
(a) is wrong because r is not any "radius", it's a real-valued
function on the set of all quadruples (t,r,theta,phi) with r>0 [note
the traditional convenient abuse of notation in which the letters play
dual role both as manifold points and functions]. This function
together with the other three (t, theta, phi) forms an R^4-valued
function called "coordinate system". It's not "of some kind" either.
(b) nothing is "assumed" about r less than or greater than 2m. One
simply seeks a solution in a certain coordinate system. Since the
solution does exist in those regions in that coordinate system, the
_conclusion_ (not _assumption_) is that both regions are valid. A
secondary immediate conclusion is that the region of validity may
actually be larger because one obviously cannot just assume that the
initial coordinate choice has no influence on the domain of the
relevant coordinate components.
(c) this looks false to me, mainly because of the Kretschmann scalar
thing. Had Crothers said "sectional curvatures" or "Riemann curvature"
instead, he would have made more sense since Riemann curvature
extending smoothly over some region sometimes implies that the metric
itself extends over that region (this is the "curvature determines the
metric" business, I forget the exact technical requirements for this
to hold). But in the end it doesn't matter since the only proof of
validity of (an extension of a piece of) a solution is that - ta-dah!!
- it satisfies the Einstein equation. Once this is verified - just
plugging in by hand will do - all the talk of the (c)-type is
irrelevancy.
So you see, practically every other sentence in Crothers' paper
requires reams of debunking like this - it's not worth it. A good
professionally written paper never requires major fumigation on such a
scale.
[...]
> To be fair, there is a genuine confusion about what historically
> constitutes the Schwarzschild solution and what constitutes the
> Hilbert solution - the two are in fact distinct,
They can't be distinct. First of all, what are these two solutions?
>From the context I gather that Crothers' "Hilbert solution" is the
usual form of the metric as commonly written these days while
Crothers' "Schwarzschild solution" is the form Schwarzschild written
in his paper. Is this correct? If yes, then it's easy to see these two
forms describe the same metric under a simple coordinate change.
[...]
> > [cut]
>
> > > > > In Crother's view, the change of t (resp. r) from timelike (resp.
> > > > > spacelike) to spacelike (resp. timelike) across the horizon means that
> > > > > the interior solution does not satisfy the prior physical constraints
> > > > > on t and r in the original statement of the problem, and therefore the
> > > > > interior solution is unphysical.
>
> > > > There are no "prior physical constraints" of the type he envisions. He
> > > > got tangled up and trapped in the logical twists of the slightly
> > > > sloppy textbook presentations I mentioned above.
>
> > > No he didn't. As I have hopefully made clear now, the matter is more
> > > subtle than that. If anything Crothers has avoided the sloppiness of
> > > the textbooks and given the correct answer to the problem as usually
> > > stated.
>
> > He got confused and made false assumptions about coordinates labelled
> > by the misleading letters "r" and "t". From these false assumptions he
> > then laboriously derives a contradiction.
>
> I don't think that these are 'false' assumptions - they are physically
> reasonable assumptions which happen to lead to a contradiction.
They may seem "reasonable" but they violate the logic of the problem
as they are just random additions. Nothing useful can be inferred from
randomly changing one's assumptions.
> I've snipped the rest because I think it mostly covers the same ground
> as above.
OK.
--
Jan Bielawski
LEJ Brouwer
> On Apr 24, 4:46 am, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
>
> > On Apr 24, 7:37 am, JanPB <film...@gmail.com> wrote:
>
> [Something was wrong with Google Groups 2 days ago - long delays]
Yes, and I am afraid I got distracted by 'real' work too, hence the
delayed response.
> > > On Apr 23, 6:13 pm, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> > > > On Apr 24, 12:52 am, JanPB <film...@gmail.com> wrote:
> > > > > On Apr 23, 4:36 am, LEJ Brouwer <intuitioni...@yahoo.com> wrote:
> 2. What is the second problem there? (You wrote "there are indeed two
> different problems...")
The first problem is to find the metric corresponding to a centrally
symmetric gravitational field outside of a point mass. The second is
to find the set of solutions with a spherical (i.e. SO(3)) symmetry.
Why do I say these are different? Well, if this were normal Galilean
spacetime, a 'centrally symmetric' anything would refer to something
with spatial rotational invariance about the origin. Applying this
intuition to GR, and assuming an index (-,+++), this means solutions
of the form:
ds^2 = A dt^2 - r^2 dOmega^2 - B dr^2
where A(r,t) and B(r,t) are positive functions, and r labels
'spheres'. This is essentially the same equation I referred to from
Landau and Lifschitz, and has a simple physical intepretation, and
leads to the exterior Schwarzschild solution. The spacelike coordinate
r labels the spheres of symmetry, as one would expect.
The problem is that the solutions with SO(3) symmetry include the
Schwarzschild interior, which looks like this:
ds^2 = -A' dr'^2 - t'^2 dOmega^2 + B' dt'^2
Here, the spherical symmetry is labelled by the timelike coordinate
t'. Now, this is very different from having the spacelike coordinate r
labelling spheres, and does not correspond to our usual physical
intuition of spherical symmetry. In particular, the 'origin' where the
point mass supposedly lies, is now at some fixed _time_ coordinate
_relative_ to the time at which the (originally radially, say)
infalling particle supposedly crossed the event horizon.
The pictures we are used to seeing which shows a particle falling in
towards what appears to be the origin are highly misleading, as the
singularity is certainly _not_ some kind of spatial origin or
'centre', and the symmetry is not a spatial spherical symmetry but
rather some difficult to intepret and highly non-intuitive temporal
SO(3) symmetry - and this is certainly not the kind of solution which
would be sought in any naive attempt to find solutions with a 'central
symmetry'. So the idea that this singularity can be interpreted in any
way as a point mass (or even a 'point' of any kind) seems to me to be
quite a stretch of the imagination. I am not saying that the interior
solution is not a solution - it is certainly a solution to the vacuum
field equations, but as you will have gathered, I personally find it
hard believe that it has any physical relationship with the exterior
solution or that radially infalling particles can fall from a
spacetime described by the exterior solution into a spacetime
described by the interior solution (i.e. the sewing between the two
spacetimes is physically invalid, even it can be done mathematically
with an appropriate choice of coordinates which are hard-to-interpret-
physically at the EH).
> > In my reading of his paper, his main assumption is that the point mass
> > lies on one of the spheres labelled by r,
>
> Well, it's just wrong.
No, it is a completely natural assumption which turns out to be
inconsistent - probably because there are no such things as point
masses in GR.
> > which is taken to be a spacelike coordinate.
>
> Also wrong to assume this a priori.
Not at all, and for the same reason. And as I said, this is how most
textbook solutions begin.
> > When one looks at the standard solution, one
> > sees that the point mass does not satisfy this condition, as r can
> > only describe the spheres in the exterior solution, which does not
> > contain the point mass.
>
> Why are you saying that "r can only describe the spheres in the
> exterior solution"? r is defined as a certain labelling of the spheres
> of symmetry.
...except that some of your 'spheres' are labelled by a time t instead
of a radius r. I hope you now see why I say we are talking about two
different problems - one which is based upon a physical intuitive
notion of spherical symmetry, and one which is essentially an abstract
mathematical one.
> In other words, a real-valued function on (an open subset
> of) spacetime. No restrictions otherwise besides the obvious
> requirements like the combined four functions (t,r,theta,phi) forming
> a homeomorphism compatible with the standard smooth structure on
> spacetime plus for purely logical reasons we must leave open the
> possibility of a restriction of the domain of the metric coordinate
> representation imposed by the Einstein equations themselves or by any
> of the derivation steps we decide to employ.
>
> > This means that the initial assumption that the point mass is
> > contained within the solution manifold is incorrect.
>
> This assumption is not being made.
...in the second problem which looks for SO(3) symmetry irrespective
of its nature.
> What's assumed is a spherically
> symmetric spacetime. (Even in Newtonian mechanics the gravitational
> field due to a point mass has a singularity there.) The usual
> derivation actually excludes the locus r=0 from the candidate
> spacetime right from the start since (t,r,theta,phi) is not a
> homeomorphism there (not 1-1 in fact). So in the standard approach to
> the derivation one solves the Einstein equations using the usual two
> candidate forms of the metric for the maximal potential domain r>0.
> Having done that, one can go back to the special case r=0 and examine
> it separately as it was excluded only by a coordinate choice - IOW, it
> was just for human convenience. (Same thing must be done for the other
> excluded locus that popped up during the derivation: r=2M.) And only
> at this point one realises that the solution so obtained does not
> extend smoothly (or even continuously) over the origin but it does
> extend smoothly over r=2M.
>
> > But this then
> > begs the question of whether it is even meaningful to talk about a
> > spherically symmetric solution about the point mass when the solution
> > does not even contain it.
>
> Why not? The only objections would be philosophical as in "where does
> all this matter go" or "does it make philosophical sense to have an
> edge of space and time inside a bounded region", etc.
Yes, these are certainly valid questions (and not just philosophical
ones if you believe the interior solution to be physically realisable)
which deserve answers.
> > If instead one looks for SO(3) invariant solutions of the vacuum field
> > equations, then we are solving a different problem,
>
> What do you mean by "different problem"? Isn't the subject of this
> thread the solution of the vacuum field equations in the spherically
> symmetric case?
>
> > but this time one
> > which has a solution containing the supposed point mass - but even in
> > that case it is not possible to find a set of coordinates continuous
> > across the event horizon where the spherical symmetry about the point
> > mass is manifest globally.
>
> No, "it is not possible to find a set of coordinates continuous across
> the event horizon" only if you insist on coordinates of a certain form
> designed to facilitate the calculations. This is a "feature" of
> certain labelling choice made for human convenience. It was merely not
> obvious from the start that this labelling choice would cause the
> coordinate lines (non-physical entities, lines of constant label
> values) to bend away from a certain surface.
It is not just for human convenience. It is always possible to choose
coordinates such that locally the metric is flat, and if we are
talking about a radially infalling particle, then we can further
choose one of those coordinates to point in the direction of motion,
and another to measure the time. Now what is it that happens at the
event horizon that causes the radial direction and the temporal
directions to magically swap places for this 'freefalling' observer?
One moment he thinks he is falling towards a spatial point, and the
next moment he is falling towards a future time? Weird and quite
unbelievable.
> > Both the interior and exterior have
> > spherical symmetry it is true, but to claim that the exterior solution
> > has a spherical symmetry with the point mass at its 'centre' does not
> > seem to be a valid conclusion given the lack of a global coordinate
> > system with manifest SO(3) symmetry throughout the manifold in which
> > this statement might make sense.
>
> Explain - as written this is incorrect. Take Kruskal-Szekeres - what's
> wrong with its symmetry properties?
There's nothing wrong with Kruskal-Szekeres (or any other similar)
coordinates except that they don't mean much physically. The
Schwarzschild (and similar) coordinates do have simple physical
intepretations - well, the exterior at least.
> > The symmetry somehow breaks down at
> > the event horizon, and this is what makes the whole matter so
> > nontrivial.
>
> Again, in what way does "symmetry somehow breaks down at the event
> horizon"?
Because the mapping to locally flat coordinates has strange properties
there (note that this is where four pieces of the Kruskal-extended
solution meet, so there is some kind of strange bi-/quadri-furcation
going on there).
Hopefully we've covered these now?
> (c) this looks false to me, mainly because of the Kretschmann scalar
> thing. Had Crothers said "sectional curvatures" or "Riemann curvature"
> instead, he would have made more sense since Riemann curvature
> extending smoothly over some region sometimes implies that the metric
> itself extends over that region (this is the "curvature determines the
> metric" business, I forget the exact technical requirements for this
> to hold). But in the end it doesn't matter since the only proof of
> validity of (an extension of a piece of) a solution is that - ta-dah!!
> - it satisfies the Einstein equation. Once this is verified - just
> plugging in by hand will do - all the talk of the (c)-type is
> irrelevancy.
This is slightly different, and is just presented as another piece of
evidence that something strange happens at the event horizon.
> So you see, practically every other sentence in Crothers' paper
> requires reams of debunking like this - it's not worth it. A good
> professionally written paper never requires major fumigation on such a
> scale.
I guess whether you are willing to put in such an effort depends upon
whether you think there is a problem with the standard solution or
not. I did go through it, and I had also read the references, so I had
some understanding of the context in which the statements were being
made.
> [...]
>
> > To be fair, there is a genuine confusion about what historically
> > constitutes the Schwarzschild solution and what constitutes the
> > Hilbert solution - the two are in fact distinct,
>
> They can't be distinct. First of all, what are these two solutions?>From the context I gather that Crothers' "Hilbert solution" is the
>
> usual form of the metric as commonly written these days while
> Crothers' "Schwarzschild solution" is the form Schwarzschild written
> in his paper. Is this correct? If yes, then it's easy to see these two
> forms describe the same metric under a simple coordinate change.
The original Schwarzschild solution only went as far as the event
horizon. See the paper by Antoci and Liebscher for details:
http://arxiv.org/abs/gr-qc/0406090
Thanks,
Sabbir.
LEJ Brouwer
> LEJ Brouwer wrote:
> > If you just look for vacuum solutions
> > with SO(3) symmetry, and make no prior assumptions about the nature of
> > r and t, then sure - it does not matter if integrating dr gives a
> > complex number, as Daryl pointed out. However, if you look for
> > solutions assuming a spherical matter distribution, and take a priori
> > the physical interpretations of r and t as spacelike and timelike
> > directions respectively (as is usually done in the textbooks), then
> > the integral of dr becoming complex does not make sense and is
> > physically incorrect, as Crothers states.
>
> The fact that you wrote this an hour after responding to my last post makes
> me think you didn't understand what I was trying to say. You still seem to
> think that one can conclude something about a solution based on the
> assumptions that went into its derivation. That's crazy talk. That's
> postmodernism. In the real, scientific world, a solution is a satisfying
> assignment. Even if you made an outright mistake in your derivation --
> getting a sign wrong or writing x^0 for the antiderivative of x^-1 or using
> complex numbers where only reals make sense -- even then, if what you end up
> with is a satisfying assignment, then it's valid. I know you understand this
> at some level, but I think at another level you don't. Otherwise you
> wouldn't continue to harp on the derivation of the Schwarzschild solution,
> as though that could possibly have any relevance to anything.
I guess what it boils down to is that I do not believe that all
coordinate choices have a meaningful physical interpretation. At a
very simplistic level, we all agree that x^2 = 9 has two solutions, +3
and -3, but I am claiming that the real problem is something like: if
the number of apples that John has is equal to the number of apples
Bob has squared, then how many apples does John have? Basically my
intuition tells me that 'background independent' GR, if I may call it
that, is too general and includes solutions which are not physically
sensible. In this particular case, my intuition is telling me that
something goes badly wrong at the event horizon.
> I'm not saying your conclusion is wrong. I'm not even saying your intuition
> is wrong. I'm just saying that your current attempt to formalize your
> intuition makes no sense, and you're going to have to find a different one.
> Perhaps what you're really concerned about is the *formation* of a black
> hole; perhaps your intuition is that an initial spherically symmetric
> configuration of matter can't dynamically evolve into something resembling
> the r<2M portion of the Schwarzschild solution. That's a potentially
> interesting line of argument, but it's not what you're arguing now. What
> you're arguing now is silly.
>
> -- Ben
Well, you are being more generous than your colleagues in that case!
Anyway, I am probably just doing a poor job of expressing in a clear
way what my intuition is telling me.
As it happens, I came across a very simple thermodynamic argument why
black holes cannot be formed from spherical collapse of ordinary
matter - thought that particular argument does not hold for photons or
gravitational waves (I do believe in the possibility of complete
gravitational collapse of the latter, but obviously not to form a
singularity). If I have some time later, I will give the details as it
was only a few lines and pretty easy to follow.
My personal opinion is that the Schwarzshild interior solution is
unphysical (that's not to say that there is not some other metric
describing the stuff which lies 'inside' the event horizon), and that
once the event horizon forms, it acts like a brick wall effectively
physically sealing off the interior from the exterior, so that
anything subsequently hitting the event horizon bounces of it both
spatially and temporally (so that it looks like particle-antiparticle
pairs are annihilating at the EH). The reasons I think this is the
case are rather involved, so I won't go into them now.
Having said, that it is of course possible that the standard view is
correct, (in which case I will insist that the singularity causes a
space-time reflection like the one described in the last paragraph!) -
I just have difficulties believing that the picture it paints can be
physically real.
Thanks,
Sabbir.
LEJ Brouwer
> LEJ Brouwer <intuitioni...@yahoo.com> wrote:
>
> [...]
>
> > In Carlip et al's view (and in the standard interpretation of general
> > relativity), the coordinates are just labels, and coordinate
> > transformations are of no physical relevance.
>
> Of course. To say otherwise is magical thinking: the belief that
> an object's properties are affected by what you name it.
That is a rather distorted interpretation of what I was saying - will
this be the basis of the strawman argument which you will subsequently
knock down?
> > In Crother et al's view,
> > coordinates can be given a physical significance a priori - e.g. if
> > the direction of t is defined to be timelike, and the direction of r
> > is defined to be spacelike when a problem is stated, then it must
> > retain these properties in the solution.
>
> How do you "define [a coordinate] to be timelike"? What does that mean?
> You can *say* "t is timelike," but why should that affect anything?
Hopefully I have clarified what I mean in my response to Jan.
> > When trying to find the metric corresponding to a centrally symmetric
> > gravitational field, the general solution can be written in the form
> > (see e.g. Landau & Lifschitz, Vol 2, p299):
> > (1) ds^2 = h(r,t) dr^2 + k(r,t) (sin^2 theta d phi^2 + d theta^2) +
> > l(r,t) dt^2 + a(r,t) dr dt
> > where r labels 'spheres' and t labels 'time' in some way. Without loss
> > of generality, coordinate transformations can be applied to bring this
> > to the form:
> > (2) ds^2 = exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta d
> > phi^2) - exp(lambda(r,t)) dr^2
>
> Not if you believe that "coordinates can be given a physical significance
> a priori"! If you believe this, you don't have any business applying
> coordinate transformations to bring a general metric to a more restricted
> form, and you *certainly* don't have any business claiming that this is
> done "without any loss of generality."
Okay, I admit that my choice of words was poor. However, (2) is indeed
what I consider to be the general case for this problem, though that
is admittedly somewhat less general than the general case which you
have in mind.
>
> In fact, the transformation you have made eliminates Painleve-Gullstrand
> coordinates
Good, because I don't think these are appropriate coordinates to use
in what is supposed to be a problem with central symmetry.
>, a nice set of coordinates which are well-behaved through
> the horizon and in which t remains a time coordinate everywhere. By
> parametrizing the metric the way you do, you've also changed h(r,t) and
> l(r,t), whose signs are arbitrary in your original metric, to positive
> functions. That's certainly *not* "without loss of generality," unless
> you allow nu and lambda to be complex.
actually h is negative and l positive (just nitpicking).
>
> This is all fine if you allow me to switch back to better coordinates
> after the field equations are solved.
What are better coordinates are of course a matter of opinion -
especially if you suspect like me that the physical world is not
necessarily quite as coordinate-independent as GR would like it to be.
> But you don't want to do that --
> you want to allow coordinate transformations before you write down the
> field equations, but not after.
>
> > where the new r and new t still label spheres and time respectively.
>
> Why? What tells you this, apart from the names you have chosen for them?
By simply looking at the expression of the metric in terms of the said
coordinates. What else?
> > The point to note at this point is that r and t are _not_ merely
> > labels - they have a definite physical attribute determined by their
> > respective spacelike and timelike natures.
>
> What physical attributes are those? How do you determine them? Or is
> this more magical thinking: you chant, "t shall be time" and that makes
> it time?
If we choose spacetime index (+,---), then if in a given coordinate
system the line element looks like this:
ds^2 = A dt^2 - r^2 dOmega^2 - B dr^2
where A and B are positive, then t parametrises time and r
parametrises radius.
>
> [...]
>
> > However, while it is true that general relativity does not care about
> > the coordinate system used, in the formulation of the problem above, a
> > physical meaning has indeed been attached to the coordinates r and t a
> > priori,
>
> How? What attaches this meaning, apart from "because I say so"?
The formulation of the problem implies this interpretation of the
coordinates.
> > The conclusion to be reached then is that analytic continuation of a
> > solution is only valid if it respects the physical constraints of the
> > original problem. If the original constraints are ignored, well, a
> > different problem is being solved, with the possibly that the
> > resulting solutions are unphysical. [If Carlip et al disagree with
> > this conclusion, then I would very much like to see a proof that the
> > interior solution does indeed solve the original problem and
> > corresponds to a physically realisable vacuum].
>
> I've answered this in the past. If you want to ask whether the solution
> is physically reasonable, set up physically reasonable initial data and
> see how they evolve. In this case, for example, you can write down an
> exact solution for a collapsing spherically symmetric shell or sphere of
> matter, with a time coordinate that has the direct physical meaning as
> the proper time as measured by an observer standing on the surface of the
> collapsing matter. In the resulting exact solution the matter passes
> through the horizon with nothing strange happening, and the solution
> evolves into an ordinary Schwarzschild black hole -- both exterior and
> interior -- typically described in Painleve-Gullstrand coordinates. You
> can find two simple examples in Adler, Bjorken, Chen, and Liu, "Simple
> analytical models of gravitational collapse," American Journal of Physics
> 73 (December 2005) 1148-1159.
>
> Steve Carlip
Yes, it is clear that such coordinates give a nice smooth evolution
from exterior to interior. I just have doubts that that evolution
corresponds to physical reality - and partly because of the
discontinuity in the transformation to Schwarzschild (or similar)
coordinates at the event horizon. Why are there no coordinate choices
in which the metric remains diagonal at every point on the path of a
radially infalling particle and which does not suffer from some kind
of discontinuity at the horizon?
The answer, "Who cares, coordinates are just labels in GR" makes me
think not that there is something wrong with the Schwarzschild
coordinates, but that there is something wrong with the background
independence assumptions of GR. As Orwell, would have said, "all
coordinate systems are equal, but some coordinate systems are more
equal than others".
And I apologise if it appears that I am just trying to wind you up -
that is not the case.
- Sabbir.
Tom Roberts
>>> Without loss
>>> of generality, coordinate transformations can be applied to bring this
>>> to the form:
>>> (2) ds^2 = exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta d
>>> phi^2) - exp(lambda(r,t)) dr^2
>
> what I consider to be the general case for this problem,
Except it isn't. Your "without loss of generality" above is wrong, and
you DID lose generality -- you fixed signs that were not originally
fixed. This, of course, is the essence of your "argument", and it is
just plain wrong.
The CORRECT form, which is truly without loss of generality is:
ds^2 = K exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta
d phi^2) - K exp(lambda(r,t)) dr^2
where K is either +1 or -1.
Simple examination of this Ansatz shows it does indeed satisfy the
requirements of spherical symmetry for either value of K.
When K=+1, the solution you get is valid only in the region r>2M. When
K=-1, the solution you get is valid only in the region 0<r<2M. When
K=-1, r is clearly the time coordinate, and there is no solution
independent of r. Your desire that there be a static solution is not
met, and you cannot impose your wishes onto the mathematics.
> If we choose spacetime index (+,---), then if in a given coordinate
> system the line element looks like this:
> ds^2 = A dt^2 - r^2 dOmega^2 - B dr^2
> where A and B are positive, then t parametrises time and r
> parametrises radius.
Yes. This is so independent of your wishes and desires. It is also
independent of how you happen to label the coordinates. But it is NOT
the general solution -- for that you must allow the signs of A and B to
be either + or - (as above, they must have the same sign, and continuity
implies they must never change sign within the region of validity of the
equation and its solution -- that is what determines the region of
validity).
>>> a
>>> physical meaning has indeed been attached to the coordinates r and t a
>>> priori,
>> How? What attaches this meaning, apart from "because I say so"?
>
> The formulation of the problem implies this interpretation of the
> coordinates.
Hmmm. The signs of dt^2 and dr^2 in the line element determine which is
timelike and which is spacelike. In limiting yourself to K=+1, you have
restricted the set of solutions without justification. Your ARBITRARY
choice to do that does not mean that the MATHEMATICS has no such
solution(s), it just means that YOU will remain ignorant of them. But
you cannot impose your ignorance on everybody else, or on the
mathematics. <shrug>
Tom Roberts
LEJ Brouwer
> Hmmm. The signs of dt^2 and dr^2 in the line element determine which is
> timelike and which is spacelike. In limiting yourself to K=+1, you have
> restricted the set of solutions without justification. Your ARBITRARY
> choice to do that does not mean that the MATHEMATICS has no such
> solution(s), it just means that YOU will remain ignorant of them. But
> you cannot impose your ignorance on everybody else, or on the
> mathematics. <shrug>
>
> Tom Roberts
You are either misunderstanding or simply ignoring what I am saying.
- Sabbir.
JanPB
Tom's answer is the only logical one given what you wrote. There must
be something else on your mind that you never wrote down - this cannot
be seen from your posts. What you wrote was perfectly straightforward
and had a perfectly straightforward answer:
You wrote:
" >>> Without loss
>>> of generality, coordinate transformations can be applied to bring this
>>> to the form:
>>> (2) ds^2 = exp(nu(r,t)) c^2 dt^2 - r^2 (d theta^2 + sin^2 theta d
>>> phi^2) - exp(lambda(r,t)) dr^2
> (2) is indeed
> what I consider to be the general case for this problem, [...]"
This is just WRONG. There was nothing else Tom could have said to
that.
Write down all the assumptions. It's impossible to guess them.
--
Jan Bielawski
LEJ Brouwer
I have already tried to explain in a response to another of your
messages what I think the two problems being solved are and the
assumptions made in each, why they are different, why one of them
fixes the signs in front of the various terms in the line element
while the other does not, and why I have reservations about the
physicality of the problem whose solution which includes the interior
region. Perhaps I should have said "without gain in generality"
instead of "without loss of generality" above. :)
Daryl McCullough
>I have already tried to explain in a response to another of your
>messages what I think the two problems being solved are and the
>assumptions made in each, why they are different, why one of them
>fixes the signs in front of the various terms in the line element
>while the other does not, and why I have reservations about the
>physicality of the problem whose solution which includes the interior
>region. Perhaps I should have said "without gain in generality"
>instead of "without loss of generality" above. :)
Here's what I think the controversy is with the interior solution.
If you adopt the convention that "t" denotes a timelike coordinate
(g_tt > 0) and "r" denotes a spacelike coordinate (g_rr < 0), then
the interior solution can be written in the following form:
(2m-t)/t dt^2 - t/(2m-t) dz^2 - (2m-t)^2 dOmega^2
This is obtained from the usual Schwarzschild line element
by making the substitutions t' = (2m-r) and z' = t, and then
dropping the primes. I used 2m-r instead of r because
"forward in time" in the interior solution means *decreasing*
values of r, rather than increasing. Also, I called the new
spatial coordinate z' instead of r' because it is *not* a
radius of any sort.
Where I believe the uneasiness is coming from is the fact
that the angular part of the line element has a coefficient
that depends on the time coordinate t, rather than the space
coordinate z.
If you have some starting point, P, and you consider the
set of all points P' that are related to P by a rotation,
then that set is a sphere of area 4 pi (2m-t)^2. So the
effective "radius" of that sphere is R(t) = (2m-t).
R here is *not* a coordinate, since it is a function of t.
So it's not to be confused with the spatial coordinate z.
When you conclude that only the exterior solution is physical,
you are implicitly assuming that the effective radius R must
be a function of the spatial coordinate. But there is no
reason to assume that.
Perhaps it becomes clearer if we drop the dimensions of spacetime
down to 3. Here's an analogous 3-D spacetime:
ds^2 = dt^2 - dz^2 - R(t)^2 dTheta^2
In this case, if P is some starting point, then the
set of all points P' that are related to P by a rotation
is a circle of circumference 2 pi R(t). If we look at
slices of constant t, then the geometry looks like this:
ds^2 = -dz^2 - R(t)^2 dTheta^2
Because the radius does not depend on the spatial coordinate z,
what this describes is *not* a sphere but a *cylinder*. The
radius of that cylinder is a function of time. The spatial
coordinate z measures distance *along* the cylinder.
In a similar way, look at the 4-D spacetime with line element
(2m-t)/t dt^2 - t/(2m-t) dz^2 - (2m-t)^2 dOmega^2
If you look at constant-time slices, the geometry looks like
this:
ds^2 = -A(t) dz^2 - R(t)^2 dOmega^2
This geometry is a 3-D analog of a cylinder. Instead of
there being a circle attached to each value of z, as is
the case with a 2-D cylinder, there is a sphere of
radius R(t) attached to each value of z.
Once again, you seem to be assuming that the effective
radius R of the spheres of symmetry must be a function
of the spatial coordinate(s). But there is no reason to
assume that. It can be a function of the temporal coordinate,
also.
--
Daryl McCullough
Ithaca, NY
LEJ Brouwer
wrote:
But this is indeed an assumption made in one of the two problems I
discussed, namely the metric of a centrally symmetric gravitational
field. And it is a very natural assumption to make in that context. I
am not sure why anyone would think otherwise (well, if it were not for
the fact that they had already read the supposedly 'correct' answer
given in the textbooks).
Moreover, since we are using the convention that spacetime has
signature (+,---), this means that coordinates (t,x,y,z) can be chose
such that locally,
ds^2 = dt^2 - dx^2 - dy^2 - dz^2
and by transforming from cartesians to spherical polars (r,theta,phi)
for the spatial coordinates, we can always find coordinates such that:
ds^2 = dt^2 - dr^2 - r^2 (dtheta^2 + sin^2(theta) dphi^2)
This is why we restrict ourselves to solutions with the signs
mentioned earlier. In the exterior solution, this form can be obtained
simply by rescaling r and t as appropriate.
In the interior solution, the metric takes the form,
ds^2 = - A dt^2 + B dr^2 + r^2 (dtheta^2 + sin^2(theta) dphi^2)
where A and B are positive. But since spacetime has signature (+,---),
r here appears to be the timelike coodinate, and t the spacelike
coordinate, but as I mentioned earlier if that is the case, then the
interior solution:
(i) does not have the (spatial) central symmetry desired, though it
does have a different and hard-to-interpret SO(3) symmetry involving
time and the 'angular' coordinates (which of course no longer have the
interpretations of angles in the usual sense)
(ii) does not have a 'centre' - the singularity is not at a spatial
point, so it cannot be the solution outside of a pointlike mass. This
is an indication that pointlike masses do not exist in GR
(iii) has a singularity which is in the temporal _future_ (i.e. larger
value of proper time) _relative_ to the time at which an infalling
particle enters the horizon, so it is not a point in any sense of the
word - not even a 'temporal' point.
I have also complained that one cannot find transformations to
coordinates in which the metric is locally flat in a continuous
manner, which indicates that the gluing of the interior solution to
the interior solution is an artificial one. This should already be
clear from the different nature of the SO(3) symmetry in both patches,
but consider also the following:
Worse still, if we take the interior metric and convert back to
cartesian coordinates and rescale r and t to get the locally flat
metric, we get:
ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
so the signature of the metric is not (+,---), but (-,+++), so it not
actually spacetime according to our original convention. How can the
_signature_ of spacetime suddenly change like this along a smooth
path? I would conclude that a particle cannot pass from the exterior
to the interior, and that according to our signature conventions for
spacetime, the so-called 'interior solution' does not exist. Changing
the coordinates to fix the signature would just give the exterior
solution again.
> Perhaps it becomes clearer if we drop the dimensions of spacetime
> down to 3. Here's an analogous 3-D spacetime:
>
> ds^2 = dt^2 - dz^2 - R(t)^2 dTheta^2
>
> In this case, if P is some starting point, then the
> set of all points P' that are related to P by a rotation
> is a circle of circumference 2 pi R(t).
It has the same U(1) symmetry that a circle has, but it is not a
circle in the usual sense.
Nice to hear from you again,
- Sabbir.
LEJ Brouwer
> In the interior solution, the metric takes the form,
>
> ds^2 = - A dt^2 + B dr^2 + r^2 (dtheta^2 + sin^2(theta) dphi^2)
My apologies - the sign should be negative in front of r^2 here, which
means that the signature is no longer obvious - so please disregard my
comments regarding that for now!
Daryl McCullough
>> Here's what I think the controversy is with the interior solution.
>> If you adopt the convention that "t" denotes a timelike coordinate
>> (g_tt > 0) and "r" denotes a spacelike coordinate (g_rr < 0), then
>> the interior solution can be written in the following form:
>>
>> (2m-t)/t dt^2 - t/(2m-t) dz^2 - (2m-t)^2 dOmega^2
>> Where I believe the uneasiness is coming from is the fact
>> that the angular part of the line element has a coefficient
>> that depends on the time coordinate t, rather than the space
>> coordinate z.
>>
>> If you have some starting point, P, and you consider the
>> set of all points P' that are related to P by a rotation,
>> then that set is a sphere of area 4 pi (2m-t)^2. So the
>> effective "radius" of that sphere is R(t) = (2m-t).
>>
>> R here is *not* a coordinate, since it is a function of t.
>> So it's not to be confused with the spatial coordinate z.
>> When you conclude that only the exterior solution is physical,
>> you are implicitly assuming that the effective radius R must
>> be a function of the spatial coordinate. But there is no
>> reason to assume that.
>
>But this is indeed an assumption made in one of the two problems I
>discussed, namely the metric of a centrally symmetric gravitational
>field.
Well, ultimately the physics is what tells us which assumptions
are reasonable, and which are not. General Relativity tells us
that we *can't* have the effective radius be a function of a
spacelike coordinate everywhere.
>Moreover, since we are using the convention that spacetime has
>signature (+,---), this means that coordinates (t,x,y,z) can be chose
>such that locally,
>
>ds^2 = dt^2 - dx^2 - dy^2 - dz^2
Yes, that's certainly true.
>and by transforming from cartesians to spherical polars (r,theta,phi)
>for the spatial coordinates, we can always find coordinates such that:
>
>ds^2 = dt^2 - dr^2 - r^2 (dtheta^2 + sin^2(theta) dphi^2)
You can always find coordinates such that the metric looks
like that *locally*. And that's true of the Schwarzschild
spacetime, also. Both inside and outside the event horizon.
>In the interior solution, the metric takes the form,
>
>ds^2 = - A dt^2 + B dr^2 + r^2 (dtheta^2 + sin^2(theta) dphi^2)
>where A and B are positive. But since spacetime has signature (+,---),
>r here appears to be the timelike coodinate, and t the spacelike
>coordinate, but as I mentioned earlier if that is the case, then the
>interior solution:
>
>(i) does not have the (spatial) central symmetry desired, though it
>does have a different and hard-to-interpret SO(3) symmetry involving
>time and the 'angular' coordinates (which of course no longer have the
>interpretations of angles in the usual sense)
That's correct. The interior solution is harder to interpret than
the exterior solution. However, *locally* the interior solution
looks like Minkowsky space. Locally, it has the metric
ds^2 = dt^2 - dx^2 - dy^2 - dz^2. The part that's hard to interpret
is how different local regions of spacetime "fit together".
>(ii) does not have a 'centre' - the singularity is not at a spatial
>point, so it cannot be the solution outside of a pointlike mass. This
>is an indication that pointlike masses do not exist in GR
>(iii) has a singularity which is in the temporal _future_ (i.e. larger
>value of proper time) _relative_ to the time at which an infalling
>particle enters the horizon, so it is not a point in any sense of the
>word - not even a 'temporal' point.
Okay.
>I have also complained that one cannot find transformations to
>coordinates in which the metric is locally flat in a continuous
>manner,
You can't transform from Schwarzschild coordinates to local
Minkowsky coordinates at the event horizon because the Schwarzschild
coordinates are singular there. But I don't see why that's a problem.
It's rare that you can describe curved spacetime using a single
nonsingular coordinate system. The more general case requires
several overlapping patches to describe the whole manifold.
>which indicates that the gluing of the interior solution to
>the interior solution is an artificial one.
It looks artificial in Schwarzschild coordinates, but not in
other coordinates (such as Kruskal).
>Worse still,
I'm not sure why you're saying "worse". There is nothing
bad about anything you've said so far. It's just that things
aren't as simple as they are in good old Minkowsky spacetime.
>if we take the interior metric and convert back to
>cartesian coordinates and rescale r and t to get the locally flat
>metric, we get:
>
>ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
No, that's not correct. You cannot convert the Schwarzschild
metric into one of that form by using a real-valued coordinate
transformation. A coordinate transformation is never going to
change the signature.
>so the signature of the metric is not (+,---), but (-,+++), so it not
>actually spacetime according to our original convention. How can the
>_signature_ of spacetime suddenly change like this along a smooth
>path?
It doesn't. The signature is the same on both sides of the
event horizon. In the exterior, t is timelike (+) and r, theta
and phi are spacelike (-). In the interior, t is spacelike (-),
r is timelike (+) and theta and phi are spacelike (-). So on
both sides, the signature is +---.
LEJ Brouwer
> >discussed, namely the metric of a centrally symmetric gravitational
> >field.
>
> Well, ultimately the physics is what tells us which assumptions
> are reasonable, and which are not. General Relativity tells us
> that we *can't* have the effective radius be a function of a
> spacelike coordinate everywhere.
What the mathematics has shown is that the interior solution does not
have spherically symmetry about a point, and it is quite unclear as to
what physical matter configuration could give rise to the properties
described by the interior solution - it is certainly not due to a
point mass. Could you explain what matter configuration the interior
solution describes?
The exterior solution does have a spatial spherical symmetry - and
this symmetry only holds as far as the event horizon. The
Schwarzschild coordinates are special because one can transform to
coordinates in which the metric is locally Minkowski by simply
rescaling r and t. Their physical intepretation is therefore clear.
The fact that Kruskal and related coordinate systems cannot be
transformed to Schwarzschild (or locally flat) coordinates at the
event horizon indicates that the event horizon itself corresponds to
the 'edge' of the solution manifold. To suggest that an infalling
particle will somehow pass beyond this edge into the interior manifold
which does not even share the same spherical symmetry as the exterior
defies reason. If the horizon were not an edge of the manifold, then
there should be a one-to-one (and certainly not one-to-four)
transformation from Kruskal coordinates to locally Minkowski
coordinates (up to a Lorentz transformation).
On the other hand, what does happen at the event horizon needs to be
investigated, bearing in mind that the exterior solution has two
sheets. My guess is that a particle is reflected off the event
horizon onto the second with space and time directions reversed (i.e.
a radially infalling particle will be reflected outwards travelling
backwards in time - i.e. it will look like a radially infalling
antiparticle, so that the overall process looks like a particle-
antiparticle annihilation occuring at the horizon).
Because the Schwarzschild coordinates are easy to interpret
physically. The fact that they break down at the horizon indicates
that the horizon is the edge of the solution manifold.
> It's rare that you can describe curved spacetime using a single
> nonsingular coordinate system. The more general case requires
> several overlapping patches to describe the whole manifold.
Yes, but that is not quite analogous to the situation here.
> >which indicates that the gluing of the interior solution to
> >the interior solution is an artificial one.
>
> It looks artificial in Schwarzschild coordinates, but not in
> other coordinates (such as Kruskal).
The passage of time and the measurement of distance are physically
meaningful, and the mapping from Kruskal coordinates to locally
Minkowski coordinates is not possible for Kruskal or Schwarzschild or
any other at the horizon.
> >Worse still,
>
> I'm not sure why you're saying "worse". There is nothing
> bad about anything you've said so far.
Surely you must be joking, Mr McCullough?
> It's just that things
> aren't as simple as they are in good old Minkowsky spacetime.
[I had already conceded that I made a sign error regarding the
signature of the metric, so I am not going to discuss it]. I do not
understand how you cannot be concerned when you have conceded that the
interior solution does not have a central spherical symmetry. All of
the textbooks and the understanding of modern general relativists is
just wrong. We had it right pre-1960, and misinterpreted the Kruskal
extension - which is not an extension but rather the sewing of two
physically distinct manifolds with boundary having an SO(3) symmetry.
The fact that we continue to accepted the standard picture without
question (and rebuke anyone who dares to question it) is merely a
reflection of our sheep-like mentality and our unfortunate lack of
critical thought. There are so many warning signs that something is
going horribly wrong, yet we ignore them simply because everyone else
is reassuring us (and each other) that it is okay.
- Sabbir.
Daryl McCullough
>What the mathematics has shown is that the interior solution does not
>have spherically symmetry about a point, and it is quite unclear as to
>what physical matter configuration could give rise to the properties
>described by the interior solution - it is certainly not due to a
>point mass. Could you explain what matter configuration the interior
>solution describes?
This has been discussed by Steven Carlip. Here's an intuitive answer:
Imagine sitting on Earth, and surrounded by a spherical shell of
distant stars. Now, imagine attaching powerful rockets to each star
in the shell so that at the same moment (according to the coordinate
system in which the Earth is at rest), the stars are all sent hurtling
towards the earth. The rockets are timed so that all the stars will
arrive simultaneously.
Now, what you can do is add up the masses of all the stars and compute
the Schwarzschild radius for that mass. At some point, the stars will
pass that radius. After that, you and the Earth and all those stars
are inside a black hole.
There still isn't any point mass anywhere except in the *future*.
The predictions of General Relativity are that all the mass inside
the spherical shell will eventually be concentrated into a single
point. That's the singularity, and it exists in the *future*, not
in any particular spatial location.
From the point of view of someone outside the spherical shell of
stars, the geometry will approach that of the Schwarzschild exterior
solution.
>The exterior solution does have a spatial spherical symmetry - and
>this symmetry only holds as far as the event horizon. The
>Schwarzschild coordinates are special because one can transform to
>coordinates in which the metric is locally Minkowski by simply
>rescaling r and t. Their physical intepretation is therefore clear.
I'm not sure what you mean by that. You can always transform
coordinates to get coordinates that are locally Minkowskian.
>The fact that Kruskal and related coordinate systems cannot be
>transformed to Schwarzschild (or locally flat) coordinates at the
>event horizon indicates that the event horizon itself corresponds to
>the 'edge' of the solution manifold.
No, it doesn't. Once again, look at this analogous case.
Suppose we have a spacetime metric that looks like this
ds^2 = X/2m dT^2 - 2m/X dX^2
This 2D spacetime metric has an "event horizon" at X=0, in the
sense that
(1) An object in freefall at rest at X > 0 will asymptotically
approach X = 0 as T --> infinity.
(2) An object that starts off in the region X < 0 can never
reach the region X > 0.
(3) The force F(X) required to keep an object stationary at a
constant value of X goes to infinity as X --> 0.
On the other hand, this metric is easily seen to be just
the Minkowski metric written in unusual accelerated coordinates:
If you let x = square-root(8m X) cosh(T/4m) and let
t = square-root(8m X) sinh(T/4m), then the line element
in terms of x and t is just
ds^2 = dt^2 - dx^2
To see this: dx = square-root(2m/X) cosh(T/4m) dX
+ square-root(X/2m) sinh(T/4m) dT
dt = square-root(2m/X) sinh(T/4m) dX
+ square-root(X/2m) cosh(T/4m) dT
dt^2 - dx^2 = -(2m/X) dX^2 + X/2m dT^2
The event horizon has disappeared by changing the coordinates,
showing that it's a *coordinate* singularity, not a curvature
singularity.
The event horizon for Schwarzschild coordinates has *exactly*
the same character. To see this, start with the full Schwarzschild
metric:
ds^2 = (r-2m)/r dt^2 - r/(r-2m) dr^2 - r^2 dOmega^2
Now, for radially infalling objects, dOmega = 0, so it
reduces to
ds^2 = (r-2m)/r dt^2 - r/(r-2m) dr^2
Now, change coordinates from r to X = r-2m:
ds^2 = X/(X+2m) dt^2 - (X+2m)/X dX^2
Now, when the particle is very close to the event horizon, so
X << 2m, we can approximate X+2m by just 2m, and this becomes,
approximately
ds^2 = X/2m dt^2 - 2m/X dX^2
The four regions of Kruskal spacetime have close analogies
in this 2D metric:
Region I, or the exterior solution of
the Schwarzschild solution, corresponds to the
region x > 0, -x < t < +x.
Region II, or the interior solution, corresponds
to the region t > 0, -t < x < +t.
Region III, or the mirror exterior solution, corresponds
to the region x < 0, -|x| < t < |x|.
Region IV, or the "white hole" solution, corresponds
to the region t < 0, -|t| < x < |t|
The separation between the regions is the
"event horizon", which is defined by X = 0,
which corresponds to the locus of points
in which x^2 = t^2.
Note: any object initially at "rest" in region IV will
eventually "fall" into Region I or Region III. So it's
like a white hole, repelling matter out of it. Any object
initially at rest in region I or region III will eventually
"fall" into region II. So it's like a black hole, drawing
all matter into it.
>To suggest that an infalling particle will somehow pass beyond
>this edge into the interior manifold which does not even share
>the same spherical symmetry as the exterior defies reason.
To suggest *otherwise* defies reason. There is no *reason* for
an infalling particle *not* to pass through the event horizon.
From the point of view of the infalling particle, there is nothing
special about the event horizon that could possibly *prevent* it
from falling through.
>If the horizon were not an edge of the manifold, then
>there should be a one-to-one (and certainly not one-to-four)
>transformation from Kruskal coordinates to locally Minkowski
>coordinates (up to a Lorentz transformation).
There is no problem transforming Kruskal coordinates to
locally Minkowski coordinates. I think what you are talking
about is the fact that there are four regions of the Kruskal
spacetime.
>On the other hand, what does happen at the event horizon needs to be
>investigated, bearing in mind that the exterior solution has two
>sheets. My guess is that a particle is reflected off the event
>horizon onto the second with space and time directions reversed (i.e.
>a radially infalling particle will be reflected outwards travelling
>backwards in time - i.e. it will look like a radially infalling
>antiparticle, so that the overall process looks like a particle-
>antiparticle annihilation occuring at the horizon).
You don't have to *guess*. You can look at what the spacetime
looks like in the neighborhood of the event horizon.
>> You can't transform from Schwarzschild coordinates to local
>> Minkowsky coordinates at the event horizon because the Schwarzschild
>> coordinates are singular there. But I don't see why that's a problem.
>
>Because the Schwarzschild coordinates are easy to interpret
>physically.
No, I would say they are definitely *not* easy to interpret,
except in the region in which r is much greater than the event
horizon, in which case it looks like Minkowsky space. Close
to the event horizon, the meaning of Schwarzschild coordinates
is very difficult to interpret.
>The fact that they break down at the horizon indicates
>that the horizon is the edge of the solution manifold.
No, it doesn't. Once again, look at the metric
ds^2 = X/2m dT^2 - 2m/X dX^2
The fact that this metric breaks down at X=0 doesn't
tell us *anything*, because this is *just* Minkowsky
spacetime written using accelerated coordinates.
Near the event horizon, Schwarzschild coordinates are
accelerated coordinates, just like X and T in the
above 2D case. Kruskal coordinates are the corresponding
well-behaved coordinates.
>> It's rare that you can describe curved spacetime using a single
>> nonsingular coordinate system. The more general case requires
>> several overlapping patches to describe the whole manifold.
>
>Yes, but that is not quite analogous to the situation here.
Yes, it certainly is. That's exactly what's going on here.
The interior solution is a perfectly good solution of the
Einstein field equations. The exterior solution is a perfectly
valid solution. It's just a matter of gluing them together.
The "glue" has to be a coordinate system that is continuous
across the event horizon. Kruskal coordinates work for that
purpose. Or we can use local Minkowski coordinates.
>The passage of time and the measurement of distance are physically
>meaningful
Yes, but t is *not* the physically measurable time, and r is
not a physically measurable distance. There is nothing physical
about the Schwarzschild coordinates.
>and the mapping from Kruskal coordinates to locally
>Minkowski coordinates is not possible for Kruskal or Schwarzschild or
>any other at the horizon.
That's plainly not true. In terms of Kruskal coordinates, the
metric looks like this:
ds^2 = 32m^3/r exp(-r/2m) (dT^2 + dR^2) - r^2 dOmega^2
where r is an implicit function of T and R.
At the event horizon r=2m, so the metric reduces to
ds^2 = 16 m^2 exp(-1) (dT^2 + dR^2) - 4 m^2 dOmega^2
There is no problem rescaling this to get Minkowski coordinates.
>> I'm not sure why you're saying "worse". There is nothing
>> bad about anything you've said so far.
>
>Surely you must be joking, Mr McCullough?
No, I'm not. You really haven't given any reason to be
suspicious of the interior solution *other* than the fact
that it is contrary to your intuitions. From the point of
view of a freefalling test particle, there is nothing
weird going on anywhere in the Schwarzschild spacetime
*except* at the singularity at r=0. There are weird
*global* properties, but nothing weird locally. And
all laws of physics are local. There is nothing local
at the event horizon to prevent someone from falling
through.
>> It's just that things
>> aren't as simple as they are in good old Minkowsky spacetime.
>
>[I had already conceded that I made a sign error regarding the
>signature of the metric, so I am not going to discuss it]. I do not
>understand how you cannot be concerned when you have conceded that the
>interior solution does not have a central spherical symmetry.
There is a pointlike singularity in the interior solution, but
it occurs in the *future*, rather than at any particular point
in space.
>All of the textbooks and the understanding of modern general
>relativists is just wrong.
>We had it right pre-1960, and misinterpreted the Kruskal
>extension - which is not an extension but rather the sewing of two
>physically distinct manifolds with boundary having an SO(3) symmetry.
Look, the whole point of Riemannian geometry is that a general
manifold is created by sewing together pieces that locally look
like sections of good old Euclidean. General Relativity says,
similarly, that every spacetime manifold can be created by
sewing together pieces that locally look like a section of
Minkowski spacetime. Any such sewing job is a legitimate
solution to General Relativity, provided that the field equations
are satisfied.
The Kruskal solution is a perfectly good solution (except possibly
at r=0). In contrast, the Schwarzschild solution is *not* a perfectly
good solution. That's because the Schwarzschild coordinates are
singular at r=2m, even though the curvature tensor is perfectly
well-defined there.
>The fact that we continue to accepted the standard picture without
>question (and rebuke anyone who dares to question it)
Why do you say that people are accepting it without question?
The reason people changed their minds about it was because
they *studied* it from many different angles, using many different
methods and many different coordinate systems, and they discovered
that they were *wrong* to think of a black hole as a "frozen star".
In contrast, the people who are saying otherwise (like you)
*haven't* studied Kruskal coordinates, haven't studied differential
geometry, haven't studied Rindler coordinates (with their apparent
"event horizon" in ordinary Minkowski spacetime).
>is merely a reflection of our sheep-like mentality
>and our unfortunate lack of critical thought.
There has been plenty of critical thought about black holes,
and it's still going on. People still don't know how black
holes interact with quantum mechanics. And certainly, the
various approaches to unifying quantum mechanics and General
Relativity may very well show that GR is *wrong* about black
holes. However, there is nothing in your arguments that say
anything about quantum effects. *If* you were arguing that
quantum effects change the picture, then I would not be
arguing against it. You haven't been arguing that. You've
been arguing about *coordinates*. Your argument assumes that
there is something special about Schwarzschild coordinates.
There isn't anything special about them. Any physically
meaningful claim *must* be expressable in whatever coordinate
system you please. The idea that an object slows to a halt
at the event horizon or worse reflects off it may have a
certain plausibility when considering Schwarzschild coordinates,
but it is completely mysterious and unmotivated looked at in
any other coordinate system.
I don't think that *you* have put very much critical thought
into it. You seem completely unable or unwilling to get past
Schwarzschild coordinates. If there is something weird happening
at the event horizon then *demonstrate* it using Kruskal coordinates.
Tom Roberts
> What the mathematics has shown is that the interior solution does not
> have spherically symmetry about a point,
This is not true. The interior Schw. solution is spherically symmetric,
but it is not static.
OK, technically you are correct about the entire manifold --
the spherical symmetry is NOT "about a point", because the
singularity must be deleted from the manifold and is not
therefore a point. But I don't think this is what you
meant at all (based on further misconceptions below).
IOW: the interior solution is spherically symmetric in the sense of an
SO(3) isometry group.
> and it is quite unclear as to
> what physical matter configuration could give rise to the properties
> described by the interior solution - it is certainly not due to a
> point mass.
Yes, it is. Or at least it is as close as one can come to such a
physical situation in GR.
> Could you explain what matter configuration the interior
> solution describes?
It is as close as one can come to a universe consisting of vacuum
surrounding a point mass. All classical theories of physics have
problems with such a "point object"....
> The exterior solution does have a spatial spherical symmetry - and
> this symmetry only holds as far as the event horizon.
Again not true. The spherical symmetry is valid throughout the entire
manifold.
[This is not "spherical symmetry about a point", but it is
spherical symmetry in the sense of an SO(3) isometry group.]
> The
> Schwarzschild coordinates are special because one can transform to
> coordinates in which the metric is locally Minkowski by simply
> rescaling r and t. Their physical intepretation is therefore clear.
This is not unique -- ALL orthogonal coordinate systems are "special" in
this way (but you got the details wrong: one must actually re-scale all
four coordinates, including the Schw. theta and phi).
You REALLY need to learn about the distinction between coordinates and
geometry.
> The fact that Kruskal and related coordinate systems cannot be
> transformed to Schwarzschild (or locally flat) coordinates at the
> event horizon indicates that the event horizon itself corresponds to
> the 'edge' of the solution manifold.
Again you are wrong, in several ways:
A) The horizon is the _boundary_ of the region of validity of either
Schw. coordinate chart, not of the manifold itself.
B) _NO_ coordinate system can be "transformed to locally flat
coordinates" ANYWHERE in the manifold, for the simple reason
that the manifold is not flat anywhere. This includes both
sets of Schw. coordinates.
C) Kruskal and related coordinate systems CAN be transformed to
local approximately-Minkowski coordinates at the horizon.
Indeed, ANY coordinates can be so transformed, ANYWHERE THEY
ARE VALID. It is the two sets of Schw. coordinates that cannot be
so transformed at the horizon, because they are not valid there.
D) Yes, no coordinate system can be transformed to Schw. coordinates
at the horizon, BECAUSE THE SCHW. COORDINATES ARE NOT VALID THERE.
This is a problem with the Schw. coordinates, not the others.
You REALLY need to get your "facts" right.
> To suggest that an infalling
> particle will somehow pass beyond this edge into the interior manifold
> which does not even share the same spherical symmetry as the exterior
> defies reason.[...]
You need to get your facts right. You have grave misconceptions about
this manifold and the Schw. coordinates.
> [... rest omitted due to failure to understand basic geometry or this
specific manifold]
It ought to be clear by now that the only "controversy" here is in LEJ
Brouwer's head. And, I suppose, in articles and textbooks that pre-dated
the current understanding of this manifold. LEJ Brouwer's confusions are
elementary, and are related to applying basic differential geometry to
the Schw. manifold.
Tom Roberts
heretic
And your intuition is right. It is incredible how GR textbooks dismiss the
event horizon as a "coordinate singularity" when it is an infinite redshift
surface in *all* coordinate systems. So hiding behind the principle of
general covariance doesn't cut it. It is not possible to find a global
diffeomorphism that makes the horizon go away, this includes Kruskal-Szekeres.
In another post the Painleve-Gullstrand solution was offered but it is not
static. So all we have, as you and others point out, is a patch of two distinct
space-times that is not continuously differentiable at the horizon in any
coordinate system.
GR Black hole believers practice a new type of mathematics, one in which
boundary conditions are selectively ignored. They choose to apply a boundary
condition at r=infinity but don't bother at r=0 (or even r=2GM/c^2). So they
end up solving two different problems and claiming they have solved the
originally posed problem.
BTW, perhaps nature creates "black holes" but how and what their properties
are is not established by the current GR mess.
heretic
The sign in front of r^2 Omega^2 is positive for both interior and
exterior solutions:
ds^2 = -(1-2GM/c^2/r)dt^2 + (1 - 2GM/c^2/r)^(-1)dr^2 + r^2 Omega^2
is the Hilbert solution 0 < r < infinity. Your original point about the
switch in timelike and spacelike coordinates stands.
heretic
> LEJ Brouwer wrote:
>> What the mathematics has shown is that the interior solution does not
>> have spherically symmetry about a point,
>
> This is not true. The interior Schw. solution is spherically symmetric,
> but it is not static.
>
> OK, technically you are correct about the entire manifold --
> the spherical symmetry is NOT "about a point", because the
> singularity must be deleted from the manifold and is not
> therefore a point. But I don't think this is what you
> meant at all (based on further misconceptions below).
>
> IOW: the interior solution is spherically symmetric in the sense of an
> SO(3) isometry group.
But it discontinously interchanges timelike and spacelike coordinates at the
event horizon. LEJ Brouwer should choose better wording of his argument,
but his argument is not about spherical symmetry.
>
>> and it is quite unclear as to
>> what physical matter configuration could give rise to the properties
>> described by the interior solution - it is certainly not due to a
>> point mass.
>
> Yes, it is. Or at least it is as close as one can come to such a
> physical situation in GR.
Proof by assertion. There is no radial curve extending from the point
to an observer outside the event horizon. In the classic Hilbert solution
the physical meaning is that (r,t) becomes (t,r) at the horizon. Using
other coordinates can't mask this fundamental transition.
>
>
>> Could you explain what matter configuration the interior
>> solution describes?
>
> It is as close as one can come to a universe consisting of vacuum
> surrounding a point mass. All classical theories of physics have
> problems with such a "point object"....
>
>
>> The exterior solution does have a spatial spherical symmetry - and
>> this symmetry only holds as far as the event horizon.
>
> Again not true. The spherical symmetry is valid throughout the entire
> manifold.
>
> [This is not "spherical symmetry about a point", but it is
> spherical symmetry in the sense of an SO(3) isometry group.]
>
>
>> The
>> Schwarzschild coordinates are special because one can transform to
>> coordinates in which the metric is locally Minkowski by simply
>> rescaling r and t. Their physical intepretation is therefore clear.
>
> This is not unique -- ALL orthogonal coordinate systems are "special" in
> this way (but you got the details wrong: one must actually re-scale all
> four coordinates, including the Schw. theta and phi).
>
> You REALLY need to learn about the distinction between coordinates and
> geometry.
You really need to stop changing the subject. The discontinuity at the
event horizon cannot be removed by any transformation. So there is no
local neighbourhood around each point on the horizon where a single piece
Minkowski map can be applied. There is a physical discontinuity at the
horizon and not merely some coordinate issue as claimed without proof by
assorted textbooks and those who religiously believe them.
>
>
>> The fact that Kruskal and related coordinate systems cannot be
>> transformed to Schwarzschild (or locally flat) coordinates at the
>> event horizon indicates that the event horizon itself corresponds to
>> the 'edge' of the solution manifold.
>
> Again you are wrong, in several ways:
> A) The horizon is the _boundary_ of the region of validity of either
> Schw. coordinate chart, not of the manifold itself.
Boundary conditions need to be applied to the solutions of the nonlinear
partial differential equations that constitute GR. Applying a single
boundary condition at r=infinity isn't mathematically (and physically)
sufficient. The outer and inner solutions are two distinct manifold
pieces in any coordinate system as evidenced by the fact that there is
no diffeomorphism which can make this infinite redshift surface go away.
> B) _NO_ coordinate system can be "transformed to locally flat
> coordinates" ANYWHERE in the manifold, for the simple reason
> that the manifold is not flat anywhere. This includes both
> sets of Schw. coordinates.
> C) Kruskal and related coordinate systems CAN be transformed to
> local approximately-Minkowski coordinates at the horizon.
Not true. Even in the contrived KS coordinates, which mix the timelike
and spacelike components, there is a discontinuous transition at the
horizon. The (r,t) to KS map is not continuously differentiable at the
event horizon. Since a single locally Minkowski manifold piece cannot
be determined in (r,t) coordinates at points on the horizon, it cannot
be found in KS coordinates also.
> Indeed, ANY coordinates can be so transformed, ANYWHERE THEY
> ARE VALID. It is the two sets of Schw. coordinates that cannot be
> so transformed at the horizon, because they are not valid there.
> D) Yes, no coordinate system can be transformed to Schw. coordinates
> at the horizon, BECAUSE THE SCHW. COORDINATES ARE NOT VALID THERE.
> This is a problem with the Schw. coordinates, not the others.
>
> You REALLY need to get your "facts" right.
Take your own advice. The event horizon is a physical singularity in
all coordinate systems. It is obviously not a point singularity but
it sure is no "coordinate singularity" as claimed by you and others.
>
>
>> To suggest that an infalling
>> particle will somehow pass beyond this edge into the interior manifold
>> which does not even share the same spherical symmetry as the exterior
>> defies reason.[...]
>
> You need to get your facts right. You have grave misconceptions about
> this manifold and the Schw. coordinates.
Actually it is up to you to show the physical meaning of a radial transition
across the horizon when the coordinate spacelike and timelike components
flip places. By Einstein's own directive you should be able to do this
in (r,t) coordinates since no coordinate system is special.
>
>
> > [... rest omitted due to failure to understand basic geometry or this
> specific manifold]
>
>
>
> It ought to be clear by now that the only "controversy" here is in LEJ
> Brouwer's head. And, I suppose, in articles and textbooks that pre-dated
> the current understanding of this manifold. LEJ Brouwer's confusions are
> elementary, and are related to applying basic differential geometry to
> the Schw. manifold.
Don't be so cocksure and ease off with the cheap ad hominems. You appear to
be completely unaware that that coordinate transforms have to be diffeomorphisms.
The physical singularity at the horizon cannot be transformed away.
Eric Gisse
[...]
>
> And your intuition is right. It is incredible how GR textbooks dismiss the
> event horizon as a "coordinate singularity" when it is an infinite redshift
> surface in *all* coordinate systems.
This is because you don't know what is being meant as a coordinate
singularity.
The Schwarzschild geometry behaves funny at r = 2GM/c^2 - g_tt goes to
zero and g_rr goes to infinity there. However, there are coordinate
systems in which the coordinates are well-behaved on both sides of the
horizon.
> So hiding behind the principle of
> general covariance doesn't cut it. It is not possible to find a global
> diffeomorphism that makes the horizon go away, this includes Kruskal-Szekeres.
> In another post the Painleve-Gullstrand solution was offered but it is not
> static. So all we have, as you and others point out, is a patch of two distinct
> space-times that is not continuously differentiable at the horizon in any
> coordinate system.
Since there isn't a coordinate transformation that makes the event
horizon go away that means it is a global property of the manifold. It
appears you understand this, just not what is being meant by
"coordinate singularity" - so you extend your misunderstanding to
conclude that there is no coordinate system that smoothy connects the
interior and exterior regions.
>
> GR Black hole believers practice a new type of mathematics, one in which
> boundary conditions are selectively ignored. They choose to apply a boundary
> condition at r=infinity but don't bother at r=0 (or even r=2GM/c^2). So they
> end up solving two different problems and claiming they have solved the
> originally posed problem.
The field equations reduce to first order differential equations in
the right coordinate system. There remains a _single_ undetermined
constant - which is determined, among other ways, by demanding that
the integrated energy density at infinity is equal to the enclosed
mass of the black hole.
What boundary conditions would you like to apply at r=0 which is a
global singularity? What boundary conditions would you like to apply
at the event horizon?
>
> BTW, perhaps nature creates "black holes" but how and what their properties
> are is not established by the current GR mess.
Really? I'll take that bet.
http://www.journals.uchicago.edu/ApJ/journal/issues/ApJ/v652n1/65488/brief/65488.abstract.html
Daryl McCullough
>Take your own advice. The event horizon is a physical singularity in
>all coordinate systems.
No, it is not. Take a look at the Krukal coordinates in the neighborhood
of the event horizon. Show me what is singular about them.
>> You need to get your facts right. You have grave misconceptions about
>> this manifold and the Schw. coordinates.
>
>Actually it is up to you to show the physical meaning of a radial transition
>across the horizon when the coordinate spacelike and timelike components
>flip places.
The fact that Schwarzschild coordinates switch spacelike and
timelike across the event horizon only means that Schwarzschild
coordinates are singular at the event horizon. They are a bad
choice of coordinates there.
Take a look at this 2D spacetime metric:
ds^2 = X/2m dT^2 - 2m/X dX^2
As you cross from X > 0 to X < 0, the coordinate
X switches from spacelike to timelike, and the
coordinate T switches from timelike to spacelike.
Does that mean that there is a physical singularity
at X=0? Obviously not, because the above metric is
obtained from the Minkowski metric
ds^2 = dt^2 - dx^2
by making the substitution
dx = square-root(8mX) cosh(T/4m)
dt = square-root(8mX) sinh(T/4m)
You cannot make conclusions about the existence of a singularity
by only looking at a single coordinate system.
>> It ought to be clear by now that the only "controversy" here is in LEJ
>> Brouwer's head. And, I suppose, in articles and textbooks that pre-dated
>> the current understanding of this manifold. LEJ Brouwer's confusions are
>> elementary, and are related to applying basic differential geometry to
>> the Schw. manifold.
>
>Don't be so cocksure and ease off with the cheap ad hominems. You
>appear to be completely unaware that that coordinate transforms have
>to be diffeomorphisms. The physical singularity at the horizon cannot
>be transformed away.
If it's really a singularity, then it can't be transformed away.
If it can be transformed away, then it isn't a real singularity.
And that's the case here. The apparent singularity at r=2m can
be transformed away through the substitution:
T = square-root(r/2m - 1) exp(r/4m) sinh(t/4m)
R = square-root(r/2m - 1) exp(r/4m) cosh(t/4m)
In terms of T,R there *is* no singularity at the event horizon.
So the singularity was transformed away, which implies that it's
not a real singularity.
Eric Gisse
> Tom Roberts wrote:
> > LEJ Brouwer wrote:
> >> What the mathematics has shown is that the interior solution does not
> >> have spherically symmetry about a point,
>
> > This is not true. The interior Schw. solution is spherically symmetric,
> > but it is not static.
>
> > OK, technically you are correct about the entire manifold --
> > the spherical symmetry is NOT "about a point", because the
> > singularity must be deleted from the manifold and is not
> > therefore a point. But I don't think this is what you
> > meant at all (based on further misconceptions below).
>
> > IOW: the interior solution is spherically symmetric in the sense of an
> > SO(3) isometry group.
>
> But it discontinously interchanges timelike and spacelike coordinates at the
> event horizon. LEJ Brouwer should choose better wording of his argument,
> but his argument is not about spherical symmetry.
Schwarzschild coordinates are a _POOR_ coordinate choice.
>
>
>
> >> and it is quite unclear as to
> >> what physical matter configuration could give rise to the properties
> >> described by the interior solution - it is certainly not due to a
> >> point mass.
>
> > Yes, it is. Or at least it is as close as one can come to such a
> > physical situation in GR.
>
> Proof by assertion. There is no radial curve extending from the point
> to an observer outside the event horizon. In the classic Hilbert solution
> the physical meaning is that (r,t) becomes (t,r) at the horizon. Using
> other coordinates can't mask this fundamental transition.
No. You misunderstand.
The radial and time coordinates swap meaning, but they _remain_ radial
and time coordinates.
[...]
>
> You really need to stop changing the subject. The discontinuity at the
> event horizon cannot be removed by any transformation. So there is no
> local neighbourhood around each point on the horizon where a single piece
> Minkowski map can be applied. There is a physical discontinuity at the
> horizon and not merely some coordinate issue as claimed without proof by
> assorted textbooks and those who religiously believe them.
This is generally known. It is not possible, in general, to reduce to
Minkowski space in the neighborhood of a point. The neighborhood is an
extra thing that comes only with metrics that play nice.
That the event horizon is a global phenomena in this manifold is
something that is known to be true.
What is _not_ true is that there is a physical discontinuity there.
You cannot be faulted for thinking this is so because physicists
argued about this for a good 40 years until Kruskal published his
coordinate system. However, you _can_ be faulted for not learning a
goddamn thing that has been known since then.
Nobody [should be] claims that the event horizon is something that can
be transformed away.
>
>
>
> >> The fact that Kruskal and related coordinate systems cannot be
> >> transformed to Schwarzschild (or locally flat) coordinates at the
> >> event horizon indicates that the event horizon itself corresponds to
> >> the 'edge' of the solution manifold.
>
> > Again you are wrong, in several ways:
> > A) The horizon is the _boundary_ of the region of validity of either
> > Schw. coordinate chart, not of the manifold itself.
>
> Boundary conditions need to be applied to the solutions of the nonlinear
> partial differential equations that constitute GR. Applying a single
> boundary condition at r=infinity isn't mathematically (and physically)
> sufficient. The outer and inner solutions are two distinct manifold
> pieces in any coordinate system as evidenced by the fact that there is
> no diffeomorphism which can make this infinite redshift surface go away.
Yes, it is mathematically sufficient. In general it is not, but the
spherical symmetry is a special thing.
The PDEs reduce, with the assistance of a timelike killing vector at
infinity and spherical symmetry, to a handful of first order partial
differential equations. Which further reduce to a bunch of first order
ordinary differential equations. Which, in the end, has all the
undetermined coefficients encapsulated in a _single_ constant.
All that is needed is _one_ boundary condition to determine the
solution to the field equations uniquely in this case.
If you don't want to take my word for it, I suggest opening a book on
general relativity or simply doing it yourself since you [appear] to
know enough to do it.
>
> > B) _NO_ coordinate system can be "transformed to locally flat
> > coordinates" ANYWHERE in the manifold, for the simple reason
> > that the manifold is not flat anywhere. This includes both
> > sets of Schw. coordinates.
> > C) Kruskal and related coordinate systems CAN be transformed to
> > local approximately-Minkowski coordinates at the horizon.
>
> Not true. Even in the contrived KS coordinates, which mix the timelike
> and spacelike components, there is a discontinuous transition at the
> horizon. The (r,t) to KS map is not continuously differentiable at the
> event horizon. Since a single locally Minkowski manifold piece cannot
> be determined in (r,t) coordinates at points on the horizon, it cannot
> be found in KS coordinates also.
Wrong.
You can always _locally_ have Minkowski space, but the definition of
local is not as you used it - in GR it is at a point, not the
neighborhood of a point.
>
> > Indeed, ANY coordinates can be so transformed, ANYWHERE THEY
> > ARE VALID. It is the two sets of Schw. coordinates that cannot be
> > so transformed at the horizon, because they are not valid there.
> > D) Yes, no coordinate system can be transformed to Schw. coordinates
> > at the horizon, BECAUSE THE SCHW. COORDINATES ARE NOT VALID THERE.
> > This is a problem with the Schw. coordinates, not the others.
>
> > You REALLY need to get your "facts" right.
>
> Take your own advice. The event horizon is a physical singularity in
> all coordinate systems. It is obviously not a point singularity but
> it sure is no "coordinate singularity" as claimed by you and others.
You are confused - it is claimed that the position r = 2GM/c^2 is a
coordinate singularity in Schwarzschild coordinates, not that the
event horizon is a coordinate artifact. You have had at least a half
century of useful thought to work from, and you are making the same
mistakes as physicists made shortly after Schwarzschild published the
results.
[...]
> Don't be so cocksure and ease off with the cheap ad hominems. You appear to
> be completely unaware that that coordinate transforms have to be diffeomorphisms.
Duh.
> The physical singularity at the horizon cannot be transformed away.
Nobody here is claiming that it can.
Dirk Van de moortel
"Eric Gisse" <jow...@gmail.com> wrote in message news:1178399218....@w5g2000hsg.googlegroups.com...
> On May 5, 12:37 pm, heretic <here...@dogma.edu> wrote:
[snip]
>> The physical singularity at the horizon cannot be transformed away.
>
> Nobody here is claiming that it can.
I think "Heretic" = Ken Tucker in disguise :-)
Dirk Vdm
JanPB
> Don't be so cocksure and ease off with the cheap ad hominems. You appear to
> be completely unaware that that coordinate transforms have to be diffeomorphisms.
...*at the intersection of the domains* of the charts under
consideration. The horizon is not in the domain of the Schwarzschild
coordinate system.
> The physical singularity at the horizon cannot be transformed away.
It's not physical and it's transformed away exactly like the origin
singularity of the standard polar coordinates in the 2D plane can be
transformed away or the singularities of the Mercator projection at
the poles can be transformed away.
--
Jan Bielawski
Tom Roberts
> It is incredible how GR textbooks dismiss the
> event horizon as a "coordinate singularity" when it is an infinite redshift
> surface in *all* coordinate systems.
Hmmm. You are confused. Yes, _IF_ a light source hovered at the horizon
it would have infinite redshift to a distant observer. But it is not
possible for a light source to hover there.
Yes, the horizon is a locus of the manifold that is completely
independent of coordinate system -- it is a geometrical property of the
manifold. But in Schwarzschild spacetime it is a _regular_ locus of the
manifold (i.e. there is no singularity there).
BTW a better description of the horizon is that it is a null trapped
surface.
> So hiding behind the principle of
> general covariance doesn't cut it. It is not possible to find a global
> diffeomorphism that makes the horizon go away, this includes
> Kruskal-Szekeres.
Of course not! No geometrical property of the manifold can be "made to
go away" by a mere coordinate transform.
But the _coordinate_ singularities of the two sets of Schw. coordinates
can be eliminated by an appropriate choice of coordinates, such as those
of Kruskal-Szerkes. Within each of their domains, there is a
diffeomorphism between each set of the Schw. coordinates and the (single
set of) K-S coordinates. This is not "global", because neither set of
Schw. coordinates is global; the K-S coordinates are global.
> In another post the Painleve-Gullstrand solution was offered but it is not
> static. So all we have, as you and others point out, is a patch of two
> distinct
> space-times that is not continuously differentiable at the horizon in any
> coordinate system.
This is simply not true. There are indeed two separate and disconnected
regions of the Schw. coordinates, but other coordinates cover the
horizon, such as Kruskal-Szerkes coordinates, the various Painleve
charts, and many more. The _MANIFOLD_ is completely regular at the
horizon; it is only the Schw. _COORDINATES_ that are singular there (and
also related coordinates).
> GR Black hole believers practice a new type of mathematics, one in which
> boundary conditions are selectively ignored.
Not true. But there are no "boundary conditions" at the horizon, for the
simple reason that is is not a boundary of the manifold -- it is a
_regular_locus_ of the manifold.
But yes, in a very real sense this manifold taught us "new mathematics",
and the field of differential geometry grew up because of it. This math,
of course, is not invalid, it's just new (back in the 1920s and 30s).
> They choose to apply a
> boundary condition at r=infinity but don't bother at r=0 (or even
> r=2GM/c^2).
Hmmm. In the initial statement of the Schw. problem, no boundary
condition is specified except at spatial infinity. It came initially as
a surprise that r=0 is singular (not a boundary), but since then we have
learned a lot about such singularities.... r=2GM/c^2 is not in any sense
a boundary of the manifold (it is a boundary of the regions of validity
for the two sets of Schw. coordinates, however).
> So they end up solving two different problems and claiming
> they have solved the originally posed problem.
Not true. It is only people who are wedded to Schw. coordinates, like
LEJ Brouwer and apparently yourself, that have "two different problems".
The actual manifold has no such difficulty, but one must use coordinates
other than Schw. to learn this.
> BTW, perhaps nature creates "black holes" but how and what their properties
> are is not established by the current GR mess.
Hmmm. The only theory we have that includes black holes is GR. At
present the general consensus in the physics community is that GR is
almost certainly a good description of black holes, except near their
singularities. For a Schw. black hole that includes the entire region
near the horizon, outside it, and inside it down to a Planck distance or
so from the singularity at the "center".
> But it discontinously interchanges timelike and spacelike coordinates at the
> event horizon. LEJ Brouwer should choose better wording of his argument,
> but his argument is not about spherical symmetry.
Only in Schw. coordinates is there such an "interchange". And even that
is a mirage -- the Schw. coordinates are not valid at the horizon, and
there are two completely different sets of Schw. coordinates, valid in
two disjoint regions: 0<r<2M for one set, and 2M<r for the other set.
The "interchange" is in the _LABELS_, and the math does not depend on
how one labels coordinates.
You seem to have the same confusions as LEJ Brouwer -- please look back
over this thread. The Schw. coordinates have no special place in the
mathematics, they are "special" only in the historical sense (in that
they were the first coordinates to be popularized for this manifold).
Note that many old physics textbooks have serious confusions about this
manifold; the math was known in the 1930s or so, but the physics
textbooks did not catch up until the 1970s (MTW and Weinberg).
Tom Roberts
Tom Roberts
> The radial and time coordinates swap meaning, but they _remain_ radial
> and time coordinates.
Not really. The Schw. labels "r" and "t" do indeed "swap" at the
horizon, but they are only "radial" and "time" in the region r>2M. In
the region 0<r<2M, r is the time coordinate (and -d/dr is future
pointing), but t is not at all "radial". For a spherically symmetric
manifold like this, the "radial" coordinate (squared) should be the
coefficient of dOmega in the line element -- that is r in this region,
not t. In this region there is no "radial" coordinate in the Schw. chart.
>> You really need to stop changing the subject. The discontinuity at the
>> event horizon cannot be removed by any transformation.
Yes. Because the Schw. coordinates are discontinuous there. <shrug>
> Heretic wrote
>> So there is no
>> local neighbourhood around each point on the horizon where a single piece
>> Minkowski map can be applied.
This is not true. One can transform form K-S coordinates to locally
approximate Minkowski coordinates at the horizon. The _MANIFOLD_ is
regular at the horizon; the Schw. _COORDINATES_ are not. <shrug>
> heretic wrote:
>> There is a physical discontinuity at the
>> horizon and not merely some coordinate issue as claimed without proof by
>> assorted textbooks and those who religiously believe them.
This is plain not true.
> [... Eric Gisse got it right]
> You can always _locally_ have Minkowski space, but the definition of
> local is not as you used it - in GR it is at a point, not the
> neighborhood of a point.
Hmmm. Usually it does mean "in a neighborhood of a given point". Of
course in any neighborhood the Minkowski coordinates and their metric
components are only APPROXIMATE (unless the manifold is flat throughout
the neighborhood).
Tom Roberts
Dirk Van de moortel
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:6e7%h.159891$Nj.11...@phobos.telenet-ops.be...
Or perhaps "Heretic" = "LEJ Brouwer" in an even
more devious disguise.
Interesting.
Dirk Vdm
Bilge
> The fact that Kruskal and related coordinate systems cannot be
> transformed to Schwarzschild (or locally flat) coordinates at the
Kruskal coordinates cannot be transformed to Schwarzschild coordinates at
the horizon because Schwarzschild coordinates are not defined at the
horiozon. What part of that do you not understand? What is your hangup with
using Schwarzschild coordinates?
[...?]
>> You can't transform from Schwarzschild coordinates to local
>> Minkowsky coordinates at the event horizon because the Schwarzschild
>> coordinates are singular there. But I don't see why that's a problem.
>
> Because the Schwarzschild coordinates are easy to interpret
> physically. The fact that they break down at the horizon indicates
> that the horizon is the edge of the solution manifold.
In other words, your criteria is ease of interpretation rather than
mathematical rigor?
[...]
>> It's rare that you can describe curved spacetime using a single
>> nonsingular coordinate system. The more general case requires
>> several overlapping patches to describe the whole manifold.
>
> Yes, but that is not quite analogous to the situation here.
Right. The difference here is that the interior and exterior Schwarzschild
solutions do not overlap. Both are singular at the horizon, which is all
the more reason to discard Schwarzschild coordinates when attempting to
describe the geometry at the horizon.
It's a Miracle
would never figure that out so I didnt look outside and that is candid.
--
My kaleidoscope art webpage:
http://community-2.webtv.net/Amused_2_Death_/Kaleidoscope/
Keep spam illegitimate, Report spam to:
http://spamcop.net/
It's a Miracle
gulp with a rubbery ending
Tom Roberts
> Tom Roberts wrote:
>> the interior solution is spherically symmetric in the sense of an
>> SO(3) isometry group.
>
> But it discontinously interchanges timelike and spacelike coordinates at
> the event horizon.
I repeat: this can be claimed ONLY for the Schw. coordinates. And in
fact it is wrong: there is no "interchange" because neither set of Schw.
coordinates is valid at the horizon; their domains to not overlap. At
most all that can be said is that the LABELS "interchange", but no
physical or mathematical properties of the manifold depend upon how
humans choose to label coordinates.
> LEJ Brouwer should choose better wording of his argument,
> but his argument is not about spherical symmetry.
His "argument" is just plain wrong, and merely illustrates his personal
misconceptions. <shrug>
>>> and it is quite unclear as to
>>> what physical matter configuration could give rise to the properties
>>> described by the interior solution - it is certainly not due to a
>>> point mass.
>>
>> Yes, it is. Or at least it is as close as one can come to such a
>> physical situation in GR.
>
> Proof by assertion.
No. See Birkhoff's theorem for a proof.
> In the classic Hilbert solution
> the physical meaning is that (r,t) becomes (t,r) at the horizon. Using
> other coordinates can't mask this fundamental transition.
THERE IS NO SUCH "TRANSITION"! The two sets of coordinates you discuss
are not valid at the horizon. Using other coordinates (e.g. K-S) AVOIDS
this problem completely. <shrug>
> The discontinuity at the
> event horizon cannot be removed by any transformation.
That discontinuity is an artifact of the coordinates you are implicitly
using. It is not present in the manifold itself, so there is no need to
"remove" it.
> So there is no
> local neighbourhood around each point on the horizon where a single piece
> Minkowski map can be applied.
This is plain and simply not true. But you need to use coordinates other
than Schw. to learn this. <shrug>
> There is a physical discontinuity at the
> horizon and not merely some coordinate issue as claimed without proof by
> assorted textbooks
This, too, is simply not true.
> [... repetition of the above mistakes omitted]
A simple counterexample to your claims is the fact that there is an
infinite set of timelike geodesics that go from the exterior region
right through the horizon and the interior region, intersecting the
singularity at r=0.
Put not your trust in coordinates, for they are merely human artifacts
used for descriptive purposes. To understand geometry and physical
processes one must look at INVARIANTS. The signature of a singularity in
GR is that one or more of the curvature invariants diverges as the
singularity is approached. In the Schwarzschild manifold there is a
singularity at r=0, but none at r=2M (the horizon) -- that is, no
curvature invariant diverges at the horizon.
[The geodesics of the 2nd previous paragraph are, of
course, completely independent of coordinates, and are
thus "invariant" in the sense of this last paragraph.]
You and LEJ Brouwer merely display elementary misunderstandings about
the role of coordinates in differential geometry, and display
misunderstandings about the Schw. manifold that have been cleared up for
almost half a century. <shrug>
Tom Roberts
carlip...@physics.ucdavis.edu
> On Apr 24, 11:41 pm, carlip-nos...@physics.ucdavis.edu wrote:
>> LEJ Brouwer <intuitioni...@yahoo.com> wrote:
[...]
>> > In Crother et al's view,
>> > coordinates can be given a physical significance a priori - e.g. if
>> > the direction of t is defined to be timelike, and the direction of r
>> > is defined to be spacelike when a problem is stated, then it must
>> > retain these properties in the solution.
>> How do you "define [a coordinate] to be timelike"? What does that mean?
>> You can *say* "t is timelike," but why should that affect anything?
> Hopefully I have clarified what I mean in my response to Jan.
Sorry, I still don't understand.
[...]
>> > The point to note at this point is that r and t are _not_ merely
>> > labels - they have a definite physical attribute determined by their
>> > respective spacelike and timelike natures.
>> What physical attributes are those? How do you determine them? Or is
>> this more magical thinking: you chant, "t shall be time" and that makes
>> it time?
> If we choose spacetime index (+,---), then if in a given coordinate
> system the line element looks like this:
> ds^2 = A dt^2 - r^2 dOmega^2 - B dr^2
> where A and B are positive, then t parametrises time and r
> parametrises radius.
But that's not a statement about the coordinates, it's a statement about
the coordinates *and the metric.* Your argument seems to be
(1) t parametrizes time when A is positive;
(2) therefore t has a "timelike nature";
(3) therefore A can't be negative, because that would violate (2).
Is it not clear that this is circular?
[...]
>> > The conclusion to be reached then is that analytic continuation of a
>> > solution is only valid if it respects the physical constraints of the
>> > original problem. If the original constraints are ignored, well, a
>> > different problem is being solved, with the possibly that the
>> > resulting solutions are unphysical. [If Carlip et al disagree with
>> > this conclusion, then I would very much like to see a proof that the
>> > interior solution does indeed solve the original problem and
>> > corresponds to a physically realisable vacuum].
>> I've answered this in the past. If you want to ask whether the solution
>> is physically reasonable, set up physically reasonable initial data and
>> see how they evolve. In this case, for example, you can write down an
>> exact solution for a collapsing spherically symmetric shell or sphere of
>> matter, with a time coordinate that has the direct physical meaning as
>> the proper time as measured by an observer standing on the surface of the
>> collapsing matter. In the resulting exact solution the matter passes
>> through the horizon with nothing strange happening, and the solution
>> evolves into an ordinary Schwarzschild black hole -- both exterior and
>> interior -- typically described in Painleve-Gullstrand coordinates. You
>> can find two simple examples in Adler, Bjorken, Chen, and Liu, "Simple
>> analytical models of gravitational collapse," American Journal of Physics
>> 73 (December 2005) 1148-1159.
> Yes, it is clear that such coordinates give a nice smooth evolution
> from exterior to interior.
Specifically, they give nice smooth evolution of a real physical system.
Furthermore, the Painleve-Gullstrand version is described in coordinates
that have clear physical meanings in terms of measurements by observers
in the system.
> I just have doubts that that evolution
> corresponds to physical reality - and partly because of the
> discontinuity in the transformation to Schwarzschild (or similar)
> coordinates at the event horizon.
I understand that you like Schwarzschild coordinates. But I don't
understand how 'I can't describe this in my favorite coordinates'
translates to 'this doesn't correspond to physical reality.'
> Why are there no coordinate choices
> in which the metric remains diagonal at every point on the path of a
> radially infalling particle and which does not suffer from some kind
> of discontinuity at the horizon?
There are. The Kruskal form of the metric is usually usually written
in terms of two null coordinates U and V, but you can define T=U+V
and R=V-U. Then the Kruskal form has "a metric [that] remains diagonal
at every point on the path of a radially infalling particle and which
does not suffer from some kind of discontinuity at the horizon."
On the other hand, I also don't understand your aversion to off-diagonal
terms in the metric. If you want to describe a rotating system in
ordinary, flat spacetime, you will find that corotating coordinates
have an off-diagonal term that goes as d\phi dt. So why should it be
such a surprise that comoving coordinates for radially collapsing matter
have a dr dt term in the metric?
Steve Carlip
LEJ Brouwer
Apologies for the long delay in response. I thank you for bringing
this scenario to my attention, as it raises some interesting questions
which I think could help to allay my earlier misgivings on the matter.
Imagine that I am initially sitting at the origin (i.e. the centre of
spherical symmetry) when the stars start accelerating radially towards
me. At the time that the stars converge at their Schwarzschild radius
to form the event horizon, then according to the form of the interior
solution, the stars don't appear to cross the event horizon and pile
in towards me because the interior of the event horizon in which I am
sitting has a spatial spherical symmetry which is not reflected in the
form of the interior Schwarzschild solution. Rather, the solution in
which I am sitting, which is interior to the event horizon, is just a
standard Schwarzschild exterior solution since it is a spherically
symmetric vacuum solution with insufficient mass to form a black hole.
However, this interior solution also has a boundary at the event
horizon. I am not sure what the boundary conditions are, though
logically it seems that nothing can enter and nothing can leave. I say
this, because once the stars 'cross' the event horizon, they do not
enter into my little sphere of tranquility, but rather head off
towards some future singularity in some region described by the
Schwarzschild interior, which happens to be at a specific point in
(star proper) time due to the artificial way in which the scenario has
been constructed wherein all stars reach/form the event horizon
simultaneously.
There therefore appear to be three separate regions in the scenario
above:
(i) The exterior of the horizon, which is just an exterior
Schwarzschild solution.
(ii) The spherical region in which I sit, which also has a
Schwarzschild exterior-like metric, but with a spherical bounding
surface which shields me from infalling stars, and also prevents
anything from leaving.
(iii) The interior Schwarzschild solution, complete with temporal
singularity, into which the stars disappear upon crossing the horizon.
This is a scenario which I am happy with, though I am guessing that
this is not the picture you are trying to describe.
In a more realistic collapse scenario where we have a continuous
spherically symmetric distribution of radially infalling matter, this
is what I think would happen:
As the matter collapses inwards, there will eventually be a high
enough mass density to form an event horizon. Any matter external to
the horizon will subsequently be 'diverted' towards the singularity
lying in its future worldline, but this interior region is not the
same as the region which contains the matter which was responsible for
the formation of the horizon. Rather, the matter which was interior to
the event horizon has a spherically symmetric distribution, and
becomes isolated from the exterior, and also from the Schwarzschild
interior containing the temporal singularity. Matter can neither enter
nor escape from this region, so it has unusual boundary conditions,
but it remains otherwise well-behaved and is without any kind of
singularity.
So this thought experiment leads me to conclude that the initial
spherically symmetric distribution undergoes a kind of phase
transition when the event horizon forms to become three separate
regions, each with spherical symmetry, and one of which contains a
future singularity (which is neither pointlike, nor a mass).
I therefore disagree with your claim that all the mass inside the
spherical shell will eventually be concentrated into a single point.
As I mentioned before, and you confirmed, the singularity is not a
'point'. The singularity exists in the (relative) *future*, not at any
particular spatial location or particular time. The mass inside the
spherical shell does not eventually become concentrated into a single
point - rather it lives to evolve happily in a little spherically
symmetric world of its own. The singularity lies in an independent,
and as I have stated before, a rather hard to physically interpret,
region and upon which all matter crossing from the exterior converges
upon.
Here is an (admittedly rather poor attempt at) representation of this
scenario:
C
x
^
|
|
A ----o<---- B
The radially infalling matter is in region B. The trapped region which
arises once the horizon (indicated by 'o') has formed is labelled A,
which contains the matter responsible for creating the horizon in the
first place. The matter crossing the horizon from B falls towards the
singularity at 'x' which lies in region C, corresponding to the
Schwarzschild interior solution. The singularity lies at a fixed time
relative to the time at which an infalling particle crosses the
horizon at 'o'. The region A becomes physically isolated from regions
B and C once the horizon has formed. As I have said before, my guess
is that at the singularity, particles are reflected back in space and
time towards the event horizon and back into the region B, so x acts
like a spacetime reflecting mirror. At a microscopic level, the
interior region A corresponds to an elementary particle, and it
appears to an external observer (i.e. in region B) that particle-
antiparticle annihilations are taking place at the event horizon,
though physically what is happening is that a single particle is being
reflected off of the singularity at 'x' which is acting as a spacetime
mirror.
Let me know if you agree with the above picture. I am not a betting
man, but I will bet that you do not. :)
- Sabbir.
Ken S. Tucker
SperM.hotmail.com> wrote:
> "Eric Gisse" <jowr...@gmail.com> wrote in messagenews:1178399218....@w5g2000hsg.googlegroups.com...
> > On May 5, 12:37 pm, heretic <here...@dogma.edu> wrote:
>
> [snip]
>
> >> The physical singularity at the horizon cannot be transformed away.
>
> > Nobody here is claiming that it can.
>
> I think "Heretic" = Ken Tucker in disguise :-)
> Dirk Vdm
Nope, heretic has gooder grammar and spelin!
But the compliment is appreciated
Ken.
PS: I 99% lurk.
Eric Gisse
It isn't a compliment, moron.
Tom Roberts
> Imagine that I am initially sitting at the origin (i.e. the centre of
> spherical symmetry) when the stars start accelerating radially towards
> me. At the time that the stars converge at their Schwarzschild radius
> to form the event horizon, then according to the form of the interior
> solution, the stars don't appear to cross the event horizon and pile
> in towards me because the interior of the event horizon in which I am
> sitting has a spatial spherical symmetry which is not reflected in the
> form of the interior Schwarzschild solution. Rather, the solution in
> which I am sitting, which is interior to the event horizon, is just a
> standard Schwarzschild exterior solution since it is a spherically
> symmetric vacuum solution with insufficient mass to form a black hole.
When those infalling stars reach the event horizon corresponding to
their total mass+energy, an horizon forms there. At all times, between
the outer edge of the stars and spatial infinity, the manifold is
Schwarzschild. This remains true as the stars continue falling inward
after the horizon, and a finite time after they cross the horizon, they
will intersect you at the center and the resulting singularity will
annihilate both you and them. You will get no warning of impending doom,
because the metric at your location remains flat until the stars arrive.
You have so many misconceptions about this, you must STUDY. I suggest:
K.Thorne, _Black_Holes_and_Time_Warps_. Then, perhaps,
you will be ready for a real textbook.
Tom Roberts
bz
@newssvr11.news.prodigy.net:
> You will get no warning of impending doom,
> because the metric at your location remains flat until the stars arrive.
>
>
would not the stars need to be falling at c (or near c) in order to 'get NO
(or little) warning'?
In other words, wouldn't you SEE them coming?
:)
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+...@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
Tom Roberts
> When those infalling stars reach the event horizon corresponding to
> their total mass+energy, an horizon forms there.
I realized this is too simplistic a description. As the stars approach
the spatial locus where the horizon will eventually be, an horizon forms
at the center and moves outward at the local speed of light, meeting the
stars at the locus of the horizon. This happens because when the stars
approach that locus, at some point in their falling it is impossible for
a light ray starting at the origin to reach the locus before the stars
reach it, so that light ray is trapped and therefore its origin is
inside the horizon. As the stars approach the locus this occurs at
successively larger distances from the center, until horizon and stars
meet at that locus. Remember any horizon is always moving with local
speed c.
> At all times, between
> the outer edge of the stars and spatial infinity, the manifold is
> Schwarzschild. This remains true as the stars continue falling inward
> after the horizon, and a finite time after they cross the horizon, they
> will intersect you at the center and the resulting singularity will
> annihilate both you and them. You will get no warning of impending doom,
> because the metric at your location remains flat until the stars arrive.
bz wrote:
> would not the stars need to be falling at c (or near c) in order to 'get NO
> (or little) warning'?
> In other words, wouldn't you SEE them coming?
Yes, if you were looking early enough, you could see them approaching.
But as I said, the metric remains unchanged until they reach you, even
though you are already inside the horizon well before they arrive. And
without foreknowledge of their total mass, there is no way for you
determine when the horizon expands around you.
Tom Roberts
Daryl McCullough
>> Imagine sitting on Earth, and surrounded by a spherical shell of
>> distant stars. Now, imagine attaching powerful rockets to each star
>> in the shell so that at the same moment (according to the coordinate
>> system in which the Earth is at rest), the stars are all sent hurtling
>> towards the earth. The rockets are timed so that all the stars will
>> arrive simultaneously.
>>
>> Now, what you can do is add up the masses of all the stars and compute
>> the Schwarzschild radius for that mass. At some point, the stars will
>> pass that radius. After that, you and the Earth and all those stars
>> are inside a black hole.
>>
>> There still isn't any point mass anywhere except in the *future*.
>> The predictions of General Relativity are that all the mass inside
>> the spherical shell will eventually be concentrated into a single
>> point. That's the singularity, and it exists in the *future*, not
>> in any particular spatial location.
>>
>> From the point of view of someone outside the spherical shell of
>> stars, the geometry will approach that of the Schwarzschild exterior
>> solution.
[Stuff deleted]
>There therefore appear to be three separate regions in the scenario
>above:
>
>(i) The exterior of the horizon, which is just an exterior
>Schwarzschild solution.
Okay.
>(ii) The spherical region in which I sit, which also has a
>Schwarzschild exterior-like metric, but with a spherical bounding
>surface which shields me from infalling stars, and also prevents
>anything from leaving.
I don't think that's correct. Maybe Steve Carlip can explain
to us what this region is like, but it certainly is not like
the Schwarzschild exterior solution. For one thing,
the metric is time-dependent. If you wait long enough, the
curvature will go to infinity (as the stars come crashing
into you).
>(iii) The interior Schwarzschild solution, complete with temporal
>singularity, into which the stars disappear upon crossing the horizon.
I believe that this is the region between the event horizon and
the spherical shell of stars. Anyone who starts off outside the
sphere of stars and falls radially inward will eventually enter
this region.
>In a more realistic collapse scenario where we have a continuous
>spherically symmetric distribution of radially infalling matter, this
>is what I think would happen:
>
>As the matter collapses inwards, there will eventually be a high
>enough mass density to form an event horizon. Any matter external to
>the horizon will subsequently be 'diverted' towards the singularity
>lying in its future worldline, but this interior region is not the
>same as the region which contains the matter which was responsible for
>the formation of the horizon.
I think that's right, although the singularity will claim
everyone in freefall, both those initially inside and initially
outside the event horizon.
>Rather, the matter which was interior to
>the event horizon has a spherically symmetric distribution, and
>becomes isolated from the exterior, and also from the Schwarzschild
>interior containing the temporal singularity. Matter can neither enter
>nor escape from this region, so it has unusual boundary conditions,
>but it remains otherwise well-behaved and is without any kind of
>singularity.
I don't think that's correct. If you are inside an inward-rushing
sphere of matter, then you will eventually get crushed by it.
Also, I'm not sure about the claim that this region being
isolated from the exterior. Nothing in this region can
affect the exterior, but I don't see why something in
the exterior couldn't affect this region.
>I therefore disagree with your claim that all the mass inside the
>spherical shell will eventually be concentrated into a single point.
What's to prevent that from happening?
>As I mentioned before, and you confirmed, the singularity is not a
>'point'. The singularity exists in the (relative) *future*, not at any
>particular spatial location or particular time. The mass inside the
>spherical shell does not eventually become concentrated into a single
>point - rather it lives to evolve happily in a little spherically
>symmetric world of its own.
Yes, but a spherically symmetric world that gets smaller all the time.
Well, you have to actually look at the math to know whether one
region is causally disconnected from another region. In particular,
you need to see if there are timelike geodesics connecting the two
regions. If so, then it's possible for objects to travel from one
region to the other.
LEJ Brouwer
wrote:
You are assuming the answer to prove it. Suppose I am the planet Earth
and all the stars start converging upon me. Whatever the metric it is
in my vicinity, it has a spherical symmetry, and will be approximately
Schwarzschild. There will be no net gravitational field acting on me
due to the stars due to the symmetry of the problem. When the event
horizon forms, nothing happens locally to change this sphercal
symmetry. We have already determined that the Schwarzschild interior
solution has an SO(3) symmetry, but it does not have the central
spherical symmetry. Therefore the interior solution cannot coincide
with the metric around me when the horizon forms. Rather, I will
remain quite oblivious of the formation of the horizon, and whatever
the metric was before horizon formation will not change greatly after
it. If it is true as claimed that the infalling stars fall into the
Schwarzschild interior, then that interior cannot coincide with my
little patch of the universe (which I labelled 'A' in my diagram), and
thus must correspond to an independent patch 'C' which only forms upon
formation of the event horizon.
>
> >(iii) The interior Schwarzschild solution, complete with temporal
> >singularity, into which the stars disappear upon crossing the horizon.
>
> I believe that this is the region between the event horizon and
> the spherical shell of stars. Anyone who starts off outside the
> sphere of stars and falls radially inward will eventually enter
> this region.
But once the stars cross the event horizon, these are not spherical
shells in the usual sense, but 'temporal' spheres, which hit the
singularity at a given time relative to passing through the event
horizon. There is no way that this can be identified with the
spherically symmetric region in which I sit. Although the assumption
that the stars continue to fall 'radially inwards' upon crossing the
horizon seems natural, it seems to me to be invalid as the
Schwarzschild interior metric does not describe a space with spatial
spherically symmetry.
> >In a more realistic collapse scenario where we have a continuous
> >spherically symmetric distribution of radially infalling matter, this
> >is what I think would happen:
>
> >As the matter collapses inwards, there will eventually be a high
> >enough mass density to form an event horizon. Any matter external to
> >the horizon will subsequently be 'diverted' towards the singularity
> >lying in its future worldline, but this interior region is not the
> >same as the region which contains the matter which was responsible for
> >the formation of the horizon.
>
> I think that's right, although the singularity will claim
> everyone in freefall, both those initially inside and initially
> outside the event horizon.
What mechanism do you propose for those initially inside to fall into
the singularity, when they are physically isolated from each other?
Again this conclusion is due to the invalid assumption that the metric
of those initially inside and the patch described by the Schwarzschild
interior can be identified. They cannot.
>
> >Rather, the matter which was interior to
> >the event horizon has a spherically symmetric distribution, and
> >becomes isolated from the exterior, and also from the Schwarzschild
> >interior containing the temporal singularity. Matter can neither enter
> >nor escape from this region, so it has unusual boundary conditions,
> >but it remains otherwise well-behaved and is without any kind of
> >singularity.
>
> I don't think that's correct. If you are inside an inward-rushing
> sphere of matter, then you will eventually get crushed by it.
There is no inward-rushing sphere in the sense that you mean - as we
have established, the interior Schwarzschild solution does not
describe a sphere converging on a spatial point, but a sphere
converging on a time relative to the time at which the horizon was
crossed. This implies that the singularity has nothing to do with the
original origin of the problem.
>
> Also, I'm not sure about the claim that this region being
> isolated from the exterior. Nothing in this region can
> affect the exterior, but I don't see why something in
> the exterior couldn't affect this region.
Anything falling in from the exterior goes into region 'C', and not
into region 'A', so nothing from region 'B' can pass into region 'A'.
It is possible, I guess that stuff in region A could end up in region
C if it manages to reach the horizon, but I suspect that the whole of
region A becomes physically disconnected from the full Schwarzschild
solution described by regions B+C as soon as the horizon forms. As I
said, I do not know what the boundary condition on region A would be,
but I am pretty sure that the boundary is impenetrable either way.
> >I therefore disagree with your claim that all the mass inside the
> >spherical shell will eventually be concentrated into a single point.
>
> What's to prevent that from happening?
This doesn't happen even in the usual case - I think we have all been
brainwashed into believing that the Schwarzschild singularity is a
spatial point when clearly it is not. As I said, it is not even a
temporal point, though in the rather artificial scenario you describe
where all stars form the event horizon at precisely the same time,
then that might appear to be the case there.
> >As I mentioned before, and you confirmed, the singularity is not a
> >'point'. The singularity exists in the (relative) *future*, not at any
> >particular spatial location or particular time. The mass inside the
> >spherical shell does not eventually become concentrated into a single
> >point - rather it lives to evolve happily in a little spherically
> >symmetric world of its own.
>
> Yes, but a spherically symmetric world that gets smaller all the time.
These are not spatial spheres in the usual sense, so the picture you
paint here is quite misleading.
The picture with the regions I describe makes sense to me. The picture
you describe where you attempt to make region A coincide with the
Schwarzschild interior (which I label 'C') is not borne out by the
form of the interior metric, which is incompatible with the spatial
symmetry in region A.
- Sabbir.
Tom Roberts
> Suppose I am the planet Earth
> and all the stars start converging upon me.
In a spherically symmetric manner.
> Whatever the metric it is
> in my vicinity, it has a spherical symmetry, and will be approximately
> Schwarzschild.
Neglecting minor deviations from spherical symmetry (sun, moon, earth
rotation, earth non-sphericity, etc.), and neglecting the atmosphere, it
will be _exactly_ Schwarzschild -- that's Birkhoff-s theorem.
I also assume you are not "the planet earth", but are an observer on its
surface (in a space suit; see above).
I also assume that by the "Schw. interior metric" you mean
Schwarzschild's solution for the interior of a spherically symmetric
uniform density star or planet. I cannot make sense of your statements
any other way. This is (bizarrely) consistent with your claim to "be"
the planet earth (though its density is by no means uniform, but ignore
that).
> There will be no net gravitational field acting on me
> due to the stars due to the symmetry of the problem. When the event
> horizon forms, nothing happens locally to change this sphercal
> symmetry.
Right. You can have no local knowledge of when the horizon expands
around you and the earth.
> We have already determined that the Schwarzschild interior
> solution has an SO(3) symmetry, but it does not have the central
> spherical symmetry.
There is no "Schw. interior solution" anywhere in this problem, except
inside the volume of the earth itself. You are on its surface, and the
metric there is the Schw. _exterior_ solution, corresponding to the mass
of the earth -- again, that is Birkhoff's theorem (given my caveats
above). The metric in this region _HAS_ spherical symmetry.
> Therefore the interior solution cannot coincide
> with the metric around me when the horizon forms.
You remain confused -- the interior solution is irrelevant. See above.
> Rather, I will
> remain quite oblivious of the formation of the horizon, and whatever
> the metric was before horizon formation will not change greatly after
> it.
True -- the metric where you are standing on the earth's surface will
not change at all until those stars reach you, at which time you will be
obliterated. Ditto for any point of the earth's interior (since you seem
hung up on the interior solution, and that's the only place it applies).
> But once the stars cross the event horizon, these are not spherical
> shells in the usual sense, but 'temporal' spheres, which hit the
> singularity at a given time relative to passing through the event
> horizon.
You are confused. This is NOT the Schw. manifold, but because it is
spherically symmetric (by my caveats above), Birkhoff's theorem applies
to its vacuum regions.
Consider the period of time between:
a) the stars cross the spatial locus where their horizon will
ultimately be.
and
b) the stars engulf you on the surface of the earth.
During this time the metric at the surface of the earth is unchanged by
the impending doom of the incoming stars.
After the incoming stars engulf you and the earth, the spatial region
where you used to be will be inside the horizon and will be vacuum, so
Birkhoff's theorem again applies to this region. This applies all the
way down to the center of the earth.
[Hmmm. I'm glossing over the difficulty of identifying
spatial regions before and after.... But then, afterwards
the entire manifold is vacuum, so it's OK.]
> There is no way that this can be identified with the
> spherically symmetric region in which I sit.
I repeat: you are confused. See above.
> Although the assumption
> that the stars continue to fall 'radially inwards' upon crossing the
> horizon seems natural,
It is well known, and is not an assumption, it is the conclusion of a
_calculation_. Oppenheimer and Snyder computed this for a quite similar
case (no earth at the center) in the 1930's.
> it seems to me to be invalid as the
> Schwarzschild interior metric does not describe a space with spatial
> spherically symmetry.
You REALLY should start doubting your ability to GUESS about such things
-- you have been wrong EVERY time you try. Moreover, the interior Schw.
region _IS_ spherically symmetric. Manifestly so -- the metric has an
SO(3) symmetry (or in the usual Schw. coordinates of this region, the
metric components are all independent of \theta and \phi).
> [... further elaboration of the above errors]
Tom Roberts
LEJ Brouwer
> LEJ Brouwer wrote:
> > Suppose I am the planet Earth
> > and all the stars start converging upon me.
>
> In a spherically symmetric manner.
As opposed to a complex banana-like formation.
> I also assume you are not "the planet earth", but are an observer on its
> surface (in a space suit; see above).
No, I _insist_ on being the planet Earth. Human observers on my
surface do not need space suits as I have a breathable atmosphere, and
the weather is particularly nice during the summer. They are welcome
to wear space suits if they like if they feel that it will improve
their street cred, though it is otherwise not necessary. T-shirts and
jeans would be just as acceptable.
> I also assume that by the "Schw. interior metric" you mean
> Schwarzschild's solution for the interior of a spherically symmetric
> uniform density star or planet.
No I don't mean that at all. Do you live in a cave?
> I cannot make sense of your statements
> any other way.
Then you must be TERRIBLY CONFUSED. You need to _STUDY_ my past
writings.
> > There will be no net gravitational field acting on me
> > due to the stars due to the symmetry of the problem. When the event
> > horizon forms, nothing happens locally to change this sphercal
> > symmetry.
>
> Right. You can have no local knowledge of when the horizon expands
> around you and the earth.
There is no horizon until the infalling stars reach their
Schwarzschild radius, at which point the horizon forms spontaneously
at that radius.
> Consider the period of time between:
> a) the stars cross the spatial locus where their horizon will
> ultimately be.
> and
> b) the stars engulf you on the surface of the earth.
This never happens.
> During this time the metric at the surface of the earth is unchanged by
> the impending doom of the incoming stars.
There is no impending doom.
> After the incoming stars engulf you and the earth
I _am_ the Earth. The stars do not engulf me at all (I assume here
that their Schwarzschild radius is greater than my radius).
> It is well known, and is not an assumption, it is the conclusion of a
> _calculation_. Oppenheimer and Snyder computed this for a quite similar
> case (no earth at the center) in the 1930's.
Their calculation is flawed.
- Sabbir.
LEJ Brouwer
> On May 15, 11:22 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> > Consider the period of time between:
> > a) the stars cross the spatial locus where their horizon will
> > ultimately be.
> > and
> > b) the stars engulf you on the surface of the earth.
>
> This never happens.
Actually, I suppose it does. Unfortunately the scenario as given is
highly artificial and obfuscates the main point - when the horizon
forms, then yes, the stars will continue falling towards me in region
'A' which, as you say, has just an ordinary Schwarzschild metric in
the vacuum regions. HOWEVER, anything that subsequently hits the
horizon from outside (i.e. stuff from region 'B' which falls in after
the horizon has been formed) will pass through to region 'C' and hit
the singularity contained therein (and bounce off it!), and will _not_
fall into to region 'A' in which I (and also now the original
infalling stars) are located. Sure the stars will engulf me, but that
is quite irrelevant to the main point, which is that the region which
contains the _singularity_ is physically distinct from the region
which contains _me_. You seem to be confused on the latter point.
- Sabbir.
Daryl McCullough
>Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> Right. You can have no local knowledge of when the horizon expands
>> around you and the earth.
>
>There is no horizon until the infalling stars reach their
>Schwarzschild radius, at which point the horizon forms spontaneously
>at that radius.
Tom is using an operational definition of being "inside the event
horizon": Event P is inside the event horizon if it is
impossible for a light signal sent from P to escape to
infinity.
Okay, so pick a spherical coordinate system centered on the Earth,
and pick two different radii R1 and R2 with R1 < R2. Suppose that
at time T1 a light signal is sent outward from radius R1. This
light signal will arrive at R2 at some time T2 with T2 > T1.
Clearly, if T1 is not too late for a light signal to escape
from radius R1, then T2 is not too late for a light signal
to escape from radius R2. So someone at radius R2 has extra
time to procrastinate before it is too late for him to escape.
Turning that around, the time at which it is too late to escape
from R1 occurs *before* the time at which it is too late to
escape from R2. So the boundary between "too late to escape"
and "not too late to escape" is a sphere that is expanding
outward at the speed of light. Tom is calling that boundary
the "event horizon".
>> Consider the period of time between:
>> a) the stars cross the spatial locus where their horizon will
>> ultimately be.
>> and
>> b) the stars engulf you on the surface of the earth.
>
>This never happens.
Why do you say that? What prevents it from happening? Are
you saying this based on General Relativity, or some alternative
theory?
Daryl McCullough
>> >There therefore appear to be three separate regions in the scenario
>> >above:
>>
>> >(i) The exterior of the horizon, which is just an exterior
>> >Schwarzschild solution.
>>
>> Okay.
>>
>> >(ii) The spherical region in which I sit, which also has a
>> >Schwarzschild exterior-like metric, but with a spherical bounding
>> >surface which shields me from infalling stars, and also prevents
>> >anything from leaving.
>>
>> I don't think that's correct. Maybe Steve Carlip can explain
>> to us what this region is like, but it certainly is not like
>> the Schwarzschild exterior solution. For one thing,
>> the metric is time-dependent. If you wait long enough, the
>> curvature will go to infinity (as the stars come crashing
>> into you).
>
>You are assuming the answer to prove it.
Okay, I'll turn around and ask you: what reason is there
for believing that the metric is time-independent, when
you have massive stars in motion?
>Suppose I am the planet Earth
>and all the stars start converging upon me. Whatever the metric it is
>in my vicinity, it has a spherical symmetry, and will be approximately
>Schwarzschild.
Why do you say that? Again, what reason is there to believe
that the metric is time-independent?
>There will be no net gravitational field acting on me
>due to the stars due to the symmetry of the problem.
I assume that you are using Newtonian physics here, because
General Relativity doesn't use "gravitational fields" it uses
a metric. So you want to reason using Newtonian physics, then
that's fine for a first cut. In Newtonian physics, it is true
that (1) the gravitational field is zero inside the shell, and
(2) after a finite amount of time, the shell of matter crushes
you.
>When the event horizon forms, nothing happens locally to change
>this spherical symmetry.
How does it follow that the matter will not crush you?
>We have already determined that the Schwarzschild interior
>solution has an SO(3) symmetry, but it does not have the central
>spherical symmetry. Therefore the interior solution cannot coincide
>with the metric around me when the horizon forms.
That was my point. The Schwarzschild interior solution does *not*
apply on the interior of the shell of matter. It applies in the
region *between* the event horizon and the shell of matter. Any
object falling radially inward that is initially outside the
shell will eventually fall into the Schwarzschild interior region.
>Rather, I will remain quite oblivious of the formation of the horizon
Yes, up until the time that the matter crushes you.
>and whatever the metric was before horizon formation will not change
>greatly after it.
Right.
>If it is true as claimed that the infalling stars fall into the
>Schwarzschild interior
I didn't claim that. I claimed that there were three regions:
1. The region inside the spherical shell of matter.
2. The Schwarzschild interior solution.
3. The Schwarzschild exterior solution.
The shell is the boundary between regions 1 and 2.
The event horizon is the boundary between regions 2 and 3.
>> >(iii) The interior Schwarzschild solution, complete with temporal
>> >singularity, into which the stars disappear upon crossing the horizon.
>>
>> I believe that this is the region between the event horizon and
>> the spherical shell of stars. Anyone who starts off outside the
>> sphere of stars and falls radially inward will eventually enter
>> this region.
>
>But once the stars cross the event horizon, these are not spherical
>shells in the usual sense, but 'temporal' spheres, which hit the
>singularity at a given time relative to passing through the event
>horizon.
>There is no way that this can be identified with the
>spherically symmetric region in which I sit.
That's why I said there are *three* regions. You are
not sitting in the Schwarzschild interior solution, you
are sitting in my region 1 above.
>> >As the matter collapses inwards, there will eventually be a high
>> >enough mass density to form an event horizon. Any matter external to
>> >the horizon will subsequently be 'diverted' towards the singularity
>> >lying in its future worldline, but this interior region is not the
>> >same as the region which contains the matter which was responsible for
>> >the formation of the horizon.
>>
>> I think that's right, although the singularity will claim
>> everyone in freefall, both those initially inside and initially
>> outside the event horizon.
>
>What mechanism do you propose for those initially inside to fall into
>the singularity, when they are physically isolated from each other?
You don't have to "fall" anywhere. The singularity will come to you
in the form of an inward-falling wall of stars.
>Again this conclusion is due to the invalid assumption that the metric
>of those initially inside and the patch described by the Schwarzschild
>interior can be identified. They cannot.
Once again, I'm *not* saying that the Earth is part of the Schwarzschild
interior solution. I'm saying that it *isn't*.
>> I don't think that's correct. If you are inside an inward-rushing
>> sphere of matter, then you will eventually get crushed by it.
>
>There is no inward-rushing sphere in the sense that you mean - as we
>have established, the interior Schwarzschild solution does not
>describe a sphere converging on a spatial point,
Once again, the Earth is *not* part of the Schwarzschild interior
solution. That's why I said there are *three* regions.
>but a sphere converging on a time relative to the time at which the
>horizon was crossed. This implies that the singularity has nothing
>to do with the original origin of the problem.
No, it's all the same singularity.
>> Also, I'm not sure about the claim that this region being
>> isolated from the exterior. Nothing in this region can
>> affect the exterior, but I don't see why something in
>> the exterior couldn't affect this region.
>
>Anything falling in from the exterior goes into region 'C', and not
>into region 'A', so nothing from region 'B' can pass into region 'A'.
I'm asking how you know that. You need to look at the geodesics
to know whether it is possible for an object to pass from one
region to another.
>> >I therefore disagree with your claim that all the mass inside the
>> >spherical shell will eventually be concentrated into a single point.
>>
>> What's to prevent that from happening?
>
>This doesn't happen even in the usual case
What do you mean by "the usual case"? Once again,
what prevents the infalling stars from reaching the Earth?
>> >As I mentioned before, and you confirmed, the singularity is not a
>> >'point'. The singularity exists in the (relative) *future*, not at any
>> >particular spatial location or particular time. The mass inside the
>> >spherical shell does not eventually become concentrated into a single
>> >point - rather it lives to evolve happily in a little spherically
>> >symmetric world of its own.
>>
>> Yes, but a spherically symmetric world that gets smaller all the time.
>
>These are not spatial spheres in the usual sense, so the picture you
>paint here is quite misleading.
You are the one who said "it lives to evolve happily in a little
spherically symmetric world of its own". What did *you* mean by it?
>> Well, you have to actually look at the math to know whether one
>> region is causally disconnected from another region. In particular,
>> you need to see if there are timelike geodesics connecting the two
>> regions. If so, then it's possible for objects to travel from one
>> region to the other.
>
>The picture with the regions I describe makes sense to me.
As I said, it is not enough to draw pictures. You have to
analyze what the geodesics look like at the boundary between
regions.
>The picture you describe where you attempt to make region A coincide
>with the Schwarzschild interior
As I have said many times, the region inside the spherical shell
is *not* the Schwarzschild interior solution.
Tom Roberts
> So the boundary between "too late to escape"
> and "not too late to escape" is a sphere that is expanding
> outward at the speed of light. Tom is calling that boundary
> the "event horizon".
This is not just me "calling it that" -- that's what the phrase "event
horizon" _means_. In this case it forms initially at the center and
expand outward at the local speed of light to meet the incoming stars at
the Schw. radius corresponding to their total mass.
Tom Roberts
Daryl McCullough
>
>Daryl McCullough wrote:
>> So the boundary between "too late to escape"
>> and "not too late to escape" is a sphere that is expanding
>> outward at the speed of light. Tom is calling that boundary
>> the "event horizon".
>
>This is not just me "calling it that" -- that's what the phrase "event
>horizon" _means_.
I didn't mean to suggest otherwise, but I think that LEJ Brouwer may be
thinking of the "event horizon" and the "Schwarzschild radius" as synonymous.
Actually, I thought that there were several alternative definitions of "event
horizon". One has to do with light signals escaping to infinity, and another
having to do with "trapped surfaces" (I'm not sure what that means, but
as I understand it, it is a more "local" definition).
>In this case it forms initially at the center and
>expand outward at the local speed of light to meet the incoming stars at
>the Schw. radius corresponding to their total mass.
Yes, I understand that.
Tom Roberts
> Actually, I thought that there were several alternative definitions of "event
> horizon". One has to do with light signals escaping to infinity, and another
> having to do with "trapped surfaces" (I'm not sure what that means, but
> as I understand it, it is a more "local" definition).
The word "horizon" is used to denote a boundary in the spacetime
manifold beyond which there is no nonspacelike path to some specified
point, worldline, or region. There are a number of different such
horizons, depending on the specific situation considered. A
"nonspacelike path" is, of course, any timelike or null trajectory or
worldline, either past- or future-directed (depending on the situation
being discussed). In Schw. spacetime, and the spacetimes considered
earlier in this thread the event horizon is the boundary inside of which
no event has a future-directed nonspacelike path to spatial infinity.
A "closed trapped surface" is a closed 2-d surface from which light
emitted in a direction outward from the surface actually travels inward.
Infinitesimally inside the above horizon is a closed trapped surface
(and others concentrically all the way down to the center). Such trapped
surfaces are easier to use in proving theorems than the above horizon.
Indeed, there is a theorem that basically says that the existence of a
closed trapped surface implies a singularity somewhere in its interior.
This gets quite complicated and detailed. The standard reference is:
Hawking and Ellis, _The_Large_Scale_Structure_of_Space-Time_.
Tom Roberts
Eric Gisse
wrote:
> In article <x6I2i.233$4Y...@newssvr19.news.prodigy.net>, Tom Roberts says...
>
>
>
> >Daryl McCullough wrote:
> >> So the boundary between "too late to escape"
> >> and "not too late to escape" is a sphere that is expanding
> >> outward at the speed of light. Tom is calling that boundary
> >> the "event horizon".
>
> >This is not just me "calling it that" -- that's what the phrase "event
> >horizon" _means_.
>
> I didn't mean to suggest otherwise, but I think that LEJ Brouwer may be
> thinking of the "event horizon" and the "Schwarzschild radius" as synonymous.
Aren't they, in _this specific case_ ?
Daryl McCullough
>stevendaryl3...@yahoo.com (Daryl McCullough) wrote:
>> In article <x6I2i.233$4Y...@newssvr19.news.prodigy.net>, Tom Roberts says...
>> >Daryl McCullough wrote:
>> >> So the boundary between "too late to escape"
>> >> and "not too late to escape" is a sphere that is expanding
>> >> outward at the speed of light. Tom is calling that boundary
>> >> the "event horizon".
>>
>> >This is not just me "calling it that" -- that's what the phrase "event
>> >horizon" _means_.
>>
>> I didn't mean to suggest otherwise, but I think that LEJ Brouwer may be
>> thinking of the "event horizon" and the "Schwarzschild radius" as synonymous.
>
>Aren't they, in _this specific case_ ?
Definitely not. In the case we are talking about, we have a
converging shell of matter. Inside the shell, there is no
matter to speak of. So the total mass involved in the
scenario is constant, M = the mass of the shell. The
corresponding Schwarzschild radius is also constant,
R = 2GM/c^2. But the event horizon is expanding outwards
at speed c. So they're not the same thing. The event
horizon is not expanding because the Schwarzshchild
radius is expanding. The relationship between them is
just this: the radius of the event horizon approaches
the Schwarzschild radius as t --> infinity.
Tom Roberts
That t is the coordinate of a distant observer. To any observer inside
r=2GM/c^2, who will be directly affected by this, the horizon races past
him at the (local) speed of light, in a finite time after any specified
point on his worldline. This happens before the stars reach him, and no
local measurement by him can determine when the horizon races past. But
as soon as the stars reach him, a finite time after the horizon passes,
he will be inevitably dragged down with them and annihilated at the
singularity which forms.
Relating this to closed trapped surfaces, throughout the region inside
the incoming stars there is no closed trapped surface, even though there
is a horizon present when the stars are close enough to r<2GM/c^2. A
closed trapped surface implies both a singularity inside and a horizon
outside, but a horizon does not imply either a c.t.s or a singularity.
Tom Roberts
Eric Gisse
wrote:
[...]
Oh.
I haven't been quite keeping track of this thread, I can only read the
same incorrect argument so many times. Didn't know y'all had moved on
to a collapsing shell.
LEJ Brouwer
wrote:
OK, but I still do not see how there can be an event horizon in what I
label region 'A', in which I sit. The only event horizon would be in
region 'C', which forms when the stars reach their Schwarzschild
radius.
> >> Consider the period of time between:
> >> a) the stars cross the spatial locus where their horizon will
> >> ultimately be.
> >> and
> >> b) the stars engulf you on the surface of the earth.
>
> >This never happens.
>
> Why do you say that? What prevents it from happening? Are
> you saying this based on General Relativity, or some alternative
> theory?
I have already retracted this statement as it was incorrect, and
summarised what the point I was making was - I am sure you have read
that by now.
- Sabbir.
Daryl McCullough
>OK, but I still do not see how there can be an event horizon in what I
>label region 'A', in which I sit. The only event horizon would be in
>region 'C', which forms when the stars reach their Schwarzschild
>radius.
All that it means for you to be sitting inside the event horizon
is that an outward-directed light signal can never escape to infinity
(to the Schwarzschild exterior solution).
LEJ Brouwer
wrote:
> LEJ Brouwer says...
>
>
>
>
>
> >> >There therefore appear to be three separate regions in the scenario
> >> >above:
>
> >> >(i) The exterior of the horizon, which is just an exterior
> >> >Schwarzschild solution.
>
> >> Okay.
>
> >> >(ii) The spherical region in which I sit, which also has a
> >> >Schwarzschild exterior-like metric, but with a spherical bounding
> >> >surface which shields me from infalling stars, and also prevents
> >> >anything from leaving.
>
> >> I don't think that's correct. Maybe Steve Carlip can explain
> >> to us what this region is like, but it certainly is not like
> >> the Schwarzschild exterior solution. For one thing,
> >> the metric is time-dependent. If you wait long enough, the
> >> curvature will go to infinity (as the stars come crashing
> >> into you).
>
> >You are assuming the answer to prove it.
>
> Okay, I'll turn around and ask you: what reason is there
> for believing that the metric is time-independent, when
> you have massive stars in motion?
One could similarly imagine a situation where the stars were hovering
just outside their Schwarzschild radius, and then allowed to gently
lower themselves to their Schwarzschild radius so that the situation
is very close to being time-independent.
> >Suppose I am the planet Earth
> >and all the stars start converging upon me. Whatever the metric it is
> >in my vicinity, it has a spherical symmetry, and will be approximately
> >Schwarzschild.
>
> Why do you say that? Again, what reason is there to believe
> that the metric is time-independent?
>
> >There will be no net gravitational field acting on me
> >due to the stars due to the symmetry of the problem.
>
> I assume that you are using Newtonian physics here, because
> General Relativity doesn't use "gravitational fields" it uses
> a metric. So you want to reason using Newtonian physics, then
> that's fine for a first cut. In Newtonian physics, it is true
> that (1) the gravitational field is zero inside the shell, and
> (2) after a finite amount of time, the shell of matter crushes
> you.
>
> >When the event horizon forms, nothing happens locally to change
> >this spherical symmetry.
>
> How does it follow that the matter will not crush you?
Okay, the stars will crush me, but anything outside them reaching the
Schwarzschild radius will not crush me more than the stars already
have as their trajectory leads them towards the singularity which is
in region 'C' which has the Schwarzschild interior metric.
> >We have already determined that the Schwarzschild interior
> >solution has an SO(3) symmetry, but it does not have the central
> >spherical symmetry. Therefore the interior solution cannot coincide
> >with the metric around me when the horizon forms.
>
> That was my point. The Schwarzschild interior solution does *not*
> apply on the interior of the shell of matter. It applies in the
> region *between* the event horizon and the shell of matter. Any
> object falling radially inward that is initially outside the
> shell will eventually fall into the Schwarzschild interior region.
I don't think this is correct. You seem to be trying to identify a
thick spherical shell (the region between the Schwarzschild radius and
the position of the stars) with part of the Schwarzschild interior
solution, which has temporal spherical symmetry, not a spatial one.
> >Rather, I will remain quite oblivious of the formation of the horizon
>
> Yes, up until the time that the matter crushes you.
>
> >and whatever the metric was before horizon formation will not change
> >greatly after it.
>
> Right.
>
> >If it is true as claimed that the infalling stars fall into the
> >Schwarzschild interior
>
> I didn't claim that. I claimed that there were three regions:
>
> 1. The region inside the spherical shell of matter.
> 2. The Schwarzschild interior solution.
> 3. The Schwarzschild exterior solution.
>
> The shell is the boundary between regions 1 and 2.
> The event horizon is the boundary between regions 2 and 3.
I still don't see how you can claim that the Schwarzschild interior
solution describes the region between the spherical shell of matter
and their Schwarzschild radius - this region has a spatial spherical
symmetry whereas the Schwarzschild interior does not.
Now I think about it more carefully, I guess that what actually must
happen is that as the spherical shell of matter moves inwards, any
excess mass above the Schwarzschild mass at any given radius will be
diverted towards its own person singularity, so that effectively the
mass of the spherical shell decreases as it approaches the origin,
until by the time it has reached the origin, there is nothing left, so
that I will not be crushed after all. The picture then becomes more
like this:
C C C C C C
x x x x x x
^ ^ ^ ^ ^ ^
| | | | | |
| | | | | |
| | | | | |
A <-o-o-o-o-o-o <-----B
where matter is shed continuously as the spherical shell continues
towards the origin having passed beyond its original Schwarzschild
radius.
- Sabbir.
Tom Roberts
> One could similarly imagine a situation where the stars were hovering
> just outside their Schwarzschild radius, and then allowed to gently
> lower themselves to their Schwarzschild radius so that the situation
> is very close to being time-independent.
To show how silly your statement is, consider just a single star --
certainly a single star headed towards earth will annihilate the planet
when it reaches it. What "magic" do you invoke to imagine that a bunch
of such stars will never reach earth?
Your claims simply do not make sense. <shrug>
Tom Roberts
LEJ Brouwer
> LEJ Brouwer wrote:
> > One could similarly imagine a situation where the stars were hovering
> > just outside their Schwarzschild radius, and then allowed to gently
> > lower themselves to their Schwarzschild radius so that the situation
> > is very close to being time-independent.
>
> To show how silly your statement is, consider just a single star --
> certainly a single star headed towards earth will annihilate the planet
> when it reaches it.
How is this related to the scenario I just described?
> What "magic" do you invoke to imagine that a bunch
> of such stars will never reach earth?
Well they might reach the Earth, but would they reach an atom?
> Your claims simply do not make sense. <shrug>
I am sorry, Tom, but that is simply not good enough. You _must_ try
harder. Now go the back of the class.
LEJ Brouwer
> On May 16, 4:53 pm, stevendaryl3...@yahoo.com (Daryl McCullough)
> wrote:
> > LEJ Brouwer says...
> > >If it is true as claimed that the infalling stars fall into the
Hi Daryl,
I am just reposting this as I didn't get a response from you with
regards to the above.
Best wishes,
Sabbir.