LEJ Brouwer wrote:
JanPB wrote:
No. It should read: "Despite this, we are still able to make a change
of coordinates in which the solution metric takes either of the two
forms:

ds^2 = -A^2 dt^2 + B^2 dr^2 + r^2 dOmega^2
or:
ds^2 = +A^2 dt^2 - B^2 dr^2 + r^2 dOmega^2".

For some mysterious reason (laziness?) most texts insist on the first
metric only (which is the wrong thing to do) and only later pull the
second metric like a rabbit out of hat which naturally looks very fishy
to the reader - exactly as you describe below.

It's probably more likely to do with the fact that there should not be
an explicit time-dependence in the solution of a manifestly
time-independent problem.

We solve the Einstein field equations, and discover the usual exterior
Schwarzschild solution. Now, Mr ****wit, our resident self-proclaimed
GR 'expert', notices that the if we let r 2m, we get another solution
of the field equations,

Actually, if we start off correctly by considering both forms of the
metric then both ranges (r2m and r2m) follow from solving the
equations. The interior is not obtained by "oh, look, I can plug in
r2m and it's still a solution" kind of thing. Some time ago the thread
you initiated made me rederive this solution with some care. I TeX'd it
and put it at http://www.mastersofcinema.org/jan/t.pdf
I hope it addresses some of these issues. I used Cartan's moving frames
as I find Christoffel symbols way too tedious.

I am sure there is nothing wrong with your derivation - the problematic
issue for me is the second of the two equations you set out to solve.

OK, you can't have both. If there is nothing wrong with my derivation
then automatically you cannot complain about the interior portion
because it's no longer pulled out of thin air. Right from the start we
know the metric can be of either of the two forms.

even though it is non-static

We've only assumed spherical symmetry (and signature 2, etc.).

Yes, but coordinate transformations can be made to bring the general
solution to a static form.

I'm losing you again - the interior is not static and it cannot be
"brought" to a static form.

[...]
In order not to upset Mr ****wit
too greatly, he might mention that Mr ****wit's metric may turn out to
be the solution to some, non-static problem, but certainly not the one
we are trying to solve. But also that Mr ****wit should take note that
his metric has a rather nasty singularity right in the middle of it, so
that the chances of it being the solution to any physically reasonable
problem are rather slim.

That's a problem of a sort of different scientific magnitude. It is
thought that this is a consequence of GR being a classical theory and
incorporating QM in it will probably resolve this.

I don't think QM is relevant if we are considering GR on its own.

Right.

I am
slightly confused that Steve Carlip and others worry (understandably)
about the original Schwarzschild solution having an edge, and therefore
disappearing off into nothingness, but in the same breath advocate an
extension to the original solution which contains a singularity into
which all infalling particles disappear into who-knows where. Why is
the same objection not raised about the presence of this singularity,
and why is the extension any better than the original?

Probably because the "edge" is avoidable by extending across the
horizon while the central singularity is unavoidable.

--
Jan Bielawski

Tom Roberts wrote:
LEJ Brouwer wrote:
I 'know' that the Schwarzschild
solution is wrong, and I also 'know' that my proposal must be either
correct, or if not completely correct at least on the right path.

The rest of us want to do physics, not whatever it is you are trying to
do. What God told you this? Why do you attempt to discuss such divine
revelations in a physics newsgroup?

Well, it is true that all knowledge ultimately stems from God, but it
is not the way in which the ideas are revealed, but the ideas
themselves which I am discussing. Having said, that I refer you to the
earlier thread entitled "Cosmogony from the Book of Genesis" in which I
give a possible physical interpretation of the opening lines of the Old
Testament.

I
can't tell you precisely how I know - it is just a very strong gut
feeling, and when I feel like this, I am usually right.

Here all you've shown is that you do not understand the MANY papers and
books that have been written about this. You merely re-hash old
objections long refuted, and old mistakes long corrected.

Maybe, maybe not.

I actually admire you a great deal. You are like a walking
encyclopaedia on gravity, yet you do not appear to be at all
pretentious or arrogant about it.

Yes, Steve Carlip is all of that.


BTW, could you please explain what you mean when you say that my
infinite cone has an 'edge'?

I assume you mean your attempt to glue the two exterior regions of the
Kruskal manifold together. The "edge" occurs when one follows an
infalling timelike geodesic -- when it reaches r=2M all of a sudden it
is impossible to compute the geodesic, because the metric is not C^2
there.

But it's the same metric - with dt - -dt and dr - -dr. My claim is
ridiculous because I am saying that the infalling particle just bounces
off the event horizon with time reversed, and I have not given any
satisfactory physical reason as to why this should happen, except to
note that the event horizon is a very strange place, so who knows what
can happen there?

Steve implied there is a boundary there, but I believe this can
be done such that the manifold is continuous there, just not smooth.
This is not a viable physical model because the Einstein field equation
must be valid everywhere, and it cannot be valid on either a boundary or
a locus where the metric is not C^2.

One can glue the two regions together there topologically.
But in doing that one must clearly distort the Kruskal
plane (i.e. the U-V coordinate plane) -- that is OK because
that can be a diffeomorphism that carries the metric
along; but at best the metric can be only C^0: for the metric
to be C^n its first n derivatives must all be equal at the
join, and the symmetry of the two exterior regions means they
must vanish; for this metric the first derivative is nonzero.

Note that on physical grounds the metric must be C^2 for two
different reasons: to satisfy the EFE, and for geodesic paths
to be C^1 (a worldline must have a 4-velocity everywhere).

["C^n" means continuously differentiable n times.]

[Hmmm. The U-V plane suppresses the two angles; I am not 100%
certain that those suppressed dimensions do not prevent
the gluing I describe; I assume that it is OK. You also
implicitly assumed this is OK.]



The trouble with physic(ist)s is not that we are "not even wrong", but
rather, from your point of view, the trouble is that we don't accept
your "strong gut feeling" as evidence of anything except the fact that
you are not doing science. shrug


Tom Roberts

There are many ways to solve a jigsaw puzzle. Just because I don't
start with the bottom left corner and work my rightwards and upwards
doesn't mean I won't solve it in the end. Besides, my method seems to
be much more efficient than yours - even if I do make more incorrect
guesses along the way. What is your definition of 'science' anyway?

- Sabbir.

That was the story of the scorpion and the frog, when once upon a time, a
scorpion asked a frog to help him to cross the river before.

Whether, the socorpion was ressuring the frog, that he would not sting him,
because as he knows, if he could he would be drown too along the river.

However, a halfway across, the scorpion did sting him, and when the sinking
frog asked why, the scorpion replied, that it is that way he does makes a
living.


--
Ahmed Ouahi, Architect
Best Regards!


"LEJ Brouwer" wrote in message
ups.com...

wrote:
We realize that we have been rather impolite to a number of people who,
as it turns out, know quite a bit more physics and mathematics than we
do. We politely ask their pardon.

I just realised that I cannot apologise to anyone on s.p.research as my
apology would be 'overly speculative'! :-)

JanPB wrote:

OK, you can't have both. If there is nothing wrong with my derivation
then automatically you cannot complain about the interior portion
because it's no longer pulled out of thin air. Right from the start we
know the metric can be of either of the two forms.

But we chose it to be of the first form. Anyway, for the sake of
argument, let's suppose that you are right - but let's also suppose
that I am right and another solution exists which only requires the two
exterior patches. We can't have both. Yours has a singularity in the
middle of it and mine doesn't. Mine leads to the prediction of
classical electrodynamics, modified Newtonian dynamics, stochastic
quantum mechanics, and the standard model gauge group.

I think I prefer my solution.

I'm losing you again - the interior is not static and it cannot be
"brought" to a static form.

You are only saying that because you are peeking at your proposed
solution. There is a priori no separation between 'interior' and
'exterior', My solution doesn't have an 'interior' which is static or
otherwise.

I don't think QM is relevant if we are considering GR on its own.

Right.

I am
slightly confused that Steve Carlip and others worry (understandably)
about the original Schwarzschild solution having an edge, and therefore
disappearing off into nothingness, but in the same breath advocate an
extension to the original solution which contains a singularity into
which all infalling particles disappear into who-knows where. Why is
the same objection not raised about the presence of this singularity,
and why is the extension any better than the original?

Probably because the "edge" is avoidable by extending across the
horizon while the central singularity is unavoidable.

But that's even worse isn't it?

--
Jan Bielawski

Cheers,

Sabbir.

"Tom Roberts" kirjoitti
igy.com...
LEJ Brouwer wrote:
I 'know' that the Schwarzschild
solution is wrong, and I also 'know' that my proposal must be either
correct, or if not completely correct at least on the right path.

The rest of us want to do physics, not whatever it is you are trying to
do. What God told you this? Why do you attempt to discuss such divine
revelations in a physics newsgroup?


I
can't tell you precisely how I know - it is just a very strong gut
feeling, and when I feel like this, I am usually right.

Here all you've shown is that you do not understand the MANY papers and
books that have been written about this. You merely re-hash old objections
long refuted, and old mistakes long corrected.


I actually admire you a great deal. You are like a walking
encyclopaedia on gravity, yet you do not appear to be at all
pretentious or arrogant about it.

Yes, Steve Carlip is all of that.


BTW, could you please explain what you mean when you say that my
infinite cone has an 'edge'?

I assume you mean your attempt to glue the two exterior regions of the
Kruskal manifold together. The "edge" occurs when one follows an infalling
timelike geodesic -- when it reaches r=2M all of a sudden it is impossible
to compute the geodesic, because the metric is not C^2 there. Steve
implied there is a boundary there, but I believe this can be done such
that the manifold is continuous there, just not smooth. This is not a
viable physical model because the Einstein field equation must be valid
everywhere, and it cannot be valid on either a boundary or a locus where
the metric is not C^2.

One can glue the two regions together there topologically.
But in doing that one must clearly distort the Kruskal
plane (i.e. the U-V coordinate plane) -- that is OK because
that can be a diffeomorphism that carries the metric
along; but at best the metric can be only C^0: for the metric
to be C^n its first n derivatives must all be equal at the
join, and the symmetry of the two exterior regions means they
must vanish; for this metric the first derivative is nonzero.

Note that on physical grounds the metric must be C^2 for two
different reasons: to satisfy the EFE, and for geodesic paths
to be C^1 (a worldline must have a 4-velocity everywhere).

["C^n" means continuously differentiable n times.]

[Hmmm. The U-V plane suppresses the two angles; I am not 100%
certain that those suppressed dimensions do not prevent
the gluing I describe; I assume that it is OK. You also
implicitly assumed this is OK.]



The trouble with physic(ist)s is not that we are "not even wrong", but
rather, from your point of view, the trouble is that we don't accept your
"strong gut feeling" as evidence of anything except the fact that you are
not doing science. shrug

Tom Roberts

I don't like that "we, physicists". You are insulting the real physicist who
are working hard to find the truth. They don't tell lies.

Henry Haapalainen

Henry Haapalainen wrote:
"Tom Roberts" kirjoitti
igy.com...
LEJ Brouwer wrote:
I 'know' that the Schwarzschild
solution is wrong, and I also 'know' that my proposal must be either
correct, or if not completely correct at least on the right path.

The rest of us want to do physics, not whatever it is you are trying to
do. What God told you this? Why do you attempt to discuss such divine
revelations in a physics newsgroup?


I
can't tell you precisely how I know - it is just a very strong gut
feeling, and when I feel like this, I am usually right.

Here all you've shown is that you do not understand the MANY papers and
books that have been written about this. You merely re-hash old objections
long refuted, and old mistakes long corrected.


I actually admire you a great deal. You are like a walking
encyclopaedia on gravity, yet you do not appear to be at all
pretentious or arrogant about it.

Yes, Steve Carlip is all of that.


BTW, could you please explain what you mean when you say that my
infinite cone has an 'edge'?

I assume you mean your attempt to glue the two exterior regions of the
Kruskal manifold together. The "edge" occurs when one follows an infalling
timelike geodesic -- when it reaches r=2M all of a sudden it is impossible
to compute the geodesic, because the metric is not C^2 there. Steve
implied there is a boundary there, but I believe this can be done such
that the manifold is continuous there, just not smooth. This is not a
viable physical model because the Einstein field equation must be valid
everywhere, and it cannot be valid on either a boundary or a locus where
the metric is not C^2.

One can glue the two regions together there topologically.
But in doing that one must clearly distort the Kruskal
plane (i.e. the U-V coordinate plane) -- that is OK because
that can be a diffeomorphism that carries the metric
along; but at best the metric can be only C^0: for the metric
to be C^n its first n derivatives must all be equal at the
join, and the symmetry of the two exterior regions means they
must vanish; for this metric the first derivative is nonzero.

Note that on physical grounds the metric must be C^2 for two
different reasons: to satisfy the EFE, and for geodesic paths
to be C^1 (a worldline must have a 4-velocity everywhere).

["C^n" means continuously differentiable n times.]

[Hmmm. The U-V plane suppresses the two angles; I am not 100%
certain that those suppressed dimensions do not prevent
the gluing I describe; I assume that it is OK. You also
implicitly assumed this is OK.]



The trouble with physic(ist)s is not that we are "not even wrong", but
rather, from your point of view, the trouble is that we don't accept your
"strong gut feeling" as evidence of anything except the fact that you are
not doing science. shrug

Tom Roberts

I don't like that "we, physicists". You are insulting the real physicist who
are working hard to find the truth. They don't tell lies.

In what way are you a physicist?


Henry Haapalainen

Tom Roberts wrote:
LEJ Brouwer wrote:
I 'know' that the Schwarzschild
solution is wrong, and I also 'know' that my proposal must be either
correct, or if not completely correct at least on the right path.

The rest of us want to do physics, not whatever it is you are trying to
do. What God told you this? Why do you attempt to discuss such divine
revelations in a physics newsgroup?


I
can't tell you precisely how I know - it is just a very strong gut
feeling, and when I feel like this, I am usually right.

Here all you've shown is that you do not understand the MANY papers and
books that have been written about this. You merely re-hash old
objections long refuted, and old mistakes long corrected.


I actually admire you a great deal. You are like a walking
encyclopaedia on gravity, yet you do not appear to be at all
pretentious or arrogant about it.

Yes, Steve Carlip is all of that.


BTW, could you please explain what you mean when you say that my
infinite cone has an 'edge'?

I assume you mean your attempt to glue the two exterior regions of the
Kruskal manifold together. The "edge" occurs when one follows an
infalling timelike geodesic -- when it reaches r=2M all of a sudden it
is impossible to compute the geodesic, because the metric is not C^2
there. Steve implied there is a boundary there, but I believe this can
be done such that the manifold is continuous there, just not smooth.

I think the manifold can be glued smoothly (one can write down a smooth
atlas on the quotient manifold) - it's the metric that has a "crease"
at the glued horizon. I posted an outline of an argument few minutes
ago. The problem is essentially with the Kruskal-Szekeres function
r(T,X) (the one given by the usual implicit equation) - it has a
nonzero slope at the horizon which would "crease" the metric there when
glued.

--
Jan Bielawski

JanPB wrote:

I think the manifold can be glued smoothly (one can write down a smooth
atlas on the quotient manifold) - it's the metric that has a "crease"
at the glued horizon. I posted an outline of an argument few minutes
ago.

Forgot to add "on the 'Any coordinate system in GR?' thread". I posted
a graph of r there.

--
Jan Bielawski

LEJ Brouwer wrote:
Why is falling into a singularity of spacetime any better than falling
off the edge of spacetime?

Because at a curvature singularity the metric is not well defined, and
the entire structure of the theory breaks down. At an edge nothing
special happens, except that you deleted a region of the complete
manifold, which is clearly artificial and unphysical.


I don't see why the incompleteness of the original solution at all
implies that the assumption of a static point particle is inconsistent.

Because the entire manifold is known to be the only spherically
symmetric static solution at large distances, but it is not static in
the region r2M.

Basically GR is nonlinear, and strange things can happen in a strong
field situation, such as inside the horizon in the Schw. manifold. Your
naive expectations and desires need not be valid in regions far removed
from your personal experience. shrug


Sure, the solution you propose is non-static, but that does not
necessarily imply that static solutions do not exist.

In this case we know there is no other solution, static or otherwise.
This is known as Birkhoff's theorem. shrug


Unless there is a
uniqueness theorem which states otherwise, like "only _maximal_
extensions" are valid. I prefer "only singularity-free solutions are
valid".

Birkhoff's theorems states that any vacuum region of any spherically
symmetric solution to the Einstein field equation is isometric to some
region of the Schwarzschild manifold (there are surely other caveats I'm
omitting).

You may prefer "singularity-free solutions", but mathematics is not
bound by your personal preferences, and in this case your preference
cannot be met. shrug


What edge? It is an infinite cone. There is a pointy bit in the middle,
but I can't say that bothers me at all.

No geodesic can cross that "pointy bit" in a self-consistent manner, and
the EFE is not valid there. Either of those would remove your construct
from consideration. See my other post for details -- the metric cannot
be C^2 at this edge, but any sensible manifold must have a metric that
is at least C^2 everywhere.


Tom Roberts

JanPB wrote:
Tom Roberts wrote:
I assume you mean your attempt to glue the two exterior regions of the
Kruskal manifold together. The "edge" occurs when one follows an
infalling timelike geodesic -- when it reaches r=2M all of a sudden it
is impossible to compute the geodesic, because the metric is not C^2
there. Steve implied there is a boundary there, but I believe this can
be done such that the manifold is continuous there, just not smooth.

I think the manifold can be glued smoothly (one can write down a smooth
atlas on the quotient manifold) - it's the metric that has a "crease"
at the glued horizon. I posted an outline of an argument few minutes
ago. The problem is essentially with the Kruskal-Szekeres function
r(T,X) (the one given by the usual implicit equation) - it has a
nonzero slope at the horizon which would "crease" the metric there when
glued.

I agree. In the last line of my statement above, I actually meant that
the metric is not smooth, as discussed in the following paragraphs of
the original post.

But as I said, since the metric can be at most C^0 at this "crease",
this is not a viable construct, for the reasons I gave.


Tom Roberts