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Misinterpretation of the radial parameter in the Schwarzschild solution?
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LEJ Brouwer
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Joined: 07 May 2005
Posts: 120

PostPosted: Mon Jul 10, 2006 1:23 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

JanPB wrote:
Quote:
LEJ Brouwer wrote:
JanPB wrote:

What do you mean "not allowed"? True, the metric for r<2M does not
satisfy the constraint you set up for yourself (namely, it's not
static) and it may contain other hidden assumptions (like r<infty) but
so what? It's still a solution to Einstein's equation and the
requirement for being static, etc., was just a simplifying (i.e., ad
hoc, random, man-made) assumption which in the end just turned out to
be an overkill. So you ended up with a more general solution covering a
larger region (the extra region - the interior - is disconnected, in
the topological sense, from the static exterior).

I refer you to the reply I gave to T.Essel's post. It is clearly not a
solution to the Einstein equation if the interior region yuo refer to
does not even physically exist.

Quote:
and (b) timelike and
spacelike directions cannot just swap places, which is what would have
to happen if matter were to fall past the horizon.

This is a mathematical feature of the Schwarzschild representation of
the metric. Nothing really swaps places, it's just that the
mathematical process of solving the relevant differential equations
mirrors one another over the two components of the disconnected domain
so the *letters* r and t end up denoting different things for r<2M
and r>2M due to what is basically an abuse of notation.

I see. And I take it that you will also remember to swap these letters
in the Einstein field equations? Why not go ahead and also swap M and
theta? In fact, why not change 'ds^2' to 'sd^2' too, just for the hell
of it? They are just letters after all.

- Sabbir.
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koobee.wublee@gmail.com
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Joined: 01 Feb 2006
Posts: 141

PostPosted: Mon Jul 10, 2006 3:52 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

"LEJ Brouwer" <intuitionist1@yahoo.com> wrote in message
news:1152494070.667050.19320@s13g2000cwa.googlegroups.com...

Quote:
[...]
Recall that in the now standard "Scharzschild exterior chart" for the
Schwarzschild vacuum, the line element expressing the Schwarzschild vacuum
solution takes the form

ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + r^2 dOmega^2,

2m < r < infty, 0 < theta < pi, -pi < phi < pi

I have often argued that physicists must learn the habit of -always-
stating coordinate ranges, since -failing- to do so can easily mislead
students, while -following- this precept can prevent embarrassing errors.

Well, essentially that's what I have been arguing all along.
[...]

Schwarzschild metric is only valid in free space where the mass density
is zero. The characterize where (r <= 2 G M / c^2), you have to find
another metric that satisfy the field equations below.

R_ij - R g_ij / 2 = 8 pi G Lm g_ij / c^4, where in this case (Lm > 0).

While forming a black hole, as you have correctly pointed out, it can
only form at eternal time. In doing so, the above equation must be
valid at some point to give another solution different from the
Schwarzschild metric, and very possibly it will prevent the formation
of a black hole or pushing the event horizon down to zero.
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Jan Bielawski
science forum Guru


Joined: 08 May 2005
Posts: 388

PostPosted: Mon Jul 10, 2006 5:59 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

LEJ Brouwer wrote:
Quote:
JanPB wrote:
LEJ Brouwer wrote:
JanPB wrote:

What do you mean "not allowed"? True, the metric for r<2M does not
satisfy the constraint you set up for yourself (namely, it's not
static) and it may contain other hidden assumptions (like r<infty) but
so what? It's still a solution to Einstein's equation and the
requirement for being static, etc., was just a simplifying (i.e., ad
hoc, random, man-made) assumption which in the end just turned out to
be an overkill. So you ended up with a more general solution covering a
larger region (the extra region - the interior - is disconnected, in
the topological sense, from the static exterior).

I refer you to the reply I gave to T.Essel's post.

I can't see it.

Quote:
It is clearly not a
solution to the Einstein equation if the interior region yuo refer to
does not even physically exist.

*If*. You haven't shown this. If 2+2=5 then all sorts of marvellous
results follow.

Quote:
and (b) timelike and
spacelike directions cannot just swap places, which is what would have
to happen if matter were to fall past the horizon.

This is a mathematical feature of the Schwarzschild representation of
the metric. Nothing really swaps places, it's just that the
mathematical process of solving the relevant differential equations
mirrors one another over the two components of the disconnected domain
so the *letters* r and t end up denoting different things for r<2M
and r>2M due to what is basically an abuse of notation.

I see. And I take it that you will also remember to swap these letters
in the Einstein field equations?
Why not go ahead and also swap M and
theta? In fact, why not change 'ds^2' to 'sd^2' too, just for the hell
of it? They are just letters after all.

No, this is incorrect. Very briefly the logic of finding this solution
goes like this. You restrict yourself to spherically symmetric static
metrics and seek solutions of this type. This forces certain
constraints on the coefficients of the metric. You denote the timelike
coordinate by t and the spacelike coordinate by r. You obtain a
solution which is only valid for r>2M (because of the assumption of t
timelike and r spacelike).

_Then_ you look at it and suddenly you notice that for r<2M the formula
still describes _a_ metric of Lorentz signature. This metric is not the
one obtained from the derivation, so to make sure it satisfies the
Einstein equation, you have to check it. Of course this is trivial
because the r<2M case is sufficiently close in purely mathematical
terms to the metric with r>2M that it's obvious its Ricci tensor is
also zero.

But note that the meaning of each coordinate in the new (r<2M) case has
nothing a priori to do with the coordinates that _happen_ to be denoted
by the same letters in the r>2M case - that's because you simply
mechanically "extended" your solution to a region not covered by the
assumptions of your derivation, where by "extended" one means simply
_mechanical copying_ of the original (r>2M) formula, retaining the
original Latin and Greek symbols and all - out of pure human laziness.
This is the abuse of notation I referred to previously.

There is no a priori physical connection between the two sets of Latin
and Greek symbols - they are the same only because we didn't feel like
using new symbols for the new metric.

The meaning of the coordinate symbols in the new metric must be deduced
from the form of the coefficients, not from the _names_ of the
coordinates! In particular, what is denoted by r is time and what is
denoted by t is a spatial coordinate because of the signs.

--
Jan Bielawski
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Sue...
science forum Guru


Joined: 08 May 2005
Posts: 2684

PostPosted: Mon Jul 10, 2006 6:26 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

JanPB wrote:
Quote:
LEJ Brouwer wrote:
JanPB wrote:
LEJ Brouwer wrote:
JanPB wrote:

What do you mean "not allowed"? True, the metric for r<2M does not
satisfy the constraint you set up for yourself (namely, it's not
static) and it may contain other hidden assumptions (like r<infty) but
so what? It's still a solution to Einstein's equation and the
requirement for being static, etc., was just a simplifying (i.e., ad
hoc, random, man-made) assumption which in the end just turned out to
be an overkill. So you ended up with a more general solution covering a
larger region (the extra region - the interior - is disconnected, in
the topological sense, from the static exterior).

I refer you to the reply I gave to T.Essel's post.

I can't see it.

It is clearly not a
solution to the Einstein equation if the interior region yuo refer to
does not even physically exist.

*If*. You haven't shown this. If 2+2=5 then all sorts of marvellous
results follow.

and (b) timelike and
spacelike directions cannot just swap places, which is what would have
to happen if matter were to fall past the horizon.

This is a mathematical feature of the Schwarzschild representation of
the metric. Nothing really swaps places, it's just that the
mathematical process of solving the relevant differential equations
mirrors one another over the two components of the disconnected domain
so the *letters* r and t end up denoting different things for r<2M
and r>2M due to what is basically an abuse of notation.

I see. And I take it that you will also remember to swap these letters
in the Einstein field equations?
Why not go ahead and also swap M and
theta? In fact, why not change 'ds^2' to 'sd^2' too, just for the hell
of it? They are just letters after all.

No, this is incorrect. Very briefly the logic of finding this solution
goes like this. You restrict yourself to spherically symmetric static
metrics and seek solutions of this type. This forces certain
constraints on the coefficients of the metric. You denote the timelike
coordinate by t and the spacelike coordinate by r. You obtain a
solution which is only valid for r>2M (because of the assumption of t
timelike and r spacelike).

_Then_ you look at it and suddenly you notice that for r<2M the formula
still describes _a_ metric of Lorentz signature. This metric is not the
one obtained from the derivation, so to make sure it satisfies the
Einstein equation, you have to check it. Of course this is trivial
because the r<2M case is sufficiently close in purely mathematical
terms to the metric with r>2M that it's obvious its Ricci tensor is
also zero.

But note that the meaning of each coordinate in the new (r<2M) case has
nothing a priori to do with the coordinates that _happen_ to be denoted
by the same letters in the r>2M case - that's because you simply
mechanically "extended" your solution to a region not covered by the
assumptions of your derivation, where by "extended" one means simply
_mechanical copying_ of the original (r>2M) formula, retaining the
original Latin and Greek symbols and all - out of pure human laziness.
This is the abuse of notation I referred to previously.

There is no a priori physical connection between the two sets of Latin
and Greek symbols - they are the same only because we didn't feel like
using new symbols for the new metric.

The meaning of the coordinate symbols in the new metric must be deduced
from the form of the coefficients, not from the _names_ of the
coordinates! In particular, what is denoted by r is time and what is
denoted by t is a spatial coordinate because of the signs.

Yes... you are correct.Where the appearance of a distant moving
clock is to be corrected then the term 'proper time' has a literal
application to the coordinate. (Special Relativity)

When a space-time is being defined to represent energy density
as volume, then the temporal component becomes imaginary.
(General Relativity and the Swartzchild solution)

The second URL below explains in nearly identical terms.

http://farside.ph.utexas.edu/teaching/em/lectures/node114.html
http://www.aoc.nrao.edu/~smyers/courses/astro12/speedoflight.html

Sue...

Quote:

--
Jan Bielawski
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koobee.wublee@gmail.com
science forum Guru Wannabe


Joined: 01 Feb 2006
Posts: 141

PostPosted: Wed Jul 12, 2006 7:24 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

"LEJ Brouwer" <intuitionist1@yahoo.com> wrote in message
news:1152183483.961504.63390@j8g2000cwa.googlegroups.com...

Quote:
The moderators of s.p.research have decided that Steve Carlip's
response has settled this issue, and refuse to allow further
discussion. I disagree, so I am continuing the thread here, and would
like to invite Steve to respond if he wishes - Sabbir.

Both Schwarzschild and Hilbert made errors. Refer to the following
article written by Schwarzschild himself.

http://arxiv.org/abs/physics/9905030

Equation (4) is only valid when the metric has a determinant of -1.
Subsequent derivation of the Schwarzschild metric has been derived
based on this error. In spherical polar cordinate, the determinant of
the metric is not -1. The retangular coordinate should not be -1
either in which Schwarzschild made another mistake in his equation
right above equation (6). Equating the spherical polar coordinate with
the rectangular one, we have

dx^2 + dy^2 + dz^2 = dr^2 + r^2 cos^2V dH^2 + r^2 dV^2

Where

** r^2 = x^2 + y^2 + z^2
** H = longitude
** V = latitude

We can also write

dx^2 + dy^2 + dz^2 = (x dx + y dy + z dz)^2 / r^2 + ((y^2 + z^2) dx^2 +
(z^2 + x^2) dy^2 + (x^2 + y^2) dz^2 - 2 x y d xdy - 2 y z dy dz - 2 z x
dz dx) / r^2

Where

** dr^2 = (x dx + y dy + z dz)^2 / r^2
** r^2 (cos^2V dH^2 + dV^2) = ((y^2 + z^2) dx^2 + (z^2 + x^2) dy^2 +
(x^2 + y^2) dz^2 - 2 x y dx dy - 2 y z dy dz - 2 z x dz dx) / r^2

This is basic geometry so far. This is flat space so far. Now, adding
the curvature in the radial direction as well as the angular direction,
we have

F dr^2 + G r^2 (cos^2V dH^2 + dV^2) = (F^2 x^2 + G y^2 + G z^2) dx^2 +
(G^2 x^2 + F y^2 + G z^2) dy^2 + (G^2 x^2 + G y^2 + F z^2) dx^2 + 2 ( F
- G) (x y dx dy + y z dy dz + z x dz dx)

Common sense says G = 1, F = 1 at r = infinity.

The spatial components of the metric in spherical polar coordinate is

** g_rr = F
** g_HH = G r^2 cos^2V
** g_VV = G r^2
** g_rH = g_rV = g_HV = 0

The spatial components of the metric in rectangular coordinate is

** g_xx = (F^2 x^2 + G y^2 + G z^2) / r^2
** g_yy = (G^2 x^2 + F y^2 + G z^2) / r^2
** g_xx = (G^2 x^2 + G y^2 + F z^2) / r^2
** g_xy = 2 (F - G) x y / r^2
** g_yz = 2 (F - G) y z / r^2
** g_zx = 2 (F - G) z x / r^2

This is not the same as what Schwarzschild wrote down. Both metrics
don't have determinants equivalent to -1. Using the more complicated
Ricci Tensor, there is no way anyone can arrive at the Schwarzschild
metric with Ricci Tensor equals to zero in free space. Ricci Tensor is
wrong. Einstein's Field Equations are total nonsense.

There are only 9 non-zero Christoffel Symbols of the 2nd kind in
spherical polar coordinate. It is a start. However, it is getting
late for me. If there is anyone interested, I can post more of the
mechanical derivations.
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Jan Bielawski
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Joined: 08 May 2005
Posts: 388

PostPosted: Wed Jul 12, 2006 6:09 pm    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

Koobee Wublee wrote:
Quote:
"LEJ Brouwer" <intuitionist1@yahoo.com> wrote in message
news:1152183483.961504.63390@j8g2000cwa.googlegroups.com...

The moderators of s.p.research have decided that Steve Carlip's
response has settled this issue, and refuse to allow further
discussion. I disagree, so I am continuing the thread here, and would
like to invite Steve to respond if he wishes - Sabbir.

Both Schwarzschild and Hilbert made errors. Refer to the following
article written by Schwarzschild himself.

http://arxiv.org/abs/physics/9905030

Equation (4) is only valid when the metric has a determinant of -1.
Subsequent derivation of the Schwarzschild metric has been derived
based on this error.

It's not an error, just a historical and - as Einstein realised before
November 1915 - unnecessary restriction on allowable coordinate
changes.

Quote:
[...]
This is not the same as what Schwarzschild wrote down. Both metrics
don't have determinants equivalent to -1. Using the more complicated
Ricci Tensor, there is no way anyone can arrive at the Schwarzschild
metric with Ricci Tensor equals to zero in free space. Ricci Tensor is
wrong. Einstein's Field Equations are total nonsense.

Study this stuff some more, and seriously, then you won't have to waste
your time on nonexisting problems.

--
Jan Bielawski
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koobee.wublee@gmail.com
science forum Guru Wannabe


Joined: 01 Feb 2006
Posts: 141

PostPosted: Thu Jul 13, 2006 4:50 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

"JanPB" <filmart@gmail.com> wrote in message
news:1152727769.721703.143900@75g2000cwc.googlegroups.com...

Quote:
Both Schwarzschild and Hilbert made errors. Refer to the following
article written by Schwarzschild himself.

http://arxiv.org/abs/physics/9905030

Equation (4) is only valid when the metric has a determinant of -1.
Subsequent derivation of the Schwarzschild metric has been derived
based on this error.

It's not an error, just a historical and - as Einstein realised before
November 1915 - unnecessary restriction on allowable coordinate
changes.

It is not an error for Einstein because he changed the rules of
mathematics. However, it remains a problem for everyone else.

Quote:
[...]
This is not the same as what Schwarzschild wrote down. Both metrics
don't have determinants equivalent to -1. Using the more complicated
Ricci Tensor, there is no way anyone can arrive at the Schwarzschild
metric with Ricci Tensor equals to zero in free space. Ricci Tensor is
wrong. Einstein's Field Equations are total nonsense.

Study this stuff some more, and seriously, then you won't have to waste
your time on nonexisting problems.

Assuming the angular distortion is unity just like the Schwarzschild
metric. The radial distortion from one field equation looks like

g_11 = 1 / (2 + k sqrt(r))

Where

** k = integration constant

Versus

g_11 = 1 / (1 + k U)

Where

** U = G M / r / c^2

I have not even bothered to calculate g_00 from another field equation.
It appears not to be (1 / g_11) though. The other two field equations
are identical because we have assumed no angular distortion in space.

You are right. I have spent a lot of time doing this, and I feel
satisfied. The field equations are utterly nonsense because of the
mathematical errors made by Schwarzschild and then Hilbert according to
Mr. Rahman.
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Tom Roberts
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Joined: 24 Mar 2005
Posts: 1399

PostPosted: Thu Jul 13, 2006 5:09 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

LEJ Brouwer wrote:
Quote:
How can spacelike and timelike directions suddenly swap places if there
is nothing unusual going on at the event horizon?

That does not happen. What does happen, when using the standard Schw.
coordinates, is that the coordinates LABELED r and t swap their
spacelike and timelike characters at the horizon. But that's OK, because
neither set of coordinates is valid at the horizon -- these TWO sets of
coordinates are valid on disjoint regions of the manifold, and how the
labels get assigned is of no physical significance whatsoever.

This is a source of confusion to students today, and was
confusing to experts until the 1960s or so.


Quote:
This is completely
unphysically,

No physical anything happens. Merely the labels of coordinates get shuffled.


Quote:
yet is accepted in a completely matter-of-fact way in all
standard textbooks. Anyone considering this objectively should
immediately smell a rat here.

Only someone who does not understand what is happening would say such
silly things.


Quote:
the main point I am making which is the fact that r<2M has no
meaning for the Schwarzschild coordinates,

You don't seem to realize that there are _TWO_SET_ of Schwarzschild
coordinates. One set is valid for r>2M and the other is valid for r<2M.
They have no intersection. Yes, the exterior coordinates are not valid
in the region r<2M. <shrug>


Quote:
so that the usual interior
solutions are invalid.

This is not true. They are valid, but are covered by a different chart
in the Schw. coordinates.


Tom Roberts
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Tom Roberts
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Joined: 24 Mar 2005
Posts: 1399

PostPosted: Thu Jul 13, 2006 5:15 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

JanPB wrote:
Quote:
The meaning of the coordinate symbols in the [interior] metric must be
deduced from the form of the coefficients, not from the _names_ of the
coordinates! In particular, what is denoted by r is time and what is
denoted by t is a spatial coordinate because of the signs.

And to make it even more confusing to the unwary, in the region r<2M the
future-pointing basis vector is -d/dr (note the minus sign) -- that is
what makes the singularity at r=0 be in the future of every timelike
path in this region (such paths enter this region at r=2M, which is
greater than 0).


Tom Roberts
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LEJ Brouwer
science forum Guru Wannabe


Joined: 07 May 2005
Posts: 120

PostPosted: Thu Jul 13, 2006 2:48 pm    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

Tom Roberts wrote:
Quote:
LEJ Brouwer wrote:
How can spacelike and timelike directions suddenly swap places if there
is nothing unusual going on at the event horizon?

That does not happen. What does happen, when using the standard Schw.
coordinates, is that the coordinates LABELED r and t swap their
spacelike and timelike characters at the horizon. But that's OK, because
neither set of coordinates is valid at the horizon -- these TWO sets of
coordinates are valid on disjoint regions of the manifold, and how the
labels get assigned is of no physical significance whatsoever.

This is a source of confusion to students today, and was
confusing to experts until the 1960s or so.


This is completely
unphysically,

No physical anything happens. Merely the labels of coordinates get shuffled.

As I said before, if you are going to 'merely' shuffle the labels of
the coordinates, then you would have to 'merely' shuffle them in the
Einstein field equations too - which means that what you are now
solving are not the Einstein field equations.

Quote:
yet is accepted in a completely matter-of-fact way in all
standard textbooks. Anyone considering this objectively should
immediately smell a rat here.

Only someone who does not understand what is happening would say such
silly things.

Quite the contrary, only someone who does not understand would _not_
say such things. The pre-1960s experts were right to be concerned.

Quote:
the main point I am making which is the fact that r<2M has no
meaning for the Schwarzschild coordinates,

You don't seem to realize that there are _TWO_SET_ of Schwarzschild
coordinates. One set is valid for r>2M and the other is valid for r<2M.
They have no intersection. Yes, the exterior coordinates are not valid
in the region r<2M. <shrug

What you still don't seem to realise is that r<2m does not physically
exist, so there is only ONE valid set of coordinates, i.e. r>2m. I
pointed this out at the start of the thread (in fact this is the whole
point of bringing the matter up), and I even went through the proof of
it in another post. Obviously its easier to just ignore these facts and
mindlessly regurgitate what the textbooks tell you, but perhaps you
could try thinking for yourself for a change. It makes a world of
difference.

Quote:
so that the usual interior
solutions are invalid.

This is not true. They are valid, but are covered by a different chart
in the Schw. coordinates.


Tom Roberts

- Sabbir.
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Dirk Van de moortel
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Joined: 01 May 2005
Posts: 3019

PostPosted: Thu Jul 13, 2006 4:12 pm    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

"Koobee Wublee" <koobee.wublee@gmail.com> wrote in message news:1152766216.377500.63490@35g2000cwc.googlegroups.com...
Quote:
"JanPB" <filmart@gmail.com> wrote in message
news:1152727769.721703.143900@75g2000cwc.googlegroups.com...

Both Schwarzschild and Hilbert made errors. Refer to the following
article written by Schwarzschild himself.

http://arxiv.org/abs/physics/9905030

Equation (4) is only valid when the metric has a determinant of -1.
Subsequent derivation of the Schwarzschild metric has been derived
based on this error.

It's not an error, just a historical and - as Einstein realised before
November 1915 - unnecessary restriction on allowable coordinate
changes.

It is not an error for Einstein because he changed the rules of
mathematics. However, it remains a problem for everyone else.

"He changed the rules of mathematics" Smile
Why do you bother breaking your mind over differential
geometry if you can't even handle baby analytic geometry?
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SRBogus.html

Dirk Vdm
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Bilge
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Joined: 30 Apr 2005
Posts: 2816

PostPosted: Thu Jul 13, 2006 11:24 pm    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

LEJ Brouwer:
Quote:
Tom Roberts wrote:

No physical anything happens. Merely the labels of coordinates
get shuffled.

As I said before, if you are going to 'merely' shuffle the labels of
the coordinates, then you would have to 'merely' shuffle them in the
Einstein field equations too - which means that what you are now
solving are not the Einstein field equations.

Don't be ridiculous. The metric has one timelike and three spacelike
coordinates. For a spacelike metric, -+++, the entry with the - sign
is timelike, regardless of what label you give it.

[...]
Quote:
Only someone who does not understand what is happening would say such
silly things.

Quite the contrary, only someone who does not understand would _not_
say such things. The pre-1960s experts were right to be concerned.

Sure, but around 1960, the issue was resolved. Anyone living in 2006
ought to know this.

[...]
Quote:
You don't seem to realize that there are _TWO_SET_ of Schwarzschild
coordinates. One set is valid for r>2M and the other is valid for r<2M.
They have no intersection. Yes, the exterior coordinates are not valid
in the region r<2M. <shrug

What you still don't seem to realise is that r<2m does not physically
exist, >so there is only ONE valid set of coordinates, i.e. r>2m. I

Since it takes an atlas with at least six charts in the standard basis
to cover a sphere are you also going to claim that only one-sixth of
a sphere exists? Of course, if you use polar coordinates, you only need
at least two charts, so now you could claim that a hemisphere can exist
unless someone claims only the standard coordinates are valid. Gee,
what a conundrum.

Quote:
pointed this out at the start of the thread (in fact this is the whole
point of bringing the matter up), and I even went through the proof of
it in another post.

If that is your point, why have you not made an attempt to prove that
``there is only ONE valid set of coordinates?'' The fact that there are
a number of charts which do not suffer the problem you have created
by making a poor choice contradicts you.

Quote:
Obviously its easier to just ignore these facts and
mindlessly regurgitate what the textbooks tell you, but perhaps you
could try thinking for yourself for a change. It makes a world of
difference.

What ``facts?'' All you have done is assert something which indicates
you have no idea what you are talking about. If your idea of ``thinking
for yourself'' is to accept non-sense just because it is recognized as
non-sense by the majority of physicists, have at it.
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LEJ Brouwer
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Joined: 07 May 2005
Posts: 120

PostPosted: Fri Jul 14, 2006 12:08 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

Bilge wrote:
Quote:
LEJ Brouwer:
Tom Roberts wrote:

No physical anything happens. Merely the labels of coordinates
get shuffled.

As I said before, if you are going to 'merely' shuffle the labels of
the coordinates, then you would have to 'merely' shuffle them in the
Einstein field equations too - which means that what you are now
solving are not the Einstein field equations.

Don't be ridiculous. The metric has one timelike and three spacelike
coordinates. For a spacelike metric, -+++, the entry with the - sign
is timelike, regardless of what label you give it.

Good grief. I despair for humanity. If dy/dx = x is solved by y = x^2 /
2 + c, and then I changed the x into a z wrote, y = z^2 / 2 + c, would
I not have to change the original problem to dy/dz = z for the new
solution to be valid? What is so ridiculous about that?

The fact that the entry with the - signature is timelike has absolutely
nothing to do with it. Or are you claiming that if a metric contains a
time-like coordinate, then this is sufficient for it to satisfy the
Einstein field equations? Because that is what it sounds like you are
arguing.

Quote:
[...]
Only someone who does not understand what is happening would say such
silly things.

Quite the contrary, only someone who does not understand would _not_
say such things. The pre-1960s experts were right to be concerned.

Sure, but around 1960, the issue was resolved. Anyone living in 2006
ought to know this.

No around 1960, the issue was fudged, shoved under the carpet, and
banished from discussion ever again. Anyone living in 2006 is expected
not to look under the carpet in case embarassing creatures which are
supposed to be extinct emerge therefrom.

Now, I have gone through the proof of why r<2m is not physical. Could
you please use your own brain to tell me why that proof is wrong? It is
only a few lines long, and should not tax even your good self unduly.

Quote:
[...]
You don't seem to realize that there are _TWO_SET_ of Schwarzschild
coordinates. One set is valid for r>2M and the other is valid for r<2M.
They have no intersection. Yes, the exterior coordinates are not valid
in the region r<2M. <shrug

What you still don't seem to realise is that r<2m does not physically
exist, >so there is only ONE valid set of coordinates, i.e. r>2m. I

Since it takes an atlas with at least six charts in the standard basis
to cover a sphere are you also going to claim that only one-sixth of
a sphere exists? Of course, if you use polar coordinates, you only need
at least two charts, so now you could claim that a hemisphere can exist
unless someone claims only the standard coordinates are valid. Gee,
what a conundrum.

We are not talking about a sphere here, and I am not arguing about
coordinate charts anyway, so your analogy is utterly worthless. It has
absolutely nothing to do with why r<2m is not physical.

Quote:
pointed this out at the start of the thread (in fact this is the whole
point of bringing the matter up), and I even went through the proof of
it in another post.

If that is your point, why have you not made an attempt to prove that
``there is only ONE valid set of coordinates?'' The fact that there are
a number of charts which do not suffer the problem you have created
by making a poor choice contradicts you.

The proof is in the papers I reference in post #1, and I have also
outlined the proof in another post. WHY DON'T YOU TRY READING IT??? If
you understand the proof, you will also understand why your 'number of
charts' argument is completely irrelevant.

Quote:
Obviously its easier to just ignore these facts and
mindlessly regurgitate what the textbooks tell you, but perhaps you
could try thinking for yourself for a change. It makes a world of
difference.

What ``facts?'' All you have done is assert something which indicates
you have no idea what you are talking about. If your idea of ``thinking
for yourself'' is to accept non-sense just because it is recognized as
non-sense by the majority of physicists, have at it.

And this from a person who has not even bothered to go through the
proof to identify what is factual and what is not. Truly pathetic.
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LEJ Brouwer
science forum Guru Wannabe


Joined: 07 May 2005
Posts: 120

PostPosted: Fri Jul 14, 2006 12:08 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

Bilge wrote:
Quote:
LEJ Brouwer:
Tom Roberts wrote:

No physical anything happens. Merely the labels of coordinates
get shuffled.

As I said before, if you are going to 'merely' shuffle the labels of
the coordinates, then you would have to 'merely' shuffle them in the
Einstein field equations too - which means that what you are now
solving are not the Einstein field equations.

Don't be ridiculous. The metric has one timelike and three spacelike
coordinates. For a spacelike metric, -+++, the entry with the - sign
is timelike, regardless of what label you give it.

Good grief. I despair for humanity. If dy/dx = x is solved by y = x^2 /
2 + c, and then I changed the x into a z wrote, y = z^2 / 2 + c, would
I not have to change the original problem to dy/dz = z for the new
solution to be valid? What is so ridiculous about that?

The fact that the entry with the - signature is timelike has absolutely
nothing to do with it. Or are you claiming that if a metric contains a
time-like coordinate, then this is sufficient for it to satisfy the
Einstein field equations? Because that is what it sounds like you are
arguing.

Quote:
[...]
Only someone who does not understand what is happening would say such
silly things.

Quite the contrary, only someone who does not understand would _not_
say such things. The pre-1960s experts were right to be concerned.

Sure, but around 1960, the issue was resolved. Anyone living in 2006
ought to know this.

No around 1960, the issue was fudged, shoved under the carpet, and
banished from discussion ever again. Anyone living in 2006 is expected
not to look under the carpet in case embarassing creatures which are
supposed to be extinct emerge therefrom.

Now, I have gone through the proof of why r<2m is not physical. Could
you please use your own brain to tell me why that proof is wrong? It is
only a few lines long, and should not tax even your good self unduly.

Quote:
[...]
You don't seem to realize that there are _TWO_SET_ of Schwarzschild
coordinates. One set is valid for r>2M and the other is valid for r<2M.
They have no intersection. Yes, the exterior coordinates are not valid
in the region r<2M. <shrug

What you still don't seem to realise is that r<2m does not physically
exist, >so there is only ONE valid set of coordinates, i.e. r>2m. I

Since it takes an atlas with at least six charts in the standard basis
to cover a sphere are you also going to claim that only one-sixth of
a sphere exists? Of course, if you use polar coordinates, you only need
at least two charts, so now you could claim that a hemisphere can exist
unless someone claims only the standard coordinates are valid. Gee,
what a conundrum.

We are not talking about a sphere here, and I am not arguing about
coordinate charts anyway, so your analogy is utterly worthless. It has
absolutely nothing to do with why r<2m is not physical.

Quote:
pointed this out at the start of the thread (in fact this is the whole
point of bringing the matter up), and I even went through the proof of
it in another post.

If that is your point, why have you not made an attempt to prove that
``there is only ONE valid set of coordinates?'' The fact that there are
a number of charts which do not suffer the problem you have created
by making a poor choice contradicts you.

The proof is in the papers I reference in post #1, and I have also
outlined the proof in another post. WHY DON'T YOU TRY READING IT??? If
you understand the proof, you will also understand why your 'number of
charts' argument is completely irrelevant.

Quote:
Obviously its easier to just ignore these facts and
mindlessly regurgitate what the textbooks tell you, but perhaps you
could try thinking for yourself for a change. It makes a world of
difference.

What ``facts?'' All you have done is assert something which indicates
you have no idea what you are talking about. If your idea of ``thinking
for yourself'' is to accept non-sense just because it is recognized as
non-sense by the majority of physicists, have at it.

And this from a person who has not even bothered to go through the
proof to identify what is factual and what is not. Truly pathetic.
Back to top
Tom Roberts
science forum Guru


Joined: 24 Mar 2005
Posts: 1399

PostPosted: Fri Jul 14, 2006 1:46 am    Post subject: Re: Misinterpretation of the radial parameter in the Schwarzschild solution? Reply with quote

LEJ Brouwer wrote:
Quote:
Tom Roberts wrote:
No physical anything happens. Merely the labels of coordinates get shuffled.

As I said before, if you are going to 'merely' shuffle the labels of
the coordinates, then you would have to 'merely' shuffle them in the
Einstein field equations too - which means that what you are now
solving are not the Einstein field equations.

This is not true. Re-labeling the coordinates is of no consequence, and
the field equation is invariant under such a shuffling. It does not
matter how you label the individual coordinates, except that it can be
confusing if you re-use a given label for different coordinates, such as
is done in the two sets of Schw. coordinates.


Quote:
What you still don't seem to realise is that r<2m does not physically
exist,

Hmmm. We are discussing the Schwarzschild solution of the Einstein field
equation, not any real, physical system. This is all _theoretical_, and
the manifold in the region r<2M is every bit as much a manifold as that
in r>2M. So is the entire "second half" of the Kruskal extension. <shrug>


Quote:
so there is only ONE valid set of coordinates, i.e. r>2m.

That is just plain wrong.


Quote:
I
pointed this out at the start of the thread (in fact this is the whole
point of bringing the matter up), and I even went through the proof of
it in another post.

Your "proof" assumed that it is possible to escape from any point in the
manifold. This assumption is not valid for the Schwarzschild spacetime.
<shrug>


Tom Roberts
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