Author 
Message 
LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Fri Jul 21, 2006 11:39 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



JanPB wrote:
Quote:  LEJ Brouwer wrote:
Daryl McCullough wrote:
That seems to be the reasoningthat only the region r >= 2m is physically
meaningful, so if the mass is anywhere, it is at r=2m.
No, the mass it at r = r0 by definition, and this corresponds to r*(r0)
= 2m. It is not at all as vague or subjective as you try to make out.
OK, define this precisely. Crothers' paper does a terrible job
explaining this point. I can understand the (unnecessary) rewriting of
the metric in the form of his equation (7):
ds^2 = (sqrt(C)alpha)/sqrt(C) dt^2  (sqrt(C)/(sqrt(C)alpha))
C'^2/(4C) dr^2  c dO^2
...where C(r) is a positive function.
What I don't understand is his constant shifting of the definitions.
Specifically: how is r0 defined? He says "Let the test particle at r0
acquire mass". So if the particle is at r0 then r0 must be the origin
because of the spherical symmetry. But a bit later he has:
"Furthermore, one can see from (13) and (14) that r0 is arbitrary [?],
i.e. the pointmass can be located at any point [??] and its location
has no intrinsic meaning". How does (13) or (14) [the formula for the
distance from r0 to another radial position r] imply that r0 is
"arbitrary"? If he means by that that coordinate value is of no
physical significance then he is stating a triviality (albeit dressed
up as a revelation), if he means that the actual position of the point
can be anywhere, then he is obviously wrong by the symmetry. What am I
missing?

Okay, Crothers' wording is not great here. Basically, 'r' parametrises
2spheres in some way (there is an arbitrariness in this choice). Given
any choice of parametrisation, the value of 'r' corresponding to the
position of the point mass, and hence the origin, must correspond to
some value of r, which is taken to be r0. So, yes he is 'stating a
triviality'. I don't agree that he has exaggerated the point because by
taking this little extra care, he has indeed come up with a
'revelation', i.e. that the event horizon and the point mass coincide.
 Sabbir.
P.S. Apologies for the appalling spelling and grammar in my last post 
it was 4am and my brain was working on impulse power! 

Back to top 


LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Fri Jul 21, 2006 10:59 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Tom Roberts wrote:
Quote:  As is well known...

Well, there's no arguing with that. You win once again.
 Sabbir. 

Back to top 


Sue... science forum Guru
Joined: 08 May 2005
Posts: 2684

Posted: Fri Jul 21, 2006 8:02 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



JanPB wrote:
snip
Quote: 
As I suspected, at the bottom of this is the confusing "switch" of
coordinates between spacelike and timelike. Even the naming conventions
("r" and "t") are powerful tricksters.

I suspect the same. Especially after seeing posters to this thread
that should know better, defend a position mathematical rigour is
not necessary to remove spatial components from a spacetime
interval. The royal invocation of the term 'timelike' is not a
substitute
for the rigourous appplication of imaginary operators to transform
from
the Lorenz gauge (anisotropic where 1/r^2 is meaningless)
back to the Coulomb gauge ( isotropic where 1/r^2 correctly
expresses the attenuation on a radial path).
[correct placement shown near bottom of page]
http://www.nrao.edu/~smyers/courses/astro12/speedoflight.html
"Gauge Transformations"
http://arxiv.org/abs/physics/0204034
"Space time"
http://farside.ph.utexas.edu/teaching/em/lectures/node113.html
Sue...


Back to top 


Jan Bielawski science forum Guru
Joined: 08 May 2005
Posts: 388

Posted: Fri Jul 21, 2006 6:34 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer wrote:
Quote:  Daryl McCullough wrote:
That seems to be the reasoningthat only the region r >= 2m is physically
meaningful, so if the mass is anywhere, it is at r=2m.
No, the mass it at r = r0 by definition, and this corresponds to r*(r0)
= 2m. It is not at all as vague or subjective as you try to make out.

OK, define this precisely. Crothers' paper does a terrible job
explaining this point. I can understand the (unnecessary) rewriting of
the metric in the form of his equation (7):
ds^2 = (sqrt(C)alpha)/sqrt(C) dt^2  (sqrt(C)/(sqrt(C)alpha))
C'^2/(4C) dr^2  c dO^2
....where C(r) is a positive function.
What I don't understand is his constant shifting of the definitions.
Specifically: how is r0 defined? He says "Let the test particle at r0
acquire mass". So if the particle is at r0 then r0 must be the origin
because of the spherical symmetry. But a bit later he has:
"Furthermore, one can see from (13) and (14) that r0 is arbitrary [?],
i.e. the pointmass can be located at any point [??] and its location
has no intrinsic meaning". How does (13) or (14) [the formula for the
distance from r0 to another radial position r] imply that r0 is
"arbitrary"? If he means by that that coordinate value is of no
physical significance then he is stating a triviality (albeit dressed
up as a revelation), if he means that the actual position of the point
can be anywhere, then he is obviously wrong by the symmetry. What am I
missing?

Jan Bielawski 

Back to top 


Tom Roberts science forum Guru
Joined: 24 Mar 2005
Posts: 1399

Posted: Fri Jul 21, 2006 5:12 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  The integrated proper separation
Integral of squareroot(g_uv dx^u dx^v)
along a path that is sometimes spacelike and sometimes timelike
yields a complex number. I'm not sure what (if anything) is the
meaning of this number,

This is a place where the semiRiemannian geometry of GR is different
from Riemannian geometry. In particular, timelike and spacelike paths
are just plain different. That integral over a spacelike path is defined
to be the path length of the path; over a timelike interval it is
defined to be the proper time duration of the path (here "path" includes
a specified pair of endpoints). Over a mixed interval as you suggest
there is no defined meaning, and I doubt there can be any useful meaning.
For geodesics, of course, there is no such mixed path, and every
geodesic path is either timelike everywhere or spacelike everywhere.
This is one aspect of why people look for "geodesically complete"
manifolds  ones that contain the entire length of every geodesic path.
A boundary or singularity is known to be present if a geodesic has a
finite limit to its affine parameter; the singularity theorems are
proved by showing a geodesic has a finite limit to its affine parameter.
As is well known, the region r>2M of Schw. spacetime is NOT geodesically
complete, and it can be extended to the complete Kruskal extension.
This, of course, includes the region r<=2M, and shows that LEJBrouwer is
just plain wrong. <shrug>
Tom Roberts 

Back to top 


LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Fri Jul 21, 2006 2:50 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



JanPB wrote:
Quote:  Daryl McCullough wrote:
JanPB says...
1. How do you justify the claim that the set r=2M is a point,
Well, what he posted before about this was the following:
OK, let's look at it. BTW, I read Crothers' paper and it's really badly
written. The reviewer (if there was one) should have pushed him much
harder.
begin quote
Here goes:
Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)
where A,B,C>0,
This assumes it's static, OK.
and applying the same transformation r* = sqrt{C} used
by Hilbert (we write r* instead of r to avoid confusion with r above),
the general solution is,
ds^2 = (1  2m/r*) dt^2  (r* / (r*  2m)) dr*^2  r*^2 dOmega^2 (eqn
7 of Crothers).
It's eq. 6 actually, OK.
Substituting sqrt{C} for r* this becomes,
ds^2 = (1  2m/sqrt{C}) dt^2  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C)
dr^2  C dOmega^2 (eqn 7 of Crothers).
This step is redundant but not wrong, just making things complex for no
reason. OK.

It's not redundant. It's very useful because it allows calculation of
the integral wrt C later on when working out the radial distance.
Quote:  [C' = dC/dr here]. This is just the usual Schwarzschild solution, but
note that r* = sqrt{C} does not need to lie in the range [0,Infty] as
Hilbert incorrectly assumed.
The form of the solution we seek forces r* to be in (2m,infty).
Now, if the coordinate associated with the point mass is r = sqrt{C} =
r0

There is actually a typo hero, which I just noticed  the sqrt{C}
should not be there, and it should just say r = r0. No harm done, so no
worries.
Quote:  (which may if you like be chosen such that r0 = 0  it makes no
difference), then we can define the 'radial distance' from the mass at
r0 to any r >= r0 to be the integral from r0 to r of sqrt{g_rr},
I assume r means r*.

No, the integral is from r = r0 to r = r >= r0. Recall that C=C(r).
Quote: 
R = int sqrt{g_rr} dr from r0 to r (eqn 11 of Crothers)
where g_rr is just the radial component of the metric:
g_rr =  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C) (eqn 12 of Crothers)
Integrating, this (I use the substitution u = sqrt{C} followed by the
substitution u = 2m sec^2 x), we find,
R = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln (sqrt{sqrt{C}}+sqrt{(sqrt{C}
 2m)}) / sqrt{2m}   sqrt{sqrt{C0}*(sqrt{C0}  2m)} + 2m ln
(sqrt{sqrt{C0}}+sqrt{(sqrt{C0}  2m)}) / sqrt{2m} 
OK.
[C0 = C(r0) here], where we have fixed the constant of integration by
applying the boundary condition that R>0 as r>r0. Since the radial
distance must be real, this means that the range of possible values for
r* = sqrt{C(r)} is given by r* >= 2m.
We already know that r*>2m if the metric is to be static.
In particular, the origin must
correspond to the minimum of this range,
Why? It is the assumption of staticity that forces 2m as the lower
bound for r* but all this implies is that no *static* spherically
symmetric solution extends all the way to the mass point.

r* = 2m is not just the position of the mass point  it is also the
position of the event horizon. If you don't realise this then you are
completely missing the point.
Quote:  The staticity
assumption was introduced just to simplify things for us humans and
nature has no obligation to obey our whims here.
i.e r*(r0) = sqtr{C0} = 2m, so
that C0 = 4m^2. But r* = 2m is the position of the event horizon, i.e.
the event horizon coincides with the point particle which is the radial
origin. Moreover, r* < 2m is unphysical.
No, at this stage we can only logically conclude far less, namely that
the assumptions we've made were too stringent (namely, staticity) to
cover the entire range.
Plugging sqrt{C0} = 2m into the above equation, we find the following
equation for the radial distance of a point at r from the origin,
R(r) = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln
(sqrt{sqrt{C}}+sqrt{(sqrt{C}  2m)} / sqrt{2m} (eqn 14 of Crothers)
Again. one cannot consider radii with r* < 2m, because these radii make
no physical sense, as they would have to have a smaller radius than
radius corresponding to the position of the point mass which defines
the origin of our space.
There is therefore NO interior solution for r* < 2m, because r* < 2m
does not physically exist.
So this is a wrong conclusion.
/end quote
So what he seems to be saying is the following: (I will use
r for his r*):
1. He's assuming a static, spherically symmetric metric,
which can always be transformed into the Schwarzchild
metric ds^2 = (1  2m/r) dt^2  (r / (r  2m)) dr^2  r^2 dOmega^2

Even if it were the case that the result only held for a static
solution (which is kind of a braindead conclusion to come as you prove
below that the above equation is sufficiently general to take into
account the nonstatic case for the purposes of my proof), I would be
satisfied with that, as it still shows that the event horizon coincides
with the mass point in that case.
Quote:  2. He defines R(r) to be the integral from r0 to r of
squareroot(g_rr) dr. This is a measure of the distance
from r0 to r. (r0 is the assumed location of the pointmass
source).
3. He notes that R(r) is only real for r >= 2m.
4. So he concludes that only r >= 2m is physically meaningful.
Amazing.

The only thing that is amazing is that you think that you insist in
allowing nonexistence points with radius less than zero, which
according to your own calculations are at imaginary distances from the
audience.
Quote:  5. So, since there is no mass at r > 2m, the mass must be at r = 2m.
Amazing.

The argument for this is actually quite rigorous as I have explained,
so there is really nothing to be amazed about.
Quote:  That seems to be the reasoningthat only the region r >= 2m is physically
meaningful, so if the mass is anywhere, it is at r=2m.

No it's not.
Quote:  Of course, the correct conclusion is that the region r < 2m *is*
physically meaningful, but r is not a spacial coordinate there,
it is a temporal coordinate.

Amazing.
Quote:  As I suspected, at the bottom of this is the confusing "switch" of
coordinates between spacelike and timelike. Even the naming conventions
("r" and "t") are powerful tricksters.

The proof makes no use of the interior solution at all. Once it has
been established that the event horizon coincides with the point mass,
there can be no interior solution.
Quote:  Here is an example of a more careful flow of derivation logic assuming
only the symmetry  this sort of care is *not* found in any of the
texts I've seen (except Hawking and Ellis who do this in a rather
abstract way)  perhaps one reason why the confusion arises. Don't
worry, it's short, I just wanted to point out what typically happens in
textbooks.
Consider the general form of a spherically symmetric metric:
ds^2 = A(t,r) dt^2 + 2B(t,r) dt dr + C(t,r) dr^2 + D(t,r) dO^2
where dO^2 = dtheta^2 + sin^2(theta) dphi^2, as usual.
All we know is that this is a metric of signature 2 and that D(t,r)>0
(because d/dtheta is spacelike  I use the +++ convention).
Because the "mixed" term dt dr is present, the signature requirement
does not translate directly into any restriction on the sign of A, B, C
yet.
Then we perform the usual two changes of variables to simplify the form
of this metric further:
Change 1. Perform the change of variables:
(t,r,theta,phi) > (t,u,theta,phi)
...where u(t,r) = sqrt(D(t,r)) (recall that D>0)
We no longer know the character of the new variable u
(spaceliketimelike) because we don't know D(t,r). As long as dD/dr is
nonzero this variable change is kosher (a diffeomorphism).
Note I'm using the letter "u" here because "r*" is too suggestive and
confusing.
So the metric simplifies to (recycling the A,B,C letter names):
ds^2 = A(t,u) dt^2 + 2B(t,u) dt du + C(t,u) du^2 + u^2 dO^2
Change 2. We get rid of the mixed term dt du by the integrating factor
trick which amounts to another change of variables:
(t,u,theta,phi) > (v,u,theta,phi)
...where v is a function of (t,u).
Again, the character of the coordinate v (spaceliketimelike) is not
known due to the mixing of t and u in an uncontrolled way.
The metric finally has the form (recycling A,B one more time):
ds^2 = A(v,u) dv^2 + B(v,u) du^2 + u^2 dO^2
And here is where most of the texts gloss over an important detail: we
still know that this metric has signature 2 but because the "mixed"
terms are gone, the signature requirement does translate now into a
specific sign constraint on A and B:
Either:
A<0 and B>0
or:
A>0 and B<0.
Both cases must be considered. In reality, they are computationally
almost identical and yield two solutions:
ds^2 = (12m/u) dv^2 + 1/(12m/u) du^2 + u^2 dO^2
...with A=(12m/u)<0 and B=1/(12m/u)>0, i.e. u>2m and v timelike, u
spacelike, static metric (thus proving Birkhoff's theorem),
and, resp.:
ds^2 = (2m/u1) dv^2  1/(2m/u1) du^2 + u^2 dO^2
...with A=2m/u1>0 and B=1/(2m/u1)<0, i.e. u<2m and u timelike, v
spacelike, nonstatic metric (still sort of proving Birkhoff in the
sense that A and B still do not depend on v here).

....after which the argument given above reconfirms that the event
horizon coincides with the point mass. All you have done is proven that
restricting to the static exterior case is no restriction at all as the
more general case reduces to this one through an appropriate choice of
coordinate transformations.
Quote:  I think just by keeping the coordinate names nonsuggestive and the
logic clean, the mystery evaporates. Unfortunately, even good texts
like d'Inverno mysteriously (and really incorrectly) write down the
general form of the simplified metric assuming A<0 and B>0 *only* (he
actually has it the other way around because he uses the +
convention). After that the interior solution  when it finally appears
much later  seems fishy.
Of course in the static case it's even simpler: the staticity forces
r>2m and when one removes this restriction, the solution for r<2m
satisfies Einstein's equation as well, except the confusion reigns
supreme because of the human reflex to keep the *letters* r and t
unchanged and subsequently falling into the "coordinate switch" trap.

Jan Bielawski 


Back to top 


LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Fri Jul 21, 2006 2:24 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  JanPB says...
1. How do you justify the claim that the set r=2M is a point,
Well, what he posted before about this was the following:
begin quote
Here goes:
Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)
where A,B,C>0, and applying the same transformation r* = sqrt{C} used
by Hilbert (we write r* instead of r to avoid confusion with r above),
the general solution is,
ds^2 = (1  2m/r*) dt^2  (r* / (r*  2m)) dr*^2  r*^2 dOmega^2 (eqn
7 of Crothers).
Substituting sqrt{C} for r* this becomes,
ds^2 = (1  2m/sqrt{C}) dt^2  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C)
dr^2  C dOmega^2 (eqn 7 of Crothers).
[C' = dC/dr here]. This is just the usual Schwarzschild solution, but
note that r* = sqrt{C} does not need to lie in the range [0,Infty] as
Hilbert incorrectly assumed.
Now, if the coordinate associated with the point mass is r = sqrt{C} =
r0 (which may if you like be chosen such that r0 = 0  it makes no
difference), then we can define the 'radial distance' from the mass at
r0 to any r >= r0 to be the integral from r0 to r of sqrt{g_rr},
R = int sqrt{g_rr} dr from r0 to r (eqn 11 of Crothers)
where g_rr is just the radial component of the metric:
g_rr =  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C) (eqn 12 of Crothers)
Integrating, this (I use the substitution u = sqrt{C} followed by the
substitution u = 2m sec^2 x), we find,
R = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln (sqrt{sqrt{C}}+sqrt{(sqrt{C}
 2m)}) / sqrt{2m}   sqrt{sqrt{C0}*(sqrt{C0}  2m)} + 2m ln
(sqrt{sqrt{C0}}+sqrt{(sqrt{C0}  2m)}) / sqrt{2m} 
[C0 = C(r0) here], where we have fixed the constant of integration by
applying the boundary condition that R>0 as r>r0. Since the radial
distance must be real, this means that the range of possible values for
r* = sqrt{C(r)} is given by r* >= 2m. In particular, the origin must
correspond to the minimum of this range, i.e r*(r0) = sqtr{C0} = 2m, so
that C0 = 4m^2. But r* = 2m is the position of the event horizon, i.e.
the event horizon coincides with the point particle which is the radial
origin. Moreover, r* < 2m is unphysical.
Plugging sqrt{C0} = 2m into the above equation, we find the following
equation for the radial distance of a point at r from the origin,
R(r) = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln
(sqrt{sqrt{C}}+sqrt{(sqrt{C}  2m)} / sqrt{2m} (eqn 14 of Crothers)
Again. one cannot consider radii with r* < 2m, because these radii make
no physical sense, as they would have to have a smaller radius than
radius corresponding to the position of the point mass which defines
the origin of our space.
There is therefore NO interior solution for r* < 2m, because r* < 2m
does not physically exist.
/end quote
So what he seems to be saying is the following: (I will use
r for his r*):
1. He's assuming a static, spherically symmetric metric,
which can always be transformed into the Schwarzchild
metric ds^2 = (1  2m/r) dt^2  (r / (r  2m)) dr^2  r^2 dOmega^2
2. He defines R(r) to be the integral from r0 to r of
squareroot(g_rr) dr. This is a measure of the distance
from r0 to r. (r0 is the assumed location of the pointmass
source).
3. He notes that R(r) is only real for r >= 2m.
4. So he concludes that only r >= 2m is physically meaningful.
5. So, since there is no mass at r > 2m, the mass must be at r = 2m.

No. The mass is at r=r0 by definition. It is true, as Jan says, that
the reality of R(r) need not imply that the minimum value r* takes is
2m  the minimum value could be larger, i.e r* >= 2n where n > m. But
this is case is even worse for you, as this would mean that there is no
event horizon at all (let alone an interior solution), as we know that
r* = 2m for the EH, which in this case would be unphysical.
Quote: 
That seems to be the reasoningthat only the region r >= 2m is physically
meaningful, so if the mass is anywhere, it is at r=2m.

No, the mass it at r = r0 by definition, and this corresponds to r*(r0)
= 2m. It is not at all as vague or subjective as you try to make out.
Quote:  Of course, the correct conclusion is that the region r < 2m *is*
physically meaningful, but r is not a spacial coordinate there,
it is a temporal coordinate. Integrating along the path of increasing
r means integrating along a timelike path up until r=2m, and then
integrating along a spacelike path thereafter.
The integrated proper separation
Integral of squareroot(g_uv dx^u dx^v)
along a path that is sometimes spacelike and sometimes timelike
yields a complex number. I'm not sure what (if anything) is the
meaning of this number, but it certainly doesn't indicate that
you have entered a physically forbidden region.

Wishful thinking at its best. Very amusing!
Quote:  
Daryl McCullough
Ithaca, NY 


Back to top 


Jan Bielawski science forum Guru
Joined: 08 May 2005
Posts: 388

Posted: Fri Jul 21, 2006 1:49 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  JanPB says...
1. How do you justify the claim that the set r=2M is a point,
Well, what he posted before about this was the following:

OK, let's look at it. BTW, I read Crothers' paper and it's really badly
written. The reviewer (if there was one) should have pushed him much
harder.
Quote:  begin quote
Here goes:
Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)
where A,B,C>0,

This assumes it's static, OK.
Quote:  and applying the same transformation r* = sqrt{C} used
by Hilbert (we write r* instead of r to avoid confusion with r above),
the general solution is,
ds^2 = (1  2m/r*) dt^2  (r* / (r*  2m)) dr*^2  r*^2 dOmega^2 (eqn
7 of Crothers).

It's eq. 6 actually, OK.
Quote:  Substituting sqrt{C} for r* this becomes,
ds^2 = (1  2m/sqrt{C}) dt^2  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C)
dr^2  C dOmega^2 (eqn 7 of Crothers).

This step is redundant but not wrong, just making things complex for no
reason. OK.
Quote:  [C' = dC/dr here]. This is just the usual Schwarzschild solution, but
note that r* = sqrt{C} does not need to lie in the range [0,Infty] as
Hilbert incorrectly assumed.

The form of the solution we seek forces r* to be in (2m,infty).
Quote:  Now, if the coordinate associated with the point mass is r = sqrt{C} =
r0 (which may if you like be chosen such that r0 = 0  it makes no
difference), then we can define the 'radial distance' from the mass at
r0 to any r >= r0 to be the integral from r0 to r of sqrt{g_rr},

I assume r means r*.
Quote:  R = int sqrt{g_rr} dr from r0 to r (eqn 11 of Crothers)
where g_rr is just the radial component of the metric:
g_rr =  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C) (eqn 12 of Crothers)
Integrating, this (I use the substitution u = sqrt{C} followed by the
substitution u = 2m sec^2 x), we find,
R = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln (sqrt{sqrt{C}}+sqrt{(sqrt{C}
 2m)}) / sqrt{2m}   sqrt{sqrt{C0}*(sqrt{C0}  2m)} + 2m ln
(sqrt{sqrt{C0}}+sqrt{(sqrt{C0}  2m)}) / sqrt{2m} 

OK.
Quote:  [C0 = C(r0) here], where we have fixed the constant of integration by
applying the boundary condition that R>0 as r>r0. Since the radial
distance must be real, this means that the range of possible values for
r* = sqrt{C(r)} is given by r* >= 2m.

We already know that r*>2m if the metric is to be static.
Quote:  In particular, the origin must
correspond to the minimum of this range,

Why? It is the assumption of staticity that forces 2m as the lower
bound for r* but all this implies is that no *static* spherically
symmetric solution extends all the way to the mass point. The staticity
assumption was introduced just to simplify things for us humans and
nature has no obligation to obey our whims here.
Quote:  i.e r*(r0) = sqtr{C0} = 2m, so
that C0 = 4m^2. But r* = 2m is the position of the event horizon, i.e.
the event horizon coincides with the point particle which is the radial
origin. Moreover, r* < 2m is unphysical.

No, at this stage we can only logically conclude far less, namely that
the assumptions we've made were too stringent (namely, staticity) to
cover the entire range.
Quote:  Plugging sqrt{C0} = 2m into the above equation, we find the following
equation for the radial distance of a point at r from the origin,
R(r) = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln
(sqrt{sqrt{C}}+sqrt{(sqrt{C}  2m)} / sqrt{2m} (eqn 14 of Crothers)
Again. one cannot consider radii with r* < 2m, because these radii make
no physical sense, as they would have to have a smaller radius than
radius corresponding to the position of the point mass which defines
the origin of our space.
There is therefore NO interior solution for r* < 2m, because r* < 2m
does not physically exist.

So this is a wrong conclusion.
Quote:  /end quote
So what he seems to be saying is the following: (I will use
r for his r*):
1. He's assuming a static, spherically symmetric metric,
which can always be transformed into the Schwarzchild
metric ds^2 = (1  2m/r) dt^2  (r / (r  2m)) dr^2  r^2 dOmega^2
2. He defines R(r) to be the integral from r0 to r of
squareroot(g_rr) dr. This is a measure of the distance
from r0 to r. (r0 is the assumed location of the pointmass
source).
3. He notes that R(r) is only real for r >= 2m.
4. So he concludes that only r >= 2m is physically meaningful.

Amazing.
Quote:  5. So, since there is no mass at r > 2m, the mass must be at r = 2m.

Amazing.
Quote:  That seems to be the reasoningthat only the region r >= 2m is physically
meaningful, so if the mass is anywhere, it is at r=2m.
Of course, the correct conclusion is that the region r < 2m *is*
physically meaningful, but r is not a spacial coordinate there,
it is a temporal coordinate.

As I suspected, at the bottom of this is the confusing "switch" of
coordinates between spacelike and timelike. Even the naming conventions
("r" and "t") are powerful tricksters.
Here is an example of a more careful flow of derivation logic assuming
only the symmetry  this sort of care is *not* found in any of the
texts I've seen (except Hawking and Ellis who do this in a rather
abstract way)  perhaps one reason why the confusion arises. Don't
worry, it's short, I just wanted to point out what typically happens in
textbooks.
Consider the general form of a spherically symmetric metric:
ds^2 = A(t,r) dt^2 + 2B(t,r) dt dr + C(t,r) dr^2 + D(t,r) dO^2
where dO^2 = dtheta^2 + sin^2(theta) dphi^2, as usual.
All we know is that this is a metric of signature 2 and that D(t,r)>0
(because d/dtheta is spacelike  I use the +++ convention).
Because the "mixed" term dt dr is present, the signature requirement
does not translate directly into any restriction on the sign of A, B, C
yet.
Then we perform the usual two changes of variables to simplify the form
of this metric further:
Change 1. Perform the change of variables:
(t,r,theta,phi) > (t,u,theta,phi)
....where u(t,r) = sqrt(D(t,r)) (recall that D>0)
We no longer know the character of the new variable u
(spaceliketimelike) because we don't know D(t,r). As long as dD/dr is
nonzero this variable change is kosher (a diffeomorphism).
Note I'm using the letter "u" here because "r*" is too suggestive and
confusing.
So the metric simplifies to (recycling the A,B,C letter names):
ds^2 = A(t,u) dt^2 + 2B(t,u) dt du + C(t,u) du^2 + u^2 dO^2
Change 2. We get rid of the mixed term dt du by the integrating factor
trick which amounts to another change of variables:
(t,u,theta,phi) > (v,u,theta,phi)
....where v is a function of (t,u).
Again, the character of the coordinate v (spaceliketimelike) is not
known due to the mixing of t and u in an uncontrolled way.
The metric finally has the form (recycling A,B one more time):
ds^2 = A(v,u) dv^2 + B(v,u) du^2 + u^2 dO^2
And here is where most of the texts gloss over an important detail: we
still know that this metric has signature 2 but because the "mixed"
terms are gone, the signature requirement does translate now into a
specific sign constraint on A and B:
Either:
A<0 and B>0
or:
A>0 and B<0.
Both cases must be considered. In reality, they are computationally
almost identical and yield two solutions:
ds^2 = (12m/u) dv^2 + 1/(12m/u) du^2 + u^2 dO^2
....with A=(12m/u)<0 and B=1/(12m/u)>0, i.e. u>2m and v timelike, u
spacelike, static metric (thus proving Birkhoff's theorem),
and, resp.:
ds^2 = (2m/u1) dv^2  1/(2m/u1) du^2 + u^2 dO^2
....with A=2m/u1>0 and B=1/(2m/u1)<0, i.e. u<2m and u timelike, v
spacelike, nonstatic metric (still sort of proving Birkhoff in the
sense that A and B still do not depend on v here).
I think just by keeping the coordinate names nonsuggestive and the
logic clean, the mystery evaporates. Unfortunately, even good texts
like d'Inverno mysteriously (and really incorrectly) write down the
general form of the simplified metric assuming A<0 and B>0 *only* (he
actually has it the other way around because he uses the +
convention). After that the interior solution  when it finally appears
much later  seems fishy.
Of course in the static case it's even simpler: the staticity forces
r>2m and when one removes this restriction, the solution for r<2m
satisfies Einstein's equation as well, except the confusion reigns
supreme because of the human reflex to keep the *letters* r and t
unchanged and subsequently falling into the "coordinate switch" trap.

Jan Bielawski 

Back to top 


Daryl McCullough science forum Guru
Joined: 24 Mar 2005
Posts: 1167

Posted: Thu Jul 20, 2006 4:27 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



JanPB says...
Quote:  1. How do you justify the claim that the set r=2M is a point,

Well, what he posted before about this was the following:
<begin quote>
Here goes:
Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)
where A,B,C>0, and applying the same transformation r* = sqrt{C} used
by Hilbert (we write r* instead of r to avoid confusion with r above),
the general solution is,
ds^2 = (1  2m/r*) dt^2  (r* / (r*  2m)) dr*^2  r*^2 dOmega^2 (eqn
7 of Crothers).
Substituting sqrt{C} for r* this becomes,
ds^2 = (1  2m/sqrt{C}) dt^2  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C)
dr^2  C dOmega^2 (eqn 7 of Crothers).
[C' = dC/dr here]. This is just the usual Schwarzschild solution, but
note that r* = sqrt{C} does not need to lie in the range [0,Infty] as
Hilbert incorrectly assumed.
Now, if the coordinate associated with the point mass is r = sqrt{C} =
r0 (which may if you like be chosen such that r0 = 0  it makes no
difference), then we can define the 'radial distance' from the mass at
r0 to any r >= r0 to be the integral from r0 to r of sqrt{g_rr},
R = int sqrt{g_rr} dr from r0 to r (eqn 11 of Crothers)
where g_rr is just the radial component of the metric:
g_rr =  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C) (eqn 12 of Crothers)
Integrating, this (I use the substitution u = sqrt{C} followed by the
substitution u = 2m sec^2 x), we find,
R = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln (sqrt{sqrt{C}}+sqrt{(sqrt{C}
 2m)}) / sqrt{2m}   sqrt{sqrt{C0}*(sqrt{C0}  2m)} + 2m ln
(sqrt{sqrt{C0}}+sqrt{(sqrt{C0}  2m)}) / sqrt{2m} 
[C0 = C(r0) here], where we have fixed the constant of integration by
applying the boundary condition that R>0 as r>r0. Since the radial
distance must be real, this means that the range of possible values for
r* = sqrt{C(r)} is given by r* >= 2m. In particular, the origin must
correspond to the minimum of this range, i.e r*(r0) = sqtr{C0} = 2m, so
that C0 = 4m^2. But r* = 2m is the position of the event horizon, i.e.
the event horizon coincides with the point particle which is the radial
origin. Moreover, r* < 2m is unphysical.
Plugging sqrt{C0} = 2m into the above equation, we find the following
equation for the radial distance of a point at r from the origin,
R(r) = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln
(sqrt{sqrt{C}}+sqrt{(sqrt{C}  2m)} / sqrt{2m} (eqn 14 of Crothers)
Again. one cannot consider radii with r* < 2m, because these radii make
no physical sense, as they would have to have a smaller radius than
radius corresponding to the position of the point mass which defines
the origin of our space.
There is therefore NO interior solution for r* < 2m, because r* < 2m
does not physically exist.
</end quote>
So what he seems to be saying is the following: (I will use
r for his r*):
1. He's assuming a static, spherically symmetric metric,
which can always be transformed into the Schwarzchild
metric ds^2 = (1  2m/r) dt^2  (r / (r  2m)) dr^2  r^2 dOmega^2
2. He defines R(r) to be the integral from r0 to r of
squareroot(g_rr) dr. This is a measure of the distance
from r0 to r. (r0 is the assumed location of the pointmass
source).
3. He notes that R(r) is only real for r >= 2m.
4. So he concludes that only r >= 2m is physically meaningful.
5. So, since there is no mass at r > 2m, the mass must be at r = 2m.
That seems to be the reasoningthat only the region r >= 2m is physically
meaningful, so if the mass is anywhere, it is at r=2m.
Of course, the correct conclusion is that the region r < 2m *is*
physically meaningful, but r is not a spacial coordinate there,
it is a temporal coordinate. Integrating along the path of increasing
r means integrating along a timelike path up until r=2m, and then
integrating along a spacelike path thereafter.
The integrated proper separation
Integral of squareroot(g_uv dx^u dx^v)
along a path that is sometimes spacelike and sometimes timelike
yields a complex number. I'm not sure what (if anything) is the
meaning of this number, but it certainly doesn't indicate that
you have entered a physically forbidden region.

Daryl McCullough
Ithaca, NY 

Back to top 


Jan Bielawski science forum Guru
Joined: 08 May 2005
Posts: 388

Posted: Thu Jul 20, 2006 1:16 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



I wrote:
Quote:  I have asked you this question
twice and haven't responded.

I meant "you haven't responded" :)

Jan Bielawski 

Back to top 


Jan Bielawski science forum Guru
Joined: 08 May 2005
Posts: 388

Posted: Thu Jul 20, 2006 1:12 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer wrote:
Quote:  JanPB wrote:
LEJ Brouwer wrote:
Daryl,
Before this discussion finally descends into a futile "oh yes I am",
"oh no you're not" type battle, I just wanted to express my
appreciation for the patience and objectivity you have shown
throughout.
Very nice of you but if you want your ideas critiqued, you should
provide a readable summary. Since your argument rests on the concept of
the set r=2M being a point, it would help if you provided a clean
argument justifying it.
I am afraid that in your case, it would not make any difference at all.
I have read many many papers, many wellwritten, many very poorly
written, and many apparently containing major logical flaws, but I
still try to take what is good from all of them, and leave the rest. To
do this efficiently requires both openmindedness and sound judgment.

Agreed. But openmindedness does not mean universal acceptance, right?
Quote:  You do not appear to particularly well endowed with either of these
qualities, and the loss is entirely yours.

I note that you still do not provide the argument. Let's leave my
humble personality out of this debate.
Quote:  A wise man can learn even from a fool. I will leave it up to you to
decide whether you are not a wise man, or I am not a fool.

Ditto. So far you have ignored two important technical questions:
1. How do you justify the claim that the set r=2M is a point,
2. What is wrong with removing a coordinate singularity by using a
diffeomorphism that's singular there. I have asked you this question
twice and haven't responded. Most recent time was here:
http://groups.google.com/group/sci.physics.relativity/msg/21ea137673105b9b?hl=en&

Jan Bielawski 

Back to top 


dda1 science forum Guru
Joined: 06 Feb 2006
Posts: 762

Posted: Wed Jul 19, 2006 10:41 pm Post subject:
Re: Sabbir Rahman  Hoax



LEJ Brouwer wrote:
Quote:  Thank you once again for your perspicacious and thoughtprovoking
critique. May I ask which part of India you are from? Y

The part that kicks your sorry ass, imbecile. I told you I'm not
Indian. You don't have to be Indian in order to detect your hoax(es) 

Back to top 


LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Wed Jul 19, 2006 10:38 pm Post subject:
Re: Sabbir Rahman  Hoax



dda1 wrote:
Quote:  LEJ Brouwer wrote:
I am afraid that in your case, it would not make any difference at all.
I have read many many papers, many wellwritten, many very poorly
written, and many apparently containing major logical flaws, but I
still try to take what is good from all of them, and leave the rest. To
do this efficiently requires both openmindedness and sound judgment.
You do not appear to particularly well endowed with either of these
qualities, and the loss is entirely yours.
A wise man can learn even from a fool. I will leave it up to you to
decide whether you are not a wise man, or I am not a fool.

Jan Bielawski
 Sabbir.
This is why your "papers" are collecting moss in arxiv, m**********r?
So will the one that you just solicited feedback, Because like all your
other ones you are not listening to criticism. Doesn't matter, the
community decides: you last one (like the ones languishing since 1997)
is also s**t.

Thank you once again for your perspicacious and thoughtprovoking
critique. May I ask which part of India you are from? Your mother back
home must be very proud of you, being a teacher with publications and
all that. 

Back to top 


dda1 science forum Guru
Joined: 06 Feb 2006
Posts: 762

Posted: Wed Jul 19, 2006 10:12 pm Post subject:
Sabbir Rahman  Hoax



LEJ Brouwer wrote:
Quote:  I am afraid that in your case, it would not make any difference at all.
I have read many many papers, many wellwritten, many very poorly
written, and many apparently containing major logical flaws, but I
still try to take what is good from all of them, and leave the rest. To
do this efficiently requires both openmindedness and sound judgment.
You do not appear to particularly well endowed with either of these
qualities, and the loss is entirely yours.
A wise man can learn even from a fool. I will leave it up to you to
decide whether you are not a wise man, or I am not a fool.

Jan Bielawski
 Sabbir.

This is why your "papers" are collecting moss in arxiv, m**********r?
So will the one that you just solicited feedback, Because like all your
other ones you are not listening to criticism. Doesn't matter, the
community decides: you last one (like the ones languishing since 1997)
is also s**t. 

Back to top 


LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Wed Jul 19, 2006 9:46 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
Daryl McCullough wrote:
In particular, where in the world are you getting your idea
that "the point mass resides at r=2m"? That makes no sense
at all.
"I refer the honourable gentleman to the answers I gave just a moment
ago..." because you are ignoring them and putting words into my mouth.
I'm not ignoring it, I'm saying that it is nonsense.

Well, I think your a very decent chap nevertheless.
Quote:  
Daryl McCullough
Ithaca, NY

 Sabbir. 

Back to top 


Google


Back to top 



The time now is Mon Oct 12, 2015 7:16 pm  All times are GMT

