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Daryl McCullough
science forum Guru
Joined: 24 Mar 2005
Posts: 1167
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 Posted: Fri Jul 14, 2006 11:26 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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LEJ Brouwer says...
| Quote: | Hi, I think the semantics are getting a little mixed up here. I am not
claiming that r is spacelike *because* of the Einstein field equation -
I am simply stating the *fact* that r is spacelike in the Einstein
field equation - and indeed in any other equation in which r is a
radial parameter. The radial direction is spacelike by definition and
the temporal direction is timelike by definition.
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If the radial direction is spacelike by definition, then that
means that r is *not* a radial coordinate in the interior region.
It's a temporal coordinate in that region. If the time direction
is timelike by definition, then that means that t is *not* a
temporal coordinate in the interior region. It's a spatial
coordinate there.
| Quote: | That's all there is to it. The fact that the radial direction is
timelike in the interior Schwarzschild solution is further evidence
that this solution is not a valid one.
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It is not true that the "radial direction is timelike in the interior
Schwarzschild solution". By definition, if a coordinate is timelike,
then it is *temporal*, not *radial*. The fact that the same variable
*name* r is used for a radial coordinate in the region outside the
event horizon, and is used for a temporal coordinate in the region
inside the event horizon has no significance.
If you prefer to restrict the name "r" to radial coordinates, and
restrict the name "t" to temporal coordinates, then you can certainly
do the following:
In the region outside the event horizon, use the metric
ds^2 = -((r-R)/r)dt^2 + (r/(r-R)) dr^2 + r^2 dOmega^2
This region is described by the coordinates (t,r,theta,phi)
where t is a time coordinate that runs from -infinity to +infinity,
r is a radial coordinate that runs from R to +infinity, and theta
and phi are angular coordinates.
In the region inside the event horizion, use the metric
ds^2 = -((R-t)/t) dt^2 + (t/(R-t)) dz^2 + (R-t)^2 dOmega^2
This region is described by the coordinates (t,z,theta,phi)
where t is a time coordinate that runs from -infinity to R,
z is a spatial coordinate that runs from -infinity to +infinity,
and theta and phi are angular coordinates. With these coordinates,
there is a catastrophe awaiting every observer in the region at
the time t=R (the singularity).
So with these new coordinates, t always refers to the temporal
coordinate, and r always refers to a radial coordinate.
What I think you are worried about is the relationship between
the coordinates in the exterior region and coordinate in the
interior region. There really *is* no relationship. They are
two separate regions of spacetime, and we use different coordinates
in the different regions.
I think people are spoiled by cartesian coordinates in Euclidean
space. Those coordinates are global; they can be used to describe
any point anywhere in the manifold. But that's not the common
case. Usually, when the manifold is curved, coordinates are only
valid in a small region, and if you travel outside of that region
you have to switch to a different coordinate system. Someone
falling towards a black hole has to switch coordinate systems
as he passes from one region to another (when he crosses the
event horizon) but that has no more significance than having
to reset your watch on Earth when you cross into a different
time zone.
--
Daryl McCullough
Ithaca, NY |
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LEJ Brouwer
science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120
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| Posted: Fri Jul 14, 2006 11:27 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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Tom Roberts wrote:
| Quote: | LEJ Brouwer wrote:
I am simply stating the *fact* that r is spacelike in the Einstein
field equation - and indeed in any other equation in which r is a
radial parameter.
Where does the symbol "r" obtain this magical power???
You _really_ need to learn the basics. Symbols have no power whatsoever,
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Some might disagree with you quite strongly. Can I recommend you take a
look at "The Temple of Man" by R A Schwaller de Lubicz to get some feel
for how thoroughly symbolism can permeate an entire civilisation?
http://www.amazon.com/gp/product/0892815701/sr=1-1/qid=1152919339/ref=pd_bbs_1/102-3323106-7127323?ie=UTF8&s=books
| Quote: | they mean merely what we designate them to mean, and in the case of GR,
as I said before, we don't know whether a given corodinate is timelike
or spacelike until we solve the field equation and examint the relevant
metric components. <shrug
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Well, I wish you luck in your research. You might need it.
| Quote: | The radial direction is spacelike by definition
Your definition of words and symbols has no power over the mathematics.
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So, you dare question the power of my words and symbols? :-)
- Sabbir. |
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Sue...
science forum Guru
Joined: 08 May 2005
Posts: 2684
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| Posted: Fri Jul 14, 2006 11:42 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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LEJ Brouwer wrote:
| Quote: | Daryl McCullough wrote:
LEJ Brouwer says...
Actually, I meant it as I said it - in the Einstein field equation, r
is spacelike and t is timelike. It is not possible therefore for r to
be timelike in the solution to the Einstein field equation, which means
that only the exterior solution is correct, and that the interior
solution is wrong.
Back up here. The Einstein field equations don't say anything about
"r" and "t". They don't say anything about which basis vectors are
spacelike and which are timelike. What they say is that
G_uv = k T_uv
where G_uv is a tensor formed from second derivatives of
the metric tensor g_uv, and T_uv is the stress-energy tensor,
and k is a constant related to Newton's constant G (there
might be a factor of pi or something).
How do you get r is spacelike and t is timelike from that?
--
Daryl McCullough
Ithaca, NY
Hi, I think the semantics are getting a little mixed up here.
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Yes.. they are problematic even in texts that should be
'authoritative'.
| Quote: | I am not
claiming that r is spacelike *because* of the Einstein field equation -
I am simply stating the *fact* that r is spacelike in the Einstein
field equation - and indeed in any other equation in which r is a
radial parameter. The radial direction is spacelike by definition and
the temporal direction is timelike by definition.
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The definition doesn't count. We need to go back to the field
equations and confirm that the temporal components are
transformed out before the interval is defined as spatial.
Likewise confirming that the spatial component is transformed
out before 'defining' the interval as temporal. The function
chi(x.t) comes to mind but I'd check here to be sure.
http://arxiv.org/abs/physics/0204034
| Quote: | That's all there is
to it. The fact that the radial direction is timelike in the interior
Schwarzschild solution is further evidence that this solution is not a
valid one.
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That is my gut feeling too. janPB came up with the idea, perhaps
in this thread, to ignore the words and look at the ingredients.
It makes perfect sense to me.
Now if the ingredients are Portland cement, the label says
Self Rising Flour and the customer is "All Smiles Bakery"
....then Houston has a problem.  )
Sue...
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Daryl McCullough
science forum Guru
Joined: 24 Mar 2005
Posts: 1167
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| Posted: Fri Jul 14, 2006 11:43 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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LEJ Brouwer says...
| Quote: | Not so - the norm of the 4-acceleration on a test particle is a local,
invariant, intrinsic quantity that diverges on the event horizon.
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What is infinite is the acceleration necessary to hover in place
right at the event horizon. What that means is that it is *impossible*
to hover at the event horizon. If you are at the event horizon, then
you're going to fall in.
But that doesn't mean that there is anything singular happening
at the event horizon. Here's an analogy: Suppose I have a spaceship
that is undergoing constant acceleration (as measured by someone
aboard the spaceship). If the proper acceleration is A, then any
object (even a light signal) that starts out a distance of 1/A or
more behind the spaceship will *never* catch up.
To someone in the spaceship, it appears that there is a horizon
a distance of 1/A below him such that anything that crosses that
horizon can never come out again. In his coordinate system, it
requires an *infinite* amount of acceleration for someone to
hover right at that horizon.
However, someone at that horizon doesn't feel anything weird
going on. The same is true of the event horizon in Schwarzschild
coordinates. The horizon seems significant to someone above
the horizon, but has no physical significance right at the horizon.
--
Daryl McCullough
Ithaca, NY |
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Jan Bielawski
science forum Guru
Joined: 08 May 2005
Posts: 388
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| Posted: Fri Jul 14, 2006 11:44 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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LEJ Brouwer wrote:
| Quote: | dda1 wrote:
LEJ Brouwer wrote:
dda1 wrote:
snipped.
Why is that you Pakis never admit to error? Is it your religion? You
afraid that you loose face?
Well, you already lost a lot of face by continuing to post, you have
been exposed as a fraud by several of us already. Why do you think that
you can't publish your stuff in any peer reviewed journal? Eh?
I don't know - maybe it's because you don't like Pakis? :)
This is why no peer reviewed journal would publish your stuff? This is
why you get kicked of moderated forums? Because we all don't like Pakis?
Well, YOU certainly don't seem to like Pakis.
A major reason for my 'stuff' not being published could be that I tend
to question knowledge that has become 'established' when the
foundations of that knowledge are dubious.
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No, no, no, no, please let go of those dreams at once. A far simpler
explanation: your papers are not good enough. There is no need to
search for grand explanations and Hollywoodesque "establishment"
conspiracies when good old plain incompetence suffices. If you don't
know how coordinate labels relate to tensor equations then there is
really nothing to add, is there?
Remember that 101 years ago a certain no-name government bureaucrat in
Switzerland was able to publish what you would call an
anti-establishment paper. So much for not being able to publish because
of discussing unpopular ideas.
--
Jan Bielawski |
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dda1
science forum Guru
Joined: 06 Feb 2006
Posts: 762
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| Posted: Fri Jul 14, 2006 11:58 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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LEJ Brouwer wrote:
| Quote: | I claim that there is a luminiferous aether, when it has been
'established' that there isn't one.
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Hmmm
1. Aether could not be detected experimentally (all experiments
atttempting to detect it failed to do so)
2. aether is not necessary as an explanation of physical reality (has
been superceeded by SR explanations)
Now, you claim that there is such a "luminiferous aether". Your only
chance is to come up with yet another experiment that detects it. Have
you done that? No? And you wonder why you are thrown out on your ass?
| Quote: | I claim that quantum theory has a classical basis, when it has already
been 'established' that quantum theory is more fundamental than
classical mechanics.
|
Let's try a simple question: can you show that quantum mechanics is
based on classical mechanics?
Did you study physics or are you the "home schooled" breed?
<rest snipped as increasingly idiotic> |
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LEJ Brouwer
science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120
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| Posted: Sat Jul 15, 2006 12:18 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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Daryl McCullough wrote:
| Quote: | LEJ Brouwer says...
Hi, I think the semantics are getting a little mixed up here. I am not
claiming that r is spacelike *because* of the Einstein field equation -
I am simply stating the *fact* that r is spacelike in the Einstein
field equation - and indeed in any other equation in which r is a
radial parameter. The radial direction is spacelike by definition and
the temporal direction is timelike by definition.
If the radial direction is spacelike by definition, then that
means that r is *not* a radial coordinate in the interior region.
It's a temporal coordinate in that region. If the time direction
is timelike by definition, then that means that t is *not* a
temporal coordinate in the interior region. It's a spatial
coordinate there.
That's all there is to it. The fact that the radial direction is
timelike in the interior Schwarzschild solution is further evidence
that this solution is not a valid one.
It is not true that the "radial direction is timelike in the interior
Schwarzschild solution". By definition, if a coordinate is timelike,
then it is *temporal*, not *radial*. The fact that the same variable
*name* r is used for a radial coordinate in the region outside the
event horizon, and is used for a temporal coordinate in the region
inside the event horizon has no significance.
If you prefer to restrict the name "r" to radial coordinates, and
restrict the name "t" to temporal coordinates, then you can certainly
do the following:
In the region outside the event horizon, use the metric
ds^2 = -((r-R)/r)dt^2 + (r/(r-R)) dr^2 + r^2 dOmega^2
This region is described by the coordinates (t,r,theta,phi)
where t is a time coordinate that runs from -infinity to +infinity,
r is a radial coordinate that runs from R to +infinity, and theta
and phi are angular coordinates.
In the region inside the event horizion, use the metric
ds^2 = -((R-t)/t) dt^2 + (t/(R-t)) dz^2 + (R-t)^2 dOmega^2
This region is described by the coordinates (t,z,theta,phi)
where t is a time coordinate that runs from -infinity to R,
z is a spatial coordinate that runs from -infinity to +infinity,
and theta and phi are angular coordinates. With these coordinates,
there is a catastrophe awaiting every observer in the region at
the time t=R (the singularity).
So with these new coordinates, t always refers to the temporal
coordinate, and r always refers to a radial coordinate.
What I think you are worried about is the relationship between
the coordinates in the exterior region and coordinate in the
interior region. There really *is* no relationship. They are
two separate regions of spacetime, and we use different coordinates
in the different regions.
I think people are spoiled by cartesian coordinates in Euclidean
space. Those coordinates are global; they can be used to describe
any point anywhere in the manifold. But that's not the common
case. Usually, when the manifold is curved, coordinates are only
valid in a small region, and if you travel outside of that region
you have to switch to a different coordinate system. Someone
falling towards a black hole has to switch coordinate systems
as he passes from one region to another (when he crosses the
event horizon) but that has no more significance than having
to reset your watch on Earth when you cross into a different
time zone.
--
Daryl McCullough
Ithaca, NY
|
Hi Daryl,
Thanks for taking the time to write that detailed explanation. I
understand your (and others') point that different coordinates can be
used on different coordinate patches, so that r and t on one patch need
not be related to r and t on the other patch. However, in this case,
both of these coordinates r and t are the same r and t that appear in
the Einstein field equations - and this surely induces a relationship
between the two coordinate systems (i.e. they must have the same
'meaning' as someone put it, as the meaning they have in the field
equation).
Another issue of concern is that the two solutions (i.e. the interior
and exterior) are not generally considered to be independent of each
other, but rather that a radially infalling particle supposedly moves
smoothly from the exterior solution to the interior solution. This is
the reason for the development of the various coordinate systems which
seem to patch the interior and exterior together. The problem is that
while everything looks fine in the new coordinates, these new
coordinates are not the 'physical' ones (i.e. radius and time) and they
unfortunately serve to mask the fact that what is physically happening
is that the radial direction *is* becoming timelike and the temporal
direction *is* becoming radial upon crossing the horizon. This is
_physically_ impossible, even though the mathematical mapping in the
transformed coordinates is smooth.
The whole matter is really quite non-trivial which is no doubt why
there has been so much confusion about it in the past. Unfortunately I
think the interpretation which survived, i.e. the particle crosses the
event horizon into the interior region, is wrong, for the reasons I
have already mentioned (most importantly because there is no interior
region). If there is no interior region, the only place the particle
can really go after it hits the event horizon is onto the other
external Kruskal sheet (and note that this has to happen smoothly as
there are no curvature singularities etc). But according to an
independent observer the orientation of time is reversed on the second
sheet relative to the first, so an external observer will actually see
what appears to be a particle-antiparticle annihilation event occurring
at t=Infty at the event horizon.
One of the reasons that I am so confident that my interpretation is
correct (besides the mathematical proofs already given by Abrams,
Antoci and Crothers) is that it has allowed me to derive classical
electrodynamics and resolve the dark matter/MOND problem essentially
from first principles in the context of classical GR. The appearance of
electrodynamics and MOND I think is pretty miraculous, and an
indication that we are on the right path. I derived the existence of a
fluid consisting of neutral spinors last year when I was not at all
concerned with MOND or dark matter, and was pretty amazed when Luc
Blanchet's paper came out in May stating that this scenario is
precisely what is required to potentially resolve the MOND/dark matter
problem.
Best wishes,
Sabbir. |
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LEJ Brouwer
science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120
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| Posted: Sat Jul 15, 2006 12:33 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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JanPB wrote:
| Quote: | LEJ Brouwer wrote:
No, no, no, no, please let go of those dreams at once. A far simpler
explanation: your papers are not good enough. There is no need to
search for grand explanations and Hollywoodesque "establishment"
conspiracies when good old plain incompetence suffices. If you don't
know how coordinate labels relate to tensor equations then there is
really nothing to add, is there?
|
Hmm, I think you are possibly confusing me with the entire population
of America. I do not claim that there is some kind of conspiracy going
on. It is just a sad state of affairs and this is just how the current
system works. As someone said, if Einstein were around today, he would
probably have a Yahoo or Hotmail email account and his submissions
would not be accepted by the arxiv. Its a shame, but that's just how
things are. Maybe my papers are not 'good enough' for whatever reason
(I have not actually submitted most of my papers for publication) but I
will still stand by my results, because I am confident that the
mathematics is sound. Nothing you have said has convinced me otherwise.
| Quote: | Remember that 101 years ago a certain no-name government bureaucrat in
Switzerland was able to publish what you would call an
anti-establishment paper. So much for not being able to publish because
of discussing unpopular ideas.
--
Jan Bielawski
|
So you seriously believe that things work in the same way now as they
did 101 years ago? Any you accuse me of wishful thinking! |
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LEJ Brouwer
science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120
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| Posted: Sat Jul 15, 2006 12:37 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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dda1 wrote:
| Quote: | LEJ Brouwer wrote:
I claim that there is a luminiferous aether, when it has been
'established' that there isn't one.
Hmmm
1. Aether could not be detected experimentally (all experiments
atttempting to detect it failed to do so)
2. aether is not necessary as an explanation of physical reality (has
been superceeded by SR explanations)
Now, you claim that there is such a "luminiferous aether". Your only
chance is to come up with yet another experiment that detects it. Have
you done that? No? And you wonder why you are thrown out on your ass?
I claim that quantum theory has a classical basis, when it has already
been 'established' that quantum theory is more fundamental than
classical mechanics.
Let's try a simple question: can you show that quantum mechanics is
based on classical mechanics?
Did you study physics or are you the "home schooled" breed?
|
I have a PhD in theoretical physics (string field theory) from MIT. My
supervisor was Barton Zwiebach.
| Quote: | rest snipped as increasingly idiotic
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I rest my case.
- Sabbir.
P.S. Perhaps the easiest prediction of my model to test is the negative
mass of antiparticles. |
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LEJ Brouwer
science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120
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| Posted: Sat Jul 15, 2006 12:44 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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Daryl McCullough wrote:
| Quote: | LEJ Brouwer says...
Not so - the norm of the 4-acceleration on a test particle is a local,
invariant, intrinsic quantity that diverges on the event horizon.
What is infinite is the acceleration necessary to hover in place
right at the event horizon. What that means is that it is *impossible*
to hover at the event horizon. If you are at the event horizon, then
you're going to fall in.
|
Daryl, this goes back to some papers by Doughty - I'll have to dig them
out again to see exactly what he says. I think his results indicate
that there is something singular about the horizon (which is not
surprising as it coincides with the 'surface' of the point mass). But
your conclusion that you cannot hover at the event horizon is probably
correct. I disagree with you that a test particle is going to fall 'in'
because there is no 'in'. Rather the test particle is going to fall
'out' - into the second exterior sheet. I recommend you take a look at
my paper where I discuss this and its consequences.
| Quote: |
But that doesn't mean that there is anything singular happening
at the event horizon. Here's an analogy: Suppose I have a spaceship
that is undergoing constant acceleration (as measured by someone
aboard the spaceship). If the proper acceleration is A, then any
object (even a light signal) that starts out a distance of 1/A or
more behind the spaceship will *never* catch up.
To someone in the spaceship, it appears that there is a horizon
a distance of 1/A below him such that anything that crosses that
horizon can never come out again. In his coordinate system, it
requires an *infinite* amount of acceleration for someone to
hover right at that horizon.
|
Sure, I agree with you.
| Quote: | However, someone at that horizon doesn't feel anything weird
going on. The same is true of the event horizon in Schwarzschild
coordinates. The horizon seems significant to someone above
the horizon, but has no physical significance right at the horizon.
|
Again, I agree with you. I only disagree about what happens next.
| Quote: | --
Daryl McCullough
Ithaca, NY |
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dda1
science forum Guru
Joined: 06 Feb 2006
Posts: 762
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| Posted: Sat Jul 15, 2006 12:50 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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LEJ Brouwer wrote:
Did you study physics or are you the "home schooled" breed?
| Quote: |
I have a PhD in theoretical physics (string field theory) from MIT. My
supervisor was Barton Zwiebach.
rest snipped as increasingly idiotic
I rest my case.
- Sabbir.
Looks like a huge waste of US taxpayer money. So let me ask you again: |
1. Aether could not be detected experimentally (all experiments
atttempting to detect it failed to do so)
2. aether is not necessary as an explanation of physical reality (has
been superceeded by SR explanations)
Now, you claim that there is such a "luminiferous aether". Your only
chance is to come up with yet another experiment that detects it. Have
you done that? No? And you wonder why you are thrown out on your ass? |
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LEJ Brouwer
science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120
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| Posted: Sat Jul 15, 2006 1:03 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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dda1 wrote:
| Quote: | LEJ Brouwer wrote:
Did you study physics or are you the "home schooled" breed?
I have a PhD in theoretical physics (string field theory) from MIT. My
supervisor was Barton Zwiebach.
rest snipped as increasingly idiotic
I rest my case.
- Sabbir.
Looks like a huge waste of US taxpayer money. So let me ask you again:
|
Yeah, even bigger than the occupation of Iraq.
| Quote: | 1. Aether could not be detected experimentally (all experiments
atttempting to detect it failed to do so)
2. aether is not necessary as an explanation of physical reality (has
been superceeded by SR explanations)
Now, you claim that there is such a "luminiferous aether". Your only
chance is to come up with yet another experiment that detects it. Have
you done that? No? And you wonder why you are thrown out on your ass?
|
My 'luminiferous aether' is none other than the physical vacuum. I
don't need an experiment to detect it because we already know it's
there. My aether does not define an absolute frame of reference that it
can be detected in the way you suggest. Read my paper and maybe you
will understand. I wouldn't put my money on that though.
Well, now you know about me, why don't you tell me a little about
yourself?
- Sabbir. |
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dda1
science forum Guru
Joined: 06 Feb 2006
Posts: 762
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| Posted: Sat Jul 15, 2006 4:16 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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LEJ Brouwer wrote:
| Quote: |
My 'luminiferous aether' is none other than the physical vacuum. I
don't need an experiment to detect it because we already know it's
there.
|
Another kook. And we paid money for an a*****le like you to get a PhD?
You" already know it's there"? This is the standard kook - fare: "it is
because I think so"
"My aether does not define an absolute frame of reference that it
| Quote: | can be detected in the way you suggest."
|
I don't suggest anything, a*****le, I just asked what proof you have.
Now we all know :NONE.
"Read my paper and maybe you
| Quote: | will understand. I wouldn't put my money on that though."
|
What "paper"? Do you have it anywhere on the web? In arxiv? If you
haven't run any experiment is worth s**t. You know at least that much
if you passed thry MIT.
What pieces of s**t we bring in this country. At least you don't bomb
trains with innocent people.
| Quote: |
Well, now you know about me, why don't you tell me a little about
yourself?
|
I teach. I publish (in peer reviewed journals) . In my free time I
debunk frauds. Like you. |
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Jan Bielawski
science forum Guru
Joined: 08 May 2005
Posts: 388
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| Posted: Sat Jul 15, 2006 4:31 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?
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dda1 wrote:
| Quote: |
What pieces of s**t we bring in this country. At least you don't bomb
trains with innocent people.
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Aw, come on, this is a bit too much!
--
Jan Bielawski |
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dda1
science forum Guru
Joined: 06 Feb 2006
Posts: 762
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Forum index
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Science and Technology
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Physics
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Relativity
