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A simple proof of the Nonexistance of Black Holes.



 
 
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  #1  
Old May 6th 05, 04:39 AM
Greysky
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Default A simple proof of the Nonexistance of Black Holes.

While Stephen Hawking recently came to the conclusion black holes are not
what he thought they were, I find it amazing he took decades to come to that
conclusion. Using a proof that relies on relativistic arguments instead of
information conservation makes things easier to comprehend. It is assumed
that a non rotating black hole will be observed to have a spherical event
horizon if observed by someone at rest with respect to the black hole. Now,
if the black hole is passing by someone at some relativistic velocity, that
observer will see that the black holes' event horizon is no longer
spherical, and that it is foreshortened in the direction of motion
according to the rules of relativity. Indeed, the faster the black hole
passes by the observer, the more flattened out the event horizon will become
in the direction of motion. Eventually, there will come a point in the
velocity curve, when the event horizon will be congruent with the
singularity producing it. At that point, the singularity will no longer be
shielded from the outside observer - it will be visible along the direction
of motion, and able to interact with the external environment.

Since this can not happen, and assuming the rules of geometrodynamics hold
for all observers in any inertial frame, there can not exist black holes of
the type currently envisioned. Damn that was simple. I shall now go on to
prove the existence of God...

Greysky

www.allocations.cc
Learn how to build a FTL radio; presently partially off-line pending
government approved modifications...


  #2  
Old May 6th 05, 04:53 AM
Starlord
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Killfile time.

"Greysky" wrote in message
m...
While



  #3  
Old May 6th 05, 04:57 AM
Tom Roberts
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Greysky wrote:
the faster the black hole
passes by the observer, the more flattened out the event horizon will become
in the direction of motion. Eventually, there will come a point in the
velocity curve, when the event horizon will be congruent with the
singularity producing it.


Not true. This is easily seen by interchanging black hole and observer.
It is also seen by noting that the existence of points of the manifold
between singularity and horizon implies that they do not intersect
(which you apparently call "congruent", you misuse this word) -- this is
independent of any motion of either black hole or observer.

And the singularity does not "produce" the horizon, they are both merely
geometric properties of the manifold....

The only possible "proof" of the nonexistence of black holes is an
exhaustive search of the universe. Good luck....


Tom Roberts
  #4  
Old May 6th 05, 05:06 AM
Double-A
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Greysky wrote:
While Stephen Hawking recently came to the conclusion black holes are

not
what he thought they were, I find it amazing he took decades to come

to that conclusion.


He didn't want to hurt the sales of the books he had already written.

Double-A

  #5  
Old May 6th 05, 05:52 AM
Mark Martin
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Default


Greysky wrote:
While Stephen Hawking recently came to the conclusion black holes are

not
what he thought they were, I find it amazing he took decades to come

to that
conclusion. Using a proof that relies on relativistic arguments

instead of
information conservation makes things easier to comprehend. It is

assumed
that a non rotating black hole will be observed to have a spherical

event
horizon if observed by someone at rest with respect to the black

hole. Now,
if the black hole is passing by someone at some relativistic

velocity, that
observer will see that the black holes' event horizon is no longer
spherical, and that it is foreshortened in the direction of motion
according to the rules of relativity. Indeed, the faster the black

hole
passes by the observer, the more flattened out the event horizon will

become
in the direction of motion. Eventually, there will come a point in

the
velocity curve, when the event horizon will be congruent with the
singularity producing it. At that point, the singularity will no

longer be
shielded from the outside observer - it will be visible along the

direction
of motion, and able to interact with the external environment.


Heh! That's a lot like saying that an observer near the event
horizon, seeing me whiz by at nearly c, will see my epidermis shorten
until my bones have to stick out of my skin.

-Mark Martin

  #6  
Old May 6th 05, 06:23 AM
N:dlzc D:aol T:com \(dlzc\)
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Default

Dear Gresky:

"Greysky" wrote in message
m...
....
Eventually, there will come a point in the velocity curve,
when the event horizon will be congruent with the singularity
producing it. At that point, the singularity will no longer be
shielded from the outside observer - it will be visible along
the direction of motion, and able to interact with the external
environment.


What is it about a singularity that you don't understand? How
big is this? What finite dimension is it expected to have? I'm
pretty sure that any finite dimension is larger than a
singularity is expected to be.

David A. Smith


  #7  
Old May 6th 05, 06:25 AM
Koobee Wublee
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Default


"Tom Roberts" wrote in message
. ..

Not true. This is easily seen by interchanging black hole and observer. It
is also seen by noting that the existence of points of the manifold
between singularity and horizon implies that they do not intersect (which
you apparently call "congruent", you misuse this word) -- this is
independent of any motion of either black hole or observer.


Therefore, you agree with me that Lorentz Transform only describes what is
observed and not necessarily what reality is.

The only possible "proof" of the nonexistence of black holes is an
exhaustive search of the universe. Good luck....


Or you can be as abstract as Mr. Hobba by closing his eyes doing the
searching. Any abstract type of thinking also produces the same equal
abstractness in imagination. However, reality checks.

So, exploring deeper in this subject of black holes, it is a common belief
that a black hole would be affected by the curvature of spacetime created by
another black hole. It is thus understood that 2 black holes have no
problems merging. In doing so, we will be enlightened with a show of
massive amount of gravitational waves. However, according to Schwarzschild
metric, a black creates a well in spacetime so deep that nothing can escape
itself. Since you should also believe in the gravitational wave not
traveling beyond the speed of light, and if the speed of light cannot escape
a black hole, how then can gravitational wave escape this black hole to
affect anything outside? Perhaps, you can resolve this paradox.



  #8  
Old May 6th 05, 08:02 AM
Greysky
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Posts: n/a
Default


"Mark Martin" wrote in message
ups.com...

Greysky wrote:
While Stephen Hawking recently came to the conclusion black holes are

not
what he thought they were, I find it amazing he took decades to come

to that
conclusion. Using a proof that relies on relativistic arguments

instead of
information conservation makes things easier to comprehend. It is

assumed
that a non rotating black hole will be observed to have a spherical

event
horizon if observed by someone at rest with respect to the black

hole. Now,
if the black hole is passing by someone at some relativistic

velocity, that
observer will see that the black holes' event horizon is no longer
spherical, and that it is foreshortened in the direction of motion
according to the rules of relativity. Indeed, the faster the black

hole
passes by the observer, the more flattened out the event horizon will

become
in the direction of motion. Eventually, there will come a point in

the
velocity curve, when the event horizon will be congruent with the
singularity producing it. At that point, the singularity will no

longer be
shielded from the outside observer - it will be visible along the

direction
of motion, and able to interact with the external environment.


Heh! That's a lot like saying that an observer near the event
horizon, seeing me whiz by at nearly c, will see my epidermis shorten
until my bones have to stick out of my skin.

Not quite. Your bones will also shorten along the direction of motion so
everything will stay in proportion. The singularity of a black hole however
will not foreshorten because of relativity. Since it is an infinitely small
point of gravity it remains a pointlike particle of gravity and relativity
allows you to look under the holes skirt, which is a no-no.

Greysky


  #9  
Old May 6th 05, 09:40 AM
Bill Hobba
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Posts: n/a
Default


"Koobee Wublee" wrote in message
news:udDee.11388$tQ.1600@fed1read06...

"Tom Roberts" wrote in message
. ..

Not true. This is easily seen by interchanging black hole and observer.

It
is also seen by noting that the existence of points of the manifold
between singularity and horizon implies that they do not intersect

(which
you apparently call "congruent", you misuse this word) -- this is
independent of any motion of either black hole or observer.


Therefore, you agree with me that Lorentz Transform only describes what is
observed and not necessarily what reality is.


Semantic gibberish, expressed in you own private language, devoid of any
objective content. Stick to standard terminology if you want to be
understood. Of course that would mean actually studying a text on
relativity - something you have no shown a willingness to do

Bill


The only possible "proof" of the nonexistence of black holes is an
exhaustive search of the universe. Good luck....


Or you can be as abstract as Mr. Hobba by closing his eyes doing the
searching. Any abstract type of thinking also produces the same equal
abstractness in imagination. However, reality checks.

So, exploring deeper in this subject of black holes, it is a common belief
that a black hole would be affected by the curvature of spacetime created

by
another black hole. It is thus understood that 2 black holes have no
problems merging. In doing so, we will be enlightened with a show of
massive amount of gravitational waves. However, according to

Schwarzschild
metric, a black creates a well in spacetime so deep that nothing can

escape
itself. Since you should also believe in the gravitational wave not
traveling beyond the speed of light, and if the speed of light cannot

escape
a black hole, how then can gravitational wave escape this black hole to
affect anything outside? Perhaps, you can resolve this paradox.





  #10  
Old May 6th 05, 09:53 AM
Dirk Van de moortel
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Posts: n/a
Default


"Bill Hobba" wrote in message ...

"Koobee Wublee" wrote in message
news:udDee.11388$tQ.1600@fed1read06...

"Tom Roberts" wrote in message
. ..

Not true. This is easily seen by interchanging black hole and observer.

It
is also seen by noting that the existence of points of the manifold
between singularity and horizon implies that they do not intersect

(which
you apparently call "congruent", you misuse this word) -- this is
independent of any motion of either black hole or observer.


Therefore, you agree with me that Lorentz Transform only describes what is
observed and not necessarily what reality is.


Semantic gibberish, expressed in you own private language, devoid of any
objective content. Stick to standard terminology if you want to be
understood. Of course that would mean actually studying a text on
relativity - something you have no shown a willingness to do

Bill


"describes what is observed and not necessarily what reality is."
What a silly thing to say - specially for a retired aerospace
engineer like Australopitheticus ;-)

Dirk Vdm


 




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