Science > Physics > A valid question which has an uncomfortable answer
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Science > Physics |
| User: |
"greysky" |
| Date: |
25 Sep 2006 02:17:36 AM |
| Object: |
A valid question which has an uncomfortable answer |
A simple non rotating black hole has a spherical event horizon which is
measured at some radius r from the singularity. Now, suppose the black hole
is observed to be moving at some significant velocity as it passes by the
observer, at say .99C. Now the black hole is not observed to be a perfect
sphere because of the relativistic foreshortening along the direction of
motion. Now to the observer the black hole appears to be a squashed hole.
The closer to the speed of light the black hole gets, the smaller value r
takes on in the direction of motion. My question is simple: What happens
when the black hole moves at C and the Schwarzschild radius along the
direction of motion equals zero? At this point is the singularity exposed
for all to see (only in the one direction of course)?
Greysky
www.allocations.cc
Learn how to build a black hole, er, a FTL radio.
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| User: "Martin Hogbin" |
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| Title: Re: A valid question which has an uncomfortable answer |
25 Sep 2006 03:15:52 AM |
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"greysky" <greysky@sbcglobal.net> wrote in message news:koLRg.4978$GR.508@newssvr29.news.prodigy.net...
A simple non rotating black hole has a spherical event horizon which is
measured at some radius r from the singularity. Now, suppose the black hole
is observed to be moving at some significant velocity as it passes by the
observer, at say .99C. Now the black hole is not observed to be a perfect
sphere because of the relativistic foreshortening along the direction of
motion. Now to the observer the black hole appears to be a squashed hole.
The closer to the speed of light the black hole gets, the smaller value r
takes on in the direction of motion. My question is simple: What happens
when the black hole moves at C ...
It cannot do this.
Martin Hogbin
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| User: "malibu" |
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| Title: Re: A valid question which has an uncomfortable answer |
25 Sep 2006 10:35:48 AM |
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Martin Hogbin wrote:
"greysky" <greysky@sbcglobal.net> wrote in message news:koLRg.4978$GR.508@newssvr29.news.prodigy.net...
A simple non rotating black hole has a spherical event horizon which is
measured at some radius r from the singularity. Now, suppose the black hole
is observed to be moving at some significant velocity as it passes by the
observer, at say .99C. Now the black hole is not observed to be a perfect
sphere because of the relativistic foreshortening along the direction of
motion. Now to the observer the black hole appears to be a squashed hole.
The closer to the speed of light the black hole gets, the smaller value r
takes on in the direction of motion. My question is simple: What happens
when the black hole moves at C ...
It cannot do this.
Martin Hogbin
Imaginary things can do anything
we imagine them to do.
Galactic centers, if that is what you
are calling 'black holes', are constrained more
than atomic centers since their speed
is much less.
John
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| User: "Painius" |
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| Title: Re: A valid question which has an uncomfortable answer |
27 Sep 2006 06:08:06 AM |
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"greysky" <greysky@sbcglobal.net> wrote in message...
news:koLRg.4978$GR.508@newssvr29.news.prodigy.net...
A simple non rotating black hole has a spherical event horizon which is
measured at some radius r from the singularity. Now, suppose the black
hole is observed to be moving at some significant velocity as it passes
by the observer, at say .99C. Now the black hole is not observed to be a
perfect sphere because of the relativistic foreshortening along the
direction of motion. Now to the observer the black hole appears to be a
squashed hole. The closer to the speed of light the black hole gets, the
smaller value r takes on in the direction of motion. My question is
simple: What happens when the black hole moves at C and the Schwarzschild
radius along the direction of motion equals zero? At this point is the
singularity exposed for all to see (only in the one direction of course)?
Greysky
www.allocations.cc
Learn how to build a black hole, er, a FTL radio.
Yes! It impinges upon the retina, and science refers to
it as a "photon". <g>
happy days and...
starry starry nights!
--
Indelibly yours,
Paine
http://www.painellsworth.net
http://www.savethechildren.org
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| User: "" |
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| Title: Re: A valid question which has an uncomfortable answer |
25 Sep 2006 03:56:40 PM |
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In sci.physics greysky <greysky@sbcglobal.net> wrote:
A simple non rotating black hole has a spherical event horizon which is
measured at some radius r from the singularity. Now, suppose the black hole
is observed to be moving at some significant velocity as it passes by the
observer, at say .99C. Now the black hole is not observed to be a perfect
sphere because of the relativistic foreshortening along the direction of
motion. Now to the observer the black hole appears to be a squashed hole.
The closer to the speed of light the black hole gets, the smaller value r
takes on in the direction of motion. My question is simple: What happens
when the black hole moves at C and the Schwarzschild radius along the
direction of motion equals zero? At this point is the singularity exposed
for all to see (only in the one direction of course)?
As others have said, you can't consistently take the speed to c while
keeping the mass nonzero. You *can*, however, take the speed to c while
simultaneously taking the rest mass to zero, in such a way that the energy
remains finite. This was first done by Aichelburg and Sexl, published in
Gen. Rel. Grav. 2 (1971) 303. The result is a spacetime with a very mild
singularity -- the curvature tensor becomes that of a "shock wave," with
delta function singularities. (These are nothing like the singularity of
a black hole.)
For the more general case of a spinning black hole, you can find a discussion
in, for example, Burinskii and Magli, gr-qc/9904012.
Note, by the way, that the definition of an event horizon is coordinate
independent, and in particular Lorentz invariant. That means that if the
singularity is behind a horizon for a black hole as seen by an observer at
rest, it will remain so for any other observer.
Steve Carlip
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| User: "Helmut Wabnig" |
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| Title: Re: A valid question which has an uncomfortable answer |
25 Sep 2006 03:00:51 AM |
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On Mon, 25 Sep 2006 07:17:36 GMT, "greysky" <greysky@sbcglobal.net>
wrote:
A simple non rotating black hole has a spherical event horizon which is
measured at some radius r from the singularity. Now, suppose the black hole
is observed to be moving at some significant velocity as it passes by the
observer, at say .99C. Now the black hole is not observed to be a perfect
sphere because of the relativistic foreshortening along the direction of
motion. Now to the observer the black hole appears to be a squashed hole.
The closer to the speed of light the black hole gets, the smaller value r
takes on in the direction of motion. My question is simple: What happens
when the black hole moves at C and the Schwarzschild radius along the
direction of motion equals zero? At this point is the singularity exposed
for all to see (only in the one direction of course)?
Greysky
www.allocations.cc
Learn how to build a black hole, er, a FTL radio.
The length contraction is not all we have to care for.
Imagine a Square moving "very" fast from left to right.
(fixed pitch font)
A----------
| |
| | -----> motion
| |
| |
B----------
^
|
We are *here* and look upwards.
The light (waves, particles) from corner B will reach us earlier than
the light from corner A. That is, we see A where it was at an earlier
time and we see corner B where it is "now". (actually we always look
into the past, may I simplify that a little).
(simplified, leaving out the horizontal lenght dilation):
What we see:
A-----------
\ \
\ \ ------>motion
\ \
\ \
B------------
The SRT formulas actually describe a rotation operation,
when we look at the mathematical formalism.
The observed horizontal length dilatation will correspond
exactly to what we see when we tilt the square counterclockwise.
Please use your imagination, ASCII graphic isn't my favorite.
w.
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| User: "Tom Roberts" |
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| Title: Re: A valid question which has an uncomfortable answer |
25 Sep 2006 08:17:55 AM |
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Helmut Wabnig wrote:
The length contraction is not all we have to care for.
Yes.
Imagine a Square moving "very" fast from left to right.
(fixed pitch font)
A----------
| |
| | -----> motion
| |
| |
B----------
^
|
We are *here* and look upwards.
The light (waves, particles) from corner B will reach us earlier than
the light from corner A. That is, we see A where it was at an earlier
time and we see corner B where it is "now". (actually we always look
into the past, may I simplify that a little).
(simplified, leaving out the horizontal lenght dilation):
What we see:
A-----------
\ \
\ \ ------>motion
\ \
\ \
B------------
Not quite -- your drawing shows the horizontal faces as unrotated; they
are also rotated. Indeed the square still looks like a square, just rotated.
Note this is due to the propagation delays of the light. If one projects
onto our coordinates, we will measure a "squashed rectangle" with no
rotation and a reduced length along the direction of motion. Note also
that the "length contraction" is fully accounted for by the rotated
image -- it remains a rotated square when including ALL effects of SR.
The SRT formulas actually describe a rotation operation,
when we look at the mathematical formalism.
The observed horizontal length dilatation will correspond
exactly to what we see when we tilt the square counterclockwise.
Yes, except it is not "tilt" but "rotate".
To the original question: it is not possible for the black hole to move
at c relative to any observer, like massive objects it can only approach c.
Tom Roberts
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| User: "Double-A" |
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| Title: Re: A valid question which has an uncomfortable answer |
25 Sep 2006 02:55:31 AM |
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greysky wrote:
A simple non rotating black hole has a spherical event horizon which is
measured at some radius r from the singularity. Now, suppose the black hole
is observed to be moving at some significant velocity as it passes by the
observer, at say .99C. Now the black hole is not observed to be a perfect
sphere because of the relativistic foreshortening along the direction of
motion. Now to the observer the black hole appears to be a squashed hole.
The closer to the speed of light the black hole gets, the smaller value r
takes on in the direction of motion. My question is simple: What happens
when the black hole moves at C and the Schwarzschild radius along the
direction of motion equals zero? At this point is the singularity exposed
for all to see (only in the one direction of course)?
Greysky
www.allocations.cc
Learn how to build a black hole, er, a FTL radio.
There is no singularity.
Double-A
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