Science > Physics > Black hole in one frame of reference, not in another?
| Topic: |
Science > Physics |
| User: |
"Todd A. Anderson" |
| Date: |
17 Feb 2004 05:02:33 PM |
| Object: |
Black hole in one frame of reference, not in another? |
If I start with an object who density is below that required for a black
hole
and accelerate that object sufficiently such that from my frame of reference
the additional relativistic mass makes the density of the object sufficient
for a black hole, what happens? In my frame of reference it would be
a black hole but in its own frame of reference it wouldn't? Is it even
possible
to do this or is the volume of the object relativistically changed such
that
the density in both frames of reference are constant?
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| User: "Old Man" |
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| Title: Re: Black hole in one frame of reference, not in another? |
17 Feb 2004 07:03:13 PM |
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"Todd A. Anderson" <drtodd@aaahawk.com.N0SPAM> wrote in message
news:c0u6ie$eld$1@news01.intel.com...
If I start with an object who density is below that required for a black
hole
and accelerate that object sufficiently such that from my frame of
reference
the additional relativistic mass makes the density of the object
sufficient
for a black hole, what happens? In my frame of reference it would be
a black hole but in its own frame of reference it wouldn't?
In the center-of-momentum frame, total energy (KE + rest mass
energy) gravitates. For a single object, the center-of-momentum
frame coincides with that of the object's rest frame. The kinetic
energy of a single object does not gravitate.
Consider two objects traveling along trajectories that eventually
meet in head on collision . As viewed in the COM frame, the
kinetic energy, as well as the rest mass energy of both objects
gravitates. In the COM, "relativistic mass" gravitates.
Kinetic energy associated with motion of the COM frame does
not gravitate. [Old Man]
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| User: "Todd A. Anderson" |
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| Title: Re: Black hole in one frame of reference, not in another? |
17 Feb 2004 07:14:29 PM |
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So, was that a yes or a no? Assume I am zooming towards another object.
From what you wrote, the kinetic energy of the incoming object gravitates.
What about the relativistic volume of the incoming object?
"Old Man" <nomail@nomail.net> wrote in message
news:0_idnYJxaPBOJK_d4p2dnA@prairiewave.com...
"Todd A. Anderson" <drtodd@aaahawk.com.N0SPAM> wrote in message
news:c0u6ie$eld$1@news01.intel.com...
If I start with an object who density is below that required for a black
hole
and accelerate that object sufficiently such that from my frame of
reference
the additional relativistic mass makes the density of the object
sufficient
for a black hole, what happens? In my frame of reference it would be
a black hole but in its own frame of reference it wouldn't?
In the center-of-momentum frame, total energy (KE + rest mass
energy) gravitates. For a single object, the center-of-momentum
frame coincides with that of the object's rest frame. The kinetic
energy of a single object does not gravitate.
Consider two objects traveling along trajectories that eventually
meet in head on collision . As viewed in the COM frame, the
kinetic energy, as well as the rest mass energy of both objects
gravitates. In the COM, "relativistic mass" gravitates.
Kinetic energy associated with motion of the COM frame does
not gravitate. [Old Man]
.
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| User: "Llanzlan Klazmon The 15th" |
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| Title: Re: Black hole in one frame of reference, not in another? |
17 Feb 2004 09:50:32 PM |
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"Todd A. Anderson" <drtodd@aaahawk.com.N0SPAM> wrote in
news:c0ue9o$j10$1@news01.intel.com:
So, was that a yes or a no? Assume I am zooming towards another
object. From what you wrote, the kinetic energy of the incoming object
gravitates. What about the relativistic volume of the incoming object?
"Old Man" <nomail@nomail.net> wrote in message
news:0_idnYJxaPBOJK_d4p2dnA@prairiewave.com...
"Todd A. Anderson" <drtodd@aaahawk.com.N0SPAM> wrote in message
news:c0u6ie$eld$1@news01.intel.com...
If I start with an object who density is below that required for a
black hole
and accelerate that object sufficiently such that from my frame of
reference
the additional relativistic mass makes the density of the object
sufficient
for a black hole, what happens? In my frame of reference it would
be a black hole but in its own frame of reference it wouldn't?
In the center-of-momentum frame, total energy (KE + rest mass
energy) gravitates. For a single object, the center-of-momentum
frame coincides with that of the object's rest frame. The kinetic
energy of a single object does not gravitate.
Consider two objects traveling along trajectories that eventually
meet in head on collision . As viewed in the COM frame, the
kinetic energy, as well as the rest mass energy of both objects
gravitates. In the COM, "relativistic mass" gravitates.
Kinetic energy associated with motion of the COM frame does
not gravitate. [Old Man]
It's an FAQ question.
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.ht
ml
Llanzlan.
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| User: "Old Man" |
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| Title: Re: Black hole in one frame of reference, not in another? |
17 Feb 2004 09:24:14 PM |
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"Todd A. Anderson" <drtodd@aaahawk.com.N0SPAM> wrote in message
news:c0ue9o$j10$1@news01.intel.com...
So, was that a yes or a no? Assume I am zooming towards another object.
From what you wrote, the kinetic energy of the incoming object gravitates.
What about the relativistic volume of the incoming object?
Definitely not. You are not in the center-of-momentum frame.
Only rest mass gravitates. Mass is invariant: regardless of
inertial reference frame whereof observed total energy, E, and
observed momentum, p, vary because of relative observer motion,
m = sqrt[ (E / c)^2 - p^2 ] = constant
In the COM frame total mass/energy, Mc^2 = E, gravitates.
Stop top-posting. Like "relativistic mass", it leads to nothing but
confusion. Get in line, and put your replies in line. [Old Man]
"Old Man" <nomail@nomail.net> wrote in message
news:0_idnYJxaPBOJK_d4p2dnA@prairiewave.com...
"Todd A. Anderson" <drtodd@aaahawk.com.N0SPAM> wrote in message
news:c0u6ie$eld$1@news01.intel.com...
If I start with an object who density is below that required for a
black
hole
and accelerate that object sufficiently such that from my frame of
reference
the additional relativistic mass makes the density of the object
sufficient
for a black hole, what happens? In my frame of reference it would be
a black hole but in its own frame of reference it wouldn't?
In the center-of-momentum frame, total energy (KE + rest mass
energy) gravitates. For a single object, the center-of-momentum
frame coincides with that of the object's rest frame. The kinetic
energy of a single object does not gravitate.
Consider two objects traveling along trajectories that eventually
meet in head on collision . As viewed in the COM frame, the
kinetic energy, as well as the rest mass energy of both objects
gravitates. In the COM, "relativistic mass" gravitates.
Kinetic energy associated with motion of the COM frame does
not gravitate. [Old Man]
.
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| User: "Edward Green" |
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| Title: Re: Black hole in one frame of reference, not in another? |
19 Feb 2004 01:34:02 AM |
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"Old Man" <nomail@nomail.net> wrote in message news:<PJadnYqwKd19R6_dRVn-gg@prairiewave.com>...
"Todd A. Anderson" <drtodd@aaahawk.com.N0SPAM> wrote in message
news:c0ue9o$j10$1@news01.intel.com...
So, was that a yes or a no? Assume I am zooming towards another object.
From what you wrote, the kinetic energy of the incoming object gravitates.
What about the relativistic volume of the incoming object?
Definitely not. You are not in the center-of-momentum frame.
Only rest mass gravitates. Mass is invariant: regardless of
inertial reference frame whereof observed total energy, E, and
observed momentum, p, vary because of relative observer motion,
m = sqrt[ (E / c)^2 - p^2 ] = constant
In the COM frame total mass/energy, Mc^2 = E, gravitates.
Must make for some interesting special cases.
For example, start with a single mass: in a frame in which the mass
is travelling towards you relativistically, duck: and while you are
ducking, both consider the possible difficulty with glibbly speaking
of frames in this manner in GR, and reflect that the solution
expressed in this frame must be analagous to that of a charged body
seen from a moving frame: rife with magnetic fields and special
effects, but really the same damn field as before.
Also reflect that in GR "frames" simply amount to change of
coordinates -- as indeed in SR, but in GR we adopt this even more
lofty, superior coordinate-independent view (even though we apparently
are forced to adopt the things to get anything done) -- and we seem
doubly foolish to expect to fool spacetime by changing them: it's the
massive bodies which set the tune, and we are but a test-particle's
worldline.
And now, on the other side of that event horizon from whose bourn from
whom no traveller returns (because you can't really duck a
relativistically closing mass), wonder what the spacetime would be
like with _two_ such bodies, having a large relative closing rate and
a small impact parameter. At what point do two bodies become one
gravitational whole? Why, just think of the possibilities!
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| User: "Jeff Relf" |
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| Title: . Momentum and predictability . |
18 Feb 2004 03:12:23 AM |
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Hi Todd A. Anderson, You asked,
" If I start with an object who density is
below that required for a black hole
and accelerate that object sufficiently such that
from my frame of reference
the additional relativistic mass
makes the density of the object
sufficient for a black hole, what happens ? "
Compared to rest mass,
relativistic mass doesn't become very significant until
a particle gets very close to the speed of light.
Other than the spin of pulsars, and certain
accelerated expansion of the universe scenarios,
nothing heavier than a nuclei ( i.e. cosmic rays )
has been observed to travel anywhere near
the speed of light .
General relativity is mostly about huge mass-energies,
while special relativity is mostly about
tiny mass-energies.
Further complicating matters,
tiny particles are notionally random,
while massive stars are much more predicable.
The greater the momentum,
the more predicable an object becomes.
And that effects the equations of physics.
.
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