observed black hole mass

From an outside observer's viewpoint, an object
falling into a black hole never passes the horizon;
it gets infinitely red shifted.

Hence, for any observer, a black hole's mass
never increases, after its creation. Why then do
astronomers routinely talk about the growth of
black holes, as they swallow nearby insects?
Does that make sense?

--
Rich

Dear RichD:

On Mar 23, 11:46*am, RichD wrote:
From an outside observer's viewpoint, an object
*falling into a black hole never passes the horizon;
it gets infinitely red shifted.

So it is seen, then it isn't seen at some point. As to assuming that
"it never passes the event horizon" is not scientific.

Hence, for any observer, a black hole's mass
never increases, after its creation.

Mass is added to that vicinity, regardless of what you think about
"real" or "apparent" ingestion.

Why then do *astronomers routinely talk about
the growth of black holes, as they swallow
nearby insects? Does that make sense?

No, you don't make sense. Have another beer.

David A. Smith

RichD wrote:
From an outside observer's viewpoint, an object
falling into a black hole never passes the horizon;
it gets infinitely red shifted.

Hence, for any observer, a black hole's mass
never increases, after its creation. Why then do
astronomers routinely talk about the growth of
black holes, as they swallow nearby insects?
Does that make sense?

No, from an observer falling into a black hole, and outside object gets
infinitely red shifted.

An outside observer will see an object disappear into a black hole, with
no problem.

Yousuf Khan

On Mar 23, 10:46*am, RichD wrote:
From an outside observer's viewpoint, an object
*falling into a black hole never passes the horizon;
it gets infinitely red shifted.

Hence, for any observer, a black hole's mass
never increases, after its creation. *Why then do
*astronomers routinely talk about the growth of
black holes, as they swallow nearby insects?
Does that make sense?

--
Rich

Define the mass of a black hole. The rigorous answer to this is the
answer to your question.

RichD wrote:
From an outside observer's viewpoint, an object
falling into a black hole never passes the horizon;
it gets infinitely red shifted.

Yes. Well, close enough.

[I make no attempt to be rigorous here. The style of the
question indicates that is not necessary. Here there be
dragons, but at this level they won't bother us.]


Hence, for any observer, a black hole's mass
never increases, after its creation. Why then do
astronomers routinely talk about the growth of
black holes, as they swallow nearby insects?

Because it does not matter.

Instead of considering the "mass of the black hole", which is inside its
horizon (r=2M for a Schwarzschild black hole), consider the mass inside
the smallest stable orbit (r=6M for a Schw. black hole). Most of the
redshifting of infalling objects occurs inside this radius, and a
distant observer can observe objects falling through r=6M without
problem (other than the usual difficulties of observing distant objects
shrouded in dust, gas, and other stars). From the standpoint of an
observer located at any radius R of a spherically-symmetric object
(R2M), only the mass inside radius R contributes to the gravitational
field.

Astronomical observations of black holes actually observe infalling
matter, not the black hole itself. They also observe an accretion disk
outside r=6M, and use it to characterize the black hole and the matter
comprising the disk. As far as this disk's dynamics goes, the mass of
the black hole is effectively the total within r=6M. That, of course,
increases as more stuff falls in.

One dragon might roar: I discussed a Schwarzschild black hole,
but astronomically observable black holes are of other types.
Details differ, but the basic argument holds for them, too.


Tom Roberts

Yousuf Khan wrote:
No, from an observer falling into a black hole, and outside object gets
infinitely red shifted.

You mean infinitely BLUEshifted. And that outside object must be
emitting EM radiation that the infalling observer can observe. As the
observer approaches the horizon and goes through, there are enormous
distortions of the visible celestial sphere.


An outside observer will see an object disappear into a black hole, with
no problem.

Not if by "disappear" you mean falls inside the horizon. But yes, the
redshifting will make an infalling object invisible for all practical
purposes (even though in principle it remains visible).


Tom Roberts

On Mar 23, 5:17 pm, Tom Roberts wrote:

Instead of considering the "mass of the black hole", which is inside its
horizon (r=2M for a Schwarzschild black hole), consider the mass inside
the smallest stable orbit (r=6M for a Schw. black hole). Most of the
redshifting of infalling objects occurs inside this radius, and a
distant observer can observe objects falling through r=6M without
problem (other than the usual difficulties of observing distant objects
shrouded in dust, gas, and other stars). From the standpoint of an
observer located at any radius R of a spherically-symmetric object
(R2M), only the mass inside radius R contributes to the gravitational
field.

So, you (plural) have never actually identified a black hole. Recall
the Schwarzschild metric below.

ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

It shows a black hole can only possibly formed in the infinite future
of the observer’s time. It is all in the mathematics. shrug

Astronomical observations of black holes actually observe infalling
matter, not the black hole itself. They also observe an accretion disk
outside r=6M, and use it to characterize the black hole and the matter
comprising the disk. As far as this disk's dynamics goes, the mass of
the black hole is effectively the total within r=6M. That, of course,
increases as more stuff falls in.

One dragon might roar: I discussed a Schwarzschild black hole,
but astronomically observable black holes are of other types.
Details differ, but the basic argument holds for them, too.

Thus, no black holes can be contemporary with any observers. As an
observer talking about observing a black hole is indeed a lie under
the Schwarzschild geometry. shrug

However, the slimy conjecture known as GR actually yield an infinite
number of solutions that are static, spherically symmetric, and
asymptotically flat where the Schwarzschild metric is actually nowhere
unique. There are still an infinite number of solutions that
degenerate to Newtonian law of gravity at large distances and do not
manifest black holes.

It is absolutely sickening that you (plural) are not willing to play
in the very mathematical confine of the $hit you (plural) have
created. *******izing observations to suit your (plural) mathemagical
can only fool the most shallow-minded. You (plural) still cannot fool
the learned scholars in physics. Do shamans showing off their
shamanistic acts in ancient primitive tribes ring any bell?

On Mar 23, 9:54*pm, wrote:
On Mar 23, 5:17 pm, Tom Roberts wrote:

Instead of considering the "mass of the black hole", which is inside its
horizon (r=2M for a Schwarzschild black hole), consider the mass inside
the smallest stable orbit (r=6M for a Schw. black hole). Most of the
redshifting of infalling objects occurs inside this radius, and a
distant observer can observe objects falling through r=6M without
problem (other than the usual difficulties of observing distant objects
shrouded in dust, gas, and other stars). From the standpoint of an
observer located at any radius R of a spherically-symmetric object
(R2M), only the mass inside radius R contributes to the gravitational
field.

So, you (plural) have never actually identified a black hole. *Recall
the Schwarzschild metric below.

ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** *U = G M / c^2 / r
** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

It shows a black hole can only possibly formed in the infinite future
of the observer’s time. *It is all in the mathematics. *shrug

Astronomical observations of black holes actually observe infalling
matter, not the black hole itself. They also observe an accretion disk
outside r=6M, and use it to characterize the black hole and the matter
comprising the disk. As far as this disk's dynamics goes, the mass of
the black hole is effectively the total within r=6M. That, of course,
increases as more stuff falls in.

* * * * One dragon might roar: I discussed a Schwarzschild black hole,
* * * * but astronomically observable black holes are of other types.
* * * * Details differ, but the basic argument holds for them, too.

Thus, no black holes can be contemporary with any observers. *As an
observer talking about observing a black hole is indeed a lie under
the Schwarzschild geometry. *shrug

As usual, stupidity from wooby is the order of the day.

The black hole never fully formes relative to an outside observer but
gets CLOSE ENOUGH rather quickly.

Find a new hobby, you braying jackass.

[snip rest]

On Mar 23, 11:11 pm, Eric Gisse wrote:

As usual, stupidity from wooby is the order of the day.

[Rest of whining crap snipped]

So, we have also learned that the college drop-out has no life as
well. Please allow me to summarize.

** Eric Gisse is a college drop-out.
** Eric Gisse has no motivation to continue his education.
** Eric Gisse has no love in his life and thus no life.
** Eric Gisse has no job.

That explains why the college drop-out sometime trolling and dumping
his $hit on these newsgroups even until 5 or 6 in the morning. It
sounds so pathetic to me. Excuse me. I need to go to cry. In the
meantime, why don’t you chew on the following.

Under the principle of relativity (discovered by Galileo) and
classical electromagnetism (Maxwell, et al), the MMX was expected to
show non-null results. Since the MMX showed null results, the
Galilean transform must be modified where the new transform must
degenerate into the Galilean transform at low speeds.

Earlier, the MMX had to utilize what Doppler effect had shown. That
is the wavelength in the propagating medium of the propagating waves
must be invariant. In doing so, there is no room for this concept to
explain the null results of the MMX. In 1887, Voigt pointed out that
instead of the invariance in wavelength, the speed of the propagating
waves must be invariant. Voigt and Larmor in 1897 or 1898 or so
(perhaps read Voigt’s work) came up with an infinite number of
transforms that would satisfy the null results of the MMX and
degenerate into the Galilean transform at low speeds.

Among the infinite numbers, Voigt went with the Voigt transform and
got a cold reception from the Einstein Dingleberries. Larmor went
with the Lorentz transform, and it was plagiarized by Einstein the
nitwit, the plagiarist, and the liar 7 to 8 years later. Since Voigt
transform does not satisfy the principle of relativity, Lorentz
transform (does satisfy the principle of relativity) is now worshipped
as a religion by the Einstein Dingleberries.

The biggest problem with the Lorentz transform is that it manifests
the twin’s paradox which is caused by the exact combination of the
principle of relativity and time dilation. To resolve the paradox,
one must show either the principle of relativity or the effect of time
dilation is false. In doing so, it would also invalidate the Lorentz
transform. Thus, to claim to have a resolution to the twin’s paradox
is like someone claiming to have invented a perpetual motion machine
again. The Einstein Dinglebeberries are so silly that their shallow
minds do not even understand their own silliness.

Any true, learned scholars in physics with training or aptitude in
scientific methodology should reject out-right the Lorentz transform.
The question if the Voigt transform as a valid and general transform
to the Galilean transform should be the subject of debate instead. Is
the principle of relativity discovered more than 400 years ago really
that sacred to allow the Einstein Dingleberries to favor the Lorentz
transform?

After you are done with that, chew on the following.

So, you (plural) have never actually identified a black hole. Recall
the Schwarzschild metric below.

ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

It shows a black hole can only possibly formed in the infinite future
of the observer’s time. It is all in the mathematics. shrug

Thus, no black holes can be contemporary with any observers. As an
observer talking about observing a black hole is indeed a lie under
the Schwarzschild geometry. shrug

However, the slimy conjecture known as GR actually yield an infinite
number of solutions that are static, spherically symmetric, and
asymptotically flat where the Schwarzschild metric is actually nowhere
unique. There are still an infinite number of solutions that
degenerate to Newtonian law of gravity at large distances and do not
manifest black holes.

It is absolutely sickening that you (plural) are not willing to play
in the very mathematical confine of the $hit you (plural) have
created. *******izing observations to suit your (plural) mathemagical
can only fool the most shallow-minded. You (plural) still cannot fool
the learned scholars in physics. Do shamans showing off their
shamanistic acts in ancient primitive tribes ring any bell?