Re: Path in Schwarzschild

On Sep 16, 9:06 pm, Tom Roberts wrote:
Koobee Wublee wrote:

Given the Schwarzschild metric, we have the following spacetime.
ds^2 = c^2 (1 - 2 U) dt^2 - dr^2 / (1 - 2 U) - r^2 dO^2
** U = G M / c^2 / r
** dO^2 = cos^2Phi dTheta^2 + dPhi^2
So, please explain why the metric element of
gravitational time dilation dominates all other element of the metric.

In the case being discussed, the first term does not merely "dominate"
the others, it is the only non-zero term in the line element (which you
call "spacetime", using your personal vocabulary). Remember please that
I specified "at rest", so dr=0, dTheta=0, and dPhi=0 for the worldlines
of the source and detector.

That is clever. You got yourself out of a seemingly checkmate jam.
Now, I have to ask you if the so precious principle of equivalence is
breaking down when the gravitational time dilation exhibits a red
shift while the transverse Doppler shift also based on time dilation
gets a blue shift in frequency.

Even if it precesses, it is actually very stable no?

As I said, it is APPROXIMATELY stable. But a truly stable orbit would
not precess at all. Yes, this precession is quite small.

Stability means it can re-attain the exact orbit sometime in the
future. In Mercury's case, is the orbit observed to be shrinking?

Does it lose
angular momentum through observation?

Why would "observation" affect the angular momentum of a planet? Somehow
I doubt that telescopes here on earth have any power over the orbit of
Mercury....

Boy, you are on a roll for a change. Yes, you are right. Any
observation should not affect the angular momentum. However, if a
physical law dictates that Mercury must lose its angular momentum, it
will. The so called observations are just intrinsic to the system.
Say I am taking a shower. Not that I would care, although I don't see
anyone observing me taking a shower, it does not mean it is not
possible to have an observer watching me taking a shower. <shrug>

Let me read into your question the better question: is the angular
momentum of Mercury conserved? Yes, its angular momentum is
APPROXIMATELY conserved, insofar as Mercury's mass can be neglected as
can the effects of all other planets. If they are taken into account,
then no, the angular momentum of mercury ALONE is not conserved.

No, I am asking if there is any observation to indicate a violation in
the conservation of angular momentum in Mercury's case.

Let me extend the scope even further: is the angular momentum of the
entire solar system conserved? Yes, to the extent that the masses of the
planets can be neglected. No, if they are taken into account (the solar
system emits gravitational radiation which breaks the conservation of
angular momentum for the solar system ALONE).

This is only a conjecture. <shrug>

[I ignore the difficulties of accurately and consistently
defining the relevant angular momenta in GR; one can do
so APPROXIMATELY and that is plenty good enough here.]

Here you go again about the Godhood of GR that only Gods themselves
can arrive at the exact solution. If that is the case, why even
bother to play Gods? GR is based on a series of mathematical nonsense
and misapplications of mathematics. You are the one who imposes this
Godliness in GR. <shrug>

Yes, but before doing that, you must decide on which model of
geodesics to embrace. [... more nonsense]

There is but a single geodesic equation, and a single metric being
discussed here. Solve for the physical situation of Mercury's orbit.

Yes, there are 4 geodesic equations. However, there should be only
one metric --- the Schwarzschild metric. What other metric do you
have in mind? And besides, the model of geodesics should be
independent of the number of geodesic equations as well as any metric
involved. <shrug>

After establishing that nothing can escape a black hole if located
inside the event horizon, you (plural) have no problems to allow the
effect of spacetime to escape the event horizon.

In order to communicate you must LEARN the vocabulary of GR, and you
must LEARN what GR actually says. What you say here is nonsense -- word
salad without meaning.

I apologize for not using the private codes of your good-old-boys
club. <shrug>

A black hole is a geometric aspect of the manifold, and in GR nothing
"escapes" from the black hole; geometry simply _IS_ and does not
"propagate" or "escape" from anywhere or anywhen.

Glad you have picked on "anywhen" --- a vocabulary not in any
dictionary but used by yours truly for more than one occasion. With a
cross pollination of vocabulary usage, I am hoping you would not fall
back on the excuse of not understanding some certain vocabulary while
confronted with serious issues that would render GR nonsensical.

Did you not criticize Dr. Van Flandern for making the claim in
observing the speed of gravity to be ten billion times the speed of
light? Did you not have claimed the speed of gravity is the speed of
light without any mathematical bases or any experimental
observations? Now, did you not have accepted the implication of the
Schwarzschild metric that time must freeze at the event horizon
relative to any observes outside the event horizon? Did you not have
claimed the speeds of light locally adjacent to each other do not have
to be the same? So, relative to any observers outside of the event
horizon, why do you impose light to freeze while the speed of gravity
not affected?

OK: tell me, in that equation for "spacetime" quoted above,
where does it indicate that "spacetime" "escapes" from
the black hole? It looks to me like one can evaluate it
anywhere and anywhen [#] and I see no "velocity" or
"propagation" or "retarded/advanced" terms in it
(remember that in electrodynamics it is the latter which
indicate propagation).

OK, you are barking up the wrong tree here. I meant whatever you have
disagreed with Dr. Van Flandern --- the speed of gravity and thus its
effect.

[#] except right at U=1/2, which we know is merely a
coordinate singularity. I ignore the subtlety that
those coordinates do not cover the manifold.

I don't understand. Now, you have deviate yourself even from the
normal usage of GR vocabulary. Why are you don't that after
criticizing me for doing so?

[... more nonsense]

That is understandable. There are no challenges from you. <shrug>

.

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