Pioneer : Anomaly Still Anonymous

"GSS" writes:
Craig Markwardt wrote:
"GSS" writes:
Craig Markwardt wrote:
"GSS" writes:

...
... Why can't we consider the observed state as a
steadily collapsing state?

Primarily, because the observations show that the globular clusters
are not presently collapsing: per the examples I cited, via Doppler
shift, the motions of the stars can be decomposed into a rotating
component and a random component (i.e. both inward- and
outward-going).


How can the observation of motions of the stars *show* that the GC is
not collapsing? ...

In particular, they show clusters where the individual stars that are
both moving inward and outward from the center of the cluster. I.e. a
rapid collapse of the entire cluster is not occurring. Computer
simulations verify that such a collapse does not occur.

... If a particular GC is steadily collapsing at present
and is bound to finally collapse in about a billion years from now,
what difference in the current observations do you visualize in that
case?
.....

You suddenly jump from a time scale of one million years to one
billion years? There is probably no direct observational signatures
presaging such an event.

OK, let us reduce the time scale to about 100 million years. If a
particular GC is steadily collapsing at present and is bound to finally
collapse in about 100 million years from now, will there be any direct
observational signatures presaging such an event. Probably none. And
that confirms my original point that current observations of the
motions of stars *cannot show* that the GC is not collapsing. Hence our
presumptions of *equilibrium state* in uncollapsed GCs may actually be
ill founded.

Since the original supposition by Zick was that the cluster would
collapse "wholesale" (i.e. presumably in free-fall), the observations
do exclude that possibility. Your question basically hinges on the
long-term dissipation processes in a cluster. Those issues have
certainly been studied very extensively, which is why I recommended
that you consult some of the excellent review articles. Apparently
you are still not interested.

.... snip ...
If a particular star, under the action of external gravitational forces
(gravitational encounter) experiences a significant deflection in its
path or experiences a significant change in its velocity, then this
gravitational encounter can be considered as significantly inelastic.

That is an incorrect definition. Consider for the moment an ideal
situation where two point masses interact totally elastically. Energy
and momentum are conserved. However, the two point masses can change
paths and velocities significantly while still obeying these
constraints. Thus while such an ideal system meets your definition,
the interaction was clearly *not* inelastic.

If you *assume* the stars to be *point masses*, then you are right. For
point masses all gravitational encounters can be considered 'elastic'.
But in reality, stars are *not* point masses and gravitational
encounters between them are *not* elastic.

Perhaps you missed the "consider for a moment" portion of my
statement. It was side argument to demonstrate the fallacy of your
supposition. You claimed that if a body was deflected significantly
or experienced a significant change in velocity, then the interaction
must be inelastic. Since your supposition is already false even for
the case of ideal point masses, there is no reason for it to be true
for extended masses.

It is true that for extended masses there will be some non-zero
internal energy dissipation. What is relevant is how *much* energy is
dissipated in an encounter. *That* determines to what degree the
encounter is elastic or inelastic.

Further examples: the earth revolves around the sun, so clearly its
straight-line path is significantly deflected, and yet.... the earth
hasn't fallen into the sun! Visual binary stars orbit each other for
billions of years... and have not coalesced! Binary neutron stars are
extended bodies, significantly deflected by their own gravities... and
yet they continue to orbit each other very stably with hardly energy
energy loss![*] Thus, your "significant deflection" test is a pure
cannard.
[*] except for the tiny amount predicted by GR of course.

Your definition leads to
a contradiction, and is thus irrelevant.

An inelastic collision is one where kinetic energy is converted to
another non-kinetic form. Thus, the relevant question is how much
energy is dissipated by the interaction into other forms.


Kindly read it again, my statement above is not any definition of
elastic or inelastic collisions. That is about the *significance* of
gravitational encounters and their 'inelastic' content.

What does "significance" mean? In the context of the question being
discussed -- energy balance in globular clusters -- the only
"significant" question is: how much energy is dissipated in
encounters? I note that you continue to avoid that question.


.... snip ...
Therefore it should be quite obvious that such gravitational inelastic
interactions must be a fairly routine phenomenon among various stars in
a globular cluster.

No it does not. To say that inelastic collisions are "fairly
routine," one would need to show:
a) the amount of energy dissipated in an interaction;
The amount of energy dissipated varies from situation to situation.
The important point here is that *all* gravitational encounters between
stars are *essentially* (as per the fundamental physical process
involved) inelastic.

If the energy loss rate in all the encounters is negligible compared
to the other effects experienced by a globular cluster, then your
statement is "essentially" irrelevant.


b) the rate of interactions;
In the *initial state* of the formation of a GC, the individual stars
are expected to possess negligible KE. During the evolution of the GC,
various stars can acquire KE only through gravitational encounters all
of which are essentially inelastic (because stars are not point
masses).

This is an erroneous statement. Your hypothetical stars can gain
kinetic energy through other means than inelastic encounters. I.e. by
"falling" in the potential of the cluster, and also by scattering
nearly elastically with other stars.

c) apply some valid distribution of impact parameters;
d) compare this dissipation rate to the some other rate to show that
it is comparatively "routine."
You did none of that.

I note that you continue to avoid addressing any of those issues in
specific.


The only degree of significance that matters for the problem at hand
is the rate of tidal energy dissipation (say) compared to the total
kinetic energy of the cluster.

In my opinion, this phenomenon is highly significant during the
transition of a GC from its initial state to the equilibrium state.
During this transition, a part of the PE is transformed into KE through
a series of gravitational interactions. Simultaneously however, a
portion of this KE keeps getting transformed into internal (pressure +
temperature) energy of the stars through the operation of differential
or tidal acceleration. The overall effect of this physical process
governing the transition of a GC from its initial state to the
equilibrium state is that the sum of PE + KE + IE remains constant
(instead of just PE + KE as is generelly presumed in most models).

Again, the burden is on you to show that the energy dissipation rate
via inelastic encounters would be significant compared to the total
energy of the cluster. You did not do that.

I note you still did not do that.

....
If the tidal effects are considered significant on the scales of
galaxies, then there is no reason why they should not be considered
significant on the scale of GC's.

....

Can you kindly define the *initial* state at the time of formation of a
GC?
Is the detailed progress of a GC from its *initial* state to the
equilibrium state well understood?

It is certainly the subject of much successful observation and
theory. Examples,

Elson, R., Hut, P., & Inagaki, S. 1987, Ann Rev Astr. Astroph., 25, 565
Meylan, G., & Heggie, D. C. 1997, A&A Reviews, 8, 1


This statement appears to be quite 'diplomatic' but does not answer my
question whether the transition of a GC from its initial state to the
equilibrium state is well understood?

I don't see it as relevant. For example, does one need to know the
detailed progress of the formation of the solar system from its
initial state in order to study its present-day equilibrium and
stability? No.

I respect your point of view. But then yiu are not studying
'evolution'.

I never claimed to study "evolution." The question was: are globular
clusters stable today? (and can they be?)

Cluster dynamics has been studied extensively. Should I transcribe
every scholarly paper onto Usenet for you? No, it's up to you at some
point to take the initiative.


.... snip ...
You stated above that the tidal forces diminish with cube of separation
distance. If they are still valid on the scale of galaxies then why not
at the scale of individual stars?

Because stars are far too small in comparison to their intracluster
separation.

Consider the approximate tidal acceleration on a body of radius R and
mass m, caused by another mass M, separated by a distance D from the
first.
a_tidal = (G M / D^2) (R/D)
The self-gravitatial acceleration at the surface of the body m is,
a_grav = (G m / R^2)

The ratio of tidal to gravitational accelerations is,
ratio = (M/m) (R/D)^3
This gives an estimate of the degree of tidal deformation of a body
compared to its own self gravity.

The tidal acceleration (which I had referred as differential
acceleration Delta_a) on a star acting over the period of gravitational
encounter tends to produce a physical deformation in the fluid
structure of the star, but the self-gravitation of the star tends to
restore its original equilibrium shape. In this process 'tidal energy'
is transferred from its system (PE+KE) energy to its internal (pressure
+ temperature) energy. If the self-gravitation is strong enough,
original shape of the star will be restored after the completion of
gravitational encounter, with the increased internal energy. This
situation corresponds to extremely low value of tidal to
self-gravitational ratio. However if the self gravitation is too weak,
the tidal acceleration will tend to produce permanent deformation as
well as transfer of tidal energy. This situation corresponds to
extremely high value of tidal to self-gravitational ratio.

Of course, the transfer of tidal energy from system energy to internal
energy of the star/structure does not depend on this *ratio* but on the
*magnitude* of tidal acceleration. Hence for assessing the 'inelastic'
nature of a particular gravitational encounter the ratios quoted by you
are irrelevant.

You are incorrect. The acceleration of any particle, including a gas
molecule in a star, is the *sum* of all individual forces acting on
the particle, divided by the particle mass. This is a basic Law of
Newton. Thus, the local gravity force is just as relevant as the
tidal force.

Yes the local gravity force is important for maintaining the dynamic
equilibrium of the fluid with a certain pressure and temperature
profile. The effect of tidal forces will be 'super-posed' and will tend
to disturb the existing dynamic equilibrium.

True.

A tidal force will cause a deformation of the star. The equipotential
surface of the star will be disturbed by a height dR, whose value is
approximately = ratio*R/2, i.e (dR/R) = ratio/2, where "ratio" is
defined above. This follows from integrating the gravitational and
tidal forces to produce potential functions.

The amount of disturbed mass will be a thin shell whose mass is about
dM ~ (dR/R)*M/3, corresponding to the two small tidal bulges, and that
mass will be raised by a height of about 2*dR. Thus, the total work
done on the gas moving from its initial spherical state to the tidally
deformed state is about,

This is too simplistic model for estimating tidal energy transfer to
the internal energy. Specifically,
(a)For a fluid mass under dynamic equilibrium, you cannot isolate the
disturbed mass to be just a thin shell of small mass. Whole fluid mass
of the star will be disturbed.

However, for the question of energy balance -- and especially for such
a tiny shell of mass -- this crude estimate of the amount of pure work
done is totally relevant. Unless you claim that stars are extremely
viscuous or highly dissipative, which they are not, the estimate will
be sufficient to an order of magnitude. Some small fraction of that
work will be turned into heat, but a large fraction will not be, when
the star is returned to its spherical shape after the encounter.

(b) During the action of tidal forces, a modified dynamic equilibrium
of the fluid mass will be continuously maintained by modifying the
pressure temperature profile of the whole fluid.
(c) Clearly, if you want to consider the combined effect of the tidal
and local gravity forces, you have to consider the 'shifting' dynamic
equilibrium of the whole fluid mass along with accompanying changes in
its pressure temperature profile. On the other hand if you want to
consider the effect of tidal forces as a 'superposed' effect, then you
have to consider the deformations purely under the action of tidal
forces first. In that case your "ratio" does not come in the picture. I
had described such a procedure for computing the tidal interaction
energy in one of my previous posts, which I am again appending below.

Your dichotomy is irrelevant. You cannot compute a pure tidal
acceleration (as you attempted to do in a previous post), and say,
"that's a big number." One as to compare it to the other relevant
forces or pressures in the star. Thus, the ratio of tidal to
gravitational forces is relevant.

Hence your subsequent derivation is not valid since you are considering
the 'deformations' under the *combined* effect of tidal and local
gravity, without considering the corresponding changes in the pressure
and temperature profile in the whole fluid mass.

Since the displacement is truly miniscule -- not even a nanometer for
a typical star in a cluster -- one can hardly expect significant
changes in the temperature of the star.

I don't question that the fluid displacement due to tides will
dissipate *some* non-zero energy, but the question is, is this amount
of energy significant compared to other loss rates for a typical star?
And the answer is, not at all. Yes, there will be a few close
encounters which significantly deform the star... but those encounters
are relatively rare.

dE = dM * g * dX
= (dR/R)*M/3 * (GM/R^2) * (2dR)
= (ratio/2)*M/3 * (GM/R^2) * (ratio*R)
~ (ratio^2) * (GM^2/R) (dropping factor of 3)
~ (ratio^2) * U_g
where U_g is the gravitational potential energy of the star. Thus the
amount of work done on the gas is proportional to the tidal "ratio"
*squared*.... which of course is *smaller* than the pure ratio.

However, work alone does not dissipate energy. When the perturbation
passes away, the star will return to its original shape. Just as it
is possible to do work on a bucket of water by raising it very gently
and then lowering it gently without producing many waves or eddies, it
is possible to deform the star gradually without much energy loss.
The "ratio" described above assumes 100% dissipation, when in fact,
only a small fraction of that tidal energy will be converted to
internal energy. Thus, as the values quoted below are upper limits,
they are still negligible.

Finally, I note that once again, you conveniently deleted the
derivation of my previous post that showed that energy dissipation
leads to a larger, not smaller, cluster when approaching equilibrium.

CM

....snip...

Is not that sad?!

The academia just accepts that there is a non-functional representation
being taught...

Well, the shame is coming.

Cry!

Lester Zick wrote:
On 20 Aug 2006 07:16:39 -0700, "Aladar"
wrote:


George Dishman wrote:
"Aladar" wrote in message
oups.com...
..
You know I was sure you would say that. I explained
it to you several times and you have had several
_years_ to go away and learn the difference between
a systematic error and a random error and what that
means for processing the data and yet here you are
making the same mistake you did last time. Well I'm
not going to waste my time trying to educate you
again, go and find out for yourself why accumulating
or averaging data has no effect on a systematic bias.

You never learned what they realy did to discover the anomaly?!

Let me run it by you, again: they found a residual frequency shift in
each of the measurement.

Right so far.

They added these anomalous shifts ...

Wrong. Adding the shift doesn't separate the
anomaly from a systematic bias because the
systematic bias appear in every record (which
is why they are called 'systematic') and
therefore it accumulates too.

There is a random much larger bias. Large number of records makes it
plausible that the random error negate each other. After that the
remaining is the systematic, which has been identified as anomaly.

So, what are you talking about?!

The whole issue is that the cumulative effect shows that it is
increasing with the distance of the photon travel. Which if equal to
the Hubble redshift then clearly proves that the Hubble effect is the
photon energy loss during the travel time.

Which in turn makes the idiotic notion of bigbangology completely
naked.

Yet, only I'm the one who is yelling: The Emperor is naked!

Well you're pretty much the only one yelling period.

"v_model is the modeled velocity of the spacecraft due
to the gravitational and other large forces discussed in
Section IV. (This velocity is outwards and hence
produces a red shift.) We have already included the
sign showing that a_P is inward. (Therefore, a_P
produces a slight blue shift on top of the larger red shift.)"

Note: there is so many way to cover the real thing: less
frequency...

The statement is clear and unambiguous, the
anomaly is a blue shift.

Looking from where?

The paragraph is quite clear, it says the outward velocity
of the craft produces a red shift. That defines the sense.
Then it says the anomaly produces a blue shift. That means
it is the opposite of the Doppler effect produced by the
speed of the craft leaving the solar system. That means the
anomaly is an increase in the received frequency compared
to what was expected.

Why don't they say so?!

They did. I realise English is not your first
language but if you cannot follow that paragraph,
get someone to help you.

George

Lester Zick
~v~~

Is not that sad?!

The academia just accepts that there is a non-functional representation
being taught...

Well, the shame is coming.

Cry!

Lester Zick wrote:
On 20 Aug 2006 07:16:39 -0700, "Aladar"
wrote:


George Dishman wrote:
"Aladar" wrote in message
oups.com...
..
You know I was sure you would say that. I explained
it to you several times and you have had several
_years_ to go away and learn the difference between
a systematic error and a random error and what that
means for processing the data and yet here you are
making the same mistake you did last time. Well I'm
not going to waste my time trying to educate you
again, go and find out for yourself why accumulating
or averaging data has no effect on a systematic bias.

You never learned what they realy did to discover the anomaly?!

Let me run it by you, again: they found a residual frequency shift in
each of the measurement.

Right so far.

They added these anomalous shifts ...

Wrong. Adding the shift doesn't separate the
anomaly from a systematic bias because the
systematic bias appear in every record (which
is why they are called 'systematic') and
therefore it accumulates too.

There is a random much larger bias. Large number of records makes it
plausible that the random error negate each other. After that the
remaining is the systematic, which has been identified as anomaly.

So, what are you talking about?!

The whole issue is that the cumulative effect shows that it is
increasing with the distance of the photon travel. Which if equal to
the Hubble redshift then clearly proves that the Hubble effect is the
photon energy loss during the travel time.

Which in turn makes the idiotic notion of bigbangology completely
naked.

Yet, only I'm the one who is yelling: The Emperor is naked!

Well you're pretty much the only one yelling period.

"v_model is the modeled velocity of the spacecraft due
to the gravitational and other large forces discussed in
Section IV. (This velocity is outwards and hence
produces a red shift.) We have already included the
sign showing that a_P is inward. (Therefore, a_P
produces a slight blue shift on top of the larger red shift.)"

Note: there is so many way to cover the real thing: less
frequency...

The statement is clear and unambiguous, the
anomaly is a blue shift.

Looking from where?

The paragraph is quite clear, it says the outward velocity
of the craft produces a red shift. That defines the sense.
Then it says the anomaly produces a blue shift. That means
it is the opposite of the Doppler effect produced by the
speed of the craft leaving the solar system. That means the
anomaly is an increase in the received frequency compared
to what was expected.

Why don't they say so?!

They did. I realise English is not your first
language but if you cannot follow that paragraph,
get someone to help you.

George

Lester Zick
~v~~

I very much agree with this one. Without a dedicated experiment,
measuring the redshift versus the distance from a craft this talk is
academic. In the regular, wrong sense of this expression = meaningless,
so usual for academia.

Yes, we can argue for ever and without a clear precise experiment we
will not get anywhere. That is why I'm so upset with the stupidities,
lining up everything to "prove the big bang". Bull****.

And there are so many other observations which all should be understood
on the same fundamental working of matter!

I came-up with something, which works. (Actually, I went back all the
way to Democritos or Lucretius.) The collision event fundation of
everything, all the massive particles and all the electromagnetic
photons works and it requires that the photon lose its energy in order
to regenerate itself.

When the very large telescopes come online they will show (already
showing, just bigbangology argues it away!) that the far away galaxies
look just like the closest ones - hence the big bang is bull****.

Thats all.

Thanks George, it was a good post.

Aladar
http://stolmarphysics.com

George Dishman wrote:
"Aladar" wrote in message
oups.com...

George Dishman wrote:
"Aladar" wrote in message
oups.com...
..
You know I was sure you would say that. I explained
it to you several times and you have had several
_years_ to go away and learn the difference between
a systematic error and a random error and what that
means for processing the data and yet here you are
making the same mistake you did last time. Well I'm
not going to waste my time trying to educate you
again, go and find out for yourself why accumulating
or averaging data has no effect on a systematic bias.

You never learned what they realy did to discover the anomaly?!

Let me run it by you, again: they found a residual frequency shift in
each of the measurement.

Right so far.

They added these anomalous shifts ...

Wrong. Adding the shift doesn't separate the
anomaly from a systematic bias because the
systematic bias appear in every record (which
is why they are called 'systematic') and
therefore it accumulates too.

There is a random much larger bias. Large number of records makes it
plausible that the random error negate each other. After that the
remaining is the systematic, which has been identified as anomaly.

So, what are you talking about?!

There is a lot of random noise. Taking an average or
accumulating records reduces the random element by
roughly the square root of the number of records, so
evaraging 100 records reduces the noise by about 10
times in comparison to the anomaly. A systematic bias
however has the same value and sign in every record
so averaging (or accumulating) doesn't change the
ratio of the anomaly to the bias. The radio beam is
such a systematic bias because the power of 8W sent
towards Earth is the same all the time. Averaging
cannot separate the anomaly from that bias and since
the 8W value is not known accurately, the effect of
cosmological redshift _cannot_ be measured from the
Pioneer Doppler data.

The whole issue is that the cumulative effect shows that it is
increasing with the distance of the photon travel. Which if equal to
the Hubble redshift then clearly proves that the Hubble effect is the
photon energy loss during the travel time.

No, it could equally be taken as confirmation of the
redshift because the craft is being 'carried along
by the Hubble flow' thus confirmation of the big bang.

Regardless, the fact is that the anomaly is 8 times
larger than the effect of the radio beam while the
Hubble redshift would be about 2000 times smaller
than the beam (hence inseparable from it even by
accumulating or averaging the data) and of course
the anomaly is a blue shift.

Which in turn makes the idiotic notion of bigbangology completely
naked.

Yet, only I'm the one who is yelling: The Emperor is naked!


"v_model is the modeled velocity of the spacecraft due
to the gravitational and other large forces discussed in
Section IV. (This velocity is outwards and hence
produces a red shift.) We have already included the
sign showing that a_P is inward. (Therefore, a_P
produces a slight blue shift on top of the larger red
shift.)"

George

On 22 Aug 2006 11:27:21 -0700, "GSS"
wrote:


Craig Markwardt wrote:
"GSS" writes:
Craig Markwardt wrote:
"GSS" writes:

...
... Why can't we consider the observed state as a
steadily collapsing state?

Primarily, because the observations show that the globular clusters
are not presently collapsing: per the examples I cited, via Doppler
shift, the motions of the stars can be decomposed into a rotating
component and a random component (i.e. both inward- and
outward-going).


How can the observation of motions of the stars *show* that the GC is
not collapsing? ...

In particular, they show clusters where the individual stars that are
both moving inward and outward from the center of the cluster. I.e. a
rapid collapse of the entire cluster is not occurring. Computer
simulations verify that such a collapse does not occur.

... If a particular GC is steadily collapsing at present
and is bound to finally collapse in about a billion years from now,
what difference in the current observations do you visualize in that
case?
.....

You suddenly jump from a time scale of one million years to one
billion years? There is probably no direct observational signatures
presaging such an event.

OK, let us reduce the time scale to about 100 million years. If a
particular GC is steadily collapsing at present and is bound to finally
collapse in about 100 million years from now, will there be any direct
observational signatures presaging such an event. Probably none. And
that confirms my original point that current observations of the
motions of stars *cannot show* that the GC is not collapsing. Hence our
presumptions of *equilibrium state* in uncollapsed GCs may actually be
ill founded.

I agree.

However, for that duration, one must be
concerned about a myriad of processes, such as: tidal disruption by
the galaxy; diruption by passage through the galactic plane;
evaporative removal of stars. Since the relaxation timescale is of
order 10^8 years, the system should remain in approximate equilibrium
while these mechanisms operate.


Whenever we treat a particular gravitational encounter as
*significant*, we must also treat that encounter to be inelastic to the
same degree of significance.

What does this statement mean?


If a particular star, under the action of external gravitational forces
(gravitational encounter) experiences a significant deflection in its
path or experiences a significant change in its velocity, then this
gravitational encounter can be considered as significantly inelastic.

That is an incorrect definition. Consider for the moment an ideal
situation where two point masses interact totally elastically. Energy
and momentum are conserved. However, the two point masses can change
paths and velocities significantly while still obeying these
constraints. Thus while such an ideal system meets your definition,
the interaction was clearly *not* inelastic.

If you *assume* the stars to be *point masses*, then you are right. For
point masses all gravitational encounters can be considered 'elastic'.
But in reality, stars are *not* point masses and gravitational
encounters between them are *not* elastic.

True. However as noted elsewhere I doubt that elastic/inelastic
collisions matter in the collapse of gc's. If they have net angular
momentum of zero internally and no opposing electrodynamic or
electrostatic forces they have to be collapsing. Simple as that.

Lester Zick
~v~~

On 22 Aug 2006 16:11:55 -0400, Craig Markwardt
wrote:


"GSS" writes:
Craig Markwardt wrote:
"GSS" writes:
Craig Markwardt wrote:
"GSS" writes:

...
... Why can't we consider the observed state as a
steadily collapsing state?

Primarily, because the observations show that the globular clusters
are not presently collapsing: per the examples I cited, via Doppler
shift, the motions of the stars can be decomposed into a rotating
component and a random component (i.e. both inward- and
outward-going).


How can the observation of motions of the stars *show* that the GC is
not collapsing? ...

In particular, they show clusters where the individual stars that are
both moving inward and outward from the center of the cluster. I.e. a
rapid collapse of the entire cluster is not occurring. Computer
simulations verify that such a collapse does not occur.

... If a particular GC is steadily collapsing at present
and is bound to finally collapse in about a billion years from now,
what difference in the current observations do you visualize in that
case?
.....

You suddenly jump from a time scale of one million years to one
billion years? There is probably no direct observational signatures
presaging such an event.

OK, let us reduce the time scale to about 100 million years. If a
particular GC is steadily collapsing at present and is bound to finally
collapse in about 100 million years from now, will there be any direct
observational signatures presaging such an event. Probably none. And
that confirms my original point that current observations of the
motions of stars *cannot show* that the GC is not collapsing. Hence our
presumptions of *equilibrium state* in uncollapsed GCs may actually be
ill founded.

Since the original supposition by Zick was that the cluster would
collapse "wholesale" (i.e. presumably in free-fall), the observations
do exclude that possibility.

Actually they can't. Absent electrostatic or electrodynamic resistance
any gravitationally self contracting object is in wholesale or free
fall collapse. Basic mechanics. There's nothing to prevent it. In a gc
you have a gravitationally self contracting object without significant
electrostatic or electrodynamic repulsion whose constituent parts have
zero net angular momentum. And nominal explanations to the contrary
have to accommodate that circumstance. Simply saying gc's are in some
kind of equilibrium state does not show what kind of equilibrium they
can be in which counters the absence of internal angular momentum. We
know electrostatic pressure counters collapse in terrestrial planets
and we know electrodynamic pressure counters collapse in stars. We
have no idea what if any repulsion does or can counter collapse in
globular clusters. And just maintaining gc's are in equilibrium does
nothing to explain what if any kind of equilibrium they can be in.

Your question basically hinges on the
long-term dissipation processes in a cluster. Those issues have
certainly been studied very extensively, which is why I recommended
that you consult some of the excellent review articles. Apparently
you are still not interested.

And I'm still waiting to see what kind of equilibrium gc's can be in.
This has nothing to do with elastic/inelastic collisions of stars in
gc's or even whether there are collisions or even close or remote
encounters for that matter. The fact is that gc's have roughly zero
net angular momentum internally and that fact alone absent any
internal repulsive resistance precipitates and continues indefinite
wholesale collapse of the structure itself.

Lester Zick
~v~~

On Tue, 22 Aug 2006 19:21:43 -0700, Lester Zick
wrote:

The fact is that gc's have roughly zero
net angular momentum internally and that fact alone absent any
internal repulsive resistance precipitates and continues indefinite
wholesale collapse of the structure itself.

You are repeating yourself, but you are *so* wrong. You have to
compute the path of each star individually, they certainly do not
"cancel out" like that. A "centre of gravity" has no real existence
and cannot act gravitationally on anything.

--
Posted via a free Usenet account from http://www.teranews.com

Lester Zick wrote:
On 22 Aug 2006 11:27:21 -0700, "GSS"
wrote:


Craig Markwardt wrote:
"GSS" writes:
Craig Markwardt wrote:
"GSS" writes:

...
... Why can't we consider the observed state as a
steadily collapsing state?

Primarily, because the observations show that the globular clusters
are not presently collapsing: per the examples I cited, via Doppler
shift, the motions of the stars can be decomposed into a rotating
component and a random component (i.e. both inward- and
outward-going).


How can the observation of motions of the stars *show* that the GC is
not collapsing? ...

In particular, they show clusters where the individual stars that are
both moving inward and outward from the center of the cluster. I.e. a
rapid collapse of the entire cluster is not occurring. Computer
simulations verify that such a collapse does not occur.

... If a particular GC is steadily collapsing at present
and is bound to finally collapse in about a billion years from now,
what difference in the current observations do you visualize in that
case?
.....

You suddenly jump from a time scale of one million years to one
billion years? There is probably no direct observational signatures
presaging such an event.

OK, let us reduce the time scale to about 100 million years. If a
particular GC is steadily collapsing at present and is bound to finally
collapse in about 100 million years from now, will there be any direct
observational signatures presaging such an event. Probably none. And
that confirms my original point that current observations of the
motions of stars *cannot show* that the GC is not collapsing. Hence our
presumptions of *equilibrium state* in uncollapsed GCs may actually be
ill founded.

I agree.

......
If you *assume* the stars to be *point masses*, then you are right. For
point masses all gravitational encounters can be considered 'elastic'.
But in reality, stars are *not* point masses and gravitational
encounters between them are *not* elastic.

True. However as noted elsewhere I doubt that elastic/inelastic
collisions matter in the collapse of gc's. If they have net angular
momentum of zero internally and no opposing electrodynamic or
electrostatic forces they have to be collapsing. Simple as that.

Lester Zick
~v~~

Let me explain 'their' point of view regarding the 'non-collapsing'
equilibrium state in a GC.
Considering an average separation between nearby stars in a GC to be
about one light year (LY) or more, the net gravitational acceleration
acting on any star is approximately towards the center of the GC.
However, since the separation between nearby stars is not *exactly*
equal, each star will therefore also experience some net transverse
acceleration (i.e. perpendicular to radial direction) which will not be
the same for all stars.

Each star under the action of radially inward gravitational
acceleration, will start moving (falling) towards the center of GC and
gain KE. But due to the presence of small transverse component, it will
acquire some angular momentum. As such this star will not pass exactly
through the center of GC but will follow an elliptical orbit. This
elliptical orbit path of the star will be governed by the controlling
relation, PE + KE = constant (which may be called system energy). Since
the separation between various stars is enormous, different stars
following their individual elliptical orbits, may 'never' experience
any close gravitational encounters with other stars. The overall size
of the elliptical orbit of any star may shrink (or contract or
collapse) only if a part of its system energy is *somehow* dissipated
out.

My point is that whenever any star experiences 'external' gravitational
acceleration (so as to change its KE) it also experiences tidal
acceleration due to its large size. The action of this tidal
acceleration is to transfer a part of its system energy to the internal
(pressure + temperature) energy. As such the controlling relation
becomes, PE + KE + IE = constant . With this dissipation of system
energy to the internal energy, the overall size of elliptical orbit of
any star will keep shrinking steadily, leading to a steady collapse of
the GC.

GSS

GSS wrote:

Let me explain 'their' point of view regarding the 'non-collapsing'
equilibrium state in a GC.

This is a fairly good explanation of my understanding
though there is a bit more I will add.

Considering an average separation between nearby stars in a GC to be
about one light year (LY) or more, the net gravitational acceleration
acting on any star is approximately towards the center of the GC.
However, since the separation between nearby stars is not *exactly*
equal, each star will therefore also experience some net transverse
acceleration (i.e. perpendicular to radial direction) which will not be
the same for all stars.

Not only that, stars will form throughout a cloud of gas
of sufficient density, not all at the same radius, so as
they fall inwards, some will pass through the centre
while others are still en route. Also stars will not all
form with no momentum even in the extreme case of a
cloud that is entirely still. The first star to form blows
the surronding gas away by radiation pressure which
both imparts momentum in opposite directions on either
side and also increases the density precipitating more
star formation.

The result is that stars will form at random times,
random radii and with random initial momentum, each
parameter being described by some statistics. You
cannot consider a model where all the stars are equally
spaced, at identical radius and with zero momentum to
be realistic.

Each star under the action of radially inward gravitational
acceleration, will start moving (falling) towards the center of GC and
gain KE. But due to the presence of small transverse component, it will
acquire some angular momentum. As such this star will not pass exactly
through the center of GC but will follow an elliptical orbit.

That would be true if all the mass of the cluster were at
the centre but even assuming spherical symmetry, as
the star falls inward through radius R, the amount of
mass in a sphere of radius R decreases and only that
mass inside the radius has an effect. That is a standard
Newtonian result, I'm sure you are aware of it.

This
elliptical orbit path of the star will be governed by the controlling
relation, PE + KE = constant (which may be called system energy). Since
the separation between various stars is enormous, different stars
following their individual elliptical orbits, may 'never' experience
any close gravitational encounters with other stars. The overall size
of the elliptical orbit of any star may shrink (or contract or
collapse) only if a part of its system energy is *somehow* dissipated
out.

Almost. The stars will experience an effect from every
other star which means the orbit will look more like a
gentle verson of a pinball machine or a random walk
with very small deviations as it passes between stars.
Overall, each pass through the centre will change the
PE+KE total slightly but any loss of KE for one star
is a gain for another, and the same for PE.

The virial theorem says this process is asymptotic to
a random distributions of velocities such that there is
a simple relationship between PE and KE though this
may take thousands of passes through the cluster.

My point is that whenever any star experiences 'external' gravitational
acceleration (so as to change its KE) it also experiences tidal
acceleration due to its large size. The action of this tidal
acceleration is to transfer a part of its system energy to the internal
(pressure + temperature) energy. As such the controlling relation
becomes, PE + KE + IE = constant . With this dissipation of system
energy to the internal energy, the overall size of elliptical orbit of
any star will keep shrinking steadily, leading to a steady collapse of
the GC.

That is true but the timescale is much longer than
the time for virialisation, so what happens is that as
the KE falls, so does the PE.

When stars collide in the centre or pass sufficiently
close, a large part of the KE becomes IE, they
become bound as either a binary or a 'blue straggler'
where the stars merge into one. Their angular
momentum is then converted into spin of the star
and lost from the motions in the cluster.

Nobody is suggesting that the process you describe
doesn't happen, only that it will not produce a disc
of stars, it will produce a shrinking sphere (unless
the gas cloud had a significant spin before the stars
formed) and that it is slow in comparison to the age
of the stars in even the oldest clusters.

The fact that we can see clusters still in existence
composed of stars more than 12 billion years old
when the time for a star to cross the cluster is of
the order of 100 thousand years shows that the
claim that the cluster would immediately collapse
cannot be correct.

George

On Wed, 23 Aug 2006 09:20:18 +0100, Ben Newsam
wrote:

On Tue, 22 Aug 2006 19:21:43 -0700, Lester Zick
wrote:

The fact is that gc's have roughly zero
net angular momentum internally and that fact alone absent any
internal repulsive resistance precipitates and continues indefinite
wholesale collapse of the structure itself.

You are repeating yourself, but you are *so* wrong.

This is puzzling. Wrong how exactly? That gc's have no net angular
momentum internally? That absent any internal repulsive resistance
zero net angular momentum precipitates and continues indefinite
wholesale collapse of the structure itself? I just can't quite make
out what you think is "so" wrong.

You have to
compute the path of each star individually,

So you're saying the aggregate of internal angular momenta for a gc is
not roughly zero regardless of how you compute it?

they certainly do not
"cancel out" like that.

They don't? Then how precisely is it that the combination for all
angular momenta is roughly zero?

A "centre of gravity" has no real existence
and cannot act gravitationally on anything.

Perhaps you should take that up with Newton. Centers of gravity are
certainly real enough in stars and planets to be calculated through
their respective centers for the purpose of gravitational attraction.

Lester Zick
~v~~