Physicists Howl at Dark Matter

Lester Zick wrote:
Yeah, George, thanks for reposting. Hadn't noticed a reply and I'd
begun to wonder what happened. Give me a day or two to ponder. - LZ

Sure. All my posts get to my local server but a very small
number don't reach Google for some reason. If there is
any part you think is wrong, please give it a second read,
I'm trying to take small steps that you will be able to agree
and work toward a mutualy understanding rather than end
up with another argument.

George

On 10 Oct 2006 00:24:32 -0700, "George Dishman"
wrote:


Lester Zick wrote:
Yeah, George, thanks for reposting. Hadn't noticed a reply and I'd
begun to wonder what happened. Give me a day or two to ponder. - LZ

Sure. All my posts get to my local server but a very small
number don't reach Google for some reason. If there is
any part you think is wrong, please give it a second read,
I'm trying to take small steps that you will be able to agree
and work toward a mutualy understanding rather than end
up with another argument.

Same here. That's why I didn't pursue what appeared to be a lack of
response.

~v~~

On 9 Oct 2006 12:07:28 -0700, "George Dishman"
wrote:


This was posted on the 29th Sept. but hasn't
made it to Google for some reason:


"Lester Zick" wrote in message
.. .
On Fri, 29 Sep 2006 00:33:49 +0100, "George Dishman"
wrote:
"Lester Zick" wrote in message
...
...
So rather than replying to your post point
by point I'd rather try to simplify what I'm trying to say and seeing
if that makes any impression because if not then I expect these posts
will degenerate just as the prior posts have and there's no reason to
go through that again.

I maintain the distribution of matter in uniform gravitational disks
varies as the square of radius and gravitational force is constant as
a function of radius because the amount of matter increases as the
square of radius while the strength of gravitation decreases as the
square of radius.

And this is what I maintain forms the ideal basis for the calculation
of velocity expectations subject to corrections for a varying density
of matter as a function of radius.

Now as I understand it Virgil maintains the strength of gravitational
force as a function of radius has to include more than just the radial
dependency for the amount of matter. However I see two difficulties
with this. For one thing it would only tend to exacerbate the effect
I'm describing by increasing the amount of gravitationally attractive
matter as a radial function of disk area. But beyond that I'm not even
sure what Virgil says is true because as delta r approaches zero this
effect would seem (at least from my perspective) to approach zero too.

Then your response to the above mainly seems to be that velocity
curves are calculated through light intensity distributions and the
integration of thin rings. However it strikes me that this reply goes
to density of matter estimates and not to the point I'm trying to make
which is that as a function of radius the strength of gravitational
attraction and the amount of matter involved in the gravitational
attraction offset one another more or less exactly in ideal terms
subject to variations in density from the ideal.

In other words it looks to me like my claim and your response are not
mutually exclusive and we could still have both together: ideal flat
velocity curve expectations based on a uniform density of matter in
disks together with variations in material density determined by light
intensity analysis and the integration of thin rings.

I would also like to comment that I consider there are three possible
geometric forms supported by this kind of ideal geometric analysis:

1) Empty space between attracting bodies in which case the strength of
gravitational attraction is inverse square as a function of radius and
velocity curves decrease accordingly;

2) Ideal disk distributions of gravitationally attractive matter in
which the amount of gravitationally attractive matter increases as the
square of radius while the strength of gravitational force decreases
as the square of radius with flat velocity curves; and

3) Ideal spherical distributions of gravitationally attractive matter
in which the amount of gravitationally attractive matter increases as
the cube of radius while the strength of gravitational force decreases
as the square of radius with velocity curves proportional to radius.

(Presumably there could also conceivably be some kind of inverse
linear distribution of matter but I just can't figure out how such a
trick could be done.)

In any event there's no sense wasting a lot of time and effort if the
thrust of what I'm suggesting here is not apparent.

OK Lester, I'll try to keep it simple too but you have
covered a lot of ground. What you are saying is fairly
apparent so if you want to continue I will too, but we
probably need to go through it in a number of smaller
steps, this post has covered a lot of ground, some of
which is correct and some is flawed. I understand what
you are trying to do, but it makes more sense if I
reply in a slightly different order so that we can
firm up some basic principles before tackling the more
contentious areas.

Your suggestion
for light density analysis only affects the density of matter involved
as far as I can tell and your comment as to the integration of thin
rings (I suppose for thin rings of different density matter) for the
calculation of velocity curves doesn't appear to bear directly on my
argument regarding the amount of matter in an ideal disk as a function
of radius squared versus the inverse square attenuation of gravitation
as a function of radius squared.

We maybe need to clear up what you mean by an "ideal"
disc. Lots of galaxies have been studied and a pattern
has been noted, the distribution of light seems to
follow an exponential distribution with radius. That
equation for a typical galaxy can be written this way:

B(r) = Bc * e^(-r/h) [1]

where E(r) is the brightness at radius r, Bc is the
brightness in the centre and h is a constant called
the scale length. For the Milky way the values I have
are 140 for Bc and 3.5 kPc for h.

That means that for every 3.5 kpc you move away from
the centre, the density reduces by a factor of 2.72.

Equation (1) is what I would call an idealised curve
and Bc and h are parameters that can be used to fit
that curve to a specific galaxy.

Do you follow that so far. If so I want to look at
your point 1) next.

This looks pretty straightforward, George, except that I'm not
conversant with "kpc" (kiloparsec?).

That's right, and the value of 140 I gave above is
in units of solar luminosity per square parsec.

Ok.

I don't dispute brightness
measures in any event although I think there might be some slight
wiggle room with respect to significance.

Certainly there is going to be a statistical spread
and structure like spiral arms and bars has some
impact but in general the exponential function has
been found to be the best basic curve that matches
many galaxies which is what I mean by idealised in
this case.

Sure.

Some wiggle also comes from the light to mass ratio.
Values might typically be between 0.5 and 2.0 so a
star of the same luminosity as the Sun might have
from half to double its mass. However, a small hot
bright star and a large cool dim star would need to
have significant other differences as well. For
example the spectra of a white dwarf and a red giant
are considerably different so we can constrain the
wiggle room from that cause.

Ok.

At least I can visualize
considerable unlit regular matter outside the galactic center which
might not be subject to quite the same dynamics as at the center.

That's all that is meant by 'dark matter'. The
conversation is about how to work out its
distribution.

Well here I'm not so sure, George. What I'm referring to is regular
matter that is unlit and I think there may be a disproportionately
greater amount away from the galactic center. But in any event when I
see the phrase "dark matter" in such contexts I generally think of a
different kind of matter altogether used to make up the discrepancy
between calculated and observed velocity curves. which can never be
lighted, fused, or whatever to produce light as in ordinary stars.

In any event by "my point 1)" are you referring to the reference to
empty space between attracting bodies and inverse square velocity
curves? Because if so this would certainly seem to be the example
analogous to the solar system which also seems to be pretty straight
forward (at least to me).

Yes but your repetition doesn't say the same as
your first statement: "the strength of gravitational
attraction is inverse square as a function of radius
and velocity curves decrease accordingly" is correct
but "inverse square velocity curves" is not correct.
The inverse square gravitational force gives an
inverse square root velocity curve.

Well I'm not trying to dot all the iotas, George, although obviously
phrasing can be important. When I refer to "inverse square velocity
curves"in the present context I'm merely referring to "velocity curves
produced by inverse square forces" such as gravitation. Too many words
to nail down every phrase precisely. As far as I know there is a
linear conjugation between r and v produced by gravitational
acceleration which results in equal swept areas over time in orbit.

If not ...

Snip alternative, what you say above is what I meant.

However I don't mean to add further complexity to what you want to say
with respect to "my point 1)". So by all means have at it and we can
discuss your remarks afterward to the extent I don't understand or
don't agree with your comments.

OK, so now think of a very thin disk, in fact just
one star thick. Applying the exponential curve to
the Milky Way tells us that the "idealised"
distribution would have 140 sun-like stars per
square parsec at the centre while at a radius of
1000 parsec it will have fallen to 105 and by 2000
parsec that will be 79 sun-like stars per square
parsec.

To work out the gravitational effect on a single
star at 1000 parsec radius, you need to add up
the effect of every other star in the disc. Those
individual effects are governed by your point 1)
so the _force_ on the individual is the sum of all
the inverse squares of the distances. That's where
the integration comes in.

Since the density varies with radius, the way I was
suggesting to approach that was to break the disc
into concentric rings. Say a circle of 0.5 parsec
radius in the centre of density 140, then a ring
between 0.5 and 1.5 of density 139.96, a ring from
1.5 to 2.5 of density 139.92 and so on. Although
the test star is at 1000 parsec, you don't stop
there but continue out to the edge of the galaxy.
In fact probably beyond 2000 parsec the contribution
is decreasing both because of the exponential and
the inverse square so will drop off rapidly.

When you add up all those gravitational forces,
you then find the expected speed by balancing the
centrifugal force against the net inward force.

Does that sound sensible Lester?

Certainly sensible as far as it goes, George, and I have no quarrel
whatsoever with your concentric ring analysis or denisty variations.
But my suggestion really just concerns your initial assumption of a
thin disk without any exponential decrease to ask what aggregate
velocities should be as a function of r. I don't suggest this to avoid
any eventual exponential decrease calculation but just to figure out
whether under the ideal initial assumptions of constant density one
can expect the amount of centripetally directed gravitationally
attractive matter to vary as the square of radius in proportion to rr
and the area of the uniform density disk.

Now as I read your final comments above I get the impression that you
aggregate the gravitational effects of all matter both within and
without the orbit of any star whose orbital velocity is measured. But
I'm not quite so sure in my own mind. In any event this seems to be
the crux of the confusion. I can't say off the top of my head why this
should be true but on the face of it my original contention was that
velocities in uniform ideal disks should remain constant as a function
of radius and velocities in comparable uniform ideal spheres should
increase directly as a function of r.

In any event I don't want to try to put too much emphasis on the
problem at this juncture. I think I understand what you're suggesting.
But I can't think of any other reason spheres of matter should rotate
more or less in unison (although obviously gaseous spheres like the
sun are subject to considerable density and uniformity variations
producing variations in rotational velocity as a function of r). At
least this is a candid reflection of my own thinking whether right or
wrong.

~v~~

On Mon, 09 Oct 2006 14:34:33 -0600, Virgil wrote:


"Lester Zick" wrote in message
. ..

I maintain the distribution of matter in uniform gravitational disks
varies as the square of radius and gravitational force is constant as
a function of radius because the amount of matter increases as the
square of radius while the strength of gravitation decreases as the
square of radius.

And this is what I maintain forms the ideal basis for the calculation
of velocity expectations subject to corrections for a varying density
of matter as a function of radius.

Now as I understand it Virgil maintains the strength of gravitational
force as a function of radius has to include more than just the radial
dependency for the amount of matter. However I see two difficulties
with this. For one thing it would only tend to exacerbate the effect
I'm describing by increasing the amount of gravitationally attractive
matter as a radial function of disk area. But beyond that I'm not even
sure what Virgil says is true because as delta r approaches zero this
effect would seem (at least from my perspective) to approach zero too.


What I say is, as usual, not what Zick says I say.

Problem is, Virgil, I can't read your writing. So I'm forced to guess
what you're saying.

What I actually say is:

Which means just what exactly?

Given a uniform circular disk of fixed radius r, the attraction towards
its center on a unit mass at an interior point at a distance a, 0 = a
r, from its center is proportional to

the integral from t = -pi/2 to t = pi/2 of
cos(t) log[(a/r cos(t) + sqrt(1 - (a/r)^2 sin(t)^2))/|1-(a/r)^2|].

For values of a r, the attraction is proportional to

the integral from t = - arcsin(r/a) to t = arcsin(r/a) of
cos(t) log[(a/r cos(t) + sqrt(1 - (a/r)^2 sin(t)^2))/|1-(a/r)^2|].


The integrand does not have any neat closed form primitive, so the
result must be expressed as an integral.

~v~~

Lester Zick wrote:
On 9 Oct 2006 12:07:28 -0700, "George Dishman"
wrote:
"Lester Zick" wrote in message
.. .
On Fri, 29 Sep 2006 00:33:49 +0100, "George Dishman"
wrote:
"Lester Zick" wrote in message
...
....

big snip of older material and stuff agreed

At least I can visualize
considerable unlit regular matter outside the galactic center which
might not be subject to quite the same dynamics as at the center.

That's all that is meant by 'dark matter'. The
conversation is about how to work out its
distribution.

Well here I'm not so sure, George. What I'm referring to is regular
matter that is unlit and I think there may be a disproportionately
greater amount away from the galactic center. But in any event when I
see the phrase "dark matter" in such contexts I generally think of a
different kind of matter altogether used to make up the discrepancy
between calculated and observed velocity curves. which can never be
lighted, fused, or whatever to produce light as in ordinary stars.

There is some evidence from other observations that
dark matter has to have unsual properties but in terms
of galactic rotation curves, a large number of asteroids
would be suitable 'unlit' matter and could be quite
conventional.

In any event by "my point 1)" are you referring to the reference to
empty space between attracting bodies and inverse square velocity
curves? Because if so this would certainly seem to be the example
analogous to the solar system which also seems to be pretty straight
forward (at least to me).

Yes but your repetition doesn't say the same as
your first statement: "the strength of gravitational
attraction is inverse square as a function of radius
and velocity curves decrease accordingly" is correct
but "inverse square velocity curves" is not correct.
The inverse square gravitational force gives an
inverse square root velocity curve.

Well I'm not trying to dot all the iotas, George, although obviously
phrasing can be important. When I refer to "inverse square velocity
curves"in the present context I'm merely referring to "velocity curves
produced by inverse square forces" such as gravitation.

OK, I misread it, I thoughtyou meant the velocity
depended on the inverse square of the radius.

Too many words
to nail down every phrase precisely. As far as I know there is a
linear conjugation between r and v produced by gravitational
acceleration which results in equal swept areas over time in orbit.

If you are referring to Kepler's second law, remember it
applies to a single given orbit. Orbits of different radii
will sweep different areas in the same time of course.

If not ...

Snip alternative, what you say above is what I meant.

However I don't mean to add further complexity to what you want to say
with respect to "my point 1)". So by all means have at it and we can
discuss your remarks afterward to the extent I don't understand or
don't agree with your comments.

OK, so now think of a very thin disk, in fact just
one star thick. Applying the exponential curve to
the Milky Way tells us that the "idealised"
distribution would have 140 sun-like stars per
square parsec at the centre while at a radius of
1000 parsec it will have fallen to 105 and by 2000
parsec that will be 79 sun-like stars per square
parsec.

To work out the gravitational effect on a single
star at 1000 parsec radius, you need to add up
the effect of every other star in the disc. Those
individual effects are governed by your point 1)
so the _force_ on the individual is the sum of all
the inverse squares of the distances. That's where
the integration comes in.

Since the density varies with radius, the way I was
suggesting to approach that was to break the disc
into concentric rings. Say a circle of 0.5 parsec
radius in the centre of density 140, then a ring
between 0.5 and 1.5 of density 139.96, a ring from
1.5 to 2.5 of density 139.92 and so on. Although
the test star is at 1000 parsec, you don't stop
there but continue out to the edge of the galaxy.
In fact probably beyond 2000 parsec the contribution
is decreasing both because of the exponential and
the inverse square so will drop off rapidly.

When you add up all those gravitational forces,
you then find the expected speed by balancing the
centrifugal force against the net inward force.

Does that sound sensible Lester?

Certainly sensible as far as it goes, George, and I have no quarrel
whatsoever with your concentric ring analysis or denisty variations.
But my suggestion really just concerns your initial assumption of a
thin disk without any exponential decrease to ask what aggregate
velocities should be as a function of r.

I may be misreading you again but I have never suggested
a disk of uniform density, that was your starting point.

I don't suggest this to avoid
any eventual exponential decrease calculation but just to figure out
whether under the ideal initial assumptions of constant density ..

Why do you say "ideal"? The idealised form in exponential
with radius. I'm happy to discuss it as a hypothetical but
it isn't relevant to real galaxies.

.. one
can expect the amount of centripetally directed gravitationally
attractive matter to vary as the square of radius in proportion to rr
and the area of the uniform density disk.

Now as I read your final comments above I get the impression that you
aggregate the gravitational effects of all matter both within and
without the orbit of any star whose orbital velocity is measured. But
I'm not quite so sure in my own mind. In any event this seems to be
the crux of the confusion.

The matter exists so you have to account for it, you
cannot just ignore it or you will get the wrong answer.

I can't say off the top of my head why this
should be true but on the face of it my original contention was that
velocities in uniform ideal disks should remain constant as a function
of radius ...

That also is what I understood you to be saying but it
isn't correct even for uniform density. Since the density
isn't uniform anyway, the question is academic but if
you want to know the answer, you need to calculate the
integral.

.. and velocities in comparable uniform ideal spheres should
increase directly as a function of r.

Ideal spheres are an entirely different matter, let's not drift
off onto a different topic. It is one of the odd aspects of
an inverse square force that a uniform spherical shell
produces the same net force at a point outside the shell
as the same total mass concentrated at the centre while
it produces no net force at all inside. That isn't true of a
ring (part of a disk).

In any event I don't want to try to put too much emphasis on the
problem at this juncture. I think I understand what you're suggesting.

All I am saying is that the rigorous way to work it out is
to add up the forces from all of the stars and you can do
that by integrating the contributions over the area of the
disk. You can't use shortcuts as you might for a sphere.

But I can't think of any other reason spheres of matter should rotate
more or less in unison ...

That's easy - viscosity and friction. The core of the Earth
apparently rotates at a slightly different speed from the
outer layers. Friction at the interface should slow one
down awhile speeding up the other until they are locked
at the same angular rate so the existence of the difference
is somewhat puzzling.

Obviously thermal convection and coriolis forces will
produce complex local motions but I think you're talking
of larger scale uniformity.

... (although obviously gaseous spheres like the
sun are subject to considerable density and uniformity variations
producing variations in rotational velocity as a function of r).

The acceleration of gravity is independent of mass so
those have no effect (other than density variations
causing convenction due to bouyancy of course).

At
least this is a candid reflection of my own thinking whether right or
wrong.

Fair enough.

George

On 11 Oct 2006 06:09:44 -0700, "George Dishman"
wrote:


Lester Zick wrote:
On 9 Oct 2006 12:07:28 -0700, "George Dishman"
wrote:
"Lester Zick" wrote in message
.. .
On Fri, 29 Sep 2006 00:33:49 +0100, "George Dishman"
wrote:
"Lester Zick" wrote in message
...
...

big snip of older material and stuff agreed

At least I can visualize
considerable unlit regular matter outside the galactic center which
might not be subject to quite the same dynamics as at the center.

That's all that is meant by 'dark matter'. The
conversation is about how to work out its
distribution.

Well here I'm not so sure, George. What I'm referring to is regular
matter that is unlit and I think there may be a disproportionately
greater amount away from the galactic center. But in any event when I
see the phrase "dark matter" in such contexts I generally think of a
different kind of matter altogether used to make up the discrepancy
between calculated and observed velocity curves. which can never be
lighted, fused, or whatever to produce light as in ordinary stars.

There is some evidence from other observations that
dark matter has to have unsual properties but in terms
of galactic rotation curves, a large number of asteroids
would be suitable 'unlit' matter and could be quite
conventional.

Yes but my understanding is that estimates for such conventional dark
matter don't yield enough to do the job.

In any event by "my point 1)" are you referring to the reference to
empty space between attracting bodies and inverse square velocity
curves? Because if so this would certainly seem to be the example
analogous to the solar system which also seems to be pretty straight
forward (at least to me).

Yes but your repetition doesn't say the same as
your first statement: "the strength of gravitational
attraction is inverse square as a function of radius
and velocity curves decrease accordingly" is correct
but "inverse square velocity curves" is not correct.
The inverse square gravitational force gives an
inverse square root velocity curve.

Well I'm not trying to dot all the iotas, George, although obviously
phrasing can be important. When I refer to "inverse square velocity
curves"in the present context I'm merely referring to "velocity curves
produced by inverse square forces" such as gravitation.

OK, I misread it, I thoughtyou meant the velocity
depended on the inverse square of the radius.

No problem. It's just difficult to qualify every characterization
exhaustively.

Too many words
to nail down every phrase precisely. As far as I know there is a
linear conjugation between r and v produced by gravitational
acceleration which results in equal swept areas over time in orbit.

If you are referring to Kepler's second law, remember it
applies to a single given orbit. Orbits of different radii
will sweep different areas in the same time of course.

You know, George, when I first read this comment I was inclined to say
"of course" but on further consideration I'm not so sure. At least for
objects in the same shape orbits. Wouldn't objects in circular orbits
one at the radius of the earth and one twice as far from the sun have
to sweep equal areas in equal times?

If not ...

Snip alternative, what you say above is what I meant.

However I don't mean to add further complexity to what you want to say
with respect to "my point 1)". So by all means have at it and we can
discuss your remarks afterward to the extent I don't understand or
don't agree with your comments.

OK, so now think of a very thin disk, in fact just
one star thick. Applying the exponential curve to
the Milky Way tells us that the "idealised"
distribution would have 140 sun-like stars per
square parsec at the centre while at a radius of
1000 parsec it will have fallen to 105 and by 2000
parsec that will be 79 sun-like stars per square
parsec.

To work out the gravitational effect on a single
star at 1000 parsec radius, you need to add up
the effect of every other star in the disc. Those
individual effects are governed by your point 1)
so the _force_ on the individual is the sum of all
the inverse squares of the distances. That's where
the integration comes in.

Since the density varies with radius, the way I was
suggesting to approach that was to break the disc
into concentric rings. Say a circle of 0.5 parsec
radius in the centre of density 140, then a ring
between 0.5 and 1.5 of density 139.96, a ring from
1.5 to 2.5 of density 139.92 and so on. Although
the test star is at 1000 parsec, you don't stop
there but continue out to the edge of the galaxy.
In fact probably beyond 2000 parsec the contribution
is decreasing both because of the exponential and
the inverse square so will drop off rapidly.

When you add up all those gravitational forces,
you then find the expected speed by balancing the
centrifugal force against the net inward force.

Does that sound sensible Lester?

Certainly sensible as far as it goes, George, and I have no quarrel
whatsoever with your concentric ring analysis or denisty variations.
But my suggestion really just concerns your initial assumption of a
thin disk without any exponential decrease to ask what aggregate
velocities should be as a function of r.

I may be misreading you again but I have never suggested
a disk of uniform density, that was your starting point.

I agree it was my starting point. But you did say "OK, so now think of
a very thin disk, in fact just one star thick" which is pretty much
the same except you go on to apply the exponential qualification right
away. Everybody uses ideal assumptions in one form or another. I just
didn't see any reason to get into density variations right away since
that didn't concern the thrust of my argument.

I don't suggest this to avoid
any eventual exponential decrease calculation but just to figure out
whether under the ideal initial assumptions of constant density ..

Why do you say "ideal"? The idealised form in exponential
with radius. I'm happy to discuss it as a hypothetical but
it isn't relevant to real galaxies.

No but it is relevant to my argument. I use the term "ideal" just as I
would the phrase "simplifying assumption". Your simplifying assumption
is somewhere beyond "2000 parsecs". Mine is "uniform density" because
my basic claim depends on area if true but density only secondarily.

.. one
can expect the amount of centripetally directed gravitationally
attractive matter to vary as the square of radius in proportion to rr
and the area of the uniform density disk.

Now as I read your final comments above I get the impression that you
aggregate the gravitational effects of all matter both within and
without the orbit of any star whose orbital velocity is measured. But
I'm not quite so sure in my own mind. In any event this seems to be
the crux of the confusion.

The matter exists so you have to account for it, you
cannot just ignore it or you will get the wrong answer.

But we seem to have gotten the wrong answer anyway. I'm just trying to
figure out how the underlying analysis might have gotten off course to
begin with.

I can't say off the top of my head why this
should be true but on the face of it my original contention was that
velocities in uniform ideal disks should remain constant as a function
of radius ...

That also is what I understood you to be saying but it
isn't correct even for uniform density. Since the density
isn't uniform anyway, the question is academic but if
you want to know the answer, you need to calculate the
integral.

Well for whatever reason I started off on what really amounted to a
casual line of reasoning with the assumption that the aggregate amount
of gravitationally attractive matter that needed to be considered lay
within the orbit of a star whose velocity was to be determined. I
don't know exactly why I assumed that but I did. And if true that
would indicate that subject to density variations we would expect
gravitational force to be roughly constant as a function of radius.
Now this is a simple assumption. It may be a false assumption but at
least it's simple and relies on a roughly symmetrical distribution of
matter around the galaxy I expect. Of course if we just had the odd
body in an outside orbit we would certainly have to calculate its
gravitational impact in addition to the matter inside the orbit. But
my line of reasoning for whatever it's worth just relied on the matter
as a function of area within the orbit.

.. and velocities in comparable uniform ideal spheres should
increase directly as a function of r.

Ideal spheres are an entirely different matter, let's not drift
off onto a different topic. It is one of the odd aspects of
an inverse square force that a uniform spherical shell
produces the same net force at a point outside the shell
as the same total mass concentrated at the centre while
it produces no net force at all inside. That isn't true of a
ring (part of a disk).

Well I'm not suggesting the two cases are identical, George, just that
there's an important connection. Where's Newton when you need him?

In any event I don't want to try to put too much emphasis on the
problem at this juncture. I think I understand what you're suggesting.

All I am saying is that the rigorous way to work it out is
to add up the forces from all of the stars and you can do
that by integrating the contributions over the area of the
disk. You can't use shortcuts as you might for a sphere.

Here we're at an impasse. I'd like to get the idea straight in my mind
one way or the other.

But I can't think of any other reason spheres of matter should rotate
more or less in unison ...

That's easy - viscosity and friction. The core of the Earth
apparently rotates at a slightly different speed from the
outer layers. Friction at the interface should slow one
down awhile speeding up the other until they are locked
at the same angular rate so the existence of the difference
is somewhat puzzling.

Obviously thermal convection and coriolis forces will
produce complex local motions but I think you're talking
of larger scale uniformity.

Yes. I've considered other sources of uniformity but they just don't
quite jibe. I know that different parts of the sun rotate at different
velocities but there you have enormous differences in density which
could explain the variations. The earth as a rocky planet has other
dynamics. But they all share the roughly cubic distribution of matter
as a function of radius. Curious to say the least.

... (although obviously gaseous spheres like the
sun are subject to considerable density and uniformity variations
producing variations in rotational velocity as a function of r).

The acceleration of gravity is independent of mass so
those have no effect (other than density variations
causing convenction due to bouyancy of course).

But the acceleration of gravity is certainly not independent of
cumulative mass so density would play a determining role.

At
least this is a candid reflection of my own thinking whether right or
wrong.

Fair enough.

Well we shall just have to see what we shall see, George.

~v~~

"Lester Zick" wrote in message
...
On 11 Oct 2006 06:09:44 -0700, "George Dishman"
wrote:


Lester Zick wrote:
On 9 Oct 2006 12:07:28 -0700, "George Dishman"
wrote:
"Lester Zick" wrote in message
.. .
On Fri, 29 Sep 2006 00:33:49 +0100, "George Dishman"
wrote:
"Lester Zick" wrote in message
...
...

big snip of older material and stuff agreed

At least I can visualize
considerable unlit regular matter outside the galactic center which
might not be subject to quite the same dynamics as at the center.

That's all that is meant by 'dark matter'. The
conversation is about how to work out its
distribution.

Well here I'm not so sure, George. What I'm referring to is regular
matter that is unlit and I think there may be a disproportionately
greater amount away from the galactic center. But in any event when I
see the phrase "dark matter" in such contexts I generally think of a
different kind of matter altogether used to make up the discrepancy
between calculated and observed velocity curves. which can never be
lighted, fused, or whatever to produce light as in ordinary stars.

There is some evidence from other observations that
dark matter has to have unsual properties but in terms
of galactic rotation curves, a large number of asteroids
would be suitable 'unlit' matter and could be quite
conventional.

Yes but my understanding is that estimates for such conventional dark
matter don't yield enough to do the job.

Yes, the searches for large bodies of ordinary matter
(perhaps brown dwarfs) using microlensing have produced
too few events and the evidence from the Bullet Cluster
reported recently both suggest matter that is not
conventional. That's what I meant by "evidence from
other observations". Galactic rotation curves only tell
us about the mass so don't distinguish between forms so
I was trying to avoid another tangent.

In any event by "my point 1)" are you referring to the reference to
empty space between attracting bodies and inverse square velocity
curves? Because if so this would certainly seem to be the example
analogous to the solar system which also seems to be pretty straight
forward (at least to me).

Yes but your repetition doesn't say the same as
your first statement: "the strength of gravitational
attraction is inverse square as a function of radius
and velocity curves decrease accordingly" is correct
but "inverse square velocity curves" is not correct.
The inverse square gravitational force gives an
inverse square root velocity curve.

Well I'm not trying to dot all the iotas, George, although obviously
phrasing can be important. When I refer to "inverse square velocity
curves"in the present context I'm merely referring to "velocity curves
produced by inverse square forces" such as gravitation.

OK, I misread it, I thought you meant the velocity
depended on the inverse square of the radius.

No problem. It's just difficult to qualify every characterization
exhaustively.

Too many words
to nail down every phrase precisely. As far as I know there is a
linear conjugation between r and v produced by gravitational
acceleration which results in equal swept areas over time in orbit.

If you are referring to Kepler's second law, remember it
applies to a single given orbit. Orbits of different radii
will sweep different areas in the same time of course.

You know, George, when I first read this comment I was inclined to say
"of course" but on further consideration I'm not so sure. At least for
objects in the same shape orbits. Wouldn't objects in circular orbits
one at the radius of the earth and one twice as far from the sun have
to sweep equal areas in equal times?

Kepler's third law: The square of the period is proportional
to the cube of the radius.

Here are the values for circular orbits of 1, 2 and 4 AU:

Radius Period Circumference Speed Area Area/time
AU yr AU AU/yr AU^2 AU^2/yr

1.00 1.00 6.28 6.28 3.14 3.14
2.00 2.83 12.57 4.44 12.57 4.44
4.00 8.00 25.13 3.14 50.27 6.28


If not ...

Snip alternative, what you say above is what I meant.

However I don't mean to add further complexity to what you want to
say
with respect to "my point 1)". So by all means have at it and we can
discuss your remarks afterward to the extent I don't understand or
don't agree with your comments.

OK, so now think of a very thin disk, in fact just
one star thick. Applying the exponential curve to
the Milky Way tells us that the "idealised"
distribution would have 140 sun-like stars per
square parsec at the centre while at a radius of
1000 parsec it will have fallen to 105 and by 2000
parsec that will be 79 sun-like stars per square
parsec.

To work out the gravitational effect on a single
star at 1000 parsec radius, you need to add up
the effect of every other star in the disc. Those
individual effects are governed by your point 1)
so the _force_ on the individual is the sum of all
the inverse squares of the distances. That's where
the integration comes in.

Since the density varies with radius, the way I was
suggesting to approach that was to break the disc
into concentric rings. Say a circle of 0.5 parsec
radius in the centre of density 140, then a ring
between 0.5 and 1.5 of density 139.96, a ring from
1.5 to 2.5 of density 139.92 and so on. Although
the test star is at 1000 parsec, you don't stop
there but continue out to the edge of the galaxy.
In fact probably beyond 2000 parsec the contribution
is decreasing both because of the exponential and
the inverse square so will drop off rapidly.

When you add up all those gravitational forces,
you then find the expected speed by balancing the
centrifugal force against the net inward force.

Does that sound sensible Lester?

Certainly sensible as far as it goes, George, and I have no quarrel
whatsoever with your concentric ring analysis or denisty variations.
But my suggestion really just concerns your initial assumption of a
thin disk without any exponential decrease to ask what aggregate
velocities should be as a function of r.

I may be misreading you again but I have never suggested
a disk of uniform density, that was your starting point.

I agree it was my starting point. But you did say "OK, so now think of
a very thin disk, in fact just one star thick" which is pretty much
the same except you go on to apply the exponential qualification right
away.

My "OK" was in response to your "So by all means
have at it ..". I think it is just another case of
us misreading each other occasionally.

Everybody uses ideal assumptions in one form or another. I just
didn't see any reason to get into density variations right away since
that didn't concern the thrust of my argument.

Well equally I haven't seen any point in discussing
uniform density when it doesn't apply. It doesn't
give the right velocity curve anyway but imagine if
it did: we could then agree that uniform density was
required but the actual density falls exponentially
hence the required dark matter would just be the
difference between the uniform and exponential
densities so I still don't see what you were trying
to prove.

I don't suggest this to avoid
any eventual exponential decrease calculation but just to figure out
whether under the ideal initial assumptions of constant density ..

Why do you say "ideal"? The idealised form in exponential
with radius. I'm happy to discuss it as a hypothetical but
it isn't relevant to real galaxies.

No but it is relevant to my argument. I use the term "ideal" just as I
would the phrase "simplifying assumption". Your simplifying assumption
is somewhere beyond "2000 parsecs". Mine is "uniform density" because
my basic claim depends on area if true but density only secondarily.

The gravitational force depends on mass and mass in
any area (for a disk) is the density. You cannot
separate the two.

.. one
can expect the amount of centripetally directed gravitationally
attractive matter to vary as the square of radius in proportion to rr
and the area of the uniform density disk.

Now as I read your final comments above I get the impression that you
aggregate the gravitational effects of all matter both within and
without the orbit of any star whose orbital velocity is measured. But
I'm not quite so sure in my own mind. In any event this seems to be
the crux of the confusion.

The matter exists so you have to account for it, you
cannot just ignore it or you will get the wrong answer.

But we seem to have gotten the wrong answer anyway. I'm just trying to
figure out how the underlying analysis might have gotten off course to
begin with.

I think it was because you assumed that the simplification
you can use for a sphere also applied to a disk.

I can't say off the top of my head why this
should be true but on the face of it my original contention was that
velocities in uniform ideal disks should remain constant as a function
of radius ...

That also is what I understood you to be saying but it
isn't correct even for uniform density. Since the density
isn't uniform anyway, the question is academic but if
you want to know the answer, you need to calculate the
integral.

Well for whatever reason I started off on what really amounted to a
casual line of reasoning with the assumption that the aggregate amount
of gravitationally attractive matter that needed to be considered lay
within the orbit of a star whose velocity was to be determined. I
don't know exactly why I assumed that but I did.

That would be true for a sphere.

And if true that
would indicate that subject to density variations we would expect
gravitational force to be roughly constant as a function of radius.

The mass inside the radius grows as the cube while
the effect falls as the square so the gravitational
acceleration would be proportional to the radius.

Now this is a simple assumption. It may be a false assumption but at
least it's simple and relies on a roughly symmetrical distribution of
matter around the galaxy I expect. Of course if we just had the odd
body in an outside orbit we would certainly have to calculate its
gravitational impact in addition to the matter inside the orbit. But
my line of reasoning for whatever it's worth just relied on the matter
as a function of area within the orbit.

Unfortunately, it is not correct.

.. and velocities in comparable uniform ideal spheres should
increase directly as a function of r.

Ideal spheres are an entirely different matter, let's not drift
off onto a different topic. It is one of the odd aspects of
an inverse square force that a uniform spherical shell
produces the same net force at a point outside the shell
as the same total mass concentrated at the centre while
it produces no net force at all inside. That isn't true of a
ring (part of a disk).

Well I'm not suggesting the two cases are identical, George, just that
there's an important connection. Where's Newton when you need him?

What you need is calculus, and he invented that
as well. :-)

In any event I don't want to try to put too much emphasis on the
problem at this juncture. I think I understand what you're suggesting.

All I am saying is that the rigorous way to work it out is
to add up the forces from all of the stars and you can do
that by integrating the contributions over the area of the
disk. You can't use shortcuts as you might for a sphere.

Here we're at an impasse. I'd like to get the idea straight in my mind
one way or the other.

Perhaps the best thing would be for you to do
some digging and find out why the shortcut works
for a sphere. Then when you consider a disk you
may see why the conditions for it to hold are not
met.

But I can't think of any other reason spheres of matter should rotate
more or less in unison ...

That's easy - viscosity and friction. The core of the Earth
apparently rotates at a slightly different speed from the
outer layers. Friction at the interface should slow one
down awhile speeding up the other until they are locked
at the same angular rate so the existence of the difference
is somewhat puzzling.

Obviously thermal convection and coriolis forces will
produce complex local motions but I think you're talking
of larger scale uniformity.

Yes. I've considered other sources of uniformity but they just don't
quite jibe. I know that different parts of the sun rotate at different
velocities but there you have enormous differences in density which
could explain the variations.

Not really. The major density variation is spherically
symmetric with the core being much denser. The whole
thing has to be close to equilibrium, just as oil forms
a layer on top of water. What keeps it mostly rotating
at similar speeds is just boring old friction.

The earth as a rocky planet has other
dynamics. But they all share the roughly cubic distribution of matter
as a function of radius. Curious to say the least.

... (although obviously gaseous spheres like the
sun are subject to considerable density and uniformity variations
producing variations in rotational velocity as a function of r).

The acceleration of gravity is independent of mass so
those have no effect (other than density variations
causing convenction due to bouyancy of course).

But the acceleration of gravity is certainly not independent of
cumulative mass so density would play a determining role.

Denser material would just sink until it reaches a
region of similar density.

Bottom line is that the atoms in the plasma of the Sun
or the mantle of the Earth are not in freefall like the
planets and are not moving nearly fast enough to be
approaching orbital speed so it is a quite different
topic and again not a sidetrack I want to pursue.

George

On Wed, 11 Oct 2006 20:08:48 +0100, "George Dishman"
wrote:


"Lester Zick" wrote in message
.. .
On 11 Oct 2006 06:09:44 -0700, "George Dishman"
wrote:


Lester Zick wrote:
On 9 Oct 2006 12:07:28 -0700, "George Dishman"
wrote:
"Lester Zick" wrote in message
.. .
On Fri, 29 Sep 2006 00:33:49 +0100, "George Dishman"
wrote:
"Lester Zick" wrote in message
...
...

big snip of older material and stuff agreed

At least I can visualize
considerable unlit regular matter outside the galactic center which
might not be subject to quite the same dynamics as at the center.

That's all that is meant by 'dark matter'. The
conversation is about how to work out its
distribution.

Well here I'm not so sure, George. What I'm referring to is regular
matter that is unlit and I think there may be a disproportionately
greater amount away from the galactic center. But in any event when I
see the phrase "dark matter" in such contexts I generally think of a
different kind of matter altogether used to make up the discrepancy
between calculated and observed velocity curves. which can never be
lighted, fused, or whatever to produce light as in ordinary stars.

There is some evidence from other observations that
dark matter has to have unsual properties but in terms
of galactic rotation curves, a large number of asteroids
would be suitable 'unlit' matter and could be quite
conventional.

Yes but my understanding is that estimates for such conventional dark
matter don't yield enough to do the job.

Yes, the searches for large bodies of ordinary matter
(perhaps brown dwarfs) using microlensing have produced
too few events and the evidence from the Bullet Cluster
reported recently both suggest matter that is not
conventional. That's what I meant by "evidence from
other observations". Galactic rotation curves only tell
us about the mass so don't distinguish between forms so
I was trying to avoid another tangent.

I'm a little confused here, George. Your comment was that from other
observations "unlit matter" could be quite conventional but is somehow
supposed to be unconventional as well? The only unconventionality I
can see is that it doesn't "light up" as predictably as conventional
theories assume.

In any event by "my point 1)" are you referring to the reference to
empty space between attracting bodies and inverse square velocity
curves? Because if so this would certainly seem to be the example
analogous to the solar system which also seems to be pretty straight
forward (at least to me).

Yes but your repetition doesn't say the same as
your first statement: "the strength of gravitational
attraction is inverse square as a function of radius
and velocity curves decrease accordingly" is correct
but "inverse square velocity curves" is not correct.
The inverse square gravitational force gives an
inverse square root velocity curve.

Well I'm not trying to dot all the iotas, George, although obviously
phrasing can be important. When I refer to "inverse square velocity
curves"in the present context I'm merely referring to "velocity curves
produced by inverse square forces" such as gravitation.

OK, I misread it, I thought you meant the velocity
depended on the inverse square of the radius.

No problem. It's just difficult to qualify every characterization
exhaustively.

Too many words
to nail down every phrase precisely. As far as I know there is a
linear conjugation between r and v produced by gravitational
acceleration which results in equal swept areas over time in orbit.

If you are referring to Kepler's second law, remember it
applies to a single given orbit. Orbits of different radii
will sweep different areas in the same time of course.

You know, George, when I first read this comment I was inclined to say
"of course" but on further consideration I'm not so sure. At least for
objects in the same shape orbits. Wouldn't objects in circular orbits
one at the radius of the earth and one twice as far from the sun have
to sweep equal areas in equal times?

Kepler's third law: The square of the period is proportional
to the cube of the radius.

Here are the values for circular orbits of 1, 2 and 4 AU:

Radius Period Circumference Speed Area Area/time
AU yr AU AU/yr AU^2 AU^2/yr

1.00 1.00 6.28 6.28 3.14 3.14
2.00 2.83 12.57 4.44 12.57 4.44
4.00 8.00 25.13 3.14 50.27 6.28

Hmm. At one earth radius F(g)=e yet at twice that radius F(g)=/=e/4?
I'm missing something here. Shouldn't the orbital velocity at twice
the radius be v/2 such that the orbital period would be 2r/(v/2) or 4
times the unit period? Just a question because this goes back a long
way.

If not ...

Snip alternative, what you say above is what I meant.

However I don't mean to add further complexity to what you want to
say
with respect to "my point 1)". So by all means have at it and we can
discuss your remarks afterward to the extent I don't understand or
don't agree with your comments.

OK, so now think of a very thin disk, in fact just
one star thick. Applying the exponential curve to
the Milky Way tells us that the "idealised"
distribution would have 140 sun-like stars per
square parsec at the centre while at a radius of
1000 parsec it will have fallen to 105 and by 2000
parsec that will be 79 sun-like stars per square
parsec.

To work out the gravitational effect on a single
star at 1000 parsec radius, you need to add up
the effect of every other star in the disc. Those
individual effects are governed by your point 1)
so the _force_ on the individual is the sum of all
the inverse squares of the distances. That's where
the integration comes in.

Since the density varies with radius, the way I was
suggesting to approach that was to break the disc
into concentric rings. Say a circle of 0.5 parsec
radius in the centre of density 140, then a ring
between 0.5 and 1.5 of density 139.96, a ring from
1.5 to 2.5 of density 139.92 and so on. Although
the test star is at 1000 parsec, you don't stop
there but continue out to the edge of the galaxy.
In fact probably beyond 2000 parsec the contribution
is decreasing both because of the exponential and
the inverse square so will drop off rapidly.

When you add up all those gravitational forces,
you then find the expected speed by balancing the
centrifugal force against the net inward force.

Does that sound sensible Lester?

Certainly sensible as far as it goes, George, and I have no quarrel
whatsoever with your concentric ring analysis or denisty variations.
But my suggestion really just concerns your initial assumption of a
thin disk without any exponential decrease to ask what aggregate
velocities should be as a function of r.

I may be misreading you again but I have never suggested
a disk of uniform density, that was your starting point.

I agree it was my starting point. But you did say "OK, so now think of
a very thin disk, in fact just one star thick" which is pretty much
the same except you go on to apply the exponential qualification right
away.

My "OK" was in response to your "So by all means
have at it ..". I think it is just another case of
us misreading each other occasionally.

Okay.

Everybody uses ideal assumptions in one form or another. I just
didn't see any reason to get into density variations right away since
that didn't concern the thrust of my argument.

Well equally I haven't seen any point in discussing
uniform density when it doesn't apply. It doesn't
give the right velocity curve anyway but imagine if
it did: we could then agree that uniform density was
required but the actual density falls exponentially
hence the required dark matter would just be the
difference between the uniform and exponential
densities so I still don't see what you were trying
to prove.

My original contention was that F(g) in plane circular galaxies should
remain constant regardless of radius because the amount of matter
increases as the square of radius in ideal terms before application of
non uniform density considerations and that this would produce flat
velocity curves rather than inverse square velocity curves (by which I
mean velocity curves produced by inverse square forces located at the
center of attraction).

I don't suggest this to avoid
any eventual exponential decrease calculation but just to figure out
whether under the ideal initial assumptions of constant density ..

Why do you say "ideal"? The idealised form in exponential
with radius. I'm happy to discuss it as a hypothetical but
it isn't relevant to real galaxies.

No but it is relevant to my argument. I use the term "ideal" just as I
would the phrase "simplifying assumption". Your simplifying assumption
is somewhere beyond "2000 parsecs". Mine is "uniform density" because
my basic claim depends on area if true but density only secondarily.

The gravitational force depends on mass and mass in
any area (for a disk) is the density. You cannot
separate the two.

True but variations in density can certainly be separated.

.. one
can expect the amount of centripetally directed gravitationally
attractive matter to vary as the square of radius in proportion to rr
and the area of the uniform density disk.

Now as I read your final comments above I get the impression that you
aggregate the gravitational effects of all matter both within and
without the orbit of any star whose orbital velocity is measured. But
I'm not quite so sure in my own mind. In any event this seems to be
the crux of the confusion.

The matter exists so you have to account for it, you
cannot just ignore it or you will get the wrong answer.

But we seem to have gotten the wrong answer anyway. I'm just trying to
figure out how the underlying analysis might have gotten off course to
begin with.

I think it was because you assumed that the simplification
you can use for a sphere also applied to a disk.

I wasn't talking about my speculation - which was only a speculation -
but about the original estimates for velocity curves which seem to
have gotten it wrong for whatever reason to begin with.

I can't say off the top of my head why this
should be true but on the face of it my original contention was that
velocities in uniform ideal disks should remain constant as a function
of radius ...

That also is what I understood you to be saying but it
isn't correct even for uniform density. Since the density
isn't uniform anyway, the question is academic but if
you want to know the answer, you need to calculate the
integral.

Well for whatever reason I started off on what really amounted to a
casual line of reasoning with the assumption that the aggregate amount
of gravitationally attractive matter that needed to be considered lay
within the orbit of a star whose velocity was to be determined. I
don't know exactly why I assumed that but I did.

That would be true for a sphere.

And definitely not for a circular disk? If so why?

And if true that
would indicate that subject to density variations we would expect
gravitational force to be roughly constant as a function of radius.

The mass inside the radius grows as the cube while
the effect falls as the square so the gravitational
acceleration would be proportional to the radius.

Are you talking about spheres here, George, or disks? Minus density
variations for disks I would expect gravitational acceleration to be a
constant function independent of radius whereas for spheres I would
expect it to be a linear function of radius.

Now this is a simple assumption. It may be a false assumption but at
least it's simple and relies on a roughly symmetrical distribution of
matter around the galaxy I expect. Of course if we just had the odd
body in an outside orbit we would certainly have to calculate its
gravitational impact in addition to the matter inside the orbit. But
my line of reasoning for whatever it's worth just relied on the matter
as a function of area within the orbit.

Unfortunately, it is not correct.

Weeell, George, I haven't made up my mind just yet. So I guess I'll
just have to go on being incorrect until I do.

.. and velocities in comparable uniform ideal spheres should
increase directly as a function of r.

Ideal spheres are an entirely different matter, let's not drift
off onto a different topic. It is one of the odd aspects of
an inverse square force that a uniform spherical shell
produces the same net force at a point outside the shell
as the same total mass concentrated at the centre while
it produces no net force at all inside. That isn't true of a
ring (part of a disk).

Well I'm not suggesting the two cases are identical, George, just that
there's an important connection. Where's Newton when you need him?

What you need is calculus, and he invented that
as well. :-)

Well of course. That's why I need him. I think there may well be more
to the inverse square gravitational apple than we've plumbed so far.

In any event I don't want to try to put too much emphasis on the
problem at this juncture. I think I understand what you're suggesting.

All I am saying is that the rigorous way to work it out is
to add up the forces from all of the stars and you can do
that by integrating the contributions over the area of the
disk. You can't use shortcuts as you might for a sphere.

Here we're at an impasse. I'd like to get the idea straight in my mind
one way or the other.

Perhaps the best thing would be for you to do
some digging and find out why the shortcut works
for a sphere. Then when you consider a disk you
may see why the conditions for it to hold are not
met.

Same considerations. I admit it's an assumption and problematic at
that. I don't admit it's wrong. In any event it lies at the basis of
my contention regardless: apart from variations in thickness and
density the amount of gravitationally attractive matter should grow as
the square of orbital radius and should exactly offset the inverse
square attentuation of gravitational force as a function of radius. At
least I've seen no convincing evidence that it wouldn't.

But I can't think of any other reason spheres of matter should rotate
more or less in unison ...

That's easy - viscosity and friction. The core of the Earth
apparently rotates at a slightly different speed from the
outer layers. Friction at the interface should slow one
down awhile speeding up the other until they are locked
at the same angular rate so the existence of the difference
is somewhat puzzling.

Obviously thermal convection and coriolis forces will
produce complex local motions but I think you're talking
of larger scale uniformity.

Yes. I've considered other sources of uniformity but they just don't
quite jibe. I know that different parts of the sun rotate at different
velocities but there you have enormous differences in density which
could explain the variations.

Not really. The major density variation is spherically
symmetric with the core being much denser. The whole
thing has to be close to equilibrium, just as oil forms
a layer on top of water. What keeps it mostly rotating
at similar speeds is just boring old friction.

I agree the whole thing is close to equilibrium but not for the same
reasons. And I agree that friction plays a huge role in the process.
But I think if gravitation were not an inverse square force as a
funtion of radius and if the amount of matter were not a cubic
function of radius the whole thing would come apart pretty quickly.
Rather like the implications of ice being less dense than liquid water
for the earth.

The earth as a rocky planet has other
dynamics. But they all share the roughly cubic distribution of matter
as a function of radius. Curious to say the least.

... (although obviously gaseous spheres like the
sun are subject to considerable density and uniformity variations
producing variations in rotational velocity as a function of r).

The acceleration of gravity is independent of mass so
those have no effect (other than density variations
causing convenction due to bouyancy of course).

But the acceleration of gravity is certainly not independent of
cumulative mass so density would play a determining role.

Denser material would just sink until it reaches a
region of similar density.

Okay.

Bottom line is that the atoms in the plasma of the Sun
or the mantle of the Earth are not in freefall like the
planets and are not moving nearly fast enough to be
approaching orbital speed so it is a quite different
topic and again not a sidetrack I want to pursue.

Well, George, I consider that atoms in the plasma of the sun and
elsewhere are actually in freefall and then ask why they do or don't
do whatever it is they do. I don't say there aren't dynamic variations
involved that operate on particular atoms and ions in isolation but I
do say they're all part of a much larger picture including galaxies
which has to be explained in exactly comparable terms overall.

~v~~

Lester Zick wrote:
On Wed, 11 Oct 2006 20:08:48 +0100, "George Dishman"
wrote:


"Lester Zick" wrote in message
.. .
On 11 Oct 2006 06:09:44 -0700, "George Dishman"
wrote:


Lester Zick wrote:
On 9 Oct 2006 12:07:28 -0700, "George Dishman"
wrote:
"Lester Zick" wrote in message
.. .
On Fri, 29 Sep 2006 00:33:49 +0100, "George Dishman"
wrote:
"Lester Zick" wrote in message
...

more snipped

There is some evidence from other observations that
dark matter has to have unsual properties but in terms
^^^^^^^^^^^^^^^^^
of galactic rotation curves, a large number of asteroids
would be suitable 'unlit' matter and could be quite
conventional.

Yes but my understanding is that estimates for such conventional dark
matter don't yield enough to do the job.

Yes, the searches for large bodies of ordinary matter
(perhaps brown dwarfs) using microlensing have produced
too few events and the evidence from the Bullet Cluster
reported recently both suggest matter that is not
conventional. That's what I meant by "evidence from
other observations". Galactic rotation curves only tell
us about the mass so don't distinguish between forms so
I was trying to avoid another tangent.

I'm a little confused here, George. Your comment was that from other
observations "unlit matter" could be quite conventional ...

No, I said _other_ observations suggested it _wasn't_
conventional.

but is somehow
supposed to be unconventional as well? The only unconventionality I
can see is that it doesn't "light up" as predictably as conventional
theories assume.

more background snipped

You know, George, when I first read this comment I was inclined to say
"of course" but on further consideration I'm not so sure. At least for
objects in the same shape orbits. Wouldn't objects in circular orbits
one at the radius of the earth and one twice as far from the sun have
to sweep equal areas in equal times?

Kepler's third law: The square of the period is proportional
to the cube of the radius.

Here are the values for circular orbits of 1, 2 and 4 AU:

Radius Period Circumference Speed Area Area/time
AU yr AU AU/yr AU^2 AU^2/yr

1.00 1.00 6.28 6.28 3.14 3.14
2.00 2.83 12.57 4.44 12.57 4.44
4.00 8.00 25.13 3.14 50.27 6.28

Hmm. At one earth radius F(g)=e yet at twice that radius F(g)=/=e/4?
I'm missing something here.

What you are missing is that none of the columns is F(g) by
which I assume you meam the foce of gravity. The force
depends on the mass but F/m = a = GM/r^2 where m is the
mass of the orbiting body and M is the mass of the star (or
whatever). The centrifugal force is a=v^2/r of course so if I
add another column you can check the value of acceleration:

Radius Period Circumference Speed Area Area/time Accn
AU yr AU AU/yr AU^2 AU^2/yr AU/yr^2

1.00 1.00 6.28 6.28 3.14 3.14 39.48
2.00 2.83 12.57 4.44 12.57 4.44 9.87
4.00 8.00 25.13 3.14 50.27 6.28 2.47

Because of the simple units I used, you need to use the
value 4*pi^2 for GM when calculating the gravitational
acceleration as a = (4*pi^") / r^2 or you can use the
centifugal acceleration a = v^2 / r. Both will give you
the same value shown in the final column.

Shouldn't the orbital velocity at twice
the radius be v/2 such that the orbital period would be 2r/(v/2) or 4
times the unit period?

No, the speed at twice the radius falls by sqrt(2) so that
the forces balance

Just a question because this goes back a long
way.

It's a good question which goes to the heart of the
disagreements.

snip 'Okay'

Everybody uses ideal assumptions in one form or another. I just
didn't see any reason to get into density variations right away since
that didn't concern the thrust of my argument.

Well equally I haven't seen any point in discussing
uniform density when it doesn't apply. It doesn't
give the right velocity curve anyway but imagine if
it did: we could then agree that uniform density was
required but the actual density falls exponentially
hence the required dark matter would just be the
difference between the uniform and exponential
densities so I still don't see what you were trying
to prove.

My original contention was that F(g) in plane circular galaxies should
remain constant regardless of radius because the amount of matter
increases as the square of radius in ideal terms before application of
non uniform density considerations and that this would produce flat
velocity curves rather than inverse square velocity curves (by which I
mean velocity curves produced by inverse square forces located at the
center of attraction).

OK but see below.

snip

.. one
can expect the amount of centripetally directed gravitationally
attractive matter to vary as the square of radius in proportion to rr
and the area of the uniform density disk.

Now as I read your final comments above I get the impression that you
aggregate the gravitational effects of all matter both within and
without the orbit of any star whose orbital velocity is measured. But
I'm not quite so sure in my own mind. In any event this seems to be
the crux of the confusion.

The matter exists so you have to account for it, you
cannot just ignore it or you will get the wrong answer.

But we seem to have gotten the wrong answer anyway. I'm just trying to
figure out how the underlying analysis might have gotten off course to
begin with.

I think it was because you assumed that the simplification
you can use for a sphere also applied to a disk.

I wasn't talking about my speculation - which was only a speculation -
but about the original estimates for velocity curves which seem to
have gotten it wrong for whatever reason to begin with.

But you calculated from a uniformly dense disk which
isn't the case so of course the calculation would appear
different. That isn't evidence that anyone got it wrong,
just that the conventional calculation started from a
different premise.

Well for whatever reason I started off on what really amounted to a
casual line of reasoning with the assumption that the aggregate amount
of gravitationally attractive matter that needed to be considered lay
within the orbit of a star whose velocity was to be determined. I
don't know exactly why I assumed that but I did.

That would be true for a sphere.

And definitely not for a circular disk? If so why?

Definitely not. To understand why, I think you
need to do the integrals or at least look them
up in a text book or on the web.

And if true that
would indicate that subject to density variations we would expect
gravitational force to be roughly constant as a function of radius.

The mass inside the radius grows as the cube while
the effect falls as the square so the gravitational
acceleration would be proportional to the radius.

Are you talking about spheres here, George, or disks?

Spheres.

Minus density
variations for disks I would expect gravitational acceleration to be a
constant function independent of radius whereas for spheres I would
expect it to be a linear function of radius.

That is right for a sphere but not for a disk. The total mass
of the disk varies as the square of the radius but the integral
over the surface doesn't produce the same acceleration as
an equal point mass at the centre.

Also, you are talking about the acceleration at the rim of
the disk whereas in galactice rotation curves we are
discussing the speed as a function of radius for a fixed
size of disk. The point in question has mass outside its
radius as well as inside.

Now this is a simple assumption. It may be a false assumption but at
least it's simple and relies on a roughly symmetrical distribution of
matter around the galaxy I expect. Of course if we just had the odd
body in an outside orbit we would certainly have to calculate its
gravitational impact in addition to the matter inside the orbit. But
my line of reasoning for whatever it's worth just relied on the matter
as a function of area within the orbit.

Unfortunately, it is not correct.

Weeell, George, I haven't made up my mind just yet. So I guess I'll
just have to go on being incorrect until I do.

It's not a question of you making up your mind, it is
a mathematical fact. You can go on being incorrect
as long as you like, it will always be incorrect.

All I am saying is that the rigorous way to work it out is
to add up the forces from all of the stars and you can do
that by integrating the contributions over the area of the
disk. You can't use shortcuts as you might for a sphere.

Here we're at an impasse. I'd like to get the idea straight in my mind
one way or the other.

Perhaps the best thing would be for you to do
some digging and find out why the shortcut works
for a sphere. Then when you consider a disk you
may see why the conditions for it to hold are not
met.

Same considerations. I admit it's an assumption and problematic at
that. I don't admit it's wrong.

Then you need to take a timeout and do the integral.

In any event it lies at the basis of
my contention regardless: apart from variations in thickness and
density the amount of gravitationally attractive matter should grow as
the square of orbital radius and should exactly offset the inverse
square attentuation of gravitational force as a function of radius. At
least I've seen no convincing evidence that it wouldn't.

The "convincing evidence" is that your assumption
is mathematically incorrect.

Not really. The major density variation is spherically
symmetric with the core being much denser. The whole
thing has to be close to equilibrium, just as oil forms
a layer on top of water. What keeps it mostly rotating
at similar speeds is just boring old friction.

I agree the whole thing is close to equilibrium but not for the same
reasons. And I agree that friction plays a huge role in the process.
But I think if gravitation were not an inverse square force as a
funtion of radius and if the amount of matter were not a cubic
function of radius the whole thing would come apart pretty quickly.

Obviously a different form of gravtitaional force would
produce a different result but that's too vague to discuss.
Note the mass isn't a cubic function of radius since the
density varies with radius.

Rather like the implications of ice being less dense than liquid water
for the earth.

That's a different matter, water is unusual in that it
has a maximum density as a function of temperature.


Bottom line is that the atoms in the plasma of the Sun
or the mantle of the Earth are not in freefall like the
planets and are not moving nearly fast enough to be
approaching orbital speed so it is a quite different
topic and again not a sidetrack I want to pursue.

Well, George, I consider that atoms in the plasma of the sun and
elsewhere are actually in freefall and then ask why they do or don't
do whatever it is they do.

They are not in freefall because they are continually
colliding with each other. That keeps their speeds
similar. Any high speed ion moving through the bulk
soon reaches the same speed distribution as every
other ion in its vicinity.

I don't say there aren't dynamic variations
involved that operate on particular atoms and ions in isolation but I
do say they're all part of a much larger picture including galaxies
which has to be explained in exactly comparable terms overall.

Nope, galaxies don't bounce off each other billions
of times per second - ions in the Sun's plasma do.

George