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The phrase 'dark matter'
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The TimeLord
science forum Guru Wannabe


Joined: 12 Jun 2005
Posts: 182

PostPosted: Thu Jul 20, 2006 7:31 am    Post subject: Re: The phrase 'dark matter' Reply with quote

On Mon, 17 Jul 2006 17:27:11 -0700, "tomgee" <tyropress@yahoo.com> wrote
in <1153182431.684397.68990@m79g2000cwm.googlegroups.com>:

[...]
Quote:
But that was based only on the notion that neutrinos had some mass. It
is only recently they have been shown to have some mass but not until
and unless they interact w/a tau neutrino.

Where is the source for that info? Based on Superkamiokande,
which verified that neutrinos could have mass, the neutrinos
don't have to interact with anything to "indicate mass".

Quote:

That means they are massless until they are transformed into particles
having mass and energy, just like my model predicted.

Model, what model? Where is it published?

[...]
Quote:
Wait now. 1st you say, "...gravitate like ordinary matter." Next you
call it "gravitational mass", which does not equate to the same thing.
To have gravitation like we know it, it must have the force of
attraction, and that force is evident only in massive objects we can
observe. For us 2b able to see them, they must have energy and the
property of time, as we know those qualities 2b. However, we cannot see
DM, so we can only guess it has gravitational attraction like visible
objects have.

That's the whole point! You can't see them, because they don't
shine, but they have gravity, thus mass and thus are matter.
The gravity is acting on things that do shine. - Think of it
this way: Newton's Second Law of Motion says that the motion
doesn't change unless a force is applied. So if the motion of
a shining star changes, there should be a force on it, even if
you can't see the thing causing the force. If the force is
not electro-magnetism, weak or strong force, then what's left?
Of course gravity. Thus there is some thing you can't see (dark)
that is generating some gravity and thus has mass (matter).
Hence it's dark matter; whatever dark matter turns out to be.

Quote:

Chew on this: How can additional gravitational attraction cause the
observed effects? Oh, sure, more attraction force will help keep

Gravity is a force. Force changes the motion of the thing it
interacts with. Duh!

Quote:
galaxies together better, yes. But how does that explain the effect
where the outer orbiting bodies move faster than can be expected? And

Because they are *not* moving slower. Duh!

Quote:
also, if DM is everywhere RM is not, how does it know what to attract

What is RM?

Quote:
and what not to attract?

Same way gravity knows what to attract. Duh!

Quote:

It is quite possible that instead of gravitational attractive forces,
they are gravitational repulsive forces. Since DM appears 2b quite

If there are gravitational repulsive forces, you will need to
show how that happens. Negative mass is not defined since "matter
holes" don't really have any physical meaning.

Quote:
opposite to RM, why should we not think that its gravita- tional force -

I'll have to take your word on that since I don't know what you
mean by "RM".

[...]
Quote:
Or, why cannot it be some force other than gravitational? It could be

Because it would show up in the spectra of the shining star we
could see.

Quote:
just "interactional", to coin a word, and not gravita- tional at all.

Gravity is interactional. So what's your point?

[...]
Quote:
At this point, we only surmise the forces to be gravitational, and it
may indeed turn out that way. But at this point, I do not see all the
questions being answered by the same attractive force of RM gravitation,
while my model proposes that it is not just the gravitational forces
that change the universe second-by-second, but also the various types of
interactions that go on between the stuff of the universe.

I am open to other ideas as long as they are well-defined and
consistent with observation. However, just saying that accepted
ideas are wrong despite their consistency with observed fact,
is like saying UFOs must exist because the government is hiding
them.

--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
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The Real Chris
science forum addict


Joined: 07 May 2006
Posts: 75

PostPosted: Thu Jul 20, 2006 7:08 pm    Post subject: Re: The phrase 'dark matter' Reply with quote

Look below for enlightenment.

The pioneer anolomy is the same.

Chris.

"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:net@nospam.com> wrote in
message news:RyXug.38482$AB3.25355@fed1read02...
Quote:
Dear The Real Chris:


The problem is, normal matter has to *not* be located where it can scatter
light passing through it. This places it in very special places,
impossible places and/or very temporary locations. We are getting the
"normal and expected" amount of light from the sources that are held in
place by Dark Matter. And closer in, we are getting normal and expected
light from those sources too.

Dust and hydrogen clouds don't work.

Junk science.

I'd like to think so too. The facts are otherwise.

David A. Smith


Galactic Rotation
The first thing to calculate before computing the distribution of galactic
rotation is the distribution of mass in the galaxy. As a first
approximation, I will take the galaxy to be a central point mass (the black
hole) of 10 million solar masses and a flat disk of uniformly distributed
matter (stars) of 1 solar mass per cubic light year. The disk will have a
diameter of 10,000 light years and the thickness of 3 light years which I
shall take as a disk of zero thickness to start off with.
I will use Newtonian mechanics throughout with a conventional value of the
gravitational constant.






Point P is R Km from the hub, the hub is M Kg and the element dR has mass
rhodR where rho is the mass per cubic Km.
So the force on the element is the sum of all the gravitational pulls from
all the other elements. Taking another arbitary point p, r Km from the Hub
and at an angle theta from P, the total force on P is :
F=GrhodRM/R^2+SUM(GrhodRrhodr/((R-rcos(theta))^2+(rSin(theta))^2))
[For all valid values of R and r]
This force is proportional to the velocity of the star at this point so the
velocity distribition can be calculated from this equation.
The results show a region of Keplerian motion distribution merging into a
region of rigid rotation like a solid disk then a further region of
Keplerian motion distribution.
This might explain the shape of the spiral arms and their winding up. These
seem to be the remains of jets from the central black hole. Without the
original existence of a central black hole the accretion disk that forms the
rest of the galaxy cannot form.
There was a programming error in the previous calculation so here are the
new results and calculations
The results
At Radius 1 The velocity is 3.6389359455973
At Radius 101 The velocity is 1.52431631615868
At Radius 201 The velocity is 1.55069372627264
At Radius 301 The velocity is 1.60870229555325
At Radius 401 The velocity is 1.67724209061256
At Radius 501 The velocity is 1.7496636322689
At Radius 601 The velocity is 1.82318147441577
At Radius 701 The velocity is 1.89647424223648
At Radius 801 The velocity is 1.96887951991093
At Radius 901 The velocity is 2.04006314709391
At Radius 1001 The velocity is 2.10986529983757
At Radius 1101 The velocity is 2.17822253859875
At Radius 1201 The velocity is 2.24512588170758
At Radius 1301 The velocity is 2.31059724104851
At Radius 1401 The velocity is 2.37467574471415
At Radius 1501 The velocity is 2.43740961568833
At Radius 1601 The velocity is 2.49885127988138
At Radius 1701 The velocity is 2.55905440181375
At Radius 1801 The velocity is 2.61807209493829
At Radius 1901 The velocity is 2.67595585871699
At Radius 2001 The velocity is 2.73275496977679
At Radius 2101 The velocity is 2.78851615784806
At Radius 2201 The velocity is 2.84328345960904
At Radius 2301 The velocity is 2.89709818197942
At Radius 2401 The velocity is 2.94999893042079
At Radius 2501 The velocity is 3.00202167297609
At Radius 2601 The velocity is 3.05319982040775
At Radius 2701 The velocity is 3.10356430885747
At Radius 2801 The velocity is 3.15314367514874
At Radius 2901 The velocity is 3.20196411690493
At Radius 3001 The velocity is 3.25004953046469
At Radius 3101 The velocity is 3.29742151932278
At Radius 3201 The velocity is 3.3440993644957
At Radius 3301 The velocity is 3.39009994558674
At Radius 3401 The velocity is 3.43543759689357
At Radius 3501 The velocity is 3.4801238756706
At Radius 3601 The velocity is 3.52416720774069
At Radius 3701 The velocity is 3.56757235538296
At Radius 3801 The velocity is 3.61033961641955
At Radius 3901 The velocity is 3.65246359614928
At Radius 4001 The velocity is 3.69393126108735
At Radius 4101 The velocity is 3.73471870669331
At Radius 4201 The velocity is 3.77478545917129
At Radius 4301 The velocity is 3.81406368734198
At Radius 4401 The velocity is 3.85243601707504
At Radius 4501 The velocity is 3.88968522742307
At Radius 4601 The velocity is 3.92536510921522
At Radius 4701 The velocity is 3.95840478075579
At Radius 4801 The velocity is 3.98549161436798
At Radius 4901 The velocity is 3.98923461011976




End of Results

Chart of Results
The Program

Option Explicit
Dim FileName As String
Dim FileNumber
Private Sub cmdCalculate_Click()
Const pi = 3.1415926
Dim Theta, dTheta, Diameter, ThetaRad, dThetaRad, rho, R, ds, dR, s, G, M,
Sum, Velocity, Force As Double
Dim ret
Diameter = Val(txtDiameter.Text)
rho = Val(txtDensity.Text)
G = Val(txtG.Text)
M = Val(txtMass.Text)
dR = 100: ds = 10: dTheta = 10
dThetaRad = dTheta * pi / 180
FileNumber = FreeFile()
Open FileName For Output As FileNumber Len = 1
For R = 1 To Diameter / 2 Step dR
For s = 1 To Diameter / 2 Step ds
For Theta = 0 To 350 Step dTheta
ret = DoEvents()
ThetaRad = Theta * pi / 180
If Not (R = s And Theta = 0) Then
Sum = Sum + rho * dR * R * dThetaRad * rho * ds * s * dThetaRad / ((R - s *
Cos(ThetaRad)) ^ 2 + (s * Sin(ThetaRad)) ^ 2)
End If
Next Theta
Next s
Force = (Sum + (rho * R * dThetaRad * dR / R ^ 2)) * G
Sum = 0
Velocity = Sqr(Force * R) / (rho * R * dThetaRad * dR)
Print #FileNumber, "At Radius "; R; " The velocity is "; Velocity
Next R
Close FileNumber
ret = MsgBox("Calculation finished", vbInformation, "Galaxy")
End Sub

Private Sub cmdExit_Click()
End
End Sub

Private Sub cmdFileName_Click()
CommonDialog1.CancelError = True
On Error GoTo ErrorHandler
Rem CommonDialog1.Flags = cdlofnh1dereadonly
CommonDialog1.Filter = "Text Files (*.txt)|*.txt"
CommonDialog1.ShowSave
FileName = CommonDialog1.FileName
Exit Sub
ErrorHandler:
MsgBox ("Error")
End Sub


End of Program
For comparison, I calculated the Keplerian velocity distribution for a
single star orbiting the central black hole. Here are the results.
At Radius 1 The velocity is 1000
At Radius 101 The velocity is 99.5037190209989
At Radius 201 The velocity is 70.5345615858598
At Radius 301 The velocity is 57.6390417704235
At Radius 401 The velocity is 49.9376169438922
At Radius 501 The velocity is 44.676705160877
At Radius 601 The velocity is 40.7908508224002
At Radius 701 The velocity is 37.7694787300249
At Radius 801 The velocity is 35.3332626668787
At Radius 901 The velocity is 33.3148302326385
At Radius 1001 The velocity is 31.6069770620507
At Radius 1101 The velocity is 30.1374387339456
At Radius 1201 The velocity is 28.8554928412381
At Radius 1301 The velocity is 27.7243486500714
At Radius 1401 The velocity is 26.7165842572632
At Radius 1501 The velocity is 25.8112866459834
At Radius 1601 The velocity is 24.9921911602031
At Radius 1701 The velocity is 24.2464322484436
At Radius 1801 The velocity is 23.5636814813137
At Radius 1901 The velocity is 22.9355385129844
At Radius 2001 The velocity is 22.3550917004948
At Radius 2101 The velocity is 21.8165952144043
At Radius 2201 The velocity is 21.3152278159744
At Radius 2301 The velocity is 20.8469099612542
At Radius 2401 The velocity is 20.4081632653061
At Radius 2501 The velocity is 19.9960011996001
At Radius 2601 The velocity is 19.6078431372549
At Radius 2701 The velocity is 19.2414460721011
At Radius 2801 The velocity is 18.8948498713306
At Radius 2901 The velocity is 18.566333001717
At Radius 3001 The velocity is 18.2543764409228
At Radius 3101 The velocity is 17.9576340436322
At Radius 3201 The velocity is 17.6749080410067
At Radius 3301 The velocity is 17.4051286546177
At Radius 3401 The velocity is 17.1473370324297
At Radius 3501 The velocity is 16.9006708854465
At Radius 3601 The velocity is 16.6643523339933
At Radius 3701 The velocity is 16.4376775728237
At Radius 3801 The velocity is 16.220008041882
At Radius 3901 The velocity is 16.0107628501689
At Radius 4001 The velocity is 15.8094122478065
At Radius 4101 The velocity is 15.6154719791128
At Radius 4201 The velocity is 15.4284983795275
At Radius 4301 The velocity is 15.2480841032965
At Radius 4401 The velocity is 15.0738543882045
At Radius 4501 The velocity is 14.9054637793553
At Radius 4601 The velocity is 14.7425932467825
At Radius 4701 The velocity is 14.5849476421374
At Radius 4801 The velocity is 14.4322534482983
At Radius 4901 The velocity is 14.2842567828501



Chart of Velocity distribution for a Keplerian System

Program

Option Explicit
Dim FileName As String
Dim FileNumber
Private Sub cmdCalculate_Click()
Dim Mp, R, G, Diameter, Ms, dR, Velocity, Force As Double
Dim ret
Diameter = Val(txtDiameter.Text)
Mp = Val(txtDensity.Text)
G = Val(txtG.Text)
Ms = Val(txtMass.Text)
dR = 100
FileNumber = FreeFile()
Open FileName For Output As FileNumber Len = 1
For R = 1 To Diameter / 2 Step dR
Force = (Ms * Mp / R ^ 2) * G
Velocity = Sqr(Force * R) / Mp
Print #FileNumber, "At Radius "; R; " The velocity is "; Velocity
Next R
Close FileNumber
ret = MsgBox("Calculation finished", vbInformation, "System")
End Sub

Private Sub cmdExit_Click()
End
End Sub

Private Sub cmdFileName_Click()
CommonDialog1.CancelError = True
On Error GoTo ErrorHandler
Rem CommonDialog1.Flags = cdlofnh1dereadonly
CommonDialog1.Filter = "Text Files (*.txt)|*.txt"
CommonDialog1.ShowSave
FileName = CommonDialog1.FileName
Exit Sub
ErrorHandler:
MsgBox ("Error")
End Sub


End of Program
Chris.
20/07/2006

For the graphs http://www.chrisscrazyideas.co.uk/
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dlzc
science forum beginner


Joined: 05 Jul 2006
Posts: 5

PostPosted: Thu Jul 20, 2006 9:46 pm    Post subject: Re: The phrase 'dark matter' Reply with quote

Dear "The Real Chris"

The Real Chris wrote:
Quote:
Look below for enlightenment.

The pioneer anolomy is the same.

The Pioneer anomaly is 90 degrees to what you are talking about
"below".

Quote:
"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:net@nospam.com> wrote in
message news:RyXug.38482$AB3.25355@fed1read02...

The first thing to calculate before computing the
distribution of galactic rotation is the distribution of
mass in the galaxy. As a first approximation, I will
take the galaxy to be a central point mass (the black
hole) of 10 million solar masses and a flat disk of
uniformly distributed matter (stars) of 1 solar mass
per cubic light year.

It would be much more representative if you had the stars be more
sparse as you got towards the rim. Might get better results also. By
the way, how do you get from "per cubic light year" to "a flat disk"?

Quote:
The disk will have a diameter of 10,000 light years
and the thickness of 3 light years which I shall take
as a disk of zero thickness to start off with.

That's probably how you did it.

Quote:
I will use Newtonian mechanics throughout with a
conventional value of the gravitational constant.

Point P is R Km from the hub, the hub is M Kg and
the element dR has mass rho[.]dR where rho is the
mass per cubic Km. So the force on the element is
the sum of all the gravitational pulls from all the
other elements.

All *which* other elements? The ones inside the radius R, or what?

Quote:
Taking another arbitary point p, r Km from the Hub
and at an angle theta from P, the total force on P is :
F=GrhodRM/R^2+SUM(GrhodRrhodr/((R-rcos(theta))^2+(rSin(theta))^2))

F=G.rho.dR.M/R^2+SUM(G.rho.dR.rho.dr/((R-r.cos(theta))^2+(r.sin(theta))^2))
theta swept with 0 on the line drawn through the central BH and P...

Quote:
[For all valid values of R and r]

requiring R <= r?

Quote:
This force is proportional to the velocity of the star at this
point so the velocity distribition can be calculated from
this equation.

Why do you say "force is proportional to velocity"?

Quote:
The results show a region of Keplerian motion distribution
merging into a region of rigid rotation like a solid disk then
a further region of Keplerian motion distribution.

.... which isn't what is seen ...

Quote:
This might explain the shape of the spiral arms and their
winding up.

Not really. The spiral arms are the result of "osculation" (my intent
is oscillating above and below plane... this may be the wrong word),
with the stars acting very much like water waves, with a very slight
difference in wave speed from individual star speeds.

Quote:
These seem to be the remains of jets from the central
black hole.

BH jets are polar emissions. Frame dragging will almost assure the
Milky Way is rotating the same way the BH is.

Quote:
Without the original existence of a central black hole the
accretion disk that forms the rest of the galaxy cannot form.

There are galaxies without central BH. Globular clusters don't have
(for the most part), central BHs.

I won't comment on your code, until I get a chance to review it, and
its undelying assumptions. I have heartburn with your assumption that
velocity correlates directly with force.

David A. Smith
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tomgee1
science forum Guru


Joined: 31 Jan 2006
Posts: 750

PostPosted: Fri Jul 21, 2006 1:09 am    Post subject: Re: The phrase 'dark matter' Reply with quote

The TimeLord wrote:
Quote:
On Mon, 17 Jul 2006 17:27:11 -0700, "tomgee" <tyropress@yahoo.com> wrote
in <1153182431.684397.68990@m79g2000cwm.googlegroups.com>:

[...]
But that was based only on the notion that neutrinos had some mass. It
is only recently they have been shown to have some mass but not until
and unless they interact w/a tau neutrino.

Where is the source for that info? Based on Superkamiokande,
which verified that neutrinos could have mass, the neutrinos
don't have to interact with anything to "indicate mass".

I've read several accounts about what was observed

in the Big Pool in '98, but the more recent one was by
a ref. from SWormly in one of his Physics News posts.

The one I refer to more often is at
http://www.phys.hawaii.edu/~jgl/nuosc_story.html.
It's very long, but it seems to cover quite a lot.

I have not read anywhere that the neuts have to
interact with anything, that I can recall. That is my own
opinion. The results are based on the apparent loss of
about 1/2 the number of neuts that fall from overhead
versus the number from the far side of the earth. Also,
the experiment suggests that earlier claims of
"oscillations" having been observed did have some
basis in fact, as the effect of that in this case was
deemed consistent with related data. Such motions
are considered confirmation that neuts have mass.

That in itself is a great discovery, but it was also found
that only certain neuts oscillate, and that they do not
do that all the time. It seems that when it happens,
Cherenkov radiation occurs. Nothing better than a
light flash to tell you something just happened, eh?

Here's where it gets complicated, I think. Neuts do
not move as fast as light does in the vacuum of space,
but they are not restricted by matter the way light is, so
in the case where light and neuts are moving in water,
light only goes at 3/4 its regular speed in vacuum, which
shows that the speed of light is affected by matter. On
the other hand, neuts are not affected by matter, and so
in water they can achieve speeds that exceed the speed
of the light In The Water. At the instant that happens, the
Cherenkov radiation occurs. Whether or not that happens
when there is no light in the same water, I don't know.

One of the neut flavors interacts with another flavor, and
that results in the oscillations, indicating that mass has been
created. I would think that the radiation is to be expected,
as my model has predicted the process where photon
radiation is created from DM and em waves. When a DM
particle is struck by an em wave, some energy is imparted
to the DM particle, and that transfer of energy transforms
the DM particle into a real matter (RM) particle that flashes
briefly very similar to the Cherenkov radiation process.

The difference is that the neuts do not interact with each
other as often as the em waves interact with the DM
particles, and so their radiation is seen rarely compared
to the rated of collisions of em waves with DM particle.

I would say that neutrinos are DM particles just like photons
are before they become transformed into visible light. In my
model, the cause for what seems to be varying rates of
collisions between em waves and DM particles is not well-
developed as yet, but I'm working on that too.
Quote:

That means they are massless until they are transformed into particles
having mass and energy, just like my model predicted.

Model, what model? Where is it published?

It is self-published but it is not available on a website. For

readers of these ngs who show an interest in reading it and
then discussing it here in these ngs, I can email it for free. I
can snailmail a copy on a 3.25" floppy disk for $4.95USD, or
mail a printed copy for $9.95USD. Or, anyone can search
my posts from as far back as 1996 to find the many tidbits of
my model that I have posted here.
Quote:

[...]
Wait now. 1st you say, "...gravitate like ordinary matter." Next you
call it "gravitational mass", which does not equate to the same thing.
To have gravitation like we know it, it must have the force of
attraction, and that force is evident only in massive objects we can
observe. For us 2b able to see them, they must have energy and the
property of time, as we know those qualities 2b. However, we cannot see
DM, so we can only guess it has gravitational attraction like visible
objects have.

That's the whole point! You can't see them, because they don't
shine, but they have gravity, thus mass and thus are matter.

We don't know that for sure yet, and to jump to that conclusion

is not wise, as there could be other processes involved we do
not yet know about. That is one possibility, yes, but it is not the
only one.
Quote:

The gravity is acting on things that do shine. - Think of it
this way: Newton's Second Law of Motion says that the motion
doesn't change unless a force is applied. So if the motion of
a shining star changes, there should be a force on it, even if
you can't see the thing causing the force. If the force is
not electro-magnetism, weak or strong force, then what's left?
Of course gravity. Thus there is some thing you can't see (dark)
that is generating some gravity and thus has mass (matter).
Hence it's dark matter; whatever dark matter turns out to be.

But there is the possibility it can be something other than gravity

as we know it. It may be repulsive instead of attractive, and if
you say it is attractive and explains the effects to your satisfac-
tion, and later it is not attractive, you will have been wrong to be
so satisfied.

Or, it could be something else other than gravitational force, as
I mention below.
Quote:

Chew on this: How can additional gravitational attraction cause the
observed effects? Oh, sure, more attraction force will help keep

Gravity is a force. Force changes the motion of the thing it
interacts with. Duh!

But gravitation is only one of the four fundamental forces. Other

forces could work to produce gravitation-like effects, duh!
Quote:

galaxies together better, yes. But how does that explain the effect
where the outer orbiting bodies move faster than can be expected? And

Because they are *not* moving slower. Duh!

Yeah, right: Duh!

also, if DM is everywhere RM is not, how does it know what to attract

What is RM?

My acronym for real matter.

and what not to attract?

Same way gravity knows what to attract. Duh!

Gravity knows to attract RM only, duh!

It is quite possible that instead of gravitational attractive forces,
they are gravitational repulsive forces. Since DM appears 2b quite

If there are gravitational repulsive forces, you will need to
show how that happens. Negative mass is not defined since "matter
holes" don't really have any physical meaning.

Both Dirac and Gamow defined them well enough for me. Now,

once you show me how gravity attracts, I'll be glad to show you
how it may repulse.
Quote:

opposite to RM, why should we not think that its gravita- tional force -

I'll have to take your word on that since I don't know what you
mean by "RM".

[...]
Or, why cannot it be some force other than gravitational? It could be

Because it would show up in the spectra of the shining star we
could see.

Why would it? What other forces show up in the light spectrum?

just "interactional", to coin a word, and not gravita- tional at all.

Gravity is interactional. So what's your point?

All forces are interactional, not just gravitational forces, that's

why I excluded them from my statement above, to make my point.
Quote:

[...]
At this point, we only surmise the forces to be gravitational, and it
may indeed turn out that way. But at this point, I do not see all the
questions being answered by the same attractive force of RM gravitation,
while my model proposes that it is not just the gravitational forces
that change the universe second-by-second, but also the various types of
interactions that go on between the stuff of the universe.

I am open to other ideas as long as they are well-defined and
consistent with observation. However, just saying that accepted
ideas are wrong despite their consistency with observed fact,
is like saying UFOs must exist because the government is hiding
them.

It is a most consistent feature of history that 99 percent or more

of "accepted ideas" have been found to be wrong. The U. no
longer revolves around us, we can no longer fall off the edge of
the world, there are no turtles holding us up, there is no martian
invasion of earth, etc., etc.
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N:dlzc D:aol T:com (dlzc)
science forum Guru


Joined: 25 Mar 2005
Posts: 2835

PostPosted: Fri Jul 21, 2006 1:50 am    Post subject: Re: The phrase 'dark matter' Reply with quote

Sigh...

"dlzc" <dlzc1@cox.net> wrote in message
news:1153431976.257552.69680@i42g2000cwa.googlegroups.com...
Quote:
Dear "The Real Chris"

The Real Chris wrote:
Look below for enlightenment.

The pioneer anolomy is the same.

The Pioneer anomaly is 90 degrees to what
you are talking about "below".

.... and is effective only on these small bodies... not on the
planets being passed.

Quote:
"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:net@nospam.com
wrote in
message news:RyXug.38482$AB3.25355@fed1read02...

The first thing to calculate before computing the
distribution of galactic rotation is the distribution of
mass in the galaxy. As a first approximation, I will
take the galaxy to be a central point mass (the black
hole) of 10 million solar masses and a flat disk of
uniformly distributed matter (stars) of 1 solar mass
per cubic light year.

It would be much more representative if you had the
stars be more sparse as you got towards the rim.
Might get better results also. By the way, how do
you get from "per cubic light year" to "a flat disk"?

The disk will have a diameter of 10,000 light years
and the thickness of 3 light years which I shall take
as a disk of zero thickness to start off with.

That's probably how you did it.

I will use Newtonian mechanics throughout with a
conventional value of the gravitational constant.

Point P is R Km from the hub, the hub is M Kg and
the element dR has mass rho[.]dR where rho is the
mass per cubic Km. So the force on the element is
the sum of all the gravitational pulls from all the
other elements.

All *which* other elements? The ones inside the
radius R, or what?

Taking another arbitary point p, r Km from the Hub
and at an angle theta from P, the total force on P is :
F=GrhodRM/R^2+SUM(GrhodRrhodr/((R-rcos(theta))^2+(rSin(theta))^2))

F=G.rho.dR.M/R^2+SUM(G.rho.dR.rho.dr/((R-r.cos(theta))^2+(r.sin(theta))^2))
theta swept with 0 on the line drawn through the
central BH and P...

[For all valid values of R and r]

requiring R <= r?

Grrr...
r <= R?

Quote:
This force is proportional to the velocity of the star at this
point so the velocity distribition can be calculated from
this equation.

Why do you say "force is proportional to velocity"?

The results show a region of Keplerian motion distribution
merging into a region of rigid rotation like a solid disk then
a further region of Keplerian motion distribution.

... which isn't what is seen ...

This might explain the shape of the spiral arms and their
winding up.

Not really. The spiral arms are the result of "osculation"
(my intent is oscillating above and below plane... this may
be the wrong word), with the stars acting very much like

.... molecule in ...

Quote:
water waves, with a very slight difference in wave speed
from individual star speeds.

These seem to be the remains of jets from the central
black hole.

BH jets are polar emissions. Frame dragging will
almost assure the Milky Way is rotating the same way
the BH is.

Without the original existence of a central black
hole the accretion disk that forms the rest of the
galaxy cannot form.

There are galaxies without central BH. Globular
clusters don't have (for the most part), central BHs.

I won't comment on your code, until I get a chance
to review it, and its undelying assumptions. I have
heartburn with your assumption that velocity
correlates directly with force.

David A. Smith
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Randy Poe
science forum Guru


Joined: 24 Mar 2005
Posts: 2485

PostPosted: Fri Jul 21, 2006 2:01 am    Post subject: Re: The phrase 'dark matter' Reply with quote

tomgee wrote:
Quote:
It is a most consistent feature of history that 99 percent or more
of "accepted ideas" have been found to be wrong.

It is a consistent feature of the internet that 87.43% of the
statistics you find people quoting are made up, and wrong.
Including this one.

- Randy
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tomgee1
science forum Guru


Joined: 31 Jan 2006
Posts: 750

PostPosted: Fri Jul 21, 2006 5:46 am    Post subject: Re: The phrase 'dark matter' Reply with quote

The TimeLord wrote:
Quote:
On Mon, 17 Jul 2006 14:11:25 -0700, "tomgee" <tyropress@yahoo.com> wrote
in <1153170685.919021.109060@35g2000cwc.googlegroups.com>:


Randy Poe wrote:
socratus wrote:
[...]
Yes, they are. DM is not dark at all, so that term used to describe it
or even define it is clearly a misnomer no matter how hard you wish upon
a star that it isn't. In fact, DM can be more accurately described as
invisible matter because we can see right through it, obviously.

Through it? How do you know? Generally stars don't resolve
below the Raleigh limit anyway in a telescope. So how would
you be able to tell that dark matter is transparent?

Because we can see through it, just like we see through light,

apparently. We are talking about two states: visible and
invisible. Those researching DM state today state that it is
invisible and can only be detected by its observed effects.

We can see "through" light, as explained in my model, because
we only see one layer of it at a time, and each layer gives us so
much information that changes with succeeding em waves, that
we see the universe similar to the way we see movies as if they
are real people up on the screens. So nothing is really trans-
parent to us, it just seems that way. We cannot see beyond our
retinal screens, we can see only the light that strikes them wave
by innumerable wave.

It appears to us that space is empty between objects, so if DM
exists, we do not "see" it the same way we do not "see" light.
They are two different things that appear to be transparent to
our eyes, but for different reasons. Light is energy as we know
it, but DM is not, since it seems to exist even where there is no
light. Recent reports show that it is everywhere objects are not.
Quote:

Physics is not all that precise in term-usage as you once imagined, eh?

That statement is moot.

Yes, it is. Sorry.

We assume it has gravitational mass, based on the observed effects, but
that conclusion is only one among others that are possible; therefore,

If dark matter is not interacting with the stars gravitationally,
then what is the underlying force? It can't be electromagnetic,
because that force would be evidenced in the spectra (Zeeman
splitting and/or synchrotron radiation).

Yes, I agree that it would be evidenced in the spectra, but how

do we know it is not? The Cherenkov radiation is evidence of
the creation of mass, and my model shows the creation of light
in a very similar way, and perhaps if we looked for it we might
find in the interactions between between other energy
exchanges such effects that may not create matter but only
act to influence it in other ways. Perhaps I'm just trying to
stay out of the same box everyone seems to have lept into
in the excitement of new discoveries.
Quote:

It can't be weak force,
since that only involves nuclear decay. It can't be the strong
force, since that only involves the nuclei. So what's left if it's
not gravity?

it is not a fact yet, as you assert. Also, to say it is call "dark"
because it does not radiate is another phony explanation, since
blackbodies do not radiate either, they only absorb radiation. For such

Blackbodies do radiate. Example = an incandescent light bulb
(3000 degK black-body radiation). Another example = the sun
(5600 degK black-body radiation).

Yes, of course blackbodies radiate. I was wrong about all

that. It was a bad day for me and I had a hard time concen-
trating on what I was doing. I do not know enough about
blackbodies to have tried to use them as an example, and I
should have known better.
Quote:

SNIP

The closest thing to a blackbody I can think of is a black hole, which

Bull. Blackbodies and black holes are completely different.
Just look at their definitions in any basic astronomy book.

That does show how little I know about blackbodies. Sorry.

It's clear you don't know about this. You should be posting this
to alt.sci.physics.new-theories, which is an NG devoted to crank
ideas.

I am grateful you said no more than that. Others would not be

so lenient. As I've said before, I am just as fallible as others,
esp. when I try to respond to everyone in a single day. I am a
skeptic, but I'm not a pessimist, and tomorrow is another day!
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