A while back I was discussing with Paul Stowe the
expression on his web site for the heating caused by
LeSagian gravity. Stowe derives a formua for heat
emission that is of the form:

(constant)x(mass)/(radius).

I did not understand why there was a radius
involved and was referred to
_Pushing Gravity_, espcially the article by Slabinski.

Well, I finally got Pushing Gravity through interlibraty
loan.

On page 191, equation (26) in the article by
Mingst and Stowe, agrees with the above formula that was
under discussion.

But on page 127, equation (19) in the article by
Slabinski has the form

(constant)x(mass).

The radius of the body is not involved.
Thus Slabinski and Mingst/Stowe disagree.

What gives?

Tom

"TC" wrote in message oups.com...
A while back I was discussing with Paul Stowe the
expression on his web site for the heating caused by
LeSagian gravity. Stowe derives a formua for heat
emission that is of the form:

(constant)x(mass)/(radius).

I did not understand why there was a radius
involved and was referred to
_Pushing Gravity_, espcially the article by Slabinski.

Well, I finally got Pushing Gravity through interlibraty
loan.

On page 191, equation (26) in the article by
Mingst and Stowe, agrees with the above formula that was
under discussion.

But on page 127, equation (19) in the article by
Slabinski has the form

(constant)x(mass).

The radius of the body is not involved.
Thus Slabinski and Mingst/Stowe disagree.

What gives?

So they have been playing with your feet to keep you busy.
That's what trolls do :-)

Dirk Vdm

TC:
A while back I was discussing with Paul Stowe the
expression on his web site for the heating caused by
LeSagian gravity. Stowe derives a formua for heat
emission that is of the form:

(constant)x(mass)/(radius).

I did not understand why there was a radius
involved and was referred to
_Pushing Gravity_, espcially the article by Slabinski.

Well, I finally got Pushing Gravity through interlibraty
loan.

On page 191, equation (26) in the article by
Mingst and Stowe, agrees with the above formula that was
under discussion.

But on page 127, equation (19) in the article by
Slabinski has the form

(constant)x(mass).

The radius of the body is not involved.
Thus Slabinski and Mingst/Stowe disagree.

What gives?


You've been duped into wasting your time trying to come up
with an argument against a well discredited idea for the sake of
trying to persudae two kooks who don't even grasp basic classical
physics why it's unphysical. You'd have a better shot trying to
convince the pope to become a rabbi at a jewish temple and his
argument for not doing so would be more plausible, even to an
athiest.

Dirk Vdm wrote:

"TC" wrote in message

[snip crap]

What gives?

So they have been playing with your feet to keep you busy.
That's what trolls do :-)

Well yeah, but don't expect your comment to influence Tom Clarke.
I can think of no one in this newsgroup who has attempted to give
more legitimacy to the extreme crank contingent than Clarke. I
suggest he approach the department chair or other faculty in the
UCF physics department, and check their level of interest toward
the "researchers" who are published by Apeiron. Who knows, maybe
Clarke can be the proud host of a colloquium of academia-bashers.


---Tim Shuba---

On 21 Apr 2005 07:53:16 -0700, "TC" wrote:

A while back I was discussing with Paul Stowe the
expression on his web site for the heating caused by
LeSagian gravity. Stowe derives a formua for heat
emission that is of the form:

(constant)x(mass)/(radius).

I did not understand why there was a radius involved and
was referred to _Pushing Gravity_, espcially the article
by Slabinski.

Ummm, actually that was to help you with the question of
'excess' heating. However, you have a good question, and
I will address it this weekend (when I have more time to
devote). Note however Slabinski's equation is not one of
specific heat (watts/m^2), as is ours, but of total heat
(watts).

Well, I finally got Pushing Gravity through interlibraty
loan.

On page 191, equation (26) in the article by Mingst and Stowe,
agrees with the above formula that was under discussion.

But on page 127, equation (19) in the article by Slabinski
has the form

(constant) x (mass).

The radius of the body is not involved. Thus Slabinski and
Mingst/Stowe disagree.

Actually, it is. But it's buried in the solid angle.

More on the weekend.

However, note the total lack of any technical content OR attempt
to address your actual question but the cynics around here...

Paul Stowe

What gives?

How does one determine the the solid angle???

Paul Stowe

Paul Stowe wrote:
On 21 Apr 2005 07:53:16 -0700, "TC" wrote:

Mingst/Stowe

On page 191, equation (26) has form
(constant)x(mass)/(radius).

page 127, equation (19) has form
Slabinksi (constant)x(mass)

... Note however Slabinski's equation is not one of
specific heat (watts/m^2), as is ours, but of total heat
(watts).

Then if Slabinski is correct, your equation should
take the form.

(constant)x(mass)/(radius^2)

.....

The radius of the body is not involved. Thus Slabinski and
Mingst/Stowe disagree.

Actually, it is. But it's buried in the solid angle.

Slabinski works out a different functional form for
heating. Is he wrong then? Does he not account for
angles correctly?

More on the weekend.

I await your explanation.

Tom

shuba wrote:
Dirk Vdm wrote:

"TC" wrote in message

[snip crap]

What gives?

So they have been playing with your feet to keep you busy.
That's what trolls do :-)

Well yeah, but don't expect your comment to influence Tom Clarke.
I can think of no one in this newsgroup who has attempted to give
more legitimacy to the extreme crank contingent than Clarke. I
suggest he approach the department chair or other faculty in the
UCF physics department, and check their level of interest toward
the "researchers" who are published by Apeiron. Who knows, maybe
Clarke can be the proud host of a colloquium of academia-bashers.

That hurts a bit.

Why do you think I got the book through
interlibrary loan instead of purchasing it? I don't want
to monetarily support them. The scary thing is that the
book came, not from a university library, but from the
Ocala public library where impressionable youngsters
interested in science might check it out and think it
real science.

I remember picking up Velikowsky's Worlds in Collision
at the library as a child and thinking - briefly - that
it might be legitimate. But I guess even then I had decent
BS filters as I didn't even finish the book.

Do you really think my attempts to convince the cranks of the
errors of their ways by talking to them as if they weren't cranks
is harmful? That it gives them legitimacy?

I really thought it was at worst just a harmless hobby.

Tom

"TC" wrote in message oups.com...

shuba wrote:
Dirk Vdm wrote:

"TC" wrote in message

[snip crap]

What gives?

So they have been playing with your feet to keep you busy.
That's what trolls do :-)

Well yeah, but don't expect your comment to influence Tom Clarke.

My comment was not really intended to influence Tom, but
rather to tease the trolls.

I can think of no one in this newsgroup who has attempted to give
more legitimacy to the extreme crank contingent than Clarke.

I think Tom did a remarkably fine job at it :-)

I
suggest he approach the department chair or other faculty in the
UCF physics department, and check their level of interest toward
the "researchers" who are published by Apeiron. Who knows, maybe
Clarke can be the proud host of a colloquium of academia-bashers.

That hurts a bit.

Why do you think I got the book through
interlibrary loan instead of purchasing it? I don't want
to monetarily support them. The scary thing is that the
book came, not from a university library, but from the
Ocala public library where impressionable youngsters
interested in science might check it out and think it
real science.

I remember picking up Velikowsky's Worlds in Collision
at the library as a child and thinking - briefly - that
it might be legitimate. But I guess even then I had decent
BS filters as I didn't even finish the book.

Do you really think my attempts to convince the cranks of the
errors of their ways by talking to them as if they weren't cranks
is harmful? That it gives them legitimacy?

I really thought it was at worst just a harmless hobby.

Keep up the good work, Tom :-)

Dirk Vdm

On 22 Apr 2005 05:43:44 -0700, "TC" wrote:

Paul Stowe wrote:
On 21 Apr 2005 07:53:16 -0700, "TC" wrote:

Mingst/Stowe

On page 191, equation (26) has form constant)x(mass)/(radius).

On page 127, equation (19) has form (constant)x(mass) [Slabinksi]

... Note however Slabinski's equation is not one of specific
heat (watts/m^2), as is ours, but of total heat (watts).

Then if Slabinski is correct, your equation should take the form.
(constant)x(mass)/(radius^2)

Fine, point to the specific step in the following that is either,

- mathematically wrong, or,
- logically inconsistent

We start will conservation and say, at equilibrium,

q_in = q_out

Where q is the power flux per unit area of the graviton field. Note
that q is the current, not omni-directional flux. For ease of
notation we'll set q_in = q & q_out = q'... Then for an attenuating
mass we say,
-ß -ß
q = q' = qe + q(1 - e )

The ß term is the total attenuation parameter. Clearly, if ß - 0
then we have,

q' = q + (q - q) = q

And, if ß - oo then,

q' = 0 + (q - 0) = q

In he first case, nothing interacts, in the second, all interacts &
is ultimately re-emitted as a secondary flux. Either way, in
equilibrium is strictly maintained!

Thus, the 'delta' or interacting component of q is (q''),

-ß
q'' = q(1 - e )

Therefore, when ß unity the Taylor series shows us that this
can be quantified (to a very high precision) by simply writing

q'' = qß

The ß term is an expression of the departure from equilibrium of
the graviton fluence. This can be found on page 190 Eq(22) and
is given as 2GM/rc^2. Now follow this through

q'' = q(2GM/rc^2)

Regrouping this we get,

q'' = (q2G/c^2)M/r

Finally, we know that within LeSage's model,

G = ¿µ^2

and

q = ¿c/4pi

Thus,

(q2G/c^2) = ([¿µ]^2/[2pi]c) = k

q'' = kM/r


Now, to Slabinski version. ...

In our analysis above we've taken a 'big picture' or macroscopic
continuum approach to the issue. In Slabinki's work he take the
microscopic or kinetic theory type approach. You could say ours
was a top down and Slabinki's a bottom up analysis. We must also
quantify Slabinki's terms and map them to their counterparts in
our approach. Slabinki defines,

N = graviton particle density (particles per unit volume)
A = test area (length squared)
A' = absorption cross-sectional area for the smallest possible
interacting material particle... (length squared)
A''= scattering cross-sectional area for the smallest possible
interacting material particle... (length squared)
ç = Solid angle sutended by A (Radians)
K = mass absorption coefficient (length squared per unit mass)
K' = mass scattering coefficient (length squared per unit mass)
m = m, m' test particles of gravitating mass
r = distance of A from m
R = Rates (R, R', R'') net, direct, scatter
£ = net decrease in graviton flux density
c = graviton mean speed
w = graviton mass

Mapping into our version.

¿ = Nwc^2
µ = Sqrt[K(K + K'[1 - Cos æ])]

Thus mapping Slabinki's Eq 19 we get,

H = (¿2piKc)m

Dimensionally this is,

kg | m^2 | kg | m kg-m^2
-------+-----+----+--- = ------
m-sec^2| kg | |sec sec^3

Converting to a per unit area (4piL^2)we get,

(¿Kc/2)m/L^2 = kg/sec^3

Note that Slabinki is evaluating the test area of a single
interacting differential particle of matter. A, A', A''
as well as ç and æ are set by this. HE IS NOT! evaluating
a macroscopic body consisting of multiple test particles.

The 'solid' angles ç and æ are affected by size. However,
for his analysis size doesn't change. Note, area is
proportional to r^2 and density to r^3, a 1/r differential.

Paul Stowe

Paul Stowe wrote:
On 22 Apr 2005 05:43:44 -0700, "TC" wrote:

Paul Stowe wrote:
On 21 Apr 2005 07:53:16 -0700, "TC" wrote:

Mingst/Stowe

On page 191, equation (26) has form constant)x(mass)/(radius).

On page 127, equation (19) has form (constant)x(mass) [Slabinksi]

... Note however Slabinski's equation is not one of specific
heat (watts/m^2), as is ours, but of total heat (watts).

Then if Slabinski is correct, your equation should take the form.
(constant)x(mass)/(radius^2)

Fine, point to the specific step in the following that is either,

- mathematically wrong, or,
- logically inconsistent

We start will conservation and say, at equilibrium,

q_in = q_out

Where q is the power flux per unit area of the graviton field. Note
that q is the current, not omni-directional flux. For ease of
notation we'll set q_in = q & q_out = q'... Then for an attenuating
mass we say,
-ß -ß
q = q' = qe + q(1 - e )

The ß term is the total attenuation parameter. Clearly, if ß - 0
then we have,

Actually this expression is true for any value of beta, you have
just added and substracted teh same thing, q e to the minus beta.

q' = q + (q - q) = q

And, if ß - oo then,

q' = 0 + (q - 0) = q

True also for any value of beta, as noted.

In he first case, nothing interacts, in the second, all interacts &
is ultimately re-emitted as a secondary flux. Either way, in
equilibrium is strictly maintained!

It's just adding and subtracting the same thing.

Thus, the 'delta' or interacting component of q is (q''),

-ß
q'' = q(1 - e )

Why "thus"? I would tend to think that this is the non-interacting
component. e to the minus beta would be the term describing
absorption and thus interaction.

Therefore, when ß unity the Taylor series shows us that this
can be quantified (to a very high precision) by simply writing

q'' = qß

We all know how Taylor series work.

The ß term is an expression of the departure from equilibrium of
the graviton fluence. This can be found on page 190 Eq(22) and
is given as 2GM/rc^2. Now follow this through

Unfortunately, I left the book at my desk. But that doesn't sound
correct. The absorbtion should just be determined by M.
r should not have anything to do with it.
Probably this is where the r term comes in in your
expressions that Slabinsky does not have in his.

q'' = q(2GM/rc^2)

Regrouping this we get,

q'' = (q2G/c^2)M/r

Finally, we know that within LeSage's model,

G = ¿µ^2

and

q = ¿c/4pi

Thus,

(q2G/c^2) = ([¿µ]^2/[2pi]c) = k

q'' = kM/r


Now, to Slabinski version. ...

In our analysis above we've taken a 'big picture' or macroscopic
continuum approach to the issue. In Slabinki's work he take the
microscopic or kinetic theory type approach. You could say ours
was a top down and Slabinki's a bottom up analysis. We must also
quantify Slabinki's terms and map them to their counterparts in
our approach. Slabinki defines,

N = graviton particle density (particles per unit volume)
A = test area (length squared)
A' = absorption cross-sectional area for the smallest possible
interacting material particle... (length squared)
A''= scattering cross-sectional area for the smallest possible
interacting material particle... (length squared)
ç = Solid angle sutended by A (Radians)
K = mass absorption coefficient (length squared per unit mass)
K' = mass scattering coefficient (length squared per unit mass)
m = m, m' test particles of gravitating mass
r = distance of A from m
R = Rates (R, R', R'') net, direct, scatter
£ = net decrease in graviton flux density
c = graviton mean speed
w = graviton mass

Mapping into our version.

¿ = Nwc^2
µ = Sqrt[K(K + K'[1 - Cos æ])]

I really wish you wouldn't use that upside down question mark. If you
mean
psi, write psi, please.

Thus mapping Slabinki's Eq 19 we get,

H = (¿2piKc)m

Heat absorbed and thus re-emitted is a constant times the mass.

Dimensionally this is,

kg | m^2 | kg | m kg-m^2
-------+-----+----+--- = ------
m-sec^2| kg | |sec sec^3

Converting to a per unit area (4piL^2)we get,

(¿Kc/2)m/L^2 = kg/sec^3

Note that Slabinki is evaluating the test area of a single
interacting differential particle of matter. A, A', A''
as well as ç and æ are set by this. HE IS NOT! evaluating
a macroscopic body consisting of multiple test particles.

A macroscopic body is comprised of test particles.
The sum of the heat input to test particles should give the
value for the macroscopic body. I guess is beta 1 there
might be shielding of the interior that would make the sum
not proportional to total mass, but you explicitly assume
beta 1.

The 'solid' angles ç and æ are affected by size.

The sum of the solid angles is the sum of the solid angles.

However,
for his analysis size doesn't change. Note, area is
proportional to r^2 and density to r^3, a 1/r differential.

This makes no sense.

Tom