From: FrediFizzx on
"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message
news:kdgi7356het3mddjjcjplvae265lsmbhc7(a)4ax.com...
> On Tue, 19 Jun 2007 22:15:43 -0700, "FrediFizzx"
> <fredifizzx(a)hotmail.com> wrote:
>
>>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message
>>news:l20h7359ctfot73g0qeng6mg9d1h3e28te(a)4ax.com...
>>> On Tue, 19 Jun 2007 14:51:53 -0700, "FrediFizzx"
>>> <fredifizzx(a)hotmail.com> wrote:
>>
>>>>I suppose the main argument is whether or not the "vacuum" is
>>>>polarizable. SI'ers are convinced that CGS buries "vacuum"
>>>>polarization
>>>>when it doesn't necessarily do so. If the "vacuum" is polarizable,
>>>>then
>>>>we will have "vacuum" capacitance and "vacuum" inductance. In order
>>>>to
>>>>have those, there must be "vacuum" charge. Which can be expressed
>>>>as
>>>>"Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must be
>>>>all >>>bound.
> Your vacuum charge is defined by hbar and c, whose definitions are not
> in dispute in either system of units.
> The product hbar*c can be shown in SI as equal to 3.16e-16 meter*volt*
> coulombs. We don't want to take the square root of such a thing, and
> call it charge. We discussed this before. We get relief by multiplying
> by Coulomb's constant, in other words re-introduce eps0, so that now
> we try hbar*c/eps0 whose square root is 5.974e-8 volt*meters. Now you
> have a cgs representation of charge in esu's voltmeters.
> The whole cgs system looks increasingly more synthetic with each
> foray.
>
>>>>SI's statement
>>>>
>>>>c^2 = 1/(eps0 mu0), (1)
>>>>
>>>>is merely shorthand for
>>>>
>>>>c^2 = lambda^2/(4pi^2 Cvac Lvac), (2)
>
> No, Eq. (1) is not shorthand for Eq. (2).

If Quantum Vacuum Charge (QVC) exists, then eq. 1 is indeed shorthand
for eq. 2.

> Eq. (1) conveys the substantive fact that c depends on two simple
> constant vacuum parameters, Their product yields c and their quotient
> yields Z.

Sure, but eq. 1 doesn't really tell us why they exist where perhaps eq.
2 gives us more insight. What it looks like to me is that eps0 and mu0
can really only exist if QVC exists.

> Your Eq. (2) is firstly a simple kinematic declaration that a
> broadcast wavelength times the broadcast frequency equals the velocity
> of transmission. which can be ascertained by drawing sine waves on
> paper. You make this velocity c, which is OK, but in the second place,
> you go too far in naming parameters Lvac and Cvac. Are they properties
> of the vacuum or of the vagaries of the broadcast transmitter?

They (Lvac and Cvac) are probably actually just a different way to look
at the electric and magnetic fields of EM radiation. Of course they are
not like capacitance and inductance of real caps and inductors.

> What is the other constraint (e.g., Z) that allows you to calculate
> Cvac and Lvac? I can make Cvac be 1 farad or any value, then solving
> for Lvac. These C and L's can't be properties of the vacuum; they must
> have the happy values that would satisfy F_brdcst.

Cvac and Lvac can't be calculated any more than k_e or k_m can be.
However, it is not hard to postulate what their values might be based on
eps0 and mu0.

Cvac = 2 eps0 lambda

Lvac = mu0 lambda/8pi^2

When converted to natural units we obtain,

Cvac = Lvac = lambda/2pi

Simply the rationalized wavelength of the EM radiation under
consideration. And a thing of beauty since they end up equal to each
other.

> There is no new information in (2) and it does not in any way resemble
> (1).

Well, I think my simple demonstration above does possibly give us some
new information.

Best,

Fred Diether
Moderator sci.physics.foundations

From: Greg Neill on
"Spirit of Truth" <juneharton(a)prodigy.net> wrote in message
news:RQoei.395$K44.183(a)newssvr13.news.prodigy.net...
>
> "Greg Neill" <gneillREM(a)OVEsympatico.ca> wrote in message
> news:466d2fd5$0$21103$9a6e19ea(a)news.newshosting.com...
> > "Spirit of Truth" <juneharton(a)prodigy.net> wrote in message
> > news:kk3bi.1304$TC1.1254(a)newssvr17.news.prodigy.net...
> >
> >> Greg, you are coping out again. The lack of simultaneity example in
there
> >> shows exactly why lack of simultaneity is false.
> >
> > Simultaneity is an observer dependent thing, and should
> > be (to any thinking individual) an obvious consequence of
> > the finite speed of light.
>
> Greg, the example has a reality of a destroyed train in one frame
> and no destroyed train in the other frame...both going into the future
> ...thus I guess you must believe in a zillion universes?

No, it does not. This is another case of your
criticizing a theory without understanding it
or its mathematics. Once again I feel that we
have no basis for a productive discussion, so
I will bow out. Have fun.


From: John C. Polasek on
On Thu, 21 Jun 2007 00:24:00 -0700, "FrediFizzx"
<fredifizzx(a)hotmail.com> wrote:

>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message
>news:kdgi7356het3mddjjcjplvae265lsmbhc7(a)4ax.com...
>> On Tue, 19 Jun 2007 22:15:43 -0700, "FrediFizzx"
>> <fredifizzx(a)hotmail.com> wrote:
>>
>>>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message
>>>news:l20h7359ctfot73g0qeng6mg9d1h3e28te(a)4ax.com...
>>>> On Tue, 19 Jun 2007 14:51:53 -0700, "FrediFizzx"
>>>> <fredifizzx(a)hotmail.com> wrote:
>>>
>>>>>I suppose the main argument is whether or not the "vacuum" is
>>>>>polarizable. SI'ers are convinced that CGS buries "vacuum"
>>>>>polarization
>>>>>when it doesn't necessarily do so. If the "vacuum" is polarizable,
>>>>>then
>>>>>we will have "vacuum" capacitance and "vacuum" inductance. In order
>>>>>to
>>>>>have those, there must be "vacuum" charge. Which can be expressed
>>>>>as
>>>>>"Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must be
>>>>>all >>>bound.
>> Your vacuum charge is defined by hbar and c, whose definitions are not
>> in dispute in either system of units.
>> The product hbar*c can be shown in SI as equal to 3.16e-16 meter*volt*
>> coulombs. We don't want to take the square root of such a thing, and
>> call it charge. We discussed this before. We get relief by multiplying
>> by Coulomb's constant, in other words re-introduce eps0, so that now
>> we try hbar*c/eps0 whose square root is 5.974e-8 volt*meters. Now you
>> have a cgs representation of charge in esu's voltmeters.
>> The whole cgs system looks increasingly more synthetic with each
>> foray.
>>
>>>>>SI's statement
>>>>>
>>>>>c^2 = 1/(eps0 mu0), (1)
>>>>>
>>>>>is merely shorthand for
>>>>>
>>>>>c^2 = lambda^2/(4pi^2 Cvac Lvac), (2)
>>
>> No, Eq. (1) is not shorthand for Eq. (2).
>
>If Quantum Vacuum Charge (QVC) exists, then eq. 1 is indeed shorthand
>for eq. 2.

It's always a red flag for me if equations can't have units affixed.
Please say what the units are for QVC, Lvac and Cvac.
You also would do us a favor by rewriting, in cgs, my SI equations in
the other note for a field of 1 million volts/meter. It should be
simple arithmetic. What is D and what is H?

>> Eq. (1) conveys the substantive fact that c depends on two simple
>> constant vacuum parameters, Their product yields c and their quotient
>> yields Z.
>
>Sure, but eq. 1 doesn't really tell us why they exist where perhaps eq.
>2 gives us more insight. What it looks like to me is that eps0 and mu0
>can really only exist if QVC exists.

No, of course one equation cannot tell what eps0 is. For that you have
the discoveries in my Dual Space book, or the permittivity paper at
http://www.dualspace.net.
It demonstrates that, in order to reproduce eps0, you need an
electron-positron pair elastically restrained in a cubic cell of size
L = 3.514x10^-14m with spring constant of
K = 2.612x10^14 N/m.
2e/L^3 makes a charge density P of 7.2x10^21 coul/m^3.
Then Eq. 14 for eps0 gives the whole picture:
eps0 = 2e*P/K = 8.854x10^-12 farad/meter
It makes the very sensible statement that permittivity is proportional
to e (for Ee = force), times charge density, and inverse to the
aforesaid spring constant K. It all works and every one of a dozen
parameters of pairspace are "forced" by logical laws without
hypothesis.
Unfortunately, the paper is not something to be speed-read nor is it,
in retrospect, very lucid.
>
>> Your Eq. (2) is firstly a simple kinematic declaration that a
>> broadcast wavelength times the broadcast frequency equals the velocity
>> of transmission. which can be ascertained by drawing sine waves on
>> paper. You make this velocity c, which is OK, but in the second place,
>> you go too far in naming parameters Lvac and Cvac. Are they properties
>> of the vacuum or of the vagaries of the broadcast transmitter?
>
>They (Lvac and Cvac) are probably actually just a different way to look
>at the electric and magnetic fields of EM radiation. Of course they are
>not like capacitance and inductance of real caps and inductors.
>
>> What is the other constraint (e.g., Z) that allows you to calculate
>> Cvac and Lvac? I can make Cvac be 1 farad or any value, then solving
>> for Lvac. These C and L's can't be properties of the vacuum; they must
>> have the happy values that would satisfy F_brdcst.
>
>Cvac and Lvac can't be calculated any more than k_e or k_m can be.
I just calculated ke (eps0?) above.
>However, it is not hard to postulate what their values might be based on
>eps0 and mu0.
>
>Cvac = 2 eps0 lambda
Farads, no? But you don't have eps0. You just have capacity in
centimeters.
>
>Lvac = mu0 lambda/8pi^2
Got to be henries. But you don't have mu0.
>When converted to natural units we obtain,
>
>Cvac = Lvac = lambda/2pi
Natural units; don't make me sick. It says farads = henries = cm's.

>Simply the rationalized wavelength of the EM radiation under
>consideration. And a thing of beauty since they end up equal to each
>other.
You should be revulsed.
>> There is no new information in (2) and it does not in any way resemble
>> (1).
>
>Well, I think my simple demonstration above does possibly give us some
>new information.
>
>Best,
>
>Fred Diether
>Moderator sci.physics.foundations
John Polasek
From: FrediFizzx on
"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message
news:5u1l7350jqkok44stf7mp6argpjiqavmdt(a)4ax.com...
> On Thu, 21 Jun 2007 00:24:00 -0700, "FrediFizzx"
> <fredifizzx(a)hotmail.com> wrote:
>
>>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message
>>news:kdgi7356het3mddjjcjplvae265lsmbhc7(a)4ax.com...
>>> On Tue, 19 Jun 2007 22:15:43 -0700, "FrediFizzx"
>>> <fredifizzx(a)hotmail.com> wrote:
>>>
>>>>"John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message
>>>>news:l20h7359ctfot73g0qeng6mg9d1h3e28te(a)4ax.com...
>>>>> On Tue, 19 Jun 2007 14:51:53 -0700, "FrediFizzx"
>>>>> <fredifizzx(a)hotmail.com> wrote:
>>>>
>>>>>>I suppose the main argument is whether or not the "vacuum" is
>>>>>>polarizable. SI'ers are convinced that CGS buries "vacuum"
>>>>>>polarization
>>>>>>when it doesn't necessarily do so. If the "vacuum" is
>>>>>>polarizable,
>>>>>>then
>>>>>>we will have "vacuum" capacitance and "vacuum" inductance. In
>>>>>>order
>>>>>>to
>>>>>>have those, there must be "vacuum" charge. Which can be expressed
>>>>>>as
>>>>>>"Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must
>>>>>>be
>>>>>>all >>>bound.
>>> Your vacuum charge is defined by hbar and c, whose definitions are
>>> not
>>> in dispute in either system of units.
>>> The product hbar*c can be shown in SI as equal to 3.16e-16
>>> meter*volt*
>>> coulombs. We don't want to take the square root of such a thing, and
>>> call it charge. We discussed this before. We get relief by
>>> multiplying
>>> by Coulomb's constant, in other words re-introduce eps0, so that now
>>> we try hbar*c/eps0 whose square root is 5.974e-8 volt*meters. Now
>>> you
>>> have a cgs representation of charge in esu's voltmeters.
>>> The whole cgs system looks increasingly more synthetic with each
>>> foray.
>>>
>>>>>>SI's statement
>>>>>>
>>>>>>c^2 = 1/(eps0 mu0), (1)
>>>>>>
>>>>>>is merely shorthand for
>>>>>>
>>>>>>c^2 = lambda^2/(4pi^2 Cvac Lvac), (2)
>>>
>>> No, Eq. (1) is not shorthand for Eq. (2).
>>
>>If Quantum Vacuum Charge (QVC) exists, then eq. 1 is indeed shorthand
>>for eq. 2.
>
> It's always a red flag for me if equations can't have units affixed.
> Please say what the units are for QVC, Lvac and Cvac.

Depends on what system of units you want to use. For SI it is, coulomb,
henry, and farad.

> You also would do us a favor by rewriting, in cgs, my SI equations in
> the other note for a field of 1 million volts/meter. It should be
> simple arithmetic. What is D and what is H?

10^6 volt/m ~= 33.3564095198 statvolt/cm in CGS. Is this for free space
EM radiation?

>>> Eq. (1) conveys the substantive fact that c depends on two simple
>>> constant vacuum parameters, Their product yields c and their
>>> quotient
>>> yields Z.
>>
>>Sure, but eq. 1 doesn't really tell us why they exist where perhaps
>>eq.
>>2 gives us more insight. What it looks like to me is that eps0 and
>>mu0
>>can really only exist if QVC exists.
>
> No, of course one equation cannot tell what eps0 is. For that you have
> the discoveries in my Dual Space book, or the permittivity paper at
> http://www.dualspace.net.
> It demonstrates that, in order to reproduce eps0, you need an
> electron-positron pair elastically restrained in a cubic cell of size

What is "restraining" the electron-positron pair? Why don't we see them
or detect them? Is it all the other e+e- "vacuum" pairs that are doing
the restraining? If so, don't you think there is some kind of BOUND
"vacuum" charge happening? Of course there would be; Quantum Vacuum
Charge.

> L = 3.514x10^-14m with spring constant of
> K = 2.612x10^14 N/m.
> 2e/L^3 makes a charge density P of 7.2x10^21 coul/m^3.
> Then Eq. 14 for eps0 gives the whole picture:
> eps0 = 2e*P/K = 8.854x10^-12 farad/meter
> It makes the very sensible statement that permittivity is proportional
> to e (for Ee = force), times charge density, and inverse to the
> aforesaid spring constant K. It all works and every one of a dozen
> parameters of pairspace are "forced" by logical laws without
> hypothesis.
> Unfortunately, the paper is not something to be speed-read nor is it,
> in retrospect, very lucid.

Well, it would be much better if you didn't ignore the rest of the
quantum objects of the Standard Model for your "vacuum".

>>> Your Eq. (2) is firstly a simple kinematic declaration that a
>>> broadcast wavelength times the broadcast frequency equals the
>>> velocity
>>> of transmission. which can be ascertained by drawing sine waves on
>>> paper. You make this velocity c, which is OK, but in the second
>>> place,
>>> you go too far in naming parameters Lvac and Cvac. Are they
>>> properties
>>> of the vacuum or of the vagaries of the broadcast transmitter?
>>
>>They (Lvac and Cvac) are probably actually just a different way to
>>look
>>at the electric and magnetic fields of EM radiation. Of course they
>>are
>>not like capacitance and inductance of real caps and inductors.
>>
>>> What is the other constraint (e.g., Z) that allows you to calculate
>>> Cvac and Lvac? I can make Cvac be 1 farad or any value, then solving
>>> for Lvac. These C and L's can't be properties of the vacuum; they
>>> must
>>> have the happy values that would satisfy F_brdcst.
>>
>>Cvac and Lvac can't be calculated any more than k_e or k_m can be.

> I just calculated ke (eps0?) above.

k_e is not eps0. In SI, k_e = 1/(4pi eps0)

>>However, it is not hard to postulate what their values might be based
>>on
>>eps0 and mu0.
>>
>>Cvac = 2 eps0 lambda

> Farads, no? But you don't have eps0. You just have capacity in
> centimeters.

??? Figure it out if the wavelength of the EM radiation is 1 meter.
Cvac ~= 1.7708376 E-11 farad in SI units.

>>Lvac = mu0 lambda/8pi^2

> Got to be henries. But you don't have mu0.

I have no idea what you mean by "But you don't have mu0 or eps0".

>>When converted to natural units we obtain,
>>
>>Cvac = Lvac = lambda/2pi

> Natural units; don't make me sick. It says farads = henries = cm's.

I can asure you that it is only you that is making yourself sick. ;-)
Most people know how to deal with this.

>>Simply the rationalized wavelength of the EM radiation under
>>consideration. And a thing of beauty since they end up equal to each
>>other.

> You should be revulsed.

It is not my fault that you can't see physics from a different
perspective. Many people have no problem with looking at physics from
different perspectives.

Best,

Fred Diether
Moderator sci.physics.foundations
http://www.vacuum-physics.com

From: anandsr21 on
On Jun 20, 1:04 am, "Greg Neill" <gneill...(a)OVEsympatico.ca> wrote:
> "Florian" <firstn...(a)lastname.net> wrote in message
>
> news:1hzyjoi.thrw70pivls0N%firstname(a)lastname.net...
>
> > Greg Neill <gneill...(a)OVEsympatico.ca> wrote:
>
> > I know. we need to invent more trick to make it work. At some point
> > there would be so many tricks (MONDtheory, dark matter, dark energy)
> > that it will certainly be time to find a better model. Right?
> > I wonder if the physicist who will find a better model is born yet?
>
> It's not "inventing more trick" if the mathematics is
> already in place in the current theory. MONDis an
> attempt to ignore theory and employ a curve-fit model
> instead. Dark matter is not a "trick" in that it is
> an empirical necessity.
>
>
MOND is not an attempt to ignore theory. It is an attempt to explain
the galactic data by thinking out of the box. We don't yet know the
reason why MOND works so well at the galactic scales. And any theory
including GR will need to come up with a reason why MOND phenomenology
works. DM would be very credible if MOND did not exist, but since it
does, GR needs to explain it.

Just to debunk some myths here:
1) MOND has exactly ONE UNIVERSAL free parameter. This parameter
has already been fixed to 1.2E-10m/s^2.

2) It has fit more than a hundred galaxies. You can search for "mond
fits" to find them. There are peer reviewed papers explaining each of
those fits. MOND is not favored by cranks, they have their own
theory.
http://www.journals.uchicago.edu/ApJ/journal/issues/ApJ/v508n1/38018/sc0.html

3) MOND had predicted the properties of LSB galaxies, before they
could be studied, so it has made more successful predictions compared
to DM theory (which cannot really predict anything).

4) Trying to fit a galaxy with DM is a purely curve fitting exercise
because there are multiple free parameters per galaxy. These
parameters have no constraints, provided by previous fits.

5) DM is only required if you don't know what is causing the higher
gravitational force.

All this is regarding galactic scales. Don't tell me about the Bullet
Cluster. I know about it and it doesn't give anything new compared to
other galaxy clusters. MOND also doesn't fit them properly.

The real problem is that we don't know what space is. How mass deforms
it? And how mass traverses space. Once we have an explanations for
these we may find the reason for MOND effects.

I do hope that LQG, which is based on a quantum transformation of GR,
will provide answers to these questions. I am sure it will take a long
time ;-).

I think that MOND effect happens due to the meeting of gravitation
effects due to a massive body and the cosmic gravitational effects.

I think that Gravitational force will never truly reduce to zero. So
that any particle will not be able to execute a straight line
trajectory. It will always execute a circle, whose radius depends on
its speed, in absence of massive objects in its vicinity. Light and
other particles moving at the speed of light will move in a circle
equal to the size of the universe and we will observe them moving in a
straight line.

Near a massive object the newtonian force will completely dominate the
cosmic force and we will see the normal newtonian drop off.

Further away the cosmic force will have more of an effect. At the MOND
scales the two forces will become equal and we will see a MONDian drop
off. Incidently MONDs force is a geometrical mean of a0 and GM/R^2.

regards,
-anandsr