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From: nuny on 18 Jan 2010 00:13
On Jan 17, 8:57 pm, jdawe <mrjd...(a)gmail.com> wrote:
> On Jan 16, 2:30 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > On 16 Jan, 02:00, jdawe <mrjd...(a)gmail.com> wrote:
>
> > > force = push + pull.
>
> > How do we distinguish push from pull?
>
> push force accelerates a mass.
>
> or
>
> pull force decelerates a mass.

So, the pull of gravity decelerates objects toward Earth's surface.
Right.


Mark L. Fergerson

From: Szczepan Bialek on 18 Jan 2010 03:44

"Rock Brentwood" <markwh04(a)yahoo.com> wrote
news:d8994184-bf44-444e-b21e-7422e48d4bc5(a)l19g2000yqb.googlegroups.com...
On Jan 15, 8:00 pm, jdawe <mrjd...(a)gmail.com> wrote:
>> force = push + pull.
>> push = solid matter + potential energy
>> pull = fluid matter + kinetic energy

>The identifications Maxwell used early on in his 1861 and 1864 papers
were:
E is potential
H is kinetic
and that in an isotropic medium (in the frame of isotropy) the
corresponding inductions would be governed by the analogue of a spring
coefficient (kappa) and of a mass coefficient (mu) and, thus, would
read:
D = kappa E
B = mu H.

In the Maxwell's words:

"According to our theory, the particles which form the partitions between
the cells constitute the matter of electricity. The motion of these
particles constitutes an electric current; the tangential force with which
the particles are pressed by the matter of the cells is electromotive force,
and the pressure oi the particles on each other corresponds to the tension
or poten- tial of the electricity." From:
http://en.wikisource.org/wiki/On_Physical_Lines_of_Force

So in the space are the two matters: electricity and magnetism (the matter
of the cells).

Is it the only solution?
I have heard that Dirac proposed the one only: the electrons.
Where can I find the details?
What is Your opinion?
S*


>That's where the coefficients and their names ultimately arose. Today,
his kappa (the dielectric coefficient) is treated synonymously as our
epsilon (the permittivity), but this identification is wrong. It's
only valid in parity-symmetric media.

>In frame of reference moving at a velocity of -G relative to the frame
of isotropy, his relations read:
D = kappa (E + G x B)
B = mu (H - G x D)
where () x () denotes the vector product.

>He DID, in fact, incorporate the G x B term in the definition of E, so
that his field-potential equation actually read:
E = -grad phi - d_t A + G x B
instead of just E = -grad phi - d_t A. So, his E is equivalent to our
(E + G x B).

>But, in his early papers he never explicitly called out B as a
separate field (though he identified it as a separate field in the
adjoining verbal language), so he only wrote B as the diglyph (mu H).
Moreover, his H is different from our H by a factor of 4 pi. So our mu
is off from his by 4 pi. We would equate mu = 4 pi (which is
normalized nowadays to 4 pi Henri per 10000km) while he would have
written mu = 1 for the equivalent ratio.

>[And note: he didn't use the letter names for the fields until around
the 1870's, but separate letter names for the individual components.
He didn't use vectors or quaternions. Rather, he used differential
forms -- both in the 1860's AND in the treatise.]

>Because he failed to fully distinguish B from H early on, he also
failed to do the proper analysis for the corresponding transformation
law. So, instead of getting
B = mu (H - G x D)
he left out the -G x D term. Only Thomson noted it, later on in the
1880's, though Maxwell apparently made an allusion to it in the verbal
language early on in his 1861 paper.

>Today, the equations would be written (as per Einstein & Laub in
1908-1909 and Minkowski in 1908) as:
D + (1/c)^2 G x H = epsilon (E + G x B)
B - (1/c)^2 G x E = mu (D - G x H).
For a medium at an index of refraction of 1 (i.e. epsilon mu = 1/c^2),
away from the Cherenkov threshold (where epsilon mu |G|^2 = 1) this is
equivalent to the "stationary" equations
D = epsilon E, B = mu H.
For media with non-trivial indices of refraction (e.g. outer space,
particularly in the earliest epochs of the universe) it is not. The
vector G identifies a universal frame of reference -- namely, the one
associated with the Big Bang metric and with the cosmic microwave
background.

>Nowadays, Maxwell's formulation would be represented by a Routhian R =
R(phi, A, E, H) rather than a Lagrangian L = L(phi, A, E, B), since he
took E and H as the independent variables. So, for an isotropic
potential-idependent Routhian R = R(E^2/2, E.B, B^2/2) (where ().()
denotes scalar product), his coefficients could be DEFINED by:
kappa = dR/d(E^2/2)
mu = dR/d(B^2/2)
along with a third coefficient, which he completely overlooked:
lambda = dR/d(B.E).
And his constitutive law for the isotropic frame would then read:
D = kappa E + lambda H
B = lambda E + mu H
which, when solved in terms of (E,B) gives the "Lagrangian" form of
the constitutive laws:
D = epsilon E + theta B
H = (1/mu) B - theta E
where
theta = mu lambda
epsilon = kappa - lambda^2 mu

>So, in the presence of the parity-violating coefficient lambda (or
theta), the dielectric coefficient kappa no longer coincides with the
permittivity epsilon.

>So, his "spring coefficient" kappa is not strictly identical to what
we presently call permittivity, though his "mass coefficient" mu
accords with our permeability mu, up to a factor of 4 pi.


From: Ste on 18 Jan 2010 04:16
On 18 Jan, 04:57, jdawe <mrjd...(a)gmail.com> wrote:
> On Jan 16, 2:30 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > On 16 Jan, 02:00, jdawe <mrjd...(a)gmail.com> wrote:
>
> > > force = push + pull.
>
> > How do we distinguish push from pull?
>
> push force accelerates a mass.
>
> or
>
> pull force decelerates a mass.

Are they not in fact two sides of the same coin? That is, pushing is
fundamentally the *same* as pulling? For example, how can one discern
the pull of gravity from the 'push of the aether'? Or perhaps less
confusingly, how can one discern the pull of a vacuum, from the push
of high-pressure gas? Is one not merely describing *relative
imbalance*?
From: jmfbahciv on 18 Jan 2010 10:20
nuny(a)bid.nes wrote:
> On Jan 17, 8:57 pm, jdawe <mrjd...(a)gmail.com> wrote:
>> On Jan 16, 2:30 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>>
>>> On 16 Jan, 02:00, jdawe <mrjd...(a)gmail.com> wrote:
>>>> force = push + pull.
>>> How do we distinguish push from pull?
>> push force accelerates a mass.
>>
>> or
>>
>> pull force decelerates a mass.
>
> So, the pull of gravity decelerates objects toward Earth's surface.
> Right.
>
>
That sure would be nice when I'm walking on ice.

/BAH
From: jdawe on 18 Jan 2010 20:57
On Jan 18, 8:16 pm, Ste <ste_ro...(a)hotmail.com> wrote:
> On 18 Jan, 04:57, jdawe <mrjd...(a)gmail.com> wrote:
>
> > On Jan 16, 2:30 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > On 16 Jan, 02:00, jdawe <mrjd...(a)gmail.com> wrote:
>
> > > > force = push + pull.
>
> > > How do we distinguish push from pull?
>
> > push force accelerates a mass.
>
> > or
>
> > pull force decelerates a mass.
>
> Are they not in fact two sides of the same coin? That is, pushing is
> fundamentally the *same* as pulling? For example, how can one discern
> the pull of gravity from the 'push of the aether'? Or perhaps less
> confusingly, how can one discern the pull of a vacuum, from the push
> of high-pressure gas? Is one not merely describing *relative
> imbalance*?

Sorry,

acceleration

or

deceleration

are opposing operands within the 'force' opposing tree but both push
\pull can generate acceleration\deceleration.

At the top of the 'force' opposing tree it should be understood there
are only 2 opposing operands:

push force 'repulses' other mass away from the mass producing the push
force.

or

pull force 'attracts' other mass toward the mass producing the pull
force.

It is only energy\matter at rest that generates push force.

It is only energy\matter in motion that generates pull force.

It should also be understood that they are always the complete
opposite and are always equally required in the universe.

A high pressure gas is fluid matter 'in motion' so it can only
generate pull force.

naturally you have 2 opposing directions forward or reverse. So
reverse pull force may look like a push to some observers.

-Josh.

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