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From: Ste on 16 Jan 2010 01:17
On 16 Jan, 04:33, "preedmont" <nos...(a)spamless.com> wrote:
> "Ste" <ste_ro...(a)hotmail.com> wrote in message
>
> news:dd68ad12-0046-499a-9391-452eb70953a0(a)a6g2000yqm.googlegroups.com...
>
> > On 16 Jan, 02:00, jdawe <mrjd...(a)gmail.com> wrote:
>
> >> force = push + pull.
>
> > How do we distinguish push from pull?
>
> push = - pull
>
> now add pull to both sides
>
> push + pull = 0
>
> therefore, force = 0

So I'll ask again, how does one *distinguish* between push and pull?
How can one tell when one is measuring push, and how can one tell when
one is measuring pull? Or are they really one and the same thing?
From: preedmont on 16 Jan 2010 09:59

"Ste" <ste_rose0(a)hotmail.com> wrote in message
news:1b611c0f-beb2-4b2e-a017-7079c480809f(a)l30g2000yqb.googlegroups.com...
> On 16 Jan, 04:33, "preedmont" <nos...(a)spamless.com> wrote:
>> "Ste" <ste_ro...(a)hotmail.com> wrote in message
>>
>> news:dd68ad12-0046-499a-9391-452eb70953a0(a)a6g2000yqm.googlegroups.com...
>>
>> > On 16 Jan, 02:00, jdawe <mrjd...(a)gmail.com> wrote:
>>
>> >> force = push + pull.
>>
>> > How do we distinguish push from pull?
>>
>> push = - pull
>>
>> now add pull to both sides
>>
>> push + pull = 0
>>
>> therefore, force = 0
>
> So I'll ask again, how does one *distinguish* between push and pull?
> How can one tell when one is measuring push, and how can one tell when
> one is measuring pull? Or are they really one and the same thing?

push = pull - ll + sh

push = not pull

not push = pull

pull - ll = push - sh

OR

pull - push = ll - sh


From: Uncle Al on 16 Jan 2010 10:21
preedmont wrote:
>
> "Ste" <ste_rose0(a)hotmail.com> wrote in message
> news:1b611c0f-beb2-4b2e-a017-7079c480809f(a)l30g2000yqb.googlegroups.com...
> > On 16 Jan, 04:33, "preedmont" <nos...(a)spamless.com> wrote:
> >> "Ste" <ste_ro...(a)hotmail.com> wrote in message
> >>
> >> news:dd68ad12-0046-499a-9391-452eb70953a0(a)a6g2000yqm.googlegroups.com...
> >>
> >> > On 16 Jan, 02:00, jdawe <mrjd...(a)gmail.com> wrote:
> >>
> >> >> force = push + pull.
> >>
> >> > How do we distinguish push from pull?
> >>
> >> push = - pull
> >>
> >> now add pull to both sides
> >>
> >> push + pull = 0
> >>
> >> therefore, force = 0
> >
> > So I'll ask again, how does one *distinguish* between push and pull?
> > How can one tell when one is measuring push, and how can one tell when
> > one is measuring pull? Or are they really one and the same thing?
>
> push = pull - ll + sh
>
> push = not pull
>
> not push = pull
>
> pull - ll = push - sh
>
> OR
>
> pull - push = ll - sh

Fermat's Little theorem less its second harmonic. Brilliant!


--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
From: Rock Brentwood on 16 Jan 2010 14:57
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.

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: Sue... on 16 Jan 2010 15:26
On Jan 16, 2:57 pm, Rock Brentwood <markw...(a)yahoo.com> wrote:

===========
[...]
> 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.

Elsewhere I have shown the equivalence of a serpent bite taken from
Newton's falling apple and the disparity between oscillators at the
top and bottom of Harvard tower. If anyone thinks you are
being overly speculative, I will try to dig up the URL to
these convincing calculations with some experimental
support. :o)

Sue...

[...]




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