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From: glird on 24 Apr 2010 17:45
On Apr 23, 11:40 am, PD <thedraperfam...(a)gmail.com> wrote:
> On Apr 23, 10:28 am, maxwell <s...(a)shaw.ca> wrote:

> > Could you supply one reference (preferably online) which MEASURES the gravitational effects on a single electron?  This effect seems very unlikely as the ratio of the EM to gravitational force on an electron is at least 10**40. > >
>
> http://adsabs.harvard.edu/abs/1977RScI...48....1W

I looked at the referenced page and found nothing about HOW the g-
force is measured. Thinking that perhaps it might be treated in
"physics today", I picked up the nearest issue and opened it to -- of
all things --
"Universal insights from few-body land" by C. H. Greene. I started
to read the article and found its main thesis ("The universal
properties of systems having short-range interactions -- be they among
cold atoms or nucleons or molecule -- connect in turn to the beautiful
but mysterious effect discovered by nuclear theorist Vitaly Efimov ...
in 1969") fascinating. On studying the article after being intrigued
by its next sentence ("Within the past four years, progress has
erupted in exploring the Efimov effect and related phenomena through
the manipulation of dilute atomic gases near a Fano-Feshbach
resonance") I read the rest of the page and the next one, studying its
Figures 1 and 2 and trying to understand what it was talking about
even though its math is WAY over (or under?) my head. Then I turned
to the next page (62, of this March, 2010 issue) and was reading the
part entitled "The Smoking Gun" and then, when I got to the place
where it said "in 2009 several experiments managed to obtain
completely convincing and unambiguous evidence -- smoking guns of
universal physics" I looked at Figure 3 across the entire top of the
page; and WHAMM!!
There, in deep blue, was a line tracing out EXACTLY THE SAME pattern
hand drawn as the structural pattern of all mono-nuclear matter-units,
from atoms to galaxies, in my 1965 book, The Nature of Matter and
Energy (Figure 41-4 on page 320). Here is one of the sentences ...
Sorry! On looking for ONE key sentence I found myself back at pg
211, and it is now an HOUR or so later and i am still too fascinated
by my own words to pick out one sentence for you, here. As to an
electron, that story is complicated.

glird
From: glird on 25 Apr 2010 18:31
On Apr 24, 6:14 pm, Don Stockbauer <donstockba...(a)hotmail.com> wrote:
> Electron's puzzles.
>
> An electron said "I'm going to be just as mysterious to those
> macroscopic humans as I can possibly be so that they spend trillons of
> hours wondering about me and my buddies."

Not quite. If you want to know what it really said, change the word
"hours" to "dollars".
To witch our scientists cleverly replied, "Where ignorance is bliss=$
$$^6, ts folly to be wise."

glird
From: Don Stockbauer on 25 Apr 2010 19:08
On Apr 25, 5:31 pm, glird <gl...(a)aol.com> wrote:
> On Apr 24, 6:14 pm, Don Stockbauer <donstockba...(a)hotmail.com> wrote:
>
> > Electron's puzzles.
>
> > An electron said "I'm going to be just as mysterious to those
> > macroscopic humans as I can possibly be so that they spend trillons of
> > hours wondering about me and my buddies."
>
>   Not quite.  If you want to know what it really said, change the word
> "hours" to "dollars".
>   To witch our scientists cleverly replied, "Where ignorance is bliss=$
> $$^6, ts folly to be wise."

You can't always buy the Truth.
From: glird on 9 May 2010 18:10
On May 8, 3:28 pm, PD <thedraperfam...(a)gmail.com> wrote:
> On May 8, 1:37 pm, "Ken S. Tucker" wrote:
> > On May 8, 10:18 am, PD wrote:
> > > On May 8, 2:48 am, "Ken S. Tucker" wrote:
><<< The spin for a given quantum state is usually represented by a vector, where the components are the expectation values of the spin operator -- not a tensor. >

><< Not quite, as far as I know, that has yet to be determined, Ken
>< I don't think so.
Yes, I know tensors, Ken, but spin is not a tensor.>

snip

>< Well Paul, I'm not sure we can simplify spin to your
level of understanding, but if you're serious we can
go to "spin connections" in Weinberg's QFT Vol.3,
Eq.(31.5.17). >

> Indeed. A spin connection is not a spin vector.
> A spin connection is a connection in a spinor bundle.
> It helps to know what the words mean, Ken.

>> The math is a bit detailed, but the concept is straight-forward, but my impression is you (Paul) enjoy science on a superficial level, so that's why I avoid math with you ... you post for some relaxation, fine with me.
Regards
Ken S. Tucker >

I find both of you reasonable and well informed. Both of you seem
logically AND mathematically adept. Even so, PD is right in saying "It
helps to know what the words mean".
Until you both know exactly what the following words mean, you can't
understand your own subjects:
mass, energy, quantum state, renormalization, expectation value,
probability, photon, point-sized, etc etc.
Here is an example: In his 1905 SR paper Einstein wrote an equation,
tau = a(), in which he said that a is a function of v; which he
expressed as phi(v). A bit later he said that phi(v) = 1. He never did
say what "a" denotes all by itself. If we let a denote acceleration,
then
a = phi(v) = 1
denotes
acceleration = d(dx/dt) = 1.
That means that for any given change in the value of v, the value of
_a_ changes by an identical amount.
However, after writing
eta = phi(v)y
and finding that phi(v) = 1, instead of realizing that this means that
the value of phi(v)y, thus of eta, CHANGES
as v does; he DELETED the symbol from his equations.
Although that did leave him with the LTE, in which eta and zeta are
independent of v or any change in it, neither he nor anyone since
seems to have realized that the latter requires that
phi(v) = 0;
i.e. that there is NO change in the value of a = phi(v) as v changes.

Regards to both of you,
glird
From: glird on 12 May 2010 15:11
On May 9, 8:24 pm, waldofj <wald...(a)verizon.net> wrote:

glird: >< In his 1905 SR paper Einstein wrote an equation,
tau a(t - vx'/(c^2 - v^2)
in which he said that a is a function of v; which he
expressed as phi(v). A bit later he said that phi(v) = 1. He never did
say what "a" dnotes all by itself. >

W: He never did because he showed it doesn't denote anything. It's
just a constant function of v which always equals one.

>< If we let a denote acceleration, >

You can't just decide to give an arbitrary meaning and then start
drawing conclusions. Well, you can but you'll just end up with
nonsense. Garbage in garbage out.

>< then a=phi(v)=1 denotes acceleration = d(dx/dt) = 1. >

Actually acceleration = (d/dt)(dx/dt)
or (d^2/dt^2)x or, since we are talking physics here,
dx/dt = v so acceleration = dv/dt
of course I realize none of this means anything to you since you
refuse to learn calculus.

>< That means that for any given change in the value of v, the value of _a_ changes by an identical amount. >

now you don't know basic physics and you're contradicting yourself.
You have set a = phi(v) = 1. One is a constant, constants don't change
(didn't you learn that in algebra class?) Constant acceleration (one
is a constant, right?) means velocity is always increasing at a
constant rate. However that last point is moot since _a_ is not
acceleration.

True. As I said, Einstein never did define what _a_ denotes.
Neither did you.

> However, after writing eta=phi(v)y and finding that phi(v) = 1, instead of realizing that this means that
the value of phi(v)y, thus of eta, CHANGES as v does; >

How can you say phi(v) is constant (one is a constant, right?) and
then say it changes?

ALMOST the same way you did when you wrote, "Constant acceleration
(one is a constant, right?) means velocity is always increasing at a
constant rate."
Btw, how do you think that E got from
"eta = a{c/sqrt(c^2-v^2)}y" to "eta = phi(v)y"?
To be explicit, even if a = phi(v) allowed him to replace a with
phi(v), how come c/sqrt(c^2-v^2) disappeared?

>< he DELETED the symbol from his equations. >

W: What's the point of writing one times something, you do realize
multiplying by one doesn't change the expression, right?

Do you realize that the only way for c/sqrt(c^2-v^2) to equal 1 is
that v = 0? If so, then why bother with transformation equations at
all?

>< Although that did leave him with the LTE, in which eta and zeta are independent of v or any change in it, neither he nor anyone since seems to have realized that the latter requires that phi(v) = 0; i.e. that there is NO change in the value of a = phi(v) as v changes. >

W: Well, zero is also a constant but phi(v) is being MULTIPLIED in
each expression. If you multiply by zero you'll just set everything to
zero, not what you want.
Or is it you don't know what phi(v) = 0 means?

Thank you for taking the time to answer me. (I knew that my wording
was misleading, and hoped that someone would correct it.) Perhaps it
would have been clearer if I'd written it this way:
Since eta and zeta are independent of v or any change in it, neither
he nor anyone since seems to have realized that the latter requires
that da/dv = 0; i.e. that there is NO change in the value of a as v
changes.

The point is this: If a = phi(v) means that a is a function of v,
and if a(v) means the same thing, and if da/dv is another way to
express that relation; then if phi(v) = da/dv = 1, and dv = 4 or 3 or
6 then da = 4 or 3 or 6 also. But if there is no change in the
relation between eta and y as v changes, as is the case in the LTE,
then that relation is NOT a function of v!

Btw, E's "proof" that "phi(v) = 1" was defective anyhow.

glird
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