Re: CODATA's Value for Hydrogen's Rydberg Constant R_H

Hi Steve, I studied your post carefully.
I find your study engrossing, and perhaps
I might even be helpful.

On May 2, 8:51 pm, "Steve Bell" <sb...@xxxxxxxxxxxx> wrote:
"Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote in messagenews:5e558bd6-ffd8-4744-9cf4-d365e96e4e87@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx



After the true 2-body effects are accounted for by using the reduced
mass of
the electron, then the main other effect that Schroedinger did not
incorporate, and that Dirac/Sommerfeld did, was relativistic
effects.
These
were only specially relativistic, though. According to current
belief,
this
is all that's needed to completely describe the unification of QM
with
relativity theory (the SR part of it) for modeling atoms,
specifically
for
deriving binding energy equations. I think there are more
complicated
relativistic effects going on in an atom, even in the simplest of
atoms,
ground state hydrogen.

Yes, if one considers magnetism as an SR effect
that should be true.

There's probably a union with geomagnetism. I believe you once thought
an
electron coasts along a geodesic?

Yes, the "Quantum Geodesic", governing orbitals.

That's was nice intuition. If true, there
ought to be a way to bring frame-dragging effects down into the atom.

LOL, I thought you wanted to eliminate subtle
GR relativistics out of the Rydberg constant.

As you probably know, Sommerfeld introduced specially relativistic effects
into Bohr's non-relativistic model. But since Sommerfeld's model gives
basically the same ground state binding energy and R_H as does Dirac's, the
data says Sommerfeld is off too. The direction both are off is "too strong"
of a binding energy. When special relativistic effects are introduced, the
electron is in a little closer, going a little faster and has a little
larger mass than Bohr. I figured since electricity's F =
(1/(4*pi*eps))*e1*e2/r^2 is so similar to gravity's F = G*m1*m2/r^2,
basically the same except for a scaling factor, if there is a differential
geometry extension of a spherically symmetric Newtonian gravitational field,
there ought to be a differential geometry extension of a spherically
symmetric Coulomb electrostatic field. So I figured out an electronic
Schwarzschild representation of a Coulomb field. But the results are even
worse. The ground state binding is even stronger than Dirac's and
Sommerfeld's. So I figured out the last step, which is to develop an
electronic Kerr representation of a Coulomb field, which as a special case
contains the electronic Schwarzschild's field. The frame dragging effects of
this electronic Kerr field are in fact magnetism. This workup unifies geo
and electro magnetism, they really are the same thing, but at different
scales. As you hinted at, it's probably magnetic effects that's causing the
binding energy to loosen up back to Bohr. The key is to find out what
electronic Kerr frame dragging effects, i.e., magnetic effects, cause this.

Posted below at ### are some diagrams and formula
I hope will be helpful, though applied to speed of
gravity, the principles involving an orbiting e-
about a p+ are very similiar.

I would characterize the problem by using a
rotating FoR, such as a phonograph turntable.

Drawing a line on the phonograph (at a finite speed)
using a stationary ruler from the circumference
through the center of rotation (think p+ there),
produces a curved line with a length "R" greater
than the radius "r" that one would find if "r" was
measured when the rotation rate is zero.

Obviously R > r thus F=qQ/R^2 < f=qQ/r^2

when using the finite speed of light "c" that
conveys electrical "force" effects.

Recall the Kerr metric has to do with a centrally
rotating mass affecting orbits.
I would call the difference F<f a magnetic effect.
Regards
Ken S. Tucker

Btw, if anyone wants to be rigorous in their use of the word "geodesic,"
that to me is tantamount to saying there is a differential geometry
representation of a Coulomb field if indeed an electron coasts on a geodesic
in this field. The orbit of an electron in Sommerfeld's theory is not
actually a geodesic. It's like only doing a specially relativistic extension
of Newtonian gravity, which doesn't get you all the way to a true
differential geometry representation.

Steve


###
Ok, I thought I'd post this (below) as
a demonstration of how the Equivalence
Principle is calculated in rotating CS's.

Speed of Gravity Theory.
According to Newton the Speed of Gravity (Gs)
is infinite. Fig.1 shows the balance of the g-force
and centrifugal force and looks like,

Centrifugal Acceleration
/|\
|
|
|
<--------(E)----- Orbital Direction of Earth
| (Circular orbit)
|
|
\|/
Solar Gravitational Acceleration Fig.1
....................(S)

Where (E) is the Earth and (S) is the Sun.

According to Einstein's GR, Gs=c, hence the
acceleration vector is directed toward the apparent
(aberated) position of the sun (s) as in this Fig.2
((seeing is believing))...

Centrifugal Acceleration
/\
/
/
/
<----(E)----- Orbital Direction of Earth
/ (Circular orbit)
/
/
\/
Solar Gravitational Acceleration Fig.2
............. (s) (S)
..........image^

IMO, the centrifugal acceleration is directly opposite
the g-acceleration, as shown in Fig.2. This is based on
the Principle of Equivalence, whereby the gravitational
acceleration and the inertial acceleration (centrifugal in
this example) precisely balance to produce a "freefall",
so that no acceleration can be detected on Earth due
to the sun's gravity.

A common misconception is to align the acceleration
vectors on (E) as in Fig.3

Centrifugal Acceleration
/|\
|
|
At |
<-<----------(E)----- Orbital Direction of Earth
/ (Circular orbit)
/
/
\/
Solar Gravitational Acceleration Fig.3
(s) (S)

This unbalanced gravitational and centrifugal
acceleration would produce a residual acceleration
and velocity accumulation "At" in the direction of
orbital velocity if it were true, and would lead to a
gross failure of Energy conservation as well as a
sense of acceleration on Earth, this doesn't happen.

But we now have a situation that needs to be
explained in terms of GR, specifically the
introduction of oblique X-Y axes relating the
sun's image and Earth as in Fig.4a,

Y-axis
/\
/
/
/
<--------(E)----- X-axis
/
/
/
\/
(s) Fig.4a

Of course, it follows, if the direction of Earth's
velocity around the Sun were to reverse then
the X-Y axes obliqueness would also reverse
and produce Fig.4b,

Y-axis
/\
\
\
\
----(E)----> X-axis
\
\
\
\/
(s) Fig.4b

Mathematically, a unit vector e1 on X and e2 on Y
produces the following results,
(assuming V=dx/dt when y=r),

Fig 4a g12 = e1.e2 = -V/c (set c=1)

Fig 4b g'12 = e'1.e'2 = -V'

and g12 = -g'12 because V = -V' . (Eqs.1)

This understanding of the non-orthogonality
of the X-Y axes is adequate to solve the speed
of gravity problem, but it does not address
magnetism. For example let's be more clear
on the definition of rotational velocity V.

Set the rotational velocity defined wrt an
inertial system to,

V12 = (x1/r)*(dx2/dt) - (x2/r)*(dx1/dt)

where x1 == x and x2 == y,

then V12 = - V21.

One can see g12 == V12 but

g12 = -V21

At this point *I presume* V12 defines g12 and thus
g12 is anti-symmetrical and, of course, non-orthogonal.

In view of this developement, we might liken the anti-
symmetrical portion of g12 to the EM field tensored
by the magnetic force q*F12 as an asymmetrical stress
on the spacetime field.

I think anyone who has played with a pair of bar
magnetics has noticed weird attactions and repulsions
as these magnetics twist in relative orientation.
Aside from exotic theory, the medium connecting
these bar magnetics is only a spacetime field.

Suppose the earth is in a circular orbit around the sun, (xy plane),
with an orbital speed V, then we should expect an aberration of
V/c in the apparent postion (image) of the Sun.

The aberration of light (and gravity when gravity
is postulated to be directed at the apparent
position of the Sun as GR suggests) is given by,

g12 = (y/r)*(dx/cdt) - (x/r)*(dy/cdt).

With x=r and y=0, g12 = -dy/cdt, and describes clockwise
motion.
If x=-r then g12 = dy/cdt and is also clockwise motion,
(anti-clockwise motion is given by g21=-g12).

The term -dy/cdt is equal to (orbital velocity)/c, = -V/c,
and accounts for aberration of light and gravity, and
g12 (and g21) account for revolution in either direction.
The quantity g12 is defined by the relation (aberration)
of light signals from the sun to the earth, and so can be
regarded as a component defining a relating spacetime
metric.

The Earth and Sun remain the same distance apart therefore,
du^i/ds = 0, where u^i is the relative 3-velocity.

There is no specific need to consider covariancy or
contravariancy details for the accuracy required, the
coordinate acceleration from the usual geodesic equation
(previously linked to in my last post),in direction x
can be written (without using "^" and "_"),

du1/ds = 0 = -1/2*g11*(g10,0 + g01,0 - g00,1)*u0*u0

Let g11~1 and u0~1, which provides,

du1/ds = 0 = 1/2*g00,1 - g10,0
(g10 is considered symmetrical herein.)

where

1/2*g00,1 = gravitational acceleration

g10,0 = inertial (centrifugal) acceleration.

Now lets calculate the coordinate acceleration in
the direction y, (these are the aberration terms
when x=r and y=0),

du2/ds = - 1/2*g21*(2*g10,0 - g00,1)

and use

g12 = (y/r)*(dx/dt ) - (x/r)*(dy/dt), (c=1) .

(since x,y is the orbital plane, for this set-up,
g12= - dy/dt = -V and g21 = V).

Break out the following,
Aberrated gravitation acceleration = 1/2*V*g00,1
(an acceleration in direction y),
This is the term that *presumeably* causes the Earth to
respond to the aberrated position of the sun, except it is
exactly balanced by inertial acceleration.

Aberrated centrifugal acceleration = - V*g10,0
(an acceleration equal in magnitude to gravitational
acceleration in direction -y)

Recall x=r, and the sum of accelerations in the
direction 'y' are,

du2/ds = 1/2*V*g00,1 - V*g10,0 = 0

These are the equal and opposite terms that nullify
the aberrated gravitational acceleration.

The above is a transparent example of the Principle of
Equivalence demonstrated using the geodesic equation
and including the aberrating effects of the rotational
metric (g12).

Frankly, I think it's a tribute to Einstein's GR that such a
complex problem can be so simplified and easily applied.
Ie. applying generally covariant equations (the geodesic)
involving gravity, inertia and rotation be expressed more
clearly, and applied so directly to find speed of gravity=c.

Regards
Ken S. Tucker
.

Relevant Pages

  • Re: Self-forces
    ... field should radiate (massless charged particle near a gravitating ... Why *would* one expect an electron to exert a force on itself? ... Once an acceleration is involved, they're not symmetric anymore, even ... a quantum theory of gravity is needed. ...
    (sci.physics)
  • Re: origin of inertia
    ... >> You are confusing force with acceleration. ... A gyro that is free to turn in all directions does not precess if it is ... Gravity exerts a force on me but no work is done because there is no ... Conservation of angular momentum changes ...
    (sci.physics)
  • Re: "Is There a Force of Gravity?"
    ... the result of an inertial acceleration in "space-time". ... Inertial and acceleration are terms that contradict each other. ... gravity CANNOT result from an inertial acceleration. ... by a single set of mathematics, as is the interchange of potential energy ...
    (sci.physics.relativity)
  • Re: "Is There a Force of Gravity?"
    ... the result of an inertial acceleration in "space-time". ... points on the Earth, perhaps London, England and Melbourne, Australia. ... gravity CANNOT result from an inertial acceleration. ... by a single set of mathematics, as is the interchange of potential energy ...
    (sci.physics)
  • Re: origin of inertia
    ... > You are confusing force with acceleration. ... All the motions are free in a gyro. ... We were talking about centripetal forces, not gravity. ... Conservation of angular momentum changes ...
    (sci.physics)