QM and electron orbits
"Andr? Michaud" wrote in message
om...
Laura wrote:
"Andrew Usher" wrote in message
om...
This message is a continuation of the discussion in the thread
'Neutrino mass'.
It is more like a reiteration of your position, already stated in that
thread.
I admit to not being formally educated in QM.
Neither am I.
But I try not to criticise things I don't understand.
I am nevertheless trying
to criticise a belief normally taught in such education.
If you're referring to the idea of the electron being "smeared" across
the
orbital, then it is you who has misunderstood.
"In a general paper on quantum mechanics, Schroedinger discusses and
rejects
the interpretation that a single quantum is somehow phyiscally "spread
out"
or "blurred" among the different parts of a superposition ."
That is what is being taught.
To my knowledge, what is being taught, in perfect accordance with
Heisenber's teachings is that the electron is not localized until
the wave function collapses. So, when in motion, it is definitely
considered in the Copenhagen school view of QM as being spread out.
Stationary states aren't subject to uncertainty. The
parameters of an electron in a stationary state can be
measured with precision. Via multiple observations
of identically prepared systems, one can measure the
distribution of degenerate states, that is, states of equal
energy and angular momentum, to unlimited accuracy.
[Old Man]
Although I
don't understand the math involved in the conventional approach, I
believe that I can understand the basics in terms of logic.
The false idea is that the Bohr-Sommerfeld orbits are an incorrect and
obsolete model of the atom. As we know, the idea of fixed orbits is
not exactly correct, but that does not make it useless.
It is very useful for chemistry and nuclear physics, but it is a model
and not meant to be taken as a true picture of the atom.
As valuable and useful as the QM model then, it would seem. No ?
The orbitals are conventionally given as time-independent
wavefunctions, and that is held to be the correct description. This
leads to the false belief that an electron's position is smeared out
over the orbital,
That's *your* false belief. The electron isn't "smeared out over the
orbital" (just as Schroedinger's cat isn't alive AND dead). It just
doesn't
have a position until one has been measured (just as you don't know if
the
cat is alive or dead until you open the chamber), and then that position
is
one that is affected by the act of measuring, and is therefore not a
true
representation of where the electron really was, if indeed it had a
position
at all.
That's why such an innate position is done away with entirely; it is
meaningless since it can't be measured without interfering with it.
and that the probability function is independent of
earlier observations. From the uncertainty principle (which states
that particles occupy h^3 in phase space), this can only be strictly
true for the 1s orbital. For all higher n, the relative uncertainty
becomes smaller, and the classical orbit becomes an increasingly
better approximation.
What difference would it make?
If none, then what is the value of your version?
This explains the solar system, for example, where the quantum numbers
are very, very large and thus quantum effects are unobservable.
Which is exactly why QM doesn't explain it.
The solar system obeys the same physical laws as the atom.
Because it looks like the classic atomic model?
You're not the first to see the similarity.
What happens inside an atom is very different from a solar system.
Is it really? A matter of interpretation possibly.
Or do you believe that the sun has a positive charge and the planets
a negative charge?
It is well known, it seems to me, that all matter making up the Sun
and planets is made up of charged particles (electrons, quarks up and
quarks down, the latter two being the charged components of the
nucleons) and that electrostatic interaction is additive and
universally obeys the inverse square law.
A very simple demonstration of this can be made when using only
the invariant masses of the charged components of nucleons instead
of their usual measured effective masses, which, if the real masses
of the constituting quarks really are in the observed range, can
only be mostly made up of relativistic inertia induced by the near
light velocities that the quarks must have to maintain the structures.
Besides, a clean set of calculated invariant masses that falls right
into the experimentally estimated range can easily be obtained from
the Coulomb inverse square law:
d = down quark
u = up quark
e = electron
Q = unit charge
k = Coulomb constant (1/(4 pi epsilon_0))
T = 1/nu_0 = 1 /6.57968391E15 Hz = 1.519829851E-16 sec
alpha_0 = Bohr radius
alpha = fine structure constant
c = speed of light
Calculating the invariant masses of stable elementary particles
m_i[d,u,e] = (k/alpha_0)(3Q / (n alpha c))^2 (n=1,2,3)
Summing up the invariant masses of the proton charged components:
m_ip = 2 m_iu + m_id
Calculating the G applicable to the invariant mass of the proton
components as a central body according to Kepler's third law
G_p = (4 pi^2 alpha_0^3) / (m_ip T^2) = 2.059446471E31 N . m^2 . kg^-2
Calculating the force at the Bohr radius from the gravitational
equation using the G applicable to the hydrogen atom:
F = (G_p m_ip m_ie) / alpha_0^2 = 8.238721758E-8 N
Which is the exact same value given by the standard Coulomb
equation using charges instead of masses:
F = (k Q^2) / alpha_0^2 = 8.238721806E-8 N
So, wouldn't it look like the same physical laws could be applying
in the atom and in the solar system ?
Some will say what is nu_0 and what has it got to do with this.
First, converting from a time basis to a distance basis, the
dimensions of the Newton (N) resolve as:
kg . m . s^-2 = J . m^-1
So, force (F) at the Bohr radius is 8.238721806E-8 J . m^-1
The energy at the Bohr radius can thus be calculated directly as
folows:
E at Bohr radius = F . alpha_0 = 4.359743805E-18 J (27.21138344 eV)
h = Planck's constant
nu_0 = E / h = 6.579683917E15 Hz
Which is the frequency of the energy induced at the Bohr radius.
Why is nu_0 significant in this context:
It is the number of times per second that the electron is deemed to
circle the proton on the Bohr orbit (hydrogen ground state) in the
Bohr model according to the de Broglie hypothesis, which allowed
calculating the traditional velocity assigned to the electron on
that orbit (his 1924 thesis).
v_0 = h / (m_e Lambda_0) = 2187691.253 m / sec
A velocity that can be confirmed from a second source here by obtaining
that traditional velocity from the length of the ground orbit multiplied
by the number of times that it theoretically circles the nucleus per
seconds according to de Broglie and the frequency of the energy induced
at the Bohr radius:
v_0 = (2 pi alpha_0) * nu_0 = 2187691.252 m / sec
So, this all seems to me a simple matter of viewpoint.
As you see, one can cause math to say or "prove" anything that
one cares to believe.
Who is right, who is wrong ?
The future will tell.
André Michaud
|