More QM Nonsense, II
Steve Bell
somewhat anew with the following. Bile says, quite justifiably so, look at the
equations about the "Zitterbeweung" and make up my own mind. I've done so at
http://www.mindspring.com/~sb635/qmnonsense.pdf
This "Darwin effect" in hydrogen still makes no "physical sense" to me. Also,
nothing in this derivation would say that the "Darwin effect" is not present
in all shells.
I would like to recount some facts. The relativistic contribution to
hydrogen's ground state binding energy, according to the series expansion of
Dirac's equation, after taking the expected value of only the relativistic
term, is 5 times larger than what the "deterministic" relativistic orbit
theory demands, per Sommerfeld. The "Darwin effect" is statistical and
"compensates for" or "negates," per the Zitterbewegung" effect, for exactly
one half of the 5-times-too-big "pure" expected relativistic effect as
predicted by QM, bringing the "net" effect on binding energy back to exactly
what Sommerfeld's *only* specially relativistic orbit model predicts (ignoring
the hyperfine splitting). Sommerfeld did not introduce spin-orbit magnetic
effects, but for the ground state, this is irrelevant because QM predicts
there is no spin-orbit coupling in the ground state.
Using the best "consistent set" of values for the electron's and proton's rest
masses and charges, and h, eps_0 and c, etc, the Dirac prediction equals
-13.5984652 eV. Converting this to an equivalent "Rydberg constant for
hydrogen," we get 10967903.3 m^-1. Using h, eps_0 and c, etc in Bohr's
nonrelativistic energy equation, we get 10967757.3 m^-1. The difference is
approximately 146 m^-1. These are the numerical facts given today's best
"consistent set" of values.
Frank Wappler recounts a value of hydrogen's Rydberg constant (R_H) as
"Indeed, a corresponding numerical coincidence can be noted
regarding the value (109677.575 +/- 0.012) cm^-1, reported by
E. R. Cohen, Phys. Rev. 88, (2), 353, (1952; received March 1951)."
The difference between this over-than 50-year old value, and what is given by
*today's* "consistent set" plugged into Bohr's nonrelativistic prediction is
only 0.2 m^-1. The value from Cohen was presumably an "experimentally
determined" value (Resnick sure described it as such), and I assume it is
"unbiased." The quoted error in the 50-year old value is +/- 1.2 m^-1, and we
see that the difference between the 50-year old value, and Bohr's
nonrelativistic prediction, using *today's* values, is essentially
statistically insignificant. The difference between Dirac's prediction and the
observed is 146 m^-1, and, assuming 1.2 m^-1 is one-sigma, 146/1.2 = about a
120 sigma difference. If you have any idea of what "statistically significant"
is, this is *gigantically* significantly different.
A significant question is how could a 50-year old value be so dead-on to a
prediction from *today's* values in Bohr's *nonrelativistic* equation? Perhaps
the older spectroscopic data was "taken as is," as has no bias, and reflects
the true physical condition in hydrogen? A true "physical condition" in
hydrogen is the relativistic effects. These effects should be, and directly
are, observable by frequency absorption and emissions. Yet the "observed" data
apparently supports a nonrelativistic signature.
Something in hydrogen is "compensating for" or "negating" the relativistic
effects. QM itself utilizes this fact in how the "Darwin effect" (which adds a
positive energy) adds to the 5-times-to"deep" *only* relativistic negative
energy contribution. But it still comes out 146 m^-1 too "deep," compared to a
50-year old "observed" value, whose data was plenty enough accurate and
precise to manifest the relativistic effects if present. If hydrogen does not
in fact "outwardly manifest" relativistic effects, this points to a
fundamental lack of "physically correct" modeling of hydrogen by QM.
----------
Steve Bell
mailto:sb...@starband.net
Astroimaging/Physics homepage: http://www.mindspring.com/~sb635
Bob Zombiewoof
> After a welcomed hiatus due to the holidays, I would like to start this thread
> somewhat anew with the following. Bile says, quite justifiably so, look at the
> equations about the "Zitterbeweung" and make up my own mind. I've done so at
>
> http://www.mindspring.com/~sb635/qmnonsense.pdf
>
> This "Darwin effect" in hydrogen still makes no "physical sense" to me. Also,
> nothing in this derivation would say that the "Darwin effect" is not present
> in all shells.
Since you don't want to actually take my suggestion and look up
"foldy-wouthuysen transformation", it isn't likely that you'll find
that the darwin term ever makes sense. It comes from the procedure
which decouples the solutions of the dirac equation for an electron
in an external potential. One of the two-component solutions has the
pauli description as a non-relativistic limit and the other
two-component
solution describes the negative energy states.
By definition, the darwin term is an effect of order, 1/m^2, in the
expansion of a unitary transformation in powers of 1/m:
\exp(iS)H\exp(-iS) = H + i[S,H] + (i^2/2!)[S,[S,H]] + ...
The expansion may be carried out to whatever level of accuracy
one wishes, with successive terms being diminished in importance
in powers of 1/m. The terms at each order what they are.
> I would like to recount some facts. The relativistic contribution to
> hydrogen's ground state binding energy, according to the series expansion of
> Dirac's equation, after taking the expected value of only the relativistic
> term, is 5 times larger than what the "deterministic" relativistic orbit
> theory demands, per Sommerfeld. The "Darwin effect" is statistical and
> "compensates for" or "negates," per the Zitterbewegung" effect, for exactly
> one half of the 5-times-too-big "pure" expected relativistic effect as
> predicted by QM, bringing the "net" effect on binding energy back to
Quantum mechanics is not a relativistic theory for precisely the
reason that a relativistic theory (e.g., using the klein-gordon eqn)
wasn't developed first. A relativistic theory must treat time and
space coordinates on equal footing. The schroedinger equation does not
do this and no one knew how to interpret the klein-gordon equation
due to the apparent non-conserved probability when interpreted as a
single particle theory (which is why dirac tried to avoid the problem
with being second order in the time coordinate by writing an equation
that was first order in the space coordinates). Zitterbewegung is
relativistic in origin and "relativistic" includes more than just
multiplying p by \gamma\beta. In particular, spin and anti-particles
are a result that are not part of the non-relativistic theory. Spin,
for example, must be included by hand into the schroedinger equation.
Anti-particles must be incorporated into the schroedinger equation
by hand (zitterbewegung).
> exactly
> what Sommerfeld's *only* specially relativistic orbit model predicts
> (ignoring the hyperfine splitting). Sommerfeld did not introduce
> spin-orbit magnetic effects, but for the ground state, this is irrelevant because QM predicts there is no spin-orbit coupling in the ground state.
For the simple reason that there is no orbital angular momentum
to couple to the spin. That doesn't seem to be a major problem,
yet because of the spin, it's a relativistic effect that only
appears in orbitals _other_ than s-states.
> Using the best "consistent set" of values for the electron's and
> proton's rest masses and charges, and h, eps_0 and c, etc, the Dirac
> prediction equals -13.5984652 eV. Converting this to an equivalent
>"Rydberg constant for hydrogen," we get 10967903.3 m^-1.
Since I already pointed out that these values are interelated,
in a complex and non-obvious way, through calculations involving
more than you are considering in your simple comparison, your
comparison is of dubious validity.
> The difference between this over-than 50-year old value, and what is given by
> *today's* "consistent set" plugged into Bohr's nonrelativistic prediction is
> only 0.2 m^-1. The value from Cohen was presumably an "experimentally
> determined" value (Resnick sure described it as such), and I assume it is
> "unbiased."
It's not unbiased. A hydrogen atom doesn't simply tell you a value
in
response to a question and it's impossible to tell what you are
comparing
to what. For example, since `c' is a defined quantity and \mu_{0} is
a defined quantity, what does it mean to use the "best values" for `c'
and \epsilon_{0}, since \epsilon_{0} = \mu_{0}c^2? What does it mean
to
use the "best values" for the electron charge and \hbar, when e is
related to \alpha and c, via \alpha = e^2/c\hbar [or, in MKSA, it's
e^2/4\pi\epsilon_{0}\hbar c, or (e^2/\hbar)(\mu_{0}/\epsilon_{0})^1/2
or e^2 Z/\hbar, where Z is the impedance of free space]? Try using
the "best values" of different eras for _measurements_ of quantities
that represent different ways of writing the same thing.
[...]
> A true "physical condition" in hydrogen is the relativistic effects.
> These effects should be, and directly are, observable by frequency
> absorption and emissions. Yet the "observed" data apparently supports
> a nonrelativistic signature.
>
> Something in hydrogen is "compensating for" or "negating" the relativistic
> effects.
No, the relativistic effects are what they are. What exactly is
the spin, if not a relativistic effect? Is something "compensating"
for the effect of the spin in the hydrogen ground state, just because
there is no spin-orbit term when L=0 and the spin-spin term that gives
the hyperfine interaction is very small? What are you calling the
hydrogen ground state? You listed the value -13.5984652 eV, but
the difference in the triplet and singlet states is 0.0000006 eV,
which affects your ground state value in the last decimal place.
> QM itself utilizes this fact in how the "Darwin effect" (which adds a
> positive energy) adds to the 5-times-to"deep" *only* relativistic
>negative energy contribution.
Quantum mechanics doesn't "utilize" the "darwin effect" to explain
anything. The darwin term comes from the relativistic theory and is
its inclusion in the non-relativistic theory is explained by the
relativistic theory and it's importance is determined by where it
appears in an expansion in powers of 1/m in the non-relativistic
reduction
of the relativistic theory.
> But it still comes out 146 m^-1 too "deep," compared to a
> 50-year old "observed" value, whose data was plenty enough accurate and
> precise to manifest the relativistic effects if present. If hydrogen
> does not in fact "outwardly manifest" relativistic effects, this points
> to a fundamental lack of "physically correct" modeling of hydrogen by QM.
Spin is an "outwardly manifest" relativistic effect. This can be
seen just by noting that the spin and the metric are obtained from the
commutator and anticommutator of the same dirac matrices. Quantum
mechanics
does not fully describe the hydrogen atom, because the spin in the
schroedinger equation comes from the non-relativistic reduction of
a fully relativistic quantum theory.
Steve Bell
>
> Steve Bell <sb...@starband.net> wrote in message news:<3E0D6D31...@starband.net>...
> > After a welcomed hiatus due to the holidays, I would like to start this thread
> > somewhat anew with the following. Bile says, quite justifiably so, look at the
> > equations about the "Zitterbeweung" and make up my own mind. I've done so at
> >
> > http://www.mindspring.com/~sb635/qmnonsense.pdf
> >
> > This "Darwin effect" in hydrogen still makes no "physical sense" to me. Also,
> > nothing in this derivation would say that the "Darwin effect" is not present
> > in all shells.
>
> Since you don't want to actually take my suggestion and look up
> "foldy-wouthuysen transformation", it isn't likely that you'll find
> that the darwin term ever makes sense. It comes from the procedure
> which decouples the solutions of the dirac equation for an electron
> in an external potential. One of the two-component solutions has the
> pauli description as a non-relativistic limit and the other
> two-component
> solution describes the negative energy states.
>
I'm sure the "essence" of the "Zitterbeweung" effect is exactly what I gave in
my PDF file. Have you had a chance to derive the expected values for the
"Schroedinger picture?
>
> Quantum mechanics is not a relativistic theory for precisely the
> reason that a relativistic theory (e.g., using the klein-gordon eqn)
> wasn't developed first.
I reject the statement that QM, in the way I refer to it, as expressed by
Dirac's energy prediction equation, is not relativistic.
> A relativistic theory must treat time and
> space coordinates on equal footing. The schroedinger equation does not
> do this and no one knew how to interpret the klein-gordon equation
> due to the apparent non-conserved probability when interpreted as a
> single particle theory (which is why dirac tried to avoid the problem
> with being second order in the time coordinate by writing an equation
> that was first order in the space coordinates). Zitterbewegung is
> relativistic in origin and "relativistic" includes more than just
> multiplying p by \gamma\beta. In particular, spin and anti-particles
> are a result that are not part of the non-relativistic theory. Spin,
> for example, must be included by hand into the schroedinger equation.
> Anti-particles must be incorporated into the schroedinger equation
> by hand (zitterbewegung).
>
> > exactly
> > what Sommerfeld's *only* specially relativistic orbit model predicts
> > (ignoring the hyperfine splitting). Sommerfeld did not introduce
> > spin-orbit magnetic effects, but for the ground state, this is irrelevant because QM predicts there is no spin-orbit coupling in the ground state.
>
> For the simple reason that there is no orbital angular momentum
> to couple to the spin. That doesn't seem to be a major problem,
> yet because of the spin, it's a relativistic effect that only
> appears in orbitals _other_ than s-states.
>
Yes, in ground state hydrogen, QM (and I mean relativistic QM) states there is
no orbital angular momentum. l = 0. The only contributions are relativistic
and Darwin. This is why the ground state is so important. It allows a critical
interpretation of QM regarding how the Darwin effect adds to the sheer
relativistic effects.
> > Using the best "consistent set" of values for the electron's and
> > proton's rest masses and charges, and h, eps_0 and c, etc, the Dirac
> > prediction equals -13.5984652 eV. Converting this to an equivalent
> >"Rydberg constant for hydrogen," we get 10967903.3 m^-1.
>
> Since I already pointed out that these values are interelated,
> in a complex and non-obvious way, through calculations involving
> more than you are considering in your simple comparison, your
> comparison is of dubious validity.
>
Absolutely not. All I did was to take the current values for h, c, etc., and
plug them into Bohr's nonrelativistic energy predicting equations for
hydrogen's ground state (and Dirac's). What else is a person suppose to do if
he/she want's to "get down" with the numbers that these energy predicting
equations tell us?
> > The difference between this over-than 50-year old value, and what is given by
> > *today's* "consistent set" plugged into Bohr's nonrelativistic prediction is
> > only 0.2 m^-1. The value from Cohen was presumably an "experimentally
> > determined" value (Resnick sure described it as such), and I assume it is
> > "unbiased."
>
> It's not unbiased. A hydrogen atom doesn't simply tell you a value
> in
> response to a question and it's impossible to tell what you are
> comparing
> to what.
No it's not. All I am doing is comparing a value obtained from spectroscopic
data to what equations predict. It really is pretty simple.
> For example, since `c' is a defined quantity and \mu_{0} is
> a defined quantity, what does it mean to use the "best values" for `c'
> and \epsilon_{0}, since \epsilon_{0} = \mu_{0}c^2? What does it mean
> to
> use the "best values" for the electron charge and \hbar, when e is
> related to \alpha and c, via \alpha = e^2/c\hbar [or, in MKSA, it's
> e^2/4\pi\epsilon_{0}\hbar c, or (e^2/\hbar)(\mu_{0}/\epsilon_{0})^1/2
> or e^2 Z/\hbar, where Z is the impedance of free space]? Try using
> the "best values" of different eras for _measurements_ of quantities
> that represent different ways of writing the same thing.
>
> [...]
>
> > A true "physical condition" in hydrogen is the relativistic effects.
> > These effects should be, and directly are, observable by frequency
> > absorption and emissions. Yet the "observed" data apparently supports
> > a nonrelativistic signature.
> >
> > Something in hydrogen is "compensating for" or "negating" the relativistic
> > effects.
>
> No, the relativistic effects are what they are. What exactly is
> the spin, if not a relativistic effect? Is something "compensating"
> for the effect of the spin in the hydrogen ground state, just because
> there is no spin-orbit term when L=0 and the spin-spin term that gives
> the hyperfine interaction is very small?
> What are you calling the
> hydrogen ground state?
If you really do not understand what hydrogen's ground state is, then we have
little common ground (pun intended) for discussion. If you don't understand
what I have very succinctly outlined in equations for hydrogen's ground state,
I can't adequately discuss things with you. I have, imo, quite adequately
described what I am calling "hydrogen's ground state."
> You listed the value -13.5984652 eV, but
> the difference in the triplet and singlet states is 0.0000006 eV,
> which affects your ground state value in the last decimal place.
>
The "error" I am talking about is some 10^3 times larger, in the fourth
decimal place.
> > QM itself utilizes this fact in how the "Darwin effect" (which adds a
> > positive energy) adds to the 5-times-to"deep" *only* relativistic
> >negative energy contribution.
>
> Quantum mechanics doesn't "utilize" the "darwin effect" to explain
> anything. The darwin term comes from the relativistic theory and is
> its inclusion in the non-relativistic theory is explained by the
> relativistic theory and it's importance is determined by where it
> appears in an expansion in powers of 1/m in the non-relativistic
> reduction
> of the relativistic theory.
>
Yes, Cohen-Tann gives a nice description of how the "net" effect of the
expected relativistic effects and the Darwin effect (and only these two, since
there is no spin-orbit effect (ignoring the hyperfine)), in hydrogen's ground
state, balance out to what Sommerfeld attributes to only relativistic effects.
> > But it still comes out 146 m^-1 too "deep," compared to a
> > 50-year old "observed" value, whose data was plenty enough accurate and
> > precise to manifest the relativistic effects if present. If hydrogen
> > does not in fact "outwardly manifest" relativistic effects, this points
> > to a fundamental lack of "physically correct" modeling of hydrogen by QM.
>
> Spin is an "outwardly manifest" relativistic effect. This can be
> seen just by noting that the spin and the metric are obtained from the
> commutator and anticommutator of the same dirac matrices. Quantum
> mechanics
> does not fully describe the hydrogen atom, because the spin in the
> schroedinger equation comes from the non-relativistic reduction of
> a fully relativistic quantum theory.
So, are you here admitting that QM does not fully describe the hydrogen atom,
even Dirac's "complete" theory? If so, please explain what you mean by that,
specifically. I have read in QM texts that QM claims there is *nothing* about
the hydrogen atom that is not modeled by QM.
It is a simple fact, it appears, that there exists in the literature of 50
years ago, a value for hydrogen's Rydberg constant that equates almost exactly
to what you get from converting Bohr's nonrelativistic energy prediction into
an inverse meters of the wavelength of the energy of the binding in the ground
state, using the best values of *today.* I can't see how this is a sheer
coincidence.
Bob Zombiewoof
> Bob Zombiewoof wrote:
> >
> > Steve Bell <sb...@starband.net> wrote in message news:<3E0D6D31...@starband.net>...
> >
> > reason that a relativistic theory (e.g., using the klein-gordon eqn)
> > wasn't developed first.
>
>
>
> I reject the statement that QM, in the way I refer to it, as
> expressed by Dirac's energy prediction equation, is not relativistic.
I don't care how you refer to it. What is "Dirac's energy prediction
equation"? Do you mean the "dirac equation"? If so, that's relativistic
quantum mechanics.
[...]
> > For the simple reason that there is no orbital angular momentum
> > to couple to the spin. That doesn't seem to be a major problem,
> > yet because of the spin, it's a relativistic effect that only
> > appears in orbitals _other_ than s-states.
> >
>
> Yes, in ground state hydrogen, QM (and I mean relativistic QM) states
> there is no orbital angular momentum. l = 0.
So does the schroedinger equation.
> The only contributions are relativistic and Darwin.
The darwin term is a relativistic correction.
[...]
> >
> > Since I already pointed out that these values are interelated,
> > in a complex and non-obvious way, through calculations involving
> > more than you are considering in your simple comparison, your
> > comparison is of dubious validity.
> >
>
>
> Absolutely not. All I did was to take the current values for h, c, etc.,
> and plug them into Bohr's nonrelativistic energy predicting equations for
> hydrogen's ground state (and Dirac's). What else is a person suppose to do if
> he/she want's to "get down" with the numbers that these energy predicting
> equations tell us?
Figure out what the equations are telling you, so you can
figure out if what you are calculating means anything, and if
so, what. I previously gave you a link to a physics today article,
which discussed some of the relationships between constants
like the rydberg constant and the electron mass.
> > It's not unbiased. A hydrogen atom doesn't simply tell
> > you a value in response to a question and it's impossible
> > to tell what you are comparing to what.
> No it's not. All I am doing is comparing a value obtained
> from spectroscopic data to what equations predict.
Exactly. You don't even know what you are comparing or
at least you haven't given any indication that you know.
I certainly can't tell from the mishmash you've posted.
> It really is pretty simple.
Then suit yourself.
> > For example, since `c' is a defined quantity and \mu_{0} is
> > a defined quantity, what does it mean to use the "best values" for `c'
> > and \epsilon_{0}, since \epsilon_{0} = \mu_{0}c^2? What does it mean
> > to
> > use the "best values" for the electron charge and \hbar, when e is
> > related to \alpha and c, via \alpha = e^2/c\hbar [or, in MKSA, it's
> > e^2/4\pi\epsilon_{0}\hbar c, or (e^2/\hbar)(\mu_{0}/\epsilon_{0})^1/2
> > or e^2 Z/\hbar, where Z is the impedance of free space]? Try using
> > the "best values" of different eras for _measurements_ of quantities
> > that represent different ways of writing the same thing.
> >
> > [...]
> > What are you calling the
> > hydrogen ground state?
>
>
> If you really do not understand what hydrogen's ground state is,
Don't be a complete imbecile. I had to point out the spin-spin
splitting of the ground state to you in a previous post.
> little common ground (pun intended) for discussion. If you don't understand
> what I have very succinctly outlined in equations for hydrogen's ground state,
> I can't adequately discuss things with you. I have, imo, quite adequately
> described what I am calling "hydrogen's ground state."
No, it's not obvious what you are calling the hydrogen ground state,
since you gave a figure to 6 decimal places and the hyperfine structure
splits the ground state such that the 6 decimal places don't make sense.
>
> > You listed the value -13.5984652 eV, but
> > the difference in the triplet and singlet states is 0.0000006 eV,
> > which affects your ground state value in the last decimal place.
> >
>
> The "error" I am talking about is some 10^3 times larger, in the fourth
> decimal place.
Since you never bother to quantify any errors nor propagate the
errors in your constants through your calculations, it's not very
clear at all what you are calling an error.
[...]
> > Quantum mechanics doesn't "utilize" the "darwin effect" to explain
> > anything. The darwin term comes from the relativistic theory and is
> > its inclusion in the non-relativistic theory is explained by the
> > relativistic theory and it's importance is determined by where it
> > appears in an expansion in powers of 1/m in the non-relativistic
> > reduction of the relativistic theory.
> >
> Yes, Cohen-Tann gives a nice description of how the "net" effect
> of the expected relativistic effects and the Darwin effect (and
> only these two, since there is no spin-orbit effect (ignoring the hyperfine)), in hydrogen's ground
What's your point?
> > Spin is an "outwardly manifest" relativistic effect. This can be
> > seen just by noting that the spin and the metric are obtained from the
> > commutator and anticommutator of the same dirac matrices. Quantum
> > mechanics
> > does not fully describe the hydrogen atom, because the spin in the
> > schroedinger equation comes from the non-relativistic reduction of
> > a fully relativistic quantum theory.
>
> So, are you here admitting that QM does not fully describe the
> hydrogen atom, even Dirac's "complete" theory?
I said quantum mechanics, not relativistic quantum mechanics.
> If so, please explain what you mean by that, specifically. I have
> read in QM texts that QM claims there is *nothing* about the hydrogen
> atom that is not modeled by QM.
They lied. You need relativistic quantum mechanics. Try
explaining the 2s->1s transition without the interaction
from the quantized radiation fields. In quantum mechanics,
that transition is strictly forbidden.
Steve Bell
>
> Steve Bell <sb...@starband.net> wrote in message news:<3E0E48CB...@starband.net>...
> > Bob Zombiewoof wrote:
> > >
> > > Steve Bell <sb...@starband.net> wrote in message news:<3E0D6D31...@starband.net>...
> > >
> > > Quantum mechanics is not a relativistic theory for precisely the
> > > reason that a relativistic theory (e.g., using the klein-gordon eqn)
> > > wasn't developed first.
> >
> >
> >
> > I reject the statement that QM, in the way I refer to it, as
> > expressed by Dirac's energy prediction equation, is not relativistic.
>
> I don't care how you refer to it. What is "Dirac's energy prediction
> equation"? Do you mean the "dirac equation"? If so, that's relativistic
> quantum mechanics.
>
The Dirac equation I refer to is at
http://sb635.mystarband.net/revpap1/page73.gif
Since Dirac used the total special relativistic energy, it includes
relativistic effects. Also, the "piece meal" representation of what Choen-Tann
calls the "Dirac equation in the weakly relativistic domain," which fits the
bill for ground state hydrogen (the electron's velocity is only about 1% c),
is found on page 1213 of his "Quantum Mechanics," vol 2. This equation is
really how I view "fitting the pieces" together for hydrogen. In it, he adds
the specially relativistic effect of time dilation, and the spin-orbit and the
Darwin term.
> [...]
> > > For the simple reason that there is no orbital angular momentum
> > > to couple to the spin. That doesn't seem to be a major problem,
> > > yet because of the spin, it's a relativistic effect that only
> > > appears in orbitals _other_ than s-states.
> > >
> >
> > Yes, in ground state hydrogen, QM (and I mean relativistic QM) states
> > there is no orbital angular momentum. l = 0.
>
> So does the schroedinger equation.
>
> > The only contributions are relativistic and Darwin.
>
> The darwin term is a relativistic correction.
>
Yes, it is, but it is separate from the "mass-velocity" relativistic effect,
which is actually best understood through time dilation.
> [...]
> > >
> > > Since I already pointed out that these values are interelated,
> > > in a complex and non-obvious way, through calculations involving
> > > more than you are considering in your simple comparison, your
> > > comparison is of dubious validity.
> > >
> >
> >
> > Absolutely not. All I did was to take the current values for h, c, etc.,
> > and plug them into Bohr's nonrelativistic energy predicting equations for
> > hydrogen's ground state (and Dirac's). What else is a person suppose to do if
> > he/she want's to "get down" with the numbers that these energy predicting
> > equations tell us?
>
> Figure out what the equations are telling you, so you can
> figure out if what you are calculating means anything, and if
> so, what. I previously gave you a link to a physics today article,
> which discussed some of the relationships between constants
> like the rydberg constant and the electron mass.
>
It's very easy to "see" what the equations are telling you. Each of these
energy equations (Bohr's, Sommerfeld's, Dirac's, etc) is an end results of
some theory. The final equation is written as a function of the electron's
mass and charge, and h and c, etc. If you think inserting the best, current
"consistent set" of these values into an equation to compute the energy the
equation predicts is an incorrect thing to do, you are not what I would call
logical, and there really is no point discussing things with you.
> > > It's not unbiased. A hydrogen atom doesn't simply tell
> > > you a value in response to a question and it's impossible
> > > to tell what you are comparing to what.
>
> > No it's not. All I am doing is comparing a value obtained
> > from spectroscopic data to what equations predict.
>
> Exactly. You don't even know what you are comparing or
> at least you haven't given any indication that you know.
> I certainly can't tell from the mishmash you've posted.
>
I know exactly what I am comparing to what, and I've already described it to
you, and if you can't follow this very simple thread of logic, then too bad.
> > It really is pretty simple.
>
> Then suit yourself.
>
> > > For example, since `c' is a defined quantity and \mu_{0} is
> > > a defined quantity, what does it mean to use the "best values" for `c'
> > > and \epsilon_{0}, since \epsilon_{0} = \mu_{0}c^2? What does it mean
> > > to
> > > use the "best values" for the electron charge and \hbar, when e is
> > > related to \alpha and c, via \alpha = e^2/c\hbar [or, in MKSA, it's
> > > e^2/4\pi\epsilon_{0}\hbar c, or (e^2/\hbar)(\mu_{0}/\epsilon_{0})^1/2
> > > or e^2 Z/\hbar, where Z is the impedance of free space]? Try using
> > > the "best values" of different eras for _measurements_ of quantities
> > > that represent different ways of writing the same thing.
> > >
> > > [...]
>
> > > What are you calling the
> > > hydrogen ground state?
> >
> >
> > If you really do not understand what hydrogen's ground state is,
>
> Don't be a complete imbecile. I had to point out the spin-spin
> splitting of the ground state to you in a previous post.
>
Oh, and by the way, you were incorrect about that, according to Cohen-Tann. On
p. 1218 he gives the hyperfine correction terms. The second term term gives
the spin-spin magnetic field interactions, that is, the correction needed for
the interaction between the dipole magnetic field set up by the intrinsic spin
of the electron and the "atom-wide" dipole magnetic field due to the spinning
of the proton. On p. 1228 on the ground state 1s level, he states: "Let us
show that the first two terms make no contribution... it can be shown that the
matrix elements of the second term (the dipole-dipole interaction) are zero
because of the spherical symmetry of the 1s state." The actual physical cause
of the hyperfine splitting is due to the interaction of the magnetic field set
up by the intrinsic spin of the electron, and the internal, non-Coulomb
(Cohen-Tann assumes uniform) field *within* the proton. He calls it the "Fermi
contact term," and he inserts this term as a "correction term" in addition to
the dipole-dipole term, which has zero-effect for the ground state.
> > little common ground (pun intended) for discussion. If you don't understand
> > what I have very succinctly outlined in equations for hydrogen's ground state,
> > I can't adequately discuss things with you. I have, imo, quite adequately
> > described what I am calling "hydrogen's ground state."
>
> No, it's not obvious what you are calling the hydrogen ground state,
> since you gave a figure to 6 decimal places and the hyperfine structure
> splits the ground state such that the 6 decimal places don't make sense.
>
The hyperfine splitting is too small to be significantly relevant to my
arguments.
> >
> > > You listed the value -13.5984652 eV, but
> > > the difference in the triplet and singlet states is 0.0000006 eV,
> > > which affects your ground state value in the last decimal place.
> > >
> >
> > The "error" I am talking about is some 10^3 times larger, in the fourth
> > decimal place.
>
> Since you never bother to quantify any errors nor propagate the
> errors in your constants through your calculations, it's not very
> clear at all what you are calling an error.
>
It is clear to me, and I hope it is clear to other "open minded" people
reading this thread. You are a QM "priest" who refuses to accept any "anti-QM"
statement. It is of little use, and it is becoming of increasingly less
interest to me, in my discussing things with you.
> [...]
> > > Quantum mechanics doesn't "utilize" the "darwin effect" to explain
> > > anything. The darwin term comes from the relativistic theory and is
> > > its inclusion in the non-relativistic theory is explained by the
> > > relativistic theory and it's importance is determined by where it
> > > appears in an expansion in powers of 1/m in the non-relativistic
> > > reduction of the relativistic theory.
> > >
> > Yes, Cohen-Tann gives a nice description of how the "net" effect
> > of the expected relativistic effects and the Darwin effect (and
> > only these two, since there is no spin-orbit effect (ignoring the hyperfine)), in hydrogen's ground
>
> What's your point?
>
The point is how to mathematically incorporate the Darwin term.
> > > Spin is an "outwardly manifest" relativistic effect. This can be
> > > seen just by noting that the spin and the metric are obtained from the
> > > commutator and anticommutator of the same dirac matrices. Quantum
> > > mechanics
> > > does not fully describe the hydrogen atom, because the spin in the
> > > schroedinger equation comes from the non-relativistic reduction of
> > > a fully relativistic quantum theory.
> >
> > So, are you here admitting that QM does not fully describe the
> > hydrogen atom, even Dirac's "complete" theory?
>
> I said quantum mechanics, not relativistic quantum mechanics.
> > If so, please explain what you mean by that, specifically. I have
> > read in QM texts that QM claims there is *nothing* about the hydrogen
> > atom that is not modeled by QM.
>
Every time you use "QM" I naturally assume you are referring to the "QM" that
incorporates relativistic effects.
> They lied. You need relativistic quantum mechanics. Try
> explaining the 2s->1s transition without the interaction
> from the quantized radiation fields. In quantum mechanics,
> that transition is strictly forbidden.
I have been comparing the relativistic Dirac binding energy equation
prediction for ground state hydrogen to the observed.
Gordon D. Pusch
> Bob Zombiewoof wrote:
[...]
> > I don't care how you refer to it. What is "Dirac's energy prediction
> > equation"? Do you mean the "dirac equation"? If so, that's relativistic
> > quantum mechanics.
> >
>
> The Dirac equation I refer to is at
>
> http://sb635.mystarband.net/revpap1/page73.gif
That is not the Dirac Equation. The Dirac Eqation is:
[\gamma^\mu (\partial_\mu - ie A_\mu) + m_0] \psi = 0.
Your alleged "Dirac equation" appears to perhaps be somebody or other's
expression for the relativistic Hydrogen energy levels --- and moreover
it is not even _Dirac's_ prediction for them, which, if one neglects
external magnetic fields and the hyperfine interaction, is:
E_{n,l} = m_0 / sqrt{ 1 + Z^2 * \alpha^2
/ [(n - l - 1/2)
+ sqrt((l + 1/2)^2 - Z^2 * \alpha^2)]^2 }
[Itzykson & Zuber, "Quantum Field Theory" eqn (2-86), p.73 (McGraw-Hill 1980)].
Do please try to get your facts (and your equations) straight before you post,
and stop tilting at windmills and straw men.
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
Steve Bell
>
> Steve Bell <sb...@starband.net> writes:
>
> > Bob Zombiewoof wrote:
> [...]
> > > I don't care how you refer to it. What is "Dirac's energy prediction
> > > equation"? Do you mean the "dirac equation"? If so, that's relativistic
> > > quantum mechanics.
> > >
> >
> > The Dirac equation I refer to is at
> >
> > http://sb635.mystarband.net/revpap1/page73.gif
>
> That is not the Dirac Equation. The Dirac Eqation is:
>
> [\gamma^\mu (\partial_\mu - ie A_\mu) + m_0] \psi = 0.
>
> Your alleged "Dirac equation" appears to perhaps be somebody or other's
> expression for the relativistic Hydrogen energy levels --- and moreover
> it is not even _Dirac's_ prediction for them, which, if one neglects
> external magnetic fields and the hyperfine interaction, is:
>
Duuh, what the hell do you think I've been talking about all this time!? It's
been hydrogen's energy levels, and if you don't realize that by now, I can
only shake my head in amusement.
> E_{n,l} = m_0 / sqrt{ 1 + Z^2 * \alpha^2
> / [(n - l - 1/2)
> + sqrt((l + 1/2)^2 - Z^2 * \alpha^2)]^2 }
>
> [Itzykson & Zuber, "Quantum Field Theory" eqn (2-86), p.73 (McGraw-Hill 1980)].
>
> Do please try to get your facts (and your equations) straight before you post,
> and stop tilting at windmills and straw men.
>
> -- Gordon D. Pusch
>
> perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
I quote from Resnick's "Quantum Physics of Atoms, etc.", p. 286: we "present
the results of Dirac's completely relativistic treatment of the hydrogen atom
energy levels, which predicts that the energies are..."
And then he gives the exact same equation I show at
http://sb635.mystarband.net/revpap1/page73.gif
I know that the words "Dirac equation" can refer to different things, but I
figured you guys were smart enough to realize which "Dirac equations" I have
been talking about. I guess not. There is another version on p. 1226 of
Cohen-Tann's "Quantum Mechanics," vol. 2. I've checked, and these two
equations give the same energy value when you plug in the values for h, c,
eps_0, etc. And of course, the one I like the best for it's "piece meal"
representation is Choen-Tann's equation, in the section entitled "The Dirac
Equation in the Weakly Relativistic Domain," p. 1213, vol 2. This equation,
too, gives the same value for hydrogen's ground state, as any "Dirac" equation
that I have been talking about, namely -1.81 x 10^-4 eV "deeper" than Bohr's
equation. You do know what Bohr's equation is, don't you?
Thinh Tran
> Yes, in ground state hydrogen, QM (and I mean relativistic QM) states there is
> no orbital angular momentum. l = 0. The only contributions are relativistic
> and Darwin. This is why the ground state is so important. It allows a critical
> interpretation of QM regarding how the Darwin effect adds to the sheer
> relativistic effects.
>
to predict the Lamb's shift (error proportional to 1/n^3 and therefore
most serious for n=1) so of course it (the Dirac equation) should not
be trusted for the ground state n=1, where l can only take the value
0.
If I may add, the Dirac equation also predicts -mc^2 for the
energy of the anti-electron. We know that this is mathematically
incorrect because energy for anti-particles are confirmed
experimentally to be positive. (I posed this question on the thread
"anti-particles" in sci.phys.research and got an answer from Joan Baez
recently.)
In brief, the Dirac equation has historical value but is
hopelessly incomplete and should not be trusted as a standard to judge
QM. I would recommend looking into the most popular version of QED
(i.e., the one formulated by Feynman) if you're interested in a
numerical critique of the QM theory.
Of course it is a good thing to know all deficiencies so that we
don't have a wrong picture of QM and somehow believe that it is
perfect in all calculations.
Thinh Tran (http://www.thinhtran.com)
Steve Bell
>
> I quote from Resnick's "Quantum Physics of Atoms, etc.", p. 286: we "present
> the results of Dirac's completely relativistic treatment of the hydrogen atom
> energy levels, which predicts that the energies are..."
>
> And then he gives the exact same equation I show at
>
> http://sb635.mystarband.net/revpap1/page73.gif
>
> I know that the words "Dirac equation" can refer to different things, but I
> figured you guys were smart enough to realize which "Dirac equations" I have
> been talking about. I guess not. There is another version on p. 1226 of
> Cohen-Tann's "Quantum Mechanics," vol. 2. I've checked, and these two
> equations give the same energy value when you plug in the values for h, c,
> eps_0, etc. And of course, the one I like the best for it's "piece meal"
> representation is Choen-Tann's equation, in the section entitled "The Dirac
> Equation in the Weakly Relativistic Domain," p. 1213, vol 2. This equation,
> too, gives the same value for hydrogen's ground state, as any "Dirac" equation
> that I have been talking about, namely -1.81 x 10^-4 eV "deeper" than Bohr's
> equation. You do know what Bohr's equation is, don't you?
>
If there are any "silent readers" of this thread (and the others I've
initiated), I would appreciate a stated consensus from each, about if you feel
I've raised valid points, or not. I would appreciate hearing from people that
haven't already responded too much. This is just a request, and of course you
can sit silent, if you wish. But I would like some constructive criticism from
people who know (and can demonstrate) something about hydrogen's ground state
energy. And I promise that if you will be civil, so will I.
Believe me, I understand how "radical" a statement it is to say that QM falls
noticeably short in its ground state energy prediction for hydrogen, and that
this has not been caught by now, if it were true. But if anyone can logically
explain (Bilge, you do not have to respond to this - I already know your
illogical opinion) about how a 50 year old value ended up essentially equaling
a prediction that you can make using *today's* current best "consistent set"
of h, c, etc, then I would like to hear it. And I hope you are appreciating
the significance that the "experimental" value was obtained over 50 years ago,
with sufficient accuracy and precision to show the "net" of the physical
effects occurring in hydrogen. There is a data point that gives some
information. If you believe Resnick, the "set" of values back in the 1985
"era" were whatever they were to give a *computed* Rydberg constant value for
hydrogen as 10968100 m^-1. This equates to an energy, using today's values of
h and c, of -13.5987094 eV. This is over -4.25 x 10^-4 eV (or about 343 m^-1)
"deeper" than the observed, well outside of the range of "experimental error"
in the observed, but still only -1.81 x 10^-4 eV different than the *then's*
Bohr energy was, using the *then* set in Bohr's equation. This is
understandable due to the fact that the equations themselves haven't changed
since their inception.
Steve Bell
>
> Steve Bell <sb...@starband.net> wrote in message news:<3E0E48CB...@starband.net>...
>
> > Yes, in ground state hydrogen, QM (and I mean relativistic QM) states there is
> > no orbital angular momentum. l = 0. The only contributions are relativistic
> > and Darwin. This is why the ground state is so important. It allows a critical
> > interpretation of QM regarding how the Darwin effect adds to the sheer
> > relativistic effects.
> >
> By combining s+l=j, the Hydrogen solution for the Dirac equation fails
> to predict the Lamb's shift (error proportional to 1/n^3 and therefore
> most serious for n=1) so of course it (the Dirac equation) should not
> be trusted for the ground state n=1, where l can only take the value
> 0.
>
I've gone over the Lamb shift issue already (please see my previous threads).
The "Dirac" equations if have been referring to take it into account.
If I may add, the Dirac equation also predicts -mc^2 for the
> energy of the anti-electron. We know that this is mathematically
> incorrect because energy for anti-particles are confirmed
> experimentally to be positive. (I posed this question on the thread
> "anti-particles" in sci.phys.research and got an answer from Joan Baez
> recently.)
Yes, negative mass is interesting. But of course, if I gave my opinions about
that, I would be ridiculed.
> In brief, the Dirac equation has historical value but is
> hopelessly incomplete and should not be trusted as a standard to judge
> QM. I would recommend looking into the most popular version of QED
> (i.e., the one formulated by Feynman) if you're interested in a
> numerical critique of the QM theory.
> Of course it is a good thing to know all deficiencies so that we
> don't have a wrong picture of QM and somehow believe that it is
> perfect in all calculations.
> Thinh Tran (http://www.thinhtran.com)
According to the texts where I get the "Dirac" equations I use, they are the
best that QM has to offer, including QED.
Steve Bell
I would like to add that I have formulated a theoretical explanation of how
hydrogen could possibly not outwardly manifest any relativistic effects. Its
basis is in the deterministic theory of "differential geometries,"
specifically Kerr geometries, and to accept it you must accept that fact that,
at the instant in time we call "the present," an electron exists as a particle
whose center of mass resides at a definite point in space, and that the
electron's center of mass also has a definite velocity. This is needed in the
differential geometry due to the total force on the electron being not only a
function of its position, but also a function of its velocity. By the way, if
this is true, it completely negates the separability of Schroedinger's
equation into independent radial and angular waves, and throws a monkey-wrench
into just about all of QM.
Thinh Tran
> Yes, negative mass is interesting. But of course, if I gave my opinions about
> that, I would be ridiculed.
Negative mass is not only interesting, but also would be a very
detrimental blow to general relativity. If I remember the logic
correctly, GR is simply wrong if it is confirmed that negative mass
exists.
>
> > In brief, the Dirac equation has historical value but is
> > hopelessly incomplete and should not be trusted as a standard to judge
> > QM. I would recommend looking into the most popular version of QED
> > (i.e., the one formulated by Feynman) if you're interested in a
> > numerical critique of the QM theory.
> > Of course it is a good thing to know all deficiencies so that we
> > don't have a wrong picture of QM and somehow believe that it is
> > perfect in all calculations.
> > Thinh Tran (http://www.thinhtran.com)
>
>
> According to the texts where I get the "Dirac" equations I use, they are the
> best that QM has to offer, including QED.
Maybe we're using the same name to refer to different things here. I'm
referring to the original equation for the free electron that Dirac
came up with in 1928, where he discovered spin and the anti-electron.
HAPPY HOLIDAYS!!!
Thinh Tran (http://www.thinhtran.com)
Thinh Tran
> Thinh Tran wrote:
> > By combining s+l=j, the Hydrogen solution for the Dirac equation fails
> > to predict the Lamb's shift (error proportional to 1/n^3 and therefore
> > most serious for n=1) so of course it (the Dirac equation) should not
> > be trusted for the ground state n=1, where l can only take the value
> > 0.
>
> I've gone over the Lamb shift issue already (please see my previous threads).
> The "Dirac" equations if have been referring to take it into account.
>
Sorry, I forgot to comment on this subject in my last posting. I read
the older thread where you mentioned the Lamb shift.
Just a minor point: The Lamb shift can also be checked at l=0
(although there is nothing to compared to). The experimentalists
simply compared the predicted energy value against the observed value
seen in the spectrum. In fact, I believe there are lots of people who
are measuring the Lamb shift for n=1, l=0 for various "Hydrogen-like"
materials and post their results on the net.
HAPPY HOLIDAYS
Thinh Tran (http://www.thinhtran.com)
Steve Bell
>
> Steve Bell <sb...@starband.net> wrote in message news:<3E0F4CCE...@starband.net>...
>
> > Yes, negative mass is interesting. But of course, if I gave my opinions about
> > that, I would be ridiculed.
>
> Negative mass is not only interesting, but also would be a very
> detrimental blow to general relativity. If I remember the logic
> correctly, GR is simply wrong if it is confirmed that negative mass
> exists.
> >
The way I view "negative mass" and its implications on motion, I just view the
accelerations produced by the curvature of spacetime, on the orbiting body, as
if the central body had a "positive mass," and reverse the direction of the
acceleration vector. This way, it does not have a detrimental blow to GR. The
result is an "outward pushing" of the velocity of the orbiting body.
> > > In brief, the Dirac equation has historical value but is
> > > hopelessly incomplete and should not be trusted as a standard to judge
> > > QM. I would recommend looking into the most popular version of QED
> > > (i.e., the one formulated by Feynman) if you're interested in a
> > > numerical critique of the QM theory.
> > > Of course it is a good thing to know all deficiencies so that we
> > > don't have a wrong picture of QM and somehow believe that it is
> > > perfect in all calculations.
> > > Thinh Tran (http://www.thinhtran.com)
> >
> >
> > According to the texts where I get the "Dirac" equations I use, they are the
> > best that QM has to offer, including QED.
>
> Maybe we're using the same name to refer to different things here. I'm
> referring to the original equation for the free electron that Dirac
> came up with in 1928, where he discovered spin and the anti-electron.
>
> HAPPY HOLIDAYS!!!
> Thinh Tran (http://www.thinhtran.com)
Probably so, but the various "Dirac" equations for the binding energy of
ground state hydrogen that I have been referring to are for a bound electron
(and have given the equations - please see my previous posts), in hydrogen's
ground state orbital, and not for a "free" electron (accept for the
"Zitterbewegung" effect equations I've cited, whose essence is understood for
the equation of the operator motion for a "free" particle, in the "Heisenberg
picture").
If you have been following my threads, what is your opinion of the "outward
manifestation" of relativistic effects in ground state hydrogen?
Gordon D. Pusch
> "Gordon D. Pusch" wrote:
>>
>> Steve Bell <sb...@starband.net> writes:
>>
>>> Bob Zombiewoof wrote:
>> [...]
>>>> I don't care how you refer to it. What is "Dirac's energy prediction
>>>> equation"? Do you mean the "dirac equation"? If so, that's relativistic
>>>> quantum mechanics.
>>>>
>>>
>>> The Dirac equation I refer to is at
>>>
>>> http://sb635.mystarband.net/revpap1/page73.gif
>>
>> That is not the Dirac Equation. The Dirac Eqation is:
>>
>> [\gamma^\mu (\partial_\mu - ie A_\mu) + m_0] \psi = 0.
>>
>> Your alleged "Dirac equation" appears to perhaps be somebody or other's
>> expression for the relativistic Hydrogen energy levels --- and moreover
>> it is not even _Dirac's_ prediction for them, which, if one neglects
>> external magnetic fields and the hyperfine interaction, is:
>
> Duuh, what the hell do you think I've been talking about all this time!? It's
> been hydrogen's energy levels, and if you don't realize that by now, I can
> only shake my head in amusement.
Then your should call it "the equation for hydrogen energy levels," not the
"Dirac equation," since the Dirac equation is:
[\gamma^\mu (\partial_\mu - ie A_\mu) + m_0] \psi = 0.
Furthermore, since your alleged "Dirac equation" is =NOT= in fact the
prediction the Dirac Equation yields for the energy levels of hydrogen,
the correct expression for which I've quoted above, you are clearly
quite confused about both the Dirac Equation, and what it predicts.
>> E_{n,l} = m_0 / sqrt{ 1 + Z^2 * \alpha^2
>> / [(n - l - 1/2)
>> + sqrt((l + 1/2)^2 - Z^2 * \alpha^2)]^2 }
>>
>> [Itzykson & Zuber, "Quantum Field Theory" eqn (2-86), p.73 (McGraw-Hill 1980)].
>>
>> Do please try to get your facts (and your equations) straight before you post,
>> and stop tilting at windmills and straw men.
>
> I quote from Resnick's "Quantum Physics of Atoms, etc.", p. 286: we "present
> the results of Dirac's completely relativistic treatment of the hydrogen atom
> energy levels, which predicts that the energies are..."
>
> And then he gives the exact same equation I show at
>
> http://sb635.mystarband.net/revpap1/page73.gif
I suggest you go back and re-read it more carefully. At best, the equation
you cite can be a leading-order _APPROXIMATION_ to the =EXACT= analytical
result I quote above. (And indeed, this has been a common conceptual error
you have made many times throughout this thread: You seem to be incapable
of distinguishing between FUNDAMENTAL results and APPROXIMATE results
obtained via perturbation theory or other means.)
> I know that the words "Dirac equation" can refer to different things,
On the contrary: Ask any physicist what the "Dirac Equation" is, and they will
immediately write down [\gamma^\mu (\partial_\mu - ie A_\mu) + m_0] \psi = 0,
not some approximate expression for the energy levels of hydrogen.
Steve Bell
>
> >
> > Duuh, what the hell do you think I've been talking about all this time!? It's
> > been hydrogen's energy levels, and if you don't realize that by now, I can
> > only shake my head in amusement.
>
> Then your should call it "the equation for hydrogen energy levels," not the
> "Dirac equation," since the Dirac equation is:
>
> [\gamma^\mu (\partial_\mu - ie A_\mu) + m_0] \psi = 0.
>
My terminology of saying a "Dirac equation" for predicting ground state
hydrogen's energy is completely appropriate.
> Furthermore, since your alleged "Dirac equation" is =NOT= in fact the
> prediction the Dirac Equation yields for the energy levels of hydrogen,
> the correct expression for which I've quoted above, you are clearly
> quite confused about both the Dirac Equation, and what it predicts.
>
> >> E_{n,l} = m_0 / sqrt{ 1 + Z^2 * \alpha^2
> >> / [(n - l - 1/2)
> >> + sqrt((l + 1/2)^2 - Z^2 * \alpha^2)]^2 }
> >>
> >> [Itzykson & Zuber, "Quantum Field Theory" eqn (2-86), p.73 (McGraw-Hill 1980)].
> >>
Could you please plug in the best "consistent set" of values of h, c, eps_0,
etc. (see NIST), and tell us what the above equation predicts? And express it
in eV, please. I have computed enough "Dirac" binding energy values, and I
really don't want to (or see a need to) do one more. If you (or whoever
derived the above equation) did it correctly, it will give the same binding
energy for hydrogen's ground state that I've been quoting for quite some time
now.
> >> Do please try to get your facts (and your equations) straight before you post,
> >> and stop tilting at windmills and straw men.
> >
> > I quote from Resnick's "Quantum Physics of Atoms, etc.", p. 286: we "present
> > the results of Dirac's completely relativistic treatment of the hydrogen atom
> > energy levels, which predicts that the energies are..."
> >
> > And then he gives the exact same equation I show at
> >
> > http://sb635.mystarband.net/revpap1/page73.gif
>
> I suggest you go back and re-read it more carefully. At best, the equation
> you cite can be a leading-order _APPROXIMATION_ to the =EXACT= analytical
> result I quote above. (And indeed, this has been a common conceptual error
> you have made many times throughout this thread: You seem to be incapable
> of distinguishing between FUNDAMENTAL results and APPROXIMATE results
> obtained via perturbation theory or other means.)
>
> > I know that the words "Dirac equation" can refer to different things,
>
> On the contrary: Ask any physicist what the "Dirac Equation" is, and they will
> immediately write down [\gamma^\mu (\partial_\mu - ie A_\mu) + m_0] \psi = 0,
> not some approximate expression for the energy levels of hydrogen.
>
> -- Gordon D. Pusch
>
Yes, I understand that, and I apologize, but I've made it clear as to which
"Dirac" binding energy equation for ground state hydrogen I was referring to.
Do you now understand what I am referring to? Although, I really would request
that you not respond. I would like to hear from other, non-biased individuals.
And if no one else responds, and if all that I've been doing is discussing
these things with you QM "priests," I'll give up on this NG. What's the point?
I will say something, and you will respond with something that you think
refutes what I've said, and then I'll respond with a refutation of your
refutation, and on and on...
Does anyone know of any "crackpot" NGs? I'd probably get more consideration
there. I absolutely admit that the ideas I profess will be claimed "crackpot"
by the mainstream QM community. I expect it, and admire this "band of
courage."
You are perhaps not the best skilled individual to ask this of, so I'll ask it
generally, do you believe there will be another "revolution" in physics? By
"revolution" I mean something like the Einstein revolution, or the Quantum
revolution initiated by Max Planck. Something truthfully revolutionary about
how we view motion and probability in the realm of the atom? But if you say
that the "revolution" will come about by the path of present-day accepted QM
itself (which includes string theory, imo), I reject that. Something else is
needed to resolve the impasse of deterministic GR and probabilistic QM.
And to Gordon D. Pusch, please do not respond. I have had enough of your
nonsense. And likewise to Bilge. Although I do want Bilge to respond to my
statement that he was incorrect in describing the spin-spin hyperfine magnetic
effect in ground state hydrogen. Or anyone else, for that matter.
Patrick Reany
Steve Bell wrote:
Take a look at the approach to the Dirac equation fostered by Hestenes at
http://modelingNTS.la.asu.edu/html/STC.html
http://modelingNTS.la.asu.edu/html/GAinQM.html
and by the Cambridge group at
http://www.mrao.cam.ac.uk/~clifford/publications/
These refernce may not gve a precise answer to your question,
but you may find them interesting anyway.
Patrick
Steve Bell
>
> Steve Bell <sb...@starband.net> wrote in message news:<3E0F4CCE...@starband.net>...
> > Thinh Tran wrote:
>
> > > By combining s+l=j, the Hydrogen solution for the Dirac equation fails
> > > to predict the Lamb's shift (error proportional to 1/n^3 and therefore
> > > most serious for n=1) so of course it (the Dirac equation) should not
> > > be trusted for the ground state n=1, where l can only take the value
> > > 0.
> >
> > I've gone over the Lamb shift issue already (please see my previous threads).
> > The "Dirac" equations if have been referring to take it into account.
> >
> Sorry, I forgot to comment on this subject in my last posting. I read
> the older thread where you mentioned the Lamb shift.
> Just a minor point: The Lamb shift can also be checked at l=0
> (although there is nothing to compared to). The experimentalists
> simply compared the predicted energy value against the observed value
> seen in the spectrum.
It is significant that you see this. There is nothing to compare to because,
QM says, technically speaking, there is no "Lamb shift," meaning a "fine
structure" splitting of the ground state level (other than the hyperfine
splitting, which is not a "Lamb shift" effect).
The "observed" value that I have been quoting for ground state hydrogen
supports the conclusion that there does not exist any "outwardly manifest"
effects of any kind (special or Darwin, or spin-orbit, or spin-spin, etc) ,
other than the "reduced mass" effects that we know must be true. I can explain
how this can "physically happen," if you wish.
> In fact, I believe there are lots of people who
> are measuring the Lamb shift for n=1, l=0 for various "Hydrogen-like"
> materials and post their results on the net.
> HAPPY HOLIDAYS
> Thinh Tran (http://www.thinhtran.com)
The important issue is what the various theories predict for ground state
hydrogen. There, due to all of QM's nonsensical "spherically symmetric"-based
conclusions of the ground state, its predictions can be analyzed with a
"critical" eye.
For example, I just pointed out that Bilge's interpretation of the spin-spin
magnetic hyperfine interaction in the ground state, is incorrect, and that it
is zero according to QM's best theoretical development. Now, let me ask you:
does the electron have a spin? Yes, it does. Does the spin set up a dipole
magnetic field? Yes, it does. Is the proton spinning? Yes, it is. Does the
proton's spin set up a dipole magnetic field? Yes, it does. Should the
electron's spin magnetic field interact with the proton's spin magnetic field,
even in the ground state, where all of these conditions are met? Yes, it
should. But QM says it doesn't. Bilge himself thought that it should. I view
this as a "physically nonsensical" prediction by QM, tied all the way back to
the use of an imaginary wave equation from the get go, and how this drives the
physically nonsensical expectations in QM, especially in the ground state,
where "spherical symmetry" can bite QM in the ass, like in its quantification
of the "hyperfine" splitting.
If the electronic field surrounding the proton can be modeled as a type of
extremely localized, very tightly curved Kerr spacetime, then the proton's
electromagnetic field is *not* spherically symmetric, but is either oblate
spheriodal or prolate spheriodal, depending on the proton's sense of direction
of its spin. This kills QM's separability of the Schroedinger's equation,
which lies at the heart and soul of all QM theory.
Steve Bell
>
>
> Take a look at the approach to the Dirac equation fostered by Hestenes at
>
> http://modelingNTS.la.asu.edu/html/STC.html
> http://modelingNTS.la.asu.edu/html/GAinQM.html
>
> and by the Cambridge group at
>
> http://www.mrao.cam.ac.uk/~clifford/publications/
>
> These refernce may not gve a precise answer to your question,
> but you may find them interesting anyway.
>
> Patrick
Thanks for these references. They imploy the "physically nonsensical" use of
the square root of minus one (i) in their developments, and I would say that
most of what they have to say is therefore "physically nonsensible."
I do find it interesting how physicists can rely on i, in such a degree, in
defining their theories of "physical reality," even though i doesn't represent
anything "physically real." I accord this due to some kind of "pandemic
brainwashing" that occurred shortly after the turn of the century, due to the
development of Schroedinger's imaginary wave equation. Imo, no equation that
contains "i" will ever be found to be "physically correct." After
Schroedinger's development, all the physicists "jumped on board," so to speak.
Schroedinger's development was certainly "brilliant," and I agree to a degree
with its deterministic predictions of quantized energy, that is, without
considering relativistic effects. But the abhorrent probability postulate of
Born is too much to bear. The electron in ground state hydrogen is undoubtedly
"jittering" in its motion, but with a gigantically smaller variance than what
Born's QM predicts. I think the observed extremely "thin" spectral lines of
hydrogen support this.
Bob Zombiewoof
The dirac equation is the not only the fundamental basis of qed,
but qed is practically an inescapable consequence of the dirac equation.
If you start with the dirac lagrangian, L = \psibar(p/ - m)\psi, and
require the lagrangian be lorentz invariant under a local gauge
transformation, \psi -> \psi\exp(iS), then the qed lagrangian follows
immediately. Since this is nothing but a restatement that the phase of
the wavefunction is unobservable, there is nothing surprising about
requiring this. Through noether's theorem, the result is a conserved
current (which is exactly the electromagnetic current density j^{u})
and the electron _must_ carry a field. Noether's theorem only appeared
in 1918 and in 1928, its importance in field theory simply wasn't
appreciated to the extent it is now.
Bob Zombiewoof
> I would like to add that I have formulated a theoretical explanation of how
> hydrogen could possibly not outwardly manifest any relativistic effects.
Since the spin is a relativistic effect, your claim can't be correct.
> basis is in the deterministic theory of "differential geometries,"
> specifically Kerr geometries, and to accept it you must accept that
> fact that, at the instant in time we call "the present," an electron
> exists as a particle whose center of mass resides at a definite point
> in space,
Calculate the electron mass from a charge distribution with the
classical electron radius. Given that the electron is known from
experiments to at least a factor 100 times smaller than this, how
do you plan to explain why its mass is 0.511 MeV?
> electron's center of mass also has a definite velocity. This is needed in the
> differential geometry due to the total force on the electron being not only a
> function of its position, but also a function of its velocity. By the way, if
> this is true, it completely negates the separability of Schroedinger's
> equation into independent radial and angular waves, and throws a monkey-wrench
> into just about all of QM.
Since the schroedinger equation is not relativistic, what's you point?
You can't explain the electron without field theory.
Gordon D. Pusch
A simple method for rating potentially revolutionary contributions to
physics.
1. A -5 point starting credit.
2. 1 point for every statement that is widely agreed on to be false.
3. 2 points for every statement that is clearly vacuous.
4. 3 points for every statement that is logically inconsistent.
5. 5 points for each such statement that is adhered to despite careful
correction.
6. 5 points for using a thought experiment that contradicts the results of
a widely accepted real experiment.
7. 5 points for each word in all capital letters (except for those with
defective keyboards).
8. 5 points for each mention of "Einstien", "Hawkins" or "Feynmann".
9. 10 points for each claim that quantum mechanics is fundamentally
misguided (without good evidence).
10. 10 points for pointing out that you have gone to school, as if this
were evidence of sanity.
11. 10 points for beginning the description of your theory by saying how
long you have been working on it.
12. 10 points for mailing your theory to someone you don't know personally
and asking them not to tell anyone else about it, for fear that your
ideas will be stolen.
13. 10 points for offering prize money to anyone who proves and/or finds
any flaws in your theory.
14. 10 points for each statement along the lines of "I'm not good at math,
but my theory is conceptually right, so all I need is for someone to
express it in terms of equations".
15. 10 points for arguing that a current well-established theory is "only a
theory", as if this were somehow a point against it.
16. 10 points for arguing that while a current well-established theory
predicts phenomena correctly, it doesn't explain "why" they occur, or
fails to provide a "mechanism".
17. 10 points for each favorable comparison of yourself to Einstein, or
claim that special or general relativity are fundamentally misguided
(without good evidence).
18. 10 points for claiming that your work is on the cutting edge of a
"paradigm shift".
19. 20 points for suggesting that you deserve a Nobel prize.
20. 20 points for each favorable comparison of yourself to Newton or claim
that classical mechanics is fundamentally misguided (without good
evidence).
21. 20 points for every use of science fiction works or myths as if they
were fact.
22. 20 points for defending yourself by bringing up (real or imagined)
ridicule accorded to your past theories.
23. 20 points for each use of the phrase "hidebound reactionary".
24. 20 points for each use of the phrase "self-appointed defender of the
orthodoxy".
25. 30 points for suggesting that a famous figure secretly disbelieved in a
theory which he or she publicly supported. (E.g., that Feynman was a
closet opponent of special relativity, as deduced by reading between
the lines in his freshman physics textbooks.)
26. 30 points for suggesting that Einstein, in his later years, was groping
his way towards the ideas you now advocate.
27. 30 points for claiming that your theories were developed by an
extraterrestrial civilization (without good evidence).
28. 30 points for allusions to a delay in your work while you spent time in
an asylum, or references to the psychiatrist who tried to talk you out
of your theory.
29. 40 points for comparing those who argue against your ideas to Nazis,
stormtroopers, or brownshirts.
30. 40 points for claiming that the "scientific establishment" is engaged
in a "conspiracy" to prevent your work from gaining its well-deserved
fame, or suchlike.
31. 40 points for comparing yourself to Galileo, suggesting that a
modern-day Inquisition is hard at work on your case, and so on.
32. 40 points for claiming that when your theory is finally appreciated,
present-day science will be seen for the sham it truly is. (30 more
points for fantasizing about show trials in which scientists who mocked
your theories will be forced to recant.)
33. 50 points for claiming you have a revolutionary theory but giving no
concrete testable predictions.
------------------------------------------------------------------------
ba...@math.ucr.edu
© 1998 John Baez
Stephen Speicher
>
> If there are any "silent readers" of this thread (and the others I've
> initiated), I would appreciate a stated consensus from each, about if you feel
> I've raised valid points, or not.
The main point which you have raised is that your antagonists are
physics "priests," and this bracketed by many sentences of blah,
blah, blah. The unintentional point which you have actually made
is that you have little understanding of physics in general, and
quantum mechanics in particular. But, don't let that fact stand
in your way of pointing out more "errors" in the field, all
overlooked by silly little physicists who lack your particular
insightful ability.
Look in the mirror that was built for you:
http://www.apa.org/journals/psp/psp7761121.html
--
Stephen
s...@speicher.com
Ignorance is just a placeholder for knowledge.
Printed using 100% recycled electrons.
-----------------------------------------------------------
Steve Bell
>
> The dirac equation is the not only the fundamental basis of qed,
> but qed is practically an inescapable consequence of the dirac equation.
> If you start with the dirac lagrangian, L = \psibar(p/ - m)\psi, and
> require the lagrangian be lorentz invariant under a local gauge
> transformation, \psi -> \psi\exp(iS), then the qed lagrangian follows
> immediately.
The inescapable use of exp(iS), dictates you are still referring to harmonic
"wave equations", per exp(ix) = cos(x) + i*sin(x). This is still imaginary
"wave mechanics" whatever you say, and is "physically nonsensible" to me. But
besides that, please do not respond to any of my posts. I really want to hear
"something anew," from unbiased individuals.
> Since this is nothing but a restatement that the phase of
> the wavefunction is unobservable, there is nothing surprising about
> requiring this. Through noether's theorem, the result is a conserved
> current (which is exactly the electromagnetic current density j^{u})
> and the electron _must_ carry a field. Noether's theorem only appeared
> in 1918 and in 1928, its importance in field theory simply wasn't
> appreciated to the extent it is now.
I really don't care about what is "observable" or not. And neither does
"nature." Please respond to the fact that you were incorrect about the
hyperfine splitting in hydrogen. Or please don't respond, I don't care. You
are a logical "idiot," one who I wish not to discuss things with anymore.
Please put me in your "kill file."
Steve Bell
>
> On Sun, 29 Dec 2002, Steve Bell wrote:
> >
> > If there are any "silent readers" of this thread (and the others I've
> > initiated), I would appreciate a stated consensus from each, about if you feel
> > I've raised valid points, or not.
>
> The main point which you have raised is that your antagonists are
> physics "priests," and this bracketed by many sentences of blah,
> blah, blah. The unintentional point which you have actually made
> is that you have little understanding of physics in general, and
> quantum mechanics in particular. But, don't let that fact stand
> in your way of pointing out more "errors" in the field, all
> overlooked by silly little physicists who lack your particular
> insightful ability.
>
> Look in the mirror that was built for you:
>
> http://www.apa.org/journals/psp/psp7761121.html
>
> --
> Stephen
> s...@speicher.com
>
Let me ask you a question: Why do you still read my posts? If you really feel
I am so wrong in what I say, why don't you just put me in your "kill file" and
be done with me? I've requested that of you in the past, and let me reiterate
it. Please put me in your "kill file" so that *you* do not respond to anymore
of my posts. I'm tired of this nonsense "puked" by you QM "priests." And yes,
I mean "priests." A "priest" to me is someone who accepts "physical truth" as
given to them by god, in their deluded egomania, without consideration of
"scientific evidence." You QM "priests" are doing just that, when you deny the
inaccurate Dirac prediction for hydrogen's ground state.
Matthew Nobes
> when you deny the
> inaccurate Dirac prediction for hydrogen's ground state.
I know I'm going to regret this, but, what the hell are
you talking about? The Dirac equation predicts the correct
spectrum for Hydrogen, up to small QED corrections of order
\alpha = 1/137.
--
Matthew Nobes
c/o Physics Dept. Simon Fraser University, 8888 University
Drive Burnaby, B.C., Canada
http://normland.phys.sfu.ca
Bob Zombiewoof
> Patrick Reany wrote:
> >
> >
> > Take a look at the approach to the Dirac equation fostered by Hestenes at
> >
> > http://modelingNTS.la.asu.edu/html/STC.html
> > http://modelingNTS.la.asu.edu/html/GAinQM.html
> >
> > and by the Cambridge group at
> >
> > http://www.mrao.cam.ac.uk/~clifford/publications/
> >
> > These refernce may not gve a precise answer to your question,
> > but you may find them interesting anyway.
> >
> > Patrick
>
>
> Thanks for these references. They imploy the "physically nonsensical" use of
> the square root of minus one (i) in their developments, and I would say that
> most of what they have to say is therefore "physically nonsensible."
>
> I do find it interesting how physicists can rely on i, in such a degree, in
> defining their theories of "physical reality," even though i doesn't represent
> anything "physically real."
Sure it does. It represents a phase angle. It differs from 1 by
a phase rotation of pi/2. exp(-kx) is a damped exponetial. exp(ikx)
is a propagating wave. cos(ix) = cosh(x). An impedance contains
a magnitude and phase. R + jwL is not the same as R + wL. "i" appears
in hermitian matrices precisely because the eigenvalues of observables
_should_ be real. The list is endless. There is nothing "unreal" about
the sqrt(-1). The term "imaginary" is a semantics issue, not a mathematical
statement about reality.
> I accord this due to some kind of "pandemic brainwashing" that
> occurred shortly after the turn of the century, due to the
> development of Schroedinger's imaginary wave equation.
What exactly is imaginary about the eigenvalues? It's exceedingly
simple to see the relation p = -i\hbar d/dx. Use it to operate on
the solutions to maxwell's wave equation. It gives you the momentum
of the wave.
Kevin Aylward
> Steve Bell <sb...@starband.net> wrote in message
> news:<3E0FC104...@starband.net>...
>>
>> I do find it interesting how physicists can rely on i, in such a
>> degree, in defining their theories of "physical reality," even
>> though i doesn't represent anything "physically real."
>
> Sure it does. It represents a phase angle. It differs from 1 by
> a phase rotation of pi/2. exp(-kx) is a damped exponetial. exp(ikx)
> is a propagating wave. cos(ix) = cosh(x). An impedance contains
> a magnitude and phase. R + jwL is not the same as R + wL. "i" appears
> in hermitian matrices precisely because the eigenvalues of observables
> _should_ be real. The list is endless. There is nothing "unreal"
> about the sqrt(-1).
I have an opposite view on numbers. *All* numbers are imaginary. I can
eat 4 apples, but I can't eat 4 "3"'s. A number is nothing more than a
mental construct.
Kevin Aylward
sa...@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
Steve Bell
Yes, I would expect this "knee jerk" reaction from QM physicists who believe
in the "standard line" that QM is "god-like" in its predictions of
experimental results. Please review my past posts ("Inaccurate QM Prediction"
and "More QM Nonsense") to qualitatively and *quantitativley* appreciate what
I have been talking about. If you, after reviewing these posts, have a
relevant contribution, I will listen to it. But let me say that you will have
to prove to me that you have read my previous posts, and that you will have to
prove to me that you yourself have "crunched the numbers" to state what you
believe is the physically correct binding energy for ground state hydrogen.
I am tired of all of this QM "nonsense." Simply "crunch the numbers" for
yourself, and tell me what *you* get! Of course, you will have to decide what
equation you will plug the best "consistent set" of values of h, c, etc.,
into, and then please tell me what equation you used for computing hydrogen's
ground state binding energy. I don't care if it's Dirac's or Bohr's or given
to you in a dream from God itself. I would be most interested in your results,
if you have somewhat of a logical description of why you used your particular
equation, and please state what the equation is, and why you believe it gives
the "physical truth."
Steve Bell
>
>
> What exactly is imaginary about the eigenvalues? It's exceedingly
> simple to see the relation p = -i\hbar d/dx. Use it to operate on
> the solutions to maxwell's wave equation. It gives you the momentum
> of the wave.
You yourself use "i". Like I've requested, please put me in your "kill file"
so you do not respond to my posts. You have nothing relevant, imo, to add to
this thread. If you do respond to anything, I want you to admit that you were
incorrect about the spin-spin hyperfine interaction in ground state hydrogen.
I readily admitted I was wrong about it being a "Zeeman effect." Can you be as
much of "a truthful man" as I am myself?
Steve Bell
Well, we all have our own views on "numbers." I think that any equational
route to some "predicting" equation, like Dirac's energy predicting equation,
which itself has no imaginary "i" in it at all, should not use "i" in its
derivation. Or let me relax that a bit, and say that I believe that "i" can be
used, but the actual "physics" of the equations can be expressed without the
use of anything "imaginary," without the use of "i" in the equations. I have
found this to be true in my professional work on narrowband stochastic
processes. This is where, to me, that QM falls "illogically apart." The use of
the square root of minus one is at the heart and soul of QM. It is, therefore
to me, a "physically inconsistent" theory. It predicts such "physically
illogical" results as the "Zitterbeweung," which is physically nonsensical, or
that as I've pointed out, that the spin-spin magnetic dipole interaction
contribution to the hyperfine splitting is zero in the ground state of
hydrogen. This is sheer "physical nonsense." And by "nonsense" I don't mean
anything necessarily derogatory, on face value. I just mean it doesn't make
"physical sense" to me, and is therefore "physical nonsense" in my opinion. I
think all physical theories of motion ("mechanics") should make physical sense
meaning that all motion, be it the very large (planet) or the very small
(electron in hydrogen), behaves according to some general "theory of motion"
that all bodies adhere to. I believe that all motion is a type of geodesic
motion in a curved spacetime. Either massively based, as in GR, or
electronically based, as in my theories of the electrostatic Coulomb field.
These "equations of motion" should be derived *deterministically* first, for
hydrogen, and then the stochasticism introduced.
Patrick Reany
Steve Bell wrote:
> Bob Zombiewoof wrote:
> >
> > The dirac equation is the not only the fundamental basis of qed,
> > but qed is practically an inescapable consequence of the dirac equation.
> > If you start with the dirac lagrangian, L = \psibar(p/ - m)\psi, and
> > require the lagrangian be lorentz invariant under a local gauge
> > transformation, \psi -> \psi\exp(iS), then the qed lagrangian follows
> > immediately.
>
> The inescapable use of exp(iS), dictates you are still referring to harmonic
> "wave equations", per exp(ix) = cos(x) + i*sin(x). This is still imaginary
> "wave mechanics" whatever you say, and is "physically nonsensible" to me. But
> besides that, please do not respond to any of my posts. I really want to hear
> "something anew," from unbiased individuals.
Do you consider yourself to be an "unbiased individual"?
You seem biased against certain mathematical formalisms
to me.
>
> > Since this is nothing but a restatement that the phase of
> > the wavefunction is unobservable, there is nothing surprising about
> > requiring this. Through noether's theorem, the result is a conserved
> > current (which is exactly the electromagnetic current density j^{u})
> > and the electron _must_ carry a field. Noether's theorem only appeared
> > in 1918 and in 1928, its importance in field theory simply wasn't
> > appreciated to the extent it is now.
>
> I really don't care about what is "observable" or not. And neither does
> "nature."
Just what DOES "nature" care about?
Patrick
Thinh Tran
fluid mechanics (i is used extensively in the analysis of
incomressible flows.)
First, I agree partly with Steve Bell. It is mathematically
possible not to use i in equations. I believe Feynman expressed more
or less the same view in his book "QED-The strange theory of light and
matter".
Second, I disagree with Steve's contention that equations with i
are not valid (i.e., unreal). Think of a case where you need to
describes two sets, one of positive and negative A's, the other of
positive and negative B's which are equivalent but different from each
others like the sets of positive and negative men and the set of
positive and negative women. In case there is an operation to change a
member of one set to a member of another set (e.g., a sex change
operation that somehow can go both way -not very realistic, I know,
but the point should be clear), it would be much simpler
mathematically to compress all information in one equation instead of
splitting them into 2 or more.
i is, in this sense, an operation (or transformation if you
will), with the following rules, which are symmetrical for A and B:
One operation (i): (A,+) => (B,+)
Two operations (i^2): (A,+) => (A,-)
Three operations (i^3): (A,+) => (B,-)
Four operations (i^4): (A,+) => (A,+)
(On the complex plane each operation corresponds to a phase angle
change -or more descriptively, a rotation- of pi/2, as pointed out by
BobZombiewolf.)
By using i I can stack more information into one equation (I know
everything with i^n where n is odd belongs to the "set of non-real
entities", therefore I can differentiate them from the "set of real
entities" with i^n, n even).
This, I think, is the main advantage of using i in mathematics,
though I agree it could create lots of confusion.
Thinh Tran (http://www.thinhtran.com)
Patrick Reany
Thinh Tran wrote:
It doesn't matter one bit what all gobbledygook one uses
at the core of his or her mathematical system, whether that
be imaginary numbers, Clifford numbers, matrices, quaternions,
spinors, vectors, or tensors. The ONLY thing that matters is
if one also has a way to map that gobbledygook into the real
numbers in such a way that predictions can be made with it
which can be tested in the lab.
Reality does NOT live within the particular mathematical
representation that is chosen. Reality only ever touches
measurements anyway. Calculations don't have to start off
in the real numbers to end up there.
Patrick
Stephen Speicher
> Stephen Speicher wrote:
> >
> > On Sun, 29 Dec 2002, Steve Bell wrote:
> > >
> > > If there are any "silent readers" of this thread (and the others I've
> > > initiated), I would appreciate a stated consensus from each, about if you feel
> > > I've raised valid points, or not.
> >
> > The main point which you have raised is that your antagonists are
> > physics "priests," and this bracketed by many sentences of blah,
> > blah, blah. The unintentional point which you have actually made
> > is that you have little understanding of physics in general, and
> > quantum mechanics in particular. But, don't let that fact stand
> > in your way of pointing out more "errors" in the field, all
> > overlooked by silly little physicists who lack your particular
> > insightful ability.
> >
> > Look in the mirror that was built for you:
> >
> > http://www.apa.org/journals/psp/psp7761121.html
> >
>
> Let me ask you a question: Why do you still read my posts?
Mainly for entertainment value.
> If you really feel
> I am so wrong in what I say, why don't you just put me in your "kill file" and
> be done with me?
Oh, don't worry, eventually all good cuckoo birds go to killfile
heaven.
> I've requested that of you in the past, and let me reiterate
> it. Please put me in your "kill file" so that *you* do not respond to anymore
> of my posts.
You are as confused about how this works as you are about
physics. You do not get to choose for me who, and when, I
"killfile" someone. You only get to choose for yourself. If you
do not want to read what I write, then don't. When you write to a
public forum you simply don't get to choose who is allowed to
respond. Heck, if only that were true, we would be rid of all you
crackpots immediately.
> I'm tired of this nonsense "puked" by you QM "priests."
>
Oh, poor boy.
> And yes,
> I mean "priests." A "priest" to me is someone who accepts "physical truth" as
> given to them by god, in their deluded egomania, without consideration of
> "scientific evidence."
Out of curiosity: The "priests" have brought us computers and MRIs;
what has the non-Priest Steve Bell given us?
Bob Zombiewoof
> Bob Zombiewoof wrote:
> >
> > The dirac equation is the not only the fundamental basis of qed,
> > but qed is practically an inescapable consequence of the dirac equation.
> > If you start with the dirac lagrangian, L = \psibar(p/ - m)\psi, and
> > require the lagrangian be lorentz invariant under a local gauge
> > transformation, \psi -> \psi\exp(iS), then the qed lagrangian follows
> > immediately.
>
>
> The inescapable use of exp(iS), dictates you are still referring to harmonic
> "wave equations", per exp(ix) = cos(x) + i*sin(x). This is still imaginary
> "wave mechanics" whatever you say, and is "physically nonsensible" to me.
Look up "hermitian operator".
> But besides that, please do not respond to any of my posts. I
> really want to hear "something anew," from unbiased individuals.
Try alt.sci* We're all biased here.
> > requiring this. Through noether's theorem, the result is a conserved
> > current (which is exactly the electromagnetic current density j^{u})
> > and the electron _must_ carry a field. Noether's theorem only appeared
> > in 1918 and in 1928, its importance in field theory simply wasn't
> > appreciated to the extent it is now.
>
>
> I really don't care about what is "observable" or not.
Or anything else that interferes with your agenda.
> And neither does
> "nature." Please respond to the fact that you were incorrect about the
> hyperfine splitting in hydrogen.
But I'm not incorrect. The splitting of the n^{th} s-orbital is
given by:
\delta_{n} = (m\alpha^2/2)[(8g_{p}/2) (Z\alpha/n)^2 (m/M_{p})]
g_{p} is the gyromagnetic moment of the proton, M_{p} is the proton nass.
[cf bjorken & drell, vol I, ppg 58-59]
> Or please don't respond, I don't care. You are a logical "idiot,"
> one who I wish not to discuss things with anymore.
Then why do you continue? Did someone hold a gun to your head?
ueb
..
> It doesn't matter one bit what all gobbledygook one uses
> at the core of his or her mathematical system, whether that
> be imaginary numbers, Clifford numbers, matrices, quaternions,
> spinors, vectors, or tensors. The ONLY thing that matters is
> if one also has a way to map that gobbledygook into the real
> numbers in such a way that predictions can be made with it
> which can be tested in the lab.
> Reality does NOT live within the particular mathematical
> representation that is chosen. Reality only ever touches
> measurements anyway. Calculations don't have to start off
> in the real numbers to end up there.
People confuse simply image with origin.
E.g. complex numbers are good for system theory, but have nothing
to do with nature. Nature knows solely real numbers - with the
little extension that time means imaginary length (and conversely).
But there are no _complex_ numbers as described with Gauss'
number plane.
Your enumeration of the "gobbledygook" is not quite correct.
Tensors (and vectors as first rank tensors) are real physical
quantities - in opposite to all the other.
Happy New Year !
Ulrich Bruchholz
www.markt-2000.de/patent
Steve Bell
>
>
> > And neither does
> > "nature." Please respond to the fact that you were incorrect about the
> > hyperfine splitting in hydrogen.
>
> But I'm not incorrect. The splitting of the n^{th} s-orbital is
> given by:
>
> \delta_{n} = (m\alpha^2/2)[(8g_{p}/2) (Z\alpha/n)^2 (m/M_{p})]
>
> g_{p} is the gyromagnetic moment of the proton, M_{p} is the proton nass.
>
> [cf bjorken & drell, vol I, ppg 58-59]
>
>
Your above equation is very similar for what Cohen-Tann gives for the
correction of the contact term, not the dipole-dipole interaction correction.
And of course it will involve the gyromagnetic moment of the proton because it
involves the proton itself. He is really quite specific about this, that there
is no dipole-dipole interaction in 1s, and shows how the contact term is
really the only effect responsible for the hyperfine splitting. According to
Cohen-Tann, you are incorrect in thinking it comes from the dipole-dipole
interaction.
Of course, QM also says the two dipole fields physically exist. But they do
not interact. If you buy that, I just don't know what to say.
> > Or please don't respond, I don't care. You are a logical "idiot,"
> > one who I wish not to discuss things with anymore.
>
> Then why do you continue? Did someone hold a gun to your head?
You are quite right, why do I continue. And I'm thinking of not continuing if
all I am doing is arguing with the likes of you. It is obvious that I will
never convince people like you that QM is seriously flawed, and if all I am
doing is talking to *you,* I quit.
Steve Bell
>
> Please allow me to express my view of i from my background, which is
> fluid mechanics (i is used extensively in the analysis of
> incomressible flows.)
> First, I agree partly with Steve Bell. It is mathematically
> possible not to use i in equations. I believe Feynman expressed more
> or less the same view in his book "QED-The strange theory of light and
> matter".
I can't believe it! Somebody actually agrees with me! Well, only partly, but
that's a start. (And according to QM, I must now be existing in a state that's
at least some proportion "correct.") Just kidding <G>.
Since "i" is so nonsensical, we could argue about it endlessly. I really would
like to hear your opinion on the bizarre predictions made by QM. That is, we
know the electron is undergoing bound motion in ground state hydrogen. The
motion must be bound, because the electron doesn't come flying out of the
atom. If it is bound, there must be a "rotational" element to its movement
through time. QM itself says the electron possesses a mass and a velocity.
This fact is used when finding the expected value of the "mass-velocity"
relativistic correction term. Yet, QM also says the electron does not
possesses any orbital angular momentum in 1s (and I understand the QM
equations, no one has to tell me why QM says this happens, I already
understand that.) Then QM says there is the "Zitterbewegung" effect. This
effect demands we view the electron as having a definite position in space,
otherwise what exactly is it that QM says is undergoing a "rapid fluctuation?
Then comes the really obnoxious QM prediction of no dipole-dipole magnetic
interaction in 1s. QM says the two dipole fields physically exist, but that
they do not interact. This is what I call "physical nonsense."
Steve Bell
>
>
> It doesn't matter one bit what all gobbledygook one uses
> at the core of his or her mathematical system, whether that
> be imaginary numbers, Clifford numbers, matrices, quaternions,
> spinors, vectors, or tensors. The ONLY thing that matters is
> if one also has a way to map that gobbledygook into the real
> numbers in such a way that predictions can be made with it
> which can be tested in the lab.
>
I completely agree with this. It is good that you see only real numbers can
show up in *both* predictions and observations.
> Reality does NOT live within the particular mathematical
> representation that is chosen. Reality only ever touches
> measurements anyway. Calculations don't have to start off
> in the real numbers to end up there.
>
> Patrick
QM uses imaginary numbers in getting to its "real only" predictions. From
these predictions, we can infer how well the theory models reality. I say QM
has made a fundamentally flawed prediction when it comes to hydrogen's ground
state energy. If you want the data why I say that, please see my past
postings.
Y.Porat
> After a welcomed hiatus due to the holidays, I would like to start this thread
> somewhat anew with the following. Bile says, quite justifiably so, look at the
> equations about the "Zitterbeweung" and make up my own mind. I've done so at
>
> http://www.mindspring.com/~sb635/qmnonsense.pdf
>
> This "Darwin effect" in hydrogen still makes no "physical sense" to me. Also,
> nothing in this derivation would say that the "Darwin effect" is not present
> in all shells.
>
> I would like to recount some facts. The relativistic contribution to
> hydrogen's ground state binding energy, according to the series expansion of
> Dirac's equation, after taking the expected value of only the relativistic
> term, is 5 times larger than what the "deterministic" relativistic orbit
> theory demands, per Sommerfeld. The "Darwin effect" is statistical and
> "compensates for" or "negates," per the Zitterbewegung" effect, for exactly
> one half of the 5-times-too-big "pure" expected relativistic effect as
> predicted by QM, bringing the "net" effect on binding energy back to exactly
> what Sommerfeld's *only* specially relativistic orbit model predicts
---------------------------
orbit shmorbit
i would like to remind you about two more QM nensense:
1 qm never went further than the element Fe (or close t it
now what is more interesting is the 'excuses' why it didnt:
a 'there is no need'!
if i were a pupil at the elementry school i would be a shamed
to bring such an excuse to my teacher
because even a child will not be such a stupid to bering such
idiotic excuses.
there is no need ??!!!
supose for an instance that Mendeleiv whould stop his pereiodic
table at the element iron
now if someone would ask him : why did you stopped it there
he would answer : 'there is no need' -'just look back
and you can aragne all the rest of it as those before!!!
now another student will ask him :
what about the Lanthanoid elements, and he would answer:
the Lanthanoids are like before- *you can arange them as before
one by one one after the other in the horizontal row.!
that was just a mathaphore to demenstrate the stupidity
of the above 'excuse.
b nonsense No 2:
what causes the em attraction is exchange of a creature called
W Boson.
tothose who dont know about that creature:
it is ctreated by a nucleid and (listen carefully anf believe
to what your ears are listening)
that w boson that is created by a single nucleid is ....
80 times heavier than the nucleid itself!!
and now the peack of crookism : it is 80 times heavier
but it is 'out of its mass shell' !!
had there been a law against fraud in physics,
the inventors and spreaders and defenders of that theory, would
been thrown to jail.
their luck is that there is no such law.
ps lucky me there is no law they invented, to trow me
into jail for exposing their fraud (:-).
but who knows croocks has power!.and may be a lot of
money behind and a lot of 'jobs' on stake depending on it ???.
all the best
Y.Porat
-------------
(ignoring
> the hyperfine splitting). Sommerfeld did not introduce spin-orbit magnetic
> effects, but for the ground state, this is irrelevant because QM predicts
> there is no spin-orbit coupling in the ground state.
>
> Using the best "consistent set" of values for the electron's and proton's rest
> masses and charges, and h, eps_0 and c, etc, the Dirac prediction equals
> -13.5984652 eV. Converting this to an equivalent "Rydberg constant for
> hydrogen," we get 10967903.3 m^-1. Using h, eps_0 and c, etc in Bohr's
> nonrelativistic energy equation, we get 10967757.3 m^-1. The difference is
> approximately 146 m^-1. These are the numerical facts given today's best
> "consistent set" of values.
>
> Frank Wappler recounts a value of hydrogen's Rydberg constant (R_H) as
>
> "Indeed, a corresponding numerical coincidence can be noted
> regarding the value (109677.575 +/- 0.012) cm^-1, reported by
> E. R. Cohen, Phys. Rev. 88, (2), 353, (1952; received March 1951)."
>
> The difference between this over-than 50-year old value, and what is given by
> *today's* "consistent set" plugged into Bohr's nonrelativistic prediction is
> only 0.2 m^-1. The value from Cohen was presumably an "experimentally
> determined" value (Resnick sure described it as such), and I assume it is
> "unbiased." The quoted error in the 50-year old value is +/- 1.2 m^-1, and we
> see that the difference between the 50-year old value, and Bohr's
> nonrelativistic prediction, using *today's* values, is essentially
> statistically insignificant. The difference between Dirac's prediction and the
> observed is 146 m^-1, and, assuming 1.2 m^-1 is one-sigma, 146/1.2 = about a
> 120 sigma difference. If you have any idea of what "statistically significant"
> is, this is *gigantically* significantly different.
>
> A significant question is how could a 50-year old value be so dead-on to a
> prediction from *today's* values in Bohr's *nonrelativistic* equation? Perhaps
> the older spectroscopic data was "taken as is," as has no bias, and reflects
> the true physical condition in hydrogen? A true "physical condition" in
> hydrogen is the relativistic effects. These effects should be, and directly
> are, observable by frequency absorption and emissions. Yet the "observed" data
> apparently supports a nonrelativistic signature.
>
> Something in hydrogen is "compensating for" or "negating" the relativistic
> effects. QM itself utilizes this fact in how the "Darwin effect" (which adds a
> positive energy) adds to the 5-times-to"deep" *only* relativistic negative
> energy contribution. But it still comes out 146 m^-1 too "deep," compared to a
> 50-year old "observed" value, whose data was plenty enough accurate and
> precise to manifest the relativistic effects if present. If hydrogen does not
> in fact "outwardly manifest" relativistic effects, this points to a
> fundamental lack of "physically correct" modeling of hydrogen by QM.
Stephen Speicher
> Bob Zombiewoof wrote:
>
> > Steve Bell wrote:
> >
> > > Or please don't respond, I don't care. You are a logical "idiot,"
> > > one who I wish not to discuss things with anymore.
> >
> > Then why do you continue? Did someone hold a gun to your head?
>
>
> You are quite right, why do I continue. And I'm thinking of not continuing if
> all I am doing is arguing with the likes of you. It is obvious that I will
> never convince people like you that QM is seriously flawed, and if all I am
> doing is talking to *you,* I quit.
>
Bye bye, and please do not slam the door on your way out.
The "Priests" are sleeping.
Bilge
Re: More QM Nonsense, II to usenet:
>Bob Zombiewoof wrote:
>>
>>
>> > And neither does
>> > "nature." Please respond to the fact that you were incorrect about the
>> > hyperfine splitting in hydrogen.
>>
>> But I'm not incorrect. The splitting of the n^{th} s-orbital is
>> given by:
>>
>> \delta_{n} = (m\alpha^2/2)[(8g_{p}/2) (Z\alpha/n)^2 (m/M_{p})]
>>
>> g_{p} is the gyromagnetic moment of the proton, M_{p} is the proton nass.
>>
>> [cf bjorken & drell, vol I, ppg 58-59]
>>
minor correction: That should be 8/3, not 8/2.
>
>
>Your above equation is very similar for what Cohen-Tann gives for the
Imagine that.
[...]
Thinh Tran
>
> Since "i" is so nonsensical, we could argue about it endlessly. I really would
> like to hear your opinion on the bizarre predictions made by QM. That is, we
> know the electron is undergoing bound motion in ground state hydrogen. The
> motion must be bound, because the electron doesn't come flying out of the
> atom. If it is bound, there must be a "rotational" element to its movement
> through time. QM itself says the electron possesses a mass and a velocity.
> This fact is used when finding the expected value of the "mass-velocity"
> relativistic correction term. Yet, QM also says the electron does not
> possesses any orbital angular momentum in 1s (and I understand the QM
> equations, no one has to tell me why QM says this happens, I already
> understand that.) Then QM says there is the "Zitterbewegung" effect. This
> effect demands we view the electron as having a definite position in space,
> otherwise what exactly is it that QM says is undergoing a "rapid fluctuation?
> Then comes the really obnoxious QM prediction of no dipole-dipole magnetic
> interaction in 1s. QM says the two dipole fields physically exist, but that
> they do not interact. This is what I call "physical nonsense."
Believe it or not, I also have thought about this problem for a long
time and -I believe- I found the reason in 2000.
In the wave equations you always see this quantity in angular
momentum-related calculations:
[l(l+1)]^(1/2)
What is it? It is simply the geometrical average of l and l+1.
What does a nice geometrical average do in a place like QM?
Because the wave equations are continuous approximation of a
fundamentally discrete (i.e. quantum) situation. Approximations always
have error. The question is when the error becomes serious.
Think of a case where you have a difference equation, but you
approximate it as a differential equation so that it is quicker to
solve.
What the sequence 1, 2, 3 of the difference equation will be?
Answer: 0 will be added to the averaging process, then depending
on the form of the PDE it will become either one of the following:
Linear average: .5, 1.5, 2.5
Geometric average:(0*1)^0.5=0, (1*2)^0.5, (2*3)^0.5
Note that l=0 is NOT the position R=0 in the spherical
coordinates used to solve the for the Hydrogen atom. So actually l=0
corresponds to the number 1 I used in the example.
For quick reference I rewrite the correct and the approximated
values for the case the number is 1:
Correct: 1
Linear average: .5
Geometric average: 0
Since the wave equations are continuous approximations it ends up
using 0 for the angular momentum of the innermost orbital, instead of
the correct value 1, which would be the case had a discrete (and more
correct) calculations been performed.
In other words: The quantity [l(l+1)]^0.5 should be simply l+1.
But of course, to do the math correctly, you must abandon the
wave equation altogether (this is why Feynman's "hocus pocus dippy
process" -his own words- works better than the wave equations.) This
analysis only pointed the problem with l=0.
The bottom line is: The wave equations are only approximations of
QM problems. In fact, they are not even good first order
approximations.
By the way -without trying to sell my book, I'm making no money
with it anyway- this is covered in the chapter "Successes and problems
of wave mechanics" in the first edition my book "The End of
Probability and The New Meaning of Quantum Physics". I'm working on an
improved edition, fixing mistakes and stuffs. If you visit my site and
link to the summary you will see many other problems with QM that I
pointed out (and supposedly solved.)
I could go on and on, but I guess I should stop somewhere.
HAPPY NEW YEAR!!!
Thinh Tran (http://www.thinhtran.com)
Thinh Tran
> The important issue is what the various theories predict for ground state
> hydrogen. There, due to all of QM's nonsensical "spherically symmetric"-based
> conclusions of the ground state, its predictions can be analyzed with a
> "critical" eye.
>
> For example, I just pointed out that Bilge's interpretation of the spin-spin
> magnetic hyperfine interaction in the ground state, is incorrect, and that it
> is zero according to QM's best theoretical development. Now, let me ask you:
> does the electron have a spin? Yes, it does. Does the spin set up a dipole
> magnetic field? Yes, it does. Is the proton spinning? Yes, it is. Does the
> proton's spin set up a dipole magnetic field? Yes, it does. Should the
> electron's spin magnetic field interact with the proton's spin magnetic field,
> even in the ground state, where all of these conditions are met? Yes, it
> should. But QM says it doesn't. Bilge himself thought that it should. I view
> this as a "physically nonsensical" prediction by QM, tied all the way back to
> the use of an imaginary wave equation from the get go, and how this drives the
> physically nonsensical expectations in QM, especially in the ground state,
> where "spherical symmetry" can bite QM in the ass, like in its quantification
> of the "hyperfine" splitting.
>
> If the electronic field surrounding the proton can be modeled as a type of
> extremely localized, very tightly curved Kerr spacetime, then the proton's
> electromagnetic field is *not* spherically symmetric, but is either oblate
> spheriodal or prolate spheriodal, depending on the proton's sense of direction
> of its spin. This kills QM's separability of the Schroedinger's equation,
> which lies at the heart and soul of all QM theory.
sb635
Somehow I missed this posting of yours. Like I wrote in an earlier
posting, the wave equations (Schrodinger's, Dirac) are simply
approximations. And even for first order approximations they are
already not good enough. They are basically linearized continuous PDE
approximations of non-linear and fundamentally discrete realities.
They had their place in history; and I think we should leave it at
that to move on.
Bob Zombiewoof
> If the electronic field surrounding the proton can be modeled as
> a type of extremely localized, very tightly curved Kerr spacetime,
The kerr metric doesn't describe spin. It describes a
rotating mass. Spin is not a rotation. The spin axis requires
a 4\pi rotation to return it to the original direction. A
mechanical angular momentum requires a 2\pi rotation. I've
already told you this.
> then the proton's electromagnetic field is *not* spherically
> symmetric, but is either oblate spheriodal or prolate spheriodal,
> depending on the proton's sense of direction of its spin.
An oblate or prolate shape would require the proton to have
an electric quadrupole moment. It does not have a quadrupole
moment. Experiments have been performed to look for this very
thing.
> This kills QM's separability of the Schroedinger's equation,
> which lies at the heart and soul of all QM theory.
No, it doesn't kill quantum mechanics. It kills the model
of the proton. But, since the proton does not have a quadrupole
moment, not to mention a dipole moment (which has also been
sought experimentally), there is no problem.
The reason the hydrogen ground state is known to be spherically
symmetric, is becuase it has no electric dipole moment, etc.
It's not simply a quantum mechanical prediction. It's an experimental
fact. Even if the hydrogen ground state were _not_ sherically
symmetric, at some precision not yet attainable, it would not
matter to quantum mechanics. Quantum mechanics describes the
deuteron just fine. The deuteron is an s-state, spin-1 particle
with a 4% d-state (L=2) admixture. Quantum mechanics is perfectly
capable of describing a wavefunction as a linear combinations
of states in a given basis.
Steve Bell
Is this all you have to say? I prove that you are incorrect in saying the
hyperfine splitting is due to the dipole-dipole interaction, and you will not
admit it. I do believe that you are quite a good mathematical physicists, but
you are also quite intellectually dishonest.
Steve Bell
>
>
> Somehow I missed this posting of yours. Like I wrote in an earlier
> posting, the wave equations (Schrodinger's, Dirac) are simply
> approximations. And even for first order approximations they are
> already not good enough. They are basically linearized continuous PDE
> approximations of non-linear and fundamentally discrete realities.
> They had their place in history; and I think we should leave it at
> that to move on.
> HAPPY NEW YEAR!!!
> Thinh Tran (http://www.thinhtran.com)
I quite agree that we should "move on" from QM. To that end, I have developed
a theory that can explain why, if true, hydrogen's ground state does not
outwardly manifest any relativistic effects. I agree, that QM is only a "first
cut" at atomic truth, but I won't stop saying that the mismatch between
experiment and QM, on this fundamental point of how relativity comes into
play, is of major significance. Imo, the mismatch is of major significance,
and points to a fundamental flaw in QM; its "first cut" at how relativistic
effects are involved is seriously flawed.
Steve Bell
>
> Steve Bell <sb...@starband.net> wrote in message
>
> > If the electronic field surrounding the proton can be modeled as
> > a type of extremely localized, very tightly curved Kerr spacetime,
>
> The kerr metric doesn't describe spin. It describes a
> rotating mass. Spin is not a rotation. The spin axis requires
> a 4\pi rotation to return it to the original direction. A
> mechanical angular momentum requires a 2\pi rotation. I've
> already told you this.
>
It is amazing to me how someone "baptized" in the "belief" of QM can come up
with such outwardly oxymoronic phrases such as "spin is not a rotation," and
not bat an eye.
> > then the proton's electromagnetic field is *not* spherically
> > symmetric, but is either oblate spheriodal or prolate spheriodal,
> > depending on the proton's sense of direction of its spin.
>
> An oblate or prolate shape would require the proton to have
> an electric quadrupole moment. It does not have a quadrupole
> moment. Experiments have been performed to look for this very
> thing.
>
This shows you really do not understand what Kerr geometries are. The central
body generating a Kerr field is *itself* totally (theoretically, perfectly)
spheroidally shaped, yet due to its rotation (and yes, it is an honest-to-god
spinning spherical ball of matter), it produces a non-spherically symmetric
external field. The Kerr field manifests "frame dragging" effects, and imo,
this is the true cause of magnetism itself.
Hayek
Steve Bell wrote:
> After a welcomed hiatus due to the holidays, I would
> like to start this thread somewhat anew with the
> following. Bile says, quite justifiably so, look at
> the equations about the "Zitterbeweung" and make
> up my own mind. I've done so at
>
> http://www.mindspring.com/~sb635/qmnonsense.pdf
[QM versus GR]
I am trying to build a consistent theory with inertia as
a pivot. You surely agree with me that inertia plays the
most important role in Newtonian Mechanics where it
pertains predictability. Newtonian Mechanics never
considered this inertia to be variable, this is
introduced by relativity. Quantum mechanics differ so
much from the 'inertial' mechanics NM & RM, and I claim
the sole reason for this is that QM does not undergo
inertia. No inertia, no predictability.
If we expand on this, it is outright silly to consider
'a relativistic' approach to QM phenomenon. Inertia does
not apply, so relativity does not apply. The Heisenberg
equation is the limit. Below that there is no inertia
whatsoever, hence no predictability is possible.
Why is it the electron does not fall in to hydrogen
nucleus ? It has no inertial mass below its lowest orbit.
When the surrounding inertia increases the size of the
orbital decreases and vice versa.
So far I concentrated on fitting this 'inertial' theory
on relativity. It works wonderfully. What do you think
its possibilities are for QM ?
Hayek
--
The small particles wave at
the big stars and get noticed.
:-)
FrediFizzx
| Bob Zombiewoof wrote:
| >
| > Steve Bell <sb...@starband.net> wrote in message
| >
| > > If the electronic field surrounding the proton can be modeled as
| > > a type of extremely localized, very tightly curved Kerr spacetime,
| >
| > The kerr metric doesn't describe spin. It describes a
| > rotating mass. Spin is not a rotation. The spin axis requires
| > a 4\pi rotation to return it to the original direction. A
| > mechanical angular momentum requires a 2\pi rotation. I've
| > already told you this.
| >
|
|
| It is amazing to me how someone "baptized" in the "belief" of QM can come
up
| with such outwardly oxymoronic phrases such as "spin is not a rotation,"
and
| not bat an eye.
Yes, this is something that also really bugs me about quantum theory.
Everywhere we see circular-like motion from stars in a galaxy to planets
around stars to electrons around nuclei then quantum theory says that spin
is not a rotation. Very difficult to swallow that one. If it is not some
kind of rotation, then what the heck is it?
| > > then the proton's electromagnetic field is *not* spherically
| > > symmetric, but is either oblate spheriodal or prolate spheriodal,
| > > depending on the proton's sense of direction of its spin.
| >
| > An oblate or prolate shape would require the proton to have
| > an electric quadrupole moment. It does not have a quadrupole
| > moment. Experiments have been performed to look for this very
| > thing.
| >
|
|
|
| This shows you really do not understand what Kerr geometries are. The
central
| body generating a Kerr field is *itself* totally (theoretically,
perfectly)
| spheroidally shaped, yet due to its rotation (and yes, it is an
honest-to-god
| spinning spherical ball of matter), it produces a non-spherically
symmetric
| external field. The Kerr field manifests "frame dragging" effects, and
imo,
| this is the true cause of magnetism itself.
|
I am glad that you decided not to quit on this discussion. For someone that
has just started to learn more about quantum theory this year, I find this
discussion fascinating. My reason for learning more about it is because of
some of the things that are being discussed here. I just can't believe
there isn't more to it or something is missing. However, it seems to me as
a beginner, that modern QED did fix some of the problems with QM. But I am
not sure yet if the fixes were "ad hoc". I am still studying and learning.
I hope that we could take name calling out of this discussion and continue
with it.
FrediFizzx
Bilge
Re: More QM Nonsense, II to usenet:
>Bilge wrote:
>>
>> Steve Bell said some stuff about
>> Re: More QM Nonsense, II to usenet:
>> >Bob Zombiewoof wrote:
>> >>
>> >>
>> >> > And neither does
>> >> > "nature." Please respond to the fact that you were incorrect about the
>> >> > hyperfine splitting in hydrogen.
>> >>
>> >> But I'm not incorrect. The splitting of the n^{th} s-orbital is
>> >> given by:
>> >>
>> >> \delta_{n} = (m\alpha^2/2)[(8g_{p}/2) (Z\alpha/n)^2 (m/M_{p})]
>> >>
>> >> g_{p} is the gyromagnetic moment of the proton, M_{p} is the proton nass.
>> >>
>> >> [cf bjorken & drell, vol I, ppg 58-59]
>> >>
>>
>> minor correction: That should be 8/3, not 8/2.
>>
>> >
>> >
>> >Your above equation is very similar for what Cohen-Tann gives for the
>>
>> Imagine that.
>>
>> [...]
>
>
>Is this all you have to say? I prove that you are incorrect in saying the
Hayek
FrediFizzx wrote:
>
> Yes, this is something that also really bugs me
> about quantum theory. Everywhere we see
> circular-like motion from stars in a galaxy to
> planets around stars to electrons around nuclei
> then quantum theory says that spin is not a
> rotation. Very difficult to swallow that one. If
> it is not some kind of rotation, then what the heck
> is it?
You are thinking too Newtonian.
For rotations you need inertia.
QM does not have inertia, or at least not in the form we
know that makes a sphere rotate.
Imagine a sphere without inertia. The slighest push
would make it 'spin' at an infinite speed. There is a
limit however, set by the Heisenberg uncertainty. But we
are looking from the other side now, where 'certainty',
read 'inertia' kicks in.
So you would not call it 'rotation', it can only accept
the right amount of energy to put it in this 'spin', and
this spin would behave more like a binary digit, in this
case not 0 or 1 but left or right. It is more like a
'state' than like a rotation.
Hayek.
Steve Bell
>
> >>
> >> >
> >> >
> >> >Your above equation is very similar for what Cohen-Tann gives for the
> >>
> >> Imagine that.
> >>
> >> [...]
> >
> >
> >Is this all you have to say? I prove that you are incorrect in saying the
>
> No, you have not.
>
Please see p. Complement BX1, p. 1120, of Choen-Tann's "Quantum Physics," vol
2. There, he proves the dipole-dipole interaction must be zero if l = 0, as it
does in the ground state. After you read it, I would think you would conclude
there is no spin-spin dipole interaction in hydrogen's 1s shell, in
contradiction to what you have previously stated.
I originally thought this "hyperfine" issue was of little importance to my
main argument, that QM is way off from predicting hydrogen's ground state
energy correctly, (and it is, given the hyperfine's small contribution), but I
thank you for forcing me to delve into the hyperfine splitting more deeply.
QM's result that no dipole-dipole interaction exists (per Cohen-Tann's very
own words) in 1s, coupled with the fact that QM admits the magnetic fields
themselves exist, is physically absurd.
Thinh Tran
> | It is amazing to me how someone "baptized" in the "belief" of QM can come
> up
> | with such outwardly oxymoronic phrases such as "spin is not a rotation,"
> and
> | not bat an eye.
>
> Yes, this is something that also really bugs me about quantum theory.
> Everywhere we see circular-like motion from stars in a galaxy to planets
> around stars to electrons around nuclei then quantum theory says that spin
> is not a rotation. Very difficult to swallow that one. If it is not some
> kind of rotation, then what the heck is it?
The reason that QM says spin is not "spinning" (in the everyday's
sense) is because it started with the point assumption (How can a
point spin?). By eliminating the point assumption, electrons, protons,
neutrons immediately take their shapes as three dimensional objects.
If you spin a spinning top it complete its spin in 1 turn, but don't
forget that this is basically a plannar spin. Now just think of the an
arbirarily shaped 3-d object spinning wildly in all sort of angle in
space. You get 3-d spinning. This intuitive dimensional analysis is
sufficient to show why the net number of spin required for these
objects to return to their starting position is about 2. (The
situation is actually different than this, but you can see the main
point of my argument. By the way a more complete argument is in my
book.)
I agree that this thread has been a great thing. In fact, I will mimic
the idea and try something similar to it in a few days to get feed
back on my common sense interpretation of the Heisenberg's uncertainty
principle.
QED is ad hoc (and Feynman was very frank about this,) IMHO, the
reason is that it still treats quantum objects as geometrical points,
which is obviously wrong. It is not difficult to see that this
point-assumption is the reason why the "self-energy" points in Feynman
diagrams blow up (with infinities) and need the "hocus pocus dippy
process" of renormalization to fix.
But since QED has fantastic numerical successes in a few cases,
it exposed the weaknesses of the wave equations as well as the de
Broglie's wave hypothesis and the Heisenberg's Uncertainty Principle
(that all these are approximated descriptions and there is no reason
for us to worship them as absolute truths.)
I believe QED is the correct pointer to a future permanent fix of
QM.
Bilge
Re: More QM Nonsense, II to usenet:
>news:3E12EB5B...@starband.net...
>| Bob Zombiewoof wrote:
>| >
>| > Steve Bell <sb...@starband.net> wrote in message
>| >
>| > > If the electronic field surrounding the proton can be modeled as
>| > > a type of extremely localized, very tightly curved Kerr spacetime,
>| >
>| > The kerr metric doesn't describe spin. It describes a
>| > rotating mass. Spin is not a rotation. The spin axis requires
>| > a 4\pi rotation to return it to the original direction. A
>| > mechanical angular momentum requires a 2\pi rotation. I've
>| > already told you this.
>| >
>|
>|
>| It is amazing to me how someone "baptized" in the "belief" of QM
>| can come up with such outwardly oxymoronic phrases such as "spin
>|is not a rotation," and not bat an eye.
>
>Yes, this is something that also really bugs me about quantum theory.
>Everywhere we see circular-like motion from stars in a galaxy to planets
>around stars to electrons around nuclei then quantum theory says that spin
>is not a rotation. Very difficult to swallow that one. If it is not some
>kind of rotation, then what the heck is it?
what spin is. When you solve the dirac hamiltonian, you get a new
internal degree of freedom, X. The dirac hamiltonian does not
commute with the angular momentum, nor does it commute with the new
degree of freedom. It _does_ commute with the sum, L + 2X,
hence it was termed "spin". It is related to the metric in terms
of dirac matricies in the following way:
[\gamma^{u},\gamma^{v}]_{+} = 2g^{uv} : [ ]_{+} denotes anti-commuator
[\gamma^{u},\gamma^{v}] = 2i\sigma^{uv}
where \sigma^{uv} are, essentially, the spin matrices. The "spin" simply
refers to an object which transforms according to the algebra, SU(2).
Let's look at some other examples of SU(2). The original nuclear model
given by either heisenberg or wigner, I forget which, treated the proton
and neutron as a single object, the nucleon. The proton was the "isospin
up" and the neutron, the "isospin down" (in case you run ito this, some
authors write this the other way). The "iso" came from the terms "isobar"
or "isotope". So, the nucleon is designated with T=1/2, the proton
Tz = +1/2 and the neutron Tz = -1/2. This probably looks rather artificial,
but it really is not, because the nuclear forces are charge independent,
although the symmetry is only approximate, because the neutron/proton mass
difference is not 0. What can you do with this? Well, for example, you can
calculate the masses of nuclei in a given isobaric multiplet (look up
isobaric mass multiplet equation). Since the symmetry is not exact, this
modelis of primary interst only in low energy nuclear physics.
What else can you do? Take the SU(2) in the SU(3)xSU(2)xU(1) of the
standard model. The SU(2) is the gauge group representing the W+/- and Z.
They are said to transform among themselves as a weak isospin triplet
(unfortunate nomenclature, since weak isospin is not the same thing as the
isospin above). Since this has the same algebra as the usual angular
momentum matrices, you can define a "z-axis" and raising and lowering
operators. The 3 possible z-projections are W+, Z, W-. The raising
opearator acting on the W-, produces a Z, for example. What can you do
with this? Well, one of the most stringent tests of the standard model
comes from this model of the weak gauge bosons as an isotriplet. Since
the bosons are merely isospin roations of each other, there should
be only 1 weak coupling constant. In nuclei, this translates into
all superallowed fermi decays have to have the same lifetime,
once the decay is corrected for kinematics. So far, every single
nucleus which decays through a superallowed fermi transition has
the exact same lifetime (i.e., "ft-value").
None of the above are geometric relations, although the terminology
used, like "rotation" and "z-axis", comes from the language of
geometry. Why is SU(2) _unlike_ a mechanical angular momentum?
For one thing, a 4\pi rotation rather than 2\pi (or 4 reflections,
rather than 2) are requred to return the oiginal object to itself.
>I am glad that you decided not to quit on this discussion. For someone
>that has just started to learn more about quantum theory this year, I find
>this discussion fascinating. My reason for learning more about it is
>because of some of the things that are being discussed here. I just can't
>believe there isn't more to it or something is missing. However, it seems
>to me as a beginner, that modern QED did fix some of the problems with QM.
>But I am not sure yet if the fixes were "ad hoc". I am still studying and
>learning.
One reason that you may get the idea that quantum mechanics is "ad hoc",
is because you see it from a relatively narrow viewpoint, e.g., the
"uncertainty principle" which makes your perspective of position and
momenta rather confusing and then see all of the comments regarding
"obsevability" as artificial. But quantum mechanics is more than just
uncertainties in p and x and ad hoc quantization. For example, in thermo-
dynamics, the classical entropy and thermodynamic variables are simply
_wrong_, except at high temperatures (gibbs paradox). There is no
classical explanation. The entropy of N particles in n_j states is
classically, N!/n1!n2!...nj!, which disagrees with experiment. Boltzmann
fixed this in a very ad hoc way by dividing by N!, but the only
justification at the time was that it resolved gibbs paradox. Quantum
mechanics, however, tells you that, for example, 2 bosons in the same
state are _indistinguishable_, not just because you can't "see" them, but
because you cannot count them. In otherwords, regardless of what
experiment you could perform, in principle, nothing would allow you to
label 1 of the two particles and then label the other so that you could be
sure you counted two particles rather than the same particle twice.
That this is true, is seen from the entropy, since it leads to the
N! removed ad hoc, by boltzmann. The entropy is nothing _but_ a
counting argument.
In qed, the unobservability of the electron wavefunction, gives
you qed. In atoms. Quantization comes from the uncertainty principle.
You can use the uncertainty principle in the electric fields, to
show that light propagates at c.
Bilge
Re: More QM Nonsense, II to usenet:
>>
>> An oblate or prolate shape would require the proton to have
>> an electric quadrupole moment. It does not have a quadrupole
>> moment. Experiments have been performed to look for this very
>> thing.
>This shows you really do not understand what Kerr geometries are. The
>central body generating a Kerr field is *itself* totally (theoretically,
>perfectly) spheroidally shaped,
The "cental body" is a singularity. It has no shape. If it had a shape,
it wouldn't be a singularity.
>yet due to its rotation (and yes, it is an honest-to-god spinning
>spherical ball of matter), it produces a non-spherically symmetric
>external field.
What's your point apart from confusing "spin" with "spinning" in
order to create a semantics argument rather than a physical one?
>The Kerr field manifests "frame dragging" effects, and imo, this
>is the true cause of magnetism itself.
Your o is wrong. Deformations are due to mechanical angular momentum,
not spin.
Bilge
Re: More QM Nonsense, II to usenet:
>
>Please see p. Complement BX1, p. 1120, of Choen-Tann's "Quantum Physics," vol
Please see ppg 58-59 of bjorken & drell in which you'll find the
derivation from the spin-spin interaction, of the expression for the
hyperfine splitting I provided.
>2. There, he proves the dipole-dipole interaction must be zero if
>l = 0, as it does in the ground state. After you read it, I would
I no longer own those books. Someone offered to exchange both
both volumes for another quantum book rather than have me throw
them in the trash. I have never encountered a quantum mechanics
book (or any other textbook, for that matter) which I despise
to the extent I despise cohen-tannoudji. Without exception, it
contains nothing which cannot be reduced by a factor of 10 in
verbosity with a net increase by a factor of 10 in clarity.
Bilge
Re: More QM Nonsense, II to usenet:
>
>sense) is because it started with the point assumption (How can a
>point spin?).
>By eliminating the point assumption, electrons, protons,
>neutrons immediately take their shapes as three dimensional objects.
>If you spin a spinning top it complete its spin in 1 turn, but don't
>forget that this is basically a plannar spin. Now just think of the an
>arbirarily shaped 3-d object spinning wildly in all sort of angle in
>space. You get 3-d spinning. This intuitive dimensional analysis is
>sufficient to show why the net number of spin required for these
>objects to return to their starting position is about 2. (The
>situation is actually different than this, but you can see the main
>point of my argument. By the way a more complete argument is in my
>book.)
[...]
>I agree that this thread has been a great thing. In fact, I will mimic
>the idea and try something similar to it in a few days to get feed
>back on my common sense interpretation of the Heisenberg's uncertainty
>principle.
You'll need to explain more than just "uncertainty", since quantum
mechanics also explains gibbs paradox.
> QED is ad hoc (and Feynman was very frank about this,) IMHO, the
>reason is that it still treats quantum objects as geometrical points,
>which is obviously wrong. It is not difficult to see that this
>point-assumption is the reason why the "self-energy" points in Feynman
>diagrams blow up (with infinities) and need the "hocus pocus dippy
>process" of renormalization to fix.
Please try using classical mechanics to explain a "finite sized"
electron. Specify the radius you use so that I can point out why it's
wrong. If you think renormalization in qed is "dippy" wait till you
try to make sense out of a classical description.
> But since QED has fantastic numerical successes in a few cases,
>it exposed the weaknesses of the wave equations as well as the de
>Broglie's wave hypothesis and the Heisenberg's Uncertainty Principle
>(that all these are approximated descriptions and there is no reason
>for us to worship them as absolute truths.)
> I believe QED is the correct pointer to a future permanent fix of
>QM.
Uh, qed is what is renormalized by the "'hocus pocus dippy process'
of renormalization", not quantum mechanics.
Steve Bell
Are you saying you disagree with his mathematical development regarding this
issue of dipole-dipole interaction in the 1s shell? If so, I can understand
why - it "mathematically proves," according to his QM, that the two dipole
fields exist, but don't outwardly express any interaction. I would like to
hear how you explain his derivation is incorrect. Sorry, I don't have your
reference. Are you sure the correction term you previously gave, from your
text, was specifically for the dipole-dipole interaction, and not for the
"Fermi contact term?" Can you give a quote from the text, as explicit about
this issue as I gave from Cohen-Tann, that directly states the dipole-dipole
interaction is not zero for the 1s shell? Assuming the "contact term" is
physically true, how would you add the two, assuming the dipole-dipole
interaction *does* contribute, and not get a "net effect" out of step with the
observed 21 cm line? How does your reference handle the "contact term"?
Steve Bell
>
> Steve Bell said some stuff about
> Re: More QM Nonsense, II to usenet:
> >Bob Zombiewoof wrote:
> >>
> >> An oblate or prolate shape would require the proton to have
> >> an electric quadrupole moment. It does not have a quadrupole
> >> moment. Experiments have been performed to look for this very
> >> thing.
>
> >This shows you really do not understand what Kerr geometries are. The
> >central body generating a Kerr field is *itself* totally (theoretically,
> >perfectly) spheroidally shaped,
>
> The "cental body" is a singularity. It has no shape. If it had a shape,
> it wouldn't be a singularity.
>
In the application of these differential geometries to the orbit problem of "a
body coasting along a geodesic," outside of the event horizon distance, you
treat the central body as an actual body with some definite spherical
diameter. To quote from Lawden's "An Introduction to Tensor Calculus, etc," in
his section on the Schwarzschild field equation solution, "We shall consider
the special case when the whole of space is devoid of matter, apart from a
spherical body with its centre at the center of the symmetry O," where "O" is
the origin (0,0,0). This doesn't change the field "outside" of the body, which
is as if all of its matter were concentrated at the origin (0,0,0).
> >yet due to its rotation (and yes, it is an honest-to-god spinning
> >spherical ball of matter), it produces a non-spherically symmetric
> >external field.
>
> What's your point apart from confusing "spin" with "spinning" in
> order to create a semantics argument rather than a physical one?
>
My point is that the spin of the central body in a differential geometry is a
physical spin of a centrally located spherical ball of matter, along a
definite "roll axis," not the "esoteric" QM "spin" of the proton.
> >The Kerr field manifests "frame dragging" effects, and imo, this
> >is the true cause of magnetism itself.
>
> Your o is wrong. Deformations are due to mechanical angular momentum,
> not spin.
>
What "deformations" are you referring to? The "deformation" in the shape of
the proton itself, or the "deformation" of the field surrounding the proton?
Steve Bell
>
> Steve Bell wrote:
>
> > After a welcomed hiatus due to the holidays, I would
> > like to start this thread somewhat anew with the
> > following. Bile says, quite justifiably so, look at
> > the equations about the "Zitterbeweung" and make
> > up my own mind. I've done so at
> >
> > http://www.mindspring.com/~sb635/qmnonsense.pdf
>
> [QM versus GR]
>
> I am trying to build a consistent theory with inertia as
> a pivot. You surely agree with me that inertia plays the
> most important role in Newtonian Mechanics where it
> pertains predictability. Newtonian Mechanics never
> considered this inertia to be variable, this is
> introduced by relativity. Quantum mechanics differ so
> much from the 'inertial' mechanics NM & RM, and I claim
> the sole reason for this is that QM does not undergo
> inertia. No inertia, no predictability.
>
If by "inertia" you mean that matter has inertia (a tendency to resist a
change in its direction of motion), I guess I don't understand why the
electron, which is certainly some amount of "matter," would not have inertia.
> If we expand on this, it is outright silly to consider
> 'a relativistic' approach to QM phenomenon. Inertia does
> not apply, so relativity does not apply. The Heisenberg
> equation is the limit. Below that there is no inertia
> whatsoever, hence no predictability is possible.
>
I certainly believe that an electron in a ground state hydrogen atom has
inertia. I also believe that there are relativistic effects "in force" in
ground state hydrogen. The "net" outwardly manifest relativistic effects are
apparently zero, though, if you accept a 50 year old "Rydberg constant for
hydrogen" as "physically correct." There is a way to explain this through the
use of modeling the electromagnetic field surrounding the proton as a type of
differential geometry.
> Why is it the electron does not fall in to hydrogen
> nucleus ? It has no inertial mass below its lowest orbit.
>
If you take a "geodesic orbit theory" approach, the reason why the electron
doesn't "spiral into" the proton in hydrogen is because its motion is
truthfully geodesic, and the total relativistic orbital energy when coasting
along a geodesic is constant. If the electron's total relativistic orbital
energy remains constant through time, there is no "emission" or "absorption"
of energy, and its orbit doesn't "degrade" into a spiral.
> When the surrounding inertia increases the size of the
> orbital decreases and vice versa.
>
> So far I concentrated on fitting this 'inertial' theory
> on relativity. It works wonderfully. What do you think
> its possibilities are for QM ?
>
> Hayek
I think there are grand possibilities of using the highly "plastic" orbit
theory of a differential geometry, with all of the "equivalence principles"
involved, for application in the world of the atom.
Steve Bell
>
>
> Yes, this is something that also really bugs me about quantum theory.
> Everywhere we see circular-like motion from stars in a galaxy to planets
> around stars to electrons around nuclei then quantum theory says that spin
> is not a rotation. Very difficult to swallow that one. If it is not some
> kind of rotation, then what the heck is it?
>
I would think a QM person would say you are "classically extrapolating
incorrectly." I certainly do not think so, though. I think an electron is in
fact a (nearly) spherical ball of matter, and "in orbit" at an instant in time
in hydrogen, with its center of matter at a definite position in space with a
definite velocity attached, and that itself has a definite "roll axis" around
which it spins, etc.
>
> I am glad that you decided not to quit on this discussion. For someone that
> has just started to learn more about quantum theory this year, I find this
> discussion fascinating. My reason for learning more about it is because of
> some of the things that are being discussed here. I just can't believe
> there isn't more to it or something is missing. However, it seems to me as
> a beginner, that modern QED did fix some of the problems with QM. But I am
> not sure yet if the fixes were "ad hoc". I am still studying and learning.
>
> I hope that we could take name calling out of this discussion and continue
> with it.
>
> FrediFizzx
I truthfully thank you for saying this. I sometimes get the feeling that no
one pays attention to what I say, and that these discussions are for nothing,
if all I am doing is talking to someone who has already made up their minds
that QM cannot possibly, in any way shape or form, be incorrect. I appreciate
that point of view though, in a "kindred" way, because that's the way I feel
about GR. It is sometimes difficult, to a degree, to "buck the trend" in QM
and absorb all the "you are an idiot" reactions you get from the QM people.
You know, though, I actually (morbidly, I guess) *like* being called an
"idiot" from someone whose future comment I hope to be able to refute. I
consider myself to be in an "intellectual competition" with the QM people, one
I hope to "win" successfully. This type of competition is not bad in any way,
and probably is the driving force behind "revolutions." It is a definite
"physics challenge" to me to refute QM. Something is going to replace QM.
Something always "replaces" our current theories. I have taken a stab at what
could replace QM, with only a very small part of the total theory addressed
(but a vitally important part - ground state hydrogen) and I completely
expect, my ideas *will* be called "crackpot" at this stage of acceptance.
*Should* they be called "crackpot"? Many friends of mine, close friends,
consider my ideas on stuff like "religion" and the like, also "of questionable
validity" so I guess I am a crackpot. I certainly meet a lot of the "crackpot"
definition per S. Spheicher. I haven't tried Spheicher's "aluminum beanie hat"
to try and block out "whatever the hell Spheicher was talking about" but if I
am being "mind invaded" by whatever (and it's not any "pets" - I don't have
any, other than the microbes that live on my body refuse), that would be
interesting too. I am definitely the kind of person who, if the mothership
landed in my backyard, I would say "Beam me up, Scotty!".
Thinh Tran
> Thinh Tran said some stuff about
> Re: More QM Nonsense, II to usenet:
> >
> >The reason that QM says spin is not "spinning" (in the everyday's
> >sense) is because it started with the point assumption (How can a
> >point spin?).
>
> No, that isn't the reason.
me ask the question again: Can a mathematical point spin?
Actually the better question to ask is: Can a mathematical point
change its orientation at a fixed location (x1, y1, z1) in space?
Ask these questions and put yourself in the position of a QM
pioneer then you may see why QM spin (i.e., something that has no
classical counterpart) is necessary as a patch to an imcomplete set of
starting axioms, and why it became permanent until further notice.
(like a software bug fix is necessary for a software that was
released too early. The fix of course had to stay until the software
is upgraded.)
> >By eliminating the point assumption, electrons, protons,
> >neutrons immediately take their shapes as three dimensional objects.
> >If you spin a spinning top it complete its spin in 1 turn, but don't
> >forget that this is basically a plannar spin. Now just think of the an
> >arbirarily shaped 3-d object spinning wildly in all sort of angle in
> >space. You get 3-d spinning. This intuitive dimensional analysis is
> >sufficient to show why the net number of spin required for these
> >objects to return to their starting position is about 2. (The
> >situation is actually different than this, but you can see the main
> >point of my argument. By the way a more complete argument is in my
> >book.)
>
> That does not work.
through?
To give you more thing to think about. Actually in my explanation
it is not spinning per se (I only use spinning as an easy analogy.)
The real effect is the set of spatial orientations of a 3-d object
with no internal symmetry.
> [...]
> >I agree that this thread has been a great thing. In fact, I will mimic
> >the idea and try something similar to it in a few days to get feed
> >back on my common sense interpretation of the Heisenberg's uncertainty
> >principle.
>
> You'll need to explain more than just "uncertainty", since quantum
> mechanics also explains gibbs paradox.
[Reply] The Gibbs paradox is simple to explain once you debunk
probability. In fact (thanks to the reminder of a previous posting by
BobZombiewoof in another thread) I will include that in the 2nd
edition of my book. Somehow I did not think about this great example
when I wrote the first edition.
>
> > QED is ad hoc (and Feynman was very frank about this,) IMHO, the
> >reason is that it still treats quantum objects as geometrical points,
> >which is obviously wrong. It is not difficult to see that this
> >point-assumption is the reason why the "self-energy" points in Feynman
> >diagrams blow up (with infinities) and need the "hocus pocus dippy
> >process" of renormalization to fix.
>
> Please try using classical mechanics to explain a "finite sized"
> electron. Specify the radius you use so that I can point out why it's
> wrong. If you think renormalization in qed is "dippy" wait till you
> try to make sense out of a classical description.
[Reply] "hocus pocus dippy process" is not my creation. It came from
the QED maestro Feynman himself.
I'm not a fan of classical physics. I agree that Newtonian
physics has no chance in solving QM problems.
>
>
> > But since QED has fantastic numerical successes in a few cases,
> >it exposed the weaknesses of the wave equations as well as the de
> >Broglie's wave hypothesis and the Heisenberg's Uncertainty Principle
> >(that all these are approximated descriptions and there is no reason
> >for us to worship them as absolute truths.)
> > I believe QED is the correct pointer to a future permanent fix of
> >QM.
>
> Uh, qed is what is renormalized by the "'hocus pocus dippy process'
> of renormalization", not quantum mechanics.
[Reply] I'm not clear what you're trying to say. Did not say QED is
not a part of QM? How do you define QM? Wave equations? Matrix? QCD?
QFD? Please clarify.
FrediFizzx
news:slrnb16vk...@radioactivex.lebesque-al.net...
"New internal degree of freedom, X," must be the key to understanding
intrinsic spin. My book shows this derivation but doesn't quite explain it
like you did. It shows that X = (hbar/2)*sigma (my book is using S instead
of X though; L+2S). I don't understand what sigma represents. This is
still a little over my head. Is there any way to describe this in words?
What exactly is meant by "new internal degree of freedom"? Another book I
have warned that the explanation of quantum spin was quite involved ;-).
But there has to be a way to describe it in words.
Well, if we take a sine wave and rotate it around in a circle twice just as
it completes one wavelength, don't we get a 4pi rotation? I can see how
angular momentum is quantized by wave phase having to match perfectly each
rotation or every other rotation. However, this doesn't explain why we
don't have 8pi rotation. Maybe 4pi is all that is necessary for 3D?
This description seems to indicate that we should be using a ensemble
interpretation. IOW, should we really be applying some of the quantum
formalism to individual particles?
| In qed, the unobservability of the electron wavefunction, gives
| you qed. In atoms. Quantization comes from the uncertainty principle.
| You can use the uncertainty principle in the electric fields, to
| show that light propagates at c.
I don't disagree with QM, QED, etc. per se. I guess "ad hoc" is really not
a good expression to use. I think that they are wonderful mathematical
tools for getting a handle on being able to calculate things in the
microscopic realm since classical physics fails. I am just not sure that
some of the interpretations are exactly complete. Even to this day, there
are some respected "experts" that have doubts that the photon is a particle.
This indicates to me that something is not quite complete.
FrediFizzx
Steve Bell
>
> ro...@radioactivex.lebesque-al.net (Bilge) wrote in message news:<slrnb1726...@radioactivex.lebesque-al.net>...
> > Thinh Tran said some stuff about
> > Re: More QM Nonsense, II to usenet:
> > >
> > >The reason that QM says spin is not "spinning" (in the everyday's
> > >sense) is because it started with the point assumption (How can a
> > >point spin?).
> >
> > No, that isn't the reason.
> [Thinh's reply]: This point can be argued forever. To save time, let
> me ask the question again: Can a mathematical point spin?
You are quite correct in thinking that a "mathematical point" can not "spin."
An absolute "point" in 3-D space has no "size" so what would be spinning? I
personally believe that the electron does have a physical size, whose radius
is greater than zero, and that the electron is a "particle complex" with a
definite, greater than zero, physical size, with a definite center-of-matter.
I believe an electron itself has physical structure, and is composed of
"truthfully" elementary particles that QM has no clue about, and never will.
Bilge
Re: More QM Nonsense, II to usenet:
>Bilge wrote:
>>
>> Steve Bell said some stuff about
>> Re: More QM Nonsense, II to usenet:
>>
>> >
>> >Please see p. Complement BX1, p. 1120, of Choen-Tann's
>> >"Quantum Physics," vol
>>
>> Please see ppg 58-59 of bjorken & drell in which you'll find the
>> derivation from the spin-spin interaction, of the expression for the
>> hyperfine splitting I provided.
>>
>> >2. There, he proves the dipole-dipole interaction must be zero if
>> >l = 0, as it does in the ground state. After you read it, I would
>>
>> I no longer own those books. Someone offered to exchange both
>> both volumes for another quantum book rather than have me throw
>> them in the trash. I have never encountered a quantum mechanics
>> book (or any other textbook, for that matter) which I despise
>> to the extent I despise cohen-tannoudji. Without exception, it
>> contains nothing which cannot be reduced by a factor of 10 in
>> verbosity with a net increase by a factor of 10 in clarity.
>
>
>Are you saying you disagree with his mathematical development regarding this
>issue of dipole-dipole interaction in the 1s shell?
by the coupling of the electron and proton spins. I have no idea what your
book is calling a dipole-dipole interaction, so I have deliberately referred
to the spins. If your book says that the there is no splitting into a
singlet and triplet state then, yes, I am disagreeing with them.
> If so, I can understand
>why - it "mathematically proves," according to his QM, that the two dipole
>fields exist, but don't outwardly express any interaction. I would like to
>hear how you explain his derivation is incorrect. Sorry, I don't have your
>reference. Are you sure the correction term you previously gave, from your
>text, was specifically for the dipole-dipole interaction, and not for the
>"Fermi contact term?" Can you give a quote from the text, as explicit about
>this issue as I gave from Cohen-Tann, that directly states the dipole-dipole
>interaction is not zero for the 1s shell?
I can give the exact text that states the origin of the splitting,
with references cited. You can try to figure out what cohen-tannoudji
have written for comparison. I really don't care to try and interpret
that, since most of it is about as transparent as tar for no particular
reason.
Equations are nominally in TeX format. If that is unfamiliar, \vec{B} is
the vector B, \vec{A}\cdot\vec{B} is the dot product of the vectors A and
B. \vec{\delta} are the pauli spin operators, etc. Where it was simpler to
superscript a group of terms rather than each term individually or to
combine terms in parentheses, I have done so for readability. X denotes
a cross product, x indicates ordinary multiplication where I had to split
an equation over multiple lines.
From Bjorken & Drell, Vol I. ppg 57-58:
---
"The hyperfine structure results from the interaction of
the proton with the electron magnetic moment[1]. This has
the effect of splitting all lines into doublets corresponding
to the two possible states of total angular momentum compounded
from the j of the electron system and the half-integer spin
of the proton. Let us compute the magnitude of this effect
for s states. For our purposes a nonrelativistic description
of the electron suffices. The interaction is of the form
H' = +(|e|/2m)\vec{\delta}\cdot\vec{B}
and \vec{B} = (g_{p}e/2M_{p})\integral d^{3}r'
x \rho(r')\nabla X (I X \nabla)/4\pi|r-r'|
Here I is the proton spin operator (I_{z} = +/-1/2) and
\rho(r') is the magnetic moment density of the proton, owing
to the fact it is not a point particle. Using the relations
\nabla x (I x \nabla) = \vec{I}\nabla^2
- (\vec{I}\cdotvec{\nabla})\vec{\nabla}
and taking the angular average for a spherically symmetric
wave function so that
\nabla_{i}\nabla_{j} => (1/3)\delta_{ij}\nabla^{2}
we find
\vec{B} = (2/3)g_{p}(e/2M_{p}) \integral d^{3}r' \rho(r')
x \vec{I}\nabla^{2}(1//4\pi|r-r'|)
= (2/3)g_{p}(e/2M_{p})\vec_{I}\rho(r)
The energy shift is then given in the nonrelativistic theory, by:
\Delta E_{n} = (\Psi H' \Psi)
= (2/3)(g_{p}e^2/4mM_{p} \vec{\delta}\cdot\vec{I}
x \integral d^{3}r \Psi_{n}^{*}\rho(r)\Psi_{n}
=~ (1/6)(g_{p}e^2/mM_{p})\vec{\delta)\cdot\vec_{I}|\Psi(0)|^2
= (1/2)m\alpha^2 [ (4/3)g_{p} (Z\alpha/n)^{2}(m/M_{p}
x \vec{\delta}\cdot\vec{I} ]
with
+1/2 triplet state
\vec{delta}\cdot\vec{I} =
-3/2 singlet state
The splitting of the nth s-state level is thus
\delta_{n} = (1/2)m\alpha^2 [(8/3)g_{p} (Z\alpha/n)^2 (m/M_p)]
and is reduced by the mass ratio (m/M_{p}) relative to the fine structure."
[1] E. Fermi, Z. Physik, 60 320 (1930); see also Bethe and
Salpeter, op. cit. p. 163
----
> Assuming the "contact term" is
>physically true, how would you add the two, assuming the dipole-dipole
>interaction *does* contribute, and not get a "net effect" out of step
>with the observed 21 cm line? How does your reference handle the
>"contact term"?
Since I don't own the text and it isn't clear from what you've
stated what it is you are talking about, I won't speculate. I
disliked every aspect of that text, but I especially disliked
its treatment of angular momentum.
Steve Bell
>
>
> I don't disagree with QM, QED, etc. per se. I guess "ad hoc" is really not
> a good expression to use. I think that they are wonderful mathematical
> tools for getting a handle on being able to calculate things in the
> microscopic realm since classical physics fails. I am just not sure that
> some of the interpretations are exactly complete. Even to this day, there
> are some respected "experts" that have doubts that the photon is a particle.
> This indicates to me that something is not quite complete.
>
> FrediFizzx
I have to keep coming back to this, that the QM equations, as complete as I
have energy predicting equations for, *do not* accurately predict hydrogen's
ground state energy. How can it be that a 50-year old experimental
quantification of hydrogen's ground state energy (ignoring the hyperfine
splitting) could possibly equal a prediction of hydrogen's ground state
energy, using *today's* consistent set of values for h, c, etc, and especially
note that the 50-year old observed equates to Bohr's *nonrelativistic* energy
prediction? Were the persons involved in putting into print, 50 years ago, the
Rydberg constant for hydrogen (50 years ago), somehow "omniscient" and knew
*today's* *current* set of "consistent values" for h, c, etc? The only
explanation that I can see is that the 50-year old value was *not* influenced
by any "bias" and gives the energy "truth" about 1s hydrogen. If the 50-year
old value is the "experimental spectroscopic truth" about ground state
hydrogen, the inescapable conclusion is that ground state hydrogen *DOES NOT*
outwardly manifest any relativistic effects. The only "out" for this,
physically as far as I can see, is that the electromagnetic field surrounding
the proton should be modeled as a type of "differential geometry." In order to
understand this, you must be well versed in Kerr geometries, and how the
intrinsic spin of the central body (even if its matter resides within a
*perfectly* spheroidal volume of space) causes a "frame dragging" in the
electromagnetic *static* field surrounding the proton. *THIS* is the
unification of QM and GR, namely that the accelerations responsible for
binding the electron to the proton are a function of the curvature of the
electromagnetic field *itself* and are not "subordinately" given by Coulomb's
relativistic (only special) potential, as in the Nordstrom metric.
Bilge
Re: More QM Nonsense, II to usenet:
>Bilge wrote:
>>
>> Steve Bell said some stuff about
>> Re: More QM Nonsense, II to usenet:
>> >Bob Zombiewoof wrote:
>> >>
>> >> An oblate or prolate shape would require the proton to have
>> >> an electric quadrupole moment. It does not have a quadrupole
>> >> moment. Experiments have been performed to look for this very
>> >> thing.
>>
>> >This shows you really do not understand what Kerr geometries are. The
>> >central body generating a Kerr field is *itself* totally (theoretically,
>> >perfectly) spheroidally shaped,
>>
>> The "cental body" is a singularity. It has no shape. If it had a shape,
>> it wouldn't be a singularity.
>>
>
>
>In the application of these differential geometries to the orbit problem of "a
>body coasting along a geodesic," outside of the event horizon distance, you
>treat the central body as an actual body with some definite spherical
>diameter. To quote from Lawden's "An Introduction to Tensor Calculus, etc," in
>his section on the Schwarzschild field equation solution, "We shall consider
>the special case when the whole of space is devoid of matter, apart from a
>spherical body with its centre at the center of the symmetry O," where "O" is
>the origin (0,0,0). This doesn't change the field "outside" of the body, which
>is as if all of its matter were concentrated at the origin (0,0,0).
precisely because it has no angular momentum.
>
>> >yet due to its rotation (and yes, it is an honest-to-god spinning
>> >spherical ball of matter), it produces a non-spherically symmetric
>> >external field.
>>
>> What's your point apart from confusing "spin" with "spinning" in
>> order to create a semantics argument rather than a physical one?
>>
>
>
>My point is that the spin of the central body in a differential geometry is a
>physical spin of a centrally located spherical ball of matter, along a
>definite "roll axis," not the "esoteric" QM "spin" of the proton.
Despite disagreeing with your description of the rotational angular
momentum for a kerr black hole, I'll just go along with that part of
it, since it really doesn't matter as far as this goes. The rotational
angular momentum is not a spin. If you rotate the symmetry axis of
kerr black hole by 2\pi, you rotate it back to its original orientation.
If you rotate a spin-1/2 by 2\pi, the wavefunction changes sign.
I've said this at least 3 times. Before introducing yet another
irrelevant analogy, deal with the fact that you need a 4\pi rotation
not at a 2\pi rotation (or 4 reflections, not 2) to return a spinor
to itself.
>> >The Kerr field manifests "frame dragging" effects, and imo, this
>> >is the true cause of magnetism itself.
>>
>> Your o is wrong. Deformations are due to mechanical angular momentum,
>> not spin.
>What "deformations" are you referring to? The "deformation" in the
>shape of the proton itself, or the "deformation" of the field surrounding
>the proton?
The proton has no deformations as may be seen by its lack of a dipole
or quadrupole moment or higher order multipolarity. The kerr metric
does have a symmetry axis as is evident from the metric, which incorporates
it explicitly as the parameter J/m, where J is the angular momentum.
In boyer-lindquist coordinates, the radius is r is parameterized by:
\rho^2 = r^2 + (a cos(\theta))^2
K = r^2 - 2Mr + a^2
The second term on the rhs of the first expression is just a deformation
parameter for a deformation along the z-axis.
But, you have more difficultoes than this. Calculate the radius of the
event horizon for a proton, not to mention the fact that quarks are
readily observed within the proton from deep inelastic scattering,
which would be precluded if the proton were a blackhole. Also, the
proton is charged, so what you _really_ want is the metric for a
kerr-newman blackhole.
Bilge
Re: More QM Nonsense, II to usenet:
>ro...@radioactivex.lebesque-al.net (Bilge) wrote in message
>news:<slrnb1726...@radioactivex.lebesque-al.net>...
>> Thinh Tran said some stuff about
>> Re: More QM Nonsense, II to usenet:
>> >
>> >The reason that QM says spin is not "spinning" (in the everyday's
>> >sense) is because it started with the point assumption (How can a
>> >point spin?).
>>
>> No, that isn't the reason.
>[Thinh's reply]: This point can be argued forever. To save time, let
>me ask the question again: Can a mathematical point spin?
> Actually the better question to ask is: Can a mathematical point
>change its orientation at a fixed location (x1, y1, z1) in space?
[...]
>> That does not work.
>[Reply] How do you know it does not work? Have you thought the problem
>through?
Did you perform the calculation I asked you to perform? If not,
then I have at least thought it through further than you have.
> To give you more thing to think about. Actually in my explanation
>it is not spinning per se (I only use spinning as an easy analogy.)
>The real effect is the set of spatial orientations of a 3-d object
>with no internal symmetry.
Since the spin comes from relativity, your explanation doesn't work.
>
>> [...]
>> >I agree that this thread has been a great thing. In fact, I will mimic
>> >the idea and try something similar to it in a few days to get feed
>> >back on my common sense interpretation of the Heisenberg's uncertainty
>> >principle.
>>
>> You'll need to explain more than just "uncertainty", since quantum
>> mechanics also explains gibbs paradox.
>
>[Reply] The Gibbs paradox is simple to explain once you debunk
Then you should have explained it, rather than just telling me
how simple it was to do so. Probability has nothing to do with
the quantum mechanical explanation. Since the entropy _only_
requires counting particles and states, I don't see how you'll
justify dividing out the permutations amoung identical rearrangements,
since you have N! distinct arrangements if the consitituents of
the ensemble are distinguishable, even in principle.
[...]
>>
>> Please try using classical mechanics to explain a "finite sized"
>> electron. Specify the radius you use so that I can point out why it's
>> wrong. If you think renormalization in qed is "dippy" wait till you
>> try to make sense out of a classical description.
>
>[Reply] "hocus pocus dippy process" is not my creation. It came from
>the QED maestro Feynman himself.
In what year?
> I'm not a fan of classical physics. I agree that Newtonian
>physics has no chance in solving QM problems.
Since everything you've advocated is straight newtonian physics,
I can't see how you can possibly say that. Adding some sort of
randomeness due to an unquantified perturbation does not alter the
newtonian aspect.
>>
>>
>> > But since QED has fantastic numerical successes in a few cases,
>> >it exposed the weaknesses of the wave equations as well as the de
>> >Broglie's wave hypothesis and the Heisenberg's Uncertainty Principle
>> >(that all these are approximated descriptions and there is no reason
>> >for us to worship them as absolute truths.)
>> > I believe QED is the correct pointer to a future permanent fix of
>> >QM.
>>
>> Uh, qed is what is renormalized by the "'hocus pocus dippy process'
>> of renormalization", not quantum mechanics.
>
>[Reply] I'm not clear what you're trying to say. Did not say QED is
>not a part of QM? How do you define QM? Wave equations? Matrix? QCD?
>QFD? Please clarify.
Plain vanilla quantum mechanics is nonrelativistic. The dirac equation
and the klein-gordon equation are the relativistic analogs. Qed is a
the result of requiring the dirac equation to be _locally_ gauge invariant
and applying noether's theorem to obtain a conserved current. The replacement
of the ordinary partial derivative by a gauge covariant derivative, is
what gives you qed.
FrediFizzx
| FrediFizzx wrote:
| >
| >
| > Yes, this is something that also really bugs me about quantum theory.
| > Everywhere we see circular-like motion from stars in a galaxy to planets
| > around stars to electrons around nuclei then quantum theory says that
spin
| > is not a rotation. Very difficult to swallow that one. If it is not
some
| > kind of rotation, then what the heck is it?
| >
|
|
| I would think a QM person would say you are "classically extrapolating
| incorrectly." I certainly do not think so, though. I think an electron is
in
| fact a (nearly) spherical ball of matter, and "in orbit" at an instant in
time
| in hydrogen, with its center of matter at a definite position in space
with a
| definite velocity attached, and that itself has a definite "roll axis"
around
| which it spins, etc.
I think maybe you are right or close to being right. My wacky idea is that
maybe there is no time in a "pure" vacuum. And bare "point" charge clears
out a pure vacuum space in the quantum vacuum then basically "puffs out" to
fill that space since there is no time in that little space. Or is like a
wavy string in that space. This last one would be more like defining a
cylindrical shaped object. I can also get the 4pi rotation with this one
and a magnetic moment. Maybe this idea also explains quantum fuzziness?
I have been on OldMan's Crackpot list for a couple of months now for
claiming the EM quantum vacuum can be modeled by inductance and capacitance
classically as an average. I don't see how anyone that believes the quantum
vacuum is polarizable around bare charge and that photons can briefly turn
into virtual charge pairs and also believes that the uncertainty principle
is correct can't see that this is inductance and capacitance. The
uncertainty principle surely allows for partially separated virtual charges
everywhere all the time in a polarizable vacuum. Separated charge in
constant motion is going to give you capacitance and inductance as a
classical average. It is even very easy to derive from w = 1/sqrt(L*C) and
c = 1/sqrt(eps0*mu0) with w being omega, angular frequency. Lvac = mu0*c/w
and Cvac = eps0*c/w or Lvac = 4*pi/c*w and Cvac = c/4*pi*w in Gaussian
units. So who is really the crackpot here? I certainly don't think I am a
crackpot for just trying to come up with different ideas. Nor would I ever
think you are a crackpot for just bucking the system in a rational way.
Although, yes, I am ready for a ride on the first warp field drive ;-). I
hope the heck they are right and that warp drives will work. I found out
that they are counting on being able to modify vacuum inductance and
capacitance (the polarizable vacuum) to work. Only they don't say that
directly most of the time.
FrediFizzx
Hayek
Steve Bell wrote:
>> I am trying to build a consistent theory with inertia
>> as a pivot. You surely agree with me that inertia
>> plays the most important role in Newtonian
>> Mechanics where it pertains predictability.
>> Newtonian Mechanics never considered this inertia
>> to be variable, this is introduced by
>> relativity. Quantum mechanics differ so much from
>> the 'inertial' mechanics NM & RM, and I claim
the sole reason for this is that QM does not undergo
>> inertia. No inertia, no predictability.
>>
>>
>
>
> If by "inertia" you mean that matter has inertia (a
tendency to resist a
> change in its direction of motion), I guess I don't
understand why the
> electron, which is certainly some amount of "matter,"
would not have inertia.
Let me elaborate.
I often refer to Mach's Principle, that states that
"mass over there causes inertia over here". By "mass
over there" Mach means all the masses surrounding us,
like stars and galaxies. "Inertia over here" is our good
old Newtonian inertia. Newton already noticed that a
bucket with water 'knows' it is at rest wrt the stars,
the famous "Newton's Bucket" or "Newton's pail"
experiment. GR confirms this by stating that time, an
other word for inertia is caused by the mass
distribution surrounding us.
http://www.xs4all.nl/~notime/inert/gravp546.html
scanned from Gravitation, Misner, Thorne, Wheeler (the MTW)
"
Nevertheless, it is a fact that Mach's Principle -that
matter there governs inertia here- and Riemann's idea
-that the geometry of space responds to physics and
participates in physics- were the two great currents of
thought which Einstein, by means of his powerful
equivalence principle, brought together into the
present-day geometric description of gravitation and motion.
"
Read the whole paragraph from the beginning :
http://www.xs4all.nl/~notime/inert/gravp543.html
"volgende" means "next".
One must realize that a clock is only an inertiameter,
and that the difference between Newtonian and
Einsteinian mechanics is only that the former assumed
inertia to be constant everywhere. Einstein was happy
to abolish constant inertia, but could not see the
connection between QM and the absence of inertia.
Since inertia is just another word for time, you could
repeat the above paragraph where you replace "time"
where it says "inertia".
Back to the "inertial (dis)connection" of QM:
Let me repeat some part of a post I made earlier
"
The connection wth QM :
I will start with an analogy :
Compare the heavy masses (stars & galaxies) which cause
our inertia, with Eagles spotting mice. If the mouse
does not move or moves slowly or a little distance, the
Eagle will not see it. If it is too small neither. Only
if the mouse is big enough, has enough speed and runs
some distance the Eagle will spot it. Mass* Speed *
Distance >= plank_constant/(2*pi). Only when this
happens, a mass will have to deal with the stars, in
this case, undergo their inertia.
No inertia means infinite speeds, and that means
particles can be at many places at the same time, with a
probability of being there. No inertia means that the
slightest force will cause infinite speeds. No inertia
means wavy character of particles, wild animal that
rages at infinite speed in its heisenberg/mach cage
dealing with its own electrical fields it generates.
That is how a photon interferes with itself.
Wheeler Dewitt found that there is no time in an empty
universe, that means no inertia, and they found the
lowest level of energy of the wave equation. Only the
lowest level, because the other levels are interactions
between the quantum-non-inertial world and the Machian
inertial world, and for that you need a whole universe.
This sort of unifies GR & QM, at least we have an idea
were to start for fitting it together.
"
The electron does not move around the nucleus, you have
a probability of finding it somewhere in or around the
nucleus. If you had a mass moving inertially, it would
nicely follow a predicted path. QM has no such paths
because then speed and position would be determined.
>
>> If we expand on this, it is outright silly to
>> consider 'a relativistic' approach to QM
>> phenomenon. Inertia does not apply, so relativity
>> does not apply. The Heisenberg equation is the
>> limit. Below that there is no inertia whatsoever,
>> hence no predictability is possible.
>
> I certainly believe that an electron in a ground
> state
hydrogen atom has
> inertia.
I believe inertia fails miserably below the Heisenberg
border. No inertia would be an excellent explanation for
"Zitterbeweging". Just imagine an object that is not
subjected to inertia in a closed cage.
I also believe that there are relativistic effects "in
force" in
> ground state hydrogen. The "net" outwardly manifest
relativistic effects are
> apparently zero, though, if you accept a 50 year old
"Rydberg constant for
> hydrogen" as "physically correct."
But of course, if the electron touches the borders of
its Heisenberg limits, then inertia kicks in. And
relativity, since relativity is also an inertial phenomenon.
If you had a universe containing a single hydrogen atom,
and further no other mass, and therefore no inertia, as
in Mach's principle/GR, the orbital of hydrogen would be
as big as the universe.
> There is a way to explain this through the use of
> modeling the electromagnetic field surrounding
the proton as a type of
> differential geometry.
Exactly compensating the relativistic effects ?
Sounds complicated.
>> Why is it the electron does not fall in to hydrogen
nucleus ? It has no inertial mass below its lowest orbit.
>
> If you take a "geodesic orbit theory" approach, the
reason why the electron
> doesn't "spiral into" the proton in hydrogen is
because its motion is
> truthfully geodesic,
That would mean predictable. It happens to be in the
nucleus too, only with a low probability. If you
consider it has no inertia to reckon with in the
orbital, you get exactly this kind of behaviour. No
inertia also means speeds greater than c, the famous
teleport capability, or the many paths Feynman
explanation.
> and the total relativistic orbital energy when coasting
> along a geodesic is constant. If the electron's total
relativistic orbital
> energy remains constant through time, there is no
"emission" or "absorption"
> of energy, and its orbit doesn't "degrade" into a spiral.
It can only be explained if the energy difference
between being in the nucleus and in the lowest orbital
is zero. And that would mean no inertia. This geodesic
explanation stinks.
>>When the surrounding inertia increases the size of the
>>orbital decreases and vice versa.
>>
>>So far I concentrated on fitting this 'inertial' theory
>>on relativity. It works wonderfully. What do you think
>>its possibilities are for QM ?
> I think there are grand possibilities of using the
highly "plastic" orbit
> theory of a differential geometry, with all of the
"equivalence principles"
> involved, for application in the world of the atom.
I think 'chiral Eotvosch" stinks too.
If there is no inertia at the lowest energy level,
there is no point measuring it there, unless you
explicitly do not want to find it. And it gets worse: no
inertia, also means no gravity. Eotvosch is about
comparing inertia to gravity.
Don't tell Uncle Al, because sadly, he does not want to
know.
Hayek.
Thinh Tran
> >> That does not work.
>
> >[Reply] How do you know it does not work? Have you thought the problem
> >through?
>
> Did you perform the calculation I asked you to perform? If not,
> then I have at least thought it through further than you have.
[Thinh's reply]: This is an unfair condition. Why should I carry out
another calculation when I already worked the problem out? A better
question would be whether I actually solved the problem or just made
foundless statements.
> Since the spin comes from relativity, your explanation doesn't work.
[reply]: I'm afraid you miss my point. Spin only appears to have come
from relativity (because you get spin with the Dirac's relativity
fix). But don't forget QM started out wrong with the point-charge
assumption.
An analogy may help: You have a software that has a big bug. You
put in a fix and get effect A. Is A caused by the fix? Answer: To a
casual user it is; but a good software engineer must assume bug+fix =>
A and therefore possibly A comes from the bug. The bug that I'm
alluding to is the point assumption.
If you've already ruled this out as a possibility, there is no
point to continue the discussion on the cause of spin.
> >[Reply] The Gibbs paradox is simple to explain once you debunk probability.
>
> Then you should have explained it, rather than just telling me
> how simple it was to do so. Probability has nothing to do with
> the quantum mechanical explanation. Since the entropy _only_
> requires counting particles and states, I don't see how you'll
> justify dividing out the permutations amoung identical rearrangements,
> since you have N! distinct arrangements if the consitituents of
> the ensemble are distinguishable, even in principle.
[Reply] I will explain it, but not right now, because I've already
issued a challenge to the scientific community to show that probabilty
and the Law of Large Number contradict each other; and the solution to
the Gibbs paradox is along the same line of logic.
But it is not true that "Probability has nothing to do with the
quantum mechanical explanation". Put yourself in the position of
someone who has to derive all of the QM distributions from scratch,
you will find that you have to go through more or less the same
procedure as Boltzmann's then added the condition of
"indistinguishability" to get things right.
Just a bit of history. Boltzmann's theory of course was
completely probabilitistic and in fact ran into trouble when Poincare
showed in 1890 in the Recurrence theorem that given enough time, the
least populous state could be occupied, thus the entropy of the grand
ensemble could decrease. Later the issue was settled, interestingly
also by Poincare, that the more populous state would be more probable,
making entropy a probabilistic statement. This conclusion, however, is
just another statement of the Law of Large Number. Thus, if
probability and the Law of Large Number are in conflict, the idea of
entropy is on very shaky ground.
Quantum distributions of course has the same conceptual problem
as the classical ensemble theory. The attempt to associate quantum
distributions with "quantum probability" as if quantum distributions
are immune to the problem of "classical probability" doesn't make the
QM foundation any less shaky.
> >[Reply] "hocus pocus dippy process" is not my creation. It came from
> >the QED maestro Feynman himself.
>
> In what year?
[Reply] Please read the chapter "loose ends", QED-The strange theory
of light and matter", Feynman, Princeton 1988 (based on a lecture
series given 1984.) In fact I'm surprised that you're not aware that
these 4 words came from Feynman.
>
> > I'm not a fan of classical physics. I agree that Newtonian
> >physics has no chance in solving QM problems.
>
> Since everything you've advocated is straight newtonian physics,
> I can't see how you can possibly say that. Adding some sort of
> randomeness due to an unquantified perturbation does not alter the
> newtonian aspect.
[Reply] FYI, it is Newtonian physics that I'm against as a QM
methodology. QM made the mistake of taking the point assumption from
Newtonian physics and translate that down to the quantum realm. The
founders forgot that the point assumption (which works for even
planets in classical physics) breaks down in the Qm realm.
So, contrary to your contention, I'm against QM because it is
"too Newtonian", not because I'm Newtonian. However, I don't think
science is a religion. I would apply whatever works, regardless of its
origin. It just so happens that the point assumption doesn't work in
QM.
> Plain vanilla quantum mechanics is nonrelativistic. The dirac equation
> and the klein-gordon equation are the relativistic analogs. Qed is a
> the result of requiring the dirac equation to be _locally_ gauge invariant
> and applying noether's theorem to obtain a conserved current. The replacement
> of the ordinary partial derivative by a gauge covariant derivative, is
> what gives you qed.
[Reply] Does this means you agree that QED is a branch of QM? Or do
you mean to say that QED is an unfortunate aberration?
Ken S. Tucker
>If you take a "geodesic orbit theory" approach, the reason why the electron
>doesn't "spiral into" the proton in hydrogen is because its motion is
>truthfully geodesic, and the total relativistic orbital energy when coasting
>along a geodesic is constant. If the electron's total relativistic orbital
>energy remains constant through time, there is no "emission" or "absorption"
>of energy, and its orbit doesn't "degrade" into a spiral.
Exactly, I found this to be true from a different approach,
(I posted sometime ago "A quantum geodesic"). I take Steve
Bell's and Thinh Tran's ideas very seriously, it is interesting
that three different (fairly) independent approaches can
converge on general agreement, and the civility Fred
suggested seems more apparent.
Steve's use of the Kerr metric is one approach, I personally
favor nonsymmetical tensors in unifying electromagnetism
and quantum theory, with GR, but that's mathematic semantics,
and may both be valid or equivalent calculation processes.
These theories are evolving and cannot be expected to
compete with Classical QM on every issue, where 50 years
of input has occurred, but QM is a theory of measurement
expectation, not a description of fundamental reality.
They potentially may explain the same thing with fewer
and more secure postulates and adhoc assumptions, just as
relativity eventually replaced aether theory.
If SR didn't come along in 1905, I'm sure aether theory
would have had many clauses, lemma's and other goodies,
so that it explained all the experiments. If SR was now
invented, it would have a difficult time competing with
entrenched and proved aether theory.
Anyway Steve, Bilge, Hayek and all, I've been reading
and re-reading this thread, but I still have one
outstanding question..
Did we agree, BOHR OBTAINED THE (very nearly) CORRECT
VALUE FOR THE RYDBERG CONSTANT, (ground state H)?
Thanks and Regards
Ken S. Tucker
Ken S. Tucker
>Thinh Tran said some stuff about
>Re: More QM Nonsense, II to usenet:
> >
> >The reason that QM says spin is not "spinning" (in the everyday's
> >sense) is because it started with the point assumption (How can a
> >point spin?).
>
> No, that isn't the reason.
>
> >By eliminating the point assumption, electrons, protons,
> >neutrons immediately take their shapes as three dimensional objects.
> >If you spin a spinning top it complete its spin in 1 turn, but don't
> >forget that this is basically a plannar spin. Now just think of the an
> >arbirarily shaped 3-d object spinning wildly in all sort of angle in
> >space. You get 3-d spinning. This intuitive dimensional analysis is
> >sufficient to show why the net number of spin required for these
> >objects to return to their starting position is about 2. (The
> >situation is actually different than this, but you can see the main
> >point of my argument. By the way a more complete argument is in my
> >book.)
>
> That does not work.
I think Tran could have explained his point better, ie,
"spinning wildly in all sort of angle", is a bit fuzzy.
Years ago I did some funny spin-stalls in aircraft, so I
recommend using a model aircraft to see his point.
Suppose you're flying straight and level due North. Then
do a 360 degree (2 pi) yaw around the vertical AND a 180
degree (pi) roll. Now you flying due North but upside down.
Note the relative spin rate difference, between yaw and roll,
roll spin = (1/2) yaw spin.
Now do this maneuver again, and you return to your original
attitude. Two rotations (4 pi) about the vertical returns the
aircraft to it's original state.
A sharp aerobatic pilot can do this on purpose, a real
pleasure to watch those smokin biplanes.
This *definition* of spin in words makes sense to me, and
I think this was Tran's argument. Does that sound reasonable
to you?
Regards
Ken S. Tucker
Hayek
Ken S. Tucker wrote:
I do not see the Kerr metric work. Ok, you have an
external inertia, and the strong attraction of the
charges of the electron-nucleus, but where in the Kerr
metric you have the Heisenberg playing field without
inertia ? And the three interacting ?
Also, your tensor approach should not only take care of
the low or nonexistant inertia of the QM world, and the
known GR world, but also where the two interact, and
that is where I think the tensor powers cease to yield
useful results.
The Wheeler-Dewitt equation, call it GR in empty space,
yields as result the lowest quantum mechanical energy
solution. Only the lowest, because the higher energy
solutions are a mix between lowest level QM and GR. The
latter GR being a full universe.
That is why I am so interested in this thread, since the
hydrogen electron is the lowest energy state.
> Anyway Steve, Bilge, Hayek and all, I've been reading
> and re-reading this thread, but I still have one
> outstanding question.. Did we agree, BOHR OBTAINED
> THE (very nearly) CORRECT VALUE FOR THE RYDBERG
> CONSTANT, (ground state H)?
My bet is on Bohr. :-)
Thinh Tran
[Comment] Thanks Ken for clarifying my point. Maybe I should learn
your method of communication. One thing I learned after participating
in sci.physics for a few months is that internet communication is very
tricky.
QM spin may have a physical meaning (I personally believe it
does). I agree that it may turn out that it is easier mathematically
to model QM objects as points then add spin as an extra property than
to start with 3-D models. Since all mathematical formulations are
precisely what they are: Formalisms, there's nothing wrong with a
point-charge formalism. However, I think physicists should be open to
a physical understanding to avoid thinking that math is all there is,
and any model different from point-charge is wrong.
There is an added benefits in thinking physical. If the
point-charge model leads to a dead end, at least we know what our next
alternative model should be.
Thinh Tran (http://www.thinhtran.com)
Thinh Tran
> Did we agree, BOHR OBTAINED THE (very nearly) CORRECT
> VALUE FOR THE RYDBERG CONSTANT, (ground state H)?
>
> Thanks and Regards
> Ken S. Tucker
[Thinh's comment] Agree 100%. What Bohr did was a spectacular achievement!
Bilge
Spin is not a 3-d rotation. How many reflections in a mirror are
required for any instantaneous position to return to itself? Since the
qusetion is rhetorical, I'll answer it for you. Two. What happens if
reflect a spinor twice? You get the negative of the original spinor.
In addition, we are not referring to rotations about two axes. We're
talking about the precession about an axis defined by a magnetic field.
I'm tired of repeating this, so pay attention and don't bother with
another analogy unless you address this first.
> Now do this maneuver again, and you return to your original
>attitude. Two rotations (4 pi) about the vertical returns the
>aircraft to it's original state.
In which case your spinor would require twice that, since you've
composed your rotation from two rotations. Read something about the
algebra SU(2) and its relationship to rotations. SU(2) is called
a "double cover", or universal covering group. It's double valued.
> A sharp aerobatic pilot can do this on purpose, a real
>pleasure to watch those smokin biplanes.
>
>This *definition* of spin in words makes sense to me, and
>I think this was Tran's argument. Does that sound reasonable
>to you?
No, not in the least, precisely because you are hung up on the semantics
rather than the physics. The reason that it makes sense to you is because
you think all of this is "ad hoc", so you've invented your own "ad hoc"
way of looking at it, which isn't correct. The same goes for thinh tran's
assertion about "point particles". The proton is not a point particle.
It's rms charge radius is just under 1 fm. It has a spin of 1/2 and
quantum mechanics deals with it just fine, as is evident from the vast
amount known about nuclei, a substantial amount of which may be deduced
using straight-forward, garden variety, quantum mechanics.
Thinh Tran
[Thinh's comment] Let's me see if I can put this in a geometrical
argument.
In 2-D spin (two dimensional spinning) there is only 1
distinguishable spinning path, I represent this with a circle.
In 3-D spin there are two independent spinning paths (i.e., two
orthogonal directions). I represent this situation with 2 identical
circles sharing the same center and perpendicular to each other. The
net effect of 3-D spin is that when an object complete 2pi you end up
on the other side of a circle (the particle turn to minus itself).
With 4pi you return to where you start. I hope this shows clearly why
4pi corresponds to all 3-D QM objects such as electron, proton,
neutron.
The advantage of this physical picture is that, considering that
there are 3 spatial directions, the following naturally arise from
geometrical considerations.
spin 1/2<=> 3-D physical objects
spin 1<=> 2-D physical objects
spin 2<=> 1-D physical objects
spin x: Partial entities (i.e. other objects)
If anyone sees a conceptual problem with this picture, please
advise.
Thinh Tran (http://www.thinhtran.com)
Y.Porat
> Ken S. Tucker wrote:
>
>
> >> orbit doesn't "degrade" into a spiral.
just for your information and physical education:
anything (to be caucious i will say nealy anything)
that can be formulated for an orbiting object:
*can be formulated as well for an object that osciates
cantileverly* from something bigger to which it is fixed
if you dont know the mechanical meaning of the concept
'cantilever' ask me and i will try to explain.
so a cantelever cannot 'fall' into its 'holder'
because it extends fron the holding (bigger) object
to further away
just a mthaphor:
if you stand on the globe and you have say your legs
fixed to it *there is no reasn why you should fall into
the earth*
or there is no reasn why your head should fall on the ground
just as simple as that
the orbiting electron of Bohr has nothing to do
with the electron structure reality.
it has a partial success only acidentally.!
IOW it is a wrong asumption. that cannot go further
than a few light elements.
so my advise: dont bet on the Bohr model
and not on the many existing paradigma for instance:
that in a heavy atom, the number of electrons
is exactly the number of protons.
if is just a friendly advise for not waisting your
time money and energy.(too much of it has been already waisted.)
all the best
Y.Porat
-----------------------
a
Hayek
Y.Porat wrote:
Not accidentally. I explain why.
> IOW it is a wrong asumption. that cannot go further
> than a few light elements.
I am perfectly aware of that.
>
> so my advise: dont bet on the Bohr model
In this case, for the lowest level I do,
for many elaborate reasons, I explained in the thread.
> and not on the many existing paradigma for instance:
> that in a heavy atom, the number of electrons
> is exactly the number of protons.
I have heared of ions, at age 16.
> if is just a friendly advise for not waisting your
> time money and energy.(too much of it has been
already waisted.)
I think debunking the weirdness of QM is a very valuable
effort.
Bilge
The disadvantage is that it's wrong. All of those are inherently
4-dimensional objects. A spin 0 is a lorentz scalar (or pseudoscalar).
It has no polarization axis. A spin 1 is a 4-vector (or pseudovector).
It has 4-polarization states, unless it's massless, in which case the
polarization must be transverse and there are only 2. Thats why it's
called a vector (or pseudovector) particle. A spin 2 has 5 polarization
states, unless it's massless in which case it must have only transverse
polarizations, in which case there are only 2. That's why it's associated
with a rank 2 tensor, for example, the graviton. For massless particles,
the time and longitudinal components vanish, which is why _propagating_
photons (and hypothetically, propagating gravitons) have only transverse
polarizations. A spin 1/2 (and 3/2, etc), is more complex, since there
exist both dirac and majorana spinors as possibilities. Electrons are
dirac particles. It's not clear whether neutrinos are majorana or dirac
particles. That's what experimental searches for neutrinoless double beta
decay are all about. Rather than explain it, I suggest using google to
find out more.
Further difficulties you have include: (1) in _no_ case, can you
specify more than one spatial axis along which to quantize the spin,
(2) particles have intrinsic parities. Please explain the difference
in spinning marbles which have positive and negative parities. I would
especially like to see a marble which is not rotating explained as
a pseudoscalar (an example from particle physics is the pion).
Hayek
Steve Bell wrote:
>
> You are perhaps not the best skilled individual to ask
> this of, so I'll ask it generally, do you believe
> there will be another "revolution" in physics? By
> "revolution" I mean something like the Einstein
> revolution, or the Quantum revolution initiated by
> Max Planck. Something truthfully revolutionary
> about how we view motion and probability in the
> realm of the atom? But if you say
that the "revolution" will come about by the path of
present-day accepted QM
> itself (which includes string theory, imo), I reject
that. Something else is
> needed to resolve the impasse of deterministic GR and
probabilistic QM.
I have been trying to tell you :
It is the abolishment of time.
And the understanding that time is inertia.
And without inertia, you get exactly this kind of
probabilistic behaviour.
With inertia you get exactly GR.
And there is lots of circumstantial evidence :
the Wheeler DeWitt equation, the Aspect - EPR
experiment, even the wave behaviour, add it all up and
the conclusion is : there is *NO* inertia below the
Heisenberg limit. Inertia gives us certainty, lack of
inertia gives us uncertainty. Inertia gives us
lightspeed, lack of inertia gives us super lightspeed.
Although I do not agree with Barbour 's platonia, I
agree with the title of his book :"The End of Time : The
Next Revolution in Physics"
The pivotal role of inertia is not understood by the
relativists-priests, nor by QM-priests.
Read my post Aplles=Whales.
http://groups.google.com/groups?&hl=en&selm=3DFC4CBF.7000806%40nospam.xs4all.nl
Hayek.
--
The small particles wave at
the big stars and get noticed.
:-)
>> that the "revolution" will come about by the path
> of present-day accepted QM itself (which includes string
> theory, imo), I reject that. Something else is
> needed to resolve the impasse of deterministic GR
> and probabilistic QM.
I have been trying to tell you :
It is the abolishment of time.
And the understanding that time is inertia.
And without inertia, you get exactly this kind of
probabilistic behaviour.
With inertia you get exactly GR.
And there is lots of circumstantial evidence :
the Wheeler DeWitt equation, the Aspect - EPR
experiment, even the wave behaviour, add it all up and
the conclusion is : there is *NO* inertia below the
Heisenberg limit. Inertia gives us certainty, lack of
inertia gives us uncertainty. Inertia gives us
lightspeed, lack of inertia gives us super lightspeed.
Although I do not agree with Barbour 's platonia, I
agree with the title of his book :"The End of Time : The
Next Revolution in Physics"
The pivotal role of inertia is not understood by the
relativist-priests, nor by QM-priests.
Read my post Apples=Whales?.
http://groups.google.com/groups?&hl=en&selm=3DFC4CBF.7000806%40nospam.xs4all.nl
It is way simpler than your Dirac-hydrogen solution.
This does not even get across : a rotating disc that
gains 20% inertial relativistic mass.
Steve Bell
I too would have to say the fact that an "experimental value" from over 50
tears ago agreeing with Bohr's prediction can not be a coincidence. But please
understand the ramifications. This means that ground state hydrogen does not
apparently outwardly manifest any relativistic effects. That's a hard thing to
accept, but up to the present, I have to accept it, right now, anyway. One of
the main reasons is I think I can explain why, or how, it happens.
Steve Bell
>
> >
> >In the application of these differential geometries to the orbit problem of "a
> >body coasting along a geodesic," outside of the event horizon distance, you
> >treat the central body as an actual body with some definite spherical
> >diameter. To quote from Lawden's "An Introduction to Tensor Calculus, etc," in
> >his section on the Schwarzschild field equation solution, "We shall consider
> >the special case when the whole of space is devoid of matter, apart from a
> >spherical body with its centre at the center of the symmetry O," where "O" is
> >the origin (0,0,0). This doesn't change the field "outside" of the body, which
> >is as if all of its matter were concentrated at the origin (0,0,0).
>
> We aren't discussing the schwarzchild metric. It's spherically symmetric,
> precisely because it has no angular momentum.
>
The symmetry of the exterior field in a Kerr differential geometry
representation, either is or is not spherical depending on whether the real,
finite sized, ball of matter emitting the field, is rotating or not rotating.
If the spherically-shaped (ball-shaped) central body is not rotating, so that
its angular momentum is zero, the Kerr field resorts to the special case
spherically-symmetric Schwarzschild field. If the totally spherically-shaped
central body is rotating, the field now is generally Kerr, and is aspherical.
But that's *the field,* not the shape of the central body, which even if it is
rotating, still maintains its theoretically perfect spherical shape.
>
> Despite disagreeing with your description of the rotational angular
> momentum for a kerr black hole, I'll just go along with that part of
> it, since it really doesn't matter as far as this goes. The rotational
> angular momentum is not a spin.
If we are talking about how the field surrounding a body is "vacuum" with all
of the matter concentrated into some spherical volume whose center is at
(0,0,0), this is "physical nonsense" when it comes to the physical rotation of
the central body generating a Kerr spacetime. The "frame dragging" effects
are a *direct* consequence that the central body is a "localized, spherically
shaped" body of matter. And I am not here saying anything about black holes. I
certainly do not think the proton is a iddy-biddy black hole.
> I've said this at least 3 times. Before introducing yet another
> irrelevant analogy, deal with the fact that you need a 4\pi rotation
> not at a 2\pi rotation (or 4 reflections, not 2) to return a spinor
> to itself.
>
Who cares about these irrelevant "QM" issues? The fact is that the electron
*is* a spinning ball of matter, and so is the proton itself (to an
approximation), that QM has no clue about. And that includes QCD with its
"nonsensical no free quark" can ever be *directly* observed bullroar.
I really detest the news "headlines" that have stated that some "quark" has
been experimentally observed." As far as I understand it, there is no such a
thing as a "free quark" and never will be.
> >> >The Kerr field manifests "frame dragging" effects, and imo, this
> >> >is the true cause of magnetism itself.
> >>
> >> Your o is wrong. Deformations are due to mechanical angular momentum,
> >> not spin.
>
> >What "deformations" are you referring to? The "deformation" in the
> >shape of the proton itself, or the "deformation" of the field surrounding
> >the proton?
>
> The proton has no deformations as may be seen by its lack of a dipole
> or quadrupole moment or higher order multipolarity. The kerr metric
> does have a symmetry axis as is evident from the metric, which incorporates
> it explicitly as the parameter J/m, where J is the angular momentum.
> In boyer-lindquist coordinates, the radius is r is parameterized by:
>
> \rho^2 = r^2 + (a cos(\theta))^2
>
> K = r^2 - 2Mr + a^2
>
> The second term on the rhs of the first expression is just a deformation
> parameter for a deformation along the z-axis.
>
Why don't you just say you were talking about the deformation of the field,
not the proton itself, directly, like I asked in my question?
> But, you have more difficultoes than this. Calculate the radius of the
> event horizon for a proton, not to mention the fact that quarks are
> readily observed within the proton from deep inelastic scattering,
If you think I am viewing the "Schwarzschild radius" of the proton as given by
its rest mass as the "size" of the proton, you don't understand what I mean by
"modeling the electromagnetic field surrounding the proton" as a Kerr field at
all. Bodies other than black holes emit Kerr fields.
> which would be precluded if the proton were a blackhole. Also, the
> proton is charged, so what you _really_ want is the metric for a
> kerr-newman blackhole.
First of all, I'm not talking about black holes at all.
I can see that it will be difficult for you accept my ideas. There really is
no need in my going further with it, with you, unless you accept that ground
state hydrogen does not outwardly manifest any relativistic effects.
Ken S. Tucker
>Ken S. Tucker said some stuff about
>Re: More QM Nonsense, II to usenet:
> >ro...@radioactivex.lebesque-al.net (Bilge) wrote in message
> >news:<slrnb1726...@radioactivex.lebesque-al.net>...
> >>Thinh Tran said some stuff about
>>>>Now just think of the an
> >> >arbirarily shaped 3-d object spinning wildly in all sort of angle in
> >> >space. You get 3-d spinning.
> >> That does not work.
> >I think Tran could have explained his point better, ie,
> >"spinning wildly in all sort of angle", is a bit fuzzy.
> > Suppose you're flying straight and level due North. Then
> >do a 360 degree (2 pi) yaw around the vertical AND a 180
> >degree (pi) roll. Now you flying due North but upside down.
> >Note the relative spin rate difference, between yaw and roll,
> >roll spin = (1/2) yaw spin.
>
> Spin is not a 3-d rotation. How many reflections in a mirror are
>required for any instantaneous position to return to itself? Since the
>qusetion is rhetorical, I'll answer it for you. Two. What happens if
>reflect a spinor twice? You get the negative of the original spinor.
>In addition, we are not referring to rotations about two axes. We're
>talking about the precession about an axis defined by a magnetic field.
>I'm tired of repeating this, so pay attention and don't bother with
>another analogy unless you address this first.
Tran defined the "3-d object" above, and introduced the analogy,
I merely understood the spin of a "3-d object" and explained it
to you, and he acknowledged an improvement in another post.
> > Now do this maneuver again, and you return to your original
> >attitude. Two rotations (4 pi) about the vertical returns the
> >aircraft to it's original state.
> In which case your spinor would require twice that, since you've
>composed your rotation from two rotations. Read something about the
>algebra SU(2) and its relationship to rotations. SU(2) is called
>a "double cover", or universal covering group. It's double valued.
> >This *definition* of spin in words makes sense to me, and
> >I think this was Tran's argument. Does that sound reasonable
> >to you?
>
> No, not in the least, precisely because you are hung up on the semantics
>rather than the physics. The reason that it makes sense to you is because
>you think all of this is "ad hoc", so you've invented your own "ad hoc"
>way of looking at it, which isn't correct. The same goes for thinh tran's
>assertion about "point particles". The proton is not a point particle.
>It's rms charge radius is just under 1 fm. It has a spin of 1/2 and
>quantum mechanics deals with it just fine, as is evident from the vast
>amount known about nuclei, a substantial amount of which may be deduced
>using straight-forward, garden variety, quantum mechanics.
Tran and Bell can argue their own theories, as you do, I'll
follow the posts, and try to be constructive on all accounts.
An improvement that replaces wave mechanical methodology,
is possible.
Regards
Ken S. Tucker
PS: Bilge, you were away for awhile, you were missed.
Thinh Tran
> Thinh Tran said some stuff about
> Re: More QM Nonsense, II to usenet:
> >[Thinh's comment] Let's me see if I can put this in a geometrical
> >argument.
> > In 2-D spin (two dimensional spinning) there is only 1
> >distinguishable spinning path, I represent this with a circle.
> > In 3-D spin there are two independent spinning paths (i.e., two
> >orthogonal directions). I represent this situation with 2 identical
> >circles sharing the same center and perpendicular to each other. The
> >net effect of 3-D spin is that when an object complete 2pi you end up
> >on the other side of a circle (the particle turn to minus itself).
> >With 4pi you return to where you start. I hope this shows clearly why
> >4pi corresponds to all 3-D QM objects such as electron, proton,
> >neutron.
> > The advantage of this physical picture is that, considering that
> >there are 3 spatial directions, the following naturally arise from
> >geometrical considerations.
> > spin 1/2<=> 3-D physical objects
> > spin 1<=> 2-D physical objects
> > spin 2<=> 1-D physical objects
> > spin x: Partial entities (i.e. other objects)
> > If anyone sees a conceptual problem with this picture, please
> >advise.
> >
>
> The disadvantage is that it's wrong. All of those are inherently
> 4-dimensional objects.
[Reply]: IMHO this is only a technicality. Of course everything is
mathematically 4-D. The question is, what is the effective total
dimension count. For example, a geometrical point that never moves
anywhere would be effectively 0-D.
I'm focusing on spatial geometry, therefore I reduce the problem
to 3-D max; and the effective total dimension count is 3-D or less.
>A spin 0 is a lorentz scalar (or pseudoscalar).
> It has no polarization axis. A spin 1 is a 4-vector (or pseudovector).
> It has 4-polarization states, unless it's massless, in which case the
> polarization must be transverse and there are only 2. Thats why it's
> called a vector (or pseudovector) particle. A spin 2 has 5 polarization
> states, unless it's massless in which case it must have only transverse
> polarizations, in which case there are only 2. That's why it's associated
> with a rank 2 tensor, for example, the graviton. For massless particles,
> the time and longitudinal components vanish, which is why _propagating_
> photons (and hypothetically, propagating gravitons) have only transverse
> polarizations. A spin 1/2 (and 3/2, etc), is more complex, since there
> exist both dirac and majorana spinors as possibilities. Electrons are
> dirac particles. It's not clear whether neutrinos are majorana or dirac
> particles. That's what experimental searches for neutrinoless double beta
> decay are all about. Rather than explain it, I suggest using google to
> find out more.
[Reply] Tensors etc. are only mathematical tools used to generalize a
description so that it meet as many external conditions as possible.
For example in fluid dynamics, the viscosity is formally a 4-order
tensor for mathematical convenience.
> Further difficulties you have include: (1) in _no_ case, can you
> specify more than one spatial axis along which to quantize the spin,
> (2) particles have intrinsic parities. Please explain the difference
> in spinning marbles which have positive and negative parities. I would
> especially like to see a marble which is not rotating explained as
> a pseudoscalar (an example from particle physics is the pion).
[Reply] Keep in mind that I'm not talking about marbles because
marbles appear are effectively two dimensional objects from a
geometric view point. We are talking about objects that are not of
perfect geometrical shape if they happen to be 3-D.
There are several natural ways to handle the parity question you
raised.
Just focusing on spatial dimensions, the simplest is that:
1. 3-D objects exist in right-hand and left-hand forms. They are
distinct because there is no rotational transformation that can make
the two identical.
2. 2-D objects also exist in righ-hand and left-hand forms.
However, since space is 3-D, these two forms can be made identical by
rotation.
Also, I would like to comment on your reply to Ken earlier. You used
the analogy of mirror reflection, i.e.
> 0 <
> 0 <
* 0 *
where the vertical triple zeroes represent the plane of symmetry.
This is not a valid analogy of the model that I proposed. So the
point is not just semantics.
In this model we are dealing with a spherical environment (hence
things appear as if they're spinning). One way to visualize this is to
shrink the mirror down to a single point and look at the left and
right images as 3-D objects approximating two circular wedges on the
opposite sides of a sphere. If there exists spherical symmetry, the
situaion would be:
> *
> 0 <
* <
I will not repeat the detail, but in a spherical environment, in
order to get from the left side situation to the right side situation
a 2pi rotational transformation is needed (note: 1pi won't do for this
case.)
But this picture is still not complete. The only way for a 3-D
object to reestablish symmetry is to get back where it came from, and
that is because there exists no real spherical symmetry in the quantum
realm (this is my main objection against carrying the point assumption
down to QM from classical physics, giving the false impression of
spherical symmetry.)
For example, in a Hydrogen atom, the center is occupied by the
proton, which is itself asymmetrical. So the more correct picture is
this:
> A *
> 0 <
* <
It should be clear now that the original orientation (left wedge)
and the orientation after 2pi rotation (right wedge) are not
symmetrical. They are anti-symmetrical because by inverting the
situation at the center, for example, all relationships are reversed.
Thus, it requires 4pi for a 3-D object to complete what I would call a
"symmetry rotation".
Last, I want to make a note on the difference between s and l. You can
visualize s as either spin or a symmetry operation to cover spherical
asymmetry (the later viewpoint shoul be taken instead of the first,
but I will not elaborate on it here.) But in either case, "spin" is
independent of R (the distance from the center, or the proton in the
case of H). On the other hand, l is a measure of R (or more correctly
l+1 is a measure of R). The two effects are therefore only
approximately equivalent. The practice of adding l and s together in
wave mechanics to form j is therefore only a first order approximation
that should suffer significant error for small l's. This error is a
contribution to the Lamb shift, though not all of it.
Thinh Tran (http://www.thinhtran.com)
Bilge
First, the difference between a propagating and a vitual photon resides
in the relationship between the polarizations along the time and long-
itudinal directions. Because the helicity of a massless particle is a
lorentz scalar, its value must be either +/-1 and the longitudinal
polarization must be identically zero for _all_ observers. A longitudinal
polarization is a mass. This, in turn forces the time polarization to be
identically zero, because the photon is not charged, i.e., charge is
locally conserved. But this is not true for virtual photons, or else there
would be no coulomb force. Virtual photons in qed carry the coulomb force.
The time component is the instantaneous coulomb potential. It must be
exactly canceled by the longitudinal component of the polarization, and
therefore _all_ four polarization states are required. I suggest a
book on qed. A very good introduction which emphasizes the physical
aspects over the formalism and is written to address the point of view
of experimentalists is "Quarks and Leptons", Halzen and Martin. It's
accessible at the upper-level undergrad/ beginning graduate physics
student, and only presumes an understanding of basic quantum mechanics
and special relativity.
[...]
>> particles. That's what experimental searches for neutrinoless double beta
>> decay are all about. Rather than explain it, I suggest using google to
>> find out more.
>[Reply] Tensors etc. are only mathematical tools used to generalize a
>description so that it meet as many external conditions as possible.
That is also incorrect. Tensors are mathematical statements about
symmetries in nature. That is why tensors are _not_ simply matrices.
(for example, the christoffel symbols which carry three indicies
could be represented as a 3-d matrix but is not a tensor because it
does not transform as a tensor). A matrix can be used to hold all
sorts of information to "meet as many external conditions as possible",
but a matrix need not have any symmetry.
>For example in fluid dynamics, the viscosity is formally a 4-order
>tensor for mathematical convenience.
Right off hand, I can't say if the viscosity is a true tensor or
not, but I'll check.
>> Further difficulties you have include: (1) in _no_ case, can you
>> specify more than one spatial axis along which to quantize the spin,
>> (2) particles have intrinsic parities. Please explain the difference
>> in spinning marbles which have positive and negative parities. I would
>> especially like to see a marble which is not rotating explained as
>> a pseudoscalar (an example from particle physics is the pion).
>
>[Reply] Keep in mind that I'm not talking about marbles because
>marbles appear are effectively two dimensional objects from a
>geometric view point. We are talking about objects that are not of
>perfect geometrical shape if they happen to be 3-D.
No, the reason that a marble has no intrinsic parity is because
it is _not_ perfectly spherical.
>There are several natural ways to handle the parity question you
>raised. Just focusing on spatial dimensions, the simplest is that:
> 1. 3-D objects exist in right-hand and left-hand forms. They are
>distinct because there is no rotational transformation that can make
>the two identical.
You can only use 3-d geometry to apply that to macroscopic objects which
have no intrinsic parity. Intrinsic parity is defined by the way an object
transforms under a reflection, i.e., coordinate inversion plus a rotation
by \pi.
> 2. 2-D objects also exist in righ-hand and left-hand forms.
>However, since space is 3-D, these two forms can be made identical by
>rotation.
That is incorrect. Right and left-handed are defined as reflections
of each other.
>
>Also, I would like to comment on your reply to Ken earlier. You used
>the analogy of mirror reflection, i.e.
> > 0 <
> > 0 <
> * 0 *
>where the vertical triple zeroes represent the plane of symmetry.
> This is not a valid analogy of the model that I proposed. So the
>point is not just semantics.
> In this model we are dealing with a spherical environment (hence
>things appear as if they're spinning). One way to visualize this is to
>shrink the mirror down to a single point and look at the left and
>right images as 3-D objects approximating two circular wedges on the
>opposite sides of a sphere. If there exists spherical symmetry, the
>situaion would be:
> > *
> > 0 <
> * <
> I will not repeat the detail, but in a spherical environment, in
>order to get from the left side situation to the right side situation
>a 2pi rotational transformation is needed (note: 1pi won't do for this
>case.)
A reflection is a coordinate inversion followed by a rotation by \pi.
You cannot shrink your object to a point, since that leaves everything
completely undetermined. You _must_ change r -> -r and rotate the
result. No continuous transformation will produce a parity transform.
> But this picture is still not complete. The only way for a 3-D
>object to reestablish symmetry is to get back where it came from, and
>that is because there exists no real spherical symmetry in the quantum
>realm (this is my main objection against carrying the point assumption
>down to QM from classical physics, giving the false impression of
>spherical symmetry.)
I have no intention of addressing anything beyond where you keep
insisting that _your_ point assumption is an assumption made by
quantum mechanics. I've repeated this enough times in responses
that you couldn't possibly have missed it and gave the proton an
example. That to which you are objecting is your own invention and
all I would be doing is choosing which of your fallacies to defend
against the other.
Thinh Tran
> You are perhaps not the best skilled individual to ask this of, so I'll ask it
> generally, do you believe there will be another "revolution" in physics? By
> "revolution" I mean something like the Einstein revolution, or the Quantum
> revolution initiated by Max Planck. Something truthfully revolutionary about
> how we view motion and probability in the realm of the atom? But if you say
> that the "revolution" will come about by the path of present-day accepted QM
> itself (which includes string theory, imo), I reject that. Something else is
> needed to resolve the impasse of deterministic GR and probabilistic QM.
[COMMENT]: I believe the next revolution in physics is right around
the corner and physics (including GR, QM) will transform completely
within 20 years. It's time for everyone to look back and reflect on
why QM must change. It has been too ridiculous for too long!!! It's
time to get back to common sense!
Those who are not ready for change will become prisoners of the
past. The good news is that they may be able to keep their jobs at
various institutions. (I personally hope that this next revolution
will not destroy lives, including the lives of those who lost the last
battle.)
Thinh Tran (http://www.thinhtran.com)
Bilge
Re: More QM Nonsense, II to usenet:
>Bilge wrote:
>> symmetric, precisely because it has no angular momentum.
>>
>
>
>
>The symmetry of the exterior field in a Kerr differential geometry
>representation, either is or is not spherical depending on whether the real,
>finite sized, ball of matter emitting the field, is rotating or not rotating.
>If the spherically-shaped (ball-shaped) central body is not rotating, so that
>its angular momentum is zero, the Kerr field resorts to the special case
>spherically-symmetric Schwarzschild field. If the totally spherically-shaped
>central body is rotating, the field now is generally Kerr, and is aspherical.
>But that's *the field,* not the shape of the central body, which even if it
shape. I specifically did not argue your point as it was irrelevant and
pointed this out. Since you cannot perform _any_ measurement on a kerr
black hole which can demonstrate the existence of a spherical "central
body" there is no physical central body which is spherical.
(If you don't want your lines snipped, wrap them at a reasonable length.
I'm not going to reformat your paragraphs any longer to conform to an
80 character line length after the insertion of a single indent level.
I'll just snip the entire line, and denote it as snipped, since that
is the least effort. If you find that to be a problem, go read the
rfc's regarding the NNRP and NNTP protocols).
[...]
>
>If we are talking about how the field surrounding a body is "vacuum" with all
>of the matter concentrated into some spherical volume whose center is at
>(0,0,0), this is "physical nonsense" when it comes to the physical rotation
>the central body generating a Kerr spacetime. The "frame dragging" effects
>are a *direct* consequence that the central body is a "localized, spherically
>shaped" body of matter. And I am not here saying anything about black holes
>certainly do not think the proton is a iddy-biddy black hole.
>
>
>> I've said this at least 3 times. Before introducing yet another
>> irrelevant analogy, deal with the fact that you need a 4\pi rotation
>> not at a 2\pi rotation (or 4 reflections, not 2) to return a spinor
>> to itself.
>>
>
>
>Who cares about these irrelevant "QM" issues?
Experiments, for one thing. What makes a neutron different from an
anti-neutron?
[...]
>I really detest the news "headlines" that have stated that some "quark" has
>been experimentally observed." As far as I understand it, there is no such a
>thing as a "free quark" and never will be.
That doesn't mean you can't observe them or that nature cares how you
think the universe should look. Since the quarks c,b,t were all
_predictions_ of the standard model with _predictions_ for their decay
modes and and what the experiments would observe, their existence is as
well substantiated as that of any other particle, such as the electron,
which is defined by experimental measurements of its properties. Since
I doubt you have ever looked at the standard model from an experimental
perspective (or even a theoretical one), your dismissal of quarks has
no basis. If I am incorrect, then you should pointout to me just what
problems you find in deep inelastic scattering experiments.
[...]
>> In boyer-lindquist coordinates, the radius is r is parameterized by:
>>
>> \rho^2 = r^2 + (a cos(\theta))^2
>>
>> K = r^2 - 2Mr + a^2
>>
>> The second term on the rhs of the first expression is just a deformation
>> parameter for a deformation along the z-axis.
>>
>
>
>Why don't you just say you were talking about the deformation of the field,
>not the proton itself, directly, like I asked in my question?
Because your distinction between the two is at best naive, if not
downright silly.
>
>If you think I am viewing the "Schwarzschild radius" of the proton as
>its rest mass as the "size" of the proton, you don't understand what
>"modeling the electromagnetic field surrounding the proton" as a Kerr
>all. Bodies other than black holes emit Kerr fields.
(wrap your lines)
If you want to put electromagnetism in geometric terms, the formalism
already exists. Look up fibre bundles.
>> which would be precluded if the proton were a blackhole. Also, the
>> proton is charged, so what you _really_ want is the metric for a
>> kerr-newman blackhole.
>
>
>First of all, I'm not talking about black holes at all.
Then don't talk about them. Just write the appropriate equations
which apply to the proton, as you see it.
>
>I can see that it will be difficult for you accept my ideas. There really is
>no need in my going further with it, with you, unless you accept that ground
>state hydrogen does not outwardly manifest any relativistic effects.
The spin is a relativistic effect. The magnetic moment is a relativistic
effect. The existence of a B-field despite the non-existence of magnetic
monopoles, is a relativistic effect. [I don't intend to discuss magnetic
monopoles. There is no experimental evidence for them and unless you
plan read something about thier existence or lack thereof in even
simple semi-classical electrodynamics and duality transformations,
don't dredge them up.]
Bilge
>Tran defined the "3-d object" above, and introduced the analogy,
>I merely understood the spin of a "3-d object" and explained it
>to you, and he acknowledged an improvement in another post.
It is still wrong, regardless of how many times it gets explained.
[...]
>Tran and Bell can argue their own theories, as you do, I'll
>follow the posts, and try to be constructive on all accounts.
The difference being that what I'm describing is supported by
experimental data and was used to make the predictions that were
responsible for 20th century technology. Try explaining a fermi
surface as a consequence of your conceptualization of spin.
>An improvement that replaces wave mechanical methodology, is possible.
Why is it necessary? There is no data to suggest anything is wrong with
it. The basic problem is that you have a very classically biased view
of waves and particles and you keep trying to pidgen-hole a quantum
description into only one of those two categories in refusing to
consider the possibity thant neither ofthose categories is adequate
to describe quantum objects. "Waves" and "particles" are semantics,
not physics. Physicists use the terms for convenience, not because
they really believe the microscopic world needs them.
Bilge
Re: More QM Nonsense, II to usenet:
>
>[COMMENT]: I believe the next revolution in physics is right around
>the corner and physics (including GR, QM) will transform completely
>within 20 years. It's time for everyone to look back and reflect on
>why QM must change. It has been too ridiculous for too long!!! It's
>time to get back to common sense!
Please describe a fermi surface and its properties without using
quantum mechanics. If that's not possible, then give an alternative
description for a transistor at the microscopic level. With no fermi
surface, you get no electron-hole pairs, since the hole is a direct
consequence of the fermi surface, which in turn is a consequence of
half-integral spin.
> Those who are not ready for change will become prisoners of the
>past. The good news is that they may be able to keep their jobs at
>various institutions.
The bad news for your thesis here is that discovering some revolutionary
new physics would be a definite enhancement to academic physicists, not
a deteriment.
Bilge
Re: More QM Nonsense, II to usenet:
>orbit shmorbit
>i would like to remind you about two more QM nensense:
>
>1 qm never went further than the element Fe (or close t it
>now what is more interesting is the 'excuses' why it didnt:
As usual, you're wrong. Quantum mechanics is used all the
time to model nuclei of any Z,N.
>a 'there is no need'!
>if i were a pupil at the elementry school i would be a shamed
>to bring such an excuse to my teacher
You should be even more ashamed of what you post.
Y.Porat
anyway the main point is it cannot go much further away.
>
>
> > IOW it is a wrong asumption. that cannot go further
> > than a few light elements.
>
>
> I am perfectly aware of that.
so ? ....... people have to make conclusions
and look for better ways.
>
>
> >
> > so my advise: dont bet on the Bohr model
>
>
> In this case, for the lowest level I do,
> for many elaborate reasons, I explained in the thread.
because as i saied, an orbiting object has some common featres with
a cantilever osciating object
but in order to go further away and get to the 'useful world'
ie to make it useful in our everyday life, you have to come as close
as possible to the 'real thing' as you note i put it in ' '
because no one yet can claim he got 'the' real thing'
we can only try at this stage to come closer and closer.
as possible.
>
>
> > and not on the many existing paradigma for instance:
> > that in a heavy atom, the number of electrons
> > is exactly the number of protons.
>
>
> I have heared of ions, at age 16.
that the 'discovery' of many fold finding of ions
(say U92+) if at all !) is done by stripping *some neutrons*
(not only protons)
so dont count even on 'ion discovery' in heavy elements.
>
> > if is just a friendly advise for not waisting your
> > time money and energy.(too much of it has been
> already waisted.)
>
> I think debunking the weirdness of QM is a very valuable
> effort.
ok we found something in common (;-)
Y.porat
-------------
> Hayek.
Y.Porat
> Thinh Tran said some stuff about
> Re: More QM Nonsense, II to usenet:
>
> >
> >[COMMENT]: I believe the next revolution in physics is right around
> >the corner and physics (including GR, QM) will transform completely
> >within 20 years. It's time for everyone to look back and reflect on
> >why QM must change. It has been too ridiculous for too long!!! It's
> >time to get back to common sense!
>
> Please describe a fermi surface and its properties without using
> quantum mechanics. If that's not possible, then give an alternative
> description for a transistor at the microscopic level. With no fermi
> surface, you get no electron-hole pairs, since the hole is a direct
> consequence of the fermi surface, which in turn is a consequence of
> half-integral spin.
i am not entirely sure that the discovery of transistors
has to be credited to the 'academy'
i have some suspicion that it was discovered by the experimental
world .might well be *accidentaly* as most useful
discoveries in our technology (no real need to bring examples (;-)
---------
>
>
> > Those who are not ready for change will become prisoners of the
> >past. The good news is that they may be able to keep their jobs at
> >various institutions.
>
> The bad news for your thesis here is that discovering some revolutionary
> new physics would be a definite enhancement to academic physicists, not
> a deteriment.
the 'deteriment' as you put it is that it might put the 'academy'
in a rediculous position --say :
how is it that so long you didnt notice the too many nonsense
you are teaching!
and how is it you alowed spending so many human reasorses
for these obvious nonsense.
do you want a little example:
see the W boson that is emmited from a nucleid
and yet it is 80 times heavier than the nucleid!
or another nosese:
some particles in prevailing theory' can be in their 'mass shell'
and than 'without their mass shell'
that is not physics that is in simple words *cheating*
all the best
Y.Porat
------------
Bilge
Re: [yank] More QM Nonsense, II to usenet:
>ro...@radioactivex.lebesque-al.net (Bilge) wrote in
>message news:<slrnb1cam...@radioactivex.lebesque-al.net>...
>>
>> quantum mechanics. If that's not possible, then give an alternative
>> description for a transistor at the microscopic level. With no fermi
>> surface, you get no electron-hole pairs, since the hole is a direct
>> consequence of the fermi surface, which in turn is a consequence of
>> half-integral spin.
>----------------
>i am not entirely sure that the discovery of transistors
>has to be credited to the 'academy'
>i have some suspicion that it was discovered by the experimental
>world .might well be *accidentaly* as most useful
>discoveries in our technology (no real need to bring examples (;-)
Steve Bell
>
> Steve Bell said some stuff about
> Re: More QM Nonsense, II to usenet:
> >Bilge wrote:
>
> >> We aren't discussing the schwarzchild metric. It's spherically
> >> symmetric, precisely because it has no angular momentum.
> >>
> >
> >
> >
> >The symmetry of the exterior field in a Kerr differential geometry
> >representation, either is or is not spherical depending on whether the real,
> >finite sized, ball of matter emitting the field, is rotating or not rotating.
> >If the spherically-shaped (ball-shaped) central body is not rotating, so that
> >its angular momentum is zero, the Kerr field resorts to the special case
> >spherically-symmetric Schwarzschild field. If the totally spherically-shaped
> >central body is rotating, the field now is generally Kerr, and is aspherical.
> >But that's *the field,* not the shape of the central body, which even if it
> >rotating, still maintains its theoretically perfect spherical shape.
>
> I already told you that the "central body" is a singularity. It has no
> shape. I specifically did not argue your point as it was irrelevant and
> pointed this out. Since you cannot perform _any_ measurement on a kerr
> black hole which can demonstrate the existence of a spherical "central
> body" there is no physical central body which is spherical.
>
You appear not to understand much about applied orbit theory. The
Schwarzschild field, which was, yes, the basis GR mathematics for black holes,
with their mathematical singularities. None-the-less it was successfully
applied to the field the Sun emits to explain a part of the observed
precession of Mercury's orbit, and the Sun is certainly not a black hole. In
fact Gravity Probe-B (I think that's its name) will be launch around the Earth
(I think it's Earth, and not into a sun orbit) to see its Kerr frame dragging
effects, and the Earth is certainly not a black hole.
Once again, I am not referring to modeling the proton as a black hole. I'm
just saying it is possible to mathematically model the electromagnetic field
surrounding the proton using a Kerr differential geometry. The fundamental
difference between my mathematics and Newman's is that I allow *all* of the
acceleration on the geodesic electron to come from the field curvature itself,
with all of the "generally relativistic *electronic* effects" tagged along, as
opposed to how Newman incorporated the central charge, and the pitiful effect
on field curvature his technique produces. It is so small, that the
relativistic Coulomb potential has to be quoted as a subsidiary equation to
get the total acceleration on the electron correctly equationally represented.
> (If you don't want your lines snipped, wrap them at a reasonable length.
> I'm not going to reformat your paragraphs any longer to conform to an
> 80 character line length after the insertion of a single indent level.
> I'll just snip the entire line, and denote it as snipped, since that
> is the least effort. If you find that to be a problem, go read the
> rfc's regarding the NNRP and NNTP protocols).
>
I wasn't aware that my lines were coming across longer that 80 characters,
thanks for telling me. I am using Netscape Messenger and it says they are
wrapped at 78 before sending, so I don't know what else to do about it. I am
not going to Outlook Express.
Are other people seeing my lines > 80?
> [...]
> >
> >If we are talking about how the field surrounding a body is "vacuum" with all
> >of the matter concentrated into some spherical volume whose center is at
> >(0,0,0), this is "physical nonsense" when it comes to the physical rotation
> >the central body generating a Kerr spacetime. The "frame dragging" effects
> >are a *direct* consequence that the central body is a "localized, spherically
> >shaped" body of matter. And I am not here saying anything about black holes
> >certainly do not think the proton is a iddy-biddy black hole.
> >
> >
> >> I've said this at least 3 times. Before introducing yet another
> >> irrelevant analogy, deal with the fact that you need a 4\pi rotation
> >> not at a 2\pi rotation (or 4 reflections, not 2) to return a spinor
> >> to itself.
> >>
> >
Say it as many times as you like.
> >
> >Who cares about these irrelevant "QM" issues?
>
> Experiments, for one thing. What makes a neutron different from an
> anti-neutron?
>
Personally, I don't think QM knows. It definitely makes a statement, but I
don't think it's correct.
> [...]
> >I really detest the news "headlines" that have stated that some "quark" has
> >been experimentally observed." As far as I understand it, there is no such a
> >thing as a "free quark" and never will be.
>
> That doesn't mean you can't observe them or that nature cares how you
> think the universe should look. Since the quarks c,b,t were all
> _predictions_ of the standard model with _predictions_ for their decay
> modes and and what the experiments would observe, their existence is as
> well substantiated as that of any other particle, such as the electron,
> which is defined by experimental measurements of its properties. Since
> I doubt you have ever looked at the standard model from an experimental
> perspective (or even a theoretical one), your dismissal of quarks has
> no basis. If I am incorrect, then you should pointout to me just what
> problems you find in deep inelastic scattering experiments.
>
Quarks are "observational nonsense." QCD predicts they will never be directly
observed in isolation.
> [...]
>
> >> In boyer-lindquist coordinates, the radius is r is parameterized by:
> >>
> >> \rho^2 = r^2 + (a cos(\theta))^2
> >>
> >> K = r^2 - 2Mr + a^2
> >>
> >> The second term on the rhs of the first expression is just a deformation
> >> parameter for a deformation along the z-axis.
> >>
> >
> >
> >Why don't you just say you were talking about the deformation of the field,
> >not the proton itself, directly, like I asked in my question?
>
> Because your distinction between the two is at best naive, if not
> downright silly.
>
What's the point in saying this again, but you really do not understand
differential geometry orbit theory.
> >
> >If you think I am viewing the "Schwarzschild radius" of the proton as
> >its rest mass as the "size" of the proton, you don't understand what
> >"modeling the electromagnetic field surrounding the proton" as a Kerr
> >all. Bodies other than black holes emit Kerr fields.
>
> (wrap your lines)
>
> If you want to put electromagnetism in geometric terms, the formalism
> already exists. Look up fibre bundles.
>
> >> which would be precluded if the proton were a blackhole. Also, the
> >> proton is charged, so what you _really_ want is the metric for a
> >> kerr-newman blackhole.
> >
> >
> >First of all, I'm not talking about black holes at all.
>
> Then don't talk about them. Just write the appropriate equations
> which apply to the proton, as you see it.
>
I have worked up the details for modeling the proton's electronic field as a
type of Schwarzschild differential geometry at
http://www.mindspring.com/~sb635/pap4.htm
Please be advised that when I wrote this paper some time ago, I felt the
correct observed binding energy for ground state hydrogen was -13.605something
eV. Obviously, I've changed my mind about that. I hope you appreciate my
confusion given all of the different values you can find for this thing in a
gazillion references. The extension of Sommerfeld's only specially
relativistic model into the realm of a full fledged Schwarzschild differential
geometry predicts even *more* binding energy that does Sommerfeld's equation,
and I now think this is in the "wrong direction," if one leaves it at only the
Schwarzschild model. By allowing the full magnetic "frame dragging" effects of
a more generalized Kerr model, one can explain how the frame dragging effects
can "negate" the Schwarschild energy, via a Kerr perturbation, and put
hydrogen back on a "nonrelativistic" outward manifestation. I haven't worked
out all the details, but I know its possible. When I do, I'll edit my paper to
include it.
But why should I even continue? There are so many learned individuals (you are
even "learned" in your "standard-line" QM philosophy) who have told me I am
wasting my time, and I get very discouraged at times. But it is my time to
waste, isn't it?
> >
> >I can see that it will be difficult for you accept my ideas. There really is
> >no need in my going further with it, with you, unless you accept that ground
> >state hydrogen does not outwardly manifest any relativistic effects.
>
> The spin is a relativistic effect. The magnetic moment is a relativistic
> effect. The existence of a B-field despite the non-existence of magnetic
> monopoles, is a relativistic effect. [I don't intend to discuss magnetic
> monopoles. There is no experimental evidence for them and unless you
> plan read something about thier existence or lack thereof in even
> simple semi-classical electrodynamics and duality transformations,
> don't dredge them up.]
Please state what you think is the best, most accurate *experimentally
determined* ground state binding energy for hydrogen, in eV, please. And
please give the source of your number. You do think it is logically correct to
ask such a question, that is, to ask what this value is, don't you?
Y.Porat
> Y.Porat said some stuff about
> Re: More QM Nonsense, II to usenet:
>
> >orbit shmorbit
> >i would like to remind you about two more QM nensense:
> >
> >1 qm never went further than the element Fe (or close t it
> >now what is more interesting is the 'excuses' why it didnt:
>
> As usual, you're wrong. Quantum mechanics is used all the
> time to model nuclei of any Z,N.
may be they tried some suggetions but heavy nucei
in qm is far from a situation that can be called 'solved'
you are completely out about that issue
it is not my claim i met the information about that situation
from members that know the situation apparently
better than you.
they solved 'thoroughly' only untill iron and then came the excuses
and as i saied you are talking with a wonderful self confidence
about things you are completely 'out'
>
> >a 'there is no need'!
> >if i were a pupil at the elementry school i would be a shamed
> >to bring such an excuse to my teacher
>
> You should be even more ashamed of what you post.
talk physics man
and substantiate your claims
hand waving does not work on thinking people.
(while i say thinking people i dont mean myself
i am just a poor crackpot (;-)
you have to convince the thinking people not the parrots.
because the parrots buy anything you sell them.
Y.Porat
---------------
Hayek
Steve Bell wrote:
>
> I too would have to say the fact that an
> "experimental value" from over 50 tears ago
> agreeing with Bohr's prediction can not be a
> coincidence. But please understand the
> ramifications. This means that ground state
> hydrogen does not apparently outwardly manifest any
> relativistic effects. That's a hard thing to accept,
> but up to the present, I have to accept it, right
> now, anyway. One of the main reasons is I think I
> can explain why, or how, it happens.
So, let me shout like Fran Dresher in "The Nanny" :
*** CHER **** I mean "SHARE !".
You know my explanation.
I started checking the QM basics on hydrogen.
Bohr first postulate (for hydrogen) is nothing but my
statement : "the electron in hydrogen is in exactly one
Heisenberg cage". I do not believe it is the centrifugal
(inertial) force that creates the orbital. It is the
"Zitterbewegung" itself. Thats why it cannot go lower.
It is in the nucleus. Only uncertainty spreads it over
the orbital. And something that starts moving without
being subjected to a force is not subjected to inertia.
And there is another conclusion to be drawn. Something
that is not subjected to inertia does not emit em waves.
I suspected for a long time that inductance is a prime
suspect to suffer from inertia. And a good candidate to
connect the electrostatic forces to inertia, formerly
mistaken for gravitation.