7 Reasons to Abandon Quantum Mechanics-And embrace this New Theory

andrewgray · 07-09-2007

Quote:

Originally Posted by Erasmus00

But to simulate something with a computer, we first need a theory to simulate.

This really isn't as bad as it seems. For example, one might define a trial blinking function:

This function time averages to 1. The blips are sinusoidal, so the derivatives are simple.

And one might also define a trial accepting function:

This function also time averages to 1. The frequency of the blips for a linearly accelerated electron would be given by De Broglie's function:

$KE_e= \frac{1}{2}h\nu_e$

Then the electric force on the electron, instead of being F(t)=eE(t), would be

$F(t) = eE(t)Accept(t)$

And the electric field given off by the charge would be:

$E(t) = k \frac{e}{r^2} Blink(t)$

where the retarded time would have to be taken into account. These two definitions would time average to Coulomb's Law very closely. With these two trial functions, a simulation for electron interference around a positively charged filament could be run, for example.

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We need a function of omega that cuts off smoothly with high frequency Fourier components. We need a function of omega that cuts off smoothly with high frequency Fourier components.

These functions would fit the bill.

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One would think, from statistical mechanics, that these electrons would end up between the ground state and the energy kt.

Will, I not sure that this makes sense, even for Schrodinger mechanics.

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I worry that perhaps local reality is being rescued at too great a cost- we sacrifice a great deal of simplicity.

The ancient Greeks once theorized that Apollo pulled the Sun across the sky behind a royal chariot. This belief was actually "correct" in the sense that it perfectly matched experimental data to the accuracy available at the time. And it "satisfied" the curiosity of the masses by giving it an official explanation. But as human history progressed, it became apparent that this theory was not based on reality and that it was indeed mythology. It stopped satisfying the curiosity of the curious, and was abandoned.

Will, QM has stopped satisfying the curiosity of the curious. In my opinion, it is "non-reality" based mythology in the same sense. And it matches experimental data in the same sense. We must move on. If the cost is high, so be it. You are welcome to continue to believe the simple explanations.

Andrew A. Gray

andrewgray · 07-09-2007

Quote:

Are you saying the magnetic field, as defined by its curl being the spherically symmetric displacement current, will be identically zero? Apart from the fact that I'm not so sure at all, how does an identically zero field have a non-zero curl?

Qfwfq,

There seems to be some confusion about spherically symmetric currents and radiation. OK, take a macroscopic problem:

Put some charge on a spherical balloon and "inhale and exhale" in a fashion to make it pulsate spherically. Does it radiate?

Andrew A. Gray

Erasmus00 · 07-09-2007

Quote:

Originally Posted by andrewgray

where the retarded time would have to be taken into account. These two definitions would time average to Coulomb's Law very closely. With these two trial functions, a simulation for electron interference around a positively charged filament could be run, for example.

But to model how the electric field works, you need to put an equation governing the field in. In this case, you have equations governing an electric field, but it gives nothing with which to simulate the relevant magnetic effects. Trying to derive equations which preserve the "sense" of the maxwell equations and at the same time allow for blinking charges seems like it might be impossible, especially as non-conservation of momentum seems to keep cropping up.

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Will, I not sure that this makes sense, even for Schrodinger mechanics.

Are you familiar with thermodynamics/statistical mechanics? In a system at temperature T, the average thermal energy will be around kT. Hence, as you cool a system you expect all the electrons to spiral into the lowest possible state.

Also, we expect any quadratic degree of freedom to participate in the specific heat of an object. For metals, this isn't true. Most of the conduction electrons don't actually participate. The answer to both questions is the pauli exclusion principle. I fail to see how your theory can answer this.
-Will

Qfwfq · 07-10-2007

No confusion Andrew...

Quote:

Originally Posted by andrewgray

Put some charge on a spherical balloon and "inhale and exhale" in a fashion to make it pulsate spherically. Does it radiate?

Of course not, and the curl of B is zero. I was talking about what you said about a radially symmetric non-zero curl, which you seemingly said wouldn't be a problem. The only confusion here appears to be between the two things, charge moving in and out, or just magically appearing and disappearing, so we should try to be clear which we're talking about.

Erasmus00 · 07-13-2007

In a discussion with a colleague, another problem with this new theory arose- a blinking charge cannot explain neutron diffraction. Is there a ready answer to this?
-Will

andrewgray · 07-16-2007

Quote:

Originally Posted by Erasmus

. . . you have equations governing an electric field, but it gives nothing with which to simulate the relevant magnetic effects.

Ah, but given the electric force from a stationary charge Q acting on a test charge q (which could be moving) is enough. Once that is established, along with the assumption that this force is independent of q's velocity, all electric and magnetic fields are defined. Following Jackson, if in the inertial frame K' the equation of motion is $\frac{d \vec p'}{dt'} = \vec F'$ , then the equation of motion in the inertial frame K, moving at velocity $\vec v$ is :

$\left ( \frac{d \vec p}{dt} \right ) _ {||} =\; \left ( \frac{d \vec p'}{dt'} \right ) _ {||} \; +\; \frac{\gamma_v}{c^2} \left [ \vec u \; \times \; \left ( \vec v \times \frac{d \vec p'}{dt'} \right ) \right ]_{||}$

and

$\left ( \frac{d \vec p}{dt} \right ) _ {\perp} =\; \gamma_v \left ( \frac{d \vec p'}{dt'} \right ) _ {\perp} \; +\; \frac{\gamma_v}{c^2} \left [ \vec u \; \times \; \left ( \vec v \times \frac{d \vec p'}{dt'} \right ) \right ]_{\perp}$

And all electric and magnetic forces follow. A computer would have no trouble with these.

Quote:

Originally Posted by Erasmus

. . . non-conservation of momentum seems to keep cropping up.

We have tunneling right? This allows a non-conservation of energy at the strictly microscopic level, right? Well, why then is it so surprising that there is no strict conservation of momentum during microscopic tunneling?

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. . . without the exclusion principle I don't think we can explain the specific heats of metals.

In a system at temperature T, the average thermal energy will be around kT. Hence, as you cool a system you expect all the electrons to spiral into the lowest possible state.

Also, we expect any quadratic degree of freedom to participate in the specific heat of an object. For metals, this isn't true. Most of the conduction electrons don't actually participate. The answer to both questions is the pauli exclusion principle. I fail to see how your theory can answer this.

You are correct. There is no Pauli exclusion principle in this New Theory. However, Boltzmann's constant k comes from ideal gases:

$\bar K = \bar {\frac{1}{2} mv^2} = \frac {3}{2}kT$

So k is defined as the average kinetic energy of an ideal gas molecule per degree Kelvin per degree of freedom. Multiplying by a mole N_o gives

$\bar E = N_o \frac{3}{2}kT = \frac {3}{2}RT$

So the specific heat of an ideal gas at constant volume becomes

$c_V \; = \; \left( \frac{ \partial \bar E }{\partial T} \right )_V \;=\; \frac {3}{2}R$

OK, fine. But for solids, my statistical mechanics text gives that the "quantum" specific heat for solids is:

$c_V \;= \; 3R \left ( \frac {\theta_E }{T} \right ) ^2 e^{-\theta_E/T}$

where θ_E is an arbitrary constant chosen to match experimental data.

So I have admitted that QM does a good job at experiment matching. I am not too concerned about QM's ability to experiment match. Again, this does not satisfy the curiosity of the curious.

The text goes on to say that "Experimentally, the specific heat approaches zero more slowly than this, indeed $c_V \propto T^3$ as $T \right 0$ . The reason for this descrepancy is . . . " (some hand waving follows).

I would expect that any application of Boltzmann's constant to electron energy levels in metals would also include some "quantum fudge factor" needed to match experimental data. This is QM's modus operandi.

For metals, my Statistical Mechanics text gives that

$c_V = \gamma T + AT^2$

where $\gamma$ and $A$ are arbitrary constants chosen to match experimental data. Se la vi.

Quote:

Originally Posted by Qfwfq

. . . and the curl of B is zero [for radial fields]. I was talking about what you said about a radially symmetric non-zero curl, which you seemingly said wouldn't be a problem. The only confusion here appears to be between the two things, charge moving in and out, or just magically appearing and disappearing, so we should try to be clear which we're talking about.

Ok, now that we have that radially pulsating charge does not radiate at the macroscopic level, it is easier to understand the microscopic level of this New Theory.

The two appropriate Maxwell equations are:

$\nabla \times \vec E - \frac{1}{c} \frac {\partial \vec B}{\partial t} \;=\; 0$

$\nabla \times \vec B - \frac{1}{c} \frac {\partial \vec E}{\partial t} \;=\; \frac{4 \pi}{c} \vec J$

If we heuristically apply these equations to the New Theory's blinking charge, then the first equation yields that since the E field is radial, then the curl of any radial function is 0, therefore there are no changing magnetic fields (actually no magnetic fields at all).

Now the second equation. For our blinking charge, there are no magnetic fields. Thus, the first term in the equation is 0. So we are left with:

$- \frac{1}{c} \frac {\partial \vec E}{\partial t} \;=\; \frac{4 \pi}{c} \vec J$

And we see that there are displacement currents. Charge, as defined by Maxwell, is where electric field lines begin (positive) and end (negative). Thus, when this New blinking charge blinks, there are areas of space that have short E field segments. For a positron, the head of this blinking E segment has negative "displacement charge" and the tail has positive "displacement charge". This yields a net "displacement charge" of zero moving outward, and we have no net charge escaping to infinity.

Qfwfq, does this answer the question?

Andrew A. Gray

andrewgray · 07-16-2007

Quote:

Originally Posted by Erasmus

In a discussion with a colleague, another problem with this new theory arose- a blinking charge cannot explain neutron diffraction. Is there a ready answer to this?

According to this New Theory, a neutron is a bound proton and electron, and an anti-neutron is a bound anti-proton and positron (and not a combination of quarks.) Perhaps for a neutron, an electron would be in some circular supersmall orbit around the proton, like some kind of femto-hydrogen.

However, with this model, it would be difficult to imagine how beams of neutrons could make their way through two slits or around a filament and interfere on their way to a detector.

I almost abandoned this New Theory because of this. But before I did, I had studied electron diffraction patterns through crystalline solids in detail. Here is that analysis:

Thin Foil Electron Diffraction Rewritten

This analysis shows that particle diffraction through crystals is probably due to "prefered paths" through the crystalline lattice and not due to "diffraction". The charged electrons would blink in such a way that they were "OFF" while they crossed the atomic planes full of nuclei. Thus, "one-particle-at-a-time" experiments could be imagined in this case, as other particles were not needed to "interfere" in any way. So the possibility of neutron diffractions became possible in this New Theory.

We imagined the neutron as being a proton and an electron. The proton's trajectory would basically decide how the neutron went through the crystal. If the proton's pulsation length were on the order of the crystalline spacing or less, one could obtain neutron diffraction patterns as the proton blinked "ON" and "OFF" while passing atomic planes. The electrons pulsation frequency would probably be so large that it would not come into play and electron-nuclei forces would time-average to an continuous attractive force, not playing into the diffraction. In other words, the proton's pulsations would be correlated to atomic planar crossings, but the electron's would not. This could leave an average neutral force, like in the case when the proton's pulsations correlated to the crossings while leaving no net force. The neutron's trajectory would like this path, and it would have no deflection. However, when the protons pulsation correlated in such a way that the net force was large, the neutron would not be "neutral", and it would not like this path. The preferred paths would correspond to maxima on the detector.

In addition, it is hard to imagine how neutrons could be very monochromatic, so we imagined a "sloppy" diffraction pattern because of this.

So we went and researched "neutron diffraction". And a"sloppy" diffraction patter is what we found:

Notice that they say that the fringe "visibility" is bad (sloppy). The minima have events in them, and the 1st order minimum is of the same intensity than the 4th order maximum.

As for calling this experiment a "double-slit" experiment, that was extremely misleading. This experiment was done using neutron trajectories through a crystalline lattice, and not through two slits.

So "one-at-a-time" neutron preferred paths through a crystalline lattice is a much better way to imagine how these patterns were formed. That is, it is advantageous to use this New Theory that "is based on reality".

Andrew A. Gray

Qfwfq · 07-17-2007

First of all:

Quote:

Originally Posted by andrewgray

Qfwfq, does this answer the question?

Thanks, it was enough to say, "OK, we're talking about a conserved charge density that flows in and out, problem does not arise..." and never more work on the charge that appears and disappears (which would not give vanishing B). No need to confuse wanting to draw a clearer distinction with ignorance.

Quote:

Originally Posted by andrewgray

We have tunneling right? This allows a non-conservation of energy at the strictly microscopic level, right?

No, not right, wrong. Where is the non-conservation? Even if the particle is observed "caught in the act" this requires an interaction such that Heisenberg's $\Delta E$ allows things to match up. RQFT is choc full of Dirac $\delta$ s for the energy-momentum requirement. In my early QM days I thought of an idea, alternative to Born-Copenhagen, which I later learned Schrödinger had entertained but then abandoned and indeed doesn't match up with many things; it would involve conservation of particle number, energy-momentum and all that follows being statistical. It is considered quite discredited. Show me what else in quantum formalism might imply what you say.

Quote:

Originally Posted by andrewgray

I would expect that any application of Boltzmann's constant to electron energy levels in metals would also include some "quantum fudge factor" needed to match experimental data. This is QM's modus operandi.

I've never learnt about some "quantum fudge factor" being QM's modus operandi. I learned the quantum formalism based on Hilbert spaces and various operators, typically self-adjoint and/or unitary, for continuous spectra (which many a quantum buff doesn't even know about) you need to add distributions and have a space that isn't self dual, much more complicated, the symmetry groups and Lie algebras, but I never heard that some "quantum fudge factor" is necessary. The only fudgy thing is renormalization which a lot of folks would gladly find a better alternative to. Why add more fudge? C'est la vie, true, but efforts are usually toward improvement.

Quote:

Originally Posted by andrewgray

According to this New Theory, a neutron is a bound proton and electron, and an anti-neutron is a bound anti-proton and positron (and not a combination of quarks.)

So weak interactions and QCD are BS too? Gosh I knew some of the people on both the Delphi and the SLAC groups and I won't believe they were lying about the top quark events that were recorded. There's a lot of stuff about hadrons and nuclei besides what Will mentions, starting from hadronic diffraction in general, DIS, not to mention low energy nuclear data. Should we perhaps discard the whole of RQFT and re-explain all those hadrons, Yukawa potential, even Regge poles and Pomeron?

Anyway I still don't see how the photoelectric effect is a problem for quantum physics to explain, even after looking through that article. So, what about polarization dependence? I remember a basic exercise in my RQFT course, which explains it fine for the Compton effect. Where does that article show it can't be explained? Saying "largely remains to be understood" isn't quite the same as what you claim, it must be more to do with the complexity.

Erasmus00 · 07-17-2007

Quote:

Originally Posted by andrewgray

You are correct. There is no Pauli exclusion principle in this New Theory. However, Boltzmann's constant k comes from ideal gases:

$\bar K = \bar {\frac{1}{2} mv^2} = \frac {3}{2}kT$

So k is defined as the average kinetic energy of an ideal gas molecule per degree Kelvin per degree of freedom. Multiplying by a mole N_o gives

$\bar E = N_o \frac{3}{2}kT = \frac {3}{2}RT$

So the specific heat of an ideal gas at constant volume becomes

$c_V \; = \; \left( \frac{ \partial \bar E }{\partial T} \right )_V \;=\; \frac {3}{2}R$

First, one has to be careful to note that k really shouldn't be defined in this way. Its better to think of k as simply a unit conversion from Temperature to energy. The reason is that the 1/2 T per degree of freedom isn't generally true- rather its mostly true. The correct statement would be 1/2 T per QUADRATIC,CLASSICAL degree of freedom.

Now, consider a solid, each atom has 6 quadratic degrees of freedom (3 potential, 3 kinetic). However, each conduction electron ALSO has 3 degrees of freedom (to lowest order we consider them free). Hence, if we have one conduction electron per atom we would get

[quote] $c_V \; = \; \left( \frac{ \partial \bar E }{\partial T} \right )_V \;=\; 9R$

The observed measurement is much closer to 3R. Why? Quantum mechanics has the usually accepted answer: the electrons obey a fermi-dirac distribution and as such they can't occupy the same energy states. Hence, the electrons near the low energy of the distribution aren't free to move, and
don't really have available states. In your theory, everything should be classical and so we have the problem of why is cv only around 3R.

Your fudge factors are less quantum fudge factors and more material science fudge factors/solid state factors. In order to actually solve a real-world (not a toy model) specific heat you have to account for all sorts of interactions that make the problem much more difficult. For a solid, your vibrations aren't really quadratic, they are only quadratic to lowest order. Hence the theta fudge factor is really a classical factor.

For a metal, we have similar problematic interactions. However, if you have a good book, I'm sure it derives the specific heat for the free electron gas and shows that it matches the data quite well (though not perfectly, because the conduction electrons aren't really a free gas). Which stat-mech book are you using,btw?
-Will

andrewgray · 07-19-2007

Quote:

Originally Posted by Qfwfq

Anyway I still don't see how the photoelectric effect is a problem for quantum physics to explain, even after looking through that article. So, what about polarization dependence?

According to QM, a "photon" is a circularly polarized thing with a certain probability of going through a polarizer. Well, if this is true, then after a successful one goes through a polarizer and strikes metal, then there should be no preferred direction for photoelectron ejection. There are no "vertically polarized photons", according to QM.

But there is a preferred direction of ejection. This clearly is the result of polarized EM fields causing transverse ejection. So the real question really becomes, "just how does a polarizer really work, anyway?"

Consider a horizontal microwave polarizing grid. It is a series of vertical wires at a spacing less than a wavelength. It might look like this:

Microwaves with random polarization strike the grid with random polarizations. The EM fields of the microwaves cause electrons in the vertical wires to oscillate up and down, as this is the only direction that they can flow. Always, with driven oscillations and re-emissions, there is a phase shift between the driving EM field and the re-emitted EM field. Thus, the vertically oscillating electrons re-emit vertically polarized EM waves with a phase shift, canceling the original unpolarized wave in the vertical direction, and leaving only horizontal polarization. One can then actually detect the horizontal motion of electrons in a detector subjected to these horizontally polarized microwaves.

Microwave circularly polarized "photons" going through the microwave polarizer with some probability? This question exposes quantum folly.

And polarizers on the microscopic level work the same way. The structure of the polarizers is such that oscillations are preferred in just one direction, just like this microwave polarizer, but on a much smaller scale.

And once the polarizer actually cancels one direction of EM oscillation and leaves the other, then electron photoejection along the remaining direction happens much more readily. Reality-based physics.

Andrew A. Gray

To be continued.

7 Reasons to Abandon Quantum Mechanics-And embrace this New Theory

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