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

[andrewgray]

OK,

Thanks for all the info.

Will, I am not denying that the beryllium reactor is causing scintillation events in the detector. All I am saying is that I doubt that the events are caused by particles going through 450 meters of dirt.

8 Billion eV protons are smashed into beryllium. These protons have huge pulsation frequencies! If we use De Broglie's approximation for the proton frequencies we get that the proton pulsations are at about

1 x 10²⁴ Hz

This to be compared to gamma ray frequencies of about

1 x 10²² Hz

So the potential for Ultra-High frequency gamma rays is serious. I would actually believe that UH-gamma rays could pass through 450 meters of dirt and cause scintillations in the detectors. But I must admit, I remain very skeptical that particles can do this.

And since the official neutrino theory seems to be a mish-mash of conjectures that will probably have to change again in the not too distant future, I think that the truly logical mind would at least be open to the possibility that these events in the detector were caused by UH-gamma rays easily going through the dirt.

Now to some of your points.

Again, you misunderstand what a background is. It is not noise, its the standard electrons/muons from the beam that hit the detector.

What I was saying was that the "background" events are what you get when the beryllium reaction beam is turned off.

Is that not true? The only reason that I call it "noise" is from the tradition in electronics of calling the signal with everything turned off, "noise". We can call it just "background" if you like. OK with me.

But you seem to be implying that the "background" originates from the beam. (??) Are you saying that the "background" comes from when the beam is ON or when it is turned OFF? Perhaps you are confused a bit.

"Background", I believe, comes from decays from the earth around the detector and possibly from UH-gamma rays from the sun, even while the beam is turned off.

getting rid of neutrinos means getting rid of conservation of momentum, conservation of angular momentum, conservation of energy, etc.

Only during microscopic tunneling. The two choices, it seems to me, are 1) the neutrino hypothesis, or 2) non-conservation of E and P during tunneling. Usually, E and P would be conserved on the time average, and macroscopically, E and P conservation would still be a law.

Andrew A. Gray

[Erasmus00]

Originally Posted by andrewgray

And since the official neutrino theory seems to be a mish-mash of conjectures that will probably have to change again in the not too distant future

Neutrino theory hasn't changes since Salam, Glashow and Weinberg wrote down their theory of the weak interaction(with the exception of the possible addition of a majorana mass,which doesn't really effect the theory). It is far from a mish-mash of conjectures, it's quite elegant.

What I was saying was that the "background" events are what you get when the beryllium reaction beam is turned off.

I was under the impression that the majority of the background was from pions and K ons in the decay pipe for the experiment. Hence, turning off the beam would remove the background.

Also, in particle physics the term background usually refers to all the extra crap the beam produces that your detectors also pick up. This is distinct from noise in the detectors, which can be made arbitrarily small by averaging over many, many events.

Only during microscopic tunneling. The two choices, it seems to me, are 1) the neutrino hypothesis, or 2) non-conservation of E and P during tunneling. Usually, E and P would be conserved on the time average, and macroscopically, E and P conservation would still be a law.

But if we get rid of the neutrinos, every time there is a weak decay we lose energy, momentum, and angular momentum. Hence, the time average doesn't lead to conservation but in fact makes the problem worse.

Also, I fear we are pulling your thread off topic- this should be about discussing your theory, right? Perhaps we should start another thread on neutrinos?
-Will

[andrewgray]

OK, good enough. Thanks for all the neutrino information.

Getting back, there were just a few loose ends in our discussion of Thermal Radiation.

Will had made a comment about reflections, which made me realize that Max Planck, the father of QM, had a self-inconsistency in his original theory of thermal (blackbody) radiation.

I am quite certain that Planck's original theory was a "quantized extension" of the Rayleigh-Jeans theory of blackbody cavities.

I am certain that Rayleigh-Jeans counted standing wave modes in a cavity and integrated the average energy:



while Planck assumed discreet energy levels and simply changed the integral to a sum:



where the energy was no longer continuous but only where .

So Planck simply "quantized" Rayleigh-Jeans. Thus, Planck's theory was a "quantum extension" of Rayleigh-Jeans. Planck incorporated all the original Rayleigh-Jeans assumptions up until he took the sum instead of a continuous integral.

But that has philosophical trouble, it seems to me. The original Rayleigh-Jeans assumptions were that the blackbody cavity was perfectly black, (and as Will has pointed out), reflections were not considered possible. However, later the derivation relies on standing waves, which requires reflections.

This self-inconsistency in Planck's original derivation seems troubling. Planck saw the data, and knew what he was looking for. He took a silly, self-inconsistent Rayleigh-Jeans theory and "quantized" it to come out with something that could match the data. I see no way around this realization. I too, used to be in awe of Planck's early derivation, but no longer.

So let's look at the big picture (or the whole picture after the whole picture is painted.) Everything emits IR at room temperature, whether it is black, white, reflective, or absorptive. Then as things get hotter, everything emits visible red, then visible white. What is the explanation for this?

1) Is it because light in a cavity can only have certain modes?
2)Or is it because one is thermally agitating the outer electron orbits, causing them to radiate?

I say, take a step back and look at the whole picture!

Andrew A. Gray

[Erasmus00]

Originally Posted by andrewgray

But that has philosophical trouble, it seems to me. The original Rayleigh-Jeans assumptions were that the blackbody cavity was perfectly black, (and as Will has pointed out), reflections were not considered possible. However, later the derivation relies on standing waves, which requires reflections.

I have tried to explain several times that we have had a miscommunication. The Rayleigh-Jeans formula was done on a reflecting cavity with a small hole in it. This is because a cavity with a small hole in it simulates the thermal spectrum of a blackbody because any light that goes into the hole will bounce around many, many times before it makes it out. Because nothing is perfectly reflective, it pretty much has to get absorbed before it finds is way back out the hole. Hence, the spectrum of radiation coming out of the hole is exactly the same as that emitted by a perfectly black body. Rayleigh-Jeans and Planck calculated the spectrum of this cavity radiation.

In more modern derivations, one often imagines a perfectly black body inside the cavity. The derivation is the same because we imagine the black body to be in thermal equilibrium with the radiation (just as much is emitted as absorbed). If one likes, the cavity can even be removed via a limit at the end. This calculation is pedagogically nice because it makes it obvious that the cavity must have the exact same spectrum as a purely black object. i.e. it makes the connection that the Rayleigh-Jeans/Planck calculations apply not just to cavities, but any perfectly black emitter.
-Will

Edit:

What is the explanation for this?

1) Is it because light in a cavity can only have certain modes?
2)Or is it because one is thermally agitating the outer electron orbits, causing them to radiate?

I say, take a step back and look at the whole picture!

I don't think anyone will argue the cause of thermal radiation to be electron transitions. However, the whole power of statistical mechanics/thermodynamics is that it applies to EVERYTHING in equilibrium. The calculation is independent of the method of creating the radiation- all that matters is the thermal equilibrium.

[andrewgray]

Will,

I see what you are saying now. However, historically, this was not the case:



http://arxiv.org/PS_cache/physics/pd.../0402064v1.pdf

So now I see that Rayleigh-Jeans and Planck did have a self-inconsistent approach, but it is "cleanup-able". Hmmm.

OK, so the point is that this New Theory has the potential to explain why thermal radiation is as it is, with a UV catastrophe and all. Planck's radiation law depends on cavities, and it does nothing for my curiosity for why my steel pipe glows red then white when I heat it. In other words, I want an explanation that matches what is really going on.

I wonder if QM addresses the thermal light emissions of solids in any way other than Planck. (??)

Doesn't QM even assume the Universe is a "blackbody" in order to calculate the "temperature" of the background radiation?

Andrew A. Gray

[andrewgray]

The Experiments

Now we will cover some of the New Experiments that will prove this New Theory correct.

1) The New Electron Interference Experiment.

This new experiment will prove that interfering electron beams in an electron microscope are actually due to preferred paths to the detector while in flight and not because of "wave function" interference at the detector.

As you recall, this New Theory has described "electron interference" as a pulsating group of electrons that go around a positively charged filament. If the pulsations of the electrons are "OFF" when they cross, then they will continue on to the detector (a maximum). If the pulsatons of the electrons are "ON" when they cross, then the huge repulsion will knock them off their paths, and they will not make it to their original detector position (a minimum). This was shown in this animation:





According to this New Theory, the electrons "interfere" while in flight, and before they strike the detector. According to QM, the electron's wave function interferes right at the detector. We should be able to take advantage of this difference to prove that this New Theory is correct.

Suppose that we increase the filament voltage so that the two sides of the beam do not overlap on the film. If the voltage gets large enough, a "gap" will appear in between the two areas where each side lands. Like this:



Thus, according to this theory, since the beams still interfere on their way to the film, fringes still appear.

However, according to QM, the wave functions do not overlap at the film, hence no interference fringes will appear.

Only one of these theories can be right. We'll see.


Andrew A. Gray

[andrewgray]

2)The New Stern Gerlach Experiment

We have already discussed this experiment in brief here:

Post 17


but for completeness we discuss it in more detail. In review, the original Stern Gerlach
Experiment looks like this:



Silver atoms are put through a huge nonuniform magnetic field, and the magnetic moments cause separations. How does this work?

Well, consider a hydrogen atom in a nonuniform magnet field:




When the electron is at the position on the left, the magnetic force, which goes as



is directed mostly outward. However, there is a vertical component. The same is true when the electron is at the right position. This force will basically cause two things:

1) A net upward force.
2) A wild nutation/precession from magnet torque.

Next consider the atom in the same scenario, but the electron orbit is going in the opposite way:



This force will basically cause two things:

1) A net downward force.
2) A wild nutation/precession from magnet torque.

In the first case, the atom will be deflected upward, and in the second case the atom will be deflected downward.

In both cases we showed in post 17 that the angular momentum L will also nutate wildly in the huge magnetic field and that the z component of the angular momentum does too. This causes radiation friction and L_z is free to change. In short, the huge magnetic field first induces L into one of the UP or DOWN states. After the atom is quickly induced into one of these UP or DOWN states, the corresponding upward or downward deflections will seem like they are "quantized", but they are not.

We wish to come up with a way to prove this. The trick would be to keep the upward or downward forces, but to minimize the huge magnet torque that is responsible for changing the values of L_z.

It turns out the the upward/downward force on the atom is given approximately by:



where is the z-component of the atom's magnetic moment which is proportional to L_z.

However, the magnetic torque is dependent on :



So we wish to minimize and maximize .

To do this, one must use a magnetic quadrapole, instead of a magnet dipole. This way, the value for itself can be small, but its z-derivative can be large. Like this:



See Beam focusing

Notice that near the center, , yet the derivitives are not. So, if we use a setup like this:



Then the final setup looks like this:




Thus, the magnetic field itself will be small, but will still be substanial. This will allow the recovery of the continuous values of L_z, proving that the magnetic moments of the atoms were not quantized.


Andrew A. Gray

[Erasmus00]

In the first experiment proposed (electron diffraction off of a point at high voltage), quantum mechanics WILL predict interference, same as your theory. The reason is simple: remember in quantum mechanics an electron will explore all paths to the detector, and even one particle can interfere with itself. The two theories may predict different interference- to test this I will need quantitative predictions from your theory.

Now, the problem I see with your quantum hall set up is that you have gradients in both the z and x direction but no field in the center. The quantum mechanical effect comes from the fact that the moment can either be aligned or anti-aligned to the magnetic field, bug the dipole field is not uniform across the beam, so different atoms will align in different directions. In short- I'm reasonably certain that quantum mechanics will predict some spread in this experiment as well. I haven't completely convinced myself yet, but I'm pretty sure.
-Will

[andrewgray]

Originally Posted by Erasmus

In the first experiment proposed, quantum mechanics WILL predict interference, same as your theory. The reason is simple: remember in quantum mechanics an electron will explore all paths to the detector, and even one particle can interfere with itself. The two theories may predict different interference- to test this I will need quantitative predictions from your theory.

I think we are agreed on this one. QM will predict interference in the overlap zone in front of the film. Here is where Ψ will overlap and interfere.



However, on the film, there aren't any places for interfering paths to overlap. QM will predict no interference right on the film, once the filament voltage gets large enough to separate the two sides.

However, this New Theory predicts that the two sides actually interfere while in-flight, and that there still will be fringes on the film, approximately the same spacing as when the two areas still overlapped. And note that the interference is around a filament wire (and not around a point).

Originally Posted by Erasmus

Now, the problem I see with your quantum hall [Stern-Gerlach] set up is that you have gradients in both the z and x direction but no field in the center. The quantum mechanical effect comes from the fact that the moment can either be aligned or anti-aligned to the magnetic field, but the dipole field is not uniform across the beam, so different atoms will align in different directions. In short- I'm reasonably certain that quantum mechanics will predict some spread in this experiment as well. I haven't completely convinced myself yet, but I'm pretty sure.

Will,

I have edited the quadrupole picture slightly to clear up the field direction along the vertical. In the beam path, the B field is vertical:



Even in the original Stern-Gerlach experiments, there are both z and x nonzero derivatives in the field. The only thing necessary for the x-changes is that they be symmetrical about the vertical, just like the original. The upper part of the quadrupole field is identical in structure to the Stern Gerlach field.

The only difference is that the vertical B field is weak while still allowing for a large z-derivative. So since one will be able to measure L_z along this vertical B field in this experiment, QM must predict quantization.

Andrew A. Gray

[andrewgray]

3)The New Bremsstrahlung Cutoff Experiment

We covered the Bremsstrahlung Cutoff Experiment briefly in post #1. In review, the setup is like this:





25 KeV electrons are blasted into a metal plate. The decelerations cause emissions of x-rays in all directions up to a cutoff frequency ν_max=E/h. We showed how the cutoff frequency resulted from pulsating electrons that have a Nyquist Cutoff Frequency. That is, if the movement frequency is much greater than the pulsation frequency, radiation at this much higher frequency cannot be generated because the electron is OFF during much of the accelerations. The radiation aliases back down to a lower frequency. (See post #1).

We seek a method to prove this. According to QM, the cutoff frequency is dependent on the max energy of an x-ray photon, produced during the interaction. According to our New Theory, the cutoff frequency is dependent of the pulsation frequency of the electrons. So we wish to change the pulsation frequency of the incident electrons without changing their energy. If the maximum frequency found is different than E/h, then the QM theory will be proven false.

Again, we seek a way to change the pulsation frequency of the incident electrons without changing their energy. To do this, we use a cyclotron instead of a ordinary voltage. There is no guarantee that a centripetal acceleration will affect electron pulsations like a linear acceleration. So our first test would involve a setup with a cyclotron like this:




According to QM, the max frequency will remain unchanged, ν_max=E/h. However, this New Theory allows for the possibility that the max frequency could be drastically reduced. Why would that be? Because in a cyclotron, the accelerations are in opposite directions with each ½ revolution. Since the electron structure is being affected, the opposing accelerations might cancel each other and drastically reduce the electron's pulsation frequency. This would in turn reduce the max Bremsstrahlung cutoff frequency. If this setup produced cutoff frequencies significantly below E/h, the photon hypothesis would indeed have to be abandoned.

Andrew A. Gray