# 7 Reasons To Abandon Quantum Mechanics-And Embrace This New Theory

### #35

Posted 13 July 2007 - 03:39 PM

-Will

### #36

Posted 16 July 2007 - 10:32 PM

. . . 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 [imath]\frac{d \vec p'}{dt'} = \vec F'[/imath], then the equation of motion in the inertial frame K, moving at velocity [imath]\vec v[/imath] is :

[math]\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 ]_{||} [/math]

and

[math]\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} [/math]

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

. . . 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?

. . . 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:

[math]\bar K = \bar {\frac{1}{2} mv^2} = \frac {3}{2}kT [/math]

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

[math]\bar E = N_o \frac{3}{2}kT = \frac {3}{2}RT[/math]

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

[math]c_V \; = \; \left( \frac{ \partial \bar E }{\partial T} \right )_V \;=\; \frac {3}{2}R[/math]

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

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

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 [imath]c_V \propto T^3[/imath] as [imath]T -> 0[/imath]. 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

[math]c_V = \gamma T + AT^2[/math]

where [imath]\gamma[/imath] and [imath]A[/imath] are arbitrary constants chosen to match experimental data.

*Se la vi*.

. . . 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:

[math]\nabla \times \vec E - \frac{1}{c} \frac {\partial \vec B}{\partial t} \;=\; 0[/math]

[math]\nabla \times \vec B - \frac{1}{c} \frac {\partial \vec E}{\partial t} \;=\; \frac{4 \pi}{c} \vec J[/math]

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:

[math]- \frac{1}{c} \frac {\partial \vec E}{\partial t} \;=\; \frac{4 \pi}{c} \vec J[/math]

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

### #37

Posted 17 July 2007 - 12:17 AM

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

**Edited by andrewgray, 16 August 2013 - 02:07 PM.**

### #38

Posted 17 July 2007 - 02:29 AM

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 . No need to confuse wanting to draw a clearer distinction with ignorance.Qfwfq, does this answer the question?

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 [imath]\Delta E[/imath] allows things to match up. RQFT is choc full of Dirac [imath]\delta[/imath]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.We have tunneling right? This allows a non-conservation of energy at the strictly microscopic level, right?

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.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.

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?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.)

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.

### #39

Posted 17 July 2007 - 09:36 AM

You are correct. There is no Pauli exclusion principle in this New Theory. However, Boltzmann's constant

*k*comes from ideal gases:

[math]\bar K = \bar {\frac{1}{2} mv^2} = \frac {3}{2}kT [/math]

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

[math]\bar E = N_o \frac{3}{2}kT = \frac {3}{2}RT[/math]

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

[math]c_V \; = \; \left( \frac{ \partial \bar E }{\partial T} \right )_V \;=\; \frac {3}{2}R[/math] [/quote]

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

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

### #40

Posted 19 July 2007 - 01:52 PM

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.

### #41

Posted 19 July 2007 - 09:09 PM

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?

First, tunneling does NOT allow for non-conservation of energy. Now, lets talk about your theory- it has a complicated time behavior (hence not symmetric in time) which leads to non-conservation of energy.

However, we should still have translational symmetry- which means we should treat momentum conservation as fundamental.

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.

We know from deep scattering that the neutron is not a proton and electron but rather seems to be made of 3 point like fractional charged objects.

-Will

### #42

Posted 20 July 2007 - 01:42 AM

But there is such a thing as coherent linear superposition. QM does not say that the field can only be circularly polarized.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.

In the angular momentum spectrum of photons there are only the two eigenvalues and this is very important in studying the interaction of photons with a simple system, because angular momentum must be conserved. For instance, selection rules for atomic spectroscopy match up with this as well as other things. Consider this and your problem with polarization begins to look a bit like the one with energy-momentum. The photon does not interact solely with the electron ejected from the photocathode.

Actually much more than three. The quarks usually mentioned for a given hadron are the overall net of quarks and antiquarks.We know from deep scattering that the neutron is not a proton and electron but rather seems to be made of 3 point like fractional charged objects.

### #43

Posted 20 July 2007 - 07:16 AM

I think electrons were not discovered to be smaller as in the quote. There was nothing detected that was smaller than the electron's electromagnetic radius. So, my thinking is that there is nothing else there. I explore that hereIt was in the latter part of the last century that electrons were discovered to be smaller than 10-15 cm from electron scattering experiments.

### #44

Posted 20 July 2007 - 09:27 AM

Actually much more than three. The quarks usually mentioned for a given hadron are the overall net of quarks and antiquarks.

Yes, but when you do deep scattering, the relevant energy scale is quite high, and the strong coupling (being asymptotically free) is small. This leads to the nucleon acting like 3 free quarks in these experiments. Feynman's so-called Parton model.

Smaller energy scattering to probe the bound states of the nucleon are much more difficult to interpret.

-Will

### #45

Posted 21 July 2007 - 03:18 PM

About Microscopic Non Strict Conservation of Energy

Beta Decay Started it all. Beta decay has an electron and anoter particle "exploding" away from each other. The two resultant particles do not comply with conservation of energy and momentum:

QM response: Invent a "massless" particle to carry away the missing energy and momentum at (nearly) the speed of light: the neutrino.

This New Theory's response: Tunneling allows strict microscopic non-conservation. (Only time averaged).

Which one to choose? Well, the QM'ers had their work cut out for them to find these "neutrinos". They built HUGE detectors that saw 2 or 3 ticks from muons or positrons PER DAY. So take note:

**Neutrino "detections" are not actually neutrino detections, they are actually detections of other particles.**There is no way to verify that "neutrinos" were actually there. But even so, this neutrino data was not even close to the numbers predicted by their theory, so

**now we have neutrino oscillation fudge!**. Yes, now we have the "neutrino oscillation" hypothesis. Since the experimental numbers did not agree with QM predictions, they simply hypothesized that they must have been changing into something else! This is the ultimate example of changing a theory to match changing data. This seems to be human nature, and change comes, (as Erasmus says), only at "great cost".

However, this New Theory seems promising, and the purely logical mind would at least look at the

**possibility**that we do not have pure microscopic conservation of energy and momentum due to tunneling. The neutrino hypotheses are still hazy enough to just let go of them.

Andrew A. Gray

P.S. To Be Continued. I realize that I have not addressed all points put forward to me. You guys are faster than me. Apologies.

### #46

Posted 22 July 2007 - 10:51 AM

But there is such a thing as coherent linear superposition. QM does not say that the field can only be circularly polarized.

Yes, QM can "make" vertically polarized light by "adding" clockwise and a counterclockwise "photons" together. However, Bohr's principle comes into play. QM says that the photoejection is a

**particle**interaction with max energy hν-Ф,

**not**2hν-Ф, implying interaction with

**one**photon. One photon does not vertical polarized light make.

QM theory needs to hypothesize the "particle" nature of light to get the

*hν*energy limit. This New Theory does not. This clearly is an advantage, as we then can restore the reality-based transverse

**E**field ejections, just like the obvious microwave example above. One final comment. Did you ever think that a "particle" collision (hypothesized as an absorption) would have the tendency to "knock" the ejected electrons forward instead of sideways? Did you wonder how the metal plate was also "knocked" sideways? If the electron was "bound" to the metal plate, and the the electron was "knocked" sideways, wouldn't it pull the plate along with it (instead of having it go off in the opposite direction?)

The above diagram is according to QM. This New Theory has

**non-acceleration resonance**instead of "photons". None of the above paradoxes apply. That is, when the pulsation frequency of the ejected photoelectron just matches the peaks of the incident wave, the electron simply moves up and down in the comoving inertial frame, going nowhere. The acceleration is over. That leaves us free to consider the incoming light as a

**wave**, as it clearly should be. The plate does not have to recoil, as this New Theory describes the photoejection simply as a transverse electrical force on the electron, and

**not**as a collision.

Andrew A. Gray

### #47

Posted 22 July 2007 - 09:06 PM

We know from deep scattering that the neutron is not a proton and electron but rather seems to be made of 3 point like fractional charged objects.

Actually much more than three. The quarks usually mentioned for a given hadron are the overall net of quarks and antiquarks.

Aunt Mary has some pain in her abdomen, so the doctors want to probe her to find out what's inside. To do this they drop a hydrogen bomb on her, and watch what comes out. "Oh, she has a tumor on her diaphram", the doctors exclaim. Pretty far fetched, right? No one would believe that watching Aunt Mary's pieces come out would reveal anything about her current abdominal condition. Blasting her apart only reveals that she can be blasted apart.

So why the surprise to find my skepticism when you probe a neutron or a proton by annihilating it with anti-protons with a GaZeV (gazillion eV's)? We see some pieces come out after the anti-proton "atomic bomb" is dropped on it, but so what? In the same way with Aunt Mary, this does not give a picture of how it was before the bomb. So how do you know that a neutron wasn't a proton and an electron if you blasted it into pieces, and you saw "some stuff" come out? You don't.

Blasting protons and neutrons into pieces and categorizing the pieces (and giving Nobels for the predictions) is all well and good, don't get me wrong. But studying the properties of the blasted neutron bits doesn't reveal that much about the original particle, in my opinion, just like blasting apart Aunt Mary with a nuke doesn't tell a psychologist much about her personality.

I am much more interested in getting nanoscale science correct at this point. Nanoscale physics is not that "mysterious" anymore, as we are starting to be able to "see" things at this dimension. In my opinion, we must get nanoscale physics correct before we can get femtoscale physics correct. And I do not believe that we have nanoscale physics correct yet. That's why the New Theory.

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.

See above.

Andrew A. Gray

### #48

Posted 22 July 2007 - 09:46 PM

I like your theory; I've followed here and on other forums.

Thanks for the encouragement.

I've always thought of an electron as a photon trapped in a repeating pattern, maybe circular.

DryLab, it seems that this New Theory does not include "photons".

Andrew A. Gray

### #49

Posted 23 July 2007 - 03:02 AM

*e*. It took me a while to figure that you must have meant your reply only in this sense.

What it does mean however is that the picture is even more complicated than three quarks just sitting there, hadroninc matter could hardly be described without RQFT. And, Andrew, DIS doesn't necessarily blow the hadron apart. In order to extract information about the electric charge of single quarks, it is enough to consider the data of events for which the lepton's interaction was hard enough (and if you don't know what that means, look it up) and that's pretty much what the "deeply inelastic" in DIS means.

And where did I mention two photons rather than a single one? You don't appear to be understanding the quantum formalism, which you are presuming to not only criticize but even refute. Not all observables are compatible, otherwise there wouldn't be Heisenberg's principle. If an interaction makes it sensible to say what the direction of the electric field vector was, then certainly it couldn't have also observed angular momentum. Those who know me well enough around here know I use different attitudes with people that are asking for help about things that are hard to understand (including QM which is freakin' wierd) and those who think they can prove something is BS, only because they know little about it. So, before trying to knock down QM, get a better understanding of it.Yes, QM can "make" vertically polarized light by "adding" clockwise and a counterclockwise "photons" together. However, Bohr's principle comes into play. QM says that the photoejection is a

particleinteraction with max energy h?-?,not2h?-?, implying interaction withonephoton. One photon does not vertical polarized light make.

What do you mean by "pure microscopic"? What do you mean by "non strict"? Without either the neutrino or some other garbage bin, we'd have to conclude that energy isHowever, this New Theory seems promising, and the purely logical mind would at least look at the

possibilitythat we do not have pure microscopic conservation of energy and momentum due to tunneling. The neutrino hypotheses are still hazy enough to just let go of them.

*consistently*disappearing. Not even average conservation.

Actually, I'd say I'm pretty darn slow, due to having other fish to fry, so you should be feeling lucky.You guys are faster than me.

### #50

Posted 23 July 2007 - 04:09 AM

Which stat-mech book are you using,btw?

Fundamentals of Statistical and Thermal Physics http://www.amazon.com/Fundamentals-Statistical-Thermal-Physics-McGraw-Hill/dp/0070518009 , by Frederick Reif.

So, what about polarization dependence? I remember a basic exercise in my RQFT course, which explains it fine for the Compton effect.

Qfwfq,

I am not familiar with a QM polarization explanation in the Compton Effect. Do you have a reference or do you care to explain? How does the Compton effect vary with polarization? I have made some predictions about this and it would be very interesting to see if they are true.

Here is how this New Theory explains the Compton Effect for changed x-ray wavelengths at different angles:

This New Theory predicts that the most likely ejection angle for the Compton electrons would be at 90

^{o}. However, since the Compton experiment uses a solid metal plate as the target, 90

^{o}ejections would be impossible (a metal vapor would be better, allowing transverse ejections). The forward direction would be the most

**un**likely for ejection, as the transverse

**E**field would again tend to eject the electrons sideways. Hence, the most likely ejection direction for a solid plate experiment would be somewhere in between 0

^{o}and 90

^{o}.

The maximum velocity would again be dependent on a

*non-acceleration resonance*, just like in the photoelectric effect. The maximum energy might not go as

*hν*, as higher harmonic resonances are probable, reducing the velocity. However, once this velocity is known, then the change in x-ray wavelength would simply be due to

**a Doppler shift**, as the receding electrons reflect the incident x-rays, shifting the frequency towards the red.

Let's try this out and see how this works. The max ejection velocities are known. The max velocity found for ejected electrons is approximately .07c for Compton's Experiment. (See Recoil Electrons From Aluminum). We will take the average velocity to be approximately .05c, for a

**rough**estimate. Also as a

**rough**estimate to see if we are in the ballpark, we will try 45

^{o}as the most likely ejection angle.

We wish to find the resulting retransmitted wavelength using the Doppler formula:

[math] \;\;\;\; \nu' \;=\; \nu \; \frac{1-\beta cos \theta}{\sqrt{1-\beta^2}}[/math]

Now the electron would continuously Doppler-shift the incident wave, from no frequency shift at the beginning of its acceleration, to a maximum shift in frequency at its final velocity (at the end of its acceleration). Thus, the retransmitted wave’s frequency shift would be a broadened spike. The center of the spike would be associated with some electron velocity in between its initial and final velocity, and not the electron’s final velocity. We will make the reasonable assumption that the average retransmission velocity of the pulsating electron is ½ its average final velocity (again for a

**rough**estimate).

So finally, we have a rough estimate for the Doppler shifted wavelengths:

Not a bad rough estimate (obviously, we could change our very rough assumptions to match the data. We will avoid this, as to not be hypocritical, but some combinations give exact results.)

However, if one changes to a metal vapor target, instead of a metal plate target, things would really change. This would allow transverse ejections, changing the character of the Doppler Shift. It would be very interesting to redo this experiment with a vapor target to see if the Compton wavelengths change. The QM prediction would stay the same.

In addition, polarizations could be considered. For Compton's experiment, if the x-rays were vertically polarized, then the ejections would no longer be in the Compton plane. This would change things. It would be easy to make predictions on how this would change things with this New Theory.

For horizontal polarizations, the ejections would tend to stay in the Compton plane. It would be interesting to do an electron velocity distribution experiment for the Compton electrons to verify these polarization predictions. QM would not make such electron velocity distribution predictions, because QM treats this phenomena as a collision instead of a transverse force ejection. Here, Qfwfq, is where I would like to see your QM Compton polarization results.

Andrew A. Gray

### #51

Posted 24 July 2007 - 01:52 AM

Look up the Klein-Nishina formula for Compton scattering and, if you're interested in the matter of polarization, you might like to go through these search results. It can be derived from a simple, loopless Feynman diagram which makes it a comparatively simple, and certainly trouble-free, computation and you can find it done in detail in Itzykson-Zuber, Quantum Field Theory 5-2-1 (Elementary Processes) with a discussion of how it relates to the non-relativistic formula, the unpolarized case and Thomson scattering.I am not familiar with a QM polarization explanation in the Compton Effect. Do you have a reference or do you care to explain? How does the Compton effect vary with polarization? I have made some predictions about this and it would be very interesting to see if they are true.