Study finds there may be multiple 'God particles'

Jun 16, 2010 by Lin Edwards report
The Standard Model. Image credit: AAAS

(PhysOrg.com) -- Recent research in the US suggests there may be five versions of the theorized Higgs boson.

The Higgs was dubbed the "" by Nobel laureate Leon Lederman because its discovery could unify our understanding of the universe and help us “know the mind of God”.

The Higgs is extremely important to the accepted theory of physics, known as the “Standard Model”, which was developed in the 1970s to incorporate everything known at the time about interactions between . The Higgs boson is thought to be the sub-atomic particle that mediates the force through which all other sub-atomic particles acquire their mass.

Scientists have been trying for five decades to detect the Higgs boson, but have so far failed. Now theoretical physicist Adam Martin and colleagues at the ’s Tevatron near Chicago in Illinois in the US have analyzed results from the DZero experiment and suggest there may be multiple versions of the Higgs boson.

The DZero experiment set up and observed collisions protons and anti-protons and was designed to examine the reason why the world is composed of normal matter rather than its opposite: . They found the collisions resulted in pairs of muons one percent more often than anti-muon particles. The could explain why matter has come to dominate over anti-matter, rather than the two annihilating each other.

This effect, called CP violation, had been seen before but not to the same degree as seen in DZero, and the degree of asymmetry found in the latest results is greater than can be accounted for by the Standard Model. The researchers said the results could be explained by the existence of five Higgs boson particles with similar masses, with one having a negative electric charge, one negative and three neutral. The theory is called the two-Higgs doublet model.

The two-Higgs doublet model is not the only possible explanation for the results, but Dr Martin said fitting a new effect in the Standard Model without disrupting its fit with other tests is difficult. The Standard Model accommodates only one Higgs doublet, and while scientists think of the Higgs as a single particle, Dr Martin said it “actually comes in a package of four”. Only one is seen because the other three are seen as W and Z bosons. Adding another Higgs doublet adds four more particles.

Many physicists have come to regard the Standard Model as incomplete since it does not explain gravity or describe dark matter. An extension to the , known as “supersymmetry,” proposes that each particle has a more massive “shadow” partner particle, effectively doubling the number of known particles. Such a scheme could accommodate the two-Higgs doublet model. So far no experimental evidence has been found for the existence of the “shadow” particles.

The search for the is one of the main aims of the Large Hadron Collider (LHC) near Geneva in Switzerland. The facility, the world’s largest particle accelerator, could also find experimental evidence for supersymmetry.

The results of the US research is published at arXiv.org.

Explore further: Tiny magnetic sensor deemed attractive

More information: Bogdan A. Dobrescu, Patrick J. Fox, Adam Martin: CP violation in B_s mixing from heavy Higgs exchange, arXiv:1005.4238, arxiv.org/abs/1005.4238

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User comments : 110

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Objectivist
4.4 / 5 (22) Jun 16, 2010
Please stop calling it the "God" particle. I'm sincerely asking you nicely.
tigger
4.7 / 5 (13) Jun 16, 2010
"one having a negative electric charge, one negative and three neutral"....

...edit...

"one having a POSITIVE electric charge, one negative and three neutral"....

KronosDeret
4.4 / 5 (14) Jun 16, 2010
agreed force of gravity can hardly be called a God. Thing fall to common center of mass.. where do I start to worship that?
Gene_H
4.3 / 5 (6) Jun 16, 2010
Please stop calling it the "God" particle. I'm sincerely asking you nicely.
This is how every propaganda ends. I'd recommend to read article here:

http://www.physor...225.html

StandingBear
1.7 / 5 (6) Jun 16, 2010
Suppose, just suppose, that the real model is three dimensional. Suppose there are more natural forces out there, and the interactions of those forces, among them the gravitational force, affect all the others
dompee
2 / 5 (7) Jun 17, 2010
Please stop calling it the "God" particle. I'm sincerely asking you nicely.
why? I think its a great nickname...what diff does the nickname make anyway, does it bother you that much? Apparently since your sincerely asking nicely, ha! The real question is who exactly are you asking? The entire physics community? Ha! Good luck with that one, I think you may have to bite the bullet and live with it, I know its asking a lot but such is life ;)

johanfprins
2.2 / 5 (6) Jun 17, 2010
I doubt the existence of any Higg's bosons: There are NO particles; ONLY waves: The core intensity of an electron wave, when viewed within its inertial refrence frame IS its mass, and the so-called "tunnelling tails" are the curvature of space around the mass-energy of the wave: A matter wave is thus nothing else but a light-wave which has been trapped within an inertial reference frame. It is for a similar reason why one can trap a light pulse within a Bose-Einstein condensate. The latter is just a macro-demonstration of what happens around a nucleus when an "orbital"-electron absorbs a light-wave. Gravity is already accounted for by the fact that matter consists of waves having inertia since their core intensities are their masses.
Are we going to waste another 100 years with nonsensical physics concepts like "wave-particle" duality, "probability amplitudes", and "complementarity"? We should rather mothball the LHD and spend the money on searching for "clean" energy generation.
Gene_H
1.3 / 5 (3) Jun 17, 2010
Particle wave duality is experimentally well proven fact, johanfprins. And experiment always goes first - physics is still an experimental science - not the theoretical babbling. Protons, grains of dust, planets and stars are all well defined particles of spherical symmetry, not the waves.
Ricochet
5 / 5 (1) Jun 17, 2010
Please, guys, remember that Physics is the study of everything, and until we know everything, we're just guessing, based on our current knowledge and training. In the scale of everything, we could be like the infant who just found his own belly button, or we could be like the teenager who is just finally starting to figure the world out.
The point is that we don't know, and until we figure out the next level of knowledge there is to be gained, we won't know if our current theories are correct, or even close. I maintain that truth can only be gained in hindsight.
Bookbinder
1 / 5 (3) Jun 17, 2010
Shorter version.
The Standard Model doesn't add up, so instead of just admitting it, and searching for the flaw in the model, we are just going to make something up. This is what children do. You physicists can do better.
typicalguy
5 / 5 (2) Jun 17, 2010
bookbinder -
I thought that's what current physics is doing? The standard model predicted various particles. These particles were searched for and discovered. The standard model predicts the higgs boson but it has not yet been discovered. Physicists are looking for that particle now. As far as flaws go, physicists inability to find the Higgs particle is paramount. If the LHC does not discover the Higgs then it will probably find some completely different physics that will take us down a road that the standard model could never have predicted. The important thing to keep in mind, any theory created with new physics discovered at the LHC would have to reduce down to the standard model in the previously tested conditions since it's an accurate description of nature for those conditions (including the real world). New physics or not, the standard model won't (and can't) be abandoned entirely.
Jigga
1 / 5 (3) Jun 17, 2010
The standard model predicts the Higgs boson but it has not yet been discovered.

You're just parroting widespread hoax or urban legend, "Higgs boson is the last missing particle in Standard Model". Standard Model actually doesn't enable/predict anything particular regarding Higgs boson, so it doesn't require it at all. There are some extensions of SM, which enable to estimate the Higgs boson mass and other parameters, but these extensions aren't part of Standard model, as they can be used in other theories as well (for example SUSY can be used both in SM, both in string theory, for example).

If some theory cannot preduct nothing about particular concept, it's invariant to this concept, it simply doesn't require this concept at all for anything - it cannot use it in any logics, any equations.

Surprisingly it seems, Standard model is actually correct in this point about Higgs boson.
Jigga
1.3 / 5 (3) Jun 17, 2010
During time, Higgs boson mass was guessed from 109+-12 GeV to 760+-21 GeV, plus two unconventional theories with 1900 GeV and 10^{18} GeV. There are so many comparably likely models - most of which contain continuous parameters whose values aren't calculable right now - that the whole interval is covered almost uniformly.

http://arxiv.org/pdf/0708.3344

The article linked is an review of more than 200(!) predictions of Higgs boson mass - all done with exact formal math... Higgs model is too vague to be considered seriously, because it has more then single formulations: It's technical derivation consists in a mere reshuffling of degrees of freedom by transforming the Higgs Lagrangian in a gauge-invariant manner. Well known "hierarchy problem" implies, that quantum corrections can make the mass of the Higgs particle arbitrarily large, since virtual particles with arbitrarily large energies are allowed in quantum mechanics, thus making such hypothesis untestable.
johanfprins
3 / 5 (6) Jun 17, 2010
Particle wave duality is experimentally well proven fact, johanfprins. And experiment always goes first - physics is still an experimental science - not the theoretical babbling.

Oh my God: Here we go again with the immature babbling of Alexa, Seneca and what have you. "Wave particle duality" is a misconception which is NOT "well proven" by experiment anywhere: Just as Cooper pairs are not well-proven by experiment. Which so-called experiment proves wave-particle duality? And PLEASE do not come with the double-slit experiment again: It does not prove duality at all: It only proves that matter MUST consist of waves: NOTHING ELSE!!
Please do not claim "experimental facts" which do not exist at all. First define what you understand as a "particle" before you claim "wave-particle duality". Physics rely on clear definitions not the type of gobblydook you are spreading around!
Jigga
1 / 5 (4) Jun 17, 2010
It only proves that matter MUST consist of waves
And waves must consist of particles... any problem?
Ethelred
not rated yet Jun 17, 2010
It only proves that matter MUST consist of waves: NOTHING ELSE!!


It does prove the matter must have wave properties anyway. I don't see it as proving the matter MUST be waves and waves only. There are other paired properties of particles in the Standard Model and those also fit the present evidence.

So I have a couple of questions for you:

Does your waves only model also handle the other paired properties?

Can it handle CP violations? As those seem real, and with my limited knowledge, I see your pure wave concept as being inherently symmetrical so I see CP violation as a problem for it.

Ethelred
johanfprins
1 / 5 (2) Jun 18, 2010
]And waves must consist of particles... any problem?

Prove it experimentally by taking a photograph of the particles! Why must all waves consist of particles? A wave within water does BUT:
A coherent laser beam does not consist of particles It is a pure coherent single wave consisting of wave-fronts moving through space driven by electromagnetic induction. No particles there at all!!
Similarly cavity radiation consists purely of standing light waves each of which cannot have less energy than a quantum amount: No particles here either! The fact that such a standing light wave can disentangle to allow quantums of light to be absorbed by the walls does not prove that the radiation within the cavity consist of moving "photon-particles".
So stop talking nonsense.
johanfprins
2 / 5 (3) Jun 18, 2010
Does your waves only model also handle the other paired properties?

Any harmonic wave exists within space AND reciprocal space; even a water wave. So when the wave increases its size in space, it decreases its size in reciprocal space: Unfortunately in quantum mechanics reciprocal space of a matter-wave has been equated with momentum of an electron owing to de Broglie's hypothesis. This is wrong since de Broglie's hypothesis is ONLY valid when you have a moving wave: NOT when it forms a time-independent standing wave. No standing wave in history has ever had momentum: Thus to assign momentum to an electron around a nucleus is daft.

I am going to make a statement and give you the opportunity to figure it out. If you cannot, let me know and I will prove it to you: "An electron-"particle" circulating around a positive charge as modelled by the Bohr model cannot have a magnetic moment at all".

CPT parameters are changes in the boundary conditions of a wave: Nothing else.
Gene_H
Jun 18, 2010
This comment has been removed by a moderator.
Gene_H
1 / 5 (2) Jun 18, 2010
Anyway, try to answer the following question:

Yes, we know, the wave at the water surface is formed with some particles. But could we see these particles, when we would some bubble sitting at the water surface, which could observe/interact with neighboring reality via surface waves only?

Or in another words: can the observer, which is using some waves for observation see just the particles, which are forming them? Can some object serve as a both subject, both mean of observation at the same moment?

I know quite well, people cannot understand the particle nature of vacuum just because of fact, these particles can be never ever seen in light waves. But this is just a feature of aether model, not its weakness. It just explains, why particles forming the waves at the water surface are observable with light waves, but the particles forming light waves cannot be seen with light waves.

Actually, without such trivial understanding we cannot move further in reality understanding.
Ethelred
not rated yet Jun 18, 2010
"An electron-"particle" circulating around a positive charge as modelled by the Bohr model cannot have a magnetic moment at all".


An electron can't circulate around a nucleus anyway. It would produce synchrotron radiation. It CAN circulate either in a circuit or in an accelerator. In which case it produces photons. When I ran across this I gave up on the planetary orbit model of the atom.

CPT parameters are changes in the boundary conditions of a wave: Nothing else.


Which implies that you don't see a way to explain the preponderance of matter vs. anti-matter. Or the muon anti-muon results and other experiments showing CPT symmetry breaking. Which means there is something going on that a wave based theory might not cover.

Ethelred
johanfprins
1 / 5 (1) Jun 18, 2010
When I ran across this I gave up on the planetary orbit model of the atom.

Good for you!! But by interpeting an electron around a nucleus in terms of the duality of position and momentum, you still maintain that the electron must have kinetic-energy! And since such an electron must stay near the nucleus, it has to change direction all the time and still radiate away energy. Just as unstable as the Bohr model.

Which implies that you don't see a way to explain the preponderance of matter vs. anti-matter.

Can you prove that both matter and antimatter were generated in equal quantities in the Big Bang? I can explain why only matter was initially generated as a single Bose-Einstein type wave. CPT symmetry breaking is NOT required since no anti-matter was created simultaneously. I have just been writing it up and will soon post it on my website. It also explains where dark matter and dark energy comes from. I will let you know as soon as it is on my website.
Gene_H
1 / 5 (3) Jun 18, 2010
CPT symmetry breaking is NOT required since no anti-matter was created simultaneously.

Both particle, both wave model of reality are completely symmetric with respect to matter-antimatter imbalance, they cannot explain the difference between matter and antimatter.

You'll need a better/deeper theory (i.e. mine, for example) to explain it without another ad-hoced assumptions, like the assumption of delayed formation of anti-matter after Big Bang.
Ethelred
not rated yet Jun 18, 2010
duality of position and momentum


Duality of wave and particle. Clearly orbits are some sort of standing wave OR it really is just a matter of uncertainty.

Can you prove that both matter and antimatter were generated in equal quantities in the Big Bang?


No. However in experiments that is what you get when energy condenses into matter. Such as a gamma ray producing an electron positron pair.

I have just been writing it up and will soon post it on my website.


Unfortunately it is likely to be way over my education. Perhaps Skeptic_Heretic will be able to follow it.

Ethelred
johanfprins
3 / 5 (2) Jun 18, 2010
Duality of wave and particle. Clearly orbits are some sort of standing wave OR it really is just a matter of uncertainty.

No standing wave EVER had momentum: "orbits" around a nucleus can only be standing waves. So there cannot be any momentum involved: What "uncertainty are you then talking about?
No. However in experiments that is what you get when energy condenses into matter. Such as a gamma ray producing an electron positron pair.

Good point. But this is assuming that there was kinetic-energy which condensed into matter. I cannot for the world of me see kinetic-energy generated within a singularity. Kinetic energy requires movement through space.
Unfortunately it is likely to be way over my education

You are clearly able to follow. I must admit that my insight is still speculative, but it makes more sense as the "hot" Bing Bang! It is more likely that it started of with a "quantum jump" which is a rapid inflation of a single, holistic matter wave!
omatumr
1 / 5 (4) Jun 18, 2010
"How many angels can dance on the head of a pin?" - A metaphor for wasting time debating topics of no practical value [Wikipedia].

Sadly that seems to describe a large part of modern theoretical physics.

With kind regards,
Oliver K. Manuel
Emeritus Professor
Nuclear & Space Studies
Former NASA Principal Investigator for Apollo
SStacey
not rated yet Jun 19, 2010

Every healthy living organism or organisation embodies 5 different qualities.
Protective Parent (+ ve): boundaries/rules, effective discipline, protection from threats, etc
Nurturing Parent (- ve): supports growth and a desire to live for something larger
Adult: problem solving, day to day running
Free child: Creativity, fun, celebration
Adapted child: Responds to requests, lives within rules

Maybe the 5 different bosons offer different forms of guidance to atoms:
A boson to ensure an atom followed its internal laws and protected it from destruction
One to support an atom's growth from lighter to heavier element and help it contribute to the larger universe
One making sure that the atom maintained itself
One that allowed the atom to participate in creativity (chaos?)
One helping the atom follow the higher universal laws or accepted bonding requests from other compatible atoms

Every atom exhibits these 5 qualities - and thus the universe is formed.
S. Stacey
creationofacouple.com
johanfprins
2 / 5 (3) Jun 19, 2010
Sadly that seems to describe a large part of modern theoretical physics.

With kind regards,
Oliver K. Manuel
Emeritus Professor
Nuclear & Space Studies
Former NASA Principal Investigator for Apollo

In fact it is worse: As Carver Mead correctly pointed out in the Spectator in 2001: In his opinion the last 7 decades of the 20th century will become known as the dark ages of theoretical physics. It all started in 1927 when The Copenhagen interpretation became accepted in Brussels. This led to quantum field theory and fudging mathematics using tricks like renormalisation. Furthermore it incorporated mistakes: For example: assuming an electric field exists around a solitary charge; assuming that a rotating single charge can cause a magnetic moment; assuming that a harmonic wave can have a phase-angle which is not determined by boundary conditions; and thus deriving that the momentum of an electron is a conservarive vector field while NO momentum field can EVER be one.
Ethelred
5 / 5 (1) Jun 19, 2010
This going to take three posts. Brevity is not always the right answer.
What "uncertainty are you then talking about?
Heisenberg's, as in the idea that the electron's position is sufficiently uncertain in comparison to the nucleus that it can't orbit since even the electron doesn't know its position exactly but the position is sufficiently tied down that the momentum is pure fluff. At least that is how I think the Standard Model treats it. Though mostly the Standard Model is just the math and not a mental model.

It seems to be the way things are done in particle physics. Do the math and don't worry about a physical model. I can't do the math and even if I could I think would still find it unsatisfying.

Continued
Ethelred
5 / 5 (1) Jun 19, 2010
Continuing

Just because I want a physical model it doesn't mean that one can ever exist that satisfies the way humans think. That is the difficult part about trying to accept what the evidence tells us. For instance I don't like the idea that the Universe will reach a state that can best be described as The End of History or perhaps Timelike Infinity. But that is what the evidence tells us at the moment so I accept it. I don't have to like it for it to be real.

But this is assuming that there was kinetic-energy which condensed into matter.


I don't see you why think that assumption is needed.

Hold that thought because there is one more to go.
Ethelred
5 / 5 (2) Jun 19, 2010
Last of the set.
I cannot for the world of me see kinetic-energy generated within a singularity. Kinetic energy requires movement through space.
Agreed but there is no space in a singularity. I don't think a singularity, that is a Point of energy, is good model for the beginning of the Universe. It should have had something like Plank length and volume due to Uncertainty. Then Space, for unknown reasons, began to expand thus allowing movement through space. This is one of the reasons that I think of Uncertainty as a real thing.
You are clearly able to follow.
By analogy and physical models. Not with math.

Ethelred

This could have been a single clean 1856 character post. But someone that hates writing thinks 1000 characters is more than enough. Its enough for sound bites.

Read my lips no new taxes.
No one ever lost money underestimating the American public.
No Cross of Gold No Crown of Thorns.
Brevity is the soul of with.

All of equal worth.
Skeptic_Heretic
3.7 / 5 (3) Jun 19, 2010
Unfortunately it is likely to be way over my education. Perhaps Skeptic_Heretic will be able to follow it.

Ethelred
I'll be happy to try to translate/explain it, that is if I fully understand it. Johan has built a very interesting framework that is self-cohesive. It's not fully explanitory, yet, but it gets stronger with each writing.

And no Alizee, W-P duality has never been proved, it has been evidenced, but not proved.
Agreed but there is no space in a singularity. I don't think a singularity, that is a Point of energy, is good model for the beginning of the Universe.
Actually, I think that's wrong. If you examine a singularity within Hawking and Einstein's frameworks, then apply Suskin's logics, you can intepret a black hole to be 3 dimensional space without time. I've been working out the formulae for this and the math keeps getting weird, then making sense, then getting weird again. It's like always being on the verge of a breakthrough, with no success.
Skeptic_Heretic
3 / 5 (2) Jun 19, 2010
Seeing a singularity as 3d space without time actually validates a lot fo our precepts that are currently operational as receiving infinity when trying to apply time based physical math to that region would result in a logical result. Effectively all information for an object contained within that singularity would be available at all points within the singularity and at all points on it's surface. Suskin was really on to something when he said that a singularity transmits and displays all relevant information twice, once within the singularity and then again on the surface of said singularity.

The drain mechanic was really pure genius and gave us grounds to use BEC research as an apt analogy to black holes. What's even more interesting is the behavior of black holes when orbiting other black holes.
Skeptic_Heretic
3 / 5 (2) Jun 19, 2010
And I apologize if the above looks alizeeish, it won't be comming up again until I have formal equations completed.
johanfprins
3.7 / 5 (3) Jun 19, 2010
Dear Ethelred and Skeptic-Heretic,
I enjoyed both of your long responses. It is such a pity that we cannot come together. Obviously I cannot address every point you have raised on this forum. A few comments:
Firstly: What is called uncertainties in pos. and mom. are the sizes of a wave within space and reciprocal space as is the case for all harmonic waves EVER!
Standard model is NOT based on harmonic waves and therefore it is "virtual physics"
The "hot" Bing Bang claims a singularity with an immense temperature: Not possible!
then apply Suskin's logics, you can intepret a black hole to be 3 dimensional space without time.

Bully for Susskind!!! The same is valid within the three-dimensional intensity distribution of a standing electron-wave. Similarly within the intensity of a REAL Bose Einstein Condensate (not condensates of "particles").
My version of BB: A single holistic matter wave inflated rapidly and only then disentangled into light and matter.
CONTINUE
johanfprins
3 / 5 (4) Jun 19, 2010
Consider ground state electron around hydrogen nucleus. It is a standing wave so time does not exist within its intensity. Harmonic motion outside 3D: "dark energy".
"Photon" light-wave comes along, resonates and is stopped in its tracks to become part of the mass of the electron-wave. The electron-wave MUST morph instantaneously (thus a "quantum jump" QJ)
The excited electron wave later finds the correct resonance and disentangles into a light wave which is emitted and it MUST then QJ back to a ground state.
Now at BB: A holistic matter wave with an intense matter density expands near instantaneously to a massive size (reason for Guth's inflation): Since it is distributed mass energy, there is no temperature. But like the excited electron in the H atom, it then disentangles to release light which now cannot escape but remains as cavity radiation. The remaining distributed mass partly disentangles into the matter we know. Some distributed mass remains as "dark matter".
Skeptic_Heretic
3 / 5 (2) Jun 19, 2010
Johan, I think you and I are working on the same math from different sides of the equation.

Ok I gave it a second read. We are working on the same math, and it all has a strong basis in information retention of wave forms. Effectively a wave never changes, it simply gains an increasing amount of information. Black holes, and elementary particles would be similacrums under this hypothesis as it isn't destroyed information but obscured information. Johan, your work isn't contrary to Quantum mechanics, it is explanitive of quantum mechanical action.
johanfprins
3.7 / 5 (3) Jun 19, 2010
@ Skeptic Heretic,
I have never said my work is contrary to quantum mechanics but it is contrary to the Copenhagen interpretation of quantum mechanics. I do have a problem with the term "information". I do not say that a wave does not change with time but that it only changes in essence when its boundary conditions are changing. Maybe I am not very bright but it seems to me as if it is assumed that a wave changes information even when its boundary conditions do not change. This is what I also gleaned from Penrose's book. Maybe I just do not understand what is meant by "information" as used by people like Penrose, Hawking, Susskind etc. My mathematical background is probably to weak. To be honest I have not yet been able to grasp what is meant by "quantum computing".
Skeptic_Heretic
3 / 5 (2) Jun 19, 2010
I'd rather not go alizee and blag the board with things I can't support. I'll keep plugging away and send over my notes once I make some reasonable progress.
TFX
not rated yet Jun 19, 2010
@ johanfprins and Skeptic Heretic. I am a not a physicist of any kind, I am however in my own right very intelligent. I have read the back and forth that you two have going here and I am more then interested. I am choosing not to share too much is I am not sure as to the kind of response I would get from the online community here. Sorry to the point, I would truly love the opportunity to read more on the research and formulations that you are working with. You do not need to share anything that would give up anything of real value but any data the you would be willing to share would be greatly appreciated.
Bloodoflamb
not rated yet Jun 19, 2010
But by interpeting an electron around a nucleus in terms of the duality of position and momentum, you still maintain that the electron must have kinetic-energy! And since such an electron must stay near the nucleus, it has to change direction all the time and still radiate away energy. Just as unstable as the Bohr model.
Please try removing the momentum term from the Schrodinger equation and correctly solving for the hydrogenic electron wavefunctions. Hint: it's not possible.
johanfprins
3 / 5 (2) Jun 20, 2010
You do not need to share anything that would give up anything of real value but any data the you would be willing to share would be greatly appreciated.

Agree! My new book should be out within two months. In the mean time you can read excerpts from it on my website.
Please try removing the momentum term from the Schrodinger equation and correctly solving for the hydrogenic electron wavefunctions. Hint: it's not possible.

Sorry to disappoint you: It is possible since Planck's constant does NOT relate to the de Broglie momentum-wavelength for a stationary electron wave. The fact is that the Schroedinger's equation is an approximation of Maxwell's equation for the electric potential. In both cases Planck's constant only defines the fact that the energy is quantised under suitable conditions. De Broglie's momentum-wavelength postulate ONLY applies to a MOVING electron-wave. A stationary e-wave does NOT have momentum; just like ANY stationary wave EVER discovered.
Skeptic_Heretic
3.7 / 5 (3) Jun 20, 2010
TFX, don't be afraid to ask questions. People only get terse with known pseudo scientitsts. We're all here to educate. Our manner of keeping to the ciriculum is just a little bit different.
Bloodoflamb
not rated yet Jun 20, 2010
Sorry to disappoint you: It is possible since Planck's constant does NOT relate to the de Broglie momentum-wavelength for a stationary electron wave. The fact is that the Schroedinger's equation is an approximation of Maxwell's equation for the electric potential. In both cases Planck's constant only defines the fact that the energy is quantised under suitable conditions. De Broglie's momentum-wavelength postulate ONLY applies to a MOVING electron-wave. A stationary e-wave does NOT have momentum; just like ANY stationary wave EVER discovered.

V(r)|psi> = i*hbar*(d/dt)|psi>

With V(r)=e^2/(4*pi*epsilon_0*r)

Find the hydrogenic wavefunctions for this situation.
MadPutz
not rated yet Jun 20, 2010
agreed force of gravity can hardly be called a God. Thing fall to common center of mass.. where do I start to worship that?


@KronosDeret - This so-called simple force of gravity is the root of why the universe has developed in the way it has since the beginning. Everything from Coca-Cola to Supernovae owes its existence to this force. That's why these critical particles are called "God particle". It has nothing to do with religious dogmas, rather just the fundamental physical science of the universe, which is all that anything godly is.
johanfprins
1 / 5 (1) Jun 20, 2010
Find the hydrogenic wavefunctions for this situation

It is because you do NOT realise that the Schroedinger equation is an approximation around the rest mass of the electron. The actual equation modelling an electron-wave (or any matter-wave for that matter) does NOT depend on the rest-mass as an INPUT, SINCE the mass-energy IS the solution!
Once you realise this, you will also realise that the del-operator in the Schroedinger equation has NOTHING to do with an electron's momentum but ONLY with the reciprocal (k-space) of the electron-wave. If you want to be stupid enough to equate this k-vector to actual momentum for a standing wave, then a standing light wave must also have momentum: You are, however, welcome to make a fool of yourself as the physics community has done for narly 100 years.

Although a standing wave is a function of k, it has NO momentum and will NEVER have momentum EVER!!! This is the case for ALL harmonic waves; and this will be so forever more!!
Bloodoflamb
not rated yet Jun 20, 2010
Once you realise this, you will also realise that the del-operator in the Schroedinger equation has NOTHING to do with an electron's momentum but ONLY with the reciprocal (k-space) of the electron-wave. If you want to be stupid enough to equate this k-vector to actual momentum for a standing wave, then a standing light wave must also have momentum: You are, however, welcome to make a fool of yourself as the physics community has done for narly 100 years.
http://en.wikipedia.org/wiki/Stone-von_Neumann_theorem
johanfprins
1 / 5 (2) Jun 20, 2010
@ Bloodoflamb,
Please stop directing arguments to websites when you are too stupid to argue on your own! Neumann, like Dirac, was a mathematician who had no inkling about what physics was all about. Let us do real physics: Show me ANY harmonic standing wave in the history of mankind which has momentum. A moving electron diffRacts: So it MUST be a harmonic moving wave. An electron around a nucleus MUST be a standing wave: So it is IMPOSSIBLE to have ANY momentum!! PLEASE STOP THIS ABSOLUTE NONSENSE THAT THERE IS MOMENTUM INVOLVED IN THE LATTER CASE!!
Bloodoflamb
not rated yet Jun 20, 2010
@ Bloodoflamb,
Please stop directing arguments to websites when you are too stupid to argue on your own! Neumann, like Dirac, was a mathematician who had no inkling about what physics was all about. Let us do real physics: Show me ANY harmonic standing wave in the history of mankind which has momentum. A moving electron diffRacts: So it MUST be a harmonic moving wave. An electron around a nucleus MUST be a standing wave: So it is IMPOSSIBLE to have ANY momentum!! PLEASE STOP THIS ABSOLUTE NONSENSE THAT THERE IS MOMENTUM INVOLVED IN THE LATTER CASE!!
I posted a link to the Stone-von Neumann theorem because it's not my job to educate you. You should educate yourself. The fact is that momentum is the generator of spatial translations and (via Stone-von Neumann) it can be shown that when working within the spatial representation the derivative operator is identically the momentum (with the appropriate hbar and i).
rah
not rated yet Jun 21, 2010
Is there any prediction of the mass of any of these particles is supposed to be? How about even a wildass guess?
johanfprins
1 / 5 (1) Jun 21, 2010

I posted a link to the Stone-von Neumann theorem because it's not my job to educate you. You should educate yourself. The fact is that momentum is the generator of spatial translations and (via Stone-von Neumann) it can be shown that when working within the spatial representation the derivative operator is identically the momentum (with the appropriate hbar and i)

I know about this theorem. It is like everything else that quantum field theory is based on, just plain claptrap. The momentum CAN NEVER BE A conservative vector field. Thus to equate momentum with the gradient of any function violates the most basic tenets of vector calculus. Stop mixing in Lagrangians with wave-fields. For a wave-field the momentum can only be perpendicular to a moving wave-front, and the phase of the wave only changes linearly along this direction. If you follow the nonsensical path Stone and von Neuamn followed, you are not working with harmonic waves, but with virtual reality.
Quasi_Intellectual
5 / 5 (4) Jun 21, 2010
A Higgs boson particle flies into a church...
Why, you ask?
*Drum-roll*
...
"Can't have mass without it."
*Ba dum-dish!*

Okay, one more:

To get to the other side!
Why did the tachyon cross the road?
*Ba dum-dish*

Thank you, thank you! I'll be here all week!
Bloodoflamb
3 / 5 (2) Jun 21, 2010
No matter what you'd like to say, the momentum operator is the linear operator on the Hilbert Space of states that produces translation. Period. End of story. Taking the x momentum operator into a spatial representation yields (hbar/i)*(d/dx). Don't like it? Tough! Fact is fact.

I don't care how much more fundamental you THINK your view of physics is. Fact is, you cannot prove that there is anything fundamentally incorrect in asserting that an electron is not purely a wave. In addition, you fundamentally cannot provide a physical reason for transition selection rules in the absence of assigning angular momentum to the electron as well as to the photon. You also cannot have photons without your abhorred quantum field theory and you cannot (in Maxwell's theory of electromagnetism) have light whose energy depends on the frequency - only on the intensity.

So, sit there and flounder in the stagnant physics of the 19th century. We'll move on without you.
Skeptic_Heretic
2.3 / 5 (3) Jun 21, 2010
I posted a link to the Stone-von Neumann theorem because it's not my job to educate you. You should educate yourself. The fact is that momentum is the generator of spatial translations and (via Stone-von Neumann) it can be shown that when working within the spatial representation the derivative operator is identically the momentum (with the appropriate hbar and i).

You have a pretty serious problem here. Neumann incorrectly assumes that Bohr's model of the atom is absolutely correct, which we know it is not. Who's talking 19th century now?
Bloodoflamb
5 / 5 (1) Jun 21, 2010
You have a pretty serious problem here. Neumann incorrectly assumes that Bohr's model of the atom is absolutely correct, which we know it is not. Who's talking 19th century now?
The Stone-von Neumann theorem has nothing to do with any specific physical situation. It has to do with the uniqueness of the commutation relations as well as the uniqueness of representation of linear operators.
Skeptic_Heretic
3 / 5 (2) Jun 21, 2010
The Stone-von Neumann theorem has nothing to do with any specific physical situation. It has to do with the uniqueness of the commutation relations as well as the uniqueness of representation of linear operators.
And how does that have nothing to do with a physical represntation composed entirely of particles (*the Bohr Interp) as opposed to standing wave mechanics? Not trying to disparage, I'm curious as to how you see this. Apologies for the 1 rank above, most people copy paste junk from their textbook as opposed to actually writing it out. I improperly prejudged your response.
Bloodoflamb
5 / 5 (1) Jun 21, 2010
The so-called standing wavefunction solutions about the hydrogen's proton are completely determined via the Schrodinger equation (or more exactly, the Dirac equation). Given that the wavefunctions are, by definition, eigenfunctions of the Hamiltonian, if we decide that standing waves must have zero momentum, then it must also be the case that the momentum operator must act as the annihilation operator on the wave function. The fact is that it does not. Every single hydrogenic electron wavefunction is a non-trivial eigenvector of the momentum operator. From this, it follows that the electron wave function DOES indeed have momentum.
Skeptic_Heretic
3 / 5 (2) Jun 21, 2010
Ok, but you're performing math that isn't relational to the actual atom itself, in order to solve the Dirac equation you have to put boundary conditions on the structure which is contrary to the HUP, this is my disagreement.
johanfprins
1 / 5 (1) Jun 21, 2010
You also cannot have photons without your abhorred quantum field theory and you cannot (in Maxwell's theory of electromagnetism) have light whose energy depends on the frequency - only on the intensity.

Any wave has a TOTAL energy which is proportional to the product of it frequency with its maximum amplitude (its intensity). This is simple high school physics! Planck's relationship for minimum energy thus limits the intensity of a light wave.

Momentum can NEVER form a conservative vector field: In contrast the result you quote DEMANDS that it must do so.

Both a photon and an electron are modelled by a harmonic wave-equation and such a wave CANNOT have momentum when it is a standing wave; as an electron around a nucleus MUST be.

The light waves within a black-body cavity are standing waves. The photons emitted by the walls entangle to form standing waves: This is demanded by the boundary conditions.

Ignoring boundary conditions as is done in QFT=voodoo.
johanfprins
1 / 5 (1) Jun 21, 2010
Given that the wavefunctions are, by definition, eigenfunctions of the Hamiltonian,


This is where the problem started. De Broglie's relationship has been assumed to be generally valid for an electron whether it is moving or not. The fact is that it is ONLY valid for an electron with momentum and such an electron MUST be modelled by a RUNNING wave. The Schroedinger equation only gives solutions within an electron's inertial refrence frame: NO RUNNING=NO MOMENTUM and NO de Broglie relationship.
You should note that the ONLY experimental proof for de Broglie's relationship is for moving electrons which diffract. It is not valid for a stationary electron since this will require a wavelength of infinte length for the electron. Although Dirac might find it acceptable, I do not; since I am trying to practice physics not "beautiful mathematis".
Yes
not rated yet Jun 21, 2010
Where I feel that the math screws up the physics is in the simple case of the lambda long antenna where we imagine the B field as a circle and the Efield as coming out of the antenna perpendicular to the Bfield. When the photons are "formed", then mysteriously the B field "circle" is cut into snippers to accomodate for the many photons going into all directions.

So the B-field lines should be imagined maybe parallel to the E-field lines, where lambda long pairs form photons. Then I can imagine the virtual scissors.
Unfortunately though mathematically possible, it is very unpractical for the math, even if in the real world it is so.

So we accomodate for math and then maybe we draw wrong conclusions when we imagine things how the math presents it to us.
JIMBO
not rated yet Jun 22, 2010
QUIT REFERRING TO THE HIGGS BOSON AS THE GODDAMN "GOD PARTICLE" !!!
THAT WAS LEON LEDERMAN'S ABSURD MONIKER, FOR WHICH HE HAS APOLOGIZED EVER SINCE. IT HAS NOTHING TO DO WITH A DEITY, JUST BOOK SALES. WRITE LIKE A SCIENCE WRITER, & DISPENSE WITH THE BS !!!!
johanfprins
1 / 5 (1) Jun 22, 2010
@Yes
Correct: A light-wave with E and B vectors doing harmonic vibrations is always emitted as per Maxwell's equations. Its size and shape of the wave is determined by the source (antenna).

ALL that Planck REALLY postulated is that a light wave with frequency (nu) can never have energy that is less than h(nu). This does not mean that a light wave with a definite frequency (nu) but with a larger total energy consist of separate photons.

It can, however, disentangle into waves with lower energies when it encounters appropriate boundary conditions which force it to do so.

In a black-body cavity the sources in the walls can only emit light-waves with the minimum energies h(nu). These light waves (or photons) encounter new boundary conditions (volume of the cavity): They inflate and entangle to form a single standing wave; as demanded by the boundary conditions. Within the cavity there are NO "photons" moving around like particles. There is thus NO wave-particle duality!!
corticalchaos
not rated yet Jun 22, 2010
it would be funny if one day it was confirmed that gravity was simply the result of a multitude of +/- electromagnetic forces acting together to attract all other collections of electromagnetic forces
johanfprins
1 / 5 (1) Jun 22, 2010
it would be funny if one day it was confirmed that gravity was simply the result of a multitude of +/- electromagnetic forces acting together to attract all other collections of electromagnetic forces

IN ESSENCE YOU ARE CORRECT: The intensity of a matter wave is NOT a probability-distribution but the mass-energy AND gravity field around this mass-distribution. In our presently accepted scientific literature this surrounding gravity field is interpreted as "tunnelling tails". This is REALLY FUNNY!!
Yes
not rated yet Jun 22, 2010
Somehow I do feel that the photons are individual h*nu "waveparticles" that can interfere with eachother to appear bigger maybe.
Standing waves need some kind of synchronization to become standing.
Entangled photons need to be 90 degrees polarized.
Do you think the wall absorbs one photon and then emits two entangled? That still makes two photons and not one with higher amplitude.
Maybe you should think one more bit about this issue.
Bloodoflamb
not rated yet Jun 22, 2010
Hydrogenic electron wavefunctions no more have identically zero momentum than does a superposition of energy eigenstates have a definite energy.
Bloodoflamb
not rated yet Jun 22, 2010
Ok, but you're performing math that isn't relational to the actual atom itself, in order to solve the Dirac equation you have to put boundary conditions on the structure which is contrary to the HUP, this is my disagreement.
[x,p]=i*hbar holds ALWAYS. The particular basis you work in does not matter.
Skeptic_Heretic
3.3 / 5 (3) Jun 22, 2010
The particular basis you work in does not matter.
It absolutely does matter. If electrons are not particles the equation doesn't hold. If electrons are orbiting particles you need to account for synch radiation which you cannot with the Planck constant.
johanfprins
1 / 5 (1) Jun 22, 2010
@Yes,
Any wave requires a specification of the source emitting it, and after having been emiitted a specification of the boundary conditions that it finds itself in. If you ignore the latter you are doing vodoo. Although the light waves being emitted and absorbed by the walls of a black-body cavity can be modelled as so-called "photons", the allowed frequencies with which they can be emitted are determined by the size of the cavity. This proves that such a light wave must immediately inflate to fulfill the boundary conditions. If it has a frequency which is NOT allowed it is not emitted because it does NOT resonate with the cavity.
@Bloodoflam
[x,p]=i*hbar holds ALWAYS. The particular basis you work in does not matter.

What holds for ANY harmonic wave with a complex amplitude is [x,k]=i. Planck's constant hbar and de Broglie's relationship has NOTHING to do with it! You will, hopefully, notice that without hbar k IS NOT the momentum of a "particle" at all.
Bloodoflamb
not rated yet Jun 22, 2010
The particular basis you work in does not matter.
It absolutely does matter. If electrons are not particles the equation doesn't hold. If electrons are orbiting particles you need to account for synch radiation which you cannot with the Planck constant.
Er... Excuse me? I'm sorry, but there's really no debate here. I can prove the canonical commutation relation using an arbitrary wavefunction, i.e. never writing down a particular basis.
johanfprins
1 / 5 (1) Jun 23, 2010
Er... Excuse me? I'm sorry, but there's really no debate here. I can prove the canonical commutation relation using an arbitrary wavefunction, i.e. never writing down a particular basis.

Exactly. This is so since the commutation relationship is between x and k and therefore it is valid for an arbitrary wavefunction. What you do is to multiply both sides of this relationship with hbar and then claiming that it is a relationship between the position and the momentum of a "particle". This is obvious nonsense since the multiplication of any mathematical expression with a suitable constant can then be used to change the interpretation of the relationship.

Just answer a simple question: Is the mass of an "electron" bound to a nucleus less than its rest mass or not?
Bloodoflamb
not rated yet Jun 23, 2010
Again, despite your silly objection of 'that's Lagrangian mechanics' (from which you can derive Hamiltonian mechanics), the momentum is the generator of translations, which can be shown to be identically (hbar/i)*(d/dx).

Rest mass energy is simply an additive constant in the Hamiltonian - one which we can ignore since energy is only DEFINED up to an additive constant and only energy DIFFERENCES are physical.
johanfprins
1 / 5 (1) Jun 23, 2010
Rest mass energy is simply an additive constant in the Hamiltonian - one which we can ignore since energy is only DEFINED up to an additive constant and only energy DIFFERENCES are physical.


WRONG! I know that this is what has been believed by physicists for nearly 100 years since it is the case for a gravitational field. It is, however, NOT the case for an electric-field. This is so since the mass of an electron is itself electric-energy: Not in the stupid manner that it is included when doing QFT, but by taking cognisance of the fact that a SOLITARY electron has NO electric-field energy in the space AROUND it: Its MASS IS its TOTAL electric-energy. Thus a bound electron has a mass-energy which is LESS than its rest mass energy and THEREFORE according to relativity it can have NO MOMENTUM. Whatever you want to believe mathematically: You cannot change physics! Similarly for nucleons: If they were not bonded in this way, we would never have had nuclear energy and atom bombs
Bloodoflamb
5 / 5 (2) Jun 23, 2010
You do not believe in quarks, if I remember correctly. If electrostatic self-interaction is the mechanism by which the electron obtains its rest mass energy, how does a neutron obtain its rest mass energy? How does a neutrino obtain its rest mass energy?

As for nucleons: how does a neutron couple to a proton via electromagnetic interactions? You cannot believe in the strong force if you do not believe in quarks as the mediator of the strong force couples only to quarks - not directly to (the composite system of) a proton or neutron.
johanfprins
1 / 5 (1) Jun 23, 2010
@Bloodoflam

I notice that you do not answer my questions!

I HAVE NOT used the term "electrostatic self-energy". The latter is BS. What I have said is that the boundary conditions which causes a solitary electron to have inertia MUST cause its intensity (since it is a wave) to be equal to its its mass-energy. Similarly for a proton. A neutron is an entanglement of a proton, electron and neutrino. The neutrino is not yet understood by anybody: It might be light energy with a magnetic-component which you will call a "spin" of 1/2. Can you prove it has mass?
As for nucleons: how does a neutron couple to a proton via electromagnetic interactions?

Probably similar to the idea Yukawa has had originally. If they resonate mass-energy, they interact with EM energy.

Again: Just answer a simple question: Is the mass of an "electron" bound to a nucleus less than its rest mass or not?
yyz
5 / 5 (1) Jun 23, 2010
"The neutrino is not yet understood by anybody....Can you prove it has mass?"

Gee, neutrino oscillations were discovered at Super-Kamiokande way back in 1998. Recent work has derived the most accurate value yet for neutrino mass: http://www.physor...658.html

Do you have a problem with neutrino flavor oscillations and quarks?

johanfprins
1 / 5 (1) Jun 23, 2010
Do you have a problem with neutrino flavor oscillations and quarks?

Yes I do: But we are now veering off from the original simple argument which requires to be settled before discussing FAR more complex matters like nuclear "particles" and their properties. I believe that we do not yet understand the latter processes whatsoever; and I do not want to be drawn into a discussion that should remain open-ended.

The question we have been arguing about is whether an electron around a nucleus has momentum or not. And the simple question I have asked is whether the energy of such an electron is less than its rest mass energy.
The only reason why I mentioned nucleons is the fact that the sum-total of their masses when they do not form a nucleus is MORE than the mass of the nucleus (unless the nucleus is radio-active). And therefore I repeat: Is the mass of an electron around a nucleus less than its rest mass or not?

How about a simple YES or NO?

Bloodoflamb
not rated yet Jun 23, 2010
No, it is not.

Does the electron have less energy than if it were unbounded? Yes. That's the definition of a bound state. It has an average potential energy equal to twice the energy level and an average kinetic energy equal to the absolute value of the energy level.

Consider a superposition of the 1s and 2s states of an electron bound to a proton:

|psi> = (1/Sqrt[2])(|1,0,0>+|2,0,0>)

Where |n,l,m> are the standard hydrogenic wavefunctions whose quantum numbers are n, l and m. You contend that all of these states have zero momentum, yes?
Skeptic_Heretic
3 / 5 (2) Jun 23, 2010
Ok, now we're at a stage where I'm intrigued by what will come of the conversation. BoL and JfP are both well studied, this should yield interesting commentary.
johanfprins
1 / 5 (1) Jun 24, 2010
I am omly responding now: My internet connection was down:
It has an average potential energy equal to twice the energy level and an average kinetic energy equal to the absolute value of the energy level.

From where do you measure this potential energy and from where this kinetic energy?

Prove to me that a 1s and a 2s state to a proton can be superposed: To form what?

Lastly, plot the actual potential energy of the charge on the electron from infinity to zero. Now draw a horizontal energy line from the value at infinity to the zero position. What is the electron's energy at infinity relative to the same inertial reference frame as the proton? It is obviously the electron's rest mass.

Now place an "electron" at this energy at position zero: It will decay to the 1s state and in the process emit em-radiation. This em radiation comes from the electron's mass-energy so that it ends up with a lower energy which is also totally mass-energy.

Continue
johanfprins
3 / 5 (2) Jun 24, 2010
Consider an electron moving with a speed v: According to Einstein its kinetic energy increases its mass-energy so that its total energy is E=mc^2.
Now apply a force and decelerate this electron. It transmits EM radiation which comes from the decrease in its mass. When the speed becomes zero, the electron is being accelerated into the opposite direction it now gains mass BUT also emits radiation. If it did not emit radiation its mass would have increased faster. Thus again the EM radiation is converted from mass energy.

So when an electron decays from a higher to a lower energy standing energy state, some of its mass converts to EM radiation. Thus when starting off as a stationary state having energy equal to rest mass (see post above), the stationary state into which it decays must have a lower mass energy. If it has kinetic energy which is not co-linear it will keep on radiating EM radiation. This is why the electron cannot have any momentum AROUND the nucleus.

Does this help?
Yes
not rated yet Jun 24, 2010
Strange.
Suppose I am going with your moving electron at speed v. For me the energy of the electron is its restmass*v^2 cause it does not move as far as I can see.
Now you in your frame do the deceleration and see your supposed photon. Now according to what you say, I who was going with the photon see an acceleration and the emission of EM (not the same EM as you see though). So is the electron now weighing less then its restmass for me while increasing speed????
Maybe the EM is produced by something else?
johanfprins
1 / 5 (1) Jun 25, 2010
Suppose I am going with your moving electron at speed v. For me the energy of the electron is its restmass*v^2 cause it does not move as far as I can see.

When it accelerates it moves relative to both reference frames: It cannot then be stationary relative to ANY inertial frame.
First consider a "neutral electron": Relative to both reference frames its total energy changes by the same amount. If not, Einstein's theory of relativity would be wrong and energy conservation violated!
Now with charge: In this case the electron does not conserve the same potential and kinetic energy as in the case of a "neutral electron". It also dissipates EM energy: Again by the same amount relative to each refrence frame: That is why the speed of light must be the same in both reference frames. But the fact is that in this case the increase in rest mass that would have been there for no charge, has been converted into EM radiation.
Continue
johanfprins
1 / 5 (1) Jun 25, 2010
I who was going with the photon see an acceleration and the emission of EM (not the same EM as you see though). So is the electron now weighing less then its restmass for me while increasing speed????

This is obviously not possible for an electron with momentum since having momentum it must have a mass which is larger than its rest mass. An electron can only decrease its mass to be less than its rest mass when the boundary conditions are such that it has to lower its energy WITHOUT HAVING MOMENTUM. This happens for electrons around a nucleus or else they will not be bound electrons. An electron around the nucleus with its total energy equal to its rest mass has the same energy as a free electron which is stationary relative to the nucleus: This energy is too high to form a bound state with the energies we are measuring.
johanfprins
1 / 5 (1) Jun 25, 2010
CHARGE-CARRIERS IN METALS:
I was hoping for the following question: Since it has not yet appeared, I will ask and answer it myself.

Question: If bound-electrons have energy less than their rest mass energy, how can charge-carriers move through a metal?

Answer: Ignoring temperature excitations, there are no "free electrons" within a metal. The electrons form standing waves between the surfaces of the metal. When applying an electric-field, the boundary conditions change: This causes the electron-waves to superpose and form wave-packets, which are not normally within the metal.

As can be found in any text book on Solid State Physics, it is the latter pseudoparticles which transport charge. When switching off the electric-field the wave packets spread to again end up as standing waves.

Thus to state that there are "free electrons" within a block of metal is physics claptrap. How about proposing an experiment to prove that the latter statement is wrong!
johanfprins
Jun 26, 2010
This comment has been removed by a moderator.
Yes
not rated yet Jun 28, 2010
How about a spark in vacuum?
johanfprins
1 / 5 (1) Jun 28, 2010
How about a spark in vacuum?

A "spark" in vacuum? What do you mean by this statement? How do you generate a spark in vacuum. "Sparks" are only generated in a "BAD" vacuum: One then generates a small plasma.

Please define what you mean by a "spark in vacuum". Only then will I be able to answer your question.
Ethelred
1 / 5 (1) Jun 28, 2010
At a guess he might be thinking of electrons, plural, crossing from the cathode to an anode in a vacuum.

Your explanation of momenta for a electrons now seems to make more sense to me. Specifically why you say that electrons don't have momentum when their mass is below rest mass. Though it seems a bit ass backwards to put it that way, in that I see electrons as not having momentum simply because they aren't moving, with respect to the atom, and their mass being lower because of the bound state in an atom.

Regarding your thinking that neutrinos don't have mass. The evidence for flavor oscillation is pretty strong at the moment which may not qualify as proof for you but it has reached the point where it should be taken into consideration.

Ethelred
frajo
1 / 5 (1) Jun 28, 2010
The evidence for flavor oscillation is pretty strong at the moment which may not qualify as proof for you but it has reached the point where it should be taken into consideration.
Which means that the standard model needs an update.
Ethelred
1 / 5 (1) Jun 28, 2010
Good, its a bit of a mess.

Though I think the idea that a replacement should be able to predict particle mass or pretty much any other constant is destined to failure.

Ethelred
johanfprins
1 / 5 (1) Jun 28, 2010
Though it seems a bit ass backwards to put it that way,..

The way you put it is also correct; since we know that any bound quantum mechanical "particle" (I hate the latter word) must have less mass than its rest mass. I have been arguing from Einstein's relativity which tells one that a moving "particle" must have a total mass-energy that is more than its rest mass; and since an electron with energy equal to its rest mass has a far higher energy than the electrons bound to the nucleus, this demands that they have energies less than their rest mass. I appreciate it that you agree that an electron around a nucleus cannot have momentum.
The evidence for flavor oscillation is pretty strong at the moment ...

I am afraid that I am highly skeptical of the standard model, but also not well-read enough to engage you in an argument about "flavours" at this point in time. So far I find it compelling to conclude that ALL physics based on QFT must be fatally flawed.
frajo
1 / 5 (1) Jun 28, 2010
Good, its a bit of a mess.
Thanks; I wouldn't dare say so.
Though I think the idea that a replacement should be able to predict particle mass or pretty much any other constant is destined to failure.
Einstein's GR didn't replace Newton's gravity neither. Thus any enhancement which encompasses the standard model as special case will do.
Ethelred
not rated yet Jun 29, 2010
Thanks; I wouldn't dare say so.
It is the Higgs that is bothering me the most at the moment. I don't see why there is any need for a specific particle to produce the mass of the other particles. With a different way to produce the mass of leptons and quarks is even more odd especially since all particles or waves have masses that equal their energy. IF there was difference in the mass energy equivalence between particles THEN it would make sense. And overturn GR which looks pretty solid.
Thus any enhancement which encompasses the standard model as special case will do.
I agree but I keep seeing real physicists expressing a desire for a:

Single Theory Which Can Produce Only One Kind Of Universe.

My point being based on the question:

Why something rather then nothing?

My thinking is there is something simply because its possible and thus must be. Since MANY different ways are possible then why insist on just one?

Ethelred
johanfprins
1 / 5 (1) Jun 29, 2010
It is the Higgs that is bothering me the most at the moment

The Higg's mechanism is based on Nambu's so-called broken symmetry concept which was inspired by the BCS model for superconduction which, in turn, is just pure poppycock. If they find the Higg's boson I will remain just as skeptical as I am about the vector bosons.
My thinking is there is something simply because its possible and thus must be. Since MANY different ways are possible then why insist on just one?

How do you know that there are many different ways possible? What you can say is that scientists are able to model the same physics in different ways. But this does not mean that all of these models are correct. It is for this reason that we must remain vigilant and not become bigots who are just blindly defending mainstream dogma: When this happens, as at present, it spells the end of physics unless we fearlessly argue new ideas and design experiments to be the final arbiters.
Yes
not rated yet Jun 29, 2010
When this happens, as at present, it spells the end of physics unless we fearlessly argue new ideas and design experiments to be the final arbiters.

There is need for clans and wars then, because I see the attitude of the regular defenders of their theories and findings not capable of setting up teams and matches.
And where do I find an objective arbiter?
The winner is the one that produces usable results.
Anyhow in this individualized world who can make a team? Everybody has his own theories that he defends jealously. I will start listening to the guy who starts supplying Amsterdam or Johannesburg with electricity from his 100 square feet backyard power plant as a result of his theories.
I might even want to buy his energy:)

As far as johanfprins with words like claptrap popyrock and some denigrating language, he has an attitude. I hope you are younger than 25, if else I wonder who wants to work with you.

I always keep saying:
"There is no such thing as a Higgs Boson."
johanfprins
1 / 5 (1) Jun 30, 2010
As far as johanfprins with words like claptrap popyrock and some denigrating language, he has an attitude.

I am 68: Yes I have developed an attitude. If you find that even those in charge of physics departments cannot (do not want to?) even understand high school physics it drives one insane: For eaxmple, Ohm's law is ONLY valid for non-zero resistivities: It only models charge-carriers when they are moving as if they have an average drift speed. It CANNOT define zero resistivity. If you can bring me a single chairman of a physics department in the world who can understand this simple fact, I will work harder on my attitude. For 10 years I have not been able to find such a person. Zero resistivity has NEVER been defined and therefore Onnes made a massive blunder when he plotted his result on superconducting mercury as resistance versus temperature. He assumed that zero voltage defined zero resistivity a la Ohm. This is NOT possible and has led to impossible models a la BCS.
johanfprins
1 / 5 (1) Jun 30, 2010
And where do I find an objective arbiter?

This the problem at present. We do not have physicists who are willing to abide by the rules that had been formulated at the founding of the Royal Society of London. You do not even find them wirthin the Royal Society anymore. See next posting

continued

johanfprins
1 / 5 (1) Jun 30, 2010
The winner is the one that produces usable results.

Supposedly so; but not anymore. When usable results are blocked from being published the end of physics is near.

Consider again superconduction: The defining fact is that as soon as it sets in, the electric-field within the material is cancelled. Why and how this happens has NEVER been explained. Just try and point this incontrovertible fact out in our "peer reviewed" literature and propose an explanation. No ways are you allowed to do so.

A referee at the Royal Society rejected such a manuscript by stating: "The description of the electromagnetism of superconductors is remarkably complicated....Being not an expert I can't say how this formulism explains the cancellation of a static electric field, but I would be very surprised if it really can't do so".

Holy Jehosepath: RoySoc asks non-experts to be referees! Can it REALLY become any worse?

Is it possible NOT to develop an attitude?

Ethelred
not rated yet Jun 30, 2010
How do you know that there are many different ways possible?
There is rather a lot math out there. I don't claim to know how many but since many systems of math and logic are self consistent it is reasonable to assume the many of them could produce a viable universe. If nothing else small differences in timing in our chaotic universe could have failed to produce our world.
What you can say is that scientists are able to model the same physics in different ways.
At least within the bounds of error at the moment yes.
But this does not mean that all of these models are correct.
For this universe certainly. And by changing just any one variable the universe would be very different. Possibly unable to support life of any kind. However if you change MORE than one balance can be achieved and life should be able to exist. Not the same life but life of some kind.

Yes this is philosophy but so is the idea that there can only be one answer that supports life.

Ethelred
johanfprins
1 / 5 (1) Jun 30, 2010
There is rather a lot math out there. ... and logic are self consistent it is reasonable to assume the many of them could produce a viable universe

When we have more than one model one should accept that any one of them could be correct, but there is no proof that more than one of them can be correct. It most probably just means that what we think is "self-consistent" is just plain wrong.

Examples from superconduction:
It is claimed that the quantum of flux trapped through a superconducting ring is consistent with doubly-charged charge-carriers. The fact is, however, that the "phase-angle" model is not viable physics at all.
In the case of radiation generated at a Josephson junction, the voltage over the junction causes an offset of qV; and a charge-carrier through the junction is accelerated by qV. Thus the energy per charge-carrier which must be radiated away is 2qV; Experimentally it is found that 2q=2e: Thus q=e. Josephsen's wrong mathematics claims that q=2e.
Ethelred
1 / 5 (1) Jul 01, 2010
When we have more than one model one should accept that any one of them could be correct, but there is no proof that more than one of them can be correct


My point is that more than one COULD BE mathematically correct and thus there could be a universe that has one and another that has the other. As both are correct there would not be a reason for them not existing.

As I said this was a matter of philosophy. The philosophy that there may be other universes with other laws as long as they are self-consistent. If so theory will not tell us how the Universe works. Only experimentation could tell which Universe a person is in.

For instance, both Einstein and Newton came up with mathematically consistent theories. But only could be correct. And yes I do know that Newton's theory was not compatible with Maxwell's equations. Its just an example.

Ethelred
johanfprins
1 / 5 (1) Jul 01, 2010
For some reason I cannot connect from my computer to physorg. I am slipping in this meassage in on another computer. As soon as I can restore contact, I will be back. It is absolutely imperative for the future of physics to stop clowns like VestaR etc. by stating logic. As soon as I can make contact again, I will be back.
Ethelred
1 / 5 (1) Jul 02, 2010
We know, relativity is incompatible with quantum mechanics.


We know no such thing. YOU think it.

Some others think that relativity should be quantized and some have made progress toward that goal. However I see no reason that gravity must be quantized. Desirable yes. But wanting something doesn't make it true. So far it is GR that has the more solid experimental foundation. Hasn't had one failure yet. Hard to beat 100% success.

For instance, since GR has gravity functioning by warping space-time just where is a need for a graviton to do this?. I mean a need besides somebody wanting it to be so.

Ethelred
frajo
2.3 / 5 (3) Jul 02, 2010
So far it is GR that has the more solid experimental foundation. Hasn't had one failure yet.
At least as long as you don't ponder over subtleties like the "missing" galactic masses or the Pioneer anomaly.
johanfprins
1 / 5 (1) Jul 02, 2010
For instance, since GR has gravity functioning by warping space-time just where is a need for a graviton to do this?. I mean a need besides somebody wanting it to be so.

The worst is that matter-waves already encompasses GR, since such a wave's intensity is already commensurate with mass and space-curvature around it. Why require "particles" like Higg's bosons, gravitons, axion's, WIMPS and what have you: If it were not of such importance one would have laughed all the time!
johanfprins
3 / 5 (2) Jul 05, 2010
At least as long as you don't ponder over subtleties like the "missing" galactic masses or the Pioneer anomaly. ]

There is a possible explanation for these "anomalies" in terms of waves ONLY. A matter wave is a region of space filled with distributed mass-energy. Thus some matter-waves might be parsecs large.

In fact, one can argue that it is probably such a single matter wave which "quantum jumped" (inflatted) into a larger matter wave during the Big Bang. All other "observable matter" and light then "preciptated" from this original single wave-entity when it became large enough.

Thus it it is possible that there could be distributed mass energy all around us which we cannot see since we cannot resonate with it. There would then be no missing mass, and I have found that the "Pioneer anomaly" can be explained when taking the gravitational pull of such a surrounding mass-distribution into account.

Interesting is it not?
Ethelred
not rated yet Jul 05, 2010
Interesting is it not?


Yes. But many other explanations are possible. A lot of people think that GR needs some tuning for instance.

The idea of a distributed matter wave may have a problem. Protons for instance seem to be very specific spatially unlike electrons. Muons have a tighter position than electrons. Oh and Muons are real enough that they have been used in Muon catalyzed fusion experiments that actually worked, within the limits of a muons life that is. They don't last long enough to make it a practical method of getting power.

The point being that larger masses have better defined positions than smaller masses. This should hold true for your matter wave ideas as it seems to fit the evidence. I just don't see how to get a mass with such vague positioning, at least with the standard particles or waves. Something of the nature of WIMPS would be called for and I thought you didn't like those.

Ethelred
johanfprins
1 / 5 (1) Jul 05, 2010
Yes. But many other explanations are possible. A lot of people think that GR needs some tuning for instance.

I agree: GR does not allow for near-instantaneous collapses and inflations. This might imply that gravity does act "instantaneously" over a distance.

The point being that larger masses have better defined positions than smaller masses.

Not generally true. A Bose-Einstein Condensate is a single holistic wave with distributed mass that has a larger mass than a proton; even though it inhabits a much larger space.

In the case of electrons, muons and protons, the GR boundary conditions probably act like restoring forces (RF). Thus, the possible shapes of the waves might be similar to those for a harmonic oscillator: electron, muon and tau: Gaussian. Proton most probably a state with n=2 and a pion a state with n=1. RF is obviously different in each case since it is determined by GR. One expects asymptotic "binding" force behaviour owing to RF's. No quarks!?
hush1
5 / 5 (1) Jul 05, 2010
In getting ahead of future published works - specifically Johann's works - I wonder if readers benefit from a Index of referenced literature the author finds questionable?

An index where, of course, all literature is referenced, yet tagged by the author as 'Useful-when-aware-of-incompatibilities' or other such helpful hints to rescue susceptible, impressionable, readership, future researchers, and the unsuspecting from expending time that otherwise hampers, rather than enhances one's scientific understanding.

Studying Ptolemy is great for exercising math, yet if taken 'seriously', hampers what all science (idealistically) strives to obtain - objectivity, or strives to avoid - bias. That point was touch on in this thread.

Although such, helpful-author-'hints', in themselves harbor potential bias, the author's own experience, acting in good faith, will rendered more 'good-caution' than 'bad-caution' to readers - the aim of sincere scientists.

Johann, is that something to consider?
johanfprins
1 / 5 (1) Jul 05, 2010
Johann, is that something to consider?

This is a difficult question: I think that we must remember that all physics is ALWAYS subject to questions. And we must consider the possibilty with open minds that what we believe today as "has been proven", might totally collapse tomorrow.

Just a single small fact is needed.: For example, the fact that BCS cannot explain how an applied electric field is cancelled within a superconductor means it cannot be correct!

I think that a study of Ptolemy's model might not "hamper" what we are wanting to obtain. It only hampers physics when such a model is accepted as "INVIOLATE HOLY TRUTH" as it was in the time of Galileo; and is now the case for BCS-model.

In physics we must learn from past mistakes, and therefore it is probably imperative that we study the past models; even those that have been discredited. They may even have been discredited for the wrong reasons: As happened to Schroedingers proposal that ALL consists of waves!

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