beejewel

Spooky idea too late for Halloween

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Continuing on from the Humpty Dumpty thread, where we were discussing the stored potential of matter, and the similarities between the gravitational field and the electrostatic fields, I proposed that the following equation be true;

 

Eq.1

[latex] \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{\phi^2}{\Phi^2}}} [/latex]

 

 

Where [latex]\phi[/latex] represents the observers potential and [latex]\Phi[/latex] represents the potential of the proton.

 

First I want to explain my reason for suggesting this to be true. I am speculatively proposing that the proton-electron are particle-antiparticle dispite their obvious difference in mass, and that the mass difference between the proton-electron is a function your (the observers) potential (this is where it gets SPOOKY.., because changing the observers potential will appear to change the mass ratio of all protons and electrons in the Universe regardless of how far away they are.

 

This rather spooky result comes from the following equation, which shows the electrons potential as a function of the observers potential.

 

Eq.2

[latex] \phi_{e}=\frac{\Phi-\phi_{gnd}}{2}\sqrt{1-\frac{\phi_{gnd}^2}{\Phi^2}} [/latex]

 

Equation 2. says that there are two particles with a combined potential (mass) of [latex]\Phi[/latex] and that their respective masses are a function of the potential [latex]\phi_{gnd}[/latex], the assumption is that ground potential is falling with passing time, so when ground potential falls to half proton potential the particles have the same mass, and presumably annihilate.

 

So why do I refer to potential instead of simply referring to the electron and protons mass?

 

My reasoning here is that the the potential in a proton-electron pair is the stored energy that went into creating the electron and the proton, and that this electrical potential between the electron-proton is the same as it's mass divided by the number of nucleons, in this case 1. So if the proton has a mass of 938 MeV * c^2, then we can legitimately drop the c's and drop the e's, and express the internal potential as simply 938 million volts.

 

Recent spectroscopic experiments have shown that the mass ratio of the proton to electron is invariant to a high degree of accuracy everywhere in the visible universe, which is expected when observations are being made from ground potential.

 

This suggests what we already know, that by changing our own potential, we change the world around us, Schroedingers cat can indeed be dead and alive, because it is our act of observing it which determines it's state.

 

Steven

Edited by beejewel

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But there is no 'absolute' EM potential.

It is a gauge field, and the universe works the same if everything is at zero potential and the electron is at one, or if everything is at 1000 eV and the electron is at 1001.

 

Or am I misunderstanding you ?

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But there is no 'absolute' EM potential. It is a gauge field, and the universe works the same if everything is at zero potential and the electron is at one, or if everything is at 1000 eV and the electron is at 1001.

Or am I misunderstanding you ?

 

I agree in theory it is arbitrary where one sets the zero point, but as far as the real world goes, it sure looks to me, as if we live within a limited potential domain, where 938 million volts is the ceiling, and 469 million volts is the "bye bye see you later" point.

 

The passage of time being the same as a drop in potential. If my hunch is correct, then we have fallen through a potential of around 8 million volts since the beginning of time, which works out to around 0.5 mV per year (That's if Google knows how old the Universe is).

 

Steven

Edited by beejewel

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[latex] \gamma=\frac{1}{\sqrt{1-\frac{v_2}{c_2}}}=\frac{1}{\sqrt{1-\frac{\phi^2}{\Phi^2}}} [/latex]

 

Just by looking at the equation I can already tell you that if it's true for all v, φ, and Φ, then it must necessarily be the case that v=±kφ and c=±kΦ. So effectively all you've done is introduce two new variables that contain the exact same information. I fail to see how that's useful. It doesn't magically become an electrostatic potential just because you've decided to label it with a Greek letter.

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It doesn't magically become an electrostatic potential just because you've decided to label it with a Greek letter.

 

Solving the equation for velocity becomes;

 

[latex]v=c(\frac{\phi}{\Phi})[/latex]

 

What this tells us is, when the observers potential is the same as the proton potential, we are in fact moving along the time dimension with velocity c, and as time passes and the observers potential falls, the velocity in time diminishes as per the function and the velocity in the x, y, z dimensions increase.

 

We can now calculate the relative velocity between two bodies by knowing their potential difference.

 

[latex]v_{rel}=c(\frac{\Delta\phi}{\Phi})[/latex]

 

Take for excample the electron at 511 keV, and the above equation tells us that the electron must be moving at 99.1% the speed of light, and it works just as well for macroscopic objects at lower speeds, try it for yourself ;)

 

 

 

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I'm telling you that you can't just decide to call a new variable "the potential." All you did was introduce a new variable. It has nothing to do with any kind of potential.

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I'm telling you that you can't just decide to call a new variable "the potential." All you did was introduce a new variable. It has nothing to do with any kind of potential.

 

Well why don't you just call it mass per nucleon?

 

So you could say there is a 1:1 relationship between the mass per nucleon and velocity as a fraction of c.

 

The theory can make predictions and it seems to work out right, ... just luck?

 

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Well why don't you just call it mass per nucleon?

 

So you could say there is a 1:1 relationship between the mass per nucleon and velocity as a fraction of c.

 

The theory can make predictions and it seems to work out right, ... just luck?

 

 

All you did was take the Lorentz factor and substitute in some different symbols. It has absolutely nothing to do with masses, nucleons, or potentials. You don't just get to insert a different symbol that means the same thing and call it a potential. It's frustrating that you're trying so hard not to understand this.

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It's frustrating that you're trying so hard not to understand this.

 

As the late Richard feynman said in one of his online lectures, "The way we come up with a new theory is we guess, then we test it against nature and if it works we call it a theory and if it doesn't its just WRONG" (something along those lines anyway)

 

Steven

 

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As the late Richard feynman said in one of his online lectures, "The way we come up with a new theory is we guess, then we test it against nature and if it works we call it a theory and if it doesn't its just WRONG" (something along those lines anyway)

 

Steven

 

 

What does that have to do with anything? You didn't come up with a theory, you came up with new symbol for velocity.

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You didn't come up with a theory, you came up with new symbol for velocity.

 

Yes, we can agree on that, [latex]m=v=\phi[/latex] , shame we can't agree that the proton and electron are a particle pair.

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First off the electron and proton cannot be particle pairs for more reasons than just its mass

 

the electron is part of the lepton family its interactions as such does not include the strong force. The electrons anti particle is the positron.

The electron is an elementary particle the best research today shows it has no subsystem. It is not made up of quarks.

 

The proton is a member of the hadron family. It is made up of quarks. Specifically 2 up and 1 down. Its interactions is all 4 forces. 99% of its rest mass is the strong force and kinetic energy of the quarks in its make up.

 

your formula also makes little sense as your using the designated symbol for the photon on the left

 

 

What precisely do you mean by proton potential its rest mass? Or observer potential?

Forgot to add the proton has an anti particle the anti proton. Both are generated everyday at LHCs so we know without a doubt they exist.

Elfmotat already pointed out the issue of merely changing symbols others have pointed out the other problems in the Humpty thread.

 

Now what constitutes a particle anti particle pair.

 

All properties of a particle and its anti particle MUST be the same except its charge.

Charge being its color charge but includes it electromagnetic charge

 

so in the case of the proton the anti proton has two up antiquarks and 1 down antiquark

 

http://en.m.wikipedia.org/wiki/Color_charge

In other words the must have the same mass, spin and be part of the same family of particles. As well as have the same interactions. The only similarity between the two is they are both 1/2 spin and have opposite electromagnetic charges.

Edited by Mordred

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Yes, we can agree on that, [latex]m=v=\phi[/latex] , shame we can't agree that the proton and electron are a particle pair.

 

Where did you get m from? Are you just giving velocity more and more symbols? Define what you mean by "particle pair," because that could mean a number of different things.

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First off the electron and proton cannot be particle pairs for more reasons than just its mass, the electron is part of the lepton family its interactions as such does not include the strong force. The electrons anti particle is the positron. The electron is an elementary particle the best research today shows it has no subsystem. It is not made up of quarks.

 

Yep, I read this on WIKI too.

 

The proton is a member of the hadron family. It is made up of quarks. Specifically 2 up and 1 down. Its interactions is all 4 forces. 99% of its rest mass is the strong force and kinetic energy of the quarks in its make up.

 

Did you verify this or figure it out from first principles?

 

your formula also makes little sense as your using the designated symbol for the photon on the left

 

I defined it, so it means what it means

 

 

What precisely do you mean by proton potential its rest mass?

 

The protons potential energy in eV divided by the number of nucleons ie. [latex]\frac{939 MeV}{1}[/latex], but as elfmotat has pointed out, it is a ratio so you can call it whatever you like.

 

Or observer potential?

 

Observers potential also referred to as ground potential, usually set to be zero in classical physics, but in my theory an observer has an upper and a lower limit to potential, so ground potential can be expressed as an absolute number, and works out to 930 million volts.

 

 

Forgot to add the proton has an anti particle the anti proton. Both are generated everyday at LHCs so we know without a doubt they exist.

Allthough we live in a domain of limited potential, particles of almost any potential energy can be created anywhere, so by colliding ions with energy twice that of the proton, it is possible to make pairs with 1876 MeV these particles are not a normal part of the furniture in our world, and therefore very short lived. We are made of pairs with a total energy of 938 MeV., and our present conciousness, for reasons I shall not speculate, is at ground potential which is 930 MV. So we are observing the world from an assymmetric point of view.

 

Now what constitutes a particle anti particle pair.

 

It seems rather indisputable that a particle pair is nothing more than a wave, with a peak, a trough, and a baseline which represents the observer. In our case the baseline is heavily biased towards the positive maximum.

 

All properties of a particle and its anti particle MUST be the same except its charge.

 

No

 

 

Charge being its color charge but includes it electromagnetic charge

so in the case of the proton the anti proton has two up antiquarks and 1 down antiquark

 

A sinewave can be disected so that the positive half has an up vector, a peak and a down vector, likewise the bottom half of the wave can have two ups and a neutral trough. A possible reason why we dont see such structure in the electron, is because it's nucleus lies outside the observers domain (in the observers future).

 

The only similarity between the two is they are both 1/2 spin and have opposite electromagnetic charges.

 

It seems likely to me that 1/2 spin particles are just half waves, but that's speculation only.

 

Just think of a Hydrogen atom as a standing wave with a positively shifted baseline, and you see the same picture I see.

 

Later I will start a new thread to explain my understanding of what causes energy to be quantised, but first we have to get over these hurdles.

 

 

 

 

Where did you get m from?

Are you just giving velocity more and more symbols?

Define what you mean by "particle pair," because that could mean a number of different things.

 

The Lorenz factor and my variation of it is a ratio, so as long as the units in the fraction are in the same units it simply doesn't matter, ie you can insert the mass per nucleon of ground potential, divided by the proton mass, if you don't know the mpn at ground potential, insert the rest and solve the equation.

 

As described above, the best way to describe a particle pair is as a wave with a peak and a trough.

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I see so you choose to ignore the related lie algebra including conservation of charge, conservation of spin, conservation of isospin conservation of flavor. The Eightfold wayen as well as the lepton and meson nonet in favor of a personal model which has no body of evidence. What part of we already can produce the antiproton at Fermilabs you didn't understand? Or for that matter the positron?

 

I suggest you buy a copy of Griffiths " Introduction to particle physics" its an excellent book the first chapter covers how particles are discovered and defined. He also covers the items I just posted

http://en.m.wikipedia.org/wiki/Quark_model

I suppose next your going to tell me quarks do not exist and the fact that the proton has quarks where the electron had none is meaningless.

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Did you verify this or figure it out from first principles?

 

!

Moderator Note

I can tell you right now that this approach is not going to fly. Your task here is to defend your idea, and that cannot be accomplished by attacking mainstream physics. You can do one or the other.

 

 

 

The theory can make predictions and it seems to work out right, ... just luck?

 

Have you shown that it works?

 

 

As the late Richard feynman said in one of his online lectures, "The way we come up with a new theory is we guess, then we test it against nature and if it works we call it a theory and if it doesn't its just WRONG" (something along those lines anyway)

 

Feynman was referring to educated guesses, not pulled from your posterior guesses. That is, one tends to derive equations. The "guessing" generally happens earlier in the process.

 

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Moderator,

 

It is far from my intention to argue with the members that the standard model is wrong. I am presenting a view, based on the principles of SR, which seems to suggest that the observers potential is limited. Solving my equation for the mass of the electron and the proton, gives a value for ground potential as follows.

 

http://www.wolframalpha.com/input/?i=0.511%3D%28%28938-x%29%2F2%29*√%281-%28x%5E2%2F938%5E2%29%29

 

It certainly came as no suprise to me that the answer equated to the mass per nucleon of Ni-62 to within 70 KeV, the nucleon with the lowest mpn, it seems reasonable that Ni-62 can not yield any energy by fission or fusion, because it is already at ground potential. All the other elements have higher potentials in their nuclei, and therfore decay towards the lowest point.

 

Further we concider living on a rock made from continuosly decaying isotopes of U, K, Th etc which are decaying below our feet emitting gamma rays, and it seems reasonable to assume that the potential is falling as time passes, ie, ground potential can not be constant.

 

I can go on to show a few more examples, but as a one man show, I can't provide proof of every experiment ever done, all I can do is present the idea which I believe has merit, and see if there are any takers who agree with the view and want to collaborate and look deeper into it.

 

Note, we don't choose our religion based on how many people believe in it, so why do it with science?

Edited by beejewel

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...it seems reasonable that Ni-62 can not yield any energy by fission or fusion, because it is already at ground potential. All the other elements have higher potentials in their nuclei, and therfore decay towards the lowest point.

 

My understanding is that the high binding energy per nucleon means that it is stable. Ni-62 will not undergo any spontaneous decay. This also means that nickel and iron isotopes are often the end products of many nuclear reactions. Stars cannot produce elements past nickel or iron.

 

 

I don't see that this means that we cannot get energy released by fission. This may be difficult to do in practice, but have you found a clear statement or better a calculation that claims this really is the case?

Edited by ajb

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ajb,

 

The chart of nucleides is usually presented with Ni-62 at the top, because it has the highest binding energy (whatever that is). I like to present it the right way around showing Ni-62 at the bottom where I believe it should be. I see both fusion and fission as decay processes.

 

I actually spent quite a bit of time and money building ion sources and particle accellerators in my own workshop, successfully fusing deuterium and producing measurable amount of neutrons. I applied for no less than three fusion related patents, so these ideas were not developed in any orifice as some of the members arrogantly suggest.

 

Regarding your question, I am 100% sure that no fission/fusion energy can be gained from nickel or iron itself, there may be a slight chance of slowing down a proton to the point where it fuses with Nickel to produce Ni-62 but in order to do that it needs to loose around 8 MeV of its kinetic energy, and I can't see any easy way to do this.

 

post-21391-0-66581500-1414923630_thumb.png

 

PS: The protons velocity with respect to an observer at ground is worked out as follows.

 

[latex]v_{p}=c(\frac{\Phi-\phi_{gnd}}{\Phi})[/latex]

 

putting in the numbers...

 

[latex]300,000,000 m/s ((938MV-930MV)/938 MV)=2,558,635 m/s[/latex]

 

This means protons simply don't stand still, when observed from our potential, the only way to slow them down is to capture them in a heavier nucleus, but how do you capture a proton travelling at 2.5 million m/s, hence the impossibly small cross section in p+p fusion (only from our potential). In the early Universe when ground potential was a lot higher, the velocity of the proton would have been significantly less, so p+p fusion would have been common place in the early Universe.

 

It's late here now, so I will no doubt read about how the Universe was much hotter in the past, but that's tomorrow :)

 

 

 

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Regarding your question, I am 100% sure that no fission/fusion energy can be gained from nickel or iron itself,

I think you are right on the fusion aspect. As nickel and iron are at the peak of the curve it will take energy to add more nucleons to them. You are probabily also right with the nickel, you can't push it closer to A= 60 by splitting it apart.

 

But anyway, this has nothing to do with your 'potential' idea. It is all explained using standard physics.

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It also has nothing to do with electrons and protons being particle pairs lol.

 

Binding energy is the amount of energy to break the nucleus up into its constituent components this is basically how stable a neutron proton configuration is. However nuclear fusion fission isn't a field I'm well versed on.

 

You mentioned studying early universe conditions

I recommend this article its one of the best Ive come across for covering nucleosynthesis.

 

http://www.wiese.itp.unibe.ch/lectures/universe.pdf Early universe particle physics

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However nuclear fusion fission isn't a field I'm well versed on.

Me neither, but for sure it is not connected to this 'potential'. It is understood okay within established physics.

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Moderator,

 

It is far from my intention to argue with the members that the standard model is wrong.

It is still out of place to ask if some aspect of it was derived from first principles; that's irrelevant. (also ironic, as you are presenting something not derived apparently from anything at all)

 

 

I am presenting a view, based on the principles of SR, which seems to suggest that the observers potential is limited. Solving my equation for the mass of the electron and the proton, gives a value for ground potential as follows.

 

http://www.wolframalpha.com/input/?i=0.511%3D((938-x)%2F2)*√(1-(x^2%2F938^2))

 

It certainly came as no suprise to me that the answer equated to the mass per nucleon of Ni-62 to within 70 KeV, the nucleon with the lowest mpn, it seems reasonable that Ni-62 can not yield any energy by fission or fusion, because it is already at ground potential. All the other elements have higher potentials in their nuclei, and therfore decay towards the lowest point.

 

Further we concider living on a rock made from continuosly decaying isotopes of U, K, Th etc which are decaying below our feet emitting gamma rays, and it seems reasonable to assume that the potential is falling as time passes, ie, ground potential can not be constant.

 

I can go on to show a few more examples, but as a one man show, I can't provide proof of every experiment ever done, all I can do is present the idea which I believe has merit, and see if there are any takers who agree with the view and want to collaborate and look deeper into it.

 

There's a whole lot missing between one example and "every experiment ever done". Since you now "know" the ground potential, can you use this to predict other masses of particles?

 

 

Note, we don't choose our religion based on how many people believe in it, so why do it with science?

 

There's a non-sequitur. Who says we do this with science?

 

 

PS: The protons velocity with respect to an observer at ground is worked out as follows.

 

[latex]v_{p}=c(\frac{\Phi-\phi_{gnd}}{\Phi})[/latex]

 

putting in the numbers...

 

[latex]300,000,000 m/s ((938MV-930MV)/938 MV)=2,558,635 m/s[/latex]

 

This means protons simply don't stand still, when observed from our potential, the only way to slow them down is to capture them in a heavier nucleus, but how do you capture a proton travelling at 2.5 million m/s, hence the impossibly small cross section in p+p fusion (only from our potential). In the early Universe when ground potential was a lot higher, the velocity of the proton would have been significantly less, so p+p fusion would have been common place in the early Universe.

 

 

OK, here's a failed prediction. Free protons, in general, don't have such a speed. Protons in a plasma in a discharge tube don't have this much energy. This is a much higher energy than the binding energy of hydrogen in a water molecule, and water molecules exist. The protons are not flying away at such a speed.

 

Solving the equation for velocity becomes;

 

[latex]v=c(\frac{\phi}{\Phi})[/latex]

 

What this tells us is, when the observers potential is the same as the proton potential, we are in fact moving along the time dimension with velocity c, and as time passes and the observers potential falls, the velocity in time diminishes as per the function and the velocity in the x, y, z dimensions increase.

 

We can now calculate the relative velocity between two bodies by knowing their potential difference.

 

[latex]v_{rel}=c(\frac{\Delta\phi}{\Phi})[/latex]

 

Take for excample the electron at 511 keV, and the above equation tells us that the electron must be moving at 99.1% the speed of light, and it works just as well for macroscopic objects at lower speeds, try it for yourself ;)

 

Since electrons can move at a range of speeds, this is another failed prediction.

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After distilling this right down over a good nights sleep, we are basically debating weather the following ratio is true.

 

[latex]\frac{v}{c}=\frac{\phi}{\Phi}[/latex]

 

Where v is the observers velocity and phi is the observers potential and Phi is the protons potential (you can call it mass per nucleon if you dont like the term potential)

 

The part that says observers velocity v is proportional to phi should agree with relativity because observers rest mass plus momentum is mc^2

 

The denominator part is the only thing new suggested by me, that c is proportional to Phi.

 

So the mass of a proton is proportional to c, well nothing new there either, it's only how we interpret it.

 

Will ponder some more and start a new thread soon..

 

Steven

 

 

 

 

 

 

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