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Spooky idea too late for Halloween


beejewel

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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

 

I don't know how to put it any more simply than I already have, about half a dozen times. If your equation is true, then [math]\phi[/math] is just another name for velocity. Calling [math]\phi[/math] a potential does not make it a potential.

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It is? Can you derive this?

 

This should be the general four-space form of the proportional term;

 

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

 

If [latex] p=\gamma m_o v [/latex] and [latex] p=m_0 U [/latex] then we should be in agreement with relativity when I say that these terms are proportional.

 

For the relative form it should become;

 

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

 

This thread is becoming long, I might start a new thread, on speculations about how I see this affecting the wave speed.

 

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This should be the general four-space form of the proportional term;

 

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

 

If [latex] p=\gamma m_o v [/latex] and [latex] p=m_0 U [/latex] then we should be in agreement with relativity when I say that these terms are proportional.

 

For the relative form it should become;

 

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

 

 

 

 

 

I see an "if" and a "should". That's not a derivation. And does not support the assertion that "rest mass plus momentum is mc^2"

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The ratios,

 

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

 

are proportional, because,

 

[latex] \frac{v \Phi}{c \phi}=0 [/latex]

 

Nothing can travel faster than light, and no potential can exceed the protons potential.

 

 

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Do you mean 1 and not 0 on the RHS of your last equation?

 

Anyway, the problem is that we don't really agree with your notion of a potential. It is so different to how we view potentials in physics. In particular, we can only measure potential differences, we have the mathematical structure of a torsor at play. A really clever guy explains some of this here. ;)

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"observers rest mass plus momentum is mc^2"

It's wrong, I make mistakes too ;)

Okay, but what about the rest of the comments/questions you've received? You seem to be selectively ignoring the parts that don't bode well for your "idea." Nobody has the slightest clue why you keep insisting that velocity is somehow equal to some kind of electrostatic potential. Your "equations" look like they came about by taking several disparate quantities you found while browsing wikipedia for a few minutes, and trying to staple them together without any real understanding of what they mean.

Ajb,

Bring me a stable fermion heavier than the proton, and I shall reconsider my case :)

Steven

What does that have to do with his question?

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we can only measure potential differences

 

Remember for this to work we have made one assumption or postulate, that the electron and proton are a pair. From that flows that the potential difference between them can never exceed 938 MV, and that comes from simple reasoning that we drop the c's and drop the e's, so we are still talking about potential as a difference.

 

So we should agree to continue debating the consequences on these terms, otherwise we are going around in circles.

 

As is pretty obvious to all, I think the above is a reasonable assumption, and the fact that we have a universe full of electrons and protons, with every proton the same as every other proton, and every electron the same as every other electron, is good evidence, further we have not found a single stable particle with more mass than a proton.

 

Just think about this for a moment...

 

 

 

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further we have not found a single stable particle with more mass than a proton.

 

Just think about this for a moment...

 

Not to give you an excuse for tap-dancing around the several outstanding objections, but this should be totally unsurprising. Mass is a form of energy, and systems tend to their lowest energy state. Something with more mass than a proton would be able to shed energy by becoming a proton, unless some other conservation law prevented it. Exactly why a neutron decays into a proton — there's nothing to stop it. So the reality is that this is not a revelation at all. We should expect never to find a heavier stable baryon than a proton (or lepton heavier than an electron) unless we also discovered some new physics

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Mass is a form of energy, and systems tend to their lowest energy state.

 

Slowly converging here, however the proton is not the lowest energy state, it is the state of the observer, which if we assume we are on the same planet (sometimes I wonder) then it's ground potential.

 

As I pointed out further up the thread, the lowest point is Ni-62 which currently lies at a potential of 930 MV towards which all elements decay towards if they can, but even ground potential is gradually falling by around 0.5 millivolts per year, so to MigL's reply above, even the proton will eventually decay.

 

Just to clarify that last statement, it is the observer that decays the proton, and not the other way around, this is a tricky part to understand, more on that later.

 

Steven

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WHAT?

 

Is it really that hard to follow this?

 

Think of the energy in every proton-electron pair as a clock spring, when it is fully wound up, it has a tension of 938 MV i.e.. it has potential to do 938 MeV worth of work.

 

It is possible for the spring to unwind, but only by falling to lower potentials, and it can only do this by converging with another proton forming a nucleus.

 

The first such merger is the fusion of H to D, then D+D to He3, then B and so on, think of these fusion processes as the governor (anchor) on the clock spring, there are no possibilities inbetween these known reactions.

 

post-21391-0-70394200-1415155600.jpeg

 

This process continues down to iron and Ni-62 (that's ground potential - where you are) the process stops temporarily there, any mass at lower potential than ground potential will actually fall UP! It floats, yes buoyancy is what we call it.

 

But as more and more of the Earths elements decay, U, Th, K etc. not to mention man's efforts to speed up this process, the core becomes heavier, and ground potential gradually falls away.

 

I know excactly how far ground potential has fallen since the beginning of time, I can work it out by knowing the mass of the proton and that mass of the electron because the relationship is:

 

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

 

All this says is that the mass of the electron is half the difference between the proton mass and the mass per nucleon at ground potential multiplied by [latex] \gamma [/latex]

 

I could simply have written,

 

[latex] \phi_e = \frac{\Delta\phi\gamma}{2}[/latex]

 

So when you do the numbers, it turns out that ground potential is 930 million volts, and proton potential is 938 million volts, therefore ground potential has fallen 8 million volts since the beginning of time.

 

Google puts the age of the universe at 13.8 Billion years, so lets divide 8 million by 13.8 billion and the drop in potential is

 

0.00057971014 Volts per year

That's of course if google knows :)

 

That's pretty clear...

 

 

Edited by beejewel
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