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The First Formula to Compute the Mass of All Particles

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posted on Apr, 3 2014 @ 04:18 AM
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The First Formula to Compute the Mass of All Particles

by John SkieSwanne


~

There are many yet unanswered questions in Particle Physics. For one:


Is there a theory that can explain the masses of particular quarks and leptons in particular generations from first principles?

source: unsolved problems in physics

I've spent years pondering upon this mass problem. The thing is, if you take a look at the standard model, especially on the mass part, you'll notice something very important: the masses of two given particles do not seem to follow a pattern relative to each other's. Which is... strongly inconvenient, to put it midly, for all of us who dream about a ToE (Theory of Everything).

Not only do these particles's mass follow different orders of magnitude (the electron is set at 0.511 MeV, while the top quark soars at 171.2 GeV), but as you go from one generation to the next, the order of "which particle is lighter than the other" itself changes. Since the beginning of particle physics, no pattern was ever found relating all these particles' masses. No formula was capable of such feat. To all of the world's eyes, it was utterly impossible; and it stayed that way since the beginning of particle physics, some one hundred and eighteen years ago.


~

Well, ladies and gents, today I finally come forward with a formula - the very first of its kind. It does the impossible - it finds order inside chaos. Today is the day I made it at least possible to compute the mass of any given, non-virtual particle of the Standard Model, based on its generation number, its charge class, and its spin.

It approximate the mass of anything from the electron to the top quark, including gluons and even neutrinos. And, here it is:



The formula is actually a group of equations, making it the most complex formula I've come to create. Let's explore its different components. I think of it a bit like a car, with different devices - a motor, a direction, a transmission, breaks - which together makes the machine a workable system.

Let's start by the core - the Lead value (L). This L is a device which lets you determine three fixed values. These 3 different values represent the mass of the tau, the top quark, and the bottom quark (the tau neutrino isn't mentioned yet, the reason being that we don't know what its mass is yet). These three values are being used by the equation as references for the fermions of Generation 2 and Generation 1. Here, L is equal to...

L = 3.785345*(1+((X^2.580742)*0.055231))

Where X is a re-shuffling device. Since mass order do not necessarily follow the rank of the particle's electric charge, X makes a conversion between the two possible. And X is equal to...

X = (3-C)-((0^C)*3)

Where C is the particle's charge class. A note about the charge: this formula was designed to use with Phoenix-I/II Theory's preons. Meaning that C is supposed to refer to the minority of preon species in the particles. In the case of an electron, with 6 majority preons but 0 minority preons, the class would be "0". In up quarks (and antiquarks), there would be 5 of majority preons of one specie and only 1 minority preon of the other specie - thus the class would be "1". Following that logic, here are the particles for each classes:

0: electron, muon, tau, positron, anti-muon, anti-tau
1: up, charm, top, anti-up, anti-charm, anti-top
2: down, strange, bottom, anti-down, anti-strange, anti-botttom
3: neutrinos, neutral bosons

I know that not much people subscribe to the Phoenix-I/II Theory, so in the case one is amongst this group, I've included a conversion formula with which one can deduce the value of C:

C = (1-C')/0.333333

Where C' is the charge of the particle - drop the "-" sign if there's one (example: "-1/3" becomes "0.333333").

Back to the topic: Once the value of L has been computed, then comes the time to find out what the value of the variable I' will be.

"I'" stands for "interval". The idea is that all fermions of the standard model are placed on an interval value relative to the Lead L. A bit like music: once you know the value, in cents, of a note, all you have to do is apply the interval value to this note to obtain the perfect forth, or a minor third. Except that here we'll be working with mass instead of cents and particles instead of notes.

To find the interval of a particle is no easy matter - in fact, this part is the hardest and it took me months to figure it out. But I did most of the job, all you need now is a scientific calculator. The Interval [I'] is equal to...

I' = ((2-(0.5*C))^(2-Gen))*((2/Gen)*I")

Where C is of course the charge class which I mentioned earlier, and Gen is the particle's generation.
In the case of an electron, the Class will be "0", and the Gen will be "1". In the case of a top quark, the Class will be "1", and the Gen will be "3". Et cetera.

Now notice there is a device called I". This is because as we hop from one generation to another, different algorithms are put into effect and/or turned off (aka, their value drop to zero). This means that the formula above can adapt itself and engage the correct algorithm, depending upon which Generation is being requested. The I" variable is part of this mechanism, and is absolutely necessary for the formula to work properly. And here's how to compute its value:

I" = I'"*((Y-C)-((0^C)*3))

Where I'" is designed to always return the value of "1", except when the particle is from Generation 3 - in which case it'll turn to zero, overriding the whole interval computation, and thus stating that to get the mass of the particle relative to the Lead Value, you don't have to do anything to the Lead Value, since it is equal to itself in the first place. I'" is equal to...

I'" = 1-(0^(3-Gen))

Now the value Y could be simply be replaced by the number 5 if you were in a hurry and didn't mind about accuracy. But its true value occillates, sometimes it can get at 4.995 or at 5.189, depending on the particle. Y give the formula its accuracy, and goes like this:

Y = (((2*(0^(Gen-1)))*Z)+((Gen-1)*10.310))-Z

Where Z is a variable equal to...

Z = 4.995*(1+(0.001801*(C^4.429988)))


(continued... )




posted on Apr, 3 2014 @ 04:19 AM
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We reached the point where we can now compute the mass of all fermions... except the neutrino, for which science didn't find a mass yet. My formula takes this into account, as I actually have placed a temporary fix. It's called the W variable. It recognizes class 3 leptons and subtract a value (which can be tweaked and corrected in the future as science progresses further) so that the result at least matches the probable mass of the electron neutrino (which is believed to be around 1.5 eV). Here is how to compute it:

W = (0^(3-c))*1.969823

Now whatever result the formula gives, one only has to square it four times (((n^2)^2)^2)^2 and et voila, the mass of the particle is returned. You might wonder why this quadruple square thing is there - the reason is that the formula itself works with representation of masses. These representations, these "avatars" are equal to the true mass square-rooted four times; this was done with the purpose of unifying all particles masses under only one order of magnitude. Once the formula is done, the number simply needs to be squared four times to give the true result.

And, last but not the least, the bosonic compensation V must be computed. Since all non-virtual particles in the standard model are with zero mass, this V device checks the spin of your particle, and cancels the formula back to zero if the spin is an integer. In other words, zero mass. It goes like this:

V = 0^(spin-0.5)

~

All of this is alot of work, but does it pay off? Well, we have the occasion to check it out - the following is a list of the results, placed in two columns, side by side for comparison. On the right column is the mass of a given particle in the real world. On the left, is the approximation which my formula makes of this said given particle.



So, this concludes my presentation of the very first formula of its kind, relating all non-virtual particles's mass from the Standard Model of Particle Physics.

~


At Time's End,

Swan



posted on Apr, 3 2014 @ 04:20 AM
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This one... I had to close my eyes before clicking "post". I'm both so excited but also so apprehensive. I know the formula is far from perfect... I just hope I won't get too flamed, lol.



edit on 3-4-2014 by swanne because: (no reason given)



posted on Apr, 3 2014 @ 04:47 AM
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This is mindless numerology. All kinds of formula without any sound theoretical justification have been devised with enough internal tweaks to derive the mass spectrum of the basic hadrons and leptons. None have convinced physicists.

Besides, you need to know that any scheme that attempts to derive the masses of quarks and leptons is doomed to failure because it assumes that quarks - like leptons - are discrete, fundamental particles, whereas I know with mathematical certainty that they are not basic but are bound states of three spin-1/2 particles (not yet discovered) bound by the same confining Meisnner Effect mechanism that generates three-quark bound states, namely, the string model of QCD in which strings are SU(3)/Z3 Nielson-Oleson vortices in the superconducting Higgs field. Subquarks, not quarks, form with leptons a unified representation of a certain subgroup of the rank-8 exceptional group E8, the symmetry group governing the unified interaction between the E8xE8' heterotic superstrings of ordinary matter. The predicted composition of up (u) and down (d) quarks is:
u = X-X-Y, d = X-Y-Y,
where X has a positive charge of 5/9 and Y has a negative charge of -4/9 in terms of the electron's charge -1.



posted on Apr, 3 2014 @ 04:55 AM
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swanne
This one... I had to close my eyes before clicking "post". I'm both so excited but also so apprehensive. I know the formula is far from perfect... I just hope I won't get too flamed, lol.



edit on 3-4-2014 by swanne because: (no reason given)


You'll only get flamed by me for setting my face on fire.

I actually got to engineering maths in uni some 20 years ago, I just passed first year by some fluke of the cosmos. I wish I had stuck with it now and not given in. I appreciate your work, but I must show my ignorance and hope you don't have any remainders left over.
Good luck with facing the internet, those maths ones can be feisty.

(edit, see above!)

edit on 3-4-2014 by Qumulys because: (no reason given)



posted on Apr, 3 2014 @ 05:24 AM
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reply to post by swanne
 



First Formula to Compute the Mass of All Particles


Good work, but the powers that be what too know how can ya make a weapon from it,???
anyhow, I guess ya know if your Formula is correct if this thread disappears down the rabbit hole



posted on Apr, 3 2014 @ 07:33 AM
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Wow, I can't even begin to get my head around ANY of that math.

I always felt that having chaos at the core of everything was kind of romantic.

I know it's not a scientific view, BUT HEY...a double dose of KUDOS to you
anyway for your original findings.

And, if it holds up, let me be the first to congratulate you on your new Nobel.

Also if you would, could you please get Neil DeGrasse Tyson's autograph for me

edit on 3-4-2014 by rival because: (no reason given)



posted on Apr, 3 2014 @ 08:30 AM
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I should point out that calculating something from first principles typically involves using only constants that represent a physical quantity in the respect that it is a constant with physical meaning that propagates. Examples being in the standard model where there are constants that represent coupling strengths and are fundamental to couplings between bosons and particles. You may say that too is just numerology, but in the standard model these appear all over the place, and their value is fixed and propagated throughout all formulations that involve those couplings.

Here I see several seeminly randomly unphysical 'tuning factors' In no way do they appear to relate to anything physical.

While it is quite a nice example of numerology what you have produced is an empirical representation that gives approximate values for the mass of each species from as you write above, using a minimal set of input parameters. The formula can also be written in a different way because compound squaring is basically just bad form. What you have is

m = c(x)^16

Where you are carefully selecting places along the polynomial to give you the different masses by using a formulated value for x. So why not write the formula as such... Also given that it is apparently 'from first principles' how do the units propagate, you have something here to the 16th power, what ever is inside that x must be unit less... so C here carries the physical mass?

point is, what you have is empirical and not physical just to reiterate

Furthermore, your data for the masses of some of the quarks have their error bounds quite exaggerated but that is just an observation and not critical.
edit on 3-4-2014 by ErosA433 because: (no reason given)



posted on Apr, 3 2014 @ 10:05 AM
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reply to post by swanne
 


Nice thread, although it's way over my math skills barrier. Hopefully an expert will have a look at this, and I wanted to be on the first page of a historic thread when you win stuff and get your face on postage stamps and things. Seriously, nice work (I think).



posted on Apr, 3 2014 @ 11:18 AM
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ErosA433
I should point out that calculating something from first principles typically involves using only constants that represent a physical quantity in the respect that it is a constant with physical meaning that propagates.

Here I see several seeminly randomly unphysical 'tuning factors' In no way do they appear to relate to anything physical.

You make a good point, and I can't argue with you - you're right, it's more empirical than physical.

It is my very first attempt. My one and only focus was to figure out the formula above all. It is indeed about tuning, more than about recognizable values.


The formula can also be written in a different way because compound squaring is basically just bad form. What you have is

m = c(x)^16

Oh. I didn't realized there was a simpler way of writing all these powers. Thanks.


what ever is inside that x must be unit less... so C here carries the physical mass?

What do you mean?



posted on Apr, 3 2014 @ 11:23 AM
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This is mindless numerology.

No. It's an operation of several tuning values so to compute the whole of the non-virtual standard model, all in one go. If you think you can do better, then please, share your formula with us all.


micpsi
The predicted composition of up (u) and down (d) quarks is:
u = X-X-Y, d = X-Y-Y,
where X has a positive charge of 5/9 and Y has a negative charge of -4/9 in terms of the electron's charge -1.


And this theory... Does it predict the mass of all particles of the standard model, including neutrinos?


edit on 3-4-2014 by swanne because: (no reason given)



posted on Apr, 3 2014 @ 11:31 AM
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reply to post by swanne
 

As George Box said, all models are wrong, but some are useful. If the model can predict some experimental results which we don't yet have, then it seems useful, and we could say this happened with the Higgs model which was experimentally verified.

For the electron neutrino mass, your chart shows the SM value as "around 1.5eV". I'm not sure if we know it that well. I think the upper bound studies say it should be less than 2.2eV, one study said it might be around 1.5 eV and other studies have suggested smaller values less than 1eV. I think more experimental results are expected by next year which might come up with a more accurate value.

So let's say hypothetically the more accurate value ends up being 0.8 eV, does that mean we should reject this "model"? The reason I put "model" in quotes is that it looks more like what analysts in my field call "curve fitting", which means generating formulae to explain historical observations, but these generally fail at making predictions of future observations if there's not some physical foundation to the way the curves are fit to the data, which as Eros suggested, such foundation doesn't seem to be present here, or if it is, perhaps you could explain it better.



posted on Apr, 3 2014 @ 11:39 AM
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Arbitrageur
So let's say hypothetically the more accurate value ends up being 0.8 eV, does that mean we should reject this "model"?

Well, As I've stated in the OP itself, I myself am still waiting for more data on the neutrino. If ever the neutrino happens to be 0.8 eV, well, all one would have to do is correct the value of

W = (0^(3-c))*1.969823

to

W = (0^(3-c))*2.669823




The reason I put "model" in quotes is that it looks more like what analysts in my field call "curve fitting", which means generating formulae to explain historical observations, but these generally fail at making predictions of future observations if there's not some physical foundation to the way the curves are fit to the data, which as Eros suggested, such foundation doesn't seem to be present here, or if it is, perhaps you could explain it better.


It seems my formula does conform to a curve fitting formula. It's what I basically intended it to be: something to account for all which we observed. Nothing more, but nothing less. I certainly can't make something as powerful as predictions of, for instance, a fourth generation of matter - it is already difficult to make a formula which fits the masses of the SM in the first place. Took me years.

And I doubt 4th generation exists anyway, at least outside of the domain of virtual particles.


edit on 3-4-2014 by swanne because: (no reason given)



posted on Apr, 3 2014 @ 01:24 PM
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but that is the point, empirical formulation and curve fitting doesn't give anything that is predictive as you just admitted. While yes you just did a very nice job of producing a formula for giving all the masses of the standard model particles (Neutrinos are likely incorrect, even in this fit)

It regardless does not hold true that the same level of prediction cannot be performed with any n-th order polynomial.

this doesn't do what it said on the tin which is a first principles derivation/formulation.

On Neutrinos

Neutrino mass scale is being determined, the estimates as presented are fair but are likely to change by up to two orders of magnitude. They are currently limited by lack of statistics as well as generally the difficulty of such a measurement



posted on Apr, 3 2014 @ 01:43 PM
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Your neutrino masses are wrong, in that all they can differ by (lightest to heaviest) is a maximum of 0.05 eV.

Does your model predict any 4th generation masses?



posted on Apr, 3 2014 @ 01:50 PM
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Reply to your question about inside the 16th power being unit-less


If you are looking at masses that are in eV/c^2 then nothing inside that power law can have a unit. otherwise it will have to some how come out as being still eV/c^2 at the end.

This is often how you atribute physical meaning to a formula. It is totally fine for it to be unitless, it would just be that the carrier of the unit in eV/c^2 would be your V parameter.



posted on Apr, 4 2014 @ 06:03 AM
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repressed
Your neutrino masses are wrong, in that all they can differ by (lightest to heaviest) is a maximum of 0.05 eV.

Um, you sure? Do you have a source I can check out.


Does your model predict any 4th generation masses?

No. It computes only what exists. We are not sure if a 4th generation exists in the first place (beyond theory).



posted on Apr, 4 2014 @ 06:18 AM
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ErosA433
This is often how you atribute physical meaning to a formula. It is totally fine for it to be unitless, it would just be that the carrier of the unit in eV/c^2 would be your V parameter.


I know you mean well, but I am not sure I understand you (I'm still a layman).

The "carrier of the unit"... is that a reference to the values inside my variable "L"? The values which I called "avatars" of the true "unit" (the unit being in eVc^2 once the ^16 operation is done)?

Also, my "V" Variable is pretty much unrelated to the rest of the formula, since its one and only purpose is to bypass the whole formula (shut it to zero) when the spin of the target particle is an integer. How does it relate to "eV/c^2", in which "V" is a reference to "Volt", not "Variable"?

regards


edit on 4-4-2014 by swanne because: (no reason given)



posted on Apr, 4 2014 @ 06:47 AM
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So does this equasion include the 97% of matter they called dark matter because that was the only way they could make the numbers add up and how does the other fudged number known as dark energy effect the equation ?

Yes they solved this gravity wave thing by finding this god particle so now we are left with a another wave that explains away the first wave.

crashing snooker balls into each other using stonger and stronger pool players will only teach you so much in the 10'[201] of a microsecond you have to view the results but lets face it you need to be wet behind the ears to think that taking a picture of the universe in the gama spectrum can take you back 14.7bn years

if this big bang does indeed exsists then it was not all matter blowing up from something thats smaller than a nat's knacker to create the universe and it was just the CPU being powered up or maybe the run button being pressed.

The double slit exsperiment gives you a clue as does DNA being computer code and that computers are getting much smarter than we are so if you add it all up then we are in something like Minecraft that has it's own set of rules that too has a type of gravity.

Knowing you are not real takes about two weeks to sink in but then it feels just the same almost as a month ago when you thought you was real and you have better hope I am right else heaven where people are not real can never exsist.



posted on Apr, 4 2014 @ 08:43 AM
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reply to post by swanne
 


Thats ok Ill see if i can come at this at a different angle.

You have a formula of the form m = v(x)^16 where x = (L/(12-l')/12) -W

Unit analysis is often used as a sanity check on equations and formulae because Units (say for example meters, seconds, kilograms) have to be representative of the quantity calculated. Thus if I produce a formula and after doing unit analysis it works out that I am getting Velocity (or i claim that it is a velocity) yet the units are kg/m then I certainly went wrong somewhere.

An example being in the equation above, mass can be presented in lb,kg or eV/c^2

So what is this magical unit of electron volts / speed of light squared!

E = mc^2 in units is [Joules] = [Kg][m/s]^2 This formula was not really discovered, it is a breakout of unit analysis

an electron volt is the energy gained by an electron accelerated through a potential of 1 volt,

[Volt] = [Joule] / [Coulomb] and given the electron charge is 1.602x10^-19 C, 1eV is defined as 1.602x10^-19 J

So how do we get a mass in units of eV/c^2?

Look at E=mc^2 and rearrange it, m = E/c^2 and bingo we have eV/c^2 and such we get that 1eV/c^2 is = 1.782x10^-36 kg


---------------
Ok so hopefully that makes sense,

Back to your original formula, m = v(x)^16 where x = (L/(12-l')/12) -W

The x parameter has to be unit-less, if i give it any unit, that unit ends up to the 16th power. thus the only physical quantity possible in this formulation is v, And it begs the question of why? and the scaling factors and solutions for each of the particles you show above, if they don't have any kind of physical units, how do they get explained not to have any physical units. Its fine to say that they are some property of the system, but thats really all i was getting at, that this is an empirical and not physical representation of the standard model particle masses. It is not a from first principle derivation.


If i take for example e=mc^2, how do we know that works?

Well if you do take an electron and accelerate it through an electric field with say 1million volts, you can make it strike a material and look at the energy it deposits. What you find once you work it all out, is that the formulation works once the inefficiencies of your detector has been unfolded.

We can also take particle anti particle pairs and accelerate them, with equal potential. (net momentum = 0, but energy = e1+e2) and what we find is that when the particle and anti-particle collide, the energy we measure emanates perpendicular to the beam, and the total energy we observe is both the physical energy of the particle given to it via acceleration, but also the rest mass energy as defined by e = mc^2.

This effect is also observed in PET scanners, in which you choose a radio isotope that emits a positron, the positron annihilates with an electron locally and two gammas of energy 511 keV are emitted back to back. You can then use timing to reconstruct where the event occurred. There is a wealth of evidence (country to some people being very ignorant) that e=mc^2 is correct.



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