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The First Formula to Compute the Mass of All Particles

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posted on Apr, 4 2014 @ 09:02 AM
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reply to post by VirusGuard
 




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 ?

If you read the post more than the title, what you would see is that the questions you asked are not at all relavent.

Individual particle mass in the standard model says NOTHING about dark matter or dark energy concentration in the universe. It could (If there was any predictive power in the formula, but since it is empirical there is almost zero predictive power) give the mass of a dark matter candidate particle, IF that particle follows the same pattern (which it doesn't have to)



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.


This statement is 100% incorrect, Who ever told you this, or if you came about it by your own deductions, it is not correct at all. The B-mode polarization observed can be used to PREDICT a higgs mass, it doesn't directly give you it. The Higgs mass predicted by the gravitational wave measurement is actually much bigger than the observed higgs at the LHC, such it begs the question, is there more than 1 Higgs?



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

You are correct though, a gamma ray spectrum can help you answer alot of things so how do we do that? OOOOOh that is right.... we already have a spectrum, because that is EXACTLY what the CMB is. And yet people still complain about sending space craft up to do exactly that.



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 big bang was an expansion OF space, not an expansion IN space, I can understand why people have trouble with this concept, all it simply means is that you cannot think of it as a point like explosion like it was anything you can imagine after seeing say... a nuke going off. It is space itself that is expanding, the matter expands within that expanding space... make sense? Suggesting the universe is a hologram... not sure the a boot up process and a spacial expansion simulation are really the same thing, no evidence presented for the universe as a hologram have ever been more than blue skies thinking. interesting, but solving nothing.



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.

On the double slit, DNA, computer code gravity and minecraft... All those things are somewhat unrelated and the tenable links between them as some kind of grand plan are lets just say fraught with logical problems. It is incorrect logic to say if B is like A then something in B should predict C which is also in A.


edit on 4-4-2014 by ErosA433 because: (no reason given)

edit on 4-4-2014 by ErosA433 because: (no reason given)




posted on Apr, 4 2014 @ 02:06 PM
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reply to post by ErosA433
 


Okay now I think I understand - Very good explanation, thanks (and a star for you)!


I would be very much inclined to agree with you. But for the sake of argument, it is my opinion that the problem you describe could, in theory, be resolved by using natural units.

If my memory serves me well, the speed of light can often be thought of as "1" when going with natural units. Thus sometimes (rarely, but sometimes), E=mc^2 is written E=m, since E=m1^2 is equal to E=m1 (and, thus, E=m). Following this logic, then, eV/c^2 could very well be written "eV" if c^2 is equal to 1. In which case, (eV)^16 would mean the squaring of the value of electronVolts only, and not of the speed of light.

Another point which, I realize, may not have been evident in my formula: notice that in the Variable L, there is a very important value: 3.785345. This is no random number. For technical reasons, this number is exactly equal to the mass of the tau particle, square-rooted 4 times (or at the 0.0625th power).

T^0.0625

=

1,777,000,000 eV ^ 0.0625

=

3.785345 eV

The formula is not exactly unitless - it uses the tau particle's mass as a reference unit (a bit like a constant) for the computation of the rest of the standard model. The whole of the variable L, in fact, returns results which are supposed to be read in electronVolts. Additionally, the simple formula

L^16

, even without the bells and whistles around, will give you the mass, in eV, of all fermions of the 3rd generation of particles (except of course the neutrino, since I excluded the variable "W").

The variable I' is indeed unitless, since it is responsible for the computation of the interval by which to divide the Lead L. Intervals are purely mathematical phenomena, with no physicality. But the L variable is based on the tau particle's mass, and its results are read in eV.

kind regards



posted on Apr, 4 2014 @ 02:16 PM
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reply to post by swanne
 


The only thing I can say about your equation is that it looks elegant and simple. For many these are the requierments for being on the right track. An other thing is that I can say to my grandchildren that I was there on ATS when this epic moment in history occured....and replied to the genius who came up with it

Cheers, have one on me,.... and give the man sitting next to you one too...



posted on Apr, 4 2014 @ 05:53 PM
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but that is the point


if i have the formula y = L^2, and L is 2 meters, it equates to a value of 4 meters^2

if you have the form L^16 and have anything that has mass in that L, that is (assuming it doesn't cancel out... which is where i am coming from) then in order for L^16 to have units of eV/c^2 then the units of L must be (eV/c^2)^(1/16)



posted on Apr, 5 2014 @ 05:44 AM
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ErosA433
if you have the form L^16 and have anything that has mass in that L, that is (assuming it doesn't cancel out... which is where i am coming from) then in order for L^16 to have units of eV/c^2 then the units of L must be (eV/c^2)^(1/16)



Why would anything cancel out? Nothing is canceled out unless I want it to cancel out (for instance the case of bosons), everything functions even though L is in eV. The list of results in my OP can vow for the fact that nothing cancels out.




edit on 5-4-2014 by swanne because: (no reason given)



posted on Apr, 5 2014 @ 05:50 AM
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reply to post by zatara
 


Hehe, Thanks mate!

But I'm really no genius. I just had alot of time on my hands, and a low tolerance for unresolved problems.



edit on 5-4-2014 by swanne because: (no reason given)



posted on Apr, 5 2014 @ 05:59 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.



But, I mean, who cares if Swanne's formula is unitless? His results clearly proves that his formula works; isn't it enough? No one else succeeded making up such a formula; is it really necessary that it has a unit?



posted on Apr, 5 2014 @ 10:44 AM
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starheart

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.



But, I mean, who cares if Swanne's formula is unitless? His results clearly proves that his formula works; isn't it enough? No one else succeeded making up such a formula; is it really necessary that it has a unit?


You miss the point, IT really isnt succeeding in anything, The formula has many free parameters that have been tuned to result in numbers close to that of the standard model. IT is an empirical formulation and little more. It is quite nice and does the task quite well, So i am not batting aside the impressive numerology, but it really isn't any form of particle physics breakthrough or success. Units are extremely important in physics. without understanding them, you dont understand anything about what an equation is saying or giving you.

And you cannot just cancel out units swanne, you cannot assign eV to something and do zero unit analysis



posted on Apr, 5 2014 @ 01:53 PM
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ErosA433
It is quite nice and does the task quite well, So i am not batting aside the impressive numerology, but it really isn't any form of particle physics breakthrough or success. Units are extremely important in physics. without understanding them, you dont understand anything about what an equation is saying or giving you.


Yes, it is a breakthrough! You can calculate the Standard Model all with one formula! Nothing like that has been done before, and no one could calculate the Standard Model - just parts of it. Regardless of its inner workings, it's still a groundbreaking calculation formula; just look at the result!


Btw, you posted your reply to swanne in your reply to me.



posted on Apr, 5 2014 @ 02:14 PM
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double post

edit on 5-4-2014 by starheart because: (no reason given)



posted on Apr, 5 2014 @ 02:16 PM
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ErosA433
And you cannot just cancel out units swanne, you cannot assign eV to something and do zero unit analysis


It seems you are misunderstanding the formula.

I simply take the (albeit reduced) mass of the tau, and I apply an operation to it. How can you say this will result in an error? It's like saying, If I put a pie of 1 kg on a scale, and I cut away one half of the pie, the scale won't display 0.5 kg but, in fact, an error.

Makes no sense.

here you say,


in order for L^16 to have units of eV/c^2 then the units of L must be (eV/c^2)^(1/16)


Or be ((eV)^0.0625)/c^2.

I don't see where the problem is??


edit on 5-4-2014 by swanne because: (no reason given)



posted on Apr, 5 2014 @ 02:25 PM
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You do not appear to understand what a first principles calculation actually means.

This is a formula to the 16th power, basically a 16th power polynomial. How many parameters/solutions does such a formula have? 16, oh and it has been tuned using a minimal input parameter to give 15 different numbers.

It is numerology, not a particle physics breakthrough. There is no physical basis or meaning behind the input parameters that give any predictive ability. IT also does not give / fit with the W and Z bosons, and yet has gamma, graviton and gluon why?


OK sooooo this is also an issue, First principles means deriving the masses without knowledge of them in the first place, more over using a very small number of input parameters (that mean something physical) and physical construct in order to deduce the masses.

Taking one of the masses and saying you apply a scaling to it, automatically means it is not a first principle derivation, and only valued for numerology.

It is possible to tune models in any way you want, or should i say numerology in any way you want. A games designer used a similar method in order to generate a procedural galaxy, with stars and planets and names for every object. Why? Because it had to fit on a 740Kb floppy disk. Impressive, yes, absolutely. Just like this is impressive. But it is as impressive as it is meaningless.



posted on Apr, 5 2014 @ 02:30 PM
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ErosA433

This is a formula to the 16th power, basically a 16th power polynomial.

This was done with the purpose of reducing all masses of the standard model under the same order of magnitude. I'ts not "numerology", it's akin to a concept in maths which is called a "modulus".


IT also does not give / fit with the W and Z bosons, and yet has gamma, graviton and gluon why?

W/Z boson can be considered, in practice, as virtual particles.



First principles means deriving the masses without knowledge of them in the first place, more over using a very small number of input parameters (that mean something physical)

Isn't a tau particle something physical?


edit on 5-4-2014 by swanne because: (no reason given)



posted on Apr, 5 2014 @ 02:58 PM
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Yes it is, but it is like saying

"OK i need knowledge of that one particle and i can calculate the rest..." so why the Tau? why not start somewhere else?

Mainstream thinkers get critisised on a day to day basis on things like the big bang, where people say "Well how did it start, how did it get there, what started it? Oh you cannot answer!"

Well, So, the same is true here, why start with the Tau to define the masses of the lower leptons for example?

Usually first principles means starting out with out prior knowledge of the problem and calculating behaviours or outcomes of a system. You may say the standard model is the same, but what the standard model does differently is that it can predict literally thousands of interactions and give you outcomes. It does so with cross referenced parameters that cannot be tuned for each individual interaction but are principles of the system.

Given that this formula apparently gives us the masses of all the quarks (I firmly say by tuned numerology only and not physics) please show how this relates to the mass of a proton and neutron.

Please Swanne, perform unit analysis on your equations and tell me what the unit of M is, you seem not to understand how units propagate, iv tried to explain it, but still it appears amiss.



posted on Apr, 5 2014 @ 04:30 PM
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ErosA433
Please Swanne, perform unit analysis on your equations and tell me what the unit of M is, you seem not to understand how units propagate, iv tried to explain it, but still it appears amiss.
I noticed that too and I understood your explanation, but maybe Swanne didn't. I'll take a crack at trying to explain it a little differently; hopefully this makes sense because I've been exposed to natural units but I don't work with them.


swanne
This was done with the purpose of reducing all masses of the standard model under the same order of magnitude. I'ts not "numerology", it's akin to a concept in maths which is called a "modulus".
When you apply a system of natural units, you have to define exactly how you are doing it. Your explanation I quoted below makes apparent reference to the Planck natural units, which does allow you to set c^2=1, but it doesn't allow you to ignore that your electron volts are raised to the sixteenth power and thus don't match the Planck units for mass under the system of natural units you appear to be trying to use.


swanne
If my memory serves me well, the speed of light can often be thought of as "1" when going with natural units. Thus sometimes (rarely, but sometimes), E=mc^2 is written E=m, since E=m1^2 is equal to E=m1 (and, thus, E=m).
True, but when you do that, you have to be careful, and understand what it means.

Natural Units

The equation c = 1 can be plugged in anywhere else. For example, Einstein's equation E = mc^2 can be rewritten in Planck units as E = m. This equation means "The energy of a particle, measured in Planck units of energy, equals the mass of the particle, measured in Planck units of mass."

Note the Planck units of energy and the Planck units of mass are not the same units when you write the equation as E=m.


Following this logic, then, eV/c^2 could very well be written "eV" if c^2 is equal to 1. In which case, (eV)^16 would mean the squaring of the value of electronVolts only, and not of the speed of light.
There are several problems with this.
First, ^2 means "squaring", but ^16 doesn't mean squaring, it's raising to the power of 16.
Second, if you want to apply the logic of natural units, you have to apply the logic of natural units. As the quote above explains, with natural units, you can deal with Planck units of mass and Planck units of energy.

The Plank mass is about 1.2209×10^19 GeV/c^2.
Note that it's not 1.2209x10^19 GeV^16/c^2

See the problem? If you set c^2=1, you still have a disagreement between eV and eV^16 units.
edit on 5-4-2014 by Arbitrageur because: clarification



posted on Apr, 6 2014 @ 05:46 AM
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reply to post by Arbitrageur
 


*sigh*

Okay, fine.

What if I were to put the formula into kg instead? Unlike eV/c^2, mass can be expressed in simple, plain kg, no planckian natural units involved.



posted on Apr, 6 2014 @ 05:57 AM
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Clarification:

I know the formula isn't perfect. My purpose here was to establish the possibility that the whole of the standard model's mass could be computed.

There are many ways to compute the mass of all particles. I've spent 5 years coming up with the first one. No, it's not perfect. But from now on we know that it's at least possible, we now have the proof that it is, and we have an idea about how we could go about it.

The formula may not please everyone, as it is a bit unconventional. That is because this formula is not ultimate - it's the first of many steps. Nothing more, but nothing less. Was the first car capable of going from 0 to 60 in 4 seconds?

Why did I posted it? Because I want mankind to give it a try too, to try and come up with a better formula. for if there is one thing I can trust mankind with, it's determination.

Mankind... never gives up.


regards


edit on 6-4-2014 by swanne because: (no reason given)



posted on Apr, 6 2014 @ 06:04 AM
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ErosA433
Well, So, the same is true here, why start with the Tau to define the masses of the lower leptons for example?

Because it was the heaviest lepton, from which could be derived, by division, the lower leptons. I went about it using division by intervals.


Given that this formula apparently gives us the masses of all the quarks (I firmly say by tuned numerology only and not physics) please show how this relates to the mass of a proton and neutron.

Don't be mean. you know very well why. It's a curve-fitting formula.

I fear I might sound blunt, but if you have come up with a better formula, please share it with us.

As I've stated, the formula isn't perfect. It's a first step, nothing more but nothing less. I know that, and I respect that. Why can't you?



edit on 6-4-2014 by swanne because: (no reason given)



posted on Apr, 6 2014 @ 06:50 AM
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swanne
What if I were to put the formula into kg instead? Unlike eV/c^2, mass can be expressed in simple, plain kg, no planckian natural units involved.
It would have to be a different formula, wouldn't it? With the formula you have, you're dealing with the exponent of 16 and I don't see how using kg units gets around that problem.

The problem isn't whether you use natural units or not, it's the exponent of 16 not matching up with either natural or SI units.


Arbitrageur
As George Box said, all models are wrong, but some are useful.
I guess I'm trying to understand if the formula is useful, and if so, exactly how. For example:



swanne
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

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.
OK so the formula can change the fudge factor to match experimental results, there is precedent for this in some cases, but...

Wouldn't be easier to just look up the mass in a table, instead of using the formulas above that will change whenever experimental results or observations change?

By the way here is the table that phycists use though this one is a little dated from 2010, but for neutrino mass it just says "mass m < 2eV (tritium decay)" which I would say is not only easier to just look up than crunching some numbers, but it's also more accurate since it only applies an upper bound, and the lower bound is not that well known except that it's not zero.

Particle Physics Booklet
pdg.lbl.gov/current/booklet.pdf


swanne
I fear I might sound blunt, but if you have come up with a better formula, please share it with us.
It would be great to have a formula, but I think you pointed out in your OP the unsolved problem:


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

If as you say it's just "curve fitting" (which with the unit problems it hasn't quite done that yet either, but for the sake of argument lets say you resolve that problem), has it brought us any closer to a solution? I don't see how. Is there a reason to use this formula instead of just looking up the value in a table or booklet?

I think folks like Eros would probably be the target audience who would use such a formula if it provided more utility than the Particle Physics Booklet, or if you think there's some other target audience who might use it, who would that be, and how would they use it? My point is if he's not finding it useful I think you should take the feedback seriously. But a formula that made some kind of dent in the unsolved problem you cited would be useful, and no we don't have one, that's why it's an unsolved problem.



posted on Apr, 6 2014 @ 01:50 PM
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reply to post by ErosA433
 


I don't understand you: yes, Swanne's formula can't make predictions. But it can, with extreme accuracy, approximate a particle mass, which NO ONE has been able to do.
The end result is: Swanne's formula works. Look at the chart:



How can you not see that no matter what the inner workings of the formula is, it WORKS!
It's like saying: because a dude doesn't know the inner workings of a computer, the computer itself doesn't work. Or, a dude doesn't know the inner workings of the Hadron Collider means that the Hadron Collider doesn't work.

Before predicting, you must have the capacity to account for something. Swanne's formula is the first one capable to account for the whole of the Standard Model. At least, give him some credit for creating something no one was capable to create.


And btw, his formula isn't numerology. It's math; unless in the US, math beyond the 6th grade is now considered numerology.



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