# All the lepton masses from G, pi, e

## Multiple poll: Check all you agree.

21 vote(s)
26.3%

19 vote(s)
23.8%

24 vote(s)
30.0%

21 vote(s)
26.3%

16 vote(s)
20.0%

24 vote(s)
30.0%

18 vote(s)
22.5%

15 vote(s)
18.8%

43 vote(s)
53.8%

29 vote(s)
36.3%
1. Oct 4, 2004

### arivero

Plus c and h, of course.

The idea is to collect here in only a thread all the approximations voiced out during the summer. Surely this is to be quarantinised in TheoryDev, but it is interesting enough to be kept open as a thread (if closed, please be free to use my weblog)

First we get Planck Mass, $$M_P$$, from G, c and h as usual.

Then we solve for $$\alpha$$, the fine structure constant, in HdV second equation (*),

$$\alpha^{-1/2}+ (1+{\alpha \over 2 \pi }) \alpha^{1/2}=e^{\pi^2 \over 4}$$

The term in parenthesis, well, is sort of a first order correction.

Then we use Nottale's remark to get the mass of the electron

$$\ln (M_P/m_e) = \alpha^{-1} \sin^2 \theta_W$$

where the square sine of Weinberg angle $$\theta_W$$ is to be rather misteriously taken at the GUT value, 3/8. I have not checked if the need of Schwinger correction above counterweights this need of a value running up to GUT scale.

Now we use HdV first equation set to get in sequence the mass of the muon, via the rather strange

$$\ln {m_\mu \over m_e}= 2\pi - 3 {1\over \pi}$$

and the mass of tau via the simpler

$$\ln {m_\tau \over m_\mu}= \pi - {1\over \pi}$$

Alternatively HdV set of equations can be presented from a quotient $$m_e m_\tau^n / m_\mu^{n+1}$$, but the original presentation hints of a hyperbolic sine, or perhaps a q-group scent.

We could try to follow towards the full mass matrix, including neutrinos, via some empirical approximations collected by Mikanata and Smirnov.

----
(*) Update: HdV has uploaded to his webpage an indication of the origin of his formula, as a 3rd term truncation of a peculiar series.

Last edited: Oct 5, 2004
2. Oct 4, 2004

### marcus

I do not imagine that this will be quarantined, these numerical relations are not your personal theory that you are developing and promoting. Rather, you are reporting on the possible follies (or wisdom) of others.

Even if this may be the work of dope-crazed hippies and Wiccan moon-worshipers, we should not be blind to the knowledge that it exists. How else shall each of us know what numerology is? At the very least, it is a very good vaccination against the temptations of numerical coincidence.

3. Oct 5, 2004

### arivero

Also it is a high barrier for guessers. I do not know anyone deriving such high precision with so small formulae (specially HdV ones have a small error). So any posting of numerology should be requested to have at least this effectivity, or to shut up. In this sense yes, it is a very effective vacunation.

BTW, while I am using here "numerology" to adhere to the broader view of this word, to me it is numerology only the use or arbitrary dimensional constants (inches/year for a famous example), while using arbitrary adimensional ones is more an issue of compacteness of infomation. Here, at least, one can use HdV relations as a mean to remember almost exactly the values of some fundamental constants; for somebody it can be easier to remember the formulae than the decimal number expressions.

Last edited: Oct 5, 2004
4. Oct 5, 2004

### Hans de Vries

Mathematics used for the Fine Structure constant formula.

Yes it should be called Mathematics rather than "Numerology"
This is especially true for the way how the formula for the
fine structure constant was generated.

It's possible to derive a function's complete Taylor expansion
from a single point and its infinitesimal small environment.
A similar "Taylor like" method was used here to derive an
entire function from a single (but highly accurate) number:

The tricky point is the start. We need to find the start of the
function presuming that a function exists and it can be expressed
as a converging series.

A successful ansatz turned out to be to express the amplitude
$\alpha^{1/2}$ as a "Gaussian like" value:

$$\alpha^{1/2}\approx e^{-{\pi^{2}/4}}$$

From here on we can develop a corrective series A by taking
successive differences so that:

$$\alpha^{1/2}\equiv Ae^{-{\pi^{2}/4}}$$

This generates the following series A:

$$A = 1+{\alpha \over (2\pi)^0 }(1+{\alpha \over (2\pi)^1 }(1+{\alpha \over (2\pi)^2 }( 1 + ........$$

or separated:

$$A = 1+{\alpha \over (2\pi)^0 }+{\alpha^2 \over (2\pi)^1 }+{\alpha^3 \over (2\pi)^3 } + ........$$

The series converges straightforward to reproduce the value
of the fine structure constant exact in all its digits:

after term 0: 0.0071918833558268
after term 1: 0.0072972279174862
after term 2: 0.0072973525456204
after term 3: 0.0072973525686533
f.s. constant 0.007297352568.(+/-24)

I hope that everybody realizes how incredible small the
possibility is that something like this happens by chance.
Say the fine structure constant is some arbitrary number
and then we get the whole series out it with an overall
precision better than one in ten billion !!

Basically the only thing I have "put in" the formula was the
ansatz $e^{-{\pi^{2}/4}}$. The rest is generated by repeatedly
subtracting the two numbers.

Regards, Hans.

Fine Structure constant article online

PS. The formula given in Alejandro's opening post represents
the series truncated after term 2.

Last edited: Oct 5, 2004
5. Oct 5, 2004

### Nereid

Staff Emeritus
Your challenge, should you choose to accept it, is to estimate the probability of finding two numbers (any two of the three lepton mass ratios) using only some 'simple' functions, pi, e (and any other 'cool' transcendentals), a few small integers (preferably <5, though a bunch of contiguous primes would also count as 'cool'), to within 0.1% (more points if 0.01%). You may also write a simple program - to generate the 'formulae' - or even an 'evolutionary' meta-program (which will all but guarrantee to find at least one formula meeting the input criteria).
Better yet, since the neutrino masses are known to have only upper limits, why not make some firm predictions (based on goat entrails, er, sorry, magic formulae)? While we're at it, how about the Higgs? the LSSP? neutralinos, axions, wimpzillas, ... the whole zoo?

6. Oct 5, 2004

### humanino

wimpzillas :surprised
I lost mine and have been unable to replace them :tongue2:

I agree with Neired : the probability that you find a relevant relation with "numerical experiments" is far from obviously non-vanishing.

7. Oct 5, 2004

### arivero

Yes, because we expect there is at least one such relation. And some of us hope it to be simple (well, a quantum group sounds "simple" to me ) and geometrical. Still, it should be good if Mathematica (TM) skilled people were able to make some symbolic experiments to find out such formulae and their probability.

8. Oct 5, 2004

### arivero

Stop, stop (I mean the verb, not the SUSY particle)... the relations I was referring to, for neutrino -thus leptonic- mixing angles, are still considered good taste in the real world. Of course, your mileage can vary.

9. Oct 5, 2004

### marcus

I expect you know that Oxford U. Press is publishing a book called Universe or Multiverse containing an article by Smolin that presents a testable scientific theory that the dimensionless fundamental constants have gone through an interation process called CNS (cosmic natural selection).

CNS would have favored convergence to constants optimized for starformation and black hole production. This is not the "Anthropic Principle" in one of its non-empirical forms---it makes testable predictions, it is falsifiable, it is not predicated on life or "consciousness", it is a simple physical theory which may in fact be wrong but seems worth testing.

I would like to understand your equation for alpha because it seems to me to be the sort of thing which might arise from an optimality condition in some iterative cosmology like CNS.

It is, as you say, mathematics. And therefore I have quoted arivero introduction

Now you have a longer equation representing successive approximations.
I would like to see this written out more explicitly.

And by the way, the value I have at hand for alpha is
0.007 297 352 533(27)

and the value which you get for alpha is
0.007 297 352 568 653...

In case you want to look at Smolin's article in that Universe or Multiverse book, the preprint is
http://arxiv.org/abs/hep-th/0407213

I do not think that your equation is necessarily in conflict with an iterative evolutionary model (although such equations in isolation, like a mysterious man dressed only in a raincoat, may scandalize and outrage propriety)

Here is the abstract on hep-th/0407213

Scientific alternatives to the anthropic principle
Lee Smolin
Comments: Contribution to "Universe or Multiverse", ed. by Bernard Carr et. al., to be published by Cambridge University Press.

"It is explained in detail why the Anthropic Principle (AP) cannot yield any falsifiable predictions, and therefore cannot be a part of science. Cases which have been claimed as successful predictions from the AP are shown to be not that. Either they are uncontroversial applications of selection principles in one universe (as in Dicke's argument), or the predictions made do not actually logically depend on any assumption about life or intelligence, but instead depend only on arguments from observed facts (as in the case of arguments by Hoyle and Weinberg). The Principle of Mediocrity is also examined and shown to be unreliable, as arguments for factually true conclusions can easily be modified to lead to false conclusions by reasonable changes in the specification of the ensemble in which we are assumed to be typical.
We show however that it is still possible to make falsifiable predictions from theories of multiverses, if the ensemble predicted has certain properties specified here. An example of such a falsifiable multiverse theory is cosmological natural selection. It is reviewed here and it is argued that the theory remains unfalsified. But it is very vulnerable to falsification by current observations, which shows that it is a scientific theory.
The consequences for recent discussions of the AP in the context of string theory are discussed."

Last edited: Oct 5, 2004
10. Oct 5, 2004

### marcus

If I apply a step of Middleschool algebra to your equation, what I get is:

$$e^{\pi^{2}/4}\alpha^{1/2} = 1+{\alpha \over (2\pi)^0 }+{\alpha^2 \over (2\pi)^1 }+{\alpha^3 \over (2\pi)^3 } + .......}$$

$$e^{\pi^{2}/4} = \alpha^{-1/2} +{\alpha^{1/2} \over (2\pi)^0 }+{\alpha^{3/2} \over (2\pi)^1 }+{\alpha^{5/2} \over (2\pi)^3 } + .......}$$

apologies for tinkering like this, just want to see how it looks when kneaded a bit

so let's imagine that instead of the number 137.0359991 (or whatever it is believed to be now) we want its square root 11.7062376... and let us define
a function

$$F(X) = X +{X^{-1} \over (2\pi)^0 }+{X^{-3} \over (2\pi)^1 }+{X^{-5} \over (2\pi)^3 } + {X^{-7} \over (2\pi)^6 }....$$

So then if we solve the equation

$$F(X) = e^{\pi^{2}/4}$$

then according to you, HdV, the solution X will turn out to
be this famous universal number 11.70623762...
which so far has only been determined by experimental measurement.

heh heh

shocking

Last edited: Oct 5, 2004
11. Oct 5, 2004

### Hans de Vries

I did use the NIST value of 0.007 297 352 568 (24)

Nist web page for alpha

Regards, Hans

12. Oct 5, 2004

### arivero

Hmm the kind of mathematics of Ramanujan, perhaps

13. Oct 5, 2004

### Nereid

Staff Emeritus
Apologies

Neutrino (and other lepton) mixing angles are, as you say, not at all like numerology, and coming at ratios from a good theoretical direction is of course a worthwhile exercise, in very good taste.

Wrt to your other post, I did some playing around a while ago, and found that it's extraordinarily easy to get some 'cool' numbers, to within 1%. For example, you can always change a really big number to something between 0 and 10 (or maybe 20 or so), simply by taking a log (natural or otherwise). $$\pi^2$$ is pretty close to 10, and various x-n, where x is a cool integer or trancendental, make a nice toolkit for fine tuning (+ or -, integer powers, etc). It's only slightly more difficult to create things that look like they're the first couple or three terms of a series. Which is why I think it's important to demand a prediction before any proposal is treated seriously. Or perhaps there's too much diamond dust in my drink.

14. Oct 5, 2004

### marcus

I see that your value is more up-to-date, and more accurate than what I was using. Thanks.

15. Oct 6, 2004

### arivero

I partly agree, but with a couple of caveats. First, that even if it is easy, it has some merit, as I have never seen extraordinary precision in most of the cranks webpages. Some of them even alter the value of the constants to fit into their theories! Some 'cool' relationships can be always found, of course. Larsson, and indirectly Lubos, brought a couple of famous ones:
TL attributes the second one to Lubos, but I have read it attributed to Feynman himself, just as an example of the kind we are discussing, random relationships. It could be interesting to know the origin of the first one to alpha, as it is a variant of KdV idea.

(Incidentaly, the third part of Wilczek "Plannk mountain" article touchs a bit the relationship between Planck scale and proton scale)

Secondly, I find in the first set of de Vries' relations a bit more of symmetry that usual. The form (q^1 - q^-1) is very geometrical, and it is very appealing that it happens between muon and tau, somehow marking the end of the number of generations. The fact of being able to find the same format for the quotient between first and second generation is also amazing as a coincidence. It lets you a lot of freedom to manipulate the relationship, for instance you can write a pair of equations
$${m_e m_\tau^2 \over m_\mu^3}=e^\pi, \; {m_e m_\tau^3 \over m_\mu^4}=e^{\frac1\pi}$$

The appeal of this format is, again, that it does not give trivial room to "predict" more generations.

PS: The Mikanata Smirnov paper I was referring before is hep-ph/0405088. They refer to a conjectured relation between Cabibbo angle and the quotient tau/muon

16. Oct 6, 2004

### marcus

Alejandro, and anyone else interested, have any of us checked the full HdV equation to see if the approximation is as good as he says?

With all respect to HdV, sometimes people make numerical error and it is good to have an independent verification.

In fact i tried to check it with my calculator and perhaps I did something wrong but I did not get such a good approximation as his post led me to expect So my calculation may be at fault. I would appreciate if someone else try it.

Here is my calculation

The inverse alpha number is 137.0359991 and I will assume that we want a formula for its square root 11.7062376...

Let us define this function

$$F(X) = X +{X^{-1} \over (2\pi)^0 }+{X^{-3} \over (2\pi)^1 }+{X^{-5} \over (2\pi)^3 } + {X^{-7} \over (2\pi)^6 } +....$$

HdV post leads me to expect that if we solve for X in the equation

$$F(X) = e^{\pi^{2}/4}$$

then the solution X will turn out to be 11.70623762...

Conversely, if I plug in this value for X then I should get (to a good approximation) the exp(pi^2/4) number, which is 11.79176...

So I expect

$$F(11.70623762...) = 11.79176...$$

But I didn't get very close to this. BTW I was avoiding roundoff error by using 137.0359991 to build even powers of 11.70623762, so there is not the most obvious explanation. Can anyone help?

17. Oct 6, 2004

### marcus

Oh oh,
I redid the calculation, using inverse alpha = 137.03599911, which I get from NIST website.
And using as the square root the approximation 11.70623762

and what do you think?
when I plug these into the function F(X) defined here, then I get 11.79176139.

But this is also equal, to the limit of the accuracy of my hand calculator, the exp(pi^2/4) number, which is also 11.79176139, or so my calculator says.

such coincidences are rare, maybe I am still doing something wrong, or the calculator battery is low. or I am dreaming this. Please someone else check the calculation

Here is the form of F(X) I used in the calculation

$$F(X) = X(1 +{X^{-2} \over (2\pi)^0 }+{X^{-4} \over (2\pi)^1 }+{X^{-6} \over (2\pi)^3 } )$$

this way all the powers of X are even and one can use 137.03599911, reducing the amount of arithmetic, then at the end there is one multiplication by 11.70623762

Last edited: Oct 6, 2004
18. Oct 6, 2004

### Hans de Vries

It's OK

X = sqrt(137.03599911) = 11.7062376154766

11.7062376154766 = X
00.0854245431237 = X^-1 /(2pi)^0
00.0000992128957 = X^-3 /(2pi)^1
00.0000000183389 = X^-5 /(2pi)^3

=================================
11.7917613898350 = exp(pi^2/4)

_3.1415926536222 = pi (measured...)
_3.1415926535897 = pi (exact)

Regards, Hans

Tip: The Windows calculator is exact to 32 digits.

Last edited: Oct 6, 2004
19. Oct 6, 2004

### marcus

so when did you first see this?

also it is maybe not so important but what do you mean by the "measured" value of pi?

20. Oct 6, 2004

### marcus

Rilke has a funny sonnet about mohammed being in a cave and
a visitor comes and shows him something. it could be like the experience of seeing this. do you read German? I will see if I can find the poem
Yes, here it is:

Mohammeds Berufung

Da aber als in sein Versteck der Hohe,
sofort Erkennbare: der Engel, trat,
aufrecht, der lautere und lichterlohe:
da tat er allen Anspruch ab und bat
bleiben zu dürfen der von seinen Reisen
innen verwirrte Kaufmann, der er war;
er hatte nie gelesen - und nun gar
ein solches Wort, zu viel für einen Weisen.

Der Engel aber, herrisch, wies und wies
ihm, was geschrieben stand auf seinem Blatte,
und gab nicht nach und wollte wieder: Lies.
Da las er: so, dass sich der Engel bog.
Und war schon einer, der gelesen hatte
und konnte und gehorchte und vollzog.

Last edited: Oct 6, 2004