# Koide Mass Formula for Neutrinos

Homework Helper

## Main Question or Discussion Point

Abstract: Since 1982 the Koide mass relation has provided an amazingly
accurate relation between the masses of the charged leptons. In
this note we show how the Koide relation can be expanded to cover
the neutrinos, and we use the relation to predict neutrino masses.

Full paper attached, this version is slightly abbreviated.

In 1982, Yoshio Koide [ http://www.arxiv.org/abs/hep-ph/0506247 ], discovered a formula relating the
masses of the charged leptons:

$$\frac{(\sqrt{m_e} + \sqrt{m_\mu} + \sqrt{m_\tau})^2} {m_e+m_\mu+m_\tau} = \frac{3}{2}.$$

Written in the above manner, this relation removes one degree of
freedom from the three charged lepton masses. In this paper, we
will first derive this relation as an eigenvalue equation, then
obtain information about the other degrees of freedom, and finally
speculatively apply the same techniques to the problem of predicting the neutrino masses.

If we suppose that the leptons are composite particles made up of
colorless combinations of colored subparticles or preons, we expect
that the three colors must be treated equally, and therefore a
natural form for a matrix operator that can cross generation
boundaries is the \emph{circulant}:

$$\Gamma(A,B,C) = \left( \begin{array}{ccc} A&B&C\\C&A&B\\B&C&A \end{array}\right).$$

where A, B and C are complex constants. Other authors have
explored these sorts of matrices in the context of neutrino masses
and mixing angles. Such matrices have eigenvectors of the form:

$$|n\ra = \left(\begin{array}{c} 1\\e^{+2in\pi/3}\\e^{-2in\pi/3}\end{array}\right), \;\;\;\;n=1,2,3.$$

If we require that the eigenvalues be real, we obtain that $A$ must
be real, and that B and C are complex conjugates. This reduces
the 6 real degrees of freedom present in the 3 complex constants
A, B and C to just 3 real degrees of freedom, the same as the
number of eigenvalues for the operators. In order to parameterize
these sorts of operators,let us write:

$$\Gamma(\mu,\eta,\delta) = \mu\left(\begin{array}{ccc} 1&\eta\exp(+i\delta)&\eta\exp(-i\delta)\\ \eta\exp(-i\delta)&1&\eta\exp(+i\delta)\\ \eta\exp(+i\delta)&\eta\exp(-i\delta)&1 \end{array}\right),$$

where we can assume eta to be non negative. Note that while
eta and delta are pure numbers, mu scales with the
eigenvalues. Then the three eigenvalues are given by:

$$\begin{array}{rcl} \Gamma(\mu,\eta,\delta)\;|n\ra &=& \lambda_n \;|n\ra,\\ &=&\mu(1+2\eta\cos(\delta+2n\pi/3))\;|n\ra. \end{array}$$

The sum of the eigenvalues are given by the trace of Gamma:
$$\lambda_1+\lambda_2+\lambda_3 = 3\mu,$$

and this allows us to calculate mu from a set of eigenvalues. The
sum of the squares of the eigenvalues are given by the trace of
Gamma^2:
$$\lambda_1^2+\lambda_2^2+\lambda_3^2 = 3\mu^2(1+2\eta^2),$$

and this gives a formula for eta^2 in terms of the eigenvalues:
$$\frac{\lambda_1^2+\lambda_2^2+\lambda_3^2} {(\lambda_1+\lambda_2+\lambda_3)^2} = \frac{1+2\eta^2}{3},$$

The value of delta is then easy to calculate.

We have restricted Gamma to a form where all its eigenvalues are
real, but it is still possible that some or all will be negative.
For situations where all the eigenvalues are non negative, it is
natural to suppose that our values are eigenvalues not of the
Gamma matrix, but instead of Gamma^2.

The masses of the charged leptons are positive, so let us compute
the square roots of the masses of the charged leptons, and find the
values for mu_1, eta_1^2 and delta_1, where the subscript
will distinguish the parameters for the masses of the charged
leptons from that of the neutral leptons.

Given the latest PDG data (MeV):
$$\begin{array}{rcccl} m_1 &=& m_e &=& 0.510998918 (44)\\ m_2 &=& m_\mu &=& 105.6583692 (94)\\ m_3 &=& m_\tau &=& 1776.99 +0.29-0.26 \end{array}$$

and ignoring, for the moment, the error bars, and keeping 7 digits
of accuracy, we obtain
$$\begin{array}{rcl} \mu_1 &=& 17.71608 \;\;{\rm MeV}^{0.5}\\ \eta_1^2 &=& 0.5000018\\ \delta_1 &=& 0.2222220 \end{array}$$

The fact that eta_1^2 is very close to 0.5 was noticed in 1982
by Yoshio Koide at a time when the
tau mass had not been experimentally measured to anywhere near
its present accuracy. That delta_1 is close to
2/9 went unnoticed until this author discovered it in
2005 and announced it here on Physics Forums.

The present constraints on the electron, muon and tau masses exclude
the possibility that \delta_1 is exactly 2/9, while on the other
hand, \eta_1^2 = 0.5 fits the data close to the middle of the
error bars. If we interpret \Gamma as a matrix of coupling
constants, \eta^2 = 1/2 is a probability. Thus if the preons are
to be spin-1/2 states, the P = (1+\cos(\theta))/2 rule for
probabilities implies that the three coupled states are
perpendicular, a situation that would be more natural for classical
waves than quantum states.

By making the assumption that \eta_1^2 is precisely 0.5, one
obtains a prediction for the mass of the tau. Since the best
measurements for the electron and muon masses are in atomic mass
units, we give the predicted tau mass in both those units and in MeV:
$$\begin{array}{rcll} m_\tau &=& 1776.968921(158) \;\; & {\rm MeV}\\ &=& 1.907654627(46) \;\; & {\rm AMU}. \end{array}$$

The error bars in the above, and in later calculations in this
paper, come from assuming that the electron and muon masses are
anywhere inside the measured mass error bars.

This gives us the opportunity to fine tune our estimate for \mu_1
and \delta_1. Since the electron and muon data are the most
exact, we assume the Koide relation and compute the tau mass from
them. Then we compute \mu_1 and then \delta_1 over the range of
electron and muon masses, obtaining:
$$\begin{array}{rcll} \delta_1 &=&.22222204715(311)\;\;\;&\textrm{from MeV data}\\ &=&.22222204717(48)\;\;\;&\textrm{from AMU data}. \end{array}$$

If \delta_1 were zero, the electron and muon would have equal
masses, while if \delta_1 were \pi/12, the electron would be
massless. Instead, \delta_1 is close to a rational fraction,
while the other terms inside the cosine are rational multiples of
pi. Using the more accurate AMU data, the difference
between \delta_1 and 2/9 is:
$$2/9 - \delta_1 = 1.7505(48)\;\times10^{-7},$$

and we can hope that a deeper theory will allow this small
difference to be computed. This difference could be written as
$$1.75\;\times 10^{-7} = \frac{4\pi}{3^{12}}\left(\alpha + \mathcal{O}(\alpha^2)\right)$$

for example.

Note that \delta_1 is close to a rational number, while the other
terms that are added to it inside the cosine are rational multiples of pi. This
distinction follows our parameterization of the eigenvalues in that
the rational fraction part comes from the operator Gamma, while
the 2n\pi/3 term comes from the eigenvectors. Rather than
depending on the details of the operator, the 2n\pi/3 depends only
on the fact that the operator has the symmetry of a circulant
matrix.

We will use m_1, m_2 and m_3 to designate the masses of the
neutrinos. The experimental situation with the neutrinos is
primitive at the moment. The only accurate measurements are from
oscillation experiments, and are for the absolute values of the
differences between squares of neutrino masses. Recent 2\sigma
data are:
$$\begin{array}{rcll} |m_2^2-m_1^2| &=& 7.92(1\pm.09) \times 10^{-5}\;\; & {\rm eV}^2\\ |m_3^2-m_2^2| &=& 2.4(1+0.21-0.26) \times 10^{-3}\;\; & {\rm eV}^2 \end{array}$$

Up to this time, attempts to apply the unaltered Koide mass formula
to the neutrinos have
failed, but these attempts have assumed that
the square roots of the neutrino masses must all be
positive. Without loss of generality, we
will assume that \mu_0 and \eta_0 are both positive, thus there
can be at most one square root mass that is negative, and it can be
only the lowest or central mass.

Of these two cases, having the central mass with a negative square
root is incompatible with the oscillation data, but we can obtain
\eta_0^2 = 1/2 with masses around:
$$\begin{array}{rcl} m_1 &=& 0.0004\;\; {\rm eV},\\ m_2 &=& 0.009\;\; {\rm eV},\\ m_3 &=& 0.05\;\; {\rm eV}. \end{array}$$

These masses approximately satisfy the squared mass differences of
as well as the Koide relation as
follows:
$$\frac{m_1+m_2+m_3}{(-\sqrt{m_1}+\sqrt{m_2}+\sqrt{m_3})^2} \;=\;\frac{2}{3}.$$

The above neutrino masses were chosen to fix the value of eta_0^2
= 1/2. As such, the fact that this can be done is of little
interest, at least until absolute measurements of the neutrino
masses are available. However, given these values, we can now
compute the mu_0 and delta_0 values for the neutrinos. The
results give that, well within experimental error:
$$\begin{array}{rcl} \delta_0 &=& \delta_1\;\; +\;\; \pi/12,\\ \mu_1 / \mu_0 &=& 3^{11}. \end{array}$$

Recalling the split between the components of the cosine in the
charged lepton mass formula, the fact that pi/12 is a rational
multiple of pi suggests that it should be related to a symmetry of
the eigenvectors rather than the operator. One possible explanation
is that in transforming from a right handed particle to a left
handed particle, the neutrino (or a portion of it) pick up a phase
difference that is a fraction of pi. As a result the neutrino
requires 12 stages to make the transition from left handed back to
left handed (with the same phase), while the electron requires only
a single stage, and it is natural for the mass hierarchy between the
charged and neutral leptons to be a power of 12-1 = 11.

We will therefore assume that the relations given above
are exact, and can then use the
measured masses of the electron and muon to write down predictions
for the absolute masses of the neutrinos. The parameterization of
the masses of the neutrinos is as follows:
$$m_n = \frac{\mu_1}{3^{11}}(1+\sqrt{2}\cos(\delta_1+\pi/12 + 2n\pi/3)),$$

where mu_1 and delta_1 are from the charged leptons.
Substituting in the the measured values for the electron and muon
masses, we obtain extremely precise predictions for the neutrino
masses:
$$\begin{array}{rcll} m_1 &=& 0.000383462480(38) \;\; & {\rm eV}\\ &=&0.4116639106(115)\times 10^{-12} \;\; & {\rm AMU} \end{array}$$

$$\begin{array}{rcll} m_2 &=& 0.00891348724(79)\;\; & {\rm eV}\\ &=&9.569022271(246)\times 10^{-12} \;\; & {\rm AMU} \end{array}$$
$$\begin{array}{rcll} m_3 &=& 0.0507118044(45) \;\; & {\rm eV}\\ &=&54.44136198(131)\times 10^{-12} \;\; & {\rm AMU} \end{array}$$

Similarly, the predictions for the differences of the squares of the
neutrino masses are:
$$\begin{array}{rcll} m_2^2-m_1^2 &=& 7.930321129(141)\times 10^{-5} \;\; & {\rm eV}^2\\ &=&.913967200(47)\times 10^{-24} \;\; & {\rm AMU}^2 \end{array}$$
$$\begin{array}{rcll} m_3^2-m_2^2 &=& 2.49223685(44)\times 10^{-3} \;\; & {\rm eV}^2\\ &=&2872.295707(138)\times 10^{-24} \;\; & {\rm AMU}^2 \end{array}$$

As with the Koide predictions for the tau mass, these predictions
for the squared mass differences are dead in the center of the error
bars. We can only hope that the future will show our calculations
to be as prescient as Koide's.

That the masses of the leptons should have these sorts of
relationships is particularly mysterious in the context of the
standard model. It is
hoped that this paper will stimulate thought among theoreticians.
Perhaps the fundamental fermions are bound states of deeper objects.

The author would like to thank Yoshio Koide and Alejandro Rivero for
their advice, encouragement and references, and Mark Mollo
(Liquafaction Corporation) for financial support.

Carl

#### Attachments

• 129.8 KB Views: 223
Last edited by a moderator:

Related Other Physics Topics News on Phys.org
Homework Helper
The above LaTex came out fairly well. The only problem is that the \rangles that should appear after each |n got deleted, likely because I use a macro for these in the original.

Since the above, I've got a new result to add to the pile. There were some mysterious 12s showing up in the above, particularly with the charged / neutral mass ratio. It turns out that the MNS mixing matrix in tribimaximal form (which is generally assumed to be the correct one in most recent reviews of neutrinos) can be put into a form where it is a 24th root of unity. To get it into this form, one multiplies it on the left by nothing other than a 3x3 matrix of eigenvectors of the circulant matrix. Here's the details:

$$\frac{1}{\sqrt{3}} \left(\begin{array}{ccc}1&e^{+2i\pi/3}&e^{-2i\pi/3}\\ 1&1&1\\ 1&e^{-2i\pi/3}&e^{+2i\pi/3}\end{array}\right) \left(\begin{array}{ccc}\sqrt{2/3}&\sqrt{1/3}&0\\ -\sqrt{1/6}&\sqrt{1/3}&-\sqrt{1/2}\\ -\sqrt{1/6}&\sqrt{1/3}&\sqrt{1/2}\end{array}\right)$$

$$=\left(\begin{array}{ccc} \sqrt{1/2}&\;\;\;\;0\;\;\;\;&-i\sqrt{1/2}\\ 0&1&0\\ \sqrt{1/2}&0&i\sqrt{1/2}\end{array}\right) = \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)^{1/24}$$

In the above, the MNS matrix is on the top right, the eigenvectors are top left. The MNS matrix transforms vectors of $$(\nu_e, \nu_\mu, \nu_\tau)$$ into vectors of $$(\nu_1,\nu_2,\nu_3)$$ where the first are what you get when you arrange a weak interaction with a charged lepton, and the second are the mass eigenstates of the neutrinos.

Some notes. First, there is one column of $$\sqrt{1/3}$$ in the MNS matrix and one row of the same in the array of eigenvectors. This implies that the (1,1,1) eigenvectors of the operator must correspond to the muon neutrino and the 2nd mass eigenstate. This is very useful information because it allows us to remove almost all the redundancy from the values of $$\delta$$ in the eigenvalue equation. That is, instead of simply finding the smallest value of $$\delta$$ that gives the three desired eigenvalues, we can instead choose the value that makes the (1,1,1) eigenvalue be the muon in the case of the charged leptons, and the middle neutrino in the case of the uncharged leptons. That is,

$$\lambda_{(1,1,1)} = \mu(1 + 2\eta^2\cos(\delta))$$

with no need to guess which value of n to choose in the $$2n\pi/3$$ that otherwise shows up in the cosine.

As a result of making this change, the new delta value for the electron can be chosen as:

$$\delta_1 = 2\pi/3 - .22222204717(48) = \frac{\pi}{180}(120 - 12.73)$$
or
$$\delta_1 = 4\pi/3 + .22222204717(48) = \frac{\pi}{180}(240 + 12.73)$$

where the two options correspond to the two solutions for cos(x) = y, and the neutrino delta is:

$$\delta_0 = 7\pi/12 - .22222204717(48) =\frac{\pi}{180}(105-12.73)$$
or
$$\delta_0 = 17\pi/12 + .22222204717(48) =\frac{\pi}{180}(75+12.73)$$

It turns out that having the angle delta_1 between 60 and 120 degrees is useful in that it implies that we can add a hidden dimension to modify the Pauli algebra's usual relationship between phase and probability so that we can match the above. This is a somewhat mysterious statement that I will expound on later. It's very simple math and not at all difficult. Accordingly, we have to choose the upper of each of the two equations, and for the angle calculation, we can use either 120-12.73 or 105-12.73, with an assumption that the electron is the simple mass relationship particle, or that the neutrino is, respectively.

Carl

Last edited:
Homework Helper
Homework Helper
My neutrino mass formula is now on arXiv in that Yoshio Koide has uploaded a paper referencing my work here:

Tribimaximal Neutrino Mixing and a Relation Between Neutrino- and Charged Lepton-Mass Spectra
Yoshio Koide
Brannen has recently pointed out that the observed charged lepton masses satisfy the relation $$m_e +m_\mu +m_\tau = {2/3} (\sqrt{m_e}+\sqrt{m_\mu}+\sqrt{m_\tau})^2$$, while the observed neutrino masses satisfy the relation $$m_{\nu 1} +m_{\nu 2} +m_{\nu 3} = {2/3} (-\sqrt{m_{\nu 1}}+\sqrt{m_{\nu 2}} +\sqrt{m_{\nu 3}})^2.$$ Stimulated by this fact, a neutrino mass matrix model, which gives a relation between the observed neutrino- and charged lepton-mass spectra, and which yields the so-called tribimaximal mixing in the neutrino sector, is proposed.
http://www.arxiv.org/abs/hep-ph/0605074

Carl

Homework Helper
Another reference to my mass formula has appeared in the literature on page 10 of this pdf file:

Koide relation [Y. Koide, Lett. Nuov. Cim., 34 (1982), 201]:
$$\frac{m_e+m_\mu+m_\tau}{(\sqrt{m_e} + \sqrt{m_\mu} + \sqrt{m_\tau})^2} = \frac{2}{3}$$
with accuracy 10^{-5}

$$tan \theta_c = \left(\frac{\sqrt{m_\mu}-\sqrt{m_e}}{2\sqrt{m_\tau}}-\sqrt{m_\mu}-\sqrt{m_e}}\right)^{1/3}$$

Both relations can be reproduced if [C A Brannen] Neutrinos
$$m_i = m_0(Z_i + Z_0)^2$$

$$\sum_i Z_i = 0$$
$$Z_0 = \sqrt{\sum_i Z_i^2/3}$$

Another representation which is closely connected to circulant symmetry
http://users.physik.tu-muenchen.de/lsoft/seminar-talks/AlexeiSmirnov06.pdf

Carl

I deal in old fashion classical terms and am awaiting judgement on a submission to PF independent research. Meanwhile I would greatly appreciate some clarification.
As far as I can understand the work mentioned on this forum, it deals only with mass and does not show any relationship between say, charge and mass, or charge and wavelength; neither does it explain why particles have their particular mass but simply predicts a mass number.
Would you please comfirm my understanding or explain where I am wrong.

Homework Helper
You are pretty much correct, the above paper just predicts masses. The one by Yoshio Koide applies a traditional vacuum expectation value Higgs field approach to the problem.

The reason I wrote the paper that way is because my papers that gave an explanation were pretty much ignored. Some of that was due to their using new math. What you end up with is a bunch of incomprehensible mathematics, followed by a formula predicting masses. It's not very convincing unless you can follow the mathematics.

So I've been concentrating on making the mathematics accessible. In many ways, what I am promoting are alternatives to doing quantum mechanics. I believe that my alternatives are easier to learn for the student, give results with less effort, and are closer to the underlying reality.

To get a hint at how the work is easier for the student, see my write up on Pauli spinors at the Wikipedia that gives a "trick" on how to find the eigenvectors of a spin operator:
http://en.wikipedia.org/wiki/Spinor#Example:_Spinors_of_the_Pauli_Spin_Matrices

Basically, the QM I work with involves a generalization of the above "trick". None of this is standard practice in the industry, which is why I wrote the wikipedia section without reference to anything heretical. I'm quite convinced that this is far superior to the usual way of doing QM.

I'm working on a new version of the MASSES2.pdf paper that will add back in the theory, but again without most of the "adult mathematics". It begins with the "density marix formulation" of quantum mechanics. I plan on submitting it to arXiv and have received a note from them apparently allowing it in the "physics" section. For an amateur, that's about as good as you can get from them.

The density matrix formulation of QM is an alternative to the usual "state vector formulation". It's based on two observations. The first is that any prediction from a state vector description of a quantum object can also be predicted from a density matrix description. Thus the two formulations are on equal grounds as far as supposing which is the fundamental.

The second observation is that just as it is possible to define a density matrix state from a state vector state, it is also possible to define a state vector state from a density matrix state. In moving from a state vector description of a situation to a density matrix description, one loses the arbitrary complex phase (i.e. the U(1) gauge freedom). Thus in going the reverse direction you must add the arbitrary phase back in, and in doing this you get insight into the nature of the complex numbers used in QM, and on gauge theory.

When you translate all that into QFT, you get insight into the nature of the quantum vacuum.

My notes for this are mostly written up. I'm hoping to publish them before the end of the month. Until then, for some introduction to density matrix theory, try http://www.DensityMatrix.com a website I started a couple weeks ago. To get a hint as to how one goes about obtaining weak hypercharge and weak isospin from density matrix calculations, my website is http://www.snuark.com [Broken] , which is devoted to the subject of the theoretical underpinnings beneath this mass theory. But these websites are just days old and nowhere near sufficient in explaining these things.

Carl

Last edited by a moderator:
Carl

Thanks for the reply. My education did not extend to the level needed to follow your work, but being dissatisfied with the inability of the experts to explain their work to the layman got me interested in interpretation.

Eventual I realise there was a need for an explanation of particle structure before Quantum physics could be explained in words. With this aim I used the experiments described in The Particle Explosion and books for beginners by Baggot, Veltman, Pais and others, to produce a formula that links mass with field force and charge. A continuation shows a possible connection with wave structure (this wave structure, I show, is an astronomically observable entity).

This is the article awaiting approval for entry into the Independent research forum which I hope we will be able to discuss and compare with your work sometime soon.

John

Homework Helper
My papers predicting the neutrino masses:

Koide Mass Formula for Neutrinos, April 7, 2006
http://brannenworks.com/MASSES.pdf
and

The Lepton Masses, May 2, 2006
http://brannenworks.com/MASSES2.pdf

predicted squared mass differences for the neutrinos of
7.93x10^{-5} eV and 2.49x10^{-5} eV. At the time
I wrote them, the best fit values were 7.92 and 2.4,
which means I was saying that the best fit experimental
numbers were high by 0.13% and low by 3.6%.

Now the latest figures from Minos (released but
not yet published, I believe) have shifted the
2.4 value up to 2.5 (see
http://www.arxiv.org/abs/hep-ph/0606060 ). So now
my formula says that the best fit experimental
numbers are high by 0.13% and 0.4%, a 9x
improvement on the second value.

Carl

Another paper that uses my neutrino masses,
but at a lower accuracy is:
http://www.arxiv.org/abs/hep-ph/0605074

Homework Helper
The Koide paper that references my formula for the neutrino masses:

Tribimaximal Neutrino Mixing and a Relation Between Neutrino- and Charged Lepton-Mass Spectra
Y. Koide
http://www.arxiv.org/abs/hep-ph/0605074

Now has two papers referencing it:

Magic Neutrino Mass Matrix and the Bjorken-Harrison-Scott Parameterization
C. S. Lam
http://www.arxiv.org/abs/hep-ph/0606220

Real Invariant Matrices and Flavour-Symmetric Mixing Variables with Emphasis on Neutrino Oscillations
P. F. Harrison, W. G. Scott, T. J. Weiler
http://www.arxiv.org/abs/hep-ph/0607335

Also, I've submitted a short abstract for the DPF2006 meeting in Hawaii:
http://www.dpf2006.org/ [Broken]

Yoshio Koide will attend the meeting, so I will likely go even if my abstract is rejected (as is likely).

Carl

Last edited by a moderator:
Homework Helper
Another paper referencing Koide's paper has come to my notice:

Real Invariant Matrices and Flavour-Symmetric Mixing Variables with Emphasis on Neutrino Oscillations
P. F. Harrison, W. G. Scott, T. J. Weiler
http://www.arxiv.org/abs/hep-ph/0607335

Carl

Homework Helper
I've started writing up a derivation of the Koide formula from first principles. The problem is that "first principles" means going back to the foundations of quantum mechanics and rebuilding the standard model from the assumption that density matrices are fundamental and spinors (and Hilbert space) are just mathematical conveniences.

The standard model is fairly complicated and deriving it is quite involved. That it has to be done with a new foundation makes it possible, but not easy. Accordingly, I'm writing a description of these new foundations of quantum mechanics in book form, with the target audience first year graduate students in physics.

The first two chapters introduce Clifford algebra in as painless and easy way as I can imagine, by analogy with the familiar Pauli spin matrices and the Dirac gamma matrices. The first chapter goes into everything that one can do, from a geometric point of view, with the algebra of the Pauli spin matrices. The chapter is mostly about what complex numbers are in a geometric context. The chapter ends with a formula showing how to reduce products of projection operators in the Pauli algebra, a calculational method that will be used repeatedly in the rest of the book.

So far, only the first chapter is complete:
http://www.brannenworks.com/dmaa.pdf

The second chapter continues the story with the Dirac algebra. The Dirac algebra allows us to investigate how particles appear in Clifford algebras. The particles of the Pauli algebra are trivial, one has spin +1/2 and spin -1/2 with the direction of orientation being the only choice. The Dirac algebra is more general and more interesting. We will discuss the various choices of operators that define the particles, and we will generalize to Clifford algebra.

Carl

Homework Helper
The Mohaptra and Smirnov review article on the neutrino masses is out:
http://staging.arjournals.annualreviews.org/doi/pdf/10.1146/annurev.nucl.56.080805.140534

My neutrino mass formula is mentioned briefly on page 599:

The Koide relation can be satisfied for neutrinos provided $$\sqrt{m_1} < 0$$ (76) implies strongly hierarchical mass spectrum, with $$m_1 = 3.9 10^{-4}eV$$ and $$\sqrt{m_1}<0$$ (76; see also Reference 77).

76. Brannen CA. http://brannenworks.com (2006)
77. Koide Y. hep-ph/0605074

This is the first published, peer reviewed mention of the formula.

Carl

Homework Helper
An article in Japanese gives more detail on the calculations for how the Koide formula is extended to the neutrinos. The various possible sign choices and how they work out with the charged leptons, neutrinos, and quarks are in a table on the bottom of page 23. A note on the coincidence with the Cabibbo angle is on the last page.

http://higgs.phys.kyushu-u.ac.jp/ppp06/slide/koide.pdf [Broken]

I presented some slides from this at the 2006 Joint Particle Physics meeting in Hawaii. The slides are here:
http://www.brannenworks.com/jpp06.pdf

Actually, the above are the slides I was tried to present. I ended up with a slightly older version that had been downloaded earlier.

Meanwhile, I'm making steady progress in describing how one obtains the Koide formula, and more, from a modification to the foundations of quantum mechanics. I'm about three chapters away from the Koide mass formulas. I've been keeping the book on the web here:
http://www.brannenworks.com/dmaa.pdf

Carl

Last edited by a moderator:
My submission has been rejected as I failed to comply with the submission rules; the length of my work makes compliance somewhat difficult. So I have decided to continue this debate without introducing my model.
What the Koide 'mass' formula does is to provide a mathematical short cut for finding the surface density of an expanding force field with a fixed quantity of matter. This can easily be checked using the following radii for the 'masses' already given: V(1) 3755164.695: V(2) 161549; V(3) 28395.06 in arbitrary units. That is to say that each neutrino is a different state (volume) of a single elementary particle.
This, of course is not a complete explanation of reality, but, it does explain what the prediction does without telling us why; that requires a different model.
John

Last edited:
Homework Helper
Some comments on that damned number

Let the complex Clifford algebra be that generated by the canonical basis vectors $$\hat{x}, \hat{y}, \hat{z}, \hat{s}, \hat{t}$$ where the signature is + + + + -, respectively.

There are a bunch of ways of writing primitive idempotents. One can convert between these. I will use as the base PI the following form:

$$(1\pm \widehat{zt})(1\pm \hat{s})(1\pm\widehat{ixyzst})/8$$

Let $$\kappa$$ be a real number. Let Z be an arbitrary element of the Clifford algebra (complex or real) that is built from the subalgebra of the Clifford algebra generated by $$\{\hat{x},\hat{y},\hat{s},\widehat{zt}\}$$. This subalgebra is used so as to define no preferred direction in 3-space. Transform the PIs as follows:
$$PI \to \exp(-\kappa Z) PI \exp(+\kappa Z)$$

where PI is any primitive idempotent (or any member of the Clifford algebra in fact). As shown in dmaa.pdf, the above transformation preserves multiplication and addition, and therefore maps complete sets of primitive idempotents to complete sets of primitive idempotents.

As in dmaa, we apply a potential energy which depends on blade, so that 1 contributes very low energy, vectors much more, bivectors even more, etc.

Now $$\widehat{ixyzst}$$ is a fixed point of the above transformation. Therefore the analysis in dmaa.pdf, which showed that the PIs would combine in pairs so as to cancel their $$\widehat{ixyzst}$$ components, still holds. But the other analyses may or may not.

As an example of how these transformations work, let $$Z = \widehat{xyzt}$$. Note that
$$\frac{d}{d\kappa}\exp(\kappa Z) = Z\exp(\kappa Z) = \exp(\kappa Z) Z.$$
The action of the transformation on the canonical basis vectors is as follows:

$$\begin{array}{rcl} \exp(-\kappa \widehat{xyzt})1\exp(+\kappa \widehat{xyzt}) &\to& 1,\\ \exp(-\kappa \widehat{xyzt})\hat{x}\exp(+\kappa \widehat{xyzt}) &\to& \hat{x}\exp(2\kappa\widehat{xyzt}),\\ \exp(-\kappa \widehat{xyzt})\hat{y}\exp(+\kappa \widehat{xyzt}) &\to& \hat{y}\exp(2\kappa\widehat{xyzt}),\\ \exp(-\kappa \widehat{xyzt})\hat{z}\exp(+\kappa \widehat{xyzt}) &\to& \hat{z}\exp(2\kappa\widehat{xyzt}),\\ \exp(-\kappa \widehat{xyzt})\hat{s}\exp(+\kappa \widehat{xyzt}) &\to& \hat{s},\\ \exp(-\kappa \widehat{xyzt})\hat{t}\exp(+\kappa \widehat{xyzt}) &\to& \hat{t}\exp(2\kappa\widehat{xyzt}), \end{array}$$

that is, the transformation multiplies all the canonical basis vectors on the right by $$\exp(2\kappa \widehat{xyzt})$$ except $$\hat{s}$$. The difference is that $$\hat{s}$$ commutes with $$\widehat{xyzt}$$.

Now $$(\widehat{xyzt})^2 = - 1$$, so we can think of $$\widehat{xyzt}$$ as an imaginary unit. The above transformation is therefore multiplication on the right by $$\exp(2 \hat{i}\kappa)$$ for most of the canonical basis vectors. Other choices for Z will give other patterns of multiplication, depending on how they commute with the canonical basis vectors.

The reason for considering transformations of a complete set of primitive idempotents is that, while these transformations preserve multiplication and addition, they do not preserve energy. Therefore, one imagines that if one understood the mass interaction better, one could write an equation for the energy as a function of the variable $$\kappa$$. Then the value of $$\kappa$$ that minimized the energy (after taking into account all possible Z values) would be a minimum of that function so we would have some sort of equation like:
$$\frac{d}{d\kappa} E(\kappa) = 0$$

Following the Koide prescription, the energy might be written as a quadratic function of $$\kappa$$. This would give some sort of linear equation for the minimizing value of $$\kappa$$.

Now my speculation is that the minimizing value of $$\kappa$$ will work out to be close to 1/9. Perhaps the proper ratio for energy between the scalar and vector components is 1/3 or 1/9. This will cause the exponential transformation to have a factor $$\exp(2i/9)$$, about the same as that damned number in Koide's equations.

Last edited:
Homework Helper
I need some scratch paper to do some calculation. Readers are invited to send me PMs rather than posting long off topic comments here.

Painleve coordinates:

$$ds^2 = -dt^2 + (dr-(2/r)^{0.5} dt)^2 + r^2 d\phi^2.$$

First step is to convert them to "Cartesian" form:

$$\begin{array}{rcl}ds^2 &=& -dt^2 + ((xdx+ydy)r^{-1}-\sqrt{2}\;r^{-0.5} dt)^2 + (y^2dx^2 + x^2dy^2 -2xy\;dx\;dy)r^{-2},\\&=&-dt^2 + dx^2 + dy^2 + 2r^{-1}dt^2 -2\sqrt{2}\;r^{-1.5}(x\;dx\;dt+y\;dy\;dt),\\I = \left(\frac{ds}{dt}\right)^2 &=&\frac{2}{r}-1 + \dot{x}^2 + \dot{y}^2-2\sqrt{2}r^{-1.5}(x\dot{x}+y\dot{y})\end{array}$$

The idea is to apply the Euler Lagrange equations to the integral over t for s:

$$s = \int \sqrt{\left(\frac{ds}{dt}\right)^2}\;dt = \int \sqrt{I}\;dt$$

to obtain the equations of motion in a Cartesian coordinate system. The resulting equations of motion (for x) are:

$$\frac{1}{2\sqrt{I}} \frac{\partial I}{\partial x} =\frac{d}{d t} \left[ \frac{1}{2\sqrt{I}} \left( \frac{\partial I}{\partial \dot{x}} \right) \right],$$ or

$$\frac{I^{-0.5}}{2} \frac{\partial I}{\partial x} =$$
$$\frac{I^{-0.5}}{2}\left( \frac{\partial^2 I}{\partial\dot{x}\partial x}\dot{x}+ \frac{\partial^2 I}{\partial\dot{x}\partial y}\dot{y}+ \frac{\partial^2 I}{\partial\dot{x}^2}\ddot{x}+ \frac{\partial^2 I}{\partial\dot{x}\partial\dot{y}}\ddot{y}\right)$$
$$-\frac{I^{-1.5}}{4}\left( \frac{\partial I}{\partial x}\frac{\partial I}{\partial\dot{x}}\dot{x}+ \frac{\partial I}{\partial y}\frac{\partial I}{\partial\dot{x}}\dot{y}+ \frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}}\ddot{x}+ \frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}}\ddot{y} \right)$$

I want to get rid of the sqrt(I) and the 4s, and rearrange the terms because we need to solve for the 2nd derivatives in terms of the 1st derivatives and constants. Moving the 2nd derivatives to the right hand side we get:

$$2I\frac{\partial I}{\partial x} + \left(\frac{\partial I}{\partial x}\frac{\partial I}{\partial\dot{x}}- 2I\frac{\partial^2 I}{\partial\dot{x}\partial x}\right)\dot{x} + \left(\frac{\partial I}{\partial y}\frac{\partial I}{\partial\dot{x}}- 2I\frac{\partial^2 I}{\partial\dot{x}\partial y}\right)\dot{y} =$$

$$\left(2I\frac{\partial^2 I}{\partial\dot{x}^2}- \frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}} \right)\ddot{x}+ \left(2I\frac{\partial^2 I}{\partial\dot{x}\partial\dot{y}}- \frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}} \right)\ddot{y}$$

Similarly, the Euler Lagrange equation in y gives:

$$2I\frac{\partial I}{\partial y} + \left(\frac{\partial I}{\partial y}\frac{\partial I}{\partial\dot{y}}- 2I\frac{\partial^2 I}{\partial\dot{y}\partial y}\right)\dot{y} + \left(\frac{\partial I}{\partial x}\frac{\partial I}{\partial\dot{y}}- 2I\frac{\partial^2 I}{\partial\dot{y}\partial x}\right)\dot{x} =$$

$$\left(2I\frac{\partial^2 I}{\partial\dot{y}^2}- \frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{y}} \right)\ddot{y}+ \left(2I\frac{\partial^2 I}{\partial\dot{y}\partial\dot{x}}- \frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{y}} \right)\ddot{x}$$

This matches the result in
https://www.physicsforums.com/showpost.php?p=1050082&postcount=25
so okay so far.

The above equations mix the 2nd deriviatives. To use them, we need to solve for $$\ddot{x}$$ and $$\ddot{y}$$. The equations are linear so they can be written in 2x2 matrix form. To solve the equations requires that we invert the matrix, which can be done provided its determinant is not zero. The determinant is:

$$\left(2I\frac{\partial^2 I}{\partial\dot{x}^2}-\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}}\right)\left(2I\frac{\partial^2 I}{\partial\dot{y}^2}-\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{y}}\right) -\left(2I\frac{\partial^2 I}{\partial\dot{x}\partial\dot{y}}-\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}}\right)\left(2I\frac{\partial^2 I}{\partial\dot{y}\partial\dot{x}}-\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{y}}\right)$$

Now to calculate the various derivatives for the Painleve coordinates.

Last edited:
Homework Helper
The first order partial derivatives of I are as follows:
$$\frac{\partial I}{\partial x} = -2r^{-3}x - \sqrt{8}r^{-1.5}\dot{x} +\sqrt{18}r^{-3.5}x(x\dot{x}+y\dot{y}),$$
$$\frac{\partial I}{\partial y} = -2r^{-3}y - \sqrt{8}r^{-1.5}\dot{y} +\sqrt{18}r^{-3.5}y(x\dot{x}+y\dot{y}),$$
$$\frac{\partial I}{\partial \dot{x}} = 2\dot{x}-\sqrt{8}r^{-1.5}x,$$
$$\frac{\partial I}{\partial \dot{y}} = 2\dot{y}-\sqrt{8}r^{-1.5}y$$

The required second order partial derivatives are:

$$\frac{\partial^2 I}{\partial x\partial\dot{x}} = -\sqrt{8}r^{-1.5} + \sqrt{18}r^{-3.5}x^2$$
$$\frac{\partial^2 I}{\partial x\partial\dot{y}} = \sqrt{18}r^{-3.5}xy$$
$$\frac{\partial^2 I}{\partial y\partial\dot{x}} = \sqrt{18}r^{-3.5}xy$$
$$\frac{\partial^2 I}{\partial y\partial\dot{y}} = -\sqrt{8}r^{-1.5} + \sqrt{18}r^{-3.5}y^2$$

$$\frac{\partial^2 I}{\partial \dot{x}^2} = 2$$
$$\frac{\partial^2 I}{\partial \dot{y}^2} = 2$$
$$\frac{\partial^2 I}{\partial \dot{x}\partial\dot{y}} = 0$$

And the value of I was:
$$I =\frac{2}{r}-1 + \dot{x}^2 + \dot{y}^2-\sqrt{8}r^{-1.5}(x\dot{x}+y\dot{y})$$

The 2nd partials are fairly simple. Substituting them into the formula for the determinant gives:

$$\left(4I-\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}}\right)\left(4I-\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{y}}\right) - \left(\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}}\right)\left(\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{y}}\right)$$

Dividing by 4, this factors to:

$$I\left(4I - \left(\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial \dot{y}}\right)^2\right)$$

which is zero when I=0 or when

$$I = \frac{1}{4}\left(\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial \dot{y}}\right)^2$$

Last edited:
Homework Helper
The determinant is evidently nonzero "almost everywhere". The usual coordinates blow up at r=2, so an interesting substitution is to put x=2, y=0. The determinant is then zero when either I=0, which gives:

$$I+1 = (\dot{x}-1)^2 + \dot{y}^2 = 1$$

At this point, $$-2 < \dot{x} < 0$$ for an object with mass, so equality can only happen with $$\dot{x}=\dot{y}=0$$. This is a light ray stuck trying to exit the black hole and modeling this does not concern me, so I will not try to avoid this division by zero.

The other case for a zero determinant happens when:

$$I = \dot{x}^2 + \dot{y}^2-2\dot{x} = 4(\dot{x}-1)^2\dot{y}^2$$

Last edited:
Homework Helper
Let's see if I can clean this up a little. The signs on the square roots are wrong, and it appears that it is easier if I rewrite things in river coordinates so:

$$I =\frac{2}{r}-1 + \dot{x}^2 + \dot{y}^2+\sqrt{8}r^{-1.5}(x\dot{x}+y\dot{y})$$

$$I = v_x^2 + v_y^2 -1$$

$$v_x = x\sqrt{2/r^3} + \dot{x}$$
$$v_y = y\sqrt{2/r^3} + \dot{y}$$

The first order partial derivatives are:

$$\frac{\partial I}{\partial x} = 2\sqrt{2}v_x\;r^{-1.5} -3\sqrt{2}(x^2v_x+xyv_y)r^{-3.5},$$

$$\frac{\partial I}{\partial y} = 2\sqrt{2}v_y\;r^{-1.5} -3\sqrt{2}(xyv_x+y^2v_y)r^{-3.5},$$

$$\frac{\partial I}{\partial \dot{x}} = 2v_x,$$
$$\frac{\partial I}{\partial \dot{y}} = 2v_y,$$

The required second order partial derivatives are:

$$\frac{\partial^2 I}{\partial x\partial\dot{x}} = 2\sqrt{2}r^{-1.5} - 3\sqrt{2}x^2r^{-3.5},$$
$$\frac{\partial^2 I}{\partial x\partial\dot{y}} = -3\sqrt{2}r^{-3.5}xy$$
$$\frac{\partial^2 I}{\partial y\partial\dot{x}} = -3\sqrt{2}r^{-3.5}xy$$
$$\frac{\partial^2 I}{\partial y\partial\dot{y}} = 2\sqrt{2}r^{-1.5} - 3\sqrt{2}y^2r^{-3.5},$$

$$\frac{\partial^2 I}{\partial \dot{x}^2} = 2$$
$$\frac{\partial^2 I}{\partial \dot{y}^2} = 2$$
$$\frac{\partial^2 I}{\partial \dot{x}\partial\dot{y}} = 0$$

Solving for the equations of motion, we find:

$$\begin{array}{rcl} \ddot{x} &=&-2x/r^3,\\ &&-4\dot{y}(x\dot{y}-y\dot{x})/r^3,\\ &&+4x/r^4,\\ &&+6x(x\dot{x}+y\dot{y})^2/r^5,\\ &&-2\sqrt{2}\dot{x}(\dot{x}^2+\dot{y}^2)/r^{1.5},\\ &&+6\sqrt{2}\dot{x}/r^{2.5},\\ &&+3\sqrt{2}\dot{x}(x\dot{x}+y\dot{y})/r^{3.5},\\ &&+4\sqrt{2}y(x\dot{y}-y\dot{x})/r^{4.5} \end{array}$$

$$\begin{array}{rcl} \ddot{y} &=&-2y/r^3,\\ &&+4\dot{x}(x\dot{y}-y\dot{x})/r^3,\\ &&+4y/r^4,\\ &&+6y(x\dot{x}+y\dot{y})^2/r^5,\\ &&-2\sqrt{2}\dot{y}(\dot{x}^2+\dot{y}^2)/r^{1.5},\\ &&+6\sqrt{2}\dot{y}/r^{2.5},\\ &&+3\sqrt{2}\dot{y}(x\dot{x}+y\dot{y})/r^{3.5},\\ &&-4\sqrt{2}x(x\dot{y}-y\dot{x})/r^{4.5} \end{array}$$

Last edited:
Homework Helper
As usual, the above has a typo, the $$(x\dot{x}+y\dot{y})$$ term should be squared, and I left off an overall factor of 2.

Organizing the terms according to powers of r we have:

$$\begin{array}{rcl|r}\ddot{x} &=&-\sqrt{2}\dot{x}(\dot{x}^2+\dot{y}^2)/r^{1.5}&1.5\\ &&+1.5\sqrt{2}\dot{x}(x\dot{x}+y\dot{y})^2/r^{3.5}&1.5\\ \hline &&-x/r^3&2.0\\ &&+3x(x\dot{x}+y\dot{y})^2/r^5&2.0\\ &&-2\dot{y}(x\dot{y}-y\dot{x})/r^3&2.0\\ \hline &&+3\sqrt{2}\dot{x}/r^{2.5}&2.5\\ &&+2\sqrt{2}y(x\dot{y}-y\dot{x})/r^{4.5}\;\;\;\;\;&2.5\\ \hline &&+2x/r^4&3.0\\ \hline \end{array}$$

$$\begin{array}{rcl|r}\ddot{y} &=&-\sqrt{2}\dot{y}(\dot{x}^2+\dot{y}^2)/r^{1.5}&1.5\\ &&+1.5\sqrt{2}\dot{y}(x\dot{x}+y\dot{y})^2/r^{3.5}&1.5\\ \hline &&-y/r^3&2.0\\ &&+3y(x\dot{x}+y\dot{y})^2/r^5&2.0\\ &&+2\dot{x}(x\dot{y}-y\dot{x})/r^3&2.0\\ \hline &&+3\sqrt{2}\dot{y}/r^{2.5}&2.5\\ &&-2\sqrt{2}x(x\dot{y}-y\dot{x})/r^{4.5}\;\;\;\;\;&2.5\\ \hline &&+2y/r^4&3.0\\ \hline \end{array}$$

The simulation is here:
http://www.gaugegravity.com/testapplet/SweetGravity.html [Broken]

Last edited by a moderator:
Homework Helper
Since the elementary particles all have spin, one might suppose that the black holes of most interest would be the ones with maximal rotation. For this reason, I will now attempt to derive equations of motion for the orbits of the Kerr metric in Doran-Cartesian coordinates. Until further notice, the following equations are taken from http://arxiv.org/abs/gr-qc/0411060. I've taken M=1. Q is the charge which seems like a useful thing to keep around.

The metric is:
$$ds^2 = -dt^2 + \left(\frac{\rho}{R} dr + \frac{\sqrt{2r}}{\rho}(dt - a \sin^2(\theta) d\phi)\right)^2 + \rho^2d\theta^2 + R^2\sin^2(\theta)d\phi^2$$

$$R = (r^2 + a^2)^{0.5},\;\;\;\rho = (r^2+a^2\cos(\theta))^{0.5}$$

These are the "coordinates of the flat background through which the river of space flows into the black hole" in the river description of black holes. They can be written in Cartesian form as:

$$ds^2 = \eta_{\mu\nu}(dx^\mu-\beta^\mu\alpha_\kappa dx^\kappa)(dx^\nu-\beta^\nu\alpha_\lambda dx^\lambda)$$

$$\beta^\mu = \frac{\beta R}{\rho}\left(0,-\frac{xr}{R\rho},-\frac{yr}{R\rho},-\frac{zR}{r\rho}\right)$$

$$\alpha_\mu dx^\mu = dt - a \sin^2(\theta) d\phi = dt + (ay/R^2)dx - (ax/R^2)dy$$

$$\beta = \sqrt{2r-Q^2}/R.$$

Now we depart from the paper. Putting the tensors into x, y, z, and t, and using a signature for eta of (-1,1,1,1) we have:

$$ds^2= -dt^2 + \left(dx-\frac{xr\sqrt{2r-Q^2}}{R^3\rho^2}(R^2dt + ay\;dx -ax\;dy)\right)^2$$
$$+ \left(dy-\frac{yr\sqrt{2r-Q^2}}{R^3\rho^2}(R^2dt + ay\;dx -ax\;dy)\right)^2$$
$$+ \left(dz-\frac{z\sqrt{2r-Q^2}}{rR\rho^2}(R^2dt + ay\;dx -ax\;dy)\right)^2$$

As before, defining $$I = (ds/dt)^2$$ we have:

$$I = -1 + \left(\dot{x}-\frac{xr\sqrt{2r-Q^2}}{R^3\rho^2}(R^2 + ay\dot{x} -ax\dot{y})\right)^2$$
$$+ \left(\dot{y}-\frac{yr\sqrt{2r-Q^2}}{R^3\rho^2}(R^2 + ay\dot{x} -ax\dot{y})\right)^2 + \left(\dot{z}-\frac{z\sqrt{2r-Q^2}}{rR\rho^2}(R^2 + ay\dot{x} -ax\dot{y})\right)^2$$

Ouch, the partial derivatives with respect to x, y, and z are going to be nasty.

Moving on, let's rewrite I as:

$$I = -1 + (v_x)^2 + (v_y)^2 + (v_z)^2$$

$$v_x = \dot{x}-\frac{xr\sqrt{2r-Q^2}}{R^3\rho^2}(R^2 + ay\dot{x} -ax\dot{y})$$

$$v_y = \dot{y}-\frac{yr\sqrt{2r-Q^2}}{R^3\rho^2}(R^2 + ay\dot{x} -ax\dot{y})$$

$$v_z = \dot{z}-\frac{z\sqrt{2r-Q^2}}{rR\rho^2}(R^2 + ay\dot{x} -ax\dot{y})$$

The partial derivatives of the v_* are messy with respect to positions, but are simple with respect to velocity. Looking at the last equation in post #17, we need only partial derivatives with respect to velocity to compute the determinant:

$$\left(2I\frac{\partial^2 I}{\partial\dot{x}^2}-\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{x}}\right)\left(2I\frac{\partial^2 I}{\partial\dot{y}^2}-\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{y}}\right) -\left(2I\frac{\partial^2 I}{\partial\dot{x}\partial\dot{y}}-\frac{\partial I}{\partial \dot{y}}\frac{\partial I}{\partial\dot{x}}\right)\left(2I\frac{\partial^2 I}{\partial\dot{y}\partial\dot{x}}-\frac{\partial I}{\partial \dot{x}}\frac{\partial I}{\partial\dot{y}}\right)$$

This time we will be doing the calculation in 3 dimensions so it will be messier. The first order partials with respect to velocity are:

$$\frac{\partial I}{\partial \dot{x}} = 2v_x - 2(xv_x+yv_y+zv_z(R^2/r^2)) ayr\sqrt{2r-Q^2}/(R^3\rho^2)$$

$$\frac{\partial I}{\partial \dot{y}} = 2v_y + 2(xv_x+yv_y+zv_z(R^2/r^2)) axr\sqrt{2r-Q^2}/(R^3\rho^2)$$

$$\frac{\partial I}{\partial \dot{z}} = 2v_z$$

Last edited:
Very interesting, CarlB. The squares and the ∏ seem to be saying geometry to me, and I note your comment It is hoped that this paper will stimulate thought among theoreticians. Perhaps the fundamental fermions are bound states of deeper objects. I have had some thoughts on certain fermions on bosons and would like to understand the electron neutrino initially. However this is not the place to discuss my idle speculations - I've printed the thread and hope to respond usefully at a later date.

Homework Helper
I thought I would add a quick calculation showing how Painleve coordinates cause the deflection of light to be 2x Newton, and giving the next order corrections. This is particularly simple to do from the equations of motion.

We will assume that the light is traveling in the +y direction at approximate a constant value of x. Thus we will have $$\dot{x}=0, \dot{y} = 1,$$ x constant, and y running from - to + infinity. The acceleration on the light is found by making these replacements into the general term for acceleration in the x direction. The result is:

$$\begin{array}{rcl|r}\ddot{x} &=&-x/r^3&2.0\\ &&+3xy^2/r^5&2.0\\ &&-2x/r^3&2.0\\ \hline &&+2\sqrt{2}xy/r^{4.5}\;\;\;\;\;&2.5\\ \hline &&+2x/r^4&3.0\\ \hline\end{array}$$

as opposed to the Newtonian acceleration $$-x/r^3$$

The total bending is the sum total of acceleration. For Newton, this is:
$$\int_{y=-\infty}^{+\infty} -x/r^3 \;dy = -\frac{1}{x}\int_{-\infty}^{+\infty}\frac{d\eta}{(1+\eta^2)^{1.5}} = -\frac{2}{x}$$

For Painleve coordinates, we apply the same integral to the more complicated Painleve acceleration. The first and last 2.0 order terms: $$-x/r^3$$ and $$-2x/r^3$$ are just like the Newtonian term, and give -2/x and -4/x. The remaining 2.0 order term is:

$$\int_{y=-\infty}^{+\infty}\frac{3xy^2}{r^5} = +\frac{3}{x}\int_{-\infty}^{+\infty}\frac{\eta^2\; d\eta}{(1+\eta^2)^{2.5}} = \frac{1}{x}\int_{-\infty}^{+\infty}\frac{d\eta}{(1+\eta^2)^{1.5}} = \frac{2}{x}$$

where the second equality follows by an integration by parts. Adding the 2.0 order terms together, we get $$-2/x+2/x-4/x = -4/x,$$ a total exactly twice the Newtonian figure.

The 2.5 and 3.0 terms give:
$$\frac{2}{x^{1.5}}\int_{-\infty}^{+\infty}\frac{\eta d\eta}{(1+\eta^2)^{2.25}} = 0$$
and
$$\frac{2}{x^2}\int_{-\infty}^{+\infty}\frac{d\eta}{(1+\eta^2)^2} = \frac{\pi}{x^2}$$

Note that these calculations ignore the deviation of the path from straight, so they are only good approximately. The first order correction to the path should be 1/x, and that should make the corrections small enough that we'll have the first order correction by ignoring the deviation from straightness. Making that jump, we have that the total deviation for Newton and Einstein are:

$$\begin{array}{l|l} -2/x&\textrm{Newton}\\ \hline -4/x+\pi/x^2&\textrm{Einstein} \end{array}$$

The above is interesting in that we know that for very small x, Einstein has to curve light a lot more than Newton. Maybe the approximation that $$\dot{y}=0$$ is breaking down, but who knows.

Last edited:
Homework Helper
I've been having some success in connecting particle physics up to Painleve gravity. Central to this is the nonlinearity in mass. Previously, I've done the Painleve coordinates for a unit mass. To show the nonlinearity in M, I need to write the Painleve acceleration with mass M.

The proper time integral has terms of the form $$(\dot{x} + x\sqrt{2M/r^3})^2$$. To get these from a term with no mass, that is, a term like $$(\dot{x} + x\sqrt{2/r^3})^2$$, I can scale by replacing all coordinates, both space and time, by themselves divided by $$M$$:

$$\begin{array}{rcl} x &\to&x/M,\\ y &\to&y/M,\\ z &\to&z/M,\\ t &\to&t/M \end{array}$$

This takes $$\dot{x} \to \dot{x}$$, and puts the $$M$$ where it belongs. It also modifies accelerations as $$a \to a M$$ because of the power of two in the denominator. The resulting equations of motion, with the gravitating mass included, are:

$$\begin{array}{rcl|r}\ddot{x} &=&-\sqrt{2M}\dot{x}(\dot{x}^2+\dot{y}^2)/r^{1.5}&1.5\\ &&+1.5\sqrt{2M}\dot{x}(x\dot{x}+y\dot{y})^2/r^{3.5}&1.5\\ \hline &&-xM/r^3&2.0\\ &&+3xM(x\dot{x}+y\dot{y})^2/r^5&2.0\\ &&-2\dot{y}M(x\dot{y}-y\dot{x})/r^3&2.0\\ \hline &&+3\sqrt{2M^3}\dot{x}/r^{2.5}&2.5\\ &&+2\sqrt{2M^3}y(x\dot{y}-y\dot{x})/r^{4.5}\;\;\;\;\;&2.5\\ \hline &&+2xM^2/r^4&3.0\\ \hline\end{array}$$

$$\begin{array}{rcl|r}\ddot{y} &=&-\sqrt{2M}\dot{y}(\dot{x}^2+\dot{y}^2)/r^{1.5}&1.5\\ &&+1.5\sqrt{2M}\dot{y}(x\dot{x}+y\dot{y})^2/r^{3.5}&1.5\\ \hline &&-yM/r^3&2.0\\ &&+3yM(x\dot{x}+y\dot{y})^2/r^5&2.0\\ &&+2\dot{x}M(x\dot{y}-y\dot{x})/r^3&2.0\\ \hline &&+3\sqrt{2M^3}\dot{y}/r^{2.5}&2.5\\ &&-2\sqrt{2M^3}x(x\dot{y}-y\dot{x})/r^{4.5}\;\;\;\;\;&2.5\\ \hline &&+2yM^2/r^4&3.0\\ \hline\end{array}$$

In short, mass enters into the terms that fall off as $$r^{-n}$$ as a multiplication by $$M^{n-1}$$.

It turns out that the accelerations (i.e. "gravitational force") of Paincare coordinates are identical to the accelerations of Schwarzschild coordinates so long as the compared test particles have no radial velocity. In particular, a stationary test particle has a particularly simple acceleration in either coordinate system:

$$\begin{array}{rcl}\ddot{x} &=&-xM/r^3\\ &&+2xM^2/r^4\\ \ddot{y} &=&-yM/r^3\\ &&+2yM^2/r^4. \end{array}$$

From a particle point of view, we can interpret this form as follows:

If the graviton were a massless particle with no other interactions, then we would expect the acceleration of a stationary test particle to have a Newtonian form. That it does not is not noticed in the usual GR theories because these theories deny the existence of a flat frame of reference in which to make the comparison with Newtonian accelerations.

We are using Painleve coordinates because they are implied by the Cambridge geometry version of gravity, which uses geometric calculus instead of tensors to describe Einstein's gravity. With this theory, we get a theory of gravity on a flat space, so comparisons with the Newtonian theory make sense. But Newtonian gravity works well only for velocities small compared to light, so we will examine the comparison in the limit of stationary test particles.

Compared to the Newtonian case, the Einstein acceleration is smaller. The discrepancy is larger the closer one gets to the origin. The force decreases with distance, but at a rate slower than we expect from a $$1/r^2$$ dependency.

The graviton is supposed to be a particle that can only attract. The fact that the gravitational force decreases more slowly than $$1/r^2$$ can be explained, in the context of a massless graviton, if we assume that, with increasing radius, the total number of gravitons is slowly increasing.

In other words, to explain this form of the gravitational force, we can assume that gravitons make gravitons. The simple assumption that gravitons stimulate the production of more gravitons, at a rate proportional to the square of their flux, is sufficient to give the Einstein acceleration, to first order. And since none of Einstein's GR equations have been tested more accurately than first order, this is a viable theory of gravity.

Perhaps the slow increase in the number of gravitons, when written as a field theory, will tend towards a minimum so that at the limit of very small accelerations (and therefore very small fluxes of gravitons) the number of gravitons is increased by an amount larger than expected. This might give the MOND gravity from particle theory.

Of course there is still a lot of stuff to do, particularly the force of gravity on moving test masses.

Last edited: