All the lepton masses from G, pi, e

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In summary, the conversation revolved around using various equations and formulae to approximate the values of fundamental constants such as the Planck Mass and the fine structure constant. The discussion also delved into the possibility of using these equations to predict the masses of leptons and other particles. Some participants raised concerns about the validity of using such numerical relations, while others argued that it could be a useful tool for remembering precise values.

Multiple poll: Check all you agree.

  • Logarithms of lepton mass quotients should be pursued.

    Votes: 21 26.6%
  • Alpha calculation from serial expansion should be pursued

    Votes: 19 24.1%
  • We should look for more empirical relationships

    Votes: 24 30.4%
  • Pythagorean triples approach should be pursued.

    Votes: 21 26.6%
  • Quotients from distance radiuses should be investigated

    Votes: 16 20.3%
  • The estimate of magnetic anomalous moment should be investigated.

    Votes: 24 30.4%
  • The estimate of Weinberg angle should be investigated.

    Votes: 18 22.8%
  • Jay R. Yabon theory should be investigate.

    Votes: 15 19.0%
  • I support the efforts in this thread.

    Votes: 43 54.4%
  • I think the effort in this thread is not worthwhile.

    Votes: 28 35.4%

  • Total voters
    79
  • #176
CarlB said:
I should exhibit a representation of SU(2) which does not satisfy the requirement that the geometrically defined squared magnitude is equivalent to the usual spinor squared magnitude (and therefore to the traces). So here it is:

[tex]\sigma_x = \left(\begin{array}{cc}0.0&0.2\\5.0&0.0\end{array}\right)[/tex]

[tex]\sigma_y = \left(\begin{array}{cc}0.0&-0.2i\\5.0i&0.0\end{array}\right)[/tex]

[tex]\sigma_z = \left(\begin{array}{cc}1.0&0.0\\0.0&-1.0\end{array}\right)[/tex]


It is more a sort of deformation of SU(2) than a representation of [the infinitesimal generators of] SU(2), isn't it? People expects selfadjointness to get unitary generators, thus unitary group, which is the thing that the U stands for, after all. Just terminology issue, but it can be a communication problem.
 
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  • #177
I'm sure I've been sloppy with my terminology. I tend to be a very calculation oriented person. Let me try again. SU(2) is the group, su(2) is the algebra.

SU(2) refers to the fact that one can represent the elements of the group by special unitary 2x2 matrices. But that is not the only representation of the group, nor are the Pauli matrices the only representation of the algebra su(2).

The set of deformed matrices are a representation of su(2) in that they satisfy the usual commutation relation:

[tex]\sigma_j\sigma_k-\sigma_k\sigma_j = 2i \epsilon_{jkm}\sigma_m[/tex].

If we wanted to describe the SU(2) that this defines, then we just take exponentials of the above. But since the above is identical to the usual su(2), we'll just get a deformation of the usual SU(2). Uh, what's the correct word for "deformation"? In this case it's something like isomorphism or homeomorphism or...

By the way, the better (i.e. geometric) way of writing the above relation is to replace i with [tex]\sigma_x \sigma_y \sigma_z[/tex].

Carl
 
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  • #178
CarlB said:
But since the above is identical to the usual su(2), we'll just get a deformation of the usual SU(2). Uh, what's the correct word for "deformation"? In this case it's something like isomorphism or homeomorphism or...
Hmm I think it is not going to be an isomorphism; supposse that in the original representation of SU(2) I build unitary elements [tex]A_i=\exp ( a_i^\mu \sigma_\mu)[/tex] from the sigmas and triplets of real numbers; this is the standard business. Now you should have an invertible map from this real triplet in in the standard representation to the real triplet components in your representation, ie a map [tex](a^\mu) \leftrightarrow (b^\mu)[/tex]. And then in order to check that you representation it really an iso-, and not a deformation, you should be sure of the details, for instance that a composition of any three elements giving the identity, [tex]A_1 A_2 A_3 = I[/tex] in the original representation should still give the identity in your mapped representation. I doubt it, because your sigmas are not idempotent anymore. So let it to be called "deformation".
 
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  • #179
CarlB said:
By the way, the better (i.e. geometric) way of writing the above relation is to replace i with [tex]\sigma_x \sigma_y \sigma_z[/tex].
Yeah, it is a very depth trick because somehow you want the product of the three sigma to have a meaning as volume element (or dual to it), the product of two as surface elements (or dual to it) and so on. But all these topics should be theme for a separate, well indexed, thread somewhere. Clifford things.
 
  • #180
arivero said:
I doubt it, because your sigmas are not idempotent anymore.

I'm not sure where you're going here. [tex]\sigma_x[/tex] is never idempotent. They square to unity, in both cases. And the idempotents you get from them, for example,

[tex](1+\sigma_x)/2.0[/tex]

is idempotent in both representations. In the usual, one obtains:

[tex]\left(\begin{array}{cc}0.5&0.5\\0.5&0.5\end{array}\right)[/tex]

In the deformed representation, one obtains:

[tex]\left(\begin{array}{cc}0.5&2.5\\0.1&0.5\end{array}\right)[/tex]

One can easily verify that both these representations of [tex](1+\sigma_x)/2.0[/tex] are, in fact, idempotent.

One can define a Clifford algebra as a vector space. On doing this, one obtains that addition is automatically identical between representations, and in this case the multiplication is identical too. So I'm pretty sure that they're both equally valid representations of su(2), and after exponentiation, SU(2).

Anyway, all this gets back to quarks and leptons in that I've discovered a beautiful representation of both the quarks and leptons that requires this as a basis for the theory.

What is going on here, in the making of the representations non unitary, is that we are preserving the usual quantum mechanics for the individual particles (i.e. the ideal generated by (1+\sigma_x) is isomorphic to the usual spin-1/2 no matter the representation), but are changing the relationship between different particles.

This shows up in the lepton mass formula because the different particles are different types (in terms of color). So you need a little more freedom than is available when you assume that the spinors (different particles) are separated from each other in the usual method of unitary representations. But they're still good su(2) reps.

It's kind of a long and involved thing and I'm splitting it up into two papers both of which are long and involved. The one having to do with representations also gets into the meaning of gauges and why Hestenes' factorization of the Dirac equation has an orientation built into it. For more on that subject, see the excellent short comment by Baylis:
http://www.arxiv.org/abs/quant-ph/0202060

Yet another way of describing this is that I am rejecting the splitting of the Banach space (or density matrix) into a Hilbert space (or spinors) because there are always multiple ways (i.e. a gauge) to do this. The Banach space is more physical because it does not depend on gauge. Along this line, but restricted to a single particle and the U(1) complex phase gauge, those who are fans of Bohmian mechanics need to read Hiley's paper which generalizes Bohmian mechanics to density matrices:
http://www.arxiv.org/abs/quant-ph/0005026

In terms of Schwinger's elegant measurement algebra, this amounts to writing the theory strictly in terms of simple measurements M(a), and rejecting the use of general measurements M(a,b). My paper shows that when one uses M(a,b), one automatically implies an orientation gauge similar to the one that Hestenes uses. And this gets back into the assumption that the vacuum is a physical state.

Carl
 
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  • #181
Maybe this is the isomorphism that is needed to explain this better:

[tex]\left(\begin{array}{cc}a&b\\c&d\end{array}\right)
\left(\begin{array}{cc}A&B\\C&D\end{array}\right) =
\left(\begin{array}{cc}aA+bC&aB+bD\\cA+dC&cB+dD\end{array}\right).[/tex]

then the isomorphism consists of multiplying the off diagonal elements by r and 1/r. Multiplication is preserved:

[tex]\left(\begin{array}{cc}a&br\\c/r&d\end{array}\right)
\left(\begin{array}{cc}A&Br\\C/r&D\end{array}\right) =
\left(\begin{array}{cc}aA+bC&(aB+bD)r\\(cA+dC)/r&cB+dD\end{array}\right).[/tex]

and addition obviously is preserved too. Also note that the transformation preserves the trace as the diagonal elements are unchanged, and it is the trace that gives probabilities in the standard density matrix approach. But what is not preserved is the natural squared magnitude of matrices. That is, the squared magnitude of the untransformed matrix is [tex]|a|^2 + |b|^2 + |c|^2 + |d|^2[/tex], but the squared magnitude of the transformed matrix gets some r activity. The geometric squared magnitude is also not preserved. In particular, the coefficients for x and y are changed in magnitude by the r.

So this gives a way of rewriting QM where the traces (and therefore the usual way of calculating QM probabilities) is preserved, but the geometric method is not. This allows one to explore cases that are outside the standard model, but with a method that includes the standard model in that the geometric probability is identical to the trace in every standard representation of the standard model. It is only when we use these unstandard representations, (such as the su(2)/SU(2) modified by r) that the predictions can be made to differ, and only then in the geometric probabilities. The trace is unchanged.

I don't think that getting into the details of the "r" transformation advances understanding much. The same trick works in the Dirac matrices, but the number of degrees of freedom available to the representation is far far larger and you can't fit into a simple description like the above "r".

The better way of seeing everything is to just look at where the (matrix) diagonal primitive idempotents and the (matrix) democratic primitive idempotents map to (in the Clifford algebra), the rest follows automatically from that. If you want to know how many representations of the Dirac algebra exist that share the same diagonal representation (i.e. the same commuting operators) as the Weyl represenstation, for example, I think you should look at it geometrically. If you restrict yourself to representations where the spinor probabilities match the geometric probabilities, and you ignore the infinite number of cases you get by rotating x into y, you end up with 96 Weyl reps that share identical diagonal representations. The representations that give different geometric probabilites that I have found are all rotations, but are rotations that use cosh and sinh instead of cos and sin. That is, they rotate different signature elements of the canonical basis elements of the Clifford algebra.

For understanding all this it helps to understand how primitive idempotents appear in Clifford algebras. That is kind of complicated, but if you've heard of Radon-Hurwitz numbers you're probably already there. The paper I'm writing up on it is very very long and has lots and lots of examples and exercises so it should be easy, if boring, to understand. Give me a week to finish it off.

Carl
 
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  • #182
CarlB said:
I'm not sure where you're going here. [tex]\sigma_x[/tex] is never idempotent. They square to unity[/tex]
Sorry, my fault. If was referring to the fact that they should square to the unity, and you are right they do (doe to the decimal notation I didn't notice it rightly). Thus OK it is the same algebra.
 
  • #183
Another paper on the Koide mass formula:

On the Koide-like Relations for the Running Masses of Charged Leptons, Neutrinos and Quarks
Zhi-zhong Xing, He Zhang
Current experimental data indicate that the Koide relation for the pole masses of charged leptons, which can be parametrized as [tex]Q^{pole}_l = 2/3[/tex], is valid up to the accuracy of [tex]O(10^{-5})[/tex]. We show that the running masses of charged leptons fail in satisfying the Koide relation (i.e., [tex]Q_l(\mu) \neq 2/3[/tex]), but the discrepancy between [tex]Q_l (\mu)[/tex] and [tex]Q^{pole}_l[/tex] is only about 0.2% at [tex]\mu=M_Z[/tex]. The Koide-like relations for the running masses of neutrinos ([tex]1/3 < Q_\nu(M_Z) < 0.6[/tex]), up-type quarks ([tex]Q_{U}(M_Z) \sim 0.89[/tex]) and down-type quarks ([tex]Q_{D}(M_Z) \sim 0.74[/tex]) are also examined from [tex]M_Z[/tex] up to the typical seesaw scale [tex]M_R \sim 10^{14}[/tex] GeV, and they are found to be nearly stable against radiative corrections. The approximate stability of [tex]Q_{U}(\mu)[/tex] and [tex]Q_{D}(\mu)[/tex] is mainly attributed to the strong mass hierarchy of quarks, while that of [tex]Q_l(\mu)[/tex] and [tex]Q_\nu(\mu)[/tex] is essentially for the reason that the lepton mass ratios are rather insensitive to radiative corrections.
http://www.arxiv.org/abs/hep-ph/0602134

Carl
 
  • #184
CarlB said:
Another paper on the Koide mass formula:

http://www.arxiv.org/abs/hep-ph/0602134

Carl

Hmm it seems that at least we have contributed to create awareness on Koide's. In this case the paper does not quote any webpage thread, but it still quotes a couple papers that redirect towards the web discussions.

As for the paper, it is more cautious that hep-ph/0601031, I hope it will be published somewhere.

Also, we missed hep-ph/0510289 because it fails to mention Koide in the abstract.
 
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  • #185
I suppose I should give the calculation that shows that, in the case of SU(2), the geometric squared magnitude gives the same number as the more usual trace.

Begin with two spinors:

[tex]|\alpha\rangle = \left(\begin{array}{c}1\\ \alpha\end{array}\right)[/tex]
[tex]|\beta\rangle = \left(\begin{array}{c}1\\ \beta\end{array}\right)[/tex]

These aren't normalized, to normalize divide by, for example, [tex]\sqrt{1+|\alpha|^2}[/tex]. Then the probability associated with transititions from one of these forms to the other is given by:

[tex]P_{\alpha\beta} = \frac{1+\alpha^*\beta+\beta^*\alpha+|\alpha\beta|^2}
{(1+|\alpha|^2)(1+|\beta|^2)}[/tex]

The density matrices associated with these spinors are (leaving off the normalization):

[tex]\rho_\alpha = \left(\begin{array}{cc}1&\alpha^*\\ \alpha&|\alpha|^2\end{array}\right)[/tex]

and similarly for beta. The products of the two density matrices is:

[tex]M_{\alpha\beta} = \rho_\alpha\rho_\beta = \left(\begin{array}{cc}1+\alpha^*\beta&\beta^*+\alpha^*|\beta|^2\\ \alpha+\beta|\alpha|^2& \beta^*\alpha+|\alpha\beta|^2\end{array}\right)[/tex]

again leaving off the normalization. Note that the trace gives the correct probability. I now write

[tex]M_{\alpha\beta} = c_1\hat{1} + c_x\sigma_x + c_y\sigma_y + c_z\sigma_z[/tex]

where c_* are complex coefficients (you can redo this with real coefficients if you wish by going to the real basis that includes bivectors like sigma_x sigma_y and the pseudoscalar sigma_x sigma_y sigma_z). The coefficients are:

[tex]c_1 = (1 + \alpha^*\beta\ + \beta^*\alpha + |\alpha\beta|^2)/2[/tex]
[tex]c_x = (\alpha+\beta|\alpha|^2+\beta^*+\alpha^*|\beta|^2)/2[/tex]
[tex]c_y = (\alpha+\beta|\alpha|^2-\beta^*-\alpha^*|\beta|^2)/2i[/tex]
[tex]c_z = (1 + \alpha^*\beta\ - \beta^*\alpha - |\alpha\beta|^2)/2[/tex]

The geometric squared magnitude is [tex]|M_{\alpha\beta}|^2_G = |c_1|^2 + |c_x|^2 + |c_y|^2 + |c_z|^2[/tex]. This can be computed easily by multiplying by the complex conjugate and adding. Parts of the 1 and the z calculations cancel each other, as do parts of the x and y calculations. What is left is equal, after a small amount of algebra, to:

[tex]|M_{\alpha\beta}|^2_G = 2(1 + \alpha^*\beta+\beta^*\alpha+|\alpha\beta|^2)(1+|\alpha|^2)(1+|\beta|^2)[/tex]

this is the same as the usual probability, after one corrects the normalization (i.e. the two things multiplying on the right), and divides by 2 to account for the fact that tr(1) = 2.

Thus we have that in the case of one spin-1/2 spinor transitioning into another, we can replace the spinor computation of the probability with a purely geometric calculation based on the geometric squared magnitude.

Now the trace is also geometrically defined in that it is the scalar part of the matrix. But the above shows that the geometric squared magnitude also gives the same result as the trace.

It should be noted that all the above calculations follow over into the various standard representations of the Dirac algebra, in particular the Weyl representation and the usual one.

We can distinguish between the trace and the geometric squared magnitude by going to representations of a Clifford algebra that are different from the usual ones. In particular, it is in the masses of the leptons that one can find evidence for the necessity of the squared magnitude, rather than the trace, as a measure of probability.

Carl
 
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  • #186
from Kea in other thread (https://www.physicsforums.com/showpost.php?p=929574&postcount=14 )

Kea said:
In this spirit:

Note that the reduced link of the electron (figure 17) is just the Hopf link, when the ends are connected up. Now taking Jones polynomials [itex]J[/itex] at 5th roots of unity for universal quantum computation, one finds that

[tex]J_{\textrm{Hopf}} = d = 1.618 \cdots[/tex]

the golden ratio. Now let a deformation parameter be

[tex]q = e^{\frac{2 \pi}{2 + d}}[/tex]

namely the [itex]SU(2)_{q}[/itex] conformal field theory expression. It is well known that the spin half rep quantum dimension is given by [itex][ 2 ]_{q} = q + q^{-1}[/itex]. Putting two of these electron graph invariants together one estimates

[tex]\alpha = 4 (q + q^{-1})^{2} = 137.08[/tex]

This is of course an extremely coarse estimate, and hence not particularly accurate. :smile:
 
  • #187
I found how to apply the Koide formula to the neutrino masses.

Several researchers have noted that the Koide formula is excluded from being applied to the masses of the neutrinos. For example, see section (III) in:
http://arxiv.org/abs/hep-ph/0601031

The experimental data that supposedly excludes the Koide formula is:

[tex]m_2^2 = m_1^2 + 8.0 \times 10^{-5}\; eV^2,\;\;\;[/tex] (1)

[tex]m_3^2 = m_2^2 + 250.0 \times 10^{-5}\; eV^2,\;\;\;[/tex] (2)

where I've left off the error bars as they are fairly small. Also note, I'm assuming the "normal" ordering, where [tex]m_1 < m_2 << m_3[/tex], which removes an absolute value from the LHS of (2).

Some time ago on this thread I published a formula for the Koide masses based on the eigenvalues of matrices of the type:

[tex]\mu\;\left( \begin{array}{ccc}1&\eta e^{+i\delta}&\eta e^{-i\delta}\\\eta e^{-i\delta}&1&\eta e^{+i\delta}\\\eta e^{+i\delta}&\eta e^{-i\delta}&1\end{array}\right)\;\;\;[/tex] (3)

The above is derived as a form for cross family operators in my latest paper in equation (217):
http://brannenworks.com/GEOPROB.pdf

For the charged leptons, one uses [tex]\eta = \sqrt{0.5}[/tex]. The three eigenvalues for a matrix of type (3) are given by:

[tex]\sqrt{m_n} = \mu(1 + 2\eta\cos(\delta + 2 n \pi/3)),\;\;\;\;\;\;n = 1,2,3\;\;\;[/tex] (4)

If one is given a set of three eigenvalues, one can determine the value of \eta by dividing the following formulas:

[tex]\left(\sum \sqrt{m_n}\right)^2 = (3 \mu)^2 = 9\mu^2,\;\;\;\;[/tex] (5)

[tex]\sum\left(\sqrt{m_n}\right)^2 = 3\mu^2(1 + 4\eta^2<\cos^2>) = \mu^2(1+2\eta^2),\;\;\;\;[/tex] (6)

where "< >" means average value (i.e. average over the three roots), and the average value of the cos^2 is 1/2. (Alternatively, you can apply trigonometry.) The above formulas are obtained by taking traces of the square root mass matrix and its square, and they allow the computation of eta^2 from sets of possible eigenvalues.


If one tries various values for [tex]m_1^2[/tex] in the neutrino squared mass difference formula, one soon finds that the Koide mass formula is not possible, assuming one assumes positive square roots for the square roots of the masses. The Koide formula has eta^2 = 1/2, and for various choices of m1, the values of eta^2 are:

[tex]\begin{array}{cc}
m_1^2&\eta^2\\
0.0&0.3753\\
1.0&0.1628\\
8.0&0.0823\\
20.0&0.0491\end{array},\;\;\;\;[/tex] (7)

Thus, eta is too low when m1=0, and increasing m1 simply makes eta even smaller. However, if we instead assume that sqrt(m1) < 0, then the Koide formula can be achieved with the following approximate neutrino masses:

[tex]m_1 = 0.00039\;eV,[/tex]
[tex]m_2 = 0.00895\;eV,[/tex]
[tex]m_3 = 0.05079\;eV,[/tex]

and the Koide mass formula is met by the relation:

[tex]\frac{(-\sqrt{m_1} + \sqrt{m_2} + \sqrt{m_3})^2}{m_1+m_2+m_3} = \frac{3}{2}[/tex]

Now so far this really hasn't shown anything. There was a free variable in the definition of the neutrino masses and I used it to satisfy the Koide relation by assuming a negative square root. However, it gets interesting when we compute the delta values for the charged leptons and compare it with the delta value for the neutral leptons. Turns out that the two angles differ, to within experimental error, by 15 degrees. That is,

[tex]\delta_0 = \delta_+ + \pi/12.[/tex]

Is this a coincidence? I doubt it. Much more on this later. By the way, if you follow the logic of my paper, you will understand why I suspect that the square roots of the masses of the charged leptons should all be taken to be negative. This transforms the delta to a value that is consistent with a hidden dimension interpretation of the modification of the spinor probability formula [tex]P = (1+\cos(\theta))/2[/tex].

This sort of transformation takes delta to 60-delta. Another obvious transformation, (that leaves the eigenvalues unchanged) is to replace delta with -delta. Together, these give four representations possible for the charged leptons and also four for the neutral leptons. I expect that one (okay, two, since you can always complement the imaginary unit) of these 16 possibilities will be picked out by the mixing matrix for neutrinos, that is, the MNS matrix. This will complete my classification of the leptons, and it should be up here tomorrow.

Carl
 
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  • #188
Carlb perhaps you have an opportunity (take it with care) to pester the bosses. It seems that ours awareness-campaing on Koide's (I say ours, but also Li & Ma, and Koide himself or course) has culminated with three pages (46-48) in the last http://arxiv.org/abs/hep-ph/0603118"
 
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  • #189
Charge Renormalization and the value of Alpha.

I came across a nice passage in Feynman's book "The Theory of
Fundamental processes" which has everything to do with our alpha
result. It shows Feynman's ideas (around 1961). In chapter 29 he
speculates that the theoretical value of alpha [itex]e_{th}^2[/itex] could be a nice
simple result from some future theory. The experimental value [itex]e_{th}^2[/itex]
is then a corrected value due to vacuum polarization.

He gives as an example [itex]e_{th}^2 = 1/141[/itex] which is getting renormalized
to [itex]e_{exp}^2 = 1/137.036[/itex]. (Just as example, no calculation) So the renor-
malized value is slightly higher as the theoretical one. (In Zee's
treatment it's the other way around)

This seems pretty much in the same ballpark as what we have:

[tex]e_{th} = e^{-\pi^2/4}, \quad e_{th}^2 = 1/139.045636[/tex]

With our empirical corrective series:

[tex]e_{exp}\ =\ \left(1 + \alpha + \frac{\alpha^2}{2\pi}\right) e_{th}[/tex]

We get a value [itex]e_{exp}^2 = 1/137.03599952[/itex] which coincides
with the two most accurate measurements:

1/137.03599911(46) From the Quantum Hall effect.
1/137.03599976(50) Magnetic Moment Anomaly
1/137.03599941(15) The two Combined

The series discussed by Feynman is the "string of loops": photon to
loop to photon to loop ... He has a formula:

[tex] e_{exp} = e_{th}/(1-Y)^{1/2}[/tex]

Where Y approaches a constant for low (photon) q's and where the
contribution of the electron positron loops is given by [itex]X=q^2Y[/itex].
There's not so much more in the book on the (divergent) series.

Has there been more work from Feynman along these lines?


Regards, Hans



P.S: Our short series can be rewritten in order to express the series in
[itex]e=e_{th}[/itex] instead of alpha. (although in a much more complicated fashion):

[tex]
e_{exp}\ = \ e\ +\ e^3\ +\ \left( 2+\frac{1}{2\pi} \right)e^5\ +\
\left( 5+\frac{6}{2\pi} \right)e^7\
+\ \left( 14+\frac{28}{2\pi}+\frac{4}{4\pi^2} \right)e^9\ +\ \left(
48+\frac{104}{2\pi}+\frac{37}{4\pi^2} \right)e^{11} \ +\ ...
[/tex]
 
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  • #190
Hans de Vries said:
Charge Renormalization and the value of Alpha.

We get a value [itex]e_{exp}^2 = 1/137.03599952[/itex] which coincides
with the two most accurate measurements:

1/137.03599911(46) From the Quantum Hall effect.
1/137.03599976(50) Magnetic Moment Anomaly
1/137.03599941(15) The two Combined

Figure 6 of eprint http://arxiv.org/abs/nucl-ex/0404013 and figure 5 of http://arxiv.org/abs/hep-ex/0512026 are historic plots of the measurements of alpha during the last twenty years. It seems the Magnetic Moment result is being favoured.
 
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  • #191
Hans de Vries said:
Charge Renormalization and the value of Alpha.

I came across a nice passage in Feynman's book "The Theory of
Fundamental processes"

...

The series discussed by Feynman is the "string of loops": photon to
loop to photon to loop ... He has a formula:

[tex] e_{exp} = e_{th}/(1-Y)^{1/2}[/tex]

Where Y approaches a constant for low (photon) q's and where the
contribution of the electron positron loops is given by [itex]X=q^2Y[/itex].
There's not so much more in the book on the (divergent) series.

Has there been more work from Feynman along these lines?

There are some approximations based on resummations of a series, I can recall for instance the "ladder approximation". On other hand, the point about renormalised series is that usually are self-generated by insertion of a finite number of counterterms that generates insertions in all orders of the expansion.

Historically, it could be that Feynman was following here, or hinting to, the "eigenvalue approach" of Adler.
 
  • #192
CarlB said:
Another unusual number, this one having to do with the ratio of the charged leptons to the neutral lepton masses.

As mentioned above, the Koide mass formula can be rewritten in a form which removes a degree of freedom from the masses. That is,

[tex]\sqrt{m_n} = \mu_\chi(1 + \sqrt{2}\cos(\delta_\chi + 2n\pi/3)),[/tex]

where [tex]\chi[/tex] is either 0 for the neutral leptons (or neutrinos) or 1 for the charged leptons (i.e. electron, muon and tau). Having done this, the previous note found that the neutrino squared mass formulas can be met, and the usual charged lepton masses can be obtained by putting [tex]\delta_0 = 0.222222047[/tex] and [tex]\delta_1 = \delta_0 + \pi/12.[/tex]

This note is to mention an unusual coincidence with the ratio [tex]\mu_1/mu_0:[/tex]

[tex]\mu_1/\mu_0 = 3^{11}.[/tex]

Now the masses are proportional to [tex]\mu^2[/tex], so the difference in scale between the charged and neutral leptons comes out to be [tex]3^{22}[/tex]. Using the best numbers for the squared neutrino mass differences, this relation is obtained to an accuracy of 0.1%, far greater than the purported accuracy of the squared neutrino mass measurements.

Assuming that the relation is exact gives a prediction for the squared neutrino masses a bit lower than the center of the current estimates. Instead of [tex]m_2^2-m_1^2 =[/tex] [tex] 8.0 \times 10^{-5} eV^2[/tex], we have, if I recall correctly, [tex]7.95 \times 10^{-5} eV^2.[/tex] I'll see if I can get back to a computer fast enough to edit in the actual numbers, unless someone computes them out.

By the way, it is also possible to put the difference between 2/9 and [tex]\delta_1[/tex] in the form of a small constant times a power of three, but I'll leave that to the reader to find as its not as simple. If you happen to find one, please post it and we will see if we happened to find the same one.

As to why the charged leptons should have a mass with such a relationship, I can only suggest that in the mysterious reaction that allows a right handed lepton to transform itself into a left handed lepton and vice versa, there might be a pathway that is 11 steps longer for the neutrinos, with each step having a probability of 1/3. For example, perhaps the neutrinos require 12 steps while the charged leptons require only one.

The reason I was looking for powers of three is because the mysterious factor 2/9 in the delta function suggested that threes were important. Also, of course, there are three colors, three generations, and three elements in the Holy Trinity.

Carl
 
  • #193
CarlB said:
[tex]\mu_1/\mu_0 = 3^{11}.[/tex]
Hmm without a model backing it, this is probably one of the most numerological formulae we have catalogued.
 
  • #194
G. Rosen

Recently I have got scent of the work of G. Rosen. Check eg SPIRES, because he does not arxiv:
http://www.slac.stanford.edu/spires/find/hep/www?rawcmd=find+a+rosen%2C+g&FORMAT=WWW&SEQUENCE=
He did some attempts at combinatorial values for Sommerfeld constant, and in 1971 he was one of the people peering at the bandwagon of using alpha to exponentiate back from Planck scale to electron scale, directly withut any renormalisation group trayectory. This was the paper http://prola.aps.org/abstract/PRD/v4/i2/p275_1

Recently he is advocating that in all the four triplets of fermions one can find a pair related by [tex]m/M=e^{2 \sqrt 2}[/tex] and that in each case the extant fermion can be adjusted from a third term depending on Barion number and Charge.
 
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  • #195
arivero said:
Hmm without a model backing it, this is probably one of the most numerological formulae we have catalogued.

I agree completely. The hope is that some of these numerical coincidences will give a clue to the underlying theory, and under that assumption, I couldn't help but mention the coincidence. And of course the Koide formula doesn't have a model backing it yet.

My feeling is that the physics of the relationship between the generations is at the same stage as when the early physicists were puzzling out the relationships between the wavelengths of light emitted by excited states of the hydrogen atom.

Carl
 
  • #196
The centrifugal force goes as
[tex]F={L \over r} {v \over r} [/tex]
and you fix the second factor using a quantum condition so that
[tex]
F={L \over r} m {c^2 \over \hbar}
[/tex]
Now, comparing with k/r^2 we have
[tex]k= L m r {c^2 \over \hbar} = L v [/tex]
so the velocity (in units c) times the required angular momentum (in units [tex]\hbar[/tex]) produces the coupling constant (in units [tex]c\hbar[/tex]) of the Coulombian force needed to meet your requisites. Er... I am telling an stupidity, or a triviality. The fact is that always [tex]k=L v[/tex], unrelated to your quantum condition.

The point I wanted to do is that your definition fixes L and v, so there are no more freedom to say that the coupling constant of the force is such or such, it is compulsorily L times v. Funny you have a method to generate coupling constants, but instead 1/137 you hit Weinberg angle.

Also there is a transition hidden somewhere. After all, your condition sets L as a function of spin s and then v as a function of L (via [tex]\gamma (\frac vc)^2 = L[/tex]). But if the coupling increases enough (above unity in bohr-schroedinger, but not sure in your model) there are spiral orbits in the allowed range, and naive relativistic quantum mechanics fails. So at highers spins we fall into a strong coupling regime probably forbidden in r.q.m.

Hans de Vries said:
We did see that the clasical velocity can be defined as:

“The velocity of a mass with spin s rotating on an orbit
with a frequency corresponding to its rest mass and an
angular momentum [itex]\sqrt{ s(s+1}\ \hbar[/itex]"


One gets a general solution for the 'classical velocity' of spin s:[tex]\beta_s \ \ \ \ \ \ = \ \ \ \ \sqrt{\sqrt{\ \ s(s+1) \ \ + \ \ ( \frac{1}{2}
\ s(s+1) \ )^2 } \ \ - \ \ \frac{1}{2} \ s(s+1) } [/tex]

Which solutions are dimensionless constants, independent
of the mass of the particle. The values of the common spins
are given below:spin 0.0: __ 0.00000000000000000000000000000000
spin 0.5: __ 0.75414143528176709788873548859945
spin 1.0: __ 0.85559967716735219296923576621118
spin 1.5: __ 0.90580479773844104117525862119228
spin 2.0: __ 0.93433577808377694874713811004304
spin inf: __ 1.00000000000000000000000000000000

Weinberg’s Electro-Weak mixing angle becomes a dimension-
less constant as well and is given in the [itex]\sin^2 \theta_W[/itex] form as:[tex] \sin^2 \theta_W
\ \ \ \ = \ \ \ \ 1 \ - \ \frac{\beta^2_f }{\beta^2_b}
\ \ \ \ = \ \ \ \ 0.22310132230086634541466926662604[/tex]

[tex] \sin^2 \theta_W
\ \ \ \ = \ \ \ \ 1 \ - \ \frac
{ \sqrt{\ \ \frac{1}{2}(\frac{1}{2}+1) \ \ + \ \ ( \frac{1}{2} \ \
\frac{1}{2}(\frac{1}{2}+1) \ )^2 } \ \ - \ \ \frac{1}{2} \
\ \frac{1}{2}(\frac{1}{2}+1) }
{\sqrt{\ \ 1(1+1) \ \ + \ \ ( \frac{1}{2} \ \
1(1+1) \ )^2 } \ \ - \ \ \frac{1}{2} \
\ 1(1+1) } [/tex]
 
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  • #197
[tex]\beta_s \ \ \ \ \ \ = \ \ \ \ \sqrt{\sqrt{\ \ s(s+1) \ \ + \ \ ( \frac{1}{2}
\ s(s+1) \ )^2 } \ \ - \ \ \frac{1}{2} \ s(s+1) } [/tex]

Which solutions are dimensionless constants, independent
of the mass of the particle. The values of the common spins
are given below:spin 0.0: __ 0.00000000000000000000000000000000
spin 0.5: __ 0.75414143528176709788873548859945
spin 1.0: __ 0.85559967716735219296923576621118
spin 1.5: __ 0.90580479773844104117525862119228
spin 2.0: __ 0.93433577808377694874713811004304
spin inf: __ 1.00000000000000000000000000000000

These dimensionless numbers related to angular momentum may also be
derived from Quantum Mechanics. In fact this does solve one of the
ad-hoc presumptions that I had to make about a “missing” relativistic
increase in energy at very small radii where v must become relativistic.

The post which summarizes various numerical coincidences with these
numbers can be found here:

https://www.physicsforums.com/showpost.php?p=382642&postcount=44Let me archieve the math of the QM derivation here below.=====================================================There are 3 steps:

1- We’ll derive a radius dependent speed (in QM) which we still need.
2- Show how QM can solve the missing relativistic mass increase.
3- Get the formula which gives us the dimensionless numbers.
- STEP 1 – We want to re-introduce speed into QM as a function of (orbital)
angular momentum and radius.Let the angular momentum be:

[tex]
\mathbf{p} \times \mathbf{r}\ =\ \sqrt{l(l+1)} \hbar
[/tex]

Entering the speed v via SR the momentum as a function of
The radius becomes:

[tex]
p\ =\ \frac{m_0 v}{\sqrt{1-v^2/c^2}}\ =\ \sqrt{l(l+1)} \frac{\hbar}{r}
[/tex]

We can write with [itex]r_c = \hbar/(mc)[/itex] as the Compton radius:

[tex]
\frac{v/c}{\sqrt{1-v^2/c^2}}\ =\ \sqrt{l(l+1)} \frac{\hbar}{m_0 c
r}\ =\ \sqrt{l(l+1)}\ \frac{r_c}{r}
[/tex]

Which we can rework to get gamma as a function of the radius:

[tex]
\gamma \ =\ \frac{1}{\sqrt{1-v^2/c^2}}\ =\ \sqrt{1+l(l+1) \frac{r_c^2}{r^2}}
[/tex]

We can now also assign a physical speed as a function of
the radius:

[tex]
v/c \ =\ \sqrt{\frac{l(l+1)}{\frac{r^2}{r_c^2}+l(l+1))}}
[/tex]

The apparent increase in mass needed to produce the angular
momentum at this speed:

[tex]
\gamma m_0\ =\ \frac{m_0}{\sqrt{1-v^2/c^2}}\ =\ m_0\sqrt{1+ l(l+1)
\frac{r_0^2}{r^2}}
[/tex]

Problem: We don’t see this radius dependent increase in energy in QM
solutions with (orbital) angular momentum. The energy is independent
of the radius.- STEP 2 –We want to show how QM can compensate for this increase with a
negative [itex] p^2_r[/itex] radial momentum term. In general we have:[tex]
E^2= p_r^2 c^2 + p_\theta^2 c^2 + p_\phi^2 c^2 + m_0^2 c^4\
[/tex]

We now want a negative square radial momentum term which
compensates for the relativistic mass increase:[tex]
E^2 - E_0^2\ \ =\ \ E^2 - m_0^2 c^4\ [/tex]

[tex]
\mbox{So we need:}\ \ \ p_r^2 c^2 \ =\ - (p_\theta^2 c^2 + p_\phi^2 c^2)\ [/tex][tex]
p_r^2 c^2 \ =\ m_0^2c^4 - E^2
\ =\ (1-\gamma^2)E_0^2
[/tex]From the expression for gamma above:

[tex]
(1-\gamma^2)\ =\ 1 - \left(\ \sqrt{1+l(l+1) \frac{r_c^2}{r^2}}\ \right)^2 = -\frac{l(l+1)}{r^2}\ r^2_c
[/tex]

The last term is exactly the term which enters the radial equation
from the theta equation when solving the Spherical Laplacian in the
Schroedinger or Dirac equations:

[tex]
-\left( \frac{l(l+1)}{r^2}r^2_c \right)\ E_0^2\ =\ -\frac{l(l+1)}{r^2} \hbar^2 c^2
[/tex]

This term you can find in any textbook which handles the derivation
of the Hydrogen solutions.
- STEP 3 -We want to know the radius [itex]r/r_c[/itex] where the physical speed v/c
is such that the orbital frequency becomes equal to the de Broglie
frequency.At one hand we now have a physical orbital frequency:[tex]
f\ =\ \frac{v}{2\pi r}\ =\ \frac{c}{2\pi r }\ \sqrt{\frac{l(l+1) }
{\frac{r^2}{r_c^2}\ +\ l(l+1)}}
[/tex]

At the other hand we have the deBroglie frequency:

[tex]
f\ =\ \frac{mc^2}{h}\ =\ \frac{mc^2}{2\pi\hbar}\ =\
\frac{c}{2\pi}\frac{1}{r_c}
[/tex]

Substituting the latter in the first:

[tex]
\frac{r}{r_c}\ =\ \sqrt{\frac{l(l+1) } {\frac{r^2}{r_c^2}\ +\
l(l+1)}}
[/tex]

Giving the quadratic equation:

[tex]
\left(\frac{r^2}{r_c^2}\right)^2\ +\
l(l+1)\left(\frac{r^2}{r_c^2}\right)\ -\ l(l+1)\ =\ 0
[/tex]

Which is solved by:

[tex]
\left(\frac{r^2}{r_c^2}\right)\ =\ -\frac{1}{2} l(l+1) \pm \
\sqrt{\left(\frac{1}{2} l(l+1) \right)^2 + l(l+1)}
[/tex]

So, finally we get our dimensionless numbers related to
quantum mechanical angular momentum:

[tex]
\frac{r}{r_c}\ =\ \sqrt{-\frac{1}{2} l(l+1)\ \ +\ \
\sqrt{\left(\frac{1}{2} l(l+1) \right)^2\ +\ l(l+1)} }
[/tex]

Regards, Hans
 
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  • #198
I was thinking, should we add to the list the suggested relationship between mass of top and fermi constant?
 
  • #199
arivero said:
I was thinking, should we add to the list the suggested relationship between mass of top and fermi constant?

Lets go. From the pdg we can get the important data: Fermi constant is measured in muon decay, being G_F=1.16637(1) 10^-5 GeV^-2 and it relates to electroweak model in the

[tex]{G_F \over \sqrt 2} = \frac 18 {g^2 \over M^2_W} [/tex]

Where the inverse of the quantity [tex]\frac 12 {g \over M_W}[/tex] is usually referred as the "electroweak vacuum", and from the above equation it has a value about 246.22 GeV.

On the other hand the top, as any other fermion, gets mass via a Yukawian term [tex]y_i\bar\psi_i h \psi_i[/tex] that under breaking generates a mass term
[tex]\bar \psi_i {y_i \over \sqrt 2} { 2 M_W \over g} \psi \equiv m_i \bar \psi_i \psi_i[/tex]

(Fermions couple to the Higgs boson via this same Yukawa-originated term
[tex]\bar \psi_i {y_i \over \sqrt 2} H \psi=\bar \psi_i {g m_i \over 2 M_W} H \psi[/tex]
and it is interesting to note that we have already met [tex]m_\mu/ m_W[/tex] before in this thread.)

(Above, h is the higgs field in unbroken coordinates, while H is the extant Higgs boson of the Minimal Standard Model.) Well, the point is that [tex]m_i^2 G_F = y_i^2 {1 \over 2 \sqrt 2 } [/tex]
so that
[tex] y_i^2= 2 \sqrt 2 G_F m_i^2 [/tex]
and in particular with [tex]m_t=172.5 \pm 2.3[/tex] we have (assuming G_F exact)

[tex]2 \sqrt 2 G_F m_t^2 = .9816 \pm 0.026[/tex]

Or

[tex]
y_t = 0.991 \pm 0.013
[/tex]

There is some theoretical justification, from infrared fixed points of the renormalisation group, to have a value of order unity, but nothing about exactly unity (In quantized Bohr-Sommerfeld relativity, a unit coupling in a coulombian interaction corresponds to the situation where the trajectory of the lowest state becomes unstable and spirals into the origin of the potential. In fact such situation could be used to define Planck's constant as coincinding with the limit angular momentum). A mass equal to the value from fermi constant, ie equal to 174.1042 GeV (\pm 0.00075), should be sort of theoretical paradise or nightmare, so it is important to refine the measure.

Perhaps it should be clarified that the sqrt(2) factors come only from two sources: the traditional normalisation of Fermi interaction gives the sqrt(2) that stands even in the final formula. The other sqrt(2) is the traditional normalisation of the vacuum and it goes both to the top yukawa coupling and, squared, to the first equation so that it cancels out at the end. The only consequence of this factor is the value of the "electroweak vacuum". Some books change normalisation so that this value is also equal to 174 GeV instead 246 but most books keep at 246 so surely there is some good argument to stay so.
 
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  • #200
Lets go. From the pdg we can get the important data: Fermi constant is measured in muon decay, being G_F=1.16637(1) 10^-5 GeV^-2 and it relates to electroweak model in the

[tex]{G_F \over \sqrt 2} = \frac 18 {g^2 \over M^2_W} [/tex]

Where the inverse of the quantity [tex]\frac 12 {g \over M_W}[/tex] is usually referred as the "electroweak vacuum", and from the above equation it has a value about 246.22 GeV.

On the other hand the top, as any other fermion, gets mass via a Yukawian term [tex]y_i\bar\psi_i h \psi_i[/tex] that under breaking generates a mass term
[tex]\bar \psi_i {y_i \over \sqrt 2} { 2 M_W \over g} \psi \equiv m_i \bar \psi_i \psi_i[/tex]

(Fermions couple to the Higgs boson via this same Yukawa-originated term
[tex]\bar \psi_i {y_i \over \sqrt 2} H \psi=\bar \psi_i {g m_i \over 2 M_W} H \psi[/tex]
and it is interesting to note that we have already met [tex]m_\mu/ m_W[/tex] before in this thread.)

(Above, h is the higgs field in unbroken coordinates, while H is the extant Higgs boson of the Minimal Standard Model.)


Well, the point is that [tex]m_i^2 G_F = y_i^2 {1 \over 2 \sqrt 2 } [/tex]
so that
[tex] y_i^2= 2 \sqrt 2 G_F m_i^2 [/tex]
and in particular with [tex]m_t=172.5 \pm 2.3[/tex] we have (assuming G_F exact)

[tex]2 \sqrt 2 G_F m_t^2 = .9816 \pm 0.026[/tex]

Or

[tex]
y_t = 0.991 \pm 0.013
[/tex]

There is some theoretical justification, from infrared fixed points of the renormalisation group, to have a value of order unity, but nothing about exactly unity (In quantized Bohr-Sommerfeld relativity, a unit coupling in a coulombian interaction corresponds to the situation where the trajectory of the lowest state becomes unstable and spirals into the origin of the potential. In fact such situation could be used to define Planck's constant as coincinding with the limit angular momentum). A mass equal to the value from fermi constant, ie equal to 174.1042 GeV (\pm 0.00075), should be sort of theoretical paradise or nightmare, so it is important to refine the measure.

Perhaps it should be clarified that the sqrt(2) factors come only from two sources: the traditional normalisation of Fermi interaction gives the sqrt(2) that stands even in the final formula. The other sqrt(2) is the traditional normalisation of the vacuum and it goes both to the top yukawa coupling and, squared, to the first equation so that it cancels out at the end. The only consequence of this factor is the value of the "electroweak vacuum". Some books change normalisation so that this value is also equal to 174 GeV instead 246 but most books keep at 246 so surely there is some good argument to stay so.
 
  • #201
arivero said:
[tex] y_i^2= 2 \sqrt 2 G_F m_i^2 [/tex]

Hmm how do we put units back in this formula? If G_F is measured in [tex](\hbar c)^3 \over GeV^2[/tex] and mass is measured in [tex]GeV \over c^2 [/tex] then the square of yukawa coupling has units, er, [tex]\hbar^3 \over c [/tex]. :confused:

Hmm it is already strange here
[tex]{G_F \over \sqrt 2} = \frac 18 {g^2 \over M^2_W} [/tex]
Because in the LHS we have again (hc)^3 / Gev^2

The point of this question is, if we agree that g and y_i are adimensional coupling constants, then we need to add the h and c to restore dimensionality. Furthermore if we postulate that the yukawa coupling of Top is unity, the formula is

[tex]1= {c \over \hbar^3} 2 \sqrt 2 G_F m_{top}^2 [/tex]

This is, if we had a theory implying the exact value of y_t=1, this theory give us a formula for the constant [tex]\hbar^3 / c [/tex]. I had expected a formula for [tex]\hbar[/tex].

Well, at least we can tell that the Area of Planck is

[tex]A_P= l_P^2 = {G_N \hbar \over c^3} = {1 \over (\hbar c) ^2} 2 \sqrt 2 G_N G_F m_{top}^2 [/tex]

or, using [tex](\hbar c)=m^2_P G_N[/tex]

[tex]G_N= {2 \sqrt 2 G_F m_{top}^2 \over m_P^4 l_P^2} =
\; 2 \sqrt 2 \; ({ m_{top} \over m_P })^2 { 1 \over m_P^2 \; l_P^2 } G_F [/tex]

But there should be more fundamental ways of defining the conversion constant (hc) because after all it is the critical value of coulombian dirac equation. This is because classically a coulombian relativistic coupling hc has minimum angular momentum h (when the orbit radius goes to cero and the orbit speed goes to c). In fact, in the formula above, the denominator [tex]m^2_X \; l^2_X[/tex] has the only goal of producing a conversion factor and it works at any scale, not compulsorily the Planck scale. It is more physical perhaps to write

[tex] m_P^2 G_N= { 2 \sqrt 2 \over m^2_X \; l^2_X} m_{top}^2 G_F [/tex]
The funny thing of this formula is that it does not use explicitly h nor c, the necessary combination is hidden in the condition of having m_X and l_X defined at the same scale. As for the powers of m and l, they come from dimensional considerations: Fermi force goes as 1/r^4, gravitation as 1/r^2 so we need a length square. Fermi force does not depend of mass, gravitations depende on product of two masses, so we need a mass square. The only ad-hocness is, as said, to ask the needed mass and the needed length to be defined at the same scale (ie the length must be compton length of the corresponding mass), even if arbitrary; in this way mass and length cancel to produce h/c factors.
 
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  • #202
Equation group (7) in page 8 of hep-ph/0604035 lists

[tex]0.22306 \pm 0.00033[/tex]

for the vector mass-based weinberg angle on-shell, so the kinematical result

[tex] \sin^2 \theta_W
\ \ \ \ = \ \ \ \ 1 \ - \ \frac{\beta^2_f }{\beta^2_b}
\ \ \ \ = \ \ \ \ 0.22310132230086634541466926662604[/tex]

is still in and for sure already for three digits. It was "(to within 0.063% or sigma 1.2)" in 2004 at the start of the thread. Now it is 0.01% and sigma 0.13 if we buy the global fit of this paper (which is from Erler, so it will most probably appear in the 2006 PDG).
 
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  • #203
Hans de Vries said:
These dimensionless numbers related to angular momentum may also be
derived from Quantum Mechanics.
...
1- We’ll derive a radius dependent speed (in QM) which we still need.
2- Show how QM can solve the missing relativistic mass increase.
3- Get the formula which gives us the dimensionless numbers.
Well, at the moment all we have is a derivation from De Broglie Quantum Mechanics, not new, not old, quantum mechanics. In fact some afirmations are very DeBroglian, as when you speak of a dependence or independence between radius and energy; in new quantum mechanics all the observables are got via averaging against the wavefunction squared, so all the three coordinates are integrated out; no observable has dependency in coordinates.

Step 1 and 2 are even valid in classical mechanics simply using L^2 instead of l(l+1)

Ah, a minor problem with the use of l(l+1), of course, is that while total angular momentum can be half-integer, orbital angular momentum as produced by quantising L=RxP according Schroedinger rules produces an operator with only integer numbers. This is the typical lore of SU(2) being the double covering of SO(3).

We want to show how QM can compensate for this increase with a
negative [itex] p^2_r[/itex] radial momentum term. In general we have:

It seems a good idea, even if I am unsure about the process in step 2. For circular orbits, we have only p_\phi different of zero, have we?

But yep, either we use some requisite relating the kinetik rotational (centrifugue term, if you wish) energy with the rest mass energy, or we look for some confining potential imposing the radius ad-hoc.
 
  • #204
arivero said:
Well, at the moment all we have is a derivation from De Broglie Quantum Mechanics, not new, not old, quantum mechanics. In fact some afirmations are very DeBroglian,

In fact it was ringing me some bell so I have revised my archives, and well, I have found your formula in De Broglie himself!

Ondes et Quanta, Note de M. Louis de Broglie, presentee par M. Jean Perrin. Seance du 10 Septembre 1923, Academie des Sciencies, pages 507 to 510. There is a footnote "au sujet de la presente Note, voir M. Brillouin, Comptes Rendus, t. 168, 1919, p 1318" . The paper is one using a small non null mass for the photon and declaring a tachionic wave over it, so it is not very popular in mainstream I guess. But after finishing with the photon, he focuses in a orbiting particle, and then in page 509 it gives

[tex]{m_0 \beta^2 c^2 \over \sqrt {1-\beta^2} }T_r= n h [/tex]

With T_r the period of revolution of the particle in the orbit. This is the justifiable part of Hans Argument. Over it we need two extra assumptions

1) fix the period from the rest mass,
[tex] T_r = { h \over m_0 c^2} [/tex]

2) Move n to be a value alike to the ones from Schroedinger wave mechanics, instead of the one from Einstein stability condition ("old" Bohr-Sommerfeld QM}.
[tex] n \to \sqrt {j (j+1)} [/tex]

One could expect at least 2) to be addressed by De Broglie in some later work but I haven't found any. I keep searching.

The j(j+1) value for L^2 appears for sure in Born version of QM, and in the article I am using (Nov. 1925, published 1926) it remarks "this result is formally reminiscent of the values of M^2 which enter the Lande g formula". Or something so, in German. The result is also general enough to do not rule out semintegers, as Schroedinger result does.

It seems (I am reading in a history report of J Brandmuller) that the substitution rule was proposed by Lande in 1923 after some previous turmoil about delaying publications for coordination with Back. The original substitution is J^2 ---> (J^2-1/4). Then Pauli in 1904 proposes a change of variables and the well known j(j-1) emerges there, empirically, with a narrow two years advantage against theoretist publications. The paper of Pauli should be Z Phys 20 p 371, and perhaps also 31 p 765.

Zur Frage der Zuordnung der Komplexstrukturterme in starken und in schwachen äußeren Feldern

The paper of pauli is restricted to subscribers of the full electronic series, I can not reach it from my campus. A commentary appears in Olivier Darrigol's http://content.cdlib.org:8088/xtf/view?docId=ft4t1nb2gv&chunk.id=0&doc.view=print, from formula 170 to 177 and ff. It seems that Pauli suggested that a derivative had been substituted by a difference (remember j has units of [tex]\hbar[/tex] so it is plausible)


:bugeye: [tex]
{1 \over j^2} \to - {d \over dj } ({1 \over j}) \to {1 \over j-1} - {1 \over j } \to {1\over j(j-1)}
[/tex] :eek:
and this hint was used by Heisenberg to fit with his own development. Schroedinger does not need it as it comes automagically from functional analysis. And De Broglie? Does he miss the point?

Some papers put l(l+1) in Sommerfeld formulae, so perhaps some edition of his Atombau corrected for it (with proof?)

A recent reference recalling the Lande g formula is hep-ph/0209068, page 7. It also discusses the semiinteger vs integer issue, but does not quote sources. Barut wrote a small biography of Lande, http://www.physik.uni-frankfurt.de/paf/paf38.html .
 
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  • #205
arivero said:
Well, at the moment all we have is a derivation from De Broglie Quantum Mechanics, not new, not old, quantum mechanics. In fact some afirmations are very DeBroglian, as when you speak of a dependence or independence between radius and energy; in new quantum mechanics all the observables are got via averaging against the wavefunction squared, so all the three coordinates are integrated out; no observable has dependency in coordinates.

Well the path-integral did bring back the idea of the moving point particle.
de Broglie was the first to use the plane wave solutions of the Klein Gordon
equation without knowing it. On the other hand, I just as well may have
some "Bohr"-ian coincidence here.


arivero said:
It seems a good idea, even if I am unsure about the process in step 2. For circular orbits, we have only p_\phi different of zero, have we?

Actually, [itex]p_\phi[/itex] attributes [itex]l^2/r^2\ \hbar[/itex] and [itex]p_\theta[/itex] attributes [itex]l/r^2\ \hbar[/itex]

So looking to the local phase behavior, the motion in a pure z-spin state
is still at an angle with the x,y plane: [itex] \psi \ =\ \pm\arctan{(1/\sqrt{l})} [/itex]
Plus or minus relates to a 50:50 superposition of the +/- x-spin and y-spin
states. This still may be related to "circular orbits" The circle at a tiled
angle crosses the x-y plane twice, at plus and minus psi.

All tiled circles combined over 360 degrees of phi then gives the
"precession picture", although it's more like happening all simultaneously.
All the circular orbits are 100% either left handed or right handed around
the z-axis and 50%-50% around the x-axis and y-axis like Stern-Gerlach
would like it to be.

I actually don't believe in moving point particles around circular orbits
but there's the correspondence between the local phase behavior and
the phase behavior of a plane wave representing a moving particle.

arivero said:
But yep, either we use some requisite relating the kinetik rotational (centrifugue term, if you wish) energy with the rest mass energy, or we look for some confining potential imposing the radius ad-hoc.

There are lots of interesting candidates for stable localized Klein-Gordon/
Dirac solutions for free particles which do not extend to infinity.
They are merely modulated by the plane wave solutions but finite in size.
(They are not super positions of on the shell plane wave solutions, these
aren't stable at all)

Some of them counter a repulsive potential just by geometry (like the
hydrogen solutions counter an attractive potential) Some cancel
radius dependent relativistic mass increase (like what I was talking
about). Yet again others can modify the rest-mass by geometry alone
and mimic exactly the wave behavior of particles with another mass.I might give some specific examples later together with the math.


Regard, Hans
 
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  • #206
arivero said:
The j(j+1) value for L^2 appears for sure in Born version of QM, and in the article I am using (Nov. 1925, published 1926) it remarks "this result is formally reminiscent of the values of M^2 which enter the Lande g formula". Or something so, in German. The result is also general enough to do not rule out semintegers, as Schroedinger result does.

I have this as a reprint in van der Waerdens book: Sources of Quantum Mechanics.
Born together with Heisenberg and Jordan: On Quantum Mechanics II

arivero said:
A recent reference recalling the Lande g formula is hep-ph/0209068, page 7. It also discusses the semiinteger vs integer issue, but does not quote sources. Barut wrote a small biography of Lande, http://www.physik.uni-frankfurt.de/paf/paf38.html .

Lots of nice links Alejandro, A good source is also "The Story of Spin"
from Tomonaga.


Regards, Hans
 
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  • #207
Hans de Vries said:
There are lots of interesting candidates

To me this should be an atractive :wink: point of the argument, it is very general old relativistic QM, so a lot of modellers can do use of it, potentially

Lots of nice links Alejandro, A good source is also "The Story of Spin"
from Tomonaga..

Thanks. The e-book of Olivier Darrigol is also a good source for spin histories. I have asked him about why De Broglie did not try the j(j+1) view, he suspects that "To him it would have been only one among many other symptoms of the failures of classical models". It is amusing because once you get the idea of using this trick the path to speeds or radiouses is clear.

(note that one of the consequences of j(j+1) is that the angular momentum is always greater than the third component of angular momentum; Born argues this is the way nature implements uncertainty principle for J).
 
  • #208
Hans de Vries said:
So, finally we get our dimensionless numbers related to
quantum mechanical angular momentum:
[tex]
... =\ \sqrt{-\frac{1}{2} l(l+1)\ \ +\ \
\sqrt{\left(\frac{1}{2} l(l+1) \right)^2\ +\ l(l+1)} }
[/tex]

A fast trivial observation is that operator-wise this is still
[tex]
\beta^2= \left<\bar \Psi \left| {-\frac{1}{2} J^2\ \ +\ \
\sqrt{J^2 + \left(\frac{1}{2} J^2 \right)^2\ } } \right| \Psi \right> [/tex]

and thus

[tex]
\sin^2 \theta = 1 - {
\left<\bar \Psi_{J=1/2} \left| {- J^2\ \ +\ \
\sqrt{4 J^2 + \left( J^2 \right)^2\ } } \right| \Psi_{J=1/2} \right>
\over
\left<\bar \Psi_{J=1} \left| {- J^2\ \ +\ \
\sqrt{4 J^2 + \left( J^2 \right)^2\ } } \right| \Psi_{J=1} \right>
}
[/tex]

(note that if we do not use natural units, we must write [tex]{1\over \hbar^2} J^2[/tex] for the algebraic operator).
 
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  • #209
arivero said:
[tex]
\sin^2 \theta = 1 - {
\left<\bar \Psi_{J=1/2} \left| {- J^2\ \ +\ \
\sqrt{4 J^2 + \left( J^2 \right)^2\ } } \right| \Psi_{J=1/2} \right>
\over
\left<\bar \Psi_{J=1} \left| {- J^2\ \ +\ \
\sqrt{4 J^2 + \left( J^2 \right)^2\ } } \right| \Psi_{J=1} \right>
}
[/tex]

(note that if we do not use natural units, we must write [tex]{1\over \hbar^2} J^2[/tex] for the algebraic operator).

Hmm of course if we had an operator Q such that
[tex]
\left| \Psi_{J=1} \right> = Q \left| \Psi_{J=1/2} \right>
[/tex]

We could even define the operatorial object

[tex]
\left(Q^+ f({1\over \hbar^2} L^2) Q \right)^{-1}
f({1\over \hbar^2} L^2)
[/tex]

Which will produce the desired quantity "M_W/M_Z" when evaluated on a solution of Dirac equation.

And of course Q is a supersymmetry generator. :smile:

The problem here is that when L is not the total angular momentum but the orbital momentum it does not commute with J (and what about Q?). But on other hand our algebraic account is independent of the starting point, it "only" serves to define f()
 
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  • #210
hyperzitterbewegung

While looking for the concept of hyperzitterbewegung, I happened to find an article very in the spirit of this thread,

E. Schönfeld
Electron and Fine-Structure Constant
Metrologia 27, p 117-125 (1990)
http://www.iop.org/EJ/article/0026-1394/27/3/002/metv27i3p117.pdf

It is cited by James G. Gilson in his preprint http://www.maths.qmul.ac.uk/~jgg/gilj.pdf , also overly speculative work I think we have already met along the thread.
 
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