# Lab Notes for a Scientific Revolution (Physics)

## December 26, 2008

### Finite Amplitudes Without +i\epsilon

To all,

I have now completed a paper at the link below, which summarizes the work I have been doing for the past two months (and in a deeper sense, for much of my adult life) to lay a foundation for understanding and calculating particle masses:

finite-amplitudes-without-i-epsilon

I have also taken the plunge and submitted this for peer review. ;-)?

The abstract is as follows:

By carefully reviewing how the invariant amplitude M is arrived at in the simplest Yang-Mills gauge group SU(2), we show how to arrive at a finite, pole-free amplitudes without having to resort to the “+i$\epsilon$ prescription.”  We first review how gauge boson mass is generated in the SU(2) action via spontaneous symmetry breaking in the standard model, and then carefully consider the formation of finite, on-shell amplitudes, without +i$\epsilon$.

Comments are welcome, and I wish everyone a happy holiday and New Year!

Jay.

## June 19, 2008

### A New Lab Note: Commutation of Linear Rest Mass with Canonical Position

It has been awhile since my last blog entry, but if you want to check out some my recent wanderings through physicsland, check out sci.physics.foundations, relativity, and research.

Here, I would like to show a rather simple calculation, which may cast a different light on how one needs to think about the canonical commutation relationship $\left[x_{j} ,p_{k} \right]=i\eta _{jk} ;\; j,k=1,2,3$.  I would very much like your comments in helping me sort this through.  You may download this in pdf form at https://jayryablon.files.wordpress.com/2008/06/linear-mass-commutator-calculation.pdf.

I.  A Known Square Mass Commutation Calculation

Consider a particle of mass m as a single particle system.  Consider canonical coordinates $x_{\mu }$, and that at least the space coordinates $x_{j} ;\; j=1,2,3$ are operators.  If we require that the mass m must commute with all operators, then we must have $\left[x_{\mu } ,m\right]=0$, and by easy extension, $\left[x_{\mu } ,m^{2} \right]=0$.  It is well known that the commutation condition $\left[x_{\mu } ,m^{2} \right]=0$, taken together with the on-shell mass relationship$m^{2} =p^{\sigma } p_{\sigma }$ and the single-particle canonical commutation relationship $\left[x_{j} ,p_{k} \right]=i\eta _{jk} ;\; j,k=1,2,3$, where $diag\left(\eta _{\mu \nu } \right)=\left(-1,+1,+1,+1\right)$ is the Minkowski tensor, leads inexorably to the commutation relationship:

$\left[x_{k} ,p_{0} \right]=-ip_{k} /p^{0} =-iv_{k}$   (1.1)

where $v_{k}$ is the particle velocity (in c=1 units) along the kth coordinate.  I leave the detailed calculation as an exercise for the reader not familiar with this calculation, and refer also to the sci.physics.research thread at http://www.physicsforums.com/archive/index.php/t-142092.html or http://groups.google.com/group/sci.physics.research/browse_frm/thread/d78cbfecf703ff6a.

I would ask for your comments on the following calculation, which is totally analogous to the calculation that leads to (1.1), but which is done using the linear mass m rather than the square mass $m^{2}$, and using the Dirac equation written as $m\psi =\gamma ^{\nu } p_{\nu } \psi$, in lieu of what is, in essence, the Klein Gordon equation $m^{2} \phi =p^{\sigma } p_{\sigma } \phi$ that leads to (1.1).

2.  Maybe New?? Linear Mass Commutation Calculation

$m\psi =\gamma ^{\nu } p_{\nu } \psi$ .  (2.1)

Require that:

$\left[x_{\mu } ,m\right]=0$   (2.2)

Continue to use the canonical commutator $\left[x_{j} ,p_{k} \right]=ig_{jk}$.  Multiply (2.1) from the left by $x_{\mu }$ noting that $\left[\gamma ^{\nu } ,x_{\mu } \right]=0$ to write:

$x_{\mu } m\psi =\gamma ^{\nu } x_{\mu } p_{\nu } \psi =\gamma ^{0} x_{\mu } p_{0} \psi +\gamma ^{j} x_{\mu } p_{j} \psi$ .  (2.3)

This separates into:

$\left\{\begin{array}{c} {x_{0} m\psi =\gamma ^{0} x_{0} p_{0} \psi +\gamma ^{j} x_{0} p_{j} \psi } \\ {x_{k} m\psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} x_{k} p_{j} \psi } \end{array}\right.$ .  (2.4)

Now, use the canonical relation $\left[x_{j} ,p_{k} \right]=i\eta _{jk}$ to commute the space (k) equation, thus:

$\begin{array}{l} {x_{k} m\psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} x_{k} p_{j} \psi =\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} \left(p_{j} x_{k} +i\eta _{jk} \right)\, \psi } \\ {=\gamma ^{0} x_{k} p_{0} \psi +\gamma ^{j} p_{j} x_{k} \psi +i\gamma _{k} \psi } \\ {=\gamma ^{0} x_{k} p_{0} \psi +mx_{k} \psi -\gamma ^{0} p_{0} x_{k} \psi +i\gamma _{k} \psi } \end{array}$ .  (2.5)

In the final line, we use Dirac’s equation written as $mx_{\mu } \psi =\gamma ^{\nu } p_{\nu } x_{\mu } \psi =\gamma ^{0} p_{0} x_{\mu } \psi +\gamma ^{j} p_{j} x_{\mu } \psi$, and specifically, the $\mu =k$ component equation $\gamma ^{j} p_{j} x_{k} \psi =mx_{k} \psi -\gamma ^{0} p_{0} x_{k} \psi$.

If we require that $\left[x_{\mu } ,m\right]=0$, which is (2.2), then (2.5) reduces easily to:

$\gamma ^{0} \left[x_{k} ,p_{0} \right]\psi =-i\gamma _{k} \psi$ ,  (2.6)

Finally, multiply from the left by $\gamma ^{0}$, and employ $\gamma ^{0} \gamma _{k} \equiv \alpha _{k}$ and $\gamma ^{0} \gamma ^{0} =1$ to write:

$\left[x_{k} ,p_{0} \right]\, \psi =-i\alpha _{k} \psi$ .  (2.7)

If we contrast (2.7) to (1.1) written as $\left[x_{k} ,p_{0} \right]\phi =-iv_{k} \phi$, we see that the velocity $p_{k} /p^{0} =v_{k}$ has been replaced by the Dirac operator $\alpha _{k}$, that is, $v_{k} \to \alpha _{k}$.

3.  Questions

Here are my first set of questions:

1)  Is the calculation leading to (2.7) correct, and is (2.7) a correct result, or have I missed something along the way?

2)  If (2.7) is correct, has anyone seen this result before?  If so where?

3)  Now use the plane wave $\psi =ue^{ip^{\sigma } x_{\sigma } }$ so that we can work with the Dirac spinors $u\left(p^{\mu } \right)$, and rewrite (2.7) as:

$\left\{\begin{array}{c} {\left(\alpha _{k} -\lambda \right)\, u=0} \\ {\lambda =i\left[x_{k} ,p_{0} \right]} \end{array}\right.$

The upper member of (3.1) is an eigenvalue equation.  Reading out this equation, I would say that the commutators $\lambda =i\left[x_{k} ,p_{0} \right]$ are the eigenvalues of the Dirac $\alpha _{k}$ matrices, which are:

${\bf \alpha }=\left(\begin{array}{cc} {0} & {{\bf \sigma }} \\ {{\bf \sigma }} & {0} \end{array}\right)$ and ${\bf \alpha }=\left(\begin{array}{cc} {-{\bf \sigma }} & {0} \\ {0} & {{\bf \sigma }} \end{array}\right)$ ,  (3.2)

in the respective Pauli/Dirac and Weyl representations, and that the u are the eigenvectors associated with these eigenvalues $\lambda =i\left[x_{k} ,p_{0} \right]$.  Am I wrong?  If not, how would one interpret this result?  Maybe the commutators $\left[x_{j} ,p_{k} \right]=i\eta _{jk}$ can be discussed in the abstract, but it seems to me that the commutators $\lambda =i\left[x_{k} ,p_{0} \right]$ can only be discussed as the eigenvalues of the matrices $\alpha _{k}$ with respect to the eigenstate vectors u.  This, it seems, would put canonical commutation into a somewhat different perspective than is usual.

Just as Dirac’s equation reveals some features that cannot be seen strictly from the Klein Gordon equation, the calculation here seems to reveal some features about the canonical commutators that the usual calculation based on $\left[x_{\mu } ,m^{2} \right]=0$ and $m^{2} =p^{\sigma } p_{\sigma }$ cannot, by itself, reveal.

I’d appreciate your thoughts on this, before I proceed downstream from here.

Thanks,

Jay.

## May 1, 2008

### Heisenberg Uncertainty and Schwinger Anomaly Continued: Draft Paper

I have been writing a paper to rigorously develop the hypothesis I presented last week, in a post linked at Heisenberg Uncertainty and Schwinger Anomaly: Two Sides of the Same Coin?.  I believe there is enough developed now, and I think enough of the kinks are now out, so you all may take a sneak preview.  Thus, I have linked my latest draft at:

Heisenberg Uncertainty and the Schwinger Anomaly

Setting aside the hypothesized connection between the magnetic anomaly and uncertainty, Sections 4 through 7, which have not been posted in any form previously, stand completely by themselves, irrespective of this hypothesis.  These sections are strictly mathematical in nature, and they provide an exact measure for how the uncertainty associated with a wavefunction varies upwards from $\hbar/2$ as a function of the potential, and the parameters of the wavefunction itself.  The wavefunction employed is completely general, and the uncertainty relation is driven by a potential $V$.

This is still under development, but this should give you a very good idea of where this is headed.

Of course, I welcome comment, as always.

Best regards,

Jay.

## April 18, 2008

### Lab Note 6: Operator Decomposition of Intrinsic Spin

I’d like to lay out a nifty little mathematical calculation which allows a “decomposition” of the intrinsic spin matrices $s^{i} ={\tfrac{1}{2}} \hbar \sigma ^{i}$ to include the position and momentum operators $x^{i}$, $p^{i}$, $i=1,2,3$.  To simplify matters, we will employ a Minkowski metric tensor with ${\rm diag}\left(\eta _{\mu \nu } \right)=\left(-1,+1,+1,+1\right)$ so that raising and lowering the space indexes $i=1,2,3$ is simple and at will, and does not entail any sign reversal.  (This lab note is available in PDF form, with a recent update including a new section 2, at intrinsic-spin-decomposition-11.)

We start with the general cross product for two three-vectors A and B.  Written in covariant (index) notation:

$\left(A\times B\right)_{i} \equiv \varepsilon _{ijk} A^{j} B^{k}$.   (1)

One can easily confirm this by taking, for example, $\left(A\times B\right)_{3} \equiv A^{1} B^{2} -A^{2} B^{1}$.  Now, let’s take the triple cross product $\left(A\times B\right)\times C$.  We can apply (1) to itself using $\left(A\times B\right)^{j} \equiv \varepsilon ^{jmn} A_{m} B_{n}$, to write:

$\left[\left(A\times B\right)\times C\right]_{i} =\varepsilon _{ijk} \left(A\times B\right)^{j} C^{k} =\varepsilon _{ijk} \varepsilon ^{jmn} A_{m} B_{n} C^{k}$.  (2)

The fact that the crossing of A and B takes precedence over crossing with C is retained in the fact that $A_{m} B_{n}$ sum with $\varepsilon ^{jmn}$, while $C^{k}$ alone sums into $\varepsilon _{ijk}$.

Let us now expand (2) for the component equation for which $i=3$.  The calculation is as such:

$\begin{array}{l} {\left[\left(A\times B\right)\times C\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} } \\ {=\varepsilon _{312} \varepsilon ^{123} A_{2} B_{3} C^{2} +\varepsilon _{312} \varepsilon ^{132} A_{3} B_{2} C^{2} +\varepsilon _{321} \varepsilon ^{231} A_{3} B_{1} C^{1} +\varepsilon _{321} \varepsilon ^{213} A_{1} B_{3} C^{1} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} -A_{3} B_{1} C^{1} -A_{3} B_{2} C^{2} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} B_{1} C^{1} -A_{3} B_{2} C^{2} -A_{3} B_{3} C^{3} } \\ {=A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} \left(B\cdot C\right)} \end{array}$,  (3)

where we have added $0=A_{3} B_{3} C^{3} -A_{3} B_{3} C^{3}$ to the fourth line.  Now in the final line, we hit an impasse, because $B_{3}$ is sandwiched between the terms we would like to form into the other dot product $A\cdot C$.  In order to complete the calculation, we must make an assumption that the $A_{i}$commute with $B_{3}$, i.e., that $\left[A_{i} ,B_{3} \right]=0$.  For now, let us make this assumption.

Therefore, we carry out the commutation in (3), and continue along to write:

$\begin{array}{l} {\left[\left(A\times B\right)\times C\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} =A_{1} B_{3} C^{1} +A_{2} B_{3} C^{2} +A_{3} B_{3} C^{3} -A_{3} \left(B\cdot C\right)} \\ {=B_{3} \left(A\cdot C\right)-A_{3} \left(B\cdot C\right)=B_{3} A_{j} C^{j} -A_{3} B_{j} C^{j} } \end{array}$.  (4)

Generalizing fully, we may now write (4) in two equivalent ways as:

$\left\{\begin{array}{c} {\left(A\times B\right)\times C=-A\left(B\cdot C\right)+B\left(A\cdot C\right)\quad \quad } \\ {\varepsilon _{ijk} \varepsilon ^{jmn} A_{m} B_{n} C^{k} =-A_{i} B_{j} C^{j} +B_{i} A_{j} C^{j} } \end{array}\right.$.  (5)

The reader will observe the well-known formula for the cross product.

Now, let’s take the cross product in which $A=x$, $B=p$ and $C={\bf \sigma }$, where x is the position operator about the center of mass, p is the momentum operator, and ${\bf \sigma }$ are the Pauli spin matrices.  We also take into account the Heisenberg canonical commutation relationship between the position and momentum operators, that is, $\left[x_{\mu } ,p_{\nu } \right]=i\hbar \delta _{\mu \nu }$.  This means that we will have to be careful at the juncture between equations (3) and (4), because the position and momentum operators along the same dimension do not commute.

So, we return to (3) with $A=x$, $B=p$ and $C={\bf \sigma }$, to write:

$\left[\left(x\times p\right)\times {\bf \sigma }\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =x_{1} p_{3} \sigma ^{1} +x_{2} p_{3} \sigma ^{2} +x_{3} p_{3} \sigma ^{3} -x_{3} \left(p\cdot {\bf \sigma }\right)$.  (6)

To take the next step, we want to place $p_{3}$ in front of the $x_{i}$.  In so doing, we can commute $p_{3}$ with $x_{i}$ for $i=1,2$.  But, for $i=3$, we must employ $x_{3} p_{3} =p_{3} x_{3} +i\hbar$.  Therefore, (6) now becomes:

$\begin{array}{l} {\left[\left(x\times p\right)\times {\bf \sigma }\right]_{3} =\varepsilon _{3jk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =p_{3} x_{1} \sigma ^{1} +p_{3} x_{2} \sigma ^{2} +\left(p_{3} x_{3} +i\hbar \right)\sigma ^{3} -x_{3} \left(p\cdot {\bf \sigma }\right)} \\ {=p_{3} \left(x\cdot {\bf \sigma }\right)-x_{3} \left(p\cdot {\bf \sigma }\right)+i\hbar \sigma _{3} =p_{3} x_{j} \sigma ^{j} -x_{3} p_{j} \sigma ^{j} +i\hbar \sigma _{3} } \end{array}$,  (7)

lowering the index on $i\hbar \sigma ^{3}$ with ${\rm diag}\left(\eta _{ij} \right)=\left(+1,+1,+1\right)$.  Now all of a sudden, $i\hbar \sigma ^{3}$ has made an unexpected appearance.  Generalizing (7), we may write:

$\left\{\begin{array}{c} {\left[\left(x\times p\right)\times {\bf \sigma }\right]=-x\left(p\cdot {\bf \sigma }\right)+p\left(x\cdot {\bf \sigma }\right)+i\hbar {\bf \sigma }\quad \quad \quad } \\ {\varepsilon _{ijk} \varepsilon ^{jmn} x_{m} p_{n} \sigma ^{k} =-x_{i} p_{j} \sigma ^{j} +p_{i} x_{j} \sigma ^{j} +i\hbar \sigma _{i} } \end{array}\right.$,  (8 )

This is the also the well-known formula for the triple-cross product, but with an additional term $i\hbar {\bf \sigma }$ emerging from the canonical commutation relationship.  In fact, moving terms, equation (8 ) gives us a way to decompose the intrinsic spin matrix so as to contain the position and momentum, and as we shall also see, orbital angular momentum operators.

First, we rewrite (8 ) as:

$\left\{\begin{array}{c} {i\hbar s=\left[\left(x\times p\right)\times s\right]+x\left(p\cdot s\right)-p\left(x\cdot s\right)\quad \quad \quad } \\ {i\hbar s_{i} =\varepsilon _{ijk} \varepsilon ^{jmn} x_{m} p_{n} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} } \end{array}\right.$,   (9)

where we have multiplied through by ${\tfrac{1}{2}} \hbar$ and then set $s_{i} \equiv {\tfrac{1}{2}} \hbar \sigma _{i}$.  This decomposes the intrinsic spin matrix into an expression involving itself, as well as the position and momentum operators.

Now, using the definition (1) but with $A=x$ and $B=p$, let’s introduce the orbital angular momentum operator :

$l^{j} \equiv \left(x\times p\right)^{j} \equiv l^{j} \equiv \varepsilon ^{jmn} x_{m} p_{n}$  (10)

It is easy to see, for example, that $l^{3} =x_{1} p_{2} -x_{2} p_{1}$.  Using (10), we now rewrite (9) as:

$\left\{\begin{array}{c} {i\hbar s=\left(l\times s\right)+x\left(p\cdot s\right)-p\left(x\cdot s\right)\quad \; \; } \\ {i\hbar s_{i} =\varepsilon _{ijk} l^{j} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} } \end{array}\right.$,   (11)

We see that part of this decomposition includes the cross-product $l\times s$ of the orbital angular momentum with the intrinsic spin.  We may also multiply the lower equation (11) through by $\varepsilon ^{mni}$ and then employ the commutation relationship $\left[s^{m} ,s^{n} \right]=i\hbar \varepsilon ^{mni} s_{i}$, to write:

$\left[s^{m} ,s^{n} \right]=\varepsilon ^{mni} \varepsilon _{ijk} l^{j} s^{k} +\varepsilon ^{mni} x_{i} p_{j} s^{j} -\varepsilon ^{mni} p_{i} x_{j} s^{j} =l^{m} s^{n} +\varepsilon ^{mni} x_{i} p_{j} s^{j} -\varepsilon ^{mni} p_{i} x_{j} s^{j}$.  (12)

Note, we have also made use of $\varepsilon ^{mni} \varepsilon _{ijk} =\delta ^{mni} _{ijk}$

Equation (11) allows us to decompose the total spin S for a Dirac field $\psi$, as follows: WORDPRESS DOES NOT LIKE THE INTEGRALS — NEED TO FIX

$\left\{\begin{array}{c} {S=\int \left(\overline{\psi }s\psi \right)\, d^{3} x =-{\tfrac{i}{\hbar }} \int \left(\overline{\psi }\left[\left(l\times s\right)+x\left(p\cdot s\right)-p\left(x\cdot s\right)\right]\, \psi \right)\, d^{3} x\quad \; \; } \\ {S_{i} =\int \left(\overline{\psi }s_{i} \psi \right)\, d^{3} x =-{\tfrac{i}{\hbar }} \int \left(\overline{\psi }\left[\varepsilon _{ijk} l^{j} s^{k} +x_{i} p_{j} s^{j} -p_{i} x_{j} s^{j} \right]\psi \right)\, d^{3} x } \end{array}\right.$ (13)

See Ohanian, H., What is spin?,  equation (18).

More to follow . . .

## March 30, 2008

### Revised Paper on Kaluza-Klein and Intrinsic Spin, Based on Spatial Isotropy

I have now prepared an updated revision of a paper demonstrating how the compact fifth dimension of Kaluza-Klein is responsible for the observed intrinsic spin of the charged leptons and their antiparticles.  The more global, underlying view, is that all intrinsic spins originate from motion through the curled up, compact $x^5$ dimension.  This latest draft is linked at:

Intrinsic Spin and the Kaluza-Klein Fifth Dimension, Rev 3.0

Thanks to some very helpful critique from Daryl M. on a thread at sci.physics.relativity, particularly post #2, I have entirely revamped section 5, which is the heart of the paper wherein we establish the existence of intrinsic spin in $x^5$, on the basis of “fitting” oscillations around a $4\pi$ loop of the compact fifth dimension (this is to maintain not only orientation but entanglement version).  From this approach, quantization of angular momentum in $x^5$ naturally emerges, it also emerges that the intrinsic $x^5$ angular momentum in the ground state is given by $(1/2) \hbar$.

In contrast to my earlier papers where I conjectured that the intrinsic spin in $x^5$ projects out into the three ordinary space dimensions by virtue of its orthogonal orientation relative to the $x^5$ plane, which was critiqued in several Usenet posts, I now have a much more direct explanation of how the intrinsic spin projects out of $x^5$ to where we observe it.

In particular, we recognize that one of the objections sometimes voiced with regards to a compactified fifth dimension is the question: how does one “bias” the vacuum toward one of four space dimensions, over the other three, by making that dimension compact?  This was at the root of some Usenet objections also raised earlier by DRL.

In this draft — and I think this will overcome many issues — we require that at least as regards intrinsic angular momentum, the square of the $J^5 = (1/2) \hbar$ obtained for the intrinsic angular momentum in $x^5$, must be isotropically shared by all four space dimensions.  That is, we require that there is to be no “bias” toward any of the four space dimensions insofar as squared intrinsic spin is concerned.  Because $J^5 = (1/2) \hbar$ emanates naturally from the five dimensional geometry, we know immediately that $\left(J^{5} \right)^{2} ={\textstyle\frac{1}{4}} \hbar ^{2}$, and then, by the isotropic requirement, that $\left(J^{1} \right)^{2} =\left(J^{2} \right)^{2} =\left(J^{3} \right)^{2} ={\textstyle\frac{1}{4}} \hbar ^{2}$ as well.  We then arrive directly at the Casimir operator $J^{2} =\left(J^{1} \right)^{2} +\left(J^{2} \right)^{2} +\left(J^{3} \right)^{2} ={\textstyle\frac{3}{4}} \hbar ^{2}$ in the usual three space dimensions, and from there, continue forward deductively.

For those who have followed this development right along, this means in the simplest terms, that rather than use “orthogonality” to get the intrinsic spin out of $x^5$ and into ordinary space, I am instead using “isotropy.”

There is also a new section 8 on positrons and Dirac’s equations which has not been posted before, and I have made other editorial changes throughout the rest of the paper.

## March 29, 2008

### Stepping Back from Kaluza-Klein: Planned Revisions

Those who have followed my Weblog are aware that I have been putting in a lot of work on Kaluza-Klein theory.  This post is to step back from the canvas, lay out the overall picture of what I am pursuing, and summarize what I plan at present to change or correct in the coming days and weeks.  This is in keeping with the concept of this Weblog as “Lab Notes,” or as a public “scientific diary.”

There are really two main aspects to this Kaluza-Klein work:

First, generally, I have found that 5-D Kaluza-Klein theory is most simply approached by starting with (classical) Lorentz force motion, and requiring the Lorentz force motion to be along the geodesics of the five dimensional geometry.  I am far from the first person who recognizes that the Lorentz force can be represented as geodesic motion in a 5-D model.  But I have found, by starting with the Lorentz force, and by requiring the 5-D electromagnetic field strength tensor to be fully antisymmetric, that all of the many “special assumptions” which are often employed in Kaluza-Klein theory energy very naturally on a completely deductive basis, with no further assumptions required.  I also believe that this approach leads to what are perhaps some new results, especially insofar as the Maxwell tensor is concerned, and insofar as QED may be considered in a non-linear context.   The latest draft of this global work on Kaluza-Klein may be seen at Kaluza-Klein Theory and Lorentz Force Geodesics.

Second, specifically, within this broader context, is the hypothesis that the fifth-dimensional “curled” motion is the direct mainspring of intrinsic spin.  More than anything else, the resistance by many physicists to Kaluza-Klein and higher-dimensional theories, rests on the simple fact that this fifth dimension — and any other higher dimensions — are thought to not be directly observable.  In simplest form, “too small” is the usual reason given for this.  Thus, if it should become possible to sustain the hypothesis that intrinsic spin is a directly-observable and universally-pervasive outgrowth of the fifth dimension, this would revitalize Kaluza-Klein as a legitimate and not accidental union of gravitation and electrodynamics, and at the same time lend credence to the higher-dimensional efforts also being undertaken by many researchers.  The latest draft paper developing with this specific line of inquiry is at Intrinsic Spin and the Kaluza-Klein Fifth Dimension.

Now, the general paper at Kaluza-Klein Theory and Lorentz Force Geodesics is very much a work in progress and there are things in this that I know need to be fixed or changed.  If you should review this, please keep in mind the following caveats:

First, sections 1-4 are superseded by the work at Intrinsic Spin and the Kaluza-Klein Fifth Dimension and have not been updated recently.

Second, sections 5-7 are still largely OK, with some minor changes envisioned.  Especially, I intend to derive the “restriction” $\Gamma^u_{55}=0$ from $F^{{\rm M} {\rm N} } =-F^{{\rm N} {\rm M} }$ rather than impose it as an ad hoc condition.

Third, sections 8-11 needs some reworking, and specifically: a) I want to start with an integration over the five-dimensional volume with a gravitational constant $G_{(5)}$ suited thereto, and relate this to the four dimensional integrals that are there at present; and b) I have serious misgivings about using a non-symmetric (torsion) energy tensor and am inclined to redevelop this to impose symmetry on the energy tensor — or at least to explore torsion versus no torsion in a way that might lead to an experimental test.  If we impose symmetry on the energy tensor, then the Maxwell tensor will be the $J^{\mu } =0$ special case of a broader tensor which includes a $J^\mu A^\nu + J^\nu A^\mu$ term and which applies, e.g., to energy flux densities (Poynting components) $T^{0k}$, k=1,2,3 for “waves” of large numbers of electrons.

Fourth, I am content with section 12, and expect it will survive the next draft largely intact.  Especially important is the covariant derivatives of the electrodynamic potentials being related to the ordinary derivatives of the gravitational potentials, which means that the way in which people often relate electrodynamic potentials to gravitational potentials in Kaluza-Klein theory is valid only in the linear approximation.  Importantly, this gives us a lever in the opposite direction, into non-linear electrodynamics.

Fifth, I expect the development of non-linear QED in section 13 to survive the next draft, but for the fact that the R=0 starting point will be removed as a consequence of my enforcing a symmetric energy tensor in sections 8-11.  Just take out all the “R=0” terms and leave the rest of the equation alone, and everything else is more or less intact.

Finally, the experiment in Section 15, if it stays, would be an experiment to test a symmetric, torsionless energy tensor against a non-symmetric energy tensor with torsion.  (Basically, metric theory versus Cartan theory.)  This is more of a “back of the envelope” section at present, but I do want to pursue specifying an experiment that will test the possible energy tensors which are available from variational principles via this Kaluza-Klein theory.

The paper at Intrinsic Spin and the Kaluza-Klein Fifth Dimension dealing specifically with the intrinsic spin hypothesis is also a work in progress, and at this time, I envision the following:

First, I will in a forthcoming draft explore positrons as well as electrons.  In compactified Kaluza-Klein, these exhibit opposite motions through $x^5$, and by developing the positron further, we can move from the Pauli spin matrices toward the Dirac $\gamma^\mu$ and Dirac’s equation.

Second, I have been engaged in some good discussion with my friend Daryl M. on a thread at sci.physics.relativity.  Though he believes I am “barking up the wrong tree,” he has provided a number of helpful comments, and especially at the bottom of post #2 where he discusses quantization in the fifth dimension using a wavelength $n \lambda = 2 \pi R$.  (I actually think that for fermions, one has to consider orientation / entanglement issues, and so to secure the correct “version,” one should use $n \lambda = 4 \pi R$ which introduces a factor of 2 which then can be turned into a half-integer spin.)  I am presently playing with some calculations based on this approach, which you will recognize as a throwback to the old Bohr models of the atom.

Third, this work of course uses $x^5 = R\phi$ to define the compact fifth dimension.  However, in obtaining $dx^5$, I have taken $R$ to be a fixed, constant radius.  In light of considering a wavelength $n \lambda = 4 \pi R$ per above, I believe it important to consider variations in $R$ rather than fixed $R$, and so, to employ $dx^5 = Rd\phi + \phi dR$.

There will likely be other changes along the way, but these are the ones which are most apparent to me at present.  I hope this gives you some perspective on where this “work in flux” is at, and where it may be headed.

Thanks for tuning in!

Jay.

## March 20, 2008

### Derivation of Heisenberg Uncertainty from Kaluza Klein Geometry

For those who have followed my Kaluza-Klein (KK) work, I believe that it is now possible to derive not only intrinsic spin, but Heisenberg uncertainty directly from a fifth, compactified dimension in Kaluza Klein.  This would put canonical quantum mechanics on a strictly Riemannian geometric foundation which — as a side benefit — unites gravitation and electromagnetism.

I need to consolidate over the next few days and will of course make a more expanded post when I am ready, but here is the basic outline.  First, take a look at:

Intrinsic Spin and the Kaluza-Klein Fifth Dimension

where I show how intrinsic spin is a consequence of the compactified fifth dimension.  This paper, at present, goes so far as to show how the Pauli spin matrices emerge from KK.

Next, go to the two page file:

Spin to Uncertainty

This shows how one can pop Heisenberg out of the spin matrices.

Finally, go to the latest draft paper on KK generally, at:

Kaluza-Klein Theory and Lorentz Force Geodesics with Non-linear QED

This lays out the full context in which I am developing this work.  Please note that the discussion on intrinsic spin in the third link is superseded by the discussion thereof in the first link.

More to follow . . .

Jay.

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