# Lab Notes for a Scientific Revolution (Physics)

## March 22, 2008

### A Possible Kaluza-Klein Experiment

It has been suggested — appropriately so — that I consider whether there might be one or more experiments which can be designed to validate or falsify some of the Klauza-Klein results which I have been posting of late. I believe that one possible experiment resides in the non-symmetric energy tensor of trace matter derived in (11.6) of my latest posted paper. Thus, I have added a new section 15 to this paper, and reposted the entire paper, with this new section 15, at Kaluza-Klein Theory and Lorentz Force Geodesics Rev. 6.0 Because this is of particular interest as it may open some new experimental windows, I have posted section 15 below as well. Please note: the specific discussion of the connection between the compactified fifth dimension, and intrinsic spin, is not updated in this paper, and the latest discusssion I have written up on this topic, is at Intrinsic Spin and the Kaluza-Klein Fifth Dimension.

Section 15: At this juncture, we have enough information to propose an experiment to validate or falsify some of the results derived thus far.  We turn for this purpose to the stress energy tensor of matter (11.6), which we raise into contravariant notation as follows:

$\kappa T^{\nu \mu } =-\kappa \left(F^{\mu \tau } F^{\nu } _{\tau } -{\textstyle\frac{1}{4}} g^{\mu \nu } F^{\sigma \tau } F_{\sigma \tau } \right)+{\textstyle\frac{\sqrt{2} }{2}} \overline{\kappa }g^{5\mu } J^{\nu } =\kappa T^{\mu \nu } _{Maxwell} +{\textstyle\frac{\sqrt{2} }{2}} \overline{\kappa }g^{5\mu } J^{\nu }$. (15.1)

The Maxwell tensor $T^{\mu \nu } _{Maxwell} =T^{\nu \mu } _{Maxwell}$ is, of course, a symmetric tensor.  But the added trace matter term $g^{5\mu } J^{\nu }$ is not necessarily symmetric, that is, there is no a priori reason why $g^{5\mu } J^{\nu }$ must be equal to $g^{5\nu } J^{\mu }$.  The origin of this non-symmetry was discussed earlier in Section 9.

With an eye toward conducting an experiment, let us now consider (15.1) in the linear approximation of (13.6) where ${\rm L}_{QCD} \approx -A^{\beta } J_{\beta } -{\textstyle\frac{1}{4}} F^{\sigma \tau } F_{\sigma \tau }$.  In the linear approximation, as used to reach (13.3), (12.11) reduces to $g^{5\mu } \approx \overline{\kappa }\left(\frac{\phi ^{5\mu } -{\textstyle\frac{1}{2}} bA^{\mu } }{1+{\textstyle\frac{1}{2}} \overline{\kappa }\phi } \right)\approx -{\textstyle\frac{1}{2}} \overline{\kappa }bA^{\mu }$, and (15.1) becomes:

$T^{\nu \mu } \approx -\left(F^{\mu \tau } F^{\nu } _{\tau } -{\textstyle\frac{1}{4}} g^{\mu \nu } F^{\sigma \tau } F_{\sigma \tau } \right)-2J^{\nu } A^{\mu } =T^{\mu \nu } _{Maxwell} -2J^{\nu } A^{\mu }$, (15.2)

where we have also used $b^{2} =8$ and $2\kappa =\overline{\kappa }^{2}$, and divided out $\kappa$.  The transpose of this non-symmetric energy tensor is:

$T^{\mu \nu } \approx -\left(F^{\mu \tau } F^{\nu } _{\tau } -{\textstyle\frac{1}{4}} g^{\mu \nu } F^{\sigma \tau } F_{\sigma \tau } \right)-2J^{\mu } A^{\nu } =T^{\mu \nu } _{Maxwell} -2J^{\mu } A^{\nu }$, (15.3)

Now, it is known that a non-symmetric energy tensor, physically, is indicative of a non-zero spin density.  In particular, using (15.2) and (15.3), the non-symmetry of the energy tensor is related to a non-zero spin density tensor $S^{\mu \nu \alpha }$ according to: [A good, basic discussion of the spin tensor is at http://en.wikipedia.org/wiki/Spin_tensor.]

$S^{\mu \nu \alpha } _{;\alpha } =T^{\mu \nu } -T^{\nu \mu } =-2J^{\mu } A^{\nu } +2J^{\nu } A^{\mu }$. (15.4)

For such a non-symmetric tensor, the “energy flux” is not identical to the “momentum density, as these differ by (15.4), for $\mu =0$, $\nu =k=1,2,3$ and vice versa.  If the spin density $S^{\mu \nu \alpha } =0$, then  in this special case, (15.4) yields:

$J^{\mu } A^{\nu } =J^{\nu } A^{\mu }$. (15.5)

So, for $S^{\mu \nu \alpha } =0$, (15.3) may be written using (15.5) as the explicitly-symmetric tensor:

$T^{\mu \nu } \approx -\left(F^{\mu \tau } F^{\nu } _{\tau } -{\textstyle\frac{1}{4}} g^{\mu \nu } F^{\sigma \tau } F_{\sigma \tau } \right)-J^{\mu } A^{\nu } -J^{\nu } A^{\mu } =T^{\mu \nu } _{Maxwell} -J^{\mu } A^{\nu } -J^{\nu } A^{\mu }$. (15.6)

Now, let’s consider a experiment which is entirely classical.  The $T^{0k}$ “Poynting” components of (15.4), (15.6) represent the energy flux across a two-dimensional area, for a flux of matter which we will take to be a stream of electrons, while the $T^{k0}$ components represent the momentum density.  The proposed experiment, then, will be to fire a stream of a very large number of electrons thereby constituting an electron “wave,” and to detect the aggregate flux of energy across a two-dimensional surface under various spin preparations, in precisely the same manner that one might test the flow of luminous energy across a surface when using light waves rather than electron waves.  Specifically, we propose in test I to fire electrons without doing anything to orient their spins, so that, statistically, the number of electrons flowing through the flux surface with positive helicity is equal to the number with negative helicity and so the spin density is zero, and (15.6) applies.  In test II, we fire electrons, but apply a magnetic field before detecting the flux, to ensure that all of the electrons are aligned to positive helicity.  In this event, the spin density, by design, is non-zero, and one of (15.2) or (15.3) will apply.  In test III, we do the same, but now apply the magnetic field to ensure that all of the electrons have negative helicity, before detecting the flux.

In the linear approximation, we take $g^{\mu \nu } \approx \eta ^{\mu \nu }$, and so the Maxwell tensor part of (15.6) is the usual:

$T^{0\kappa } _{Maxwell} =-F^{0\tau } F^{k} _{\tau } =E\times B$. (15.7)

Therefore, for test I, where $S^{\mu \nu \alpha } =0$, (15.6) applies and the Poynting vector is:

$T^{0k} _{I} \approx T^{0k} _{Maxwell} -J^{0} A^{k} -J^{k} A^{0} =E\times B-\rho \, A-\phi \, J$. (15.8)

where we employ the current density four-vector $J^{\mu } =\left(\rho ,J_{x} ,J_{y} ,J_{z} \right)=\left(\rho ,J\right)$ and the vector potential $A^{\mu } \equiv \left(\phi ,A_{x} ,A_{y} ,A_{z} \right)=\left(\phi ,A\right)$.  Via (15.5), $\rho \, A=\phi \, J$, for test I.  Referring to (15.8), we regard the term $\phi \, J$ with the electrostatic current density $J$ to contribute to the “energy flux” and the term $\rho \, A$ with the charge density $\rho$ to contribute to the “momentum density,” that is, we use the four vector current density $J^{\mu }$, rather than the vector potential $A^{\mu }$, to establish whether we are speaking of “flux” versus “density.”

In tests II and III, we note from (15.4), $S^{\mu \nu \alpha } _{;\alpha } =-2J^{\mu } A^{\nu } +2J^{\nu } A^{\mu }$, that for $S^{\mu \nu \alpha } >0$ (positive spin density), $\partial _{\alpha } S^{\mu \nu \alpha } >0$, and so $J^{\nu } A^{\mu } >J^{\mu } A^{\nu }$.  For $S^{\mu \nu \alpha } <0$ (negative spin density), $J^{\nu } A^{\mu } .  Thus, we identify (15.2) with a positive, and (15.3) with a negative helicity electron beam.  Thus, in test II, the “energy flux” should be observed to be:

$T^{k0} _{II} \approx T^{0k} _{Maxwell} -2J^{\nu } A^{0} =E\times B-2\phi \, J$. (15.9)

and in test III, the “momentum density” should be observed to be:

$T^{0k} _{III} \approx T^{0k} _{Maxwell} -2J^{0} A^{k} =E\times B-2\rho \, A$, (15.10)

If we orient the test so that the electrons are fired along the z axis, and detected to flow through the x-y plane, then:

$T^{03} _{I} \approx -F^{0\tau } F^{3} _{\tau } -J^{0} A^{3} -J^{3} A^{0} =E_{x} B_{y} -E_{y} B_{x} -\rho A_{z} -\phi J_{z}$, (15.11)

$T^{03} _{II} \approx -F^{0\tau } F^{3} _{\tau } -J^{0} A^{3} -J^{3} A^{0} =E_{x} B_{y} -E_{y} B_{x} -2\phi J_{z}$, and (15.12)

$T^{03} _{III} \approx -F^{0\tau } F^{3} _{\tau } -J^{0} A^{3} -J^{3} A^{0} =E_{x} B_{y} -E_{y} B_{x} -2\rho A_{z}$. (15.13)

One may, if desired, use $F^{\mu \nu } =A^{\nu ,\mu } -A^{\mu ,\nu }$ to reformulate the electromagnetic field terms into potential terms, via $E=-\nabla \phi -\partial A/\partial t$ and $B=\nabla \times A$.  With firing along the z-axis, this is $E_{z} =-\partial \phi /\partial z-\partial A_{z} /\partial t$ and $B_{z} =\partial A_{x} /\partial y-\partial A_{y} /\partial x$.

Validation (or falsification) of energy fluxes specified by (15.11) through (15.13) under the various spin density preparations I, II and III, would then serve as a test of the trace matter tensor (15.1), and the steps which were undertaken to derive this tensor in the first instance.