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%TCIDATA{Created=Thu Nov 21 07:20:12 1996}
%TCIDATA{LastRevised=Tue Dec 03 18:21:08 1996}

\input{tcilatex}
\begin{document}

\date{Report: URI 520P}
\title{Spacetime and Matter}
\author{R. A. McCorkle \\
%EndAName
Physics Department\\
University of Rhode Island\\
311C East Hall\\
Kingston, RI 02881}
\maketitle

\begin{abstract}
A spacetime theory of matter, based on general relativity, is shown to
provide, as a special case, standard electroweak theory.
\end{abstract}

\section{Introduction}

Although the necessity of employing quantum fields in order to satisfy the
cluster decomposition principle and Lorentz invariance has been made clear,
there is still no understanding as to if and why quantum fields are basic
and what, if anything, they have to do with elementarity \cite{wberg}.
Moreover, despite the overwhelming success of the standard model \cite
{smodel1}, it exists as a composite that is not fundamental: no insight is
given into the composition and range of matter. Numerous loose ends of the
theory are well known \cite{swberg}: why just quarks and leptons, why three
generations, why the direct correspondence between quark and lepton flavors,
why chiral assymmetry, why particle masses are so small and have a large
range of values, why SU(3)xSU(2)xSU(1), what is the relation of inertia to
the Higgs field, what is the nature of the Higgs sector, how many Higgs
fields, the large number of free parameters required, and how to include and
unify gravity with the other forces.

\section{Conceptual basis}

The theory of general relativity \cite{einstein} considers spacetime to be a
continuum and there is no experimental evidence to the contrary. Our
extensive experience with energy and momentum conservation lead to the
conclusion that spacetime is homogeneous and isotropic. For a theory
constructed out of spacetime, there is no loss of generality in restricting
consideration of the energy momentum tensor to the perfect fluid form since
it is the most general form $T_{\mu \nu }$ can then take \cite{wald}. Two
scalars, pressure $p$ and energy density $\varepsilon $, then characterize
this tensor: 
\begin{equation}
T_{\nu }^{\mu }=(\varepsilon +p)u^{\mu }u_{\nu }+p\delta _{\nu }^{\mu },
\label{1}
\end{equation}
where $u^{\mu }=dx^{\mu }/d\tau $ with $u^{\mu }u_{\mu }=-1$, proper time
being related to the fundamental arclength by $ds^{2}=-d\tau ^{2}$. Greek
indices range from 0 to 3, the metric tensor $g_{\mu \nu }$ having signature 
$(-+++)$ with $G_{\mu \nu }=8\pi GT_{\mu \nu }$ being the field equations,
and rationalized natural units, $\hbar =c=1$ and $e^{2}/4\pi \simeq 1/137$,
are used. Based on both the invariance of action and a direct derivation by
perturbational analysis, the requisite equation of state may be argued to be
(see Appendix A) 
\begin{equation}
p=-\varepsilon /3.  \label{2}
\end{equation}
Then the energy momentum tensor is given in terms of a single timelike
vector, 
\begin{equation}
T_{\nu }^{\mu }=f^{\mu }f_{\nu }-f^{2}\delta _{\nu }^{\mu }/2  \label{3}
\end{equation}
where $f^{2}=\varepsilon +p$ and $f^{\mu }=fu^{\mu }$.

The group structure of spacetime, due to the timelike congruence associated
with $u^{\mu }$, is described through the contracted Bianchi identities $%
T_{\nu ;\mu }^{\mu }=0$: 
\begin{equation}
f_{;\mu }^{\mu }f_{\nu }+f^{\mu }f_{\nu ;\mu }-ff_{,\nu }=0.  \label{4}
\end{equation}
Since $f_{;\nu }^{\mu }=f_{,\nu }u^{\mu }+fu_{;\nu }^{\mu }$, the
irreduciable decomposition of the gradient of the four-velocity, 
\begin{equation}
u_{\mu ;\nu }=\frac{\Theta }{3}h_{\mu \nu }-u_{\nu }a_{\mu }+\sigma _{\mu
\nu }+\varpi _{\mu \nu },  \label{5}
\end{equation}
into a volume expansion $\Theta =u_{;\sigma }^{\sigma }$, an acceleration $%
a_{\nu }=\dot{u}_{\nu }=u_{\nu ;\sigma }u^{\sigma }$, a symmetric trace-free
shear, $\sigma _{\mu \nu }=u_{(\mu ;\nu )}+\dot{u}_{(\mu }u_{\nu )}-\Theta
h_{\mu \nu }/3$, and a skew-symmetric vorticity, $\varpi _{\mu \nu }=u_{[\mu
;\nu ]}+\dot{u}_{[\mu }u_{\nu ]}$, (with the orthogonality properties $%
0=h_{\mu \nu }u^{\nu }=a_{\nu }u^{\nu }=\sigma _{\mu \nu }u^{\nu }=\varpi
_{\mu \nu }u^{\nu }$ where $h_{\mu \nu }=g_{\mu \nu }+u_{\mu }u_{\nu }$ is
the projection tensor onto the spacelike hypersurface orthogonal to $u^{\nu
} $) allows a group-theoretic irreducible expression of the fluid dynamics: 
\begin{equation}
f_{\mu ;\nu }=\xi k_{\mu \nu }+f_{\mu }a_{\nu }-f_{\nu }a_{\mu }+\zeta _{\mu
\nu }  \label{6}
\end{equation}
for which $\xi =f_{;\sigma }^{\sigma }/2=-f^{\nu }f_{,\nu }/f=f\Theta /3$, $%
k_{\mu \nu }=g_{\mu \nu }+2f_{\mu }f_{\nu }/f^{2}$, and $\zeta _{\mu \nu
}=f(\sigma _{\mu \nu }+\varpi _{\mu \nu })$ with $\zeta _{(\mu \nu
)}=f\sigma _{\mu \nu }$ and $\zeta _{[\mu \nu ]}=f\varpi _{\mu \nu }$. Round
brackets indicate symmetrized indices; square brackets, skew-symmetrized
indices. The acceleration may be expressed as 
\begin{equation}
a_{\nu }=f_{,\nu }/f-\xi f_{\nu }/f^{2}.  \label{7}
\end{equation}

The excitation modes $\xi k_{\mu \nu }$ (expansion), $f_{\mu }a_{\nu
}-f_{\nu }a_{\mu }$ (Fermi-Walker), and $\zeta _{\mu \nu }$ (shear and
rotation) are mutually orthogonal: $k^{\mu \nu }(f_{\mu }a_{\nu }-f_{\nu
}a_{\mu })=k^{\mu \nu }\zeta _{\mu \nu }=\zeta ^{\mu \nu }(f_{\mu }a_{\nu
}-f_{\nu }a_{\mu })$. Also, $f^{\nu }a_{\nu }=f^{\nu }\zeta _{\mu \nu }=$ $%
f^{\nu }\zeta _{\nu \mu }=\zeta _{.\nu }^{\nu }=0$. Thus, the expansion,
Fermi-Walker and shear/rotation excitation modes involving $\xi $ and the
linearly independent components of $a_{\nu }$ and $\zeta _{\mu \nu }$ are
one, three, and eight parameter functions respectively. The shear/rotation
excitation is composed of the antisymmetric rotation motions $\zeta _{[\mu
\nu ]}=f\varpi _{\mu \nu }$ consisting of three independent components and
the symmetric shear $\zeta _{(\mu \nu )}=f\sigma _{\mu \nu }$ with five
independent components. Solution of the first order equations for each mode
results in an integration parameter $(y^{\nu })$ which will be referred to
as the excitation core.

Excitations of combined expansion and Fermi-Walker modes (hereafter referred
to as EW excitations) are of interest in that they will be seen to provide
leptons and their electroweak interactions. Development of the dynamical
equations for this combination, based on the orthonormal tetrad $(\mathbf{u},%
\mathbf{s},\mathbf{q},\mathbf{r})$ incorporating the velocity $\mathbf{u}$,
the unit vector associated with the acceleration, $\mathbf{s}=\mathbf{a}/a$,
and two symmetrically disposed spacelike vectors $\mathbf{q}$ and $\mathbf{r}
$ is given in Appendix B.

\section{A Different Representation}

A representation change that is Lorentz invariant extended by parity \cite
{ryder} in the local Lorentz frame is now made. In the local Lorentzian
(indices for which are taken from the early part of the Greek alphabet and
enclosed in parenthesis) with metric $\eta _{(\alpha )(\beta
)}=diag(-1,1,1,1)$, the Dirac matrices are utilized in Weyl representation
and are designated by $\gamma _{\alpha }$: 
\[
\gamma _{\alpha }=\left( 
\begin{array}{cc}
0 & \mathbf{\sigma } \\ 
\mathbf{\bar{\sigma}} & 0
\end{array}
\right) 
\]
with $\mathbf{\sigma =}(\sigma _{\alpha })=(-I,\left\{ \sigma _{i}\right\} )$
and $\mathbf{\bar{\sigma}=(}\bar{\sigma}_{\alpha })=(-I,-\left\{ \sigma
_{i}\right\} )$ where Latin indices from the middle of the alphabet range
from 1 to 3 and $\left\{ \sigma _{i}\right\} $ are the Pauli matrices. With $%
\chi _{a}=\left( 
\begin{array}{c}
\chi _{a1} \\ 
\chi _{a2}
\end{array}
\right) $ and $\phi _{a}=\left( 
\begin{array}{c}
\phi _{a1} \\ 
\phi _{a2}
\end{array}
\right) $ being two-component matrices and $\psi _{a}=\left( 
\begin{array}{c}
\chi _{a} \\ 
\phi _{a}
\end{array}
\right) $ with $a$ being either 1 or 2, it follows that $\gamma
^{5}=\varepsilon ^{\alpha \beta \delta \lambda }\gamma _{\alpha }\gamma
_{\beta }\gamma _{\delta }\gamma _{\lambda }/4!=\left( 
\begin{array}{cc}
-I & 0 \\ 
0 & I
\end{array}
\right) $, $\psi _{aL}=P_{L}\psi _{a}=\left( 
\begin{array}{c}
\chi _{a} \\ 
0
\end{array}
\right) $ and $\psi _{aR}=P_{R}\psi _{a}=\left( 
\begin{array}{c}
0 \\ 
\phi _{a}
\end{array}
\right) $with $P_{L}=(1-\gamma ^{5})/2$, $P_{R}=(1+\gamma ^{5})/2$. $\bar{%
\psi}_{a}=\psi _{a}^{\dagger }\gamma ^{0}$ where $\psi _{a}^{\dagger }$ is
the conjugate transpose of $\psi _{a}$. With $e_{(\alpha )}^{\mu }(\mathbf{x}%
)\equiv e_{(\alpha )}^{\mu }$ serving as tetrad (i.e., $g^{\mu \nu }(\mathbf{%
x})=e_{(\alpha )}^{\mu }e_{(\beta )}^{\nu }\eta ^{(\alpha )(\beta )}$), then
the $\gamma $-matrices in general coordinates become $\gamma ^{\nu }(\mathbf{%
x})=e_{(\alpha )}^{\nu }$ $\gamma ^{\alpha }=\gamma ^{\nu }$ and $\gamma
\cdot u=\gamma ^{\nu }u_{\nu }=\gamma ^{\alpha }u_{(\alpha )}$, for example.
In particular, the attendant properties $\gamma _{\mu }\gamma ^{\nu }+\gamma
^{\nu }\gamma _{\mu }=-2I\delta _{\mu }^{\nu }$ and $\gamma ^{5}\gamma ^{\nu
}=-\gamma ^{\nu }\gamma ^{5}$, $\gamma ^{5}\gamma ^{5}=I$ follow.

Transformation to an irreduciable representation satisfying the above stated
requirements is then obtained by taking 
\begin{equation}
(\gamma \cdot u-\gamma ^{5}\gamma \cdot s)/4=\psi _{1L}\bar{\psi}_{1L}+\psi
_{1R}\bar{\psi}_{1R}  \label{8}
\end{equation}
and 
\begin{equation}
-(i\gamma \cdot q+\gamma ^{5}\gamma \cdot r)/4=\psi _{2L}\bar{\psi}%
_{2L+}\psi _{2R}\bar{\psi}_{2R}.  \label{9}
\end{equation}
Right multiplication of both equations by $\gamma ^{\nu },$ taking the trace
and utilizing the trace properties $Tr(\gamma ^{\mu }\gamma ^{\nu
})=-4g^{\mu \nu }$ and $Tr(\gamma ^{5}\gamma ^{\mu }\gamma ^{\nu })=0$,
gives 
\begin{equation}
u^{\nu }=-\bar{\psi}_{1}\gamma ^{\nu }\psi _{1}  \label{10}
\end{equation}
and 
\begin{equation}
q^{\nu }=-i\bar{\psi}_{2}\gamma ^{\nu }\psi _{2},  \label{11}
\end{equation}
respectively. Similarly, right multiplication of each by $\gamma ^{\nu
}\gamma ^{5}$ and taking the trace gives 
\begin{equation}
s^{\nu }=\bar{\psi}_{1}\gamma ^{\nu }\gamma ^{5}\psi _{1}  \label{12}
\end{equation}
and 
\begin{equation}
r^{\nu }=\bar{\psi}_{2}\gamma ^{\nu }\gamma ^{5}\psi _{2}.  \label{13}
\end{equation}

Then, $(\gamma \cdot u+\gamma ^{5}\gamma \cdot s)(\gamma \cdot u-\gamma
^{5}\gamma \cdot s)=0$ (since $\gamma \cdot u\gamma \cdot u=-\gamma \cdot
s\gamma \cdot s=I$ and $\gamma \cdot u\gamma \cdot s+\gamma \cdot s\gamma
\cdot u=0$) gives 
\[
(\gamma \cdot u+\gamma ^{5}\gamma \cdot s)(\psi _{1L}\bar{\psi}_{1L}+\psi
_{1R}\bar{\psi}_{1R})=0. 
\]
Right multiplication of this latter expression by $\psi _{1L}$ and noting
that $\bar{\psi}_{1L}\psi _{1L}=0$ whereas $\bar{\psi}_{1L}\psi _{1R}\neq 0$
gives $(\gamma \cdot u+\gamma ^{5}\gamma \cdot s)\psi _{1R}=0$, and
similarly, $(\gamma \cdot u+\gamma ^{5}\gamma \cdot s)\psi _{1L}=0;$ that is 
\begin{equation}
\gamma \cdot u\psi _{1}=-\gamma ^{5}\gamma \cdot s\psi _{1}.  \label{14}
\end{equation}
Likewise, $\gamma \cdot q\psi _{2}=-i\gamma ^{5}\gamma \cdot r\psi _{2}$.
Furthermore, since $P_{L}\gamma ^{5}=-P_{L}$ , $P_{R}\gamma ^{5}=P_{R}$ and $%
\gamma ^{5}\psi _{iL}=-\psi _{iL}$, $\gamma ^{5}\psi _{iR}=\psi _{iR}$, it
follows that $\gamma \cdot (u-s)\psi _{1L}/4=\gamma \cdot u\psi _{1L}/2=\psi
_{1R}\bar{\psi}_{1R}\psi _{1L}$ and $\gamma \cdot (u+s)\psi _{1R}/4=\gamma
\cdot u\psi _{1R}/2=\psi _{1L}\bar{\psi}_{1L}\psi _{1R}$. Thus $\gamma \cdot
u\gamma \cdot u\psi _{1L}/2=\psi _{1L}/2=\gamma \cdot u\psi _{1R}\bar{\psi}%
_{1R}\psi _{1L}=2\psi _{1L}\bar{\psi}_{1L}\psi _{1R}\bar{\psi}_{1R}\psi
_{1L}=2\psi _{1L}\left| \bar{\psi}_{1L}\psi _{1R}\right| ^{2}$ since $\bar{%
\psi}_{1L}\psi _{1R}=(\bar{\psi}_{1R}\psi _{1L})^{*}$, giving $\left| \bar{%
\psi}_{1L}\psi _{1R}\right| ^{2}=1/4.$ With $\bar{\psi}_{1L}\psi _{1R}=1/2$,
it follows that $\gamma \cdot u\psi _{1L}=\psi _{1R}$ and $\gamma \cdot
u\psi _{1R}=\psi _{1L}$, that is 
\begin{equation}
\gamma \cdot u\psi _{1}=\psi _{1}.  \label{15}
\end{equation}
Again, in a similar manner, with $\bar{\psi}_{2L}\psi _{2R}=1/2$, it follows
that 
\begin{equation}
\gamma \cdot q\psi _{2}=i\psi _{2}=-i\gamma ^{5}\gamma \cdot r\psi _{2}.
\label{16}
\end{equation}

Now, directly 
\[
P_{R}\gamma \cdot (q-ir)\gamma \cdot (u+s)/16=i\psi _{2R}\bar{\psi}_{2R}\psi
_{1L}\bar{\psi}_{1L}. 
\]
Taking the trace of this equation and using the trace properties of the $%
\gamma $-matrices and the orthogonality of the vectors $(u,s,q,r)$ gives $%
\left| \bar{\psi}_{2R}\psi _{1L}\right| ^{2}=0$. Also, 
\[
P_{L}\gamma \cdot (q+ir)\gamma \cdot (u-s)/16=i\psi _{2L}\bar{\psi}_{2L}\psi
_{1R}\bar{\psi}_{1R}, 
\]
and taking the trace then gives $\left| \bar{\psi}_{2L}\psi _{1R}\right|
^{2}=0$.

Then, $\gamma \cdot (u+s)\psi _{2R}/4=\psi _{1L}\bar{\psi}_{1L}\psi _{2R}=0$%
, that is, 
\begin{equation}
\gamma \cdot u\psi _{2R}=-\gamma \cdot s\psi _{2R}.  \label{17}
\end{equation}
Also, $\gamma \cdot (u-s)\psi _{2L}/4=\psi _{1R}\bar{\psi}_{1R}\psi _{2L}=0$
giving 
\begin{equation}
\gamma \cdot u\psi _{2L}=\gamma \cdot s\psi _{2L}.  \label{18}
\end{equation}
However, $\gamma \cdot q\gamma \cdot u=-\gamma \cdot u\gamma \cdot q$ and $%
\gamma \cdot q\gamma \cdot s=-\gamma \cdot s\gamma \cdot q$ implies that $%
\gamma \cdot q\gamma \cdot u\psi _{2R}=-\gamma \cdot u\gamma \cdot q\psi
_{2R}=-\gamma \cdot q\gamma \cdot s\psi _{2R}=\gamma \cdot s\gamma \cdot
q\psi _{2R}$. But $\gamma \cdot q\psi _{2R}=i\psi _{2L}$ then gives 
\begin{equation}
-\gamma \cdot u\psi _{2L}=\gamma \cdot s\psi _{2L}.  \label{19}
\end{equation}
It then follows that $\gamma \cdot s\psi _{2L}=\gamma \cdot u\psi _{2L}=0$
and similarly, $\gamma \cdot s\psi _{2R}=\gamma \cdot u\psi _{2R}=0,$that is 
\begin{equation}
\gamma \cdot u\psi _{2}=\gamma \cdot s\psi _{2}=0.  \label{20}
\end{equation}
In a like manner, 
\begin{equation}
\gamma \cdot q\psi _{1}=\gamma \cdot r\psi _{1}=0.  \label{21}
\end{equation}

\section{Fundamentals of Quantum Field Theory and the Standard Model}

In both classical particle theory as well as in relativistic quantum field
theory the momentum of a particle is so defined that it is not conserved. In
classical theory, this effect shows up, for example, as runaway solutions
when radiation effects are included \cite{dirac}. To address this
difficulty, a new complex was introduced by Teitelboim \cite{teitelboim} in
order to separate the particle-field system into two dynamically independent
subsystems: the particle plus attached field (the complex), and the
radiation field. Momentum conservation for the complex is observed. This
latter entity corresponds to defining particle momentum using both the
energy momentum tensor and pseudotensor so as to properly treat radiative
field effects \cite{mtw}. In quantum field theory, radiative corrections are
necessary \cite{itzykson} and fundamental to the formulation is the fact
that it is impossible to construct the theory with a fixed number of
particles \cite{wberg}. Both classical and quantum field theory do not
define momentum inclusive of the pseudotensor. \cite{footnote}

In accord with standard theory, the momentum associated with the i$^{th}$
excitation is obtained by integration of the i$^{th}$ contribution to the
energy momentum over a spacelike hypersurface, 
\begin{equation}
p_{i}^{\nu }=\int T_{i}^{\nu \sigma }d\Sigma _{\sigma }=\int
f_{i}^{2}(u^{\nu }u^{\sigma }-\frac{1}{2}g^{\nu \sigma })(-u_{\sigma }dV)=%
\frac{3}{2}\int f_{i}^{2}u^{\nu }dV=\int d\Omega (\eta _{i}^{\nu })
\label{22}
\end{equation}
where in a coordinate basis $d\Omega =dx^{\sigma }d\Sigma _{\sigma
}=-u_{\sigma }dx^{\sigma }dV$ is the measure of spacetime and $\eta
_{i}^{\nu }=\frac{3}{2}f_{i}^{2}u^{\nu }u^{\sigma }\partial _{\sigma }$ ,
the spacelike surface being $d\Sigma _{\nu }=-u_{\nu }dV$ where $%
dV=u^{\sigma }d\Sigma _{\sigma }$.

Now, by Stoke's theorem \cite{schutz}, $\int_{\partial U}d\Omega (\eta
_{i}^{\nu })\mid _{\partial U}=\int_{U}\mathcal{L}_{\eta _{i}^{\nu }}d\Omega 
$ where $\mathcal{L}_{\eta ^{\nu }}$ indicates the Lie derivative and $%
\partial U$ is the boundry of $U$. The integration over spacetime is to be
taken over all time history of the i$^{th}$ excitation up until the time of
interest and over all spacelike surface influenced by the excitation \cite
{goldstein}. But 
\[
(\mathcal{L}_{\eta _{i}^{\nu }}d\Omega )_{\mu }=(-u_{\mu }dV)_{;\sigma }%
\frac{3}{2}f_{i}^{2}u^{\nu }u^{\sigma }+(-u_{\sigma }dV)(\frac{3}{2}%
f_{i}^{2}u^{\nu }u^{\sigma })_{;\mu }. 
\]
Then, using $dV_{;\sigma }u^{\sigma }=d\dot{V}=\Theta dV$ and previous
expressions for $(f_{i}^{2})_{,\nu }$ and u$_{;\mu }^{\nu }$ (from equs. (%
\ref{5}) and (\ref{7})) gives 
\begin{equation}
(\mathcal{L}_{\eta _{i}^{\nu }}d\Omega )_{\mu }=\frac{1}{2}f_{i}^{2}\left\{
3a(u^{\nu }s_{\mu }-s^{\nu }u_{\mu })+\Theta \delta _{\mu }^{\nu }+3(\sigma
_{.\mu }^{\nu }+\omega _{.\mu }^{\nu })\right\} dV.  \label{23}
\end{equation}
Thereby, 
\begin{equation}
p_{i}^{\nu }=\frac{1}{2}\int f_{i}^{2}\left\{ 3a(u^{\nu }s_{\sigma }-s^{\nu
}u_{\sigma })+\Theta \delta _{\sigma }^{\nu }+3(\sigma _{.\sigma }^{\nu
}+\omega _{.\sigma }^{\nu })\right\} dx^{\sigma }dV.  \label{24}
\end{equation}

Further consideration is given to type EW excitations: $\sigma _{.\mu }^{\nu
}=\omega _{.\mu }^{\nu }=0$. (It is always possible to take $\omega _{\mu
\nu }=0$ or $\sigma _{\mu \nu }=0$ for stationary, axially symmetric vacuum
solutions \cite{chinea}.) Writing the Kronecker function as $\delta _{\mu
}^{\nu }=-u^{\nu }u_{\mu }+s^{\nu }s_{\mu }+q^{\nu }q_{\mu }+r^{\nu }r_{\mu
}=-u^{\nu }u_{\mu }+h_{\mu }^{\nu }$ and using $dx^{\nu }dV=u^{\nu }d\Omega
+h_{\sigma }^{\nu }dx^{\sigma }dV$ gives 
\begin{equation}
p_{i}^{\nu }=\int \frac{f_{i}^{2}}{2}(3as^{\nu }+\Theta u^{\nu })d\Omega
+\int \frac{f_{i}^{2}}{2}\left\{ \Theta (q^{\nu }\tilde{q}+r^{\nu }\tilde{r}%
+s^{\nu }\tilde{s})+3au^{\nu }\tilde{s}\right\} dV,  \label{25}
\end{equation}
having employed the one-forms $\tilde{u}=u_{\sigma }dx^{\sigma },\tilde{s}%
=s_{\sigma }dx^{\sigma },\tilde{q}=q_{\sigma }dx^{\sigma },$ and $\tilde{r}%
=r_{\sigma }dx^{\sigma }$. Then, using $dV=-d\Omega /\tilde{u}$ allows
writing 
\begin{equation}
p_{i}^{\nu }=\int \frac{f_{i}^{2}}{2\tilde{u}}\left\{ (\Theta \tilde{u}-3a%
\tilde{s})u^{\nu }+(3a\tilde{u}-\Theta \tilde{s})s^{\nu }-\Theta (\tilde{q}%
q^{\nu }+\tilde{r}r^{\nu })\right\} d\Omega .  \label{26}
\end{equation}

Now, define an external space for excitation i by (see Appendix C for the
approximate model utilized here) 
\begin{equation}
\left| \int_{\infty }^{\left\{ x^{\nu }\right\} }\tilde{a}\right| =\left|
\int_{\infty }^{\left\{ x^{\nu }\right\} }\frac{1}{2}d\left[ \ln \left(
1+2m_{i}\mu _{i}\right) \right] \right| \leq 1,  \label{27}
\end{equation}
since it fails to be accurate to extend beyond this limit toward an
excitation core the asymptotically flat spacetime existing at large
distances from the core \cite{mtw2}. Here, the limit $\infty $ designates
the asymptotically flat region far from the $i^{th}$-excitation core.
Thereby, external space for excitation i is given by 
\begin{equation}
\frac{1}{2}\left[ \ln \left( 1+2m_{i}\mu _{i}\right) \right] _{\infty
}^{\left\{ x^{\nu }\right\} }=\frac{1}{2}\ln \left( 1+2m_{i}\mu _{i}(x^{\nu
})\right) \leq 1,  \label{28}
\end{equation}
which is satisfied for any distance from core position $\mathsf{r}_{i}\geq
m_{i}/3=m_{i0}\xi _{0}^{2}/6\pi \simeq 4\times 10^{-34}cm$. This is the
minimum radius that may be taken for a point particle when representing
spacetime behavior as the interaction of point particles with interactions
in standard form, as developed below. Then, writing the momentum integral as
two integrals, 
\[
\mathbf{p}_{i}=\frac{3}{2}\int f_{i}^{2}\mathbf{u}dV= 
\]
\begin{equation}
\frac{3}{2}\int_{in}f_{i}^{2}\mathbf{u}dV+\int_{ex}\frac{f_{i}^{2}}{2\tilde{u%
}}\left\{ (\Theta \tilde{u}-3a\tilde{s})\mathbf{u}+(3a\tilde{u}-\Theta 
\tilde{s})\mathbf{s}-\Theta (\tilde{q}\mathbf{q}+\tilde{r}\mathbf{r}%
)\right\} d\Omega ,  \label{29}
\end{equation}
where the designation of an integral over external space by ex is for values
of $\left\{ x^{\nu }\right\} $ compatible with the above restriction and
internal space (borrowing from standard terminology) is over the remaining
spacetime. In practice, radial distances generally taken to comprise
external space are determined by collision energies in scattering events and
are strongly in compliance with the above established crtieria. It is seen
that the asymptotically Lorentzian rest frame far from the core of an
excitation, with metric $g_{L\alpha \beta }$ (indices for which are taken
from the early part of the Greek alphabet without parentheses) may be
utilized throughout external space for all energy ranges of practical
interest. This frame will subsequently be referred to as the extended
Lorentzian.

Notice that $\mathbf{p}_{i}$ is parametrically dependent on the relative
position of all other excitations with respect to the $i^{th}$ excitation,
but is not a function of $\mathbf{x}$ since integration is over $\mathbf{x}$
. Defining 
\begin{equation}
\mathbf{P}_{i}=\frac{3}{2}\int_{in}f_{i}^{2}\mathbf{u}dV  \label{30}
\end{equation}
and recognizing that $\mathbf{p}_{i}$ resides \cite{mtw3} in the extended
Lorentzian region of spacetime gives 
\begin{equation}
p_{i}^{\alpha }=P_{i}^{\alpha }+\int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{2%
\tilde{u}}\left\{ (\Theta \tilde{u}-3a\tilde{s})u^{\alpha }+(3a\tilde{u}%
-\Theta \tilde{s})s^{\alpha }-\Theta (\tilde{q}q^{\alpha }+\tilde{r}%
r^{\alpha })\right\} d\Omega (\mathbf{x}),  \label{31}
\end{equation}
with the understanding that Euclidean spatial coordinates are now employed.
(One way of evaluating $P_{i}^{\alpha }$ is to project $\mathbf{u}$ onto the
extended Lorentzian basis, $u^{\alpha }=e_{\mu }^{\alpha }(\mathbf{x})u^{\mu
}$, and then integrate; $P^{\alpha }{}_{i}=\frac{3}{2}\int_{in}f_{i}^{2}e_{%
\mu }^{\alpha }(\mathbf{x})u^{\mu }dV.$) Multiplying this expression by $%
\psi _{a}(\mathbf{x})\exp (i\mathbf{p}_{i}\cdot \left[ \mathbf{x}-\mathbf{x}%
_{i}\right] \mathbf{)}d^{4}p_{i}/(2\pi )^{4}$ and integrating over all
momentum space for which $d^{4}p_{i}$ is a measure followed by integration
over all $\mathbf{x}$ gives 
\[
i\partial _{i}^{\alpha }\psi _{a}(\mathbf{x}_{i})=P_{i}^{\alpha }\psi _{a}(%
\mathbf{x}_{i})+ 
\]
\begin{equation}
\psi _{a}(\mathbf{x}_{i})\int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{2\tilde{u}}%
\left\{ (\Theta \tilde{u}-3a\tilde{s})u^{\alpha }+(3a\tilde{u}-\Theta \tilde{%
s})s^{\alpha }-\Theta (\tilde{q}q^{\alpha }+\tilde{r}r^{\alpha })\right\}
d\Omega (\mathbf{x}),  \label{32}
\end{equation}
where $\partial _{i}^{\alpha }=\frac{\partial }{\partial x_{i\alpha }}$.
(Due to the orthogonality $(\Phi _{\mathbf{p}},\Phi _{\mathbf{p}^{\prime
}})=\int \Phi _{\mathbf{p}}^{\dagger }\Phi _{\mathbf{p}}d^{4}x/(2\pi
)^{4}=\delta ^{4}(\mathbf{p}-\mathbf{p}^{\prime })$ of plane waves $\Phi _{%
\mathbf{p}}(\mathbf{x)=}\exp (i\mathbf{p}\cdot \mathbf{x)}$, and the
convolution theorem, the spectral power density in momentum space for $\psi
_{a}(\mathbf{x}_{i})$ becomes $\left| (\psi _{a},\Phi _{\mathbf{p}})\right|
^{2}$ yielding the interpretation of this latter quantity as the probability
that the state described by $\psi _{a}$ is found in state $\Phi _{\mathbf{p}%
} $ with momentum $\mathbf{p}$.\cite{wberg2})

For $a=1$, the delta-function property of $f_{i}^{2}$ (see Appendix C) is
used to place $\psi _{1}(\mathbf{x}_{i})$ inside the integral: $%
f_{i}^{2}\psi _{1}(\mathbf{x}_{i})=f_{i}^{2}\psi _{1}(\mathbf{x})$.
Contracting on the left with the Dirac matrices, $\gamma _{d\alpha }$, where
the subscript $d$ has been inserted for clarity, then gives 
\[
i\gamma _{d}\cdot \partial _{i}\psi _{1}(\mathbf{x}_{i})=\gamma _{d}\cdot
P_{i}\psi _{1}(\mathbf{x}_{i})+ 
\]
\begin{equation}
i\gamma \int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{2\tilde{u}}\left\{ (\Theta 
\tilde{u}-3a\tilde{s})\gamma \cdot u+(3a\tilde{u}-\Theta \tilde{s})\gamma
\cdot s\right\} \psi _{1}(\mathbf{x})d\Omega (\mathbf{x}),  \label{33}
\end{equation}
where it has been recognized that $\gamma ^{\alpha }(\mathbf{x})=\gamma
_{d}^{\alpha }$ throughout external space and that $\gamma \cdot q\psi
_{1}=\gamma \cdot r\psi _{1}=0$. Then, using $\gamma \cdot s\psi
_{1}=-\gamma ^{5}\gamma \cdot u\psi _{1}$ allows writing 
\[
\gamma \cdot u\psi _{1}=(3\gamma \cdot u-\gamma \cdot s)(3+\gamma ^{5})\psi
_{1}/8 
\]
and 
\[
\gamma \cdot s\psi _{1}=(3\gamma \cdot s-\gamma \cdot u)(3+\gamma ^{5})\psi
_{1}/8. 
\]
Using these results to replace $\gamma \cdot s\psi _{1}$ and $\gamma \cdot
u\psi _{1}$in the integral above gives 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{1L}(\mathbf{x}_{i})=\gamma _{d}\cdot
P_{i}\psi _{1L}(\mathbf{x}_{i})+\frac{g\gamma _{d}^{\alpha }Z_{\alpha }}{%
2\cos \theta _{w}}\psi _{1L}(\mathbf{x}_{i})  \label{34}
\end{equation}
and 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{1R}(\mathbf{x}_{i})=\gamma _{d}\cdot
P_{i}\psi _{1R}(\mathbf{x}_{i})+\frac{g\gamma _{d}^{\alpha }Z_{\alpha }}{%
\cos \theta _{w}}\psi _{1R}(\mathbf{x}_{i})  \label{35}
\end{equation}
where the definition 
\begin{equation}
\frac{gZ_{\alpha }}{\cos \theta _{w}}=\int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{4%
\tilde{u}}\left\{ (\Theta \tilde{u}-3a\tilde{s})(3u_{\alpha }-s_{\alpha
})+(3a\tilde{u}-\Theta \tilde{s})(3s_{\alpha }-u_{\alpha })\right\} d\Omega (%
\mathbf{x})  \label{36}
\end{equation}
has been employed and the delta function properties of $f_{i}^{2}$ have been
used to remove $\psi _{1L}$ and $\psi _{1R}$ from the integral.

In a similar manner, for $a=2$, utilizing the delta function properties of $%
f_{i}^{2}$, and contracting on the left with $\gamma _{d\alpha }$ gives 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{2}(\mathbf{x}_{i})=\gamma _{d}\cdot
P_{i}\psi _{2}(\mathbf{x}_{i})+\int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{2\tilde{%
u}}\left\{ -\Theta (\tilde{q}\gamma \cdot q+\tilde{r}\gamma \cdot r)\right\}
\psi _{2}(\mathbf{x})d\Omega (\mathbf{x}),  \label{37}
\end{equation}
where it has been recognized that $\gamma \cdot u\psi _{2}=\gamma \cdot
s\psi _{2}=0$. Using $\gamma \cdot r\psi _{2}=i\gamma ^{5}\gamma \cdot q\psi
_{2}$ then gives 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{2}(\mathbf{x}_{i})=\gamma _{d}\cdot
P_{i}\psi _{2}(\mathbf{x}_{i})-\int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{2\tilde{%
u}}\Theta (\tilde{q}+i\tilde{r}\gamma ^{5})\gamma \cdot q\psi _{2}(\mathbf{x}%
)d\Omega (\mathbf{x}),  \label{38}
\end{equation}
that is, 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{2L}(\mathbf{x}_{i})=\gamma _{d}\cdot
P_{i}\psi _{2L}(\mathbf{x}_{i})-\left\{ \gamma _{d}^{\alpha }\int_{ex}\frac{%
f_{i}^{2}(\mathbf{x})}{2\tilde{u}}\Theta (\tilde{q}+i\tilde{r})q_{\alpha
}d\Omega (\mathbf{x})\right\} \psi _{2L}(\mathbf{x}_{i})  \label{39}
\end{equation}
and 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{2R}(\mathbf{x}_{i})=\gamma _{d}\cdot
P_{i}\psi _{2R}(\mathbf{x}_{i})-\left\{ \gamma _{d}^{\alpha }\int_{ex}\frac{%
f_{i}^{2}(\mathbf{x})}{2\tilde{u}}\Theta (\tilde{q}-i\tilde{r})q_{\alpha
}d\Omega (\mathbf{x})\right\} \psi _{2R}(\mathbf{x}_{i}),  \label{40}
\end{equation}
the delta function properties of $f_{i}^{2}$ having been used to remove $%
\psi _{2L}$ and $\psi _{2R}$ from the integral.

Now, 
\begin{equation}
\gamma _{d}\cdot P_{i}\psi (\mathbf{x}_{i})=\frac{3}{2}\int_{in}f_{i}^{2}%
\gamma _{d}\cdot u\psi (\mathbf{x})dV=\frac{3}{2}\int_{in}f_{i}^{2}(\gamma
\cdot u-\breve{\gamma}\cdot u)\psi (\mathbf{x})dV  \label{41}
\end{equation}
where $\breve{\gamma}\cdot u=\gamma \cdot u-\gamma _{d}^{\alpha }u_{\alpha
}= $ $\gamma \cdot u-\gamma ^{\mu }e_{\mu }^{(\alpha )}e_{\alpha }^{\nu
}u_{\nu }$, having used $\gamma _{d}^{\alpha }=\gamma ^{\mu }e_{\mu
}^{(\alpha )}$ and $u_{\alpha }=$ $e_{\alpha }^{\nu }u_{\nu }$. At the
boundry between internal and external space, $e_{\mu }^{(\alpha )}(\mathbf{x}%
_{b})e_{\alpha }^{\nu }(\mathbf{x}_{b})=\delta _{\mu }^{\nu }$, and in
general, to first order, $e_{\mu }^{(\alpha )}(\mathbf{x})e_{\alpha }^{\nu }(%
\mathbf{x})\cong \delta _{\mu }^{\nu }(1+\zeta (\mathbf{x}))$. Then $\breve{%
\gamma}\cdot u\cong -(\gamma \cdot u)\zeta (\mathbf{x})$ and 
\begin{equation}
\gamma _{d}\cdot P_{i}\psi (\mathbf{x}_{i})\cong \frac{3}{2}%
\int_{in}f_{i}^{2}\gamma \cdot u\psi (\mathbf{x})dV+\frac{3}{2}%
\int_{in}f_{i}^{2}(\gamma \cdot u)\zeta (\mathbf{x})\psi (\mathbf{x})dV,
\label{42}
\end{equation}
neglecting higher order terms. Then, with the notation 
\begin{equation}
gW_{\alpha }^{1}=\int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{\tilde{u}}\Theta 
\tilde{q}q_{\alpha }d\Omega (\mathbf{x}),  \label{43}
\end{equation}
\begin{equation}
gW_{\alpha }^{2}=\int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{-\tilde{u}}\Theta 
\tilde{r}q_{\alpha }d\Omega (\mathbf{x}),  \label{44}
\end{equation}
\begin{equation}
gW_{\alpha }^{3}=\int_{ex}\frac{f_{i}^{2}(\mathbf{x})}{\tilde{u}}\Theta 
\tilde{r}r_{\alpha }d\Omega (\mathbf{x}),  \label{45}
\end{equation}
\begin{equation}
M_{i}=\frac{3}{2}\int_{in}f_{i}^{2}dV,  \label{46}
\end{equation}
and 
\begin{equation}
\eta _{i}=\frac{3}{2}\int_{in}f_{i}^{2}\zeta dV  \label{47}
\end{equation}
it follows, since $\gamma \cdot u\psi _{1}(\mathbf{x})=\psi _{1}(\mathbf{x})$
and $\gamma \cdot u\psi _{2}(\mathbf{x})=0$, that 
\begin{equation}
\gamma _{d}\cdot P_{i}\psi _{1}(\mathbf{x}_{i})=M_{i}\psi _{1}(\mathbf{x}%
_{i})+\eta _{i}\psi _{1}(\mathbf{x}_{i})  \label{48}
\end{equation}
and 
\begin{equation}
\gamma _{d}\cdot P_{i}\psi _{2}(\mathbf{x}_{i})=0,  \label{49}
\end{equation}
giving 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{1L}(\mathbf{x}_{i})=M_{i}\psi _{1R}(%
\mathbf{x}_{i})+\frac{g\gamma _{d}\cdot Z}{2\cos \theta _{w}}\psi _{1L}(%
\mathbf{x}_{i})+\eta _{i}\psi _{1R}(\mathbf{x}_{i}),  \label{50}
\end{equation}
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{1R}(\mathbf{x}_{i})=M_{i}\psi _{1L}(%
\mathbf{x}_{i})+\frac{g\gamma _{d}\cdot Z}{\cos \theta _{w}}\psi _{1R}(%
\mathbf{x}_{i})+\eta _{i}\psi _{1L}(\mathbf{x}_{i}),  \label{51}
\end{equation}
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{2L}(\mathbf{x}_{i})=-\frac{g\gamma
_{d}\cdot (W^{1}-iW^{2})}{2}\psi _{2L}(\mathbf{x}_{i}),  \label{52}
\end{equation}
and 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{2R}(\mathbf{x}_{i})=-\frac{g\gamma
_{d}\cdot (W^{1}+iW^{2})}{2}\psi _{2R}(\mathbf{x}_{i}).  \label{53}
\end{equation}
Noting that $\gamma _{d}\cdot Z\psi _{2}=\gamma _{d}\cdot W^{1}\psi
_{1}=\gamma _{d}\cdot W^{2}\psi _{1}=0$ allows writing, for the linearly
independent combinations $(\psi _{1L}-\psi _{2L})/\sqrt{2}=\psi _{u}$ and $%
(\psi _{1L}+\psi _{2L})/\sqrt{2}=\psi _{lL},$ 
\[
i\gamma _{d}\cdot \partial _{i}\left( 
\begin{array}{c}
\psi _{u} \\ 
\psi _{lL}
\end{array}
\right) =\frac{g}{2}\gamma _{d}^{\alpha }\left( 
\begin{array}{cc}
Z_{\alpha }/\cos \theta _{w} & W_{\alpha }^{1}-iW_{\alpha }^{2} \\ 
W_{\alpha }^{1}+iW_{\alpha }^{2} & -Z_{\alpha }/\cos \theta _{w}
\end{array}
\right) \left( 
\begin{array}{c}
\psi _{u} \\ 
\psi _{lL}
\end{array}
\right) + 
\]
\begin{equation}
\frac{g\gamma _{d}^{\alpha }}{\sqrt{2}}\left( 
\begin{array}{c}
0 \\ 
iW_{\alpha }^{2}\psi _{2L}+Z_{\alpha }\psi _{1L}/\cos \theta _{w}
\end{array}
\right) +\frac{M_{i}+\eta _{i}}{\sqrt{2}}\left( 
\begin{array}{c}
\psi _{1R} \\ 
\psi _{1R}
\end{array}
\right) .  \label{54}
\end{equation}

Designating 
\begin{equation}
g^{\prime }B_{\alpha }=g(W_{\alpha }^{3}-\frac{Z_{\alpha }}{\cos \theta _{w}}%
),  \label{55}
\end{equation}
and noting that $-i\gamma _{d}^{\alpha }W_{\alpha }^{2}\psi _{2L}=\gamma
_{d}^{\alpha }W_{\alpha }^{3}\psi _{2L}$ (since $\gamma \cdot q\psi
_{2}=-i\gamma ^{5}\gamma \cdot r\psi _{2}$) gives 
\begin{equation}
\frac{g\gamma _{d}^{\alpha }}{\sqrt{2}}\left( iW_{\alpha }^{2}\psi _{2L}+%
\frac{Z_{\alpha }}{\cos \theta _{w}}\psi _{1L}\right) =g\gamma _{d}^{\alpha
}(-W_{\alpha }^{3}+\frac{Z_{\alpha }}{\cos \theta _{w}})\psi
_{lL}=-g^{\prime }\gamma _{d}^{\alpha }B_{\alpha }\psi _{lL},  \label{56}
\end{equation}
which allows writing 
\[
i\gamma _{d}\cdot \partial _{i}\left( 
\begin{array}{c}
\psi _{u} \\ 
\psi _{lL}
\end{array}
\right) =\frac{g}{2}\gamma _{d}^{\alpha }\left( 
\begin{array}{cc}
Z_{\alpha }/\cos \theta _{w} & W_{\alpha }^{1}-iW_{\alpha }^{2} \\ 
W_{\alpha }^{1}+iW_{\alpha }^{2} & -Z_{\alpha }/\cos \theta _{w}
\end{array}
\right) \left( 
\begin{array}{c}
\psi _{u} \\ 
\psi _{lL}
\end{array}
\right) - 
\]
\begin{equation}
\left( 
\begin{array}{c}
0 \\ 
g^{\prime }\gamma _{d}^{\alpha }B_{\alpha }\psi _{lL}
\end{array}
\right) +\frac{M_{i}+\eta _{i}}{\sqrt{2}}\left( 
\begin{array}{c}
\psi _{1R} \\ 
\psi _{1R}
\end{array}
\right) .  \label{57}
\end{equation}

Now, defining $\psi _{l}=\psi _{1R}+\psi _{lL}$ gives 
\[
i\gamma _{d}\cdot \partial _{i}\left( 
\begin{array}{c}
\psi _{u} \\ 
\psi _{l}
\end{array}
\right) =\frac{g}{2}\gamma _{d}^{\alpha }\left( 
\begin{array}{cc}
Z_{\alpha }/\cos \theta _{w} & W_{\alpha }^{1}-iW_{\alpha }^{2} \\ 
W_{\alpha }^{1}+iW_{\alpha }^{2} & -Z_{\alpha }/\cos \theta _{w}
\end{array}
\right) P_{L}\left( 
\begin{array}{c}
\psi _{u} \\ 
\psi _{l}
\end{array}
\right) - 
\]
\begin{equation}
\gamma _{d}^{\alpha }\left\{ g^{\prime }B_{\alpha }\psi _{lL}-\frac{%
gZ_{\alpha }}{\cos \theta _{w}}\psi _{1R}\right\} \left( 
\begin{array}{c}
0 \\ 
1
\end{array}
\right) +\frac{M_{i}+\eta _{i}}{\sqrt{2}}\left( 
\begin{array}{c}
\psi _{1R} \\ 
\sqrt{2}\psi _{1L}+\psi _{1R}
\end{array}
\right) .  \label{58}
\end{equation}
The three components $\psi _{1R,}$ $\psi _{1L,}\psi _{2L,}$ give a complete
physical description just as three members of the tetrad $(u,s,q,r)$ can be
used to construct the fourth member by Gram-Schmidt orthogonalization \cite
{letaw}. Invariant properties under the Lorentz group as well as the
representations with respect to the electroweak gauge groups determine the
combinations that are called particles.

Recalling that $\gamma _{d}^{\alpha }W_{\alpha }^{3}\psi _{1R}=0$, then 
\begin{equation}
\gamma _{d}^{\alpha }\left\{ g^{\prime }B_{\alpha }\psi _{lL}-\frac{%
gZ_{\alpha }}{\cos \theta _{w}}\psi _{1R}\right\} =\gamma _{d}^{\alpha
}g^{\prime }B_{\alpha }(\psi _{lL}+\psi _{1R})=\gamma _{d}^{\alpha
}g^{\prime }B_{\alpha }\psi _{l}.  \label{59}
\end{equation}
For comparison purposes, define 
\begin{equation}
G_{e}\phi =G_{e}\left( 
\begin{array}{c}
\phi _{1} \\ 
\phi _{2}
\end{array}
\right) =\frac{M_{i}+\eta _{i}}{\sqrt{2}}\left( 
\begin{array}{c}
1 \\ 
1
\end{array}
\right)  \label{60}
\end{equation}
where $G_{e}$ is Fermi's constant, and 
\begin{equation}
\psi _{A}=\left( 
\begin{array}{c}
\psi _{u} \\ 
\psi _{l}
\end{array}
\right) ,  \label{61}
\end{equation}
with $\psi _{AL}=P_{L}\psi _{A}$. It follows that 
\[
G_{e}\psi _{lR}\phi +G_{e}\phi ^{\dagger }\psi _{AL}\left( 
\begin{array}{c}
0 \\ 
1
\end{array}
\right) =G_{e}\left( 
\begin{array}{c}
\psi _{lR}\phi _{1} \\ 
\psi _{lR}\phi _{2}+\psi _{u}\phi _{1}^{*}+\psi _{lL}\phi _{2}^{*}
\end{array}
\right) 
\]
\begin{equation}
=\frac{M_{i}+\eta _{i}}{\sqrt{2}}\left( 
\begin{array}{c}
\psi _{lR} \\ 
\psi _{l}+\psi _{u}
\end{array}
\right) =\frac{M_{i}+\eta _{i}}{\sqrt{2}}\left( 
\begin{array}{c}
\psi _{1R} \\ 
\sqrt{2}\psi _{1L}+\psi _{1R}
\end{array}
\right) .  \label{62}
\end{equation}
Thereby, 
\[
i\gamma _{d}\cdot \partial _{i}\psi _{A}=\frac{g}{2}\gamma _{d}^{\alpha
}\left( W_{\alpha }^{1}\sigma _{1}+W_{\alpha }^{2}\sigma _{2}+\frac{%
Z_{\alpha }}{\cos \theta _{w}}\sigma _{3}\right) \psi _{AL}- 
\]
\begin{equation}
\left( 
\begin{array}{c}
0 \\ 
g^{\prime }\gamma _{d}^{\alpha }B_{\alpha }\psi _{l}
\end{array}
\right) +G_{e}\psi _{lR}\phi +G_{e}\phi ^{\dagger }\psi _{AL}\left( 
\begin{array}{c}
0 \\ 
1
\end{array}
\right) ,  \label{63}
\end{equation}
or equivalently, with $\mathbf{W}_{\alpha }\mathbf{\cdot \sigma }\doteq
W_{\alpha }^{i}\sigma _{i}$ ($i$ running from 1 to 3), 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{AL}=\frac{g}{2}\gamma _{d}^{\alpha }%
\mathbf{W}_{\alpha }\mathbf{\cdot \sigma }\psi _{AL}-\frac{g^{\prime }}{2}%
\gamma _{d}^{\alpha }B_{\alpha }\psi _{AL}+G_{e}\psi _{lR}\phi ,  \label{64}
\end{equation}
and 
\begin{equation}
i\gamma _{d}\cdot \partial _{i}\psi _{AR}=-g^{\prime }\gamma _{d}^{\alpha
}B_{\alpha }\psi _{AR}+G_{e}\phi ^{\dagger }\psi _{AL},  \label{65}
\end{equation}
where $\psi _{AR}=\psi _{1R}$. Then the normal form of standard electroweak
theory follows by going to the unitary gauge \cite{cheng}.

It is seen that the scalar field $\eta (\mathbf{x})$, identifiable as the
Higgs field in the standard model, arises from the departure from flatness
of the metric in the vicinity of the core of an excitation and thus the
association of the Higgs field with particle mass generation is evident.

In particular, since in the unitary gauge, $G_{e}\phi \rightarrow
(M_{i}+\eta _{i})\left( 
\begin{array}{c}
0 \\ 
1
\end{array}
\right) $, the doublet $\psi _{A}$ contains a massless component and a
massive component of mass $M_{i}$. For evaluation of the expression for
mass, poles in the integrand may be handled in the standard way (see
Appendix C); 
\begin{equation}
f_{i}^{2}dV=\frac{2\pi ^{2}}{\xi _{0}}\left( \frac{\Sigma }{\xi _{0}^{2}}+%
\frac{m_{i0}r}{\pi }\right) \delta (r)\sin \theta drd\theta .  \label{66}
\end{equation}
Then, with $\Sigma \delta (r)=(J^{2}\cos ^{2}\theta )\delta (r)$, 
\[
\int_{in}f_{i}^{2}dV=\frac{2\pi ^{2}J^{2}}{\xi _{0}^{3}}\int_{0^{+}}^{L}%
\delta (r)dr\int_{0}^{\pi }\cos ^{2}\theta \sin \theta d\theta +\frac{2\pi
m_{i0}}{\xi _{0}}\int_{0^{+}}^{L}r\delta (r)dr\int_{0}^{\pi }\sin \theta
d\theta 
\]
\begin{equation}
=\frac{2\pi ^{2}J^{2}}{3\xi _{0}^{3}}+2\epsilon m_{i0}\ln (1+\frac{L^{2}}{%
\epsilon ^{2}\xi _{0}^{2}}),  \label{67}
\end{equation}
where the regularized $\delta $-function, $\delta _{\epsilon }(r)=\epsilon
\xi _{0}/\pi (r^{2}+\epsilon ^{2}\xi _{0}^{2})$, is used for evaluation: $%
\int_{0^{+}}^{L}\delta (r)dr=\frac{1}{2}$ and 
\[
\int_{0^{+}}^{L}r\delta (r)dr=\frac{\epsilon \xi _{0}}{2\pi }\ln (1+\frac{%
L^{2}}{\epsilon ^{2}\xi _{0}^{2}}).
\]
Thereby, upon inclusion of the core contribution, $-m_{i0}$, 
\begin{equation}
M_{i}=\frac{3}{2}\int f_{i}^{2}dV\cong -m_{i0}+\frac{\pi ^{2}J^{2}}{\xi
_{0}^{3}}+3\epsilon m_{i0}\ln (\frac{L^{2}}{\epsilon ^{2}\xi _{0}^{2}}),
\label{69}
\end{equation}
where $L\gg \epsilon \xi _{0}=4\epsilon \times 10^{-33}cm$. For $j=\sqrt{3}/2
$ it follows, to within the accuracy of the present model, that with $m_{i0}$
being the extreme mass $\sqrt{2\pi j}/\xi _{0}\simeq 2.33/\xi _{0}$ (or near
extreme), then $m_{i0}=\pi ^{2}J^{2}/\xi _{0}^{3}$ since $J=j/m_{i0}$ then
gives $m_{i0}=(\pi j)^{2/3}/\xi _{0}\simeq 1.95/\xi _{0}$. The hierarchy
problem is thus addressed with a low mass Fermion emerging on the length
scale L; 
\begin{equation}
M_{i}\simeq 3\epsilon m_{i0}\ln (\frac{L^{2}}{\epsilon ^{2}\xi _{0}^{2}}).
\label{70}
\end{equation}
The ''infinite'' negative mass of the core is compensated by the positive
mass surrounding the core with a slight excess emerging as the effective
mass of the Fermion. Then, writing $L=M^{-1},$%
\begin{equation}
M_{i}(M)\simeq 3\epsilon m_{i0}\left\{ \ln (\frac{1}{M_{x}^{2}\epsilon
^{2}\xi _{0}^{2}})+\ln (\frac{M_{x}^{2}}{M^{2}})\right\} ,  \label{71}
\end{equation}
where $M_{x}$ is a ''unification'' mass scale. Comparison with the standard
model, for which Fermion mass, at mass scale $M$, is 
\begin{eqnarray*}
M_{f}(M) &\simeq &M_{f}(M_{x})\left\{ 1-\frac{5\alpha (M_{x})}{6\pi }\ln (%
\frac{M_{x}^{2}}{M^{2}})\right\} ^{-27/40}\left\{ 1+\frac{\alpha (M_{x})}{%
\pi }\ln (\frac{M_{x}^{2}}{M^{2}})\right\} ^{9/40} \\
&\simeq &M_{f}(M_{x})\left\{ 1+\frac{63\alpha (M_{x})}{80\pi }\ln (\frac{%
M_{x}^{2}}{M^{2}})\right\} 
\end{eqnarray*}
and $\alpha (M_{x})$ is the running coupling constant of $U(1)\times SU(2)$
theory at the mass scale $M_{x}\simeq 300$ $Gev$, gives 
\begin{equation}
M_{f}(M_{x})\simeq 3\epsilon m_{i0}\ln (\frac{1}{M_{x}^{2}\epsilon ^{2}\xi
_{0}^{2}}).  \label{72}
\end{equation}
With $M_{f}(M_{x})\simeq 1Gev$, since $\xi _{0}^{-1}\simeq 10^{19}Gev/2$ and 
$m_{i0}\simeq 10^{19}Gev$, it follows that $\epsilon \simeq 1.9\times
10^{-22}$ and that 
\begin{equation}
\alpha ^{-1}(M_{x})\simeq \frac{63}{80\pi }\ln (\frac{1}{M_{x}^{2}\epsilon
^{2}\xi _{0}^{2}})\simeq 43.  \label{73}
\end{equation}

\appendix 

\section{Equation of State}

The equation of state of the (Hausdorf) spacetime continuum is sought. Since
a process for establishing spacetime is outside the realm of experimental
investigation, a physically equivalent process whereby spacetime, and
consequently the energy density and pressure of spacetime, could be
established is needed. The reliability of such an equivalent process can
only be established based on comparison of the results obtained with actual
physical situations. Necessarily, a physically equivalent concept of
pregeometry is needed. Consistent with spacetime being the collection of
events, an event being a localization of spacetime, pregeometry requires the
absence of events, that is, localizability. An intuitive way to effect the
delocalization of events is to let each spacetime point go over into a light
cone of unselected direction. Then the delocalization of a spacetime point
appears physically degenerate with a massless particle of zero energy
occupying a null geodesic of unselected direction: the absence of spacetime
could be degenerate with a Petroff Type-O flat space. Gravitational
interaction of null currents provides a mechanism for conversion of these
currents into a perfect fluid state: Chandrasekhar and Xanthopoulos \cite
{chandrasekhar} showed ''the transformation of massless particles describing
null trajectories into a perfect fluid in which the stream lines describe
time-like trajectories'', thereby providing ''the first example of an
induced transformation of a massless into a massive particle''.

The gravitational fields established by null currents, which are here
considered to generate the matter density and pressure that characterize
spacetime, are plane-fronted gravitational waves which account for the
interactions of the currents and are obtainable exactly from consideration
of a finite mass $m$ moving at speed $v$ in the limit as $v\rightarrow 1$
and $m\rightarrow 0$ for which the energy $e$ is held fixed with $m=e\sqrt{%
1-v^{2}}$.\cite{bonnor} Furthermore, for this case, linearized calculations
give solutions identical with exact solutions. Then, a model for
self-interaction of a null current with an ambient may be obtained from
consideration of the linear interaction of a mass $m$ in relative motion
with respect to the ambient and for which the above stated limit is taken.

In order to relate the energy density and pressure of spacetime, the energy
content of a small sample volume is taken to be physically degenerate with
the energy $e$ of a null current: $e$ is given by the local energy density
of spacetime, $\varepsilon $, times the size of the small volume $V/N$,
there being a large number $N$ of such volumes in a finite volume $V$. The
equation of state is obtained by evaluation of fluid pressure from the
volume derivative, at constant entropy, of the energy of the fluid: $%
p=-(\partial E/\partial V)_{s}$, the basic expression of which, based on a
statistical model, is the derivative of the ensemble-averaged energy with
respect to volume \cite{landau-st}, 
\begin{equation}
p=-\partial \overline{E/\partial V.}  \label{a-one}
\end{equation}

The ensemble-averaged energy is obtained by considering the small sample
volume to be a test mass $m$ with velocity $v$ and energy $e$ in a perfect
fluid which may be treated in the Newtonian limit \cite{14chand},
perturbations of which \cite{3weinberg}, written in Fourier space, give for
the relation between perturbation density and potential 
\begin{equation}
\left( \omega ^{2}-v_{s}^{2}k^{2}\right) \widetilde{\rho }=\varepsilon k^{2}%
\widetilde{\phi }  \label{a-two}
\end{equation}
with the gravitational potential given by 
\begin{equation}
\widetilde{\phi }=-4\pi G\left[ \widetilde{\rho }+e\delta (\omega -%
\overrightarrow{k}\cdot \overrightarrow{v})/2\pi \right] /k^{2},
\label{a-three}
\end{equation}
transformed quantities being indicated by a tilde, each component having a
dependence $\exp i(\overrightarrow{k}\cdot \overrightarrow{x}-\omega t)$, $%
\varepsilon $ being the matter density of the fluid, $\rho $ the density
response to the test mass, and $v_{s}=\sqrt{(\partial p/\partial \varepsilon
)_{s}}$ the sound speed of the medium. Thereby, the response is found to be 
\begin{equation}
\widetilde{\phi }=e\Omega ^{2}\delta (\omega -\overrightarrow{k}\cdot 
\overrightarrow{v})\left( \omega ^{2}-v_{s}^{2}k^{2}\right) /2\pi
\varepsilon k^{2}\left( v_{s}^{2}k^{2}-\omega ^{2}-\Omega ^{2}\right)
\label{a-four}
\end{equation}
with $\Omega ^{2}=4\pi G\varepsilon $.

Sine the post-Newtonian approximation \cite{15einstein} is needed to treat a
test mass, the self-energy of the test mass is the sum of the rest and
kinetic contribution, $m\sqrt{1-v^{2}}=e$, with the total correlation
energy. In the statistical model, the fluid consists of a large number $N$
of such masses in a volume $V$, the ensemble average being 
\begin{equation}
\overline{E}=Ne+\left( N^{2}/2V\right) \int e\phi (\overrightarrow{x}%
)\,d^{3}x.  \label{a-five}
\end{equation}
This formulation of the correlation energy is exact in that the
self-consistent correlation function \cite{landau-st} is identically unity
for a perfect fluid. Then, use of the above expression for $\widetilde{\phi }
$ and Fourier inversion gives 
\begin{equation}
\int e\phi \,d^{3}x=e^{2}\left( v_{s}^{2}-v^{2}\right) /\varepsilon ,
\label{a-six}
\end{equation}
and, since $\varepsilon V=Ne$, then 
\begin{equation}
\overline{E}=\varepsilon V+\varepsilon V\left( v_{s}^{2}-v^{2}\right) /2.
\label{a-seven}
\end{equation}
The pressure is then given by 
\begin{equation}
p=-\partial \bar{E}/\partial V=-\varepsilon \left( 1+\frac{v_{s}^{2}}{2}-%
\frac{v^{2}}{2}\right) .  \label{aa-seven}
\end{equation}
Thereby $v_{s}^{2}=\partial p/\partial \varepsilon =(v^{2}-2)/3$ and it
follows that $p=-$ $\varepsilon (2-v^{2})/3$. Thus, $v\rightarrow 1$ gives 
\begin{equation}
p=-\varepsilon /3.  \label{a-eight}
\end{equation}
This is the required equation of state. (It is of interest to notice that
the limit $v\rightarrow 1$ gives $v_{s}^{2}=1/3$ and, with the dispersion
relation for density perturbations \cite{3weinberg} $\omega
^{2}=k^{2}v_{s}^{2}-\Omega ^{2}$, then $\omega ^{2}\rightarrow
-k^{2}/3-\Omega ^{2}$, revealing that in the absence of matter ($\Omega
\rightarrow 0$), flat-space (pregeometry) is unstable: $\func{Im}(\omega
)\rightarrow k/\sqrt{3}$, with infinite exponential growth rate for
vanishingly small perturbations. Also, the above equation of state is
recognized as that of the static Einstein universe, applicable to the
flat-space equivalent of pregeometry.)

Perhaps a stronger case can be made for the above equation of state by
observing that it can be obtained by requiring that the action remain
stationary at zero from a state of no matter to a state with matter present.
The Lagrangian density, being the sum of the geometrical contribution $\sqrt{%
-g}R/16\pi G$ and the matter contribution $-\sqrt{-g}\varepsilon $ \cite
{38soper}, becomes $\sqrt{-g}(-\varepsilon -3p)$ upon use of the field
equations requirement that $-R=8\pi GT_{\sigma }^{\sigma }=8\pi
G(3p-\varepsilon )$. The equation of state then follows from the zero
stationary value for the action.

\section{EW Dynamics}

Dynamics for EW type excitations may be developed using a standard
formulation \cite{6ellis} based on contraction of Ricci's identity, $u_{\mu
;\lambda ;\nu }-u_{\mu ;\nu ;\lambda }=R_{\mu \sigma \nu \lambda }u^{\sigma
} $ with $u^{\lambda }$. The Riemann tensor is treated in terms of its
irreducible parts: the Weyl tensor $C_{\mu \sigma \nu \lambda }$, the
traceless part of the Ricci tensor $R_{\mu \nu }-g_{\mu \nu }R/4$, and the
curvature scalar $R$. For these excitations, the only non-vanishing part of
the Weyl tensor is the electric component: $E_{\mu \nu }=C_{\mu \sigma \nu
\lambda }u^{\sigma }u^{\lambda }$. The field equations give $R_{\mu \nu
}=8\pi Gf^{2}h_{\mu \nu }$; the Ricci tensor is the metric tensor in the
hyperspace orthogonal to $u^{\nu }$ scaled by $8\pi Gf^{2}=R/3$. Then with $%
R_{\mu \sigma \nu \lambda }u^{\sigma }u^{\lambda }=E_{\mu \nu }$, standard
algebra gives 
\begin{equation}
a_{\mu ;\nu }=E_{\mu \nu }+\frac{\Lambda }{3}ah_{\mu \nu }+\frac{\Theta }{3}%
u_{\mu }a_{\nu }-a_{\mu }a_{\nu }-\dot{a}_{\mu }u_{\nu },  \label{b-two}
\end{equation}
where $\Lambda =a_{;\sigma }^{\sigma }/a$. Self-contraction of this result
and recognition that $E_{\mu \nu }$ is traceless and that $u^{\nu }$ and $%
a^{\nu }$ are orthogonal gives the Raychaudhuri equation for this case; 
\begin{equation}
a_{;\sigma }^{\sigma }=\dot{\Theta}+\Theta ^{2}/3.  \label{b-three}
\end{equation}
Contraction of Ricci's identity with $g^{\mu \nu }h^{\kappa \lambda }$ and
use of the field equations and Raychaudhuri's equation then gives $\dot{%
\Theta}u_{\nu }=-\Theta _{,\nu }$.

Using the orthonormal tetrad $(u^{\nu },s^{\nu },q^{\nu },r^{\nu })$
comprised of $u^{\nu }$, the unit vector $s^{\nu }=a^{\nu }/a$ associated
with the acceleration, and two symmetrically disposed spacelike vectors $%
q^{\nu }$, $r^{\nu }$, it follows that 
\begin{equation}
s_{\mu ;\nu }=\left( a^{-1}E_{.\nu }^{\sigma }+\Lambda \delta _{\nu
}^{\sigma }/3\right) p_{\sigma \mu }+\frac{\Theta }{3}u_{\mu }s_{\nu
}-u_{\nu }\dot{s}_{\mu }  \label{b-four}
\end{equation}
where $p_{\nu }^{\mu }=$ $\delta _{\nu }^{\mu }+u^{\mu }u_{\nu }-s^{\mu
}s_{\nu }=q^{\mu }q_{\nu }+r^{\mu }r_{\nu }$ is the projection tensor onto
the $(q^{\nu },r^{\nu })$ spacelike surface. Orthonormality and the above
equation then show 
\begin{equation}
r_{;\lambda }^{\sigma }s_{\sigma }h_{\nu }^{\lambda }=-\frac{\Lambda }{3}%
r_{\nu }-\frac{E_{\lambda \nu }}{a}r^{\lambda }.  \label{b-five}
\end{equation}
Also, $r_{;\nu }^{\sigma }u_{\sigma }=-r^{\sigma }u_{\sigma ;\nu }$ and
using decomposition for $u_{\sigma ;\nu }$ gives 
\begin{equation}
r_{;\nu }^{\sigma }u_{\sigma }=-\Theta r_{\nu }/3.  \label{b-one}
\end{equation}
Equivalent equations result for $q^{\nu }$.

Now, since $\dot{r}^{\sigma }u_{\sigma }=-r^{\sigma }\dot{u}_{\sigma }=0$
and $\dot{r}^{\sigma }r_{\sigma }=0$, by expansion $\dot{r}^{\sigma
}=As^{\sigma }+Bq^{\sigma }$ and, likewise, $\dot{q}^{\sigma }=Cs^{\sigma
}+Dr^{\sigma }$. But since $\dot{r}^{\sigma }q_{\sigma }+r^{\sigma }\dot{q}%
_{\sigma }=0$, then $B=-D$ and the symmetric disposition of $q^{\nu }$ and $%
r^{\nu }$ renders $B=-D=0$ and $A=C$. Then (\ref{b-one}) and $\dot{r}%
^{\sigma }=As^{\sigma }$ are satisfied by 
\begin{equation}
r_{;\nu }^{\sigma }=-As^{\sigma }u_{\nu }+\frac{\Theta }{3}r_{\nu }u^{\sigma
},  \label{b-six}
\end{equation}
with a similar expression for $q_{;\nu }^{\sigma }$. Immediately, (\ref
{b-six}) renders the left side of (\ref{b-five}) zero, showing that $%
E_{\lambda \nu }r^{\lambda }/a=-\Lambda r_{\nu }/3$ and, equivalently, $%
E_{\lambda \nu }q^{\lambda }/a=-\Lambda q_{\nu }/3$. By direct expansion, $%
\dot{a}_{\nu }=a^{2}u_{\nu }+\dot{a}s_{\nu }-aA(q_{\nu }+r_{\nu })$.
Self-contraction gives $\dot{a}^{2}=-a^{4}+$ $\dot{a}^{2}+2a^{2}A^{2}$.
Taking $A=-a/\sqrt{2}$ results in 
\begin{equation}
r_{;\nu }^{\sigma }=a\frac{s^{\sigma }}{\sqrt{2}}u_{\nu }+\frac{\Theta }{3}%
r_{\nu }u^{\sigma }  \label{b-seven}
\end{equation}
\begin{equation}
q_{;\nu }^{\sigma }=a\frac{s^{\sigma }}{\sqrt{2}}u_{\nu }+\frac{\Theta }{3}%
q_{\nu }u^{\sigma }  \label{b-eight}
\end{equation}
and, thereby, $\dot{r}^{\nu }=\dot{q}^{\nu }=-a^{\nu }/\sqrt{2}$.
Furthermore, since $\dot{a}_{\nu }=\dot{a}s_{\nu }+a\dot{s}_{\nu }$, using $%
A=-a/\sqrt{2}$ in the above expression for $\dot{a}_{\nu }$ gives 
\begin{equation}
\dot{s}_{\nu }=au_{\nu }+a(q_{\nu }+r_{\nu })/\sqrt{2}.  \label{b-nine}
\end{equation}
(\ref{b-four}) now simplifies: 
\begin{equation}
s_{\mu ;\nu }=\frac{\Theta }{3}u_{\mu }s_{\nu }-u_{\nu }\dot{s}_{\mu }.
\label{b-ten}
\end{equation}
In particular, from (\ref{b-nine}) and (\ref{b-ten}), $a=s_{;\sigma
}^{\sigma }$. Thereby, $a_{;\sigma }^{\sigma }=as_{;\sigma }^{\sigma }$ $%
+a_{,\sigma }s^{\sigma }=a^{2}+a^{\prime }$ where $a^{\prime }=a_{,\sigma
}s^{\sigma }$, and (\ref{b-three}) becomes 
\begin{equation}
a^{\prime }+a^{2}=\dot{\Theta}+\Theta ^{2}/3.  \label{b-eleven}
\end{equation}
Evaluating Lie derivatives gives $\pounds _{r}q=\pounds _{s}q=\pounds
_{s}r=0 $, showing that $(\mathbf{r},\mathbf{s})$, $(\mathbf{s},\mathbf{q})$
and $(\mathbf{q},\mathbf{r})$ are surface forming.

Contraction of $E^{\mu \sigma }$ with $\delta _{\sigma }^{\nu }=-u^{\nu
}u_{\sigma }+s^{\nu }s_{\sigma }+q^{\nu }q_{\sigma }+r^{\nu }r_{\sigma }$
and noting that $E^{\mu \sigma }u_{\sigma }=0$ gives $E^{\mu \nu }=E^{\mu
\sigma }s_{\sigma }s^{\nu }-\Lambda a(q^{\mu }q^{\nu }+r^{\mu }r^{\nu })/3$.
Since $E_{.\sigma }^{\sigma }$ vanishes by symmetry, then $E^{\mu \sigma
}s_{\sigma }s_{\mu }=2\Lambda a/3$. Thereby, the expansion $E^{\mu \sigma
}s_{\sigma }=E^{\nu \sigma }s_{\sigma }(-u^{\mu }u_{\nu }+s^{\mu }s_{\nu
}+q^{\mu }q_{\nu }+r^{\mu }r_{\nu })=2\Lambda as^{\mu }/3$ follows, giving 
\begin{equation}
E^{\mu \nu }=\Lambda a(2s^{\mu }s^{\nu }-q^{\mu }q^{\nu }-r^{\mu }r^{\nu
})/3;  \label{b-twelve}
\end{equation}
the electric component of the Weyl tensor lies in the spacelike hypersurface
orthogonal to $\mathbf{u}$ and is diagonal with respect to the tetrad
coordinates.

These motions are three space rotations as is seen by transformation from $(%
\mathbf{q,r)}$ to $(\mathbf{w},\mathbf{v)}$ given by $\sqrt{2}\mathbf{w}=%
\mathbf{q+r}$ and $\sqrt{2}\mathbf{v}=\mathbf{q-r}$. Then, 
\begin{equation}
w_{;\nu }^{\sigma }=a^{\sigma }u_{\nu }+\frac{\Theta }{3}w_{\nu }u^{\sigma }
\label{b-thirteen}
\end{equation}
\begin{equation}
v_{;\nu }^{\sigma }=\frac{\Theta }{3}v_{\nu }u^{\sigma }  \label{b-fourteen}
\end{equation}
and, thereby, $\dot{w}^{\nu }=-a^{\nu }$ and $\dot{v}^{\nu }=0$.
Additionally, $\dot{s}_{\nu }=a(u_{\nu }+w_{\nu })$.

\section{Approximate EW Model}

Although exact solutions are available, an approximate, but simple and
illustrative model of an EW excitation may be realized using the Kerr family
of solutions. Vaidya \cite{vaidya} presented the external solutions for the
Kerr metric in the Einstein universe for which the energy density and
pressure are given by 
\begin{equation}
\varepsilon =-3p=\frac{3}{4\xi _{0}^{2}S^{2}}(1+2m\mu )  \label{c-1}
\end{equation}
with $S$ being the scale factor, $\xi _{0}=\sqrt{2\pi G}\simeq 4\times
10^{-33}cm$, and $m$ and $\mu $ are specified below. For\texttt{\ }$%
S^{2}\rightarrow \infty $, the solution becomes the Kerr metric in Minkowski
space. This latter metric is written here in Boyer-Lindquist (S-type)
coordinates \cite{boyer} as 
\begin{equation}
ds^{2}=-\left( \pm \sqrt{1+2m\mu }dt\pm \frac{2m\mu J\sin ^{2}\theta }{\sqrt{%
1+2m\mu }}d\phi \right) ^{2}+\frac{\Sigma }{\Delta }dr^{2}+\Sigma d\theta
^{2}+\frac{\Delta \sin ^{2}\theta }{1+2m\mu }d\phi ^{2},  \label{c-2}
\end{equation}
where $J=j/m_{0}$, $\Delta =r^{2}+J^{2}+2mr,m=m_{0}\xi _{0}^{2}/2\pi ,\Sigma
=r^{2}+J^{2}\cos ^{2}\theta $ and $\mu =r/\Sigma $ with $j$ being the
angular momentum of the singularity in units of $\hbar $ and $-m_{0}$ being
the negative mass of the singularity, the negative r side of the disk having
been taken and a relabeling having been carried-out so that r values are
taken to be positive. The extreme case, $(m_{0}\xi _{0})^{2}=2\pi j$, is of
interest. (Metrics of the Kerr-Schild form are practical in that they can be
inclusive of radiation fields \cite{kramer}.) The invariant measure of this
space is $d\Omega =\sqrt{-g}d^{4}x=\Sigma \sin \theta dtdrd\theta d\phi .$

Choosing a sign for $\tilde{u}=u_{\sigma }dx^{\sigma }$ allows taking 
\begin{equation}
\tilde{u}=-\sqrt{1+2m\mu }dt-\frac{2m\mu J\sin ^{2}\theta }{\sqrt{1+2m\mu }}%
d\phi ,  \label{c-5}
\end{equation}
the associated velocity being $\left\{ u^{\nu }\right\} =(1/\sqrt{1+2m\mu }%
,0,0,0)$ and 
\begin{equation}
R_{\mu \nu }=4\xi _{0}^{2}f^{2}h_{\mu \nu }=4\xi _{0}^{2}f^{2}diag(0,\Sigma
/\Delta ,\Sigma ,\Delta \sin ^{2}\theta /\left[ 1+2m\mu \right] ).
\label{c-4}
\end{equation}
As $r\rightarrow \infty $, $\left\{ g_{\mu \nu }\right\} \rightarrow
diag(-1,1,r^{2},r^{2}\sin ^{2}\theta )=\left\{ g_{L\alpha \beta }\right\} $,
a Lorentzian metric.

Transformation of the above C-type coordinates to global coordinates (a
globally synchronizable embedding) by $t=\tau \sqrt{1+\rho ^{2}}$ and $%
r=\rho \tau $ gives 
\[
ds^{2}=-\left\{ (1+\rho ^{2})(1+2m\mu )-\frac{\rho ^{2}\Sigma }{\Delta }%
\right\} d\tau ^{2}-2\rho \tau (1+2m\mu -\frac{\Sigma }{\Delta })d\tau d\rho
- 
\]
\[
4mJ\mu \sqrt{1+\rho ^{2}}\sin ^{2}\theta d\tau d\phi +\tau ^{2}\left\{ \frac{%
\Sigma }{\Delta }-\frac{\rho ^{2}(1+2m\mu )}{1+\rho ^{2}}\right\} d\rho ^{2}-%
\frac{4mJ\mu \tau \rho }{\sqrt{1+\rho ^{2}}}\sin ^{2}\theta d\rho d\phi 
\]
\begin{equation}
+\Sigma d\theta ^{2}+\sin ^{2}\theta (\rho ^{2}\tau ^{2}+J^{2}-2m\mu
J^{2}\sin ^{2}\theta )d\phi ^{2}  \label{c-12}
\end{equation}
where now $\Sigma =\rho ^{2}\tau ^{2}+J^{2}\cos ^{2}\theta $ and $\Delta
=\rho ^{2}\tau ^{2}+J^{2}+2m\rho \tau $. Far from the singularity, $\rho
^{2}\tau ^{2}\gg J^{2}$, the metric goes over to the FRW negative curvature
empty space value; 
\begin{equation}
ds^{2}=-d\tau ^{2}+\tau ^{2}(\frac{d\rho ^{2}}{1+\rho ^{2}}+\rho ^{2}d\theta
^{2}+\rho ^{2}\sin ^{2}\theta d\phi ^{2}).  \label{c-13}
\end{equation}
In the same limit, the field equations reduce to $\ddot{S}=0$ , yielding a
constant $\dot{S}=1-\epsilon $, and 
\begin{equation}
\mu =3(\left[ 1-\epsilon \right] ^{2}-1)/4\xi _{0}^{2}S^{2}\rightarrow 0,
\label{c-14}
\end{equation}
revealing $\epsilon $ as the physical limit corresponding to 0. Then $\mu
\rightarrow -3\epsilon /2\xi _{0}^{2}\tau ^{2}$ at large radii and the
identification $-3\epsilon /2\xi _{0}^{2}\tau ^{2}=3/4\xi _{0}^{2}S^{2}$
follows, giving $S^{2}=-\tau ^{2}/2\epsilon \rightarrow \infty $ as $%
\epsilon \rightarrow 0$. (It is shown above in section IV that $\epsilon
\sim 10^{-22}$: the present theory reveals finite physical limits for both
the 0 and the $\infty $ of standard theory, an example of the latter being $%
m_{0}\sim 1/\xi _{0}\sim 10^{19}Gev$, the mass of an excitation core which
is the ''bare mass'' of standard renormalizable field theory.) Thereby, 
\begin{equation}
f^{2}=-\frac{\epsilon }{\xi _{0}^{2}\tau ^{2}}(1+2m\mu ),  \label{c-15}
\end{equation}
which represents the ''renormalized'' but structured vacuum. Explicitly, in
C-type coordinates, 
\begin{equation}
f^{2}=-\frac{\epsilon }{\xi _{0}^{2}(t^{2}-r^{2})}(1+\frac{2mr}{%
r^{2}+J^{2}\cos ^{2}\theta }).  \label{c-16}
\end{equation}
Then, since $(\ln f),_{\nu }=a_{\nu }+\Theta u_{\nu }/3$, it follows that 
\begin{equation}
\Theta =-3(\ln f),_{\sigma }u^{\sigma }=-3(\ln f),_{t}u^{t}=\frac{3t}{%
(t^{2}-r^{2})\sqrt{1+2m\mu }}.  \label{c17}
\end{equation}
Thereby, with $f^{2}=\varepsilon +p=2\varepsilon /3$, 
\begin{equation}
\tilde{a}=d(\ln f)-\Theta \tilde{u}/3=\frac{1}{2}d\{\ln (1+2m\mu )\}+\frac{%
r\,dr+2m\mu Jt\sin ^{2}\theta \cdot d\phi /(1+2m\mu )}{t^{2}-r^{2}}.
\label{c25}
\end{equation}
For evaluation of integrals with $\tilde{a}$ in the integrand, the poles due
to $(t^{2}-r^{2})^{-1}$ may be handled in the standard way by letting 
\[
\frac{1}{(t^{2}-r^{2})}\rightarrow i\pi \delta (t^{2}-r^{2})\rightarrow i\pi
S(t)\frac{\delta (t-r)}{r}\rightarrow -\pi ^{2}S(t)\delta (t-r)\delta (r), 
\]
giving, for example, for the external space integral 
\begin{equation}
\int_{ex}\tilde{a}=\int_{ex}\frac{1}{2}d\{\ln (1+2m\mu )\}.  \label{c26}
\end{equation}

Now, the above solutions represent a single excitation at the origin of a
FRW negative curvature universe expanding at light speed. In order to extend
this model to the case in which there are, for example, N excitations of
EW-type, let the metric go to a multicenter form: \cite{hawking} 
\begin{equation}
f^{2}\rightarrow -\frac{\epsilon }{\xi _{0}^{2}}\sum_{i=1}^{N}\frac{%
(1+2m_{i}\mu _{i})}{\tau _{i}^{2}}=\sum_{i=1}^{N}f_{i}^{2}  \label{c-28}
\end{equation}
where $\mu _{i}=\mathsf{r}_{i}/(\mathsf{r}_{i}^{2}+J_{i}^{2}\cos ^{2}\theta
_{i})$, \textsf{r}$_{i}=\left| \vec{r}-\vec{r}_{i}\right| $, the C-type
coordinates of the i$^{th}$excitation being $\vec{r}_{i}=(r_{i},\theta
_{i},\phi _{i})$. Obviously the parameters $m_{i}$ and $J_{i}$ pertain to
the i$^{th}$ excitation and $\tau _{i}^{2}=t^{2}-\mathsf{r}_{i}^{2}$. Then 
\begin{equation}
\Sigma \rightarrow \left( \sum_{i=1}^{N}\frac{m_{i}\mathsf{r}_{i}}{\tau
_{i}^{2}}\right) \left( \sum_{i=1}^{N}\frac{m_{i}\mu _{i}}{\tau _{i}^{2}}%
\right) ^{-1},  \label{c-29}
\end{equation}
giving $\Sigma \rightarrow \Sigma _{i}=\mathsf{r}_{i}^{2}+J_{i}^{2}\cos
^{2}\theta _{i}$ as $\tau _{i}^{2}\rightarrow 0.$

For evaluation of integrals with $f_{i}^{2}$ in the integrand, poles due to $%
f_{i}^{2}$ may again be handled in the standard way by letting 
\begin{equation}
\frac{1}{(t^{2}-\mathsf{r}_{i}^{2})}\rightarrow i\pi \delta (t^{2}-\mathsf{r}%
_{i}^{2})\rightarrow i\pi S(t)\frac{\delta (t-\mathsf{r}_{i})}{\mathsf{r}_{i}%
}\rightarrow -\pi ^{2}S(t)\delta (t-\mathsf{r}_{i})\delta (\mathsf{r}_{i}).
\label{c-34}
\end{equation}
Thus, near the origin (for the $i^{th}$-particle at the origin) 
\begin{equation}
f_{i}^{2}d\Omega =\frac{\pi ^{2}\epsilon }{\xi _{0}^{2}}S(t)\delta
(t-r)\delta (r)(1+2m_{i}\mu )\Sigma \sin \theta dtdrd\theta d\phi ,
\label{c-37}
\end{equation}
and, since $-\delta (r)/\tilde{u}\rightarrow \delta (r)/dt$ and $\delta
(0^{+})=1/\pi \epsilon \xi _{0}$, 
\[
f_{i}^{2}dV=\frac{f_{i}^{2}d\Omega }{-\tilde{u}}\rightarrow \frac{2\pi
^{3}\epsilon }{\xi _{0}^{2}}S(t)\delta (t)\delta (r)\left( \Sigma
+2m_{i}r\right) \sin \theta drd\theta 
\]
\begin{equation}
=\frac{2\pi ^{2}}{\xi _{0}}\left( \frac{\Sigma }{\xi _{0}^{2}}+\frac{m_{i0}r%
}{\pi }\right) \delta (r)\sin \theta drd\theta .  \label{c-38}
\end{equation}

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