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\author{\copyright\ Copyright 2002--2005 David E. Rutherford \\
All Rights Reserved \\ \\
E-mail: \\
http://www.softcom.net/users/der555/quantum.pdf}
\title{Proposed Quantum Mechanical Connection}
\date{February 21, 2005}
\begin{document}
\maketitle
This is a proposed connection between my theory \cite{mypaper1}
and relativistic quantum mechanics (RQM). In order to emphasize
the relationship between the two, I have attempted to derive the
major equations of RQM (Klein-Gordon, Dirac, Proca) from my
equations. Note, however, that my knowledge of quantum mechanics
is nominal. I have learned only enough to attempt to show a
possible relationship between my equations and those of RQM. Also,
the equations I've derived here from my theory are meant only to
show possible similarities between my theory and RQM. They do not
necessarily represent the way in which my theory best deals with
quantum mechanical situations.
The scalar electric potential $\phi$ \cite{mypaper2} at the event
$P(x, y, z, t)$, or $P$, due to a stationary `point' charge $q$ at
the origin, in SI units is
\begin{equation}\label{1}
\phi = \frac{1}{4\pi\epsilon_0}\frac{q}{s}
\end{equation}
where $s$ is the spacetime interval \cite{mypaper3} from $P$ to
the origin
\begin{equation}\label{2}
s = \sqrt{x^2 + y^2 + z^2 + c^2 t^2}
\end{equation}
where $c$ is the speed of light in vacuo. For a stationary charge
density $\rho$, we have the four-dimensional Poisson's equation
\cite{mypaper4}
\begin{equation}\label{3}
\nabla^2\phi + \frac{\partial^2\phi}{c^2\partial t^2} =
-\frac{\rho}{\epsilon_0}
\end{equation}
where
\begin{equation}
\nabla^2 = \frac{\partial^2}{\partial x^2} +
\frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}
\end{equation}
Setting $r^2 = x^2 + y^2 + z^2$, where $r$ is the spatial distance
between $P$ and the origin, and $a^2 = c^2 t^2$ in (\ref{2}),
where $a$ is an unspecified (nonzero) constant, we can write
(\ref{1}) as
\begin{equation}\label{4}
\phi = \frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{r^2 + a^2}}
\end{equation}
I intend now to show that (\ref{4}) satisfies (\ref{3}). Inserting
(\ref{4}) into the left-hand side of (\ref{3}), we have, for the
first term,
\begin{align}\label{5}
\nabla^2\phi =
\,&\frac{q}{4\pi\epsilon_0}\nabla^2\left(\frac{1}{\sqrt{r^2 +
a^2}}\right)\\ =
\,&-\frac{q}{4\pi\epsilon_0}\left(\frac{3a^2}{(r^2 +
a^2)^{5/2}}\right) \nonumber
\end{align}
and for the second term,
\begin{equation}\label{6}
\frac{\partial^2\phi}{c^2\partial t^2} = 0
\end{equation}
Adding (\ref{5}) and (\ref{6}) we have
\begin{equation}\label{7}
\nabla^2\phi + \frac{\partial^2\phi}{c^2\partial t^2} =
-\frac{q}{4\pi\epsilon_0}\left(\frac{3a^2}{(r^2 +
a^2)^{5/2}}\right)
\end{equation}
Equation (\ref{7}) is well-behaved and, in general, nonzero
everywhere.
Now, in order to show that (\ref{4}) satisfies (\ref{3}), we must
show that the right-hand sides of (\ref{7}) and (\ref{3}) are
equivalent, in other words, we must show that
\begin{equation}\label{7a}
-\frac{q}{4\pi\epsilon_0}\left(\frac{3a^2}{(r^2 +
a^2)^{5/2}}\right) = -\frac{\rho}{\epsilon_0}
\end{equation}
To do this, we start by noting that integrating $\rho dV$ over all
space, where $dV$ is the volume element, must result in the
magnitude of the charge $q$. Solving (\ref{7a}) for $\rho$, we get
\begin{equation}\label{7b}
\rho = \frac{3a^2q}{4\pi(r^2 + a^2)^{5/2}}
\end{equation}
In order to show that (\ref{7a}) is true, we must be able to
substitute the right-hand side of (\ref{7b}) for $\rho$ in our
integration of $\rho dV$, and obtain the result $q$. Carrying out
the integration, using (\ref{7b}) and the volume element $dV =
4\pi r^2dr$, we find
\begin{equation}\label{7c}
\int_\textrm{all space}{\rho dV} =
\int_0^\infty{\frac{3a^2q}{4\pi(r^2 + a^2)^{5/2}}\,4\pi r^2dr} = q
\end{equation}
showing that (\ref{4}) does indeed satisfy (\ref{3}). Note that,
since the right-hand side of (\ref{7b}) is nonzero everywhere, the
charge density $\rho$ on the left-hand side of (\ref{7b}) must,
also, be nonzero everywhere.
At $r = 0$, (\ref{7}) becomes
\begin{align}\label{8}
\nabla^2\phi + \frac{\partial^2\phi}{c^2\partial t^2} =
\,&-\frac{3q}{4\pi\epsilon_0 a^3}\\
= \,&-\frac{3}{a^2}\,\frac{q}{4\pi\epsilon_0 a} \nonumber
\end{align}
On the right-hand side of (\ref{8}), we see that $q/4\pi\epsilon_0
a$ is equivalent to the scalar potential $\phi$ from (\ref{4}) at
$r = 0$. Thus, setting $k^2 = 3/a^2$, we can write the right-hand
side of (\ref{8}) as
\begin{equation}\label{11}
-\frac{3}{a^2}\,\frac{q}{4\pi\epsilon_0 a} = -k^2\phi
\end{equation}
for $a > 0$. Using (\ref{11}), we can now write (\ref{8})
as
\begin{equation}\label{12} \nabla^2\phi +
\frac{\partial^2\phi}{c^2\partial t^2} = -k^2\phi
\end{equation}
or
\begin{equation}\label{14}
\left(\partial^2 + k^2\right)\phi = 0
\end{equation}
where
\begin{equation}\label{15}
\partial^2 = \nabla^2 + \frac{\partial^2}{c^2\partial t^2}
\end{equation}
with general solution
\begin{equation}\label{13}
\phi = A\,\textrm{cos}({\bf{k}} \cdot {\bf{x}}) +
B\,\textrm{sin}({\bf{k}} \cdot {\bf{x}})
\end{equation}
where $A$ and $B$ are arbitrary constants, ${\bf{k}} =
(k_x,k_y,k_z,k_t)$, ${\bf{x}} = (x,y,z,ct)$, and $k^2 = k_x^2 +
k_y^2 + k_z^2 + k_t^2$.\footnote{An equally valid solution to
(\ref{12}) is $\phi = A\,\textrm{exp}({\bf{k}} \cdot {\bf{x}}) +
B\,\textrm{exp}(-{\bf{k}} \cdot {\bf{x}})$ with the basis vectors
${\bf{e}}_\mu$ \cite{mypaper5} satisfying the relations
${\bf{e}}_\mu{\bf{e}}_\nu = -{\bf{e}}_4$ for $\mu = \nu$, where
$\mu, \nu = 1, 2, 3, 4$.}
If $a < 0$, (\ref{12}) becomes
\begin{equation}\label{16}
\nabla^2\phi + \frac{\partial^2\phi}{c^2\partial t^2} = k^2\phi
\end{equation}
or
\begin{equation}\label{14a}
\left(\partial^2 - k^2\right)\phi = 0
\end{equation}
with general solution
\begin{equation}\label{17}
\phi = C\,\textrm{exp}({\bf{k}} \cdot {\bf{x}}) +
D\,\textrm{exp}(-{\bf{k}} \cdot {\bf{x}})
\end{equation}
where $C$ and $D$ are arbitrary constants.\footnote{Here, in
(\ref{17}), the basis vectors ${\bf{e}}_\mu$ \cite{mypaper5}
satisfy the relations ${\bf{e}}_\mu{\bf{e}}_\nu = {\bf{e}}_4$ for
$\mu = \nu$, where $\mu, \nu = 1, 2, 3, 4$.}
We can form still another solution to (\ref{16}) if we multiply
the two solutions (\ref{13}) and (\ref{17}) together. In order to
differentiate between the two, I will refer to solution (\ref{13})
as $\phi_+$, since $a > 0$, and solution (\ref{17}) as $\phi_-$,
since $a < 0$. Note that the ${\bf{k}}$'s in (\ref{13}) and
(\ref{17}) need not be equivalent. Allowing for this possibility,
I will replace the ${\bf{k}}$ associated with (\ref{17}) with
${\bf{p}} = (p_x,p_y,p_z,p_t)$, where $p^2 = p_x^2 + p_y^2 + p_z^2
+ p_t^2$. The combined potential $\phi$ is, thus,
\begin{equation}\label{18}
\phi = \phi_+\phi_- = (A\,\textrm{cos}({\bf{k}} \cdot {\bf{x}}) +
B\,\textrm{sin}({\bf{k}} \cdot {\bf{x}}))\,
(C\,\textrm{exp}({\bf{p}} \cdot {\bf{x}}) +
D\,\textrm{exp}(-{\bf{p}} \cdot {\bf{x}}))
\end{equation}
Expanding the right-hand side of (\ref{18}) produces the sum $\phi
= \phi_1 + \phi_2 + \phi_3 + \phi_4$, where
\begin{align}\label{19}
\phi_1 &= A_1\,\textrm{exp}({\bf{p}} \cdot {\bf{x}})\,
\textrm{cos}({\bf{k}} \cdot {\bf{x}}) \\
\phi_2 &= A_2\,\textrm{exp}({\bf{p}} \cdot {\bf{x}})\,
\textrm{sin}({\bf{k}} \cdot {\bf{x}}) \nonumber\\
\phi_3 &= A_3\,\textrm{exp}(-{\bf{p}}
\cdot {\bf{x}})\, \textrm{cos}({\bf{k}} \cdot {\bf{x}}) \nonumber\\
\phi_4 &= A_4\,\textrm{exp}(-{\bf{p}} \cdot {\bf{x}})
\,\textrm{sin}({\bf{k}} \cdot {\bf{x}}) \nonumber
\end{align}
and $A_1 = AC$, $A_2 = BC$, $A_3 = AD$ and $A_4 = BD$. If we set
$A = B = C = D = 1$ and insert (\ref{18}) into the left-hand side
of (\ref{16}), we get
\begin{equation}\label{16a}
\nabla^2\phi + \frac{\partial^2\phi}{c^2\partial t^2} =
\left({\bf{p}}^2 - {\bf{k}}^2\right)\phi
\end{equation}
Then, due to the multiplication rules for the orthonormal basis
vectors ${\bf{e}}_\mu$ \cite{mypaper5}, we can write
$\left({\bf{p}}^2 - {\bf{k}}^2\right) = \left({\bf{p}} -
{\bf{k}}\right)^2$ so that (\ref{16a}) becomes
\begin{equation}\label{16b}
\nabla^2\phi + \frac{\partial^2\phi}{c^2\partial t^2} =
\left({\bf{p}} - {\bf{k}}\right)^2\phi
\end{equation}
Therefore, (\ref{18}) is also a solution to (\ref{16}) with $k =
{\bf{p}} - {\bf{k}}$. Furthermore, it can be shown that $k =
{\bf{p}} + {\bf{k}}$, $k = -{\bf{p}} - {\bf{k}}$, and $k =
-{\bf{p}} + {\bf{k}}$ also satisfy $k^2 = {\bf{p}}^2 - {\bf{k}}^2$
and, therefore, $k^2 = ({\bf{p}} - {\bf{k}})^2$. Note, also, that
if ${\bf{p}} < {\bf{k}}$, then $k^2 = {\bf{p}}^2 - {\bf{k}}^2 <
0$, thus, (\ref{18}) is also a solution to (\ref{12}). This would
seem to suggest that whether $k^2$ is positive, negative, or zero
depends on whether ${\bf{p}}$ is greater than, less than, or equal
to ${\bf{k}}$, respectively.
Evidently, from (\ref{18}), a particle is \emph{not} a localized
entity, in the classical sense (i.e., it's not a little round
ball) - it extends over all space. It does, however, have a
localized energy maximum (or minimum) at $r = 0$. The localized
energy maximum at $r = 0$ is associated with the
$\textrm{exp}(-{\bf{p}} \cdot {\bf{x}})$ terms, and the localized
energy minimum is associated with the $\textrm{exp}({\bf{p}} \cdot
{\bf{x}})$ terms. I hesitate, however, to disregard the
$\textrm{exp}({\bf{p}} \cdot {\bf{x}})$ terms on the basis that
they `blow up' as $r \rightarrow \infty$. I suspect they describe
physically valid quantities - possibly antiparticles. The terms
$\textrm{cos}({\bf{k}} \cdot {\bf{x}})$ and $\textrm{sin}({\bf{k}}
\cdot {\bf{x}})$ may describe the field of the particle or
antiparticle described by $\textrm{exp}(-{\bf{p}} \cdot {\bf{x}})$
or $\textrm{exp}({\bf{p}} \cdot {\bf{x}})$.\footnote{In my paper
\cite{mypaper1} the field of a particle (antiparticle) travelling
`forward' in time is its associated antiparticle (particle)
travelling (relatively) `backward' in time, thus, the
particle/field `pair' form an electric `dipole', since every
charged particle is accompanied by its field \cite{mypaper6}. In
this case, $\phi_+ = A\,\textrm{cos}({\bf{k}} \cdot {\bf{x}}) +
B\,\textrm{sin}({\bf{k}} \cdot {\bf{x}})$ in (\ref{18}) might
better be replaced with $\phi_+ = A\,\textrm{exp}({\bf{k}} \cdot
{\bf{x}}) + B\,\textrm{exp}(-{\bf{k}} \cdot {\bf{x}})$ with the
basis vectors ${\bf{e}}_\mu$ \cite{mypaper5} satisfying the
relations ${\bf{e}}_\mu{\bf{e}}_\nu = -{\bf{e}}_4$ for $\mu =
\nu$, where $\mu, \nu = 1, 2, 3, 4$. In addition, if $k = 0$ in
(\ref{12}) or (\ref{16}), or if ${\bf{p}} - {\bf{k}} = 0$ in
(\ref{16b}), the implication is that there are no electric
monopoles.}
It would seem natural to attribute the sinusoidal parts of
(\ref{18}) to the wave aspect, and the exponential parts to the
particle aspect, of matter. In order to make the correspondence
with the equations of RQM, I will associate ${\bf{p}}$ with the
momentum four-vector ${\bf{p}} = (p_x,p_y,p_z,E/c)$ (in which case
we would need to divide by $\hbar$) and ${\bf{k}}$ with the wave
four-vector ${\bf{k}} = (k_x,k_y,k_z,\omega/c)$.
Using these definitions for ${\bf{p}}$ and ${\bf{k}}$, if we set
$k = ({\bf{p}}/\hbar) - {\bf{k}} = 0$, after rearranging, we
arrive at
\begin{equation}\label{21a}
{\bf{p}} = \hbar{\bf{k}}
\end{equation}
which contains the Planck-Einstein-de Broglie
relations.\footnote{Thus the Planck-Einstein-de Broglie relations
imply no electric monopoles.} Analogously, if $k =
({\bf{p}}/\hbar) - {\bf{k}}
0$, then
\begin{equation}\label{21b}
{\bf{p}} > \hbar{\bf{k}}
\end{equation}
Combining (\ref{21a}) and (\ref{21b}), we get
\begin{equation}\label{21d}
{\bf{p}} \geq \hbar{\bf{k}}
\end{equation}
which leads to the Heisenberg uncertainty relations $\Delta
{\bf{p}} \Delta {\bf{x}} \geq \hbar$.\footnote{$\Delta {\bf{p}}
\Delta {\bf{x}}$ signifies the product of the uncertainties in the
corresponding components of ${\bf{p}}$ and ${\bf{x}}$, i.e.
$\Delta {\bf{p}} \Delta {\bf{x}} \geq \hbar$ signifies $\Delta p_x
\Delta x \geq \hbar$, $\Delta p_y \Delta y \geq \hbar$, $\Delta
p_z \Delta z \geq \hbar$, and $\Delta E \Delta t \geq \hbar$.} If
$k = ({\bf{p}}/\hbar) - {\bf{k}} < 0$, then
\begin{equation}\label{21c}
{\bf{p}} < \hbar{\bf{k}}
\end{equation}
The meaning of (\ref{21c}) is not clear at this time, however, one
implication is that $\Delta {\bf{p}} \Delta {\bf{x}} < \hbar$,
providing ${\bf{p}} < \hbar{\bf{k}}$.
If we set $k^2 = m^2c^2/\hbar^2$, in (\ref{14a}), where $m$ is the
mass of the particle, equation (\ref{14a}) becomes
\begin{equation}\label{14b}
\left(\partial^2 - \left(\frac{mc}{\hbar}\right)^2\right)\phi = 0
\end{equation}
This is the \emph{Klein-Gordon equation}, up to a sign, which
describes a particle with spin 0 and mass $m$.
If we specify that $k^2 = m^2c^2/\hbar^2 = 3/a^2$, the value of
$a$ would be $a = \pm\hbar\sqrt{3}/mc$. Note, however, that this
value for $a$ arises from the time interval $t$, \emph{not} a
spatial interval, since $a = ct$.
To get the Dirac equation, we start with (\ref{14b}), multiply by
$\hbar^2$, then, due to the multiplication rules for the
orthonormal basis vectors ${\bf{e}}_\mu$ \cite{mypaper5}
(remembering that $\partial$ is a four-vector \cite{mypaper7}), we
can factor, to obtain
\begin{equation}\label{16d}
\left(\hbar\partial + mc\right)\left(\hbar\partial - mc\right)\phi
= 0
\end{equation}
Eliminating the first term in parentheses, by convention, we then
have
\begin{equation}\label{16e}
\left(\hbar\partial - mc\right)\phi = 0
\end{equation}
This is the \emph{Dirac equation}, up to a sign, which describes a
particle with spin 1/2 and mass $m$.
Equation (\ref{14}) can easily be made to describe a particle with
spin 1 and mass $m$, by simply replacing the scalar potential
$\phi$ with the potential four-vector $A_\mu$, and setting $k^2 =
m^2c^2/\hbar^2$, thus,
\begin{equation}\label{20}
\left(\partial^2 + \left(\frac{mc}{\hbar}\right)^2\right)A_\mu = 0
\end{equation}
This is the \emph{Proca equation}, up to a sign, where the term
$\partial_\nu(\partial_\mu A_\nu)$ from Maxwell's equations has
been cancelled by the time component of the electric field
\cite{mypaper8}, thus it is unnecessary to invoke the Lorenz
condition $\partial_\nu A_\nu = 0$. The free photon is described
by setting $m = 0$ in (\ref{20}).
We can incorporate angular momentum and spin \cite{mypaper9} by
replacing $\phi$, $\phi_+$, and $\phi_-$ in (\ref{18}) with
\begin{equation}\label{21}
\phi' = \phi'_+\phi'_- = (A\,\textrm{cos}({\bf{k}}{\bf{x}}) +
B\,\textrm{sin}({\bf{k}}{\bf{x}}))\, (C\,\textrm{exp}({\bf{p}}
{\bf{x}} / \hbar) + D\,\textrm{exp}(-{\bf{p}}{\bf{x}} / \hbar))
\end{equation}
where ${\bf{p}}$ is the momentum four-vector, ${\bf{k}}$ is the
wave four-vector, and ${\bf{k}}{\bf{x}}$ and ${\bf{p}}{\bf{x}}$
are four-vector products \cite{mypaper10}.
I intend to elaborate on the above in the future as I extend my
knowledge of quantum mechanics.
\begin{thebibliography}{999}
\bibitem{mypaper1} See ``New Transformation
Equations and the Electric Field Four-vector'', at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper2} See ``New Transformation
Equations and the Electric Field Four-vector'', Section 11.1, at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper3} See ``New Transformation
Equations and the Electric Field Four-vector'', Section 2.2, at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper4} See ``New Transformation
Equations and the Electric Field Four-vector'', Section 12.3, at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper5} See ``New Transformation
Equations and the Electric Field Four-vector'', Section 4.1.1, at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper6} See ``New Transformation
Equations and the Electric Field Four-vector'', Conclusions, at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper7} See ``New Transformation
Equations and the Electric Field Four-vector'', Section 4.1.3, at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper8} See ``New Transformation
Equations and the Electric Field Four-vector'', Section 12.1, at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper9} See ``New Transformation
Equations and the Electric Field Four-vector'', Section 20, at
http://www.softcom.net/users/der555/newtransform.pdf.
\bibitem{mypaper10} See ``New Transformation
Equations and the Electric Field Four-vector'', Section 4.1.2, at
http://www.softcom.net/users/der555/newtransform.pdf.
\end{thebibliography}
\end{document}
--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf
"Energy Density Correction"
http://www.softcom.net/users/der555/enerdens.pdf
"Proposed Quantum Mechanical Connection"
http://www.softcom.net/users/der555/quantum.pdf
"Biot-Savart's Companion"
http://www.softcom.net/users/der555/biotcomp.pdf
.
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