| Topic: |
Science > Physics |
| User: |
"David Rutherford" |
| Date: |
04 Nov 2004 01:37:57 AM |
| Object: |
Biot-Savart's Companion (11/3/04) |
\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{graphicx}
\topmargin=-3.5pc \oddsidemargin=-1pc \textheight=55pc
\textwidth=41pc
\newcommand{\A}{\bf{A}}
\newcommand{\B}{\bf{B}}
\newcommand{\E}{\bf{E}}
\newcommand{\F}{\bf{F}}
\newcommand{\I}{\bf{I}}
\newcommand{\rhat}{\hat{\bf{r}}}
\newcommand{\vv}{\bf{v}}
\newcommand{\el}{\bf{l}}
\newcommand{\p}{\partial}
\author{\copyright\ Copyright 2004 David E. Rutherford \\
All Rights Reserved \\ \\
E-mail: \\
http://www.softcom.net/users/der555/biotcomp.pdf}
\title{\bf Biot-Savart's Companion}
\date{November 3, 2004}
\begin{document}
\maketitle
\section{The Companion Law}
The Biot-Savart law states that the element of magnetic field
$d\B$ produced by a short segment $d{\el}$ of wire of arbitrary
shape carrying a steady line current $I$, in SI units, is
\begin{equation}\label{dB}
d{\B} = {\mu_0 \over 4 \pi}\,{I d{\el} \times \rhat \over r^2}
\end{equation}
with magnitude
\begin{equation}\label{magdB}
|d{\B}| = dB = {\mu_0 \over 4 \pi}\,{I dl \sin\theta \over r^2}
\end{equation}
where $\theta$ is the angle between d{\el} and $\rhat$.
I believe the Biot-Savart law has a companion law that, to my
knowledge, has gone undiscovered until now. My companion laws to
(\ref{dB}) and (\ref{magdB}) introduce a `scalar' field $H$ whose
element $dH$ has similar form to (\ref{dB}) and
(\ref{magdB}).\footnote{See
http://www.softcom.net/users/der555/newtransform.pdf. The field
$H$ is actually part of my generalized electric field (however, it
is not referred to as $H$, there). Note that, here, I am using a
three-dimensional, non-relativistic treatment, as opposed to the
four-dimensional, relativistic treatment used at the above link.}
The companion law is derived from $\nabla \cdot {\A}$, where
${\A}$ is the vector potential due to a `point' charge $q'$ moving
with constant velocity ${\vv}'$.\footnote{$\nabla \cdot {\A}$ is
physical and, in general, nonzero in my theory.} We first note
that ${\A} = {\vv}'\phi/c^2$, where $\phi = q'/(4 \pi \epsilon_0
r)$ is the scalar electric potential, due to $q'$, at a field
point $P$ a distance $r$ from $q'$.\footnote{It is important to
note that the scalar potential $\phi$, here, is the
three-dimensional part of my four-dimensional potential at
http://www.softcom.net/users/der555/newtransform.pdf, since this
is a three-dimensional treatment.} Thus, after substitution we get
\begin{equation}\label{divA}
\nabla \cdot {\A} = \nabla \cdot \left({{\vv}'\phi \over
c^2}\right) = -\,{q' \over 4 \pi \epsilon_0 c^2}\,{{\vv}' \cdot
\rhat \over r^2}
\end{equation}
where $\rhat$ is a unit vector pointing from $q'$ to $P$.
If we refer to $\nabla \cdot {\A}$ here as $H$, and note that
$\epsilon_0\mu_0 = 1/c^2$, we can write (\ref{divA}) as
\begin{equation}\label{H1}
H = -\,{\mu_0 q' \over 4\pi}\, {{\vv}'\! \cdot \rhat \over r^2}
\end{equation}
Since ${\vv}'\! \cdot \rhat$ is a scalar quantity (in
three-dimensional space), we could also write (\ref{H1}) as
\begin{equation}\label{H2}
H = -\,{\mu_0 q' \over 4\pi}\, {v'\! \cos\theta \over r^2}
\end{equation}
where $v'$ is the magnitude of ${\vv}'$ and $\theta$ is the angle
between ${\vv}'$ and $\rhat$.
The element $dH$ at $P$ due to a `point' charge $dq'$ is
\begin{equation}\label{dH1}
dH = -\,{\mu_0 dq' \over 4 \pi}\, {{\vv}' \cdot \rhat \over r^2}
\end{equation}
or
\begin{equation}\label{dH2}
dH = -\,{\mu_0 dq' \over 4 \pi}\, {v'\! \cos\theta \over r^2}
\end{equation}
Now consider a small segment $d{\el}$ of wire carrying a steady
current $I$, within which the `point' charge $dq'$ is moving with
velocity $\vv'$ parallel to $d{\el}$. In terms of the current $I$,
substituting $dq' = I dt$ and ${\vv}' = d{\el}/dt$, (\ref{dH1})
and (\ref{dH2}) become
\begin{equation}\label{dH3}
dH = -\,{\mu_0 \over 4 \pi}\, {I d{\el} \cdot \rhat \over r^2}
\end{equation}
or
\begin{equation}\label{magdH3}
dH = -\,{\mu_0 \over 4 \pi}\, {I dl \cos\theta \over r^2}
\end{equation}
where $dl$ is the magnitude of $d{\el}$. The equations (\ref{dH3})
and (\ref{magdH3}) are my companions to (\ref{dB}) and
(\ref{magdB}), respectively.
We can find the total field $H$ at $P$ by integrating (\ref{dH3})
along the wire, so that
\begin{equation}
H = -\,{\mu_0 I \over 4 \pi} \int{{d{\el} \cdot \rhat \over r^2}}
\end{equation}
\section{Force on a Test Charge in the Field $H$}
The \emph{additional force} ${\F}_a$ on a test charge $q$ at point
$P$ moving with velocity $\vv$ in the field $H$ at $P$
is\footnote{See http://www.softcom.net/users/der555/actreact.pdf.}
\begin{equation}\label{F}
{\F}_a = -\,q{\vv}H
\end{equation}
If $H$ at $P$ is due to a `point' charge $q'$ moving with constant
velocity ${\vv}'$, we can substitute (\ref{H1}) or (\ref{H2}) into
(\ref{F}), to obtain
\begin{equation}
{\F}_a = -\,q {\vv} \left(-\,{\mu_0 q' \over 4\pi}\, {{\vv}'\!
\cdot \rhat \over r^2}\right) = {\mu_0 q q' \over 4 \pi}\, {{\vv}
\left({\vv}'\! \cdot \rhat \right) \over r^2}
\end{equation}
or
\begin{equation}
{\F}_a = {\mu_0 q q' \over 4 \pi}\, {{\vv} \left(v'\!
\cos\theta\right) \over r^2}
\end{equation}
respectively.
For a steady line current $I$, the element of force $d{\F}_a$ on
$q$ due to a short segment $d{\el}$ of wire, using (\ref{dH3}), is
\begin{equation}\label{dF}
d{\F}_a = -\,q {\vv} dH = {\mu_0 q {\vv} \over 4 \pi}\, {I d{\el}
\cdot \rhat \over r^2}
\end{equation}
The total force ${\F}_a$ on $q$ can be found by integrating
(\ref{dF}) along the wire, resulting in
\begin{equation}
{\F}_a = {\mu_0 q I {\vv} \over 4 \pi} \int{{d{\el} \cdot \rhat
\over r^2}}
\end{equation}
It is interesting to note that the magnetic force ${\F}_m = q
{\vv} \times {\B}$ changes the \emph{direction} of the velocity of
a test particle, but not the magnitude of its velocity. In
contrast, my additional force ${\F}_a = -\,q {\vv} H$ changes the
\emph{magnitude} of the velocity, but not its direction. These two
forces go hand-in-hand, along with the electric force ${\F}_e = q
{\E}$, to complete the three-dimensional, non-relativistic
equations of motion of a test particle.\footnote{See
http://www.softcom.net/users/der555/actreact.pdf.}
These equations predict new physics, but the most interesting
prediction involves the time component of the force. Analyzing the
time component of the force is outside of the three-dimensional
treatment here, however due to its implications I would like to
mention it.
The time component of the force on $q$, due to $q'$,
is\footnote{See http://www.softcom.net/users/der555/actreact.pdf.}
\begin{equation}
F_t = q\left(\frac{1}{c}\,{\bf{v}} \cdot {\E} + c\,\nabla \cdot
{\A}\right)
\end{equation}
The last term is the time component of the additional force
\begin{equation}
F_{ta} = qc\,\nabla \cdot {\A}
\end{equation}
or
\begin{equation}\label{Ft}
F_{ta} = qcH
\end{equation}
Substituting (\ref{H1}) into (\ref{Ft}), we obtain
\begin{equation}\label{Fta}
F_{ta} = -\,{\mu_0 q q' c \over 4\pi}\, {{\vv}'\! \cdot \rhat
\over r^2}
\end{equation}
For a steady line current $I$, using the same methods as above, we
find that the element of force $dF_{ta}$ on $q$ due to a short
segment $d{\el}$ of wire, is
\begin{equation}\label{dFta}
dF_{ta} = -\,{\mu_0 q c \over 4 \pi}\, {I d{\el} \cdot \rhat \over
r^2}
\end{equation}
The total time component of the additional force $F_{ta}$ on $q$
can be found by integrating (\ref{dFta}) along the wire, to obtain
\begin{equation}\label{Ftint}
F_{ta} = -\,{\mu_0 q c I \over 4 \pi} \int{{d{\el} \cdot \rhat
\over r^2}}
\end{equation}
If we extend Newton's Second Law ${\F} = m {\bf{a}}$ to include
the time component of the force $F_t$, that is $F_t = m a_t$,
where $a_t$ is the time component of the acceleration, then we can
write the time component of the additional force as $F_{ta} = m
a_{ta}$, where $a_{ta}$ is the acceleration due to the time
component of the additional force $F_{ta}$.\footnote{There are
terms in addition to the time rate of change of four-momentum in
my four-dimensional, relativistic treatment at
http://www.softcom.net/users/der555/newtransform.pdf. These terms
are considered to be relativistic, however, and are not considered
here.} Thus we can now write the time component of the
acceleration $a_{ta}$ of $q$, due to $F_{ta}$, as
\begin{equation}\label{at}
a_{ta} = -\,{\mu_0 q c I \over 4 \pi m} \int{{d{\el} \cdot \rhat
\over r^2}}
\end{equation}
where $m$ is the mass of $q$.
Note that $c$ in $F_{ta}$ and $dF_{ta}$ is the time component of
the velocity of $q$, thus $q$ is \emph{stationary} in
space.\footnote{I'm focusing, here, on the case where $q$ is
stationary in $F_t$. There might also be an acceleration in time
due to $(q/c)\,{\bf{v}} \cdot {\E}$ if $q$ is not stationary in
space.} Also note that, for a nonzero $F_{ta}$ (or $dF_{ta}$),
there is a \emph{nonzero} acceleration in time. In other words,
(\ref{at}) describes a means of transporting $q$ in time.
\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
.
|
|
| User: "Dave" |
|
| Title: Re: Biot-Savart's Companion (11/3/04) |
05 Nov 2004 06:10:07 AM |
|
|
is there anything in here worth finding a tool to make this crud readable?
on a quick scan it looks like old already discreditted stuff just going
around in a new wrapper.
"David Rutherford" <> wrote in message
news:VLGdnY0sm7TPQRTcRVn-rQ@softcom.net...
\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{graphicx}
\topmargin=-3.5pc \oddsidemargin=-1pc \textheight=55pc
\textwidth=41pc
\newcommand{\A}{\bf{A}}
\newcommand{\B}{\bf{B}}
\newcommand{\E}{\bf{E}}
\newcommand{\F}{\bf{F}}
\newcommand{\I}{\bf{I}}
\newcommand{\rhat}{\hat{\bf{r}}}
\newcommand{\vv}{\bf{v}}
\newcommand{\el}{\bf{l}}
\newcommand{\p}{\partial}
\author{\copyright\ Copyright 2004 David E. Rutherford \\
All Rights Reserved \\ \\
E-mail: \\
http://www.softcom.net/users/der555/biotcomp.pdf}
\title{\bf Biot-Savart's Companion}
\date{November 3, 2004}
\begin{document}
\maketitle
\section{The Companion Law}
The Biot-Savart law states that the element of magnetic field
$d\B$ produced by a short segment $d{\el}$ of wire of arbitrary
shape carrying a steady line current $I$, in SI units, is
\begin{equation}\label{dB}
d{\B} = {\mu_0 \over 4 \pi}\,{I d{\el} \times \rhat \over r^2}
\end{equation}
with magnitude
\begin{equation}\label{magdB}
|d{\B}| = dB = {\mu_0 \over 4 \pi}\,{I dl \sin\theta \over r^2}
\end{equation}
where $\theta$ is the angle between d{\el} and $\rhat$.
I believe the Biot-Savart law has a companion law that, to my
knowledge, has gone undiscovered until now. My companion laws to
(\ref{dB}) and (\ref{magdB}) introduce a `scalar' field $H$ whose
element $dH$ has similar form to (\ref{dB}) and
(\ref{magdB}).\footnote{See
http://www.softcom.net/users/der555/newtransform.pdf. The field
$H$ is actually part of my generalized electric field (however, it
is not referred to as $H$, there). Note that, here, I am using a
three-dimensional, non-relativistic treatment, as opposed to the
four-dimensional, relativistic treatment used at the above link.}
The companion law is derived from $\nabla \cdot {\A}$, where
${\A}$ is the vector potential due to a `point' charge $q'$ moving
with constant velocity ${\vv}'$.\footnote{$\nabla \cdot {\A}$ is
physical and, in general, nonzero in my theory.} We first note
that ${\A} = {\vv}'\phi/c^2$, where $\phi = q'/(4 \pi \epsilon_0
r)$ is the scalar electric potential, due to $q'$, at a field
point $P$ a distance $r$ from $q'$.\footnote{It is important to
note that the scalar potential $\phi$, here, is the
three-dimensional part of my four-dimensional potential at
http://www.softcom.net/users/der555/newtransform.pdf, since this
is a three-dimensional treatment.} Thus, after substitution we get
\begin{equation}\label{divA}
\nabla \cdot {\A} = \nabla \cdot \left({{\vv}'\phi \over
c^2}\right) = -\,{q' \over 4 \pi \epsilon_0 c^2}\,{{\vv}' \cdot
\rhat \over r^2}
\end{equation}
where $\rhat$ is a unit vector pointing from $q'$ to $P$.
If we refer to $\nabla \cdot {\A}$ here as $H$, and note that
$\epsilon_0\mu_0 = 1/c^2$, we can write (\ref{divA}) as
\begin{equation}\label{H1}
H = -\,{\mu_0 q' \over 4\pi}\, {{\vv}'\! \cdot \rhat \over r^2}
\end{equation}
Since ${\vv}'\! \cdot \rhat$ is a scalar quantity (in
three-dimensional space), we could also write (\ref{H1}) as
\begin{equation}\label{H2}
H = -\,{\mu_0 q' \over 4\pi}\, {v'\! \cos\theta \over r^2}
\end{equation}
where $v'$ is the magnitude of ${\vv}'$ and $\theta$ is the angle
between ${\vv}'$ and $\rhat$.
The element $dH$ at $P$ due to a `point' charge $dq'$ is
\begin{equation}\label{dH1}
dH = -\,{\mu_0 dq' \over 4 \pi}\, {{\vv}' \cdot \rhat \over r^2}
\end{equation}
or
\begin{equation}\label{dH2}
dH = -\,{\mu_0 dq' \over 4 \pi}\, {v'\! \cos\theta \over r^2}
\end{equation}
Now consider a small segment $d{\el}$ of wire carrying a steady
current $I$, within which the `point' charge $dq'$ is moving with
velocity $\vv'$ parallel to $d{\el}$. In terms of the current $I$,
substituting $dq' = I dt$ and ${\vv}' = d{\el}/dt$, (\ref{dH1})
and (\ref{dH2}) become
\begin{equation}\label{dH3}
dH = -\,{\mu_0 \over 4 \pi}\, {I d{\el} \cdot \rhat \over r^2}
\end{equation}
or
\begin{equation}\label{magdH3}
dH = -\,{\mu_0 \over 4 \pi}\, {I dl \cos\theta \over r^2}
\end{equation}
where $dl$ is the magnitude of $d{\el}$. The equations (\ref{dH3})
and (\ref{magdH3}) are my companions to (\ref{dB}) and
(\ref{magdB}), respectively.
We can find the total field $H$ at $P$ by integrating (\ref{dH3})
along the wire, so that
\begin{equation}
H = -\,{\mu_0 I \over 4 \pi} \int{{d{\el} \cdot \rhat \over r^2}}
\end{equation}
\section{Force on a Test Charge in the Field $H$}
The \emph{additional force} ${\F}_a$ on a test charge $q$ at point
$P$ moving with velocity $\vv$ in the field $H$ at $P$
is\footnote{See http://www.softcom.net/users/der555/actreact.pdf.}
\begin{equation}\label{F}
{\F}_a = -\,q{\vv}H
\end{equation}
If $H$ at $P$ is due to a `point' charge $q'$ moving with constant
velocity ${\vv}'$, we can substitute (\ref{H1}) or (\ref{H2}) into
(\ref{F}), to obtain
\begin{equation}
{\F}_a = -\,q {\vv} \left(-\,{\mu_0 q' \over 4\pi}\, {{\vv}'\!
\cdot \rhat \over r^2}\right) = {\mu_0 q q' \over 4 \pi}\, {{\vv}
\left({\vv}'\! \cdot \rhat \right) \over r^2}
\end{equation}
or
\begin{equation}
{\F}_a = {\mu_0 q q' \over 4 \pi}\, {{\vv} \left(v'\!
\cos\theta\right) \over r^2}
\end{equation}
respectively.
For a steady line current $I$, the element of force $d{\F}_a$ on
$q$ due to a short segment $d{\el}$ of wire, using (\ref{dH3}), is
\begin{equation}\label{dF}
d{\F}_a = -\,q {\vv} dH = {\mu_0 q {\vv} \over 4 \pi}\, {I d{\el}
\cdot \rhat \over r^2}
\end{equation}
The total force ${\F}_a$ on $q$ can be found by integrating
(\ref{dF}) along the wire, resulting in
\begin{equation}
{\F}_a = {\mu_0 q I {\vv} \over 4 \pi} \int{{d{\el} \cdot \rhat
\over r^2}}
\end{equation}
It is interesting to note that the magnetic force ${\F}_m = q
{\vv} \times {\B}$ changes the \emph{direction} of the velocity of
a test particle, but not the magnitude of its velocity. In
contrast, my additional force ${\F}_a = -\,q {\vv} H$ changes the
\emph{magnitude} of the velocity, but not its direction. These two
forces go hand-in-hand, along with the electric force ${\F}_e = q
{\E}$, to complete the three-dimensional, non-relativistic
equations of motion of a test particle.\footnote{See
http://www.softcom.net/users/der555/actreact.pdf.}
These equations predict new physics, but the most interesting
prediction involves the time component of the force. Analyzing the
time component of the force is outside of the three-dimensional
treatment here, however due to its implications I would like to
mention it.
The time component of the force on $q$, due to $q'$,
is\footnote{See http://www.softcom.net/users/der555/actreact.pdf.}
\begin{equation}
F_t = q\left(\frac{1}{c}\,{\bf{v}} \cdot {\E} + c\,\nabla \cdot
{\A}\right)
\end{equation}
The last term is the time component of the additional force
\begin{equation}
F_{ta} = qc\,\nabla \cdot {\A}
\end{equation}
or
\begin{equation}\label{Ft}
F_{ta} = qcH
\end{equation}
Substituting (\ref{H1}) into (\ref{Ft}), we obtain
\begin{equation}\label{Fta}
F_{ta} = -\,{\mu_0 q q' c \over 4\pi}\, {{\vv}'\! \cdot \rhat
\over r^2}
\end{equation}
For a steady line current $I$, using the same methods as above, we
find that the element of force $dF_{ta}$ on $q$ due to a short
segment $d{\el}$ of wire, is
\begin{equation}\label{dFta}
dF_{ta} = -\,{\mu_0 q c \over 4 \pi}\, {I d{\el} \cdot \rhat \over
r^2}
\end{equation}
The total time component of the additional force $F_{ta}$ on $q$
can be found by integrating (\ref{dFta}) along the wire, to obtain
\begin{equation}\label{Ftint}
F_{ta} = -\,{\mu_0 q c I \over 4 \pi} \int{{d{\el} \cdot \rhat
\over r^2}}
\end{equation}
If we extend Newton's Second Law ${\F} = m {\bf{a}}$ to include
the time component of the force $F_t$, that is $F_t = m a_t$,
where $a_t$ is the time component of the acceleration, then we can
write the time component of the additional force as $F_{ta} = m
a_{ta}$, where $a_{ta}$ is the acceleration due to the time
component of the additional force $F_{ta}$.\footnote{There are
terms in addition to the time rate of change of four-momentum in
my four-dimensional, relativistic treatment at
http://www.softcom.net/users/der555/newtransform.pdf. These terms
are considered to be relativistic, however, and are not considered
here.} Thus we can now write the time component of the
acceleration $a_{ta}$ of $q$, due to $F_{ta}$, as
\begin{equation}\label{at}
a_{ta} = -\,{\mu_0 q c I \over 4 \pi m} \int{{d{\el} \cdot \rhat
\over r^2}}
\end{equation}
where $m$ is the mass of $q$.
Note that $c$ in $F_{ta}$ and $dF_{ta}$ is the time component of
the velocity of $q$, thus $q$ is \emph{stationary} in
space.\footnote{I'm focusing, here, on the case where $q$ is
stationary in $F_t$. There might also be an acceleration in time
due to $(q/c)\,{\bf{v}} \cdot {\E}$ if $q$ is not stationary in
space.} Also note that, for a nonzero $F_{ta}$ (or $dF_{ta}$),
there is a \emph{nonzero} acceleration in time. In other words,
(\ref{at}) describes a means of transporting $q$ in time.
\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
.
|
|
|
|
|
Related Articles |
|
|
