Energy Density Correction (10/14/03)
David Rutherford
http://www.softcom.net/users/der555/enerdens.pdf
-----------------------------------------------
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\author{\copyright\ Copyright 2002-2003 David E. Rutherford \\
All Rights Reserved \\ \\
E-mail: druth...@softcom.net \\
http://www.softcom.net/users/der555/enerdens.pdf}
\title{Energy Density Correction}
\date{October 14, 2003}
\begin{document}
\maketitle
The energy density $u$ of a continuous charge distribution,
according to popular belief, in SI units, is
\begin{equation}\label{1}
u = \frac{\epsilon_0}{2}\,E^2
\end{equation}
where $\epsilon_0$ is the permittivity constant and $E$ is the
conventional electric field strength. I intend to show, using
conventional terminology, that this should instead be
\begin{equation}\label{2}
u = \epsilon_0 E^2
\end{equation}
In order to simplify, I'll start with the energy of a point charge
distribution. The energy of the distribution is just the work
required to assemble the distribution. Let's start with a pair of
identical charged particles, $q_1$ and $q_2$, at $+\infty$ and
$-\infty$, respectively. The work $W_{12}$ required to bring $q_1$
in from $+\infty$ to the origin against the field of $q_2$, is
\begin{equation}\label{3}
W_{12} = \frac{1}{4\pi\epsilon_0}\frac{q_1\,q_2}{r_{12}}
\end{equation}
where $(1/4\pi\epsilon_0)(q_2/r_{12})$ is the potential at the
location of $q_1$ due to $q_2$, $r_{12}$ is the distance between
$q_1$ and $q_2$ after we're through bringing in $q_1$. But, since
$q_2$ is not in the picture yet ($r_{12} = \infty$), the work
required to bring $q_1$ in is zero, i.e., $W_{12} = 0$.
The work $W_{21}$ necessary to bring in $q_2$ is
\begin{equation}\label{4}
W_{21} = \frac{1}{4\pi\epsilon_0}\frac{q_2\,q_1}{r_{21}}
\end{equation}
However, as we bring in $q_2$, we must do work on $q_1$ against
the field of $q_2$ in order to keep it in place at the
origin.\footnote{This contradicts the conventional definition of
work in which work can only be done on a body that undergoes a
displacement, however, I believe this definition is incomplete.
Please refer to Section 19 in
http://www.softcom.net/users/der555/newtransform.pdf for my
definition of work density.} The work necessary to keep $q_1$ in
place, is the same as the work that would be required to bring
$q_1$ in against the field of $q_2$, had we brought $q_2$ in
first. But this is the same as (\ref{3}) for $r_{12} = r_{21}$. So
the total magnitude of the work $W$ required to assemble the two
particles is
\begin{equation}\label{5}
W = W_{12} + W_{21} =
\frac{1}{4\pi\epsilon_0}\left(\frac{q_1\,q_2}{r_{12}} +
\frac{q_2\,q_1}{r_{21}}\right)
\end{equation}
or
\begin{equation}\label{6}
W = \frac{1}{4\pi\epsilon_0}\sum_{i=1}^2\sum_{j=1}^2\frac{q_i
q_j}{r_{ij}}
\end{equation}
for $i \neq j$.
For three particles, we have to add the work required to bring in
a third particle $q_3$ from infinity against the fields of both
$q_1$ and $q_2$
\begin{equation}\label{7}
W_{31} + W_{32} =
\frac{1}{4\pi\epsilon_0}\left(\frac{q_3\,q_1}{r_{31}} +
\frac{q_3\,q_2}{r_{32}}\right)
\end{equation}
to the work required to assemble $q_1$ and $q_2$ from (\ref{5}).
But again we have to include the work required to keep $q_1$ and
$q_2$ in position as we bring in $q_3$. The work $W_{13}$ and
$W_{23}$ required to keep $q_1$ and $q_2$ in position is
\begin{equation}\label{8}
W_{13} = \frac{1}{4\pi\epsilon_0}\frac{q_1\,q_3}{r_{13}}
\end{equation}
and
\begin{equation}\label{9}
W_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2\,q_3}{r_{23}}
\end{equation}
respectively. So the total work $W$ required to assemble the three
particles, from (\ref{5}), (\ref{7}), (\ref{8}) and (\ref{9}), is
\begin{equation}\label{10}
W = W_{12} + W_{13} + W_{21} + W_{23} + W_{31} + W_{32}
\end{equation}
or
\begin{equation}\label{11}
W = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1\,q_2}{r_{12}} +
\frac{q_1\,q_3}{r_{13}} + \frac{q_2\,q_1}{r_{21}} +
\frac{q_2\,q_3}{r_{23}} + \frac{q_3\,q_1}{r_{31}} +
\frac{q_3\,q_2}{r_{32}}\right)
\end{equation}
which we can write as
\begin{equation}\label{12}
W = \frac{1}{4\pi\epsilon_0}\sum_{i=1}^3\sum_{j=1}^3\frac{q_i
q_j}{r_{ij}}
\end{equation}
for $i \neq j$. So for any number of particles $n$, we can
apparently write
\begin{equation}\label{13}
W = \frac{1}{4\pi\epsilon_0}\sum_{i=1}^n\sum_{j=1}^n\frac{q_i
q_j}{r_{ij}}
\end{equation}
for $i \neq j$. We can also write this as
\begin{equation}\label{14}
W = \sum_{i=1}^n q_i
\left(\sum_{j=1}^n\frac{1}{4\pi\epsilon_0}\frac{q_j}{r_{ij}}\right)
\end{equation}
for $i \neq j$. The term in brackets is the potential
$\phi({\bf{r}}_i)$ at the point ${\bf{r}}_i$ (the position of
$q_i$) due to all other charges. Thus, we can write (\ref{14}) as
\begin{equation}\label{15}
W = \sum_{i=1}^n q_i\phi({\bf{r}}_i)
\end{equation}
For a volume charge density $\rho$, (\ref{15}) becomes
\begin{equation}\label{16}
W = \int{\rho\phi\,dV}
\end{equation}
where $dV$ is the volume element. Using Poisson's equation
$\nabla^2\phi = -\rho/\epsilon_0$ we can write (\ref{16}) as
\begin{equation}\label{16a}
W = -\epsilon_0\int{\phi\nabla^2\phi\,dV}
\end{equation}
Integration by parts then leads to
\begin{equation}\label{17}
W = \epsilon_0\int{|\nabla\phi|^2\,dV}
\end{equation}
and, since $E = |\nabla\phi|$, we get
\begin{equation}\label{17a}
W = \epsilon_0\int{E^2 dV}
\end{equation}
This represents the energy stored in the electric field,
therefore, we can interpret $\epsilon_0E^2$ as the energy density
$u$, or
\begin{equation}\label{18}
u = \epsilon_0 E^2
\end{equation}
Clearly, this is in contradiction to the conventional physics
claim that the energy density is $u = (\epsilon_0/2)E^2$. The
extra energy density is due to the extra work that must be done on
the charges already assembled in order to keep them in place
against the field of each new charge as we add it to the
configuration.
\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
John C. Polasek
<druth...@softcom.net> wrote:
>You can see the pdf version of this at
>
>http://www.softcom.net/users/der555/enerdens.pdf
>
>-----------------------------------------------
>
snip
>--
>Dave Rutherford
>"New Transformation Equations and the Electric Field Four-vector"
>http://www.softcom.net/users/der555/newtransform.pdf
>
snip
>
In your pdf, just after Eq. 4, the work to hold q1 in place as we
bring up q2 is zero! It doesn't move. Don't double count the
potentials.
Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).
Uncle Al
[snip 170 lines of LaTeX drool]
Hey git - NOBODY posts LaTeX to Usenet.
There are very capable freeware and shareware LaTeX viewers. LaTeX
viewers typically install 7000-9000 files on your hard drive - most of
them being less than a kilobyte. It is an intensely effective way to
waste cluster storage space and slow response as your HD stepper
grinds away. Even ZTREE coughed trying to swallow the partition.
You'd think LaTeX was inventd by Microcrap.
After installing a LaTeX reader in its own partition, the most
effective next step is
FORMAT [parition] /u /q
If you think you have big balls re LaTeX posting, any pimply loser can
convert MS Word to LaTeX with Word2Tex,
The most recent version is particularly adept with tables. Tech
support is awesome. Remember that the LatTeX "X" is a German "h" at
the back of your throat, so you don't sound like a complete idiot
while performing like one.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
David Rutherford
John C. Polasek wrote:
>
> In your pdf, just after Eq. 4, the work to hold q1 in place as we
> bring up q2 is zero! It doesn't move. Don't double count the
> potentials.
Read the sentence that preceeds the one you are referring to.
"However, as we bring in q2, we must do work on q1 against the field
of q2 in order to keep it in place at the origin."
Now look at the little "1" in between the two lines. That number refers
to a footnote at the bottom of the page. Go to the bottom of the page
and read the footnote.
You'll find a reference to my paper at
http://www.softcom.net/users/der555/newtransform.pdf
where, in Section 19 you'll find _my_ definition of work (density), as I
stated in the footnote. My definition is a correction of the
conventional definition of work which, in my opinion, is incomplete.
--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
Old Man
news:voon28h...@corp.supernews.com...
>
>
> John C. Polasek wrote:
> >
> > In your pdf, just after Eq. 4, the work to hold q1 in place as we
> > bring up q2 is zero! It doesn't move. Don't double count the
> > potentials.
>
> Read the sentence that preceeds the one you are referring to.
>
> "However, as we bring in q2, we must do work on q1 against the field
> of q2 in order to keep it in place at the origin."
>
> Now look at the little "1" in between the two lines. That number refers
> to a footnote at the bottom of the page. Go to the bottom of the page
> and read the footnote.
>
> You'll find a reference to my paper at
>
> http://www.softcom.net/users/der555/newtransform.pdf
>
> where, in Section 19 you'll find _my_ definition of work (density), as I
> stated in the footnote. My definition is a correction of the
> conventional definition of work which, in my opinion, is incomplete.
Clearly, Polasek has found Rutherford's rotten core. [Old Man]
> Dave Rutherford