| Topic: |
Science > Physics |
| User: |
"David Rutherford" |
| Date: |
13 Nov 2003 01:46:30 PM |
| Object: |
Charge is not conserved in classical EM |
The classical EM equation for conservation of charge is
div J + @\rho/@t = 0
where J is the current density, @/@t is the partial derivative with
respect to t, and \rho is the charge density.
I will show now that this does _not_ imply conservation of charge.
Since current I can be written
I = \oint_S{J.da} (1)
for a closed surface. But
I = dq/dt = (d/dt)\int_V{\rho dV} = \int_V{(@\rho/@t)dV} (2)
and
\oint_S{J.da} = \int_V{(div J)dV} (3)
so, plugging (2) and (3) into (1) and rearranging terms, we have
\int_V{(div J)dV} = \int_V{(@\rho/@t)dV}
Since this applies to any volume, we can say
div J = @\rho/@t
or
div J - @\rho/@t = 0
This is the _correct_ continuity equation for local charge conservation.
It is _not_ the same as the classical continuity equation, which is
div J + @\rho/@t = 0
therefore, the classical continuity equation does _not_ result in local
conservation of charge.
Fortunately, my theory
http://www.softcom.net/users/der555/newtransform.pdf
gives div J - @\rho/@t = 0, so it results in local conservation of charge.
--
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
.
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| User: "Mark" |
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| Title: Re: Charge is not conserved in classical EM |
14 Nov 2003 03:56:38 PM |
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David Rutherford <drutherford@softcom.net> writes:
Since current I can be written
I = \oint_S{J.da} (1)
for a closed surface.
I = dq/dt = (d/dt)\int_V{\rho dV} = \int_V{(@\rho/@t)dV} (2)
I = -dq/dt, since da points OUTward from the closed surface; while
q is how much charges is IN the region.
The outflux of current from a closed region DEcreases the charge
in the region, not INcreases it. So, it's I = -dq/dt, or
more precisely:
I(dV) = -dq(V)/dt
where I(dV) is the current along the boundary dV of region V.
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| User: "Randy Poe" |
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| Title: Re: Charge is not conserved in classical EM |
13 Nov 2003 03:56:05 PM |
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David Rutherford <drutherford@softcom.net> wrote in message news:<vr7nlcoffvgqf6@corp.supernews.com>...
The classical EM equation for conservation of charge is
div J + @\rho/@t = 0
where J is the current density, @/@t is the partial derivative with
respect to t, and \rho is the charge density.
I will show now that this does _not_ imply conservation of charge.
Since current I can be written
I = \oint_S{J.da} (1)
for a closed surface. But
I = dq/dt
Wait a minute. Suppose you had a net flow of current
out of a volume, e.g. J pointing out everywhere, so
J.da is positive over the whole surface. Then clearly
the amount of charge inside the volume is decreasing,
not increasing. That is, the surface integral is
positive if and only if the net dq/dt inside the
volume is negative.
Maxwell's right, you're wrong.
I think you're simultaneously mixing and matching several
different meanings of I and dq/dt.
- Randy
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
14 Nov 2003 03:15:35 PM |
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Randy Poe wrote:
David Rutherford <drutherford@softcom.net> wrote in message news:<vr7nlcoffvgqf6@corp.supernews.com>...
The classical EM equation for conservation of charge is
div J + @\rho/@t = 0
where J is the current density, @/@t is the partial derivative with
respect to t, and \rho is the charge density.
I will show now that this does _not_ imply conservation of charge.
Since current I can be written
I = \oint_S{J.da} (1)
for a closed surface. But
I = dq/dt
Wait a minute. Suppose you had a net flow of current
out of a volume, e.g. J pointing out everywhere, so
J.da is positive over the whole surface. Then clearly
the amount of charge inside the volume is decreasing,
not increasing. That is, the surface integral is
positive if and only if the net dq/dt inside the
volume is negative.
Maxwell's right, you're wrong.
I think you're simultaneously mixing and matching several
different meanings of I and dq/dt.
Which of my equations requires a sign change, in your opinion. After you
decide, email all of the math and physics textbook publishers and tell
them they need to rewrite their books to reflect your change of definition.
--
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
.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Charge is not conserved in classical EM |
17 Nov 2003 08:51:43 AM |
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David Rutherford wrote:
Randy Poe wrote:
David Rutherford <drutherford@softcom.net> wrote in message news:<vr7nlcoffvgqf6@corp.supernews.com>...
The classical EM equation for conservation of charge is
div J + @\rho/@t = 0
where J is the current density, @/@t is the partial derivative with
respect to t, and \rho is the charge density.
I will show now that this does _not_ imply conservation of charge.
Since current I can be written
I = \oint_S{J.da} (1)
for a closed surface. But
I = dq/dt
Wait a minute. Suppose you had a net flow of current
out of a volume, e.g. J pointing out everywhere, so
J.da is positive over the whole surface. Then clearly
the amount of charge inside the volume is decreasing,
not increasing. That is, the surface integral is
positive if and only if the net dq/dt inside the
volume is negative.
Maxwell's right, you're wrong.
I think you're simultaneously mixing and matching several
different meanings of I and dq/dt.
Which of my equations requires a sign change, in your opinion.
The last one above. It should read
I = - dq/dt.
After you
decide, email all of the math and physics textbook publishers and tell
them they need to rewrite their books to reflect your change of definition.
They already have it right. It's simpy you who can't read correctly. You
have to be careful to which current (which direction) and which charge
the formula is referring.
Bye,
Bjoern
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| User: "Randy Poe" |
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| Title: Re: Charge is not conserved in classical EM |
16 Nov 2003 07:46:59 PM |
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On Fri, 14 Nov 2003 13:15:35 -0800, David Rutherford
<drutherford@softcom.net> wrote:
Randy Poe wrote:
I = dq/dt
Wait a minute. Suppose you had a net flow of current
out of a volume, e.g. J pointing out everywhere, so
J.da is positive over the whole surface. Then clearly
the amount of charge inside the volume is decreasing,
not increasing. That is, the surface integral is
positive if and only if the net dq/dt inside the
volume is negative.
Maxwell's right, you're wrong.
I think you're simultaneously mixing and matching several
different meanings of I and dq/dt.
Which of my equations requires a sign change, in your opinion.
Perhaps the one where I said "wait a minute" and pointed out to you
how the signs were inconsistent in the way you were interpreting "I"
and "dq/dt". Did you actually read what I wrote? Or what anybody else
wrote? You quoted it. The answer to your question is above. If I is
positive, it means charge is flowing out of the volume. It means q is
decreasing. Therefore using "I" to mean charge flowing out of the
volume and "q" meaning charge in the volume is not consistent with "I
= dq/dt". Which you'd know if you read my post, as that was the entire
content of my post.
Read it again.
- Randy
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
16 Nov 2003 08:57:47 PM |
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Randy Poe wrote:
On Fri, 14 Nov 2003 13:15:35 -0800, David Rutherford
<drutherford@softcom.net> wrote:
Randy Poe wrote:
I = dq/dt
Wait a minute. Suppose you had a net flow of current
out of a volume, e.g. J pointing out everywhere, so
J.da is positive over the whole surface. Then clearly
the amount of charge inside the volume is decreasing,
not increasing. That is, the surface integral is
positive if and only if the net dq/dt inside the
volume is negative.
Maxwell's right, you're wrong.
I think you're simultaneously mixing and matching several
different meanings of I and dq/dt.
Which of my equations requires a sign change, in your opinion.
Perhaps the one where I said "wait a minute" and pointed out to you
how the signs were inconsistent in the way you were interpreting "I"
and "dq/dt". Did you actually read what I wrote? Or what anybody else
wrote? You quoted it. The answer to your question is above. If I is
positive, it means charge is flowing out of the volume. It means q is
decreasing. Therefore using "I" to mean charge flowing out of the
volume and "q" meaning charge in the volume is not consistent with "I
= dq/dt". Which you'd know if you read my post, as that was the entire
content of my post.
Read it again.
Read
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm=vrcv53qlokkvab%40corp.supernews.com
--
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
.
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| User: "Randy Poe" |
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| Title: Re: Charge is not conserved in classical EM |
17 Nov 2003 10:00:30 AM |
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David Rutherford <drutherford@softcom.net> wrote in message news:<vrgdvicvnpaka5@corp.supernews.com>...
Randy Poe wrote:
On Fri, 14 Nov 2003 13:15:35 -0800, David Rutherford
<drutherford@softcom.net> wrote:
Randy Poe wrote:
I = dq/dt
Wait a minute. S
Which of my equations requires a sign change, in your opinion.
Perhaps the one where I said "wait a minute" and pointed out to you
how the signs were inconsistent in the way you were interpreting "I"
and "dq/dt".
Read
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm=vrcv53qlokkvab%40corp.supernews.com
There you correctly write I = -dQ/dt.
But you incorrectly add a negative sign to the
divergence theorem, claiming it is "necessary for the
conservation of fluid".
The divergence theorem is pure geometry. Despite
your assertion that there should be a negative sign
there, there shouldn't be. That makes no physical
sense.
- Randy
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Charge is not conserved in classical EM |
17 Nov 2003 08:56:13 AM |
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David Rutherford wrote:
Randy Poe wrote:
On Fri, 14 Nov 2003 13:15:35 -0800, David Rutherford
<drutherford@softcom.net> wrote:
Randy Poe wrote:
I = dq/dt
Wait a minute. Suppose you had a net flow of current
out of a volume, e.g. J pointing out everywhere, so
J.da is positive over the whole surface. Then clearly
the amount of charge inside the volume is decreasing,
not increasing. That is, the surface integral is
positive if and only if the net dq/dt inside the
volume is negative.
Maxwell's right, you're wrong.
I think you're simultaneously mixing and matching several
different meanings of I and dq/dt.
Which of my equations requires a sign change, in your opinion.
Perhaps the one where I said "wait a minute" and pointed out to you
how the signs were inconsistent in the way you were interpreting "I"
and "dq/dt". Did you actually read what I wrote? Or what anybody else
wrote? You quoted it. The answer to your question is above. If I is
positive, it means charge is flowing out of the volume. It means q is
decreasing. Therefore using "I" to mean charge flowing out of the
volume and "q" meaning charge in the volume is not consistent with "I
= dq/dt". Which you'd know if you read my post, as that was the entire
content of my post.
Read it again.
Read
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm=vrcv53qlokkvab%40corp.supernews.com
The comment after the very first equation
"\int_V{(div v)dV} = \oint_S{v.da} (1)
The physical meaning of (1) is that the total amount of fluid flowing
out through the closed surface S, per unit time, equals the total amount
of fluid flowing _into_ the volume V enclosed by the surface, from
sources inside V."
is wrong. This equation simply says that the integral over the sources
inside a volume is equal to the flow through the surface of the volume.
The equation has nothing to do with "flowing into" or "out of" - it is
about the *total* flow. Therefore your change of sign in equation (2)
makes no sense at all - it is mathematical nonsense! (err, BTW: you know
that da is a vector, right? And you know that it conventionally points
outward, right?)
Bye,
Bjoern
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| User: "Bruce Scott TOK" |
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| Title: Re: Charge is not conserved in classical EM |
14 Nov 2003 04:53:30 AM |
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Randy Poe wrote:
|> David Rutherford <drutherford@softcom.net> wrote in message news:<vr7nlcoffvgqf6@corp.supernews.com>...
|> > Since current I can be written
|> >
|> > I = \oint_S{J.da} (1)
|> >
|> > for a closed surface. But
|> >
|> > I = dq/dt
[...]
|> I think you're simultaneously mixing and matching several
|> different meanings of I and dq/dt.
I think he simply missed a sign. Surprising enough he could set up the
integrals correctly, but the div J gives the current flowing _out_ of
the volume so he should set
d rho/dt = - div J
which of course gives the correct equation.
It is an easy mistake to make; I made it the first time I wrote up my
lecture notes; but it is really stupid to then charge off to Usenet and
broadcast to all posterity that you've just discovered "new physics"
that overthrows 150 years of work :-))))
--
cu,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
14 Nov 2003 03:14:45 PM |
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Bruce Scott TOK wrote:
Randy Poe wrote:
|> David Rutherford <drutherford@softcom.net> wrote in message news:<vr7nlcoffvgqf6@corp.supernews.com>...
|> > Since current I can be written
|> >
|> > I = \oint_S{J.da} (1)
|> >
|> > for a closed surface. But
|> >
|> > I = dq/dt
[...]
|> I think you're simultaneously mixing and matching several
|> different meanings of I and dq/dt.
I think he simply missed a sign.
Where? Please be specific.
Surprising enough he could set up the
integrals correctly, but the div J gives the current flowing _out_ of
the volume so he should set
d rho/dt = - div J
which of course gives the correct equation.
It is an easy mistake to make; I made it the first time I wrote up my
lecture notes; but it is really stupid to then charge off to Usenet and
broadcast to all posterity that you've just discovered "new physics"
that overthrows 150 years of work :-))))
First you demean me, then you try to claim credit after the fact.
Typical behavior from a physicist.
--
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
.
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| User: "Big Bird" |
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| Title: Re: Charge is not conserved in classical EM |
13 Nov 2003 05:07:06 PM |
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David Rutherford <drutherford@softcom.net> wrote in message news:<vr7nlcoffvgqf6@corp.supernews.com>...
The classical EM equation for conservation of charge is
div J + @\rho/@t = 0
where J is the current density, @/@t is the partial derivative with
respect to t, and \rho is the charge density.
This is a continuity equation -- it asserts conservation of a
quantity. It does NOT have to be charge -- the exact same equation can
be used for the conservation of mass in a similar context, for
example.
As it is written down here in a differential way, it makes the
approximation that charge is a continuous quantity; this is OK aslong
as we are talking about large numbers of charged particles. Since it
is understood that charge is really quantized at the lowest level, one
might want to go to an integral form for sufficiently small volumes:
\int div J + \int @\rho/@t = 0
I + dq/dt = 0
This can accomodate quantized "q" as long as we allow for quantized
"I". Note the sign: for "I" to point out of the volume means that the
total "q" in the volume is decreasing; in other words if the total
amount of charge in a volume increases then there is a current into
that volume and if it decreases then there is a current out of it.
Hence conservation of charge: charge cannot suddenly appear or
disappear.
The same can be done with any other quantity: if the amount of
gelatinous matter in a volume (say a plastic bag) decreases, then
there must be gummy bears taken from that bag. Conservation of gummy
bears.
I will show now that this does _not_ imply conservation of charge.
It DEFINES conservation of charge, to be precise.
[...]
Since current I can be written
I = \oint_S{J.da} (1)
for a closed surface. But
I = dq/dt =
Absolutely false: you are asserting here that a current INTO a volume
leads to a DECREASE of charge inside that volume. That is nonsense.
I am *suspecting* that you took this equation from some electrical
engineering text. EE folks tend to count currents sometimes in the
opposite direction from the way physicists handle it, (as they always
only deal with negative charges, namely electrons).
To a physicist an electron going form A to B constitutes a *negative*
current from A to B, i.e. a positive current from B to A. Thus if you
put more electrons into a volume, the charge inside *decreases* (i.e.
becomes more negative). To EE folks, the charge sometimes "increases"
because there are more charge carriers in the volume.
Both are fine views as long as you don't blindly take an equation from
the one area and an equation from the other and imagine you can plug
them into each other. That will invariably lead to sign errors like
the one you made right up there.
That sign error propagates all the way through to the point where you
get
div J = @\rho/@t
instead of the correct
div J = - @\rho/@t
----
Now comes the problematic part: Your error is a trivial one, that
every physics freshman would be embarrassed about. Yet you have the
FAT EGO to presume that you are right and *all* *physicists* *who*
*lived* *since* *Maxwell* somehow never noticed that charge isn't
actually conserved.
In other words you find it perfectly acceptable to insult millions of
smart men but you will invariably balk at the expression "FAT EGO"
that I used in the previous sentence since it is somehow "needlessly
insulting". You will call me all kinds of names (including a claim
that it is somehow *my* ego that is inflated) because you imagine that
anything insultig handed to you is less justified than the coarse vile
insult you just spewed at a large segment of the human population.
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
14 Nov 2003 03:28:00 PM |
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Big Bird wrote:
David Rutherford <drutherford@softcom.net> wrote in message news:<vr7nlcoffvgqf6@corp.supernews.com>...
The classical EM equation for conservation of charge is
div J + @\rho/@t = 0
where J is the current density, @/@t is the partial derivative with
respect to t, and \rho is the charge density.
This is a continuity equation -- it asserts conservation of a
quantity.
No it doesn't. I showed you that
div J + @\rho/@t =/= 0
It does NOT have to be charge -- the exact same equation can
be used for the conservation of mass in a similar context, for
example.
As it is written down here in a differential way, it makes the
approximation that charge is a continuous quantity; this is OK aslong
as we are talking about large numbers of charged particles. Since it
is understood that charge is really quantized at the lowest level, one
might want to go to an integral form for sufficiently small volumes:
\int div J + \int @\rho/@t = 0
I + dq/dt = 0
So you are saying that I = -dq/dt?
Tell me what physics book you got that definition from.
To a physicist an electron going form A to B constitutes a *negative*
current from A to B, i.e. a positive current from B to A. Thus if you
put more electrons into a volume, the charge inside *decreases* (i.e.
becomes more negative). To EE folks, the charge sometimes "increases"
because there are more charge carriers in the volume.
Both are fine views as long as you don't blindly take an equation from
the one area and an equation from the other and imagine you can plug
them into each other. That will invariably lead to sign errors like
the one you made right up there.
That sign error propagates all the way through to the point where you
get
div J = @\rho/@t
instead of the correct
div J = - @\rho/@t
Exactly where, in your opinion, does my sign error originate? Please be
specific.
--
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
.
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| User: "Big Bird" |
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| Title: Re: Charge is not conserved in classical EM |
14 Nov 2003 09:42:57 PM |
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David Rutherford <drutherford@softcom.net> wrote in message news:<vrahu46fjmsi74@corp.supernews.com>...
Big Bird wrote:
This is a continuity equation -- it asserts conservation of a
quantity.
No it doesn't. I showed you that
You showed no such thing.
This has been pointed out and explained to you by several people.
You are a liar.
As it is written down here in a differential way, it makes the
approximation that charge is a continuous quantity; this is OK aslong
as we are talking about large numbers of charged particles. Since it
is understood that charge is really quantized at the lowest level, one
might want to go to an integral form for sufficiently small volumes:
\int div J + \int @\rho/@t = 0
I + dq/dt = 0
So you are saying that I = -dq/dt?
Tell me what physics book you got that definition from.
This is not a "definition", it is the result of a single, simple
integration.
I just derived it right there in front of your eyes. Contrary to
brainless symbol-manipulation mechanisms like you, I don't need to get
trivial things like this from a book: everybody with more than two
braincells can see me derive it right up there in the part you quoted.
Exactly where, in your opinion, does my sign error originate? Please be
specific.
I *was* specific.
I pointed out *exactly* where you are missing a minus sign.
If you are expecting people to help you out by explaining the basics
of calculus to you, they you really need to stop insulting them like
this.
.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Charge is not conserved in classical EM |
17 Nov 2003 08:59:46 AM |
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David Rutherford wrote:
Big Bird wrote:
David Rutherford <drutherford@softcom.net> wrote in message news:<vr7nlcoffvgqf6@corp.supernews.com>...
The classical EM equation for conservation of charge is
div J + @\rho/@t = 0
where J is the current density, @/@t is the partial derivative with
respect to t, and \rho is the charge density.
This is a continuity equation -- it asserts conservation of a
quantity.
No it doesn't. I showed you that
div J + @\rho/@t =/= 0
By a false premise.
It does NOT have to be charge -- the exact same equation can
be used for the conservation of mass in a similar context, for
example.
As it is written down here in a differential way, it makes the
approximation that charge is a continuous quantity; this is OK aslong
as we are talking about large numbers of charged particles. Since it
is understood that charge is really quantized at the lowest level, one
might want to go to an integral form for sufficiently small volumes:
\int div J + \int @\rho/@t = 0
I + dq/dt = 0
So you are saying that I = -dq/dt?
Yes, for the definitions: I = current flowing out of the surface and q =
charge contained in the volume.
Tell me what physics book you got that definition from.
Try Jackson, Electrodynamics. It should be in there.
(in lots of places one finds the equation I = + dq/dt, right, but only
if I and q are *not* defined as above!!!)
[snip rest]
Bye,
Bjoern
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
13 Nov 2003 02:53:39 PM |
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David Rutherford wrote:
Since current I can be written
I = \oint_S{J.da} (1)
for a closed surface.
This should be
"Current I can be written
I = \oint_S{J.da} (1)
for a closed surface."
--
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
.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Charge is not conserved in classical EM |
14 Nov 2003 06:49:46 AM |
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David Rutherford wrote:
The classical EM equation for conservation of charge is
div J + @\rho/@t = 0
where J is the current density, @/@t is the partial derivative with
respect to t, and \rho is the charge density.
I will show now that this does _not_ imply conservation of charge.
Since current I can be written
I = \oint_S{J.da} (1)
for a closed surface. But
I = dq/dt
You go wrong exactly here. If q denotes the charge inside the surface,
and I denotes the current going out of the surface, then obviously
I = - dq/dt,
and this gives the usual continuity equation.
= (d/dt)\int_V{\rho dV} = \int_V{(@\rho/@t)dV} (2)
and
\oint_S{J.da} = \int_V{(div J)dV} (3)
so, plugging (2) and (3) into (1) and rearranging terms, we have
\int_V{(div J)dV} = \int_V{(@\rho/@t)dV}
Since this applies to any volume, we can say
div J = @\rho/@t
or
div J - @\rho/@t = 0
This is the _correct_ continuity equation for local charge conservation.
Wrong. See above.
It is _not_ the same as the classical continuity equation, which is
div J + @\rho/@t = 0
therefore, the classical continuity equation does _not_ result in local
conservation of charge.
Wrong. It does. You only used the wrong relationship between current and
time derivation of charge.
Fortunately, my theory
http://www.softcom.net/users/der555/newtransform.pdf
gives div J - @\rho/@t = 0, so it results in local conservation of charge.
Wrong.
Bye,
Bjoern
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
14 Dec 2003 05:16:08 PM |
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David Rutherford wrote:
Fortunately, my theory
http://www.softcom.net/users/der555/newtransform.pdf
gives div J - @\rho/@t = 0, so it results in local conservation of charge.
I forgot to mention that it also demands no electric monopoles, which is
also a prediction of my theory.
--
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
.
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| User: "Franz Heymann" |
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| Title: Re: Charge is not conserved in classical EM |
15 Dec 2003 04:02:27 PM |
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"David Rutherford" <drutherford@softcom.net> wrote in message
news:vtprlvftkcj06@corp.supernews.com...
David Rutherford wrote:
Fortunately, my theory
http://www.softcom.net/users/der555/newtransform.pdf
gives div J - @\rho/@t = 0, so it results in local conservation of
charge.
I forgot to mention that it also demands no electric monopoles, which is
also a prediction of my theory.
Electrons and quarks all are electric monopoles, so your theory is patently
nonsense
Franz Heymann
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| User: "S. Enterprize Company" |
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| Title: Re: Charge is not conserved in classical EM |
15 Dec 2003 05:31:15 PM |
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"David Rutherford" <drutherford@softcom.net> wrote in message
news:vtprlvftkcj06@corp.supernews.com...
David Rutherford wrote:
Fortunately, my theory
http://www.softcom.net/users/der555/newtransform.pdf
gives div J - @\rho/@t = 0, so it results in local conservation of
charge.
I forgot to mention that it also demands no electric monopoles, which is
also a prediction of my theory.
Electrons and quarks all are electric monopoles, so your theory is patently
nonsense
No, electrons aren't electric monopoles.
Franz Heymann
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
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| User: "Franz Heymann" |
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| Title: Re: Charge is not conserved in classical EM |
17 Dec 2003 04:38:40 PM |
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"S. Enterprize Company" <smart1234@aol.com> wrote in message
news:20031215183115.01560.00000607@mb-m20.aol.com...
"David Rutherford" <drutherford@softcom.net> wrote in message
news:vtprlvftkcj06@corp.supernews.com...
David Rutherford wrote:
Fortunately, my theory
http://www.softcom.net/users/der555/newtransform.pdf
gives div J - @\rho/@t = 0, so it results in local conservation of
charge.
I forgot to mention that it also demands no electric monopoles, which
is
also a prediction of my theory.
Electrons and quarks all are electric monopoles, so your theory is
patently
nonsense
No, electrons aren't electric monopoles.
Well, they are certainly not electric dipoles, so what are they then?
Franz
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| User: "S. Enterprize Company" |
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| Title: Re: Charge is not conserved in classical EM |
17 Dec 2003 07:28:15 PM |
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"S. Enterprize Company" <smart1234@aol.com> wrote in message
news:20031215183115.01560.00000607@mb-m20.aol.com...
"David Rutherford" <drutherford@softcom.net> wrote in message
news:vtprlvftkcj06@corp.supernews.com...
David Rutherford wrote:
Fortunately, my theory
http://www.softcom.net/users/der555/newtransform.pdf
gives div J - @\rho/@t = 0, so it results in local conservation of
charge.
I forgot to mention that it also demands no electric monopoles, which
is
also a prediction of my theory.
Electrons and quarks all are electric monopoles, so your theory is
patently
nonsense
No, electrons aren't electric monopoles.
Well, they are certainly not electric dipoles, so what are they then?
Franz
There is an anti-matter pole and a matter pole to an electron. What makes
charge or the attractiveness of repulsiveness of particles EXTERNALLY. The
Helix Spiral Spinning Fields of each particle. Why is a proton postively
charged compared to an electron? Because of the size and strength of this
field. Example:
electron ( - )
( ) matter side
( )
(((()))
(((((()))))) Smart Neutral Particle
(((()))
( )
( | ) anti- matter side
|
|
proton (+) |
( | )
( v )
( )
((((((())))))
(((((((((())))))))))
((((((()))))))
( )
( )
( )
The electron is pulled into the proton helix spiral field causing the
attractiveness.
It's like a large black hole sucking in the smaller particle.
With like charges the fields and strengths are the same size so they bounce
off of each other, thus causing the repulsion.
Charge in this sence is just the size and strength of the Helix Spiral Field
difference.
The INTERNAL charge of the particle has to do with the capacitance of the
"Smarks" ( like (quarks)) and this works together with the inductance or
magnetic field the helix spiral field entrances that causes EM waves around the
particle.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
.
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| User: "S. Enterprize Company" |
|
| Title: Re: Charge is not conserved in classical EM |
15 Dec 2003 05:36:52 PM |
|
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"David Rutherford" <drutherford@softcom.net> wrote in message
news:vtprlvftkcj06@corp.supernews.com...
David Rutherford wrote:
Fortunately, my theory
http://www.softcom.net/users/der555/newtransform.pdf
gives div J - @\rho/@t = 0, so it results in local conservation of
charge.
I forgot to mention that it also demands no electric monopoles, which is
also a prediction of my theory.
Electrons and quarks all are electric monopoles, so your theory is patently
nonsense
No, electrons aren't electric monopoles.
Franz Heymann
Electrons are composed of _positive_ charged positrons and negative charged
sub-electrons with a Smart Neutral particle preventing their annihilation.
You still don't even know what charge is at the sub-atomic level.
Smart's Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
.
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| User: "David Rutherford" |
|
| Title: Re: Charge is not conserved in classical EM |
15 Nov 2003 01:25:31 PM |
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If v represents the flow of an imcompressible fluid, then the divergence
theorem says
\int_V{(div v)dV} = \oint_S{v.da} (1)
The physical meaning of (1) is that the total amount of fluid flowing
out through the closed surface S, per unit time, equals the total amount
of fluid flowing _into_ the volume V enclosed by the surface, from
sources inside V.
However, for conservation of fluid, the amount of fluid flowing out
through the surface S, per unit time, must equal the total amount of
fluid flowing _out_of_ the volume V enclosed by the surface. So, for the
conservation of fluid, we must have
\int_V{(div v)dV} = -\oint_S{v.da} (2)
In terms of current density J, (2) becomes
\int_V{(div J)dV} = -\oint_S{J.da} (3)
Current I is
I = \oint_S{J.da}
= d(Q_outside)/dt
= -d(Q_inside)/dt
= -(d/dt)(\int_V{\rho dV})
= -\int_V{(@\rho/@t)dV}
where Q_inside is the total charge inside V, and Q_outside is the total
charge outside V, @/@t is the partial derivative with respect to t, and
\rho is the charge density. So we can write
-\oint_S{J.da} = \int_V{(@\rho/@t)dV} (4)
Inserting (4) into (3), we get
\int_V{(div J)dV} = \int_V{(@\rho/@t)dV} (5)
and, since (5) applies to any volume, we have
div J = @\rho/@t
or
div J - @\rho/@t = 0 (6)
Equation (6) is, therefore, the _correct_ equation for conservation of
charge. The equation for conservation of charge in classical EM is
div J + @\rho/@t = 0 (7)
Clearly, since div J = -@\rho/@t, in classical EM, you get div J -
@\rho/@t =/= 0. So in classical EM, charge is _not_ conserved.
--
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
.
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| User: "Alfred Einstead" |
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| Title: Re: Charge is not conserved in classical EM |
21 Nov 2003 07:03:20 AM |
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David Rutherford <drutherford@softcom.net> wrote:
If v represents the flow of an imcompressible fluid, then the divergence
theorem says
\int_V{(div v)dV} = \oint_S{v.da} (1)
If v represents the flow of an incompressible fluid, i.e.,
div v = 0
then the "Divergence Theorem"
Integral_V (div v) dV = Integral_{dV} dA.v
says that
Integral_{dV} dA.v = 0;
i.e., that the fluid has no net influx or outflux from a
closed surface dV bounding a volume V.
.
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
22 Nov 2003 11:10:07 AM |
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Alfred Einstead wrote:
David Rutherford <drutherford@softcom.net> wrote:
If v represents the flow of an imcompressible fluid, then the divergence
theorem says
\int_V{(div v)dV} = \oint_S{v.da} (1)
If v represents the flow of an incompressible fluid, i.e.,
div v = 0
then the "Divergence Theorem"
Integral_V (div v) dV = Integral_{dV} dA.v
says that
Integral_{dV} dA.v = 0;
i.e., that the fluid has no net influx or outflux from a
closed surface dV bounding a volume V.
In this case, \int_V{(div v)dV} = -\oint_S{v.da} would be true, also
(or, using your equation, Integral_V (div v) dV = -Integral_{dV} dA.v).
--
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
.
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| User: "Alfred Einstead" |
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| Title: Re: Charge is not conserved in classical EM |
24 Nov 2003 05:25:49 PM |
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David Rutherford <drutherford@softcom.net> wrote:
Integral_V (div v) dV = Integral_{dV} dA.v
says that
Integral_{dV} dA.v = 0;
i.e., that the fluid has no net influx or outflux from a
closed surface dV bounding a volume V.
In this case [...] Integral_V (div v) dV = -Integral_{dV} dA.v).
Integral_V (div v) dV = +Integral_{dV} dA.v.
dA points outward, not inward.
Example:
Integral_{x=0..1,y=0..1,z=0..1) (u_x(x,y,z)+v_y(x,y,z)+w_z(x,y,z) dx dy dz
= Integral_{y=0..1,z=0..1) (Integral_{x=0..1) u_x(x,y,z) dx) dy dz
+ Integral_{z=0..1,x=0..1) (Integral_{y=0..1) v_y(x,y,z) dx) dy dz
+ Integral_{x=0..1,y=0..1) (Integral_{z=0..1) w_z(x,y,z) dx) dy dz
= Integral_{y=0..1,z=0..1) (u(1,y,z)-u(0,y,z)) dy dz
+ Integral_{z=0..1,x=0..1) (v(x,1,z)-v(x,0,z)) dz dx
+ Integral_{x=0..1,y=0..1) (w(x,y,1)-w(x,y,0)) dx dy
= Integral_{y=0..1,z=0..1) (u,v,w)|x=1 . ((1,0,0) dy dz)
+ Integral_{y=0..1,z=0..1) (u,v,w)|x=0 . ((-1,0,0) dy dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=1 . ((0,1,0) dx dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=0 . ((0,-1,0) dx dz)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=1 . ((0,0,1) dx dy)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=0 . ((0,0,-1) dx dy)
= Integral_{dV} (u,v,w) . dA;
(1,0,0) dy dz points OUTward at x = 1, (-1,0,0) dy dz points OUTward
at x = 0, etc. They are outward normals. The sign is +, not -.
.
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
07 Dec 2003 07:49:27 PM |
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Alfred Einstead wrote:
David Rutherford <drutherford@softcom.net> wrote:
Integral_V (div v) dV = Integral_{dV} dA.v
says that
Integral_{dV} dA.v = 0;
i.e., that the fluid has no net influx or outflux from a
closed surface dV bounding a volume V.
In this case [...] Integral_V (div v) dV = -Integral_{dV} dA.v).
Integral_V (div v) dV = +Integral_{dV} dA.v.
dA points outward, not inward.
Example:
Integral_{x=0..1,y=0..1,z=0..1) (u_x(x,y,z)+v_y(x,y,z)+w_z(x,y,z) dx dy dz
= Integral_{y=0..1,z=0..1) (Integral_{x=0..1) u_x(x,y,z) dx) dy dz
+ Integral_{z=0..1,x=0..1) (Integral_{y=0..1) v_y(x,y,z) dx) dy dz
+ Integral_{x=0..1,y=0..1) (Integral_{z=0..1) w_z(x,y,z) dx) dy dz
= Integral_{y=0..1,z=0..1) (u(1,y,z)-u(0,y,z)) dy dz
+ Integral_{z=0..1,x=0..1) (v(x,1,z)-v(x,0,z)) dz dx
+ Integral_{x=0..1,y=0..1) (w(x,y,1)-w(x,y,0)) dx dy
= Integral_{y=0..1,z=0..1) (u,v,w)|x=1 . ((1,0,0) dy dz)
+ Integral_{y=0..1,z=0..1) (u,v,w)|x=0 . ((-1,0,0) dy dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=1 . ((0,1,0) dx dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=0 . ((0,-1,0) dx dz)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=1 . ((0,0,1) dx dy)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=0 . ((0,0,-1) dx dy)
= Integral_{dV} (u,v,w) . dA;
(1,0,0) dy dz points OUTward at x = 1, (-1,0,0) dy dz points OUTward
at x = 0, etc. They are outward normals. The sign is +, not -.
I'll use Integral_S{v.dS}, where dS is the surface element, instead of
Integral_{dV} (u,v,w) . dA or \oint_S{v.da}. It means the same thing (if
S is a closed surface), but I think it's a little less confusing.
Integral_S{v.dS} is equivalent to Integral_V{(div v)dV} evaluated
_OUTside_ V, not INside V, since the normals to S point OUTward,
not INward, so actually,
Integral_V{(div v)dV} OUTside V = Integral_S{v.dS}
But we need to evaluate Integral_V{(div v)dV} _INside_ V, so we would
need to evaluate either Integral_S{v.dS} using INward normals or,
equivalently, -Integral_S{v.dS} using OUTward normals. If we assume
Integral_S{v.dS} to mean Integral_S{v.dS} using OUTward normals, we can
write
Integral_V{(div v)dV} INside V = -Integral_S{v.dS}
In terms of the current density J, this becomes
Integral_V{(div J)dV} INside V = -Integral_S{J.dS}
But
Integral_S{J.dS} = I
= d(Q_outside)/dt
= -d(Q_inside)/dt
= -(d/dt)(Integral_V{\rho dV}) INside V
= -Integral_V{(@\rho/@t)dV} INside V
so we can write
Integral_V{(div J)dV} INside V = Integral_V{(@\rho/@t)dV} INside V
and, since this applies to any volume V, we can say
div J = @\rho/@t
In the classical equation, Integral_V{(div J)dV} is _assumed_ to be
evaluated INside V (which it actually isn't). So classically, based on
this incorrect assumption,
Integral_V{(div J)dV} INside V = Integral_S{J.dS}
= -Integral_V{(@\rho/@t)dV} INside V
or
div J = -@\rho/@t
which are clearly untrue, since the _actual_ Integral_V{(div J)dV}
INside V is negative, and -Integral_V{(@\rho/@t)dV} INside V is
positive, for a net flow of charge out of V.
--
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
.
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| User: "Alfred Einstead" |
|
| Title: Re: Charge is not conserved in classical EM |
09 Dec 2003 07:19:25 PM |
|
|
David Rutherford <drutherford@softcom.net> wrote:
Alfred Einstead wrote:
Example:
Integral_{x=0..1,y=0..1,z=0..1) (u_x(x,y,z)+v_y(x,y,z)+w_z(x,y,z) dx dy dz
= Integral_{y=0..1,z=0..1) (Integral_{x=0..1) u_x(x,y,z) dx) dy dz
+ Integral_{z=0..1,x=0..1) (Integral_{y=0..1) v_y(x,y,z) dx) dy dz
+ Integral_{x=0..1,y=0..1) (Integral_{z=0..1) w_z(x,y,z) dx) dy dz
= Integral_{y=0..1,z=0..1) (u(1,y,z)-u(0,y,z)) dy dz
+ Integral_{z=0..1,x=0..1) (v(x,1,z)-v(x,0,z)) dz dx
+ Integral_{x=0..1,y=0..1) (w(x,y,1)-w(x,y,0)) dx dy
= Integral_{y=0..1,z=0..1) (u,v,w)|x=1 . ((1,0,0) dy dz)
+ Integral_{y=0..1,z=0..1) (u,v,w)|x=0 . ((-1,0,0) dy dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=1 . ((0,1,0) dx dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=0 . ((0,-1,0) dx dz)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=1 . ((0,0,1) dx dy)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=0 . ((0,0,-1) dx dy)
= Integral_{dV} (u,v,w) . dA;
I'll use Integral_S{v.dS}, where dS is the surface element, instead of
Integral_{dV} (u,v,w) . dA
i.e., in this example:
Integral_{y=0..1,z=0..1) (Integral_{x=0..1) u_x(x,y,z) dx) dy dz
+ Integral_{z=0..1,x=0..1) (Integral_{y=0..1) v_y(x,y,z) dx) dy dz
+ Integral_{x=0..1,y=0..1) (Integral_{z=0..1) w_z(x,y,z) dx) dy dz
= Integral_{y=0..1,z=0..1) (u(1,y,z)-u(0,y,z)) dy dz
+ Integral_{z=0..1,x=0..1) (v(x,1,z)-v(x,0,z)) dz dx
+ Integral_{x=0..1,y=0..1) (w(x,y,1)-w(x,y,0)) dx dy
= Integral_{y=0..1,z=0..1) (u,v,w)|x=1 . ((1,0,0) dy dz)
+ Integral_{y=0..1,z=0..1) (u,v,w)|x=0 . ((-1,0,0) dy dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=1 . ((0,1,0) dx dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=0 . ((0,-1,0) dx dz)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=1 . ((0,0,1) dx dy)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=0 . ((0,0,-1) dx dy)
Integral_S{v.dS} is equivalent to Integral_V{(div v)dV} evaluated
_OUTside_ V
i.e., in this example, outside of [0,1]x[0,1]x[0,1] or:
integral for
x over (-infinity, 0) union (1, +infinity)
y over (-infinity, 0) union (1, +infinity)
z over (-infinity, 0) union (1, +infinity)
of
u_x(x,y,z) + v_y(x,y,z) + w_z(x,y,z) dx dy dz
(8 triple integrals in all, one for each combination of
intervals).
-- which is, generally, ill-defined and also completely
unrelated to anything being discussed here.
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| User: "David Rutherford" |
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| Title: Re: Charge is not conserved in classical EM |
11 Dec 2003 07:46:22 PM |
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Alfred Einstead wrote:
David Rutherford <drutherford@softcom.net> wrote:
Alfred Einstead wrote:
Example:
Integral_{x=0..1,y=0..1,z=0..1) (u_x(x,y,z)+v_y(x,y,z)+w_z(x,y,z) dx dy dz
= Integral_{y=0..1,z=0..1) (Integral_{x=0..1) u_x(x,y,z) dx) dy dz
+ Integral_{z=0..1,x=0..1) (Integral_{y=0..1) v_y(x,y,z) dx) dy dz
+ Integral_{x=0..1,y=0..1) (Integral_{z=0..1) w_z(x,y,z) dx) dy dz
= Integral_{y=0..1,z=0..1) (u(1,y,z)-u(0,y,z)) dy dz
+ Integral_{z=0..1,x=0..1) (v(x,1,z)-v(x,0,z)) dz dx
+ Integral_{x=0..1,y=0..1) (w(x,y,1)-w(x,y,0)) dx dy
= Integral_{y=0..1,z=0..1) (u,v,w)|x=1 . ((1,0,0) dy dz)
+ Integral_{y=0..1,z=0..1) (u,v,w)|x=0 . ((-1,0,0) dy dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=1 . ((0,1,0) dx dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=0 . ((0,-1,0) dx dz)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=1 . ((0,0,1) dx dy)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=0 . ((0,0,-1) dx dy)
= Integral_{dV} (u,v,w) . dA;
I'll use Integral_S{v.dS}, where dS is the surface element, instead of
Integral_{dV} (u,v,w) . dA
My v in Integral_S{v.dS}, is not the same as your v in (u,v,w). My v is
a vector, your v in Integral_{dV} (u,v,w) . dA is the y-component of a
vector.
i.e., in this example:
Integral_{y=0..1,z=0..1) (Integral_{x=0..1) u_x(x,y,z) dx) dy dz
+ Integral_{z=0..1,x=0..1) (Integral_{y=0..1) v_y(x,y,z) dx) dy dz
+ Integral_{x=0..1,y=0..1) (Integral_{z=0..1) w_z(x,y,z) dx) dy dz
I'm assuming you meant
+ Integral_{z=0..1,x=0..1) (Integral_{y=0..1) v_y(x,y,z) dy) dx dz
+ Integral_{x=0..1,y=0..1) (Integral_{z=0..1) w_z(x,y,z) dz) dx dy
in the last two lines.
= Integral_{y=0..1,z=0..1) (u(1,y,z)-u(0,y,z)) dy dz
+ Integral_{z=0..1,x=0..1) (v(x,1,z)-v(x,0,z)) dz dx
+ Integral_{x=0..1,y=0..1) (w(x,y,1)-w(x,y,0)) dx dy
= Integral_{y=0..1,z=0..1) (u,v,w)|x=1 . ((1,0,0) dy dz)
+ Integral_{y=0..1,z=0..1) (u,v,w)|x=0 . ((-1,0,0) dy dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=1 . ((0,1,0) dx dz)
+ Integral_{z=0..1,x=0..1) (u,v,w)|y=0 . ((0,-1,0) dx dz)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=1 . ((0,0,1) dx dy)
+ Integral_{x=0..1,y=0..1) (u,v,w)|z=0 . ((0,0,-1) dx dy)
Integral_S{v.dS} is equivalent to Integral_V{(div v)dV} evaluated
_OUTside_ V
i.e., in this example, outside of [0,1]x[0,1]x[0,1] or:
integral for
x over (-infinity, 0) union (1, +infinity)
y over (-infinity, 0) union (1, +infinity)
z over (-infinity, 0) union (1, +infinity)
of
u_x(x,y,z) + v_y(x,y,z) + w_z(x,y,z) dx dy dz
(8 triple integrals in all, one for each combination of
intervals).
-- which is, generally, ill-defined and also completely
unrelated to anything being discussed here.
I meant that Integral_V{(div v)dV} is evaluated _OUTside_ V in the sense
that Integral_S{v.dS} is evaluated 'outside' of S, due to the 'outward'
pointing normals.
By the way, I noticed something interesting on this website,
http://scienceworld.wolfram.com/physics/ContinuityEquation.html
On the website, the continuity equation for current in integral form is
I = integral_S{J.dS} = -(@/@t)integral_V{\rho dV} (1)
Do you agree with (1)? If not, why not?
The integral form of the continuity equation for a compressible fluid
(which is the type of fluid that's used to model conservation of
charge), with J = \rho u and dS instead of da, is
integral_S{J.dS} = -integral_V{(div J)dV} (4)
Note the _negative_ sign on the right-hand side of (4).
Do you agree with (4)? If not, why not?
If you combine (1) and (4), you get
integral_V{(div J)dV} = (@/@t)integral_V{\rho dV}
Do you agree with this? If not, why not?
Since the last equation applies to any volume, it can also be written as
div J = @\rho/@t
Do you agree with this? If not, why not?
The classical equation for local charge conservation is
div J = -@\rho/@t
--
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
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Charge is not conserved in classical EM |
17 Nov 2003 09:02:49 AM |
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David Rutherford wrote:
If v represents the flow of an imcompressible fluid, then the divergence
theorem says
\int_V{(div v)dV} = \oint_S{v.da} (1)
The physical meaning of (1) is that the total amount of fluid flowing
out through the closed surface S, per unit time, equals the total amount
of fluid flowing _into_ the volume V enclosed by the surface, from
sources inside V.
Nonsense. The physical meaning of (1) is that the total amount of
sources inside the volume equals the total amount of fluid flowing
through the surface. "out" and "into" isn't contained explicitly in this
formula - it gives the *total* flow.
However, for conservation of fluid, the amount of fluid flowing out
through the surface S, per unit time, must equal the total amount of
fluid flowing _out_of_ the volume V enclosed by the surface. So, for the
conservation of fluid, we must have
\int_V{(div v)dV} = -\oint_S{v.da} (2)
Total nonsense, based on your misunderstanding above.
(BTW, in which direction does da point?)
[snip rest]
Bye,
Bjoern
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