A Question about Maxwell first equation, rho = DIV D
Paul Stowe
Given Maxwell definition of the point form of Gauss's Law, namely:
rho = DIV D
By its definition, this must be a physical constant. My question is,
has it ever been quantified or measured? If so, then this value when
divided by elemental charge (e), must yield the definitive volume of
the electron and proton. Since I can find no reference to this value
I am assuming that it has not. But wouldn't logic dictate that given
this definition, and a finite value for e, there can't exist point
charges? This would, in turn, make such an assumption of point charges
mute?
Thanks,
Paul Stowe
Charles H Bennett - Alpena
Wrong on practically all counts. The equation is not a definition, it is
a statement of the value of DIV D, saying simply that the divergence of
the field in a certain volume is equal to the charge contained in that
volume.
I assume you mean "moot", not "mute".
Grant Bennett <cben...@northland.lib.mi.us>
Paul Stowe
In <Pine.SUN.3.91.97051...@hostserver.merit.edu> Charles H
Bennett - Alpena <cben...@northland.lib.mi.us> writes:
>
>Wrong on practically all counts. The equation is not a definition, it is
>a statement of the value of DIV D, saying simply that the divergence of
>the field in a certain volume is equal to the charge contained in that
>volume.
Last time I checked rho = DIV D was a definition.
Let's see, DIV is difined to be THE LIMIT as volume goes to zero of the
fluctuation if a property crossing perpendicular to the surface area
enclosing that volume. Limit theorem is, as the name implies, a limiting
value...
As used in standard applications we do use your definition above, which
is written as:
-
/
|
rho = <|> D dS / Volume
|
/
-
Which, from a strict mathematical standpoint is not divergence except
at the limit.
>
>I assume you mean "moot", not "mute".
>
Yes, on this I chose the wrong word...
Paul Stowe
Drizzt Do'Urden
But wouldn't logic dictate that given
> this definition, and a finite value for e, there can't exist point
> charges? This would, in turn, make such an assumption of point charges
> mute?
As far as the assumption of point charges go, you are absolutely correct.
There is no basis. It's kind of like in an elementary physics course,
where they use Acme massless rope and Acme frictionless pulleys. None of
the stuff is real, but it makes the calculations a lot easier. If we had
to go around using stuff we could realistically mesure with our hands and
eyes, nothing would ever get done. I believe you are right in the
interpretation of the numbers. The assumption is only there to derive the
equation, however, not to be used in the calculations.
Bill Rowe
In article <5lignc$m...@dfw-ixnews9.ix.netcom.com>, pst...@ix.netcom.com
says...
>
>
>
>Given Maxwell definition of the point form of Gauss's Law, namely:
>
> rho = DIV D
>
>By its definition, this must be a physical constant. My question is,
>has it ever been quantified or measured? If so, then this value when
>divided by elemental charge (e), must yield the definitive volume of
>the electron and proton.
No, this doesn't follow. To see why it is probably better to look at Gauss's
Law in integral form rather than differential form.
In integral form Gauss's Law is q/e = surface integral of E . dS over the
entire surface. This is independent of the surface shape or size. It is only
dependent on the amount of charge contained within the specified volume. One
could assume a sphere of radius r or 10r and the integral would still
evaluate to the same value.
--
"Against stupidity, the Gods themselves contend in vain"
To reply remove the *nospam*
Paul Stowe
In <5ltafe$s...@hacgate2.hac.com> ghostbit@sprynet.*nospam*.com (Bill Rowe)
writes:
I know this form well, but unless I have completely fouled up my
understanding
of math definitions, DIV is the value that is taken at the limit. It
is a
unique finite value in this regard. A mundane example is the density
of a gas
like air. This density fluctuates slightly due to the statistical
variation in molecular content of the enclosed volume element. This
magnitude of this will
of course be dependent on the volume chosen to be enclosed. But the
divergence value is not (given an equilibrium condition).
Paul Stowe
Jim Carr
pst...@ix.netcom.com(Paul Stowe) writes:
>
>Given Maxwell definition of the point form of Gauss's Law, namely:
>
> rho = DIV D
>
>By its definition, this must be a physical constant.
It states a physical relationship that is equivalent to Gauss' Law
in integral form and to Coulomb's Law. It is not a "constant"
since there are variable dependences you have not written down
and D itself requires boundary conditions to be fully specified.
>My question is,
>has it ever been quantified or measured?
It was originally defined empirically, so it is fair to say that
this equation was quantified before it was written down.
>If so, then this value when
>divided by elemental charge (e), must yield the definitive volume of
>the electron and proton.
Which value? Where would a volume come in?
--
James A. Carr <j...@scri.fsu.edu> | "What are those, Daddy?"
http://www.scri.fsu.edu/~jac/ | Young girl at Smithsonian
Supercomputer Computations Res. Inst. | exhibit of objects and letters
Florida State, Tallahassee FL 32306 | left at the Vietnam Memorial.
Jim Carr
Charles H Bennett - Alpena <cben...@northland.lib.mi.us> writes:
}
} Wrong on practically all counts. The equation is not a definition, it is
} a statement of the value of DIV D, saying simply that the divergence of
} the field in a certain volume is equal to the charge contained in that
} volume.
Alpena? Cool. My mother grew up in Alpena, as did a good friend
on the faculty here.
pst...@ix.netcom.com(Paul Stowe) writes:
>
>Last time I checked rho = DIV D was a definition.
It is the statement (definition if you like) of an equation (that
is, a relationship) that describes Coulomb's Law.
>Let's see, DIV is difined to be THE LIMIT as volume goes to zero of the
>fluctuation if a property crossing perpendicular to the surface area
>enclosing that volume.
Fluctuation?
Note that the volume is at the point (unspecified in the equation
above) where you do the evaluation of the vector operator and rho.
There need not (and often is not) any charge in *that* volume.
>As used in standard applications we do use your definition above, which
>is written as:
>
> -
> /
> |
> rho = <|> D dS / Volume
> |
> /
> -
You have made some assumptions about rho here that cannot be guaranteed
(and left out some vector details), but it is true that the surface
integral gives you the charge enclosed, which is equivalent to Coulomb's
Law and has been tested to extreme accuracy as the most stringent test
on the mass of the photon. Note that you cannot integrate D until
you know it, which requires solving the diff.eq. with BCs.
Paul Stowe
In <5m543g$s3r$1...@news.fsu.edu> j...@ibms48.scri.fsu.edu (Jim Carr) writes:
>
>Charles H Bennett - Alpena <cben...@northland.lib.mi.us> writes:
>}
>} Wrong on practically all counts. The equation is not a definition, it is
>} a statement of the value of DIV D, saying simply that the divergence of
>} the field in a certain volume is equal to the charge contained in that
>} volume.
>
> Alpena? Cool. My mother grew up in Alpena, as did a good friend
> on the faculty here.
>
>pst...@ix.netcom.com(Paul Stowe) writes:
>>
>>Last time I checked rho = DIV D was a definition.
>
> It is the statement (definition if you like) of an equation (that
> is, a relationship) that describes Coulomb's Law.
I think perhaps you mis-spoke yourself, Coulomb's Law (perhaps you
meant Gauss's Law) defines the force relationship between charges
and the resulting electrical field.
One reference for this is "Engineering Electro-Magnetics", William
Hayt Jr., McGraw-Hill, 1974 Chapter 2.
Divergence on the other hand is a field operator that results in a
scalar quantity of a field vector property. It is indeed a
definition just as Grad and Curl are.
>>Let's see, DIV is difined to be THE LIMIT as volume goes to zero of the
>>fluctuation if a property crossing perpendicular to the surface area
>>enclosing that volume.
>
> Fluctuation?
Let's see if I can explain this. Consider the illustration below:
_________________________________
| | | * |
| * | * | * |
| * *| | * |
| | * | * * |
---------------------------------
| * | * | |
| * | | * |
| * | * | |
| * | * | * |
---------------------------------
| * | | * |
| | * | |
| * | * | * |
| * | * | * |
---------------------------------
each box represents a sub volume element of air and each
"*" a molecule. The illustration is a snapshot of the
region at time zero. Since each molecule is moving with
an average velocity of c (sonic velocity), they will of
course leave one box and enter another. So at time dt
later this situation will change.
As is illustrated, the "average concentration is three
molecules per box, but obviously each box will "fluctuate"
in how many molecules it actually contains. Divergence
is a measure of this effect. This is why the Divergence
properties of an "incompressible" medium is zero.
In that case there is no spacing between particles so as
one "moves out" another one "moves in" to take its place,
net effect, zero...
> Note that the volume is at the point (unspecified in the equation
> above) where you do the evaluation of the vector operator and rho.
> There need not (and often is not) any charge in *that* volume.
>
>>As used in standard applications we do use your definition above,
which
>>is written as:
>>
>> -
>> /
>> |
>> rho = <|> D dS / Volume
>> |
>> /
>> -
>
> You have made some assumptions about rho here that cannot be
guaranteed
> (and left out some vector details), but it is true that the surface
> integral gives you the charge enclosed, which is equivalent to
Coulomb's
> Law and has been tested to extreme accuracy as the most stringent
test
> on the mass of the photon. Note that you cannot integrate D until
> you know it, which requires solving the diff.eq. with BCs.
Centered around an arbitrary point yes, and as can be seen in the
above illustration, if the volume actually shrinks to zero, the
concept of Divergence is meaningless. This is because we have gone
beyond the bounds of validity for "Continuum Mechanics" on which
the property is defined.
Paul Stowe
Jim Carr
Note well, netcom spammers: just because I reply to a post from
a netcom site does not mean I want your junk e-mail. If you get
my address from this newsgroup, I assume you know I do not want
such mail and you are sending it to harras me in violation of
the CDA. You have been warned.
j...@ibms48.scri.fsu.edu (Jim Carr) writes:
|
| pst...@ix.netcom.com(Paul Stowe) writes:
| >
| >Last time I checked rho = DIV D was a definition.
|
| It is the statement (definition if you like) of an equation (that
| is, a relationship) that describes Coulomb's Law.
pst...@ix.netcom.com(Paul Stowe) writes:
>
>I think perhaps you mis-spoke yourself, Coulomb's Law (perhaps you
>meant Gauss's Law) defines the force relationship between charges
>and the resulting electrical field.
No, I meant Coulomb's Law. That equation of Maxwell's is the
Coulomb's Law equation in the form of the differential equation
that gives the field that gives Coulomb's force law. That is
why it is there. If it is not there, you don't have a Coulomb
force, and if it is different you don't get 1/r^2.
>Divergence on the other hand is a field operator that results in a
>scalar quantity of a field vector property. It is indeed a
>definition just as Grad and Curl are.
There is a definition of the divergence operator. When you
use that in an equation you get a differential equation. If
you solve that equation, you get the field D. That field will
produce a force on a charge. That force is the Coulomb force.
The 1/r^2 dependence of the Coulomb force is a result of this
form of the differential equation. When I went to school, you
could find this as an example in a calculus text (Thomas), not
just in physics books.
| >Let's see, DIV is difined to be THE LIMIT as volume goes to zero of the
| >fluctuation if a property crossing perpendicular to the surface area
| >enclosing that volume.
|
| Fluctuation?
>Let's see if I can explain this. Consider the illustration below:
> ...
>each box represents a sub volume element of air and each
>"*" a molecule.
Hmmm. For a second there I thought you had a really strange
calculus teacher. Instead of fluctuation (which never appears
in your example), why not just say conserved current?
Anyway, your example is a scalar field, not a vector field.
>Centered around an arbitrary point yes, and as can be seen in the
>above illustration, if the volume actually shrinks to zero, the
>concept of Divergence is meaningless.
Sorry, but you clearly do not understand the meaning of the
limit in the definition of differential operators.
>This is because we have gone
>beyond the bounds of validity for "Continuum Mechanics" on which
>the property is defined.
You might also review the meaning of continuum.
Paul Stowe
In <5m8ebf$kg8$1...@news.fsu.edu> j...@ibms48.scri.fsu.edu (Jim Carr)
writes:
>j...@ibms48.scri.fsu.edu (Jim Carr) writes:
>|
>| pst...@ix.netcom.com(Paul Stowe) writes:
>| >
>| >Last time I checked rho = DIV D was a definition.
>|
>| It is the statement (definition if you like) of an equation (that
>| is, a relationship) that describes Coulomb's Law.
>
>pst...@ix.netcom.com(Paul Stowe) writes:
>>
>>I think perhaps you mis-spoke yourself, Coulomb's Law (perhaps you
>>meant Gauss's Law) defines the force relationship between charges
>>and the resulting electrical field.
>
> No, I meant Coulomb's Law. That equation of Maxwell's is the
> Coulomb's Law equation in the form of the differential equation
> that gives the field that gives Coulomb's force law. That is
> why it is there. If it is not there, you don't have a Coulomb
> force, and if it is different you don't get 1/r^2.
>
It is Maxwell's first equation that give charge in a region of field
defined by a volume element which is the minimum attainable, while
still retaining the continuum properties. I shall quote from the
reference mentioned earlier ( "Engineering Electro-Magnetics", William
Hayt Jr., McGraw-Hill, 1974) page 80:
"This is the first of Maxwell's four equations as they apply to
electro-statics and steady magnetic fields, and it states that
the electric flux per unit volume leaving a vanishingly small
volume unit is exactly equal to the volume charge density
there. This equation is aptly called the point form of Gauss's
law. ... Maxwell's first equation is also described as the
differential form of Gauss's law, and conversely, Gauss's law
is recognized as the integral form of Maxwell's first equation"
>>Divergence on the other hand is a field operator that results in a
>>scalar quantity of a field vector property. It is indeed a
>>definition just as Grad and Curl are.
>
> There is a definition of the divergence operator. When you
> use that in an equation you get a differential equation. If
> you solve that equation, you get the field D. That field will
> produce a force on a charge. That force is the Coulomb force.
> The 1/r^2 dependence of the Coulomb force is a result of this
> form of the differential equation. When I went to school, you
> could find this as an example in a calculus text (Thomas), not
> just in physics books.
On page 24 of "Theoretical Physics", G. Joos, Dover 1958 we
get the definition of Divergence:
-
/
Div = Lim as volume -> 0 of <|> dS / volume
/
-
and it is using the velocity vector as the property upon which
it operate in the example.
This is equivalent to writing:
Div = nabla = (d/dx + d/dy + d/dz)
understanding that these are each d/di is a partial derivative...
>
>| >Let's see, DIV is difined to be THE LIMIT as volume goes to zero of
>| >the fluctuation if a property crossing perpendicular to the surface
>| >area enclosing that volume.
>|
>| Fluctuation?
>
>>Let's see if I can explain this. Consider the illustration below:
>> ...
>>each box represents a sub volume element of air and each
>>"*" a molecule.
>
> Hmmm. For a second there I thought you had a really strange
> calculus teacher. Instead of fluctuation (which never appears
> in your example), why not just say conserved current?
>
No, take any single box, (which is the volume of concern) at time
t=0 let it contains three molecules. At time t=1 it contains four
having gain one from an adjacent region. At time t=2 it contains
two molecules as two have moved into adjacent regions ...etc.
The question answered by Divergence is, Given an average velocity
of the molecules for example, what is the rate of change in the
molecular contents of the region? The answer returned is the frequency
or rate of oscillation i.e. Div v = nu
If this is not true then what does Div v mean? it's units are 1/sec.
> Anyway, your example is a scalar field, not a vector field.
Yes the measure of density is a scalar property of a vector field.
>>Centered around an arbitrary point yes, and as can be seen in the
>>above illustration, if the volume actually shrinks to zero, the
>>concept of Divergence is meaningless.
>
> Sorry, but you clearly do not understand the meaning of the
> limit in the definition of differential operators.
Good, in the example under discussion, what is the divergence of
the momentum content of air (at STP) when the volume truly is zero?
>>This is because we have gone
>>beyond the bounds of validity for "Continuum Mechanics" on which
>>the property is defined.
>
> You might also review the meaning of continuum.
OK, let's look at "Continuum Mechanics", T. J. Chung, Prentice
Hall 1988. On page 1&2 we find:
"To distinguish the continuum or macroscopic model from a
microscopic one, we may list a number of criteria. ... A
concept of fundamental importance here is that of mean free
path, which can be defined as the average distance that a
molecule travels between successive collisions with other
molecules. The ratio of the mean free path L to the
characteristic length S of the physical boundaries of interest,
called the Knudsen number Kn, may be used to determine the
dividing line between macroscopic and microscopic models."
Bottom line, the limit of validity of the continuum model is
when L/S < 1 period. If our boxes become smaller that L
we simply can't use the continuum mathematics.
Paul Stowe
Force Ten
>> A mundane example is the density of a gas like air. This density fluctuates slightly due to the statistical variation in molecular content of the enclosed volume element. This magnitude of this will of course be dependent on the volume chosen to be enclosed. But the divergence value is not (given an equilibrium condition). <<
Your example is well chosen to illustrate the problem in your approach.
The use of integral and differential calculus itself is appropriate only
when dealing with continuous functions. The density of air, for
example, is continous only when looking at volume elements that are
large compared to molecular sizes. You wouldn't, for example, expect to
be able to learn anything useful about the density inside an atomic
nucleus by studying density fluctuations in air. The same applies to
the radius of an electron.
Jim Carr
j...@ibms48.scri.fsu.edu (Jim Carr) writes:
|
| pst...@ix.netcom.com(Paul Stowe) writes:
| >
| >I think perhaps you mis-spoke yourself, Coulomb's Law (perhaps you
| >meant Gauss's Law) defines the force relationship between charges
| >and the resulting electrical field.
|
| No, I meant Coulomb's Law. That equation of Maxwell's is the
| Coulomb's Law equation in the form of the differential equation
| that gives the field that gives Coulomb's force law. That is
| why it is there. If it is not there, you don't have a Coulomb
| force, and if it is different you don't get 1/r^2.
pst...@ix.netcom.com(Paul Stowe) writes:
>
>It is Maxwell's first equation that give charge in a region of field
>defined by a volume element which is the minimum attainable, while
>still retaining the continuum properties.
It would be more precise to say that the first equation that relates
the electric field to the presence of a charged source. That is,
given a charge distribution, solving that equation with suitable
boundary conditions gives you the field. That field gives you
Coulomb's Law. How do you think you get Coulomb's Law out of
Maxwell's equations?
Your choice of words in the sentence above is very imprecise
and confused, mixing things related to mathematics up with
things that have physical content.
>I shall quote from the
>reference mentioned earlier ( "Engineering Electro-Magnetics", William
>Hayt Jr., McGraw-Hill, 1974) page 80:
>
> "This is the first of Maxwell's four equations as they apply to
> electro-statics and steady magnetic fields, and it states that
> the electric flux per unit volume leaving a vanishingly small
> volume unit is exactly equal to the volume charge density
> there. This equation is aptly called the point form of Gauss's
> law. ... Maxwell's first equation is also described as the
> differential form of Gauss's law, and conversely, Gauss's law
> is recognized as the integral form of Maxwell's first equation"
All true, and not contradicting anything I wrote above. Now, what
property does the solution to this equation have? It produces a
field that gives a 1/r^2 force between charged particles.
>On page 24 of "Theoretical Physics", G. Joos, Dover 1958 we
>get the definition of Divergence:
>
> -
> /
> Div = Lim as volume -> 0 of <|> dS / volume
> /
> -
>
>and it is using the velocity vector as the property upon which
>it operate in the example.
>
>This is equivalent to writing:
>
> Div = nabla = (d/dx + d/dy + d/dz)
>
>understanding that these are each d/di is a partial derivative...
You are still making some serious errors concerning the
omission of vectors in your equations above, and have not
defined "S". Nothing you write above contradicts anything
I wrote about the divergence, particularly the property that
the solution to the differential equation has. It is in
Thomas, and I assume it is in the book you cite. There is
more to physics than quoting those books; you need to know
what those words mean in practice. Look up what sort of
fields obey Gauss' Law. Fields that produce 1/r^2 Coulomb
forces. That is why they are equivalent.
Further, you must surely know that you cannot apply continuum
mathematics with limits to small countable numbers of particles
in a discrete space.
| Anyway, your example is a scalar field, not a vector field.
>Yes the measure of density is a scalar property of a vector field.
You were taking the divergence of a scalar field as far as I could
see from your description, and doing so in a case where a limit
cannot be taken and thus your definition does not apply. But
this is all irrelevant to the fact that the empirically defined
Coulomb's Law gives you fields that obey Gauss' Law, and this
is where Gauss' Law came from and why it is included in the
set of Maxwell's equations.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Paul Stowe
In <5mdg85$ckl$1...@news.fsu.edu> j...@ibms48.scri.fsu.edu (Jim Carr)
writes:
>
>j...@ibms48.scri.fsu.edu (Jim Carr) writes:
>
>>pst...@ix.netcom.com(Paul Stowe) writes:
>>
>>It is Maxwell's first equation that give charge in a region of field
>>defined by a volume element which is the minimum attainable, while
>>still retaining the continuum properties.
>
>the electric field to the presence of a charged source. That is,
>given a charge distribution, solving that equation with suitable
>boundary conditions gives you the field. That field gives you
>Coulomb's Law. How do you think you get Coulomb's Law out of
>Maxwell's equations?
>
>Your choice of words in the sentence above is very imprecise
>and confused, mixing things related to mathematics up with
>things that have physical content.
>>I shall quote from the reference mentioned earlier ( "Engineering
>>Electro-Magnetics", William Hayt Jr., McGraw-Hill, 1974) page 80:
>>
>> "This is the first of Maxwell's four equations as they apply to
>> electro-statics and steady magnetic fields, and it states that
>> the electric flux per unit volume leaving a vanishingly small
>> volume unit is exactly equal to the volume charge density
>> there. This equation is aptly called the point form of Gauss's
>> law. ... Maxwell's first equation is also described as the
>> differential form of Gauss's law, and conversely, Gauss's law
>> is recognized as the integral form of Maxwell's first equation"
>
>All true, and not contradicting anything I wrote above. Now, what
>property does the solution to this equation have? It produces a
>field that gives a 1/r^2 force between charged particles.
Div D yield a charge density, nothing more. We can then apply
Gauss's law to "enclosed" the specific region of field (i.e. multiply
by a selected volume) where we have defined this density to obtain
an enclosed "total charge". This then can be input into
E = Q/4pi epsion R^2 a->
(your 1/R^2 relationship) where a-> is the vector designator, to get
the "electric field" at a point R distance from where Div D was
derived. However, I fail to see how this defines Div D as Coulomb's
Law.
However, I think at this point we have immersed ourselves so much into
the forest that we see only trees, not the forest anymore. The point
of the original post was to ask if this value, Div D (for an elemental
charge e) had been defined. Since this is a well understood
mathematical operator, itsvalue should be a fixed constant value (given
that e is a fixed constant value). This is a logical conclusion based
on the assumed validity of the mathematical definition of Divergence.
>>On page 24 of "Theoretical Physics", G. Joos, Dover 1958 we
>>get the definition of Divergence:
>>
>> -
>> /
>> Div = Lim as volume -> 0 of <|> dS / volume
>> /
>> -
>>
>>and it is using the velocity vector as the property upon which
>>it operate in the example.
>
>>This is equivalent to writing:
>>
>> Div = nabla = (d/dx + d/dy + d/dz)
>>
>>understanding that these are each d/di is a partial derivative...
>
>You are still making some serious errors concerning the
>omission of vectors in your equations above, and have not
>defined "S".
I now see what your saying, I left out the i,j,k components or the
vector designators. I simply assumed this was understood.
Also, yes I left out Joos's definition that S as area.
>Nothing you write above contradicts anything
>I wrote about the divergence, particularly the property that
>the solution to the differential equation has. It is in
>Thomas, and I assume it is in the book you cite. There is
>more to physics than quoting those books; you need to know
>what those words mean in practice. Look up what sort of
>fields obey Gauss' Law. Fields that produce 1/r^2 Coulomb
>forces. That is why they are equivalent.
>
>Further, you must surely know that you cannot apply continuum
>mathematics with limits to small countable numbers of particles
>in a discrete space.
Interesting, I pointed this out in the last post. The illustrations
were to visualize the process. I understand that CM can not be
used at or below the "grain" (mean free path). Which is the main
basis for a non point charge argument.
>>>Anyway, your example is a scalar field, not a vector field.
>>
>>Yes the measure of density is a scalar property of a vector field.
>>
>You were taking the divergence of a scalar field as far as I could
>see from your description, and doing so in a case where a limit
>cannot be taken and thus your definition does not apply. But
>this is all irrelevant to the fact that the empirically defined
>Coulomb's Law gives you fields that obey Gauss' Law, and this
>is where Gauss' Law came from and why it is included in the
>set of Maxwell's equations.
The point was to illustrate your question on my use of the word
"fluctuation". I should have used the vector sum of the momentum
of the particles found in each region. This would resolve this
quibble.
Paul Stowe
Jim Carr
I am having trouble understanding your fixation on looking at it
this way, since this is not how Maxwell's equations are used to
obtain electromagnetic effects. Normally one thinks of the
problem where you are given charges and currents and use those
to determine the fields, which then produce forces on charges
and currents.
You can use the equations to invert this process, but only because
the field D obeys this differential equation. It is not the case
that any arbitrary field will obey this equation, only ones that
correspond to a Coulomb-law force.
>We can then apply
>Gauss's law to "enclosed" the specific region of field (i.e. multiply
>by a selected volume) where we have defined this density to obtain
>an enclosed "total charge".
No, you cannot just do a mulitiplication. You must do the integral.
You can simplify the problem in some cases, but not for arbitrary
rho -- and you see concerned with arbitrary rho from any old D.
But for simple cases it is OK but when you write
>This then can be input into
^^^^^^^^^^
>
> E = Q/4pi epsion R^2 a->
>
>(your 1/R^2 relationship) where a-> is the vector designator, to get
>the "electric field" at a point R distance from where Div D was
>derived.
you are writing nonsense. You already have E. That is what you
used to find $\rho({\vec r}) = Q \delta({\vec r})$.
>However, I fail to see how this defines Div D as Coulomb's Law.
Because F = qE *is* Coulomb's law, only because E satisfies
that first differential equation.
The divergence equation gives you fields with this property.
I ask again: how do you think you get Coulomb's Law for point
charges out of Maxwell's equations? You use div D = rho.
Why is div D = rho one of Maxwell's equations? Because it
produces the empirically defined fields.
> ... the original post was to ask if this value, Div D (for an elemental
>charge e) had been defined. Since this is a well understood
>mathematical operator, itsvalue should be a fixed constant value (given
>that e is a fixed constant value).
If you have a point charge, you do get a simple relationship.
But it is not a *constant* since many rho are possible.
Bill Rowe
Paul Stowe:
Given Maxwell definition of the point form of Gauss's Law, namely:
rho = DIV D
By its definition, this must be a physical constant. My question is, has
it ever been quantified or measured? If so, then this value when divided
by elemental charge (e), must yield the definitive volume of the electron
and proton.
Bill Rowe:
No, this doesn't follow. To see why it is probably better to look at
Gauss's Law in integral form rather than differential form.
In integral form Gauss's Law is q/e = surface integral of E . dS over the
entire surface. This is independent of the surface shape or size.
Paul Stowe:
I know this form well, but unless I have completely fouled up my
understanding of math definitions, DIV is the value that is taken at the
limit. It is a unique finite value in this regard. A mundane example is
the density of a gas like air. This density fluctuates slightly due to the
statistical variation in molecular content of the enclosed volume element.
This magnitude of this will of course be dependent on the volume chosen
to be enclosed. But the divergence value is not (given an equilibrium
condition).
----
At this point I am confused. Your last comments show you are aware the div
is independent volume contained by the surface. Yet you started this thread
by suggesting div D could be used to define a volume for an electon. Why do
you expect something that is independent of volume to be useful to define a
volume? This makes no sense.