| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
03 Apr 2006 09:38:18 PM |
| Object: |
Magnetic potential |
How does the fact that the divergence of a magnetic field B is zero
allow one to construct a vector potential A such that the curl of A is
B? What is the construction?
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| User: "FrediFizzx" |
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| Title: Re: Magnetic potential |
04 Apr 2006 01:52:25 AM |
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<gmarkowsky@gmail.com> wrote in message
news:1144118298.009324.18580@i40g2000cwc.googlegroups.com...
How does the fact that the divergence of a magnetic field B is zero
allow one to construct a vector potential A such that the curl of A is
B? What is the construction?
In addition to what Timo said, think of it this way. The B field is
rotational and A is the irrotational part of the magnetic field that is
rotating. I think in cgs units, A has the dimensions of charge/length.
The same as the scalar potential V. In SI units it seems to get screwy.
The dimension of A would be momentum/charge.
FrediFizzx
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.vacuum-physics.com
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| User: "Timo Nieminen" |
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| Title: Re: Magnetic potential |
03 Apr 2006 11:30:20 PM |
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On Tue, 3 Apr 2006 wrote:
How does the fact that the divergence of a magnetic field B is zero
allow one to construct a vector potential A such that the curl of A is
B? What is the construction?
What is div(curl(F)), where F is some vector field? See, eg, the inside
front cover of Jackson. Since div(B) = 0 = div(curl(A)), there must be
some A such that curl(A) = B. Since div is a differential operator, A is
not unique.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Henning Makholm" |
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| Title: Re: Magnetic potential |
04 Apr 2006 07:47:26 AM |
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Scripsit Timo Nieminen <timo@physics.uq.edu.au>
Since div(B) = 0 = div(curl(A)), there must be some A such that
curl(A) = B.
Why?
--
Henning Makholm "Monarki, er ikke noget materielt ... Borger!"
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| User: "" |
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| Title: Re: Magnetic potential |
04 Apr 2006 08:56:35 AM |
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Scripsit Timo Nieminen <t...@physics.uq.edu.au>
Since div(B) = 0 = div(curl(A)), there must be some A such that
curl(A) = B.
Why?
Yeah, where does the A come from? I can see that once we have it,
div(curl(A)) = 0.
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| User: "" |
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| Title: Re: Magnetic potential |
04 Apr 2006 11:28:02 AM |
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wrote:
Scripsit Timo Nieminen <t...@physics.uq.edu.au>
Since div(B) = 0 = div(curl(A)), there must be some A such that
curl(A) = B.
Why?
Yeah, where does the A come from? I can see that once we have it,
div(curl(A)) = 0.
From the differential geometry point of view, it's just a consequence
of the fact R^3 has zero cohomology. Think of B as being a 1-form on
R^3. Then div(B) = 0 is equivalent to:
d *B = 0
where * is the Hodge star operator.
Since *B is closed it is therefore exact so that there exists some
1-form A such that:
*B = dA
so that B = *dA since ** = 1 in R^3. That's just the same as B =
curl(A).
Michael
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| User: "Timo Nieminen" |
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| Title: Re: Magnetic potential |
04 Apr 2006 02:48:54 PM |
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On Tue, 4 Apr 2006 wrote:
Scripsit Timo Nieminen <t...@physics.uq.edu.au>
Since div(B) = 0 = div(curl(A)), there must be some A such that
curl(A) = B.
Why?
Yeah, where does the A come from? I can see that once we have it,
div(curl(A)) = 0.
Seeing as there's been a thorough reply already, I'll confine myself to a
basic example. Consider the differential operator d/dx. If we have
dA/dx = dB/dx
then, integrating both sides, we have A = B + constant, and it's possible
to choose B so that the constant is zero.
Don't think of curl(A) as "the curl of A", think of it as C, so you have
div(B) = 0 = div(C).
If C = curl(A), you know that div(C) = 0 for any vector field A. This
means that div(B) = div(C) for any (twice-differentiable?) vector field A.
One (well, actually, infinitely many) of this infinity of choices will
give you B=curl(A).
Or, you can think of it as following directly from Helmholtz's theorem.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "" |
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| Title: Re: Magnetic potential |
04 Apr 2006 04:29:46 PM |
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Or, you can think of it as following directly from Helmholtz's theorem.
Thank you, this is the answer I was after.
Greg
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| User: "Edward Green" |
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| Title: Re: Magnetic potential |
04 Apr 2006 05:17:35 PM |
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Timo Nieminen wrote:
On Tue, 4 Apr 2006 wrote:
Scripsit Timo Nieminen <t...@physics.uq.edu.au>
Since div(B) = 0 = div(curl(A)), there must be some A such that
curl(A) = B.
Why?
Yeah, where does the A come from? I can see that once we have it,
div(curl(A)) = 0.
Seeing as there's been a thorough reply already
Ah, if only I understood the elegance of arguments with "cohomology"!
I'll confine myself to a
basic example. Consider the differential operator d/dx. If we have
dA/dx = dB/dx
then, integrating both sides, we have A = B + constant, and it's possible
to choose B so that the constant is zero.
Don't think of curl(A) as "the curl of A", think of it as C, so you have
div(B) = 0 = div(C).
If C = curl(A), you know that div(C) = 0 for any vector field A. This
means that div(B) = div(C) for any (twice-differentiable?) vector field A.
One (well, actually, infinitely many) of this infinity of choices will
give you B=curl(A).
Now, sir -- one is hampered in suggesting any flaw in this argument by
the knowledge that the result is certainly correct, but would you allow
such an argument if you didn't know it came out right in the end!?
May one not be troubled that we have not merely
dA = dB
but in fact
dA = 0 = dB,
and fear we are talking nonsense, even in elementary calculus?
Ok. It apparently is still true in the calculus of a single variable
that we may infer from this condition that
A = B + constant
trivially, since A and B are both constants. But is it obvious what
the correct extension is of the results for dA = dB in the calculus of
a single variable is in the case of vector fields with
something-like-the-differentials equal, and further both equal to zero?
I am reminded of an off-hand assertion once seen that "d/dt x.x =
2x.(dx/dt)" where x is a vector. It may be true, but if it doesn't make
one stop and say "Just a damn minute, how do we know we can do that?",
then one is not thinking.
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| User: "Timo Nieminen" |
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| Title: Re: Magnetic potential |
04 Apr 2006 05:41:37 PM |
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On Wed, 4 Apr 2006, Edward Green wrote:
Timo Nieminen wrote:
On Tue, 4 Apr 2006 wrote:
Scripsit Timo Nieminen <t...@physics.uq.edu.au>
Since div(B) = 0 = div(curl(A)), there must be some A such that
curl(A) = B.
Why?
Yeah, where does the A come from? I can see that once we have it,
div(curl(A)) = 0.
Seeing as there's been a thorough reply already
Ah, if only I understood the elegance of arguments with "cohomology"!
I'll confine myself to a
basic example. Consider the differential operator d/dx. If we have
dA/dx = dB/dx
then, integrating both sides, we have A = B + constant, and it's possible
to choose B so that the constant is zero.
Don't think of curl(A) as "the curl of A", think of it as C, so you have
div(B) = 0 = div(C).
If C = curl(A), you know that div(C) = 0 for any vector field A. This
means that div(B) = div(C) for any (twice-differentiable?) vector field A.
One (well, actually, infinitely many) of this infinity of choices will
give you B=curl(A).
Now, sir -- one is hampered in suggesting any flaw in this argument by
the knowledge that the result is certainly correct, but would you allow
such an argument if you didn't know it came out right in the end!?
May one not be troubled that we have not merely
dA = dB
but in fact
dA = 0 = dB,
and fear we are talking nonsense, even in elementary calculus?
Ok. It apparently is still true in the calculus of a single variable
that we may infer from this condition that
A = B + constant
trivially, since A and B are both constants. But is it obvious what
the correct extension is of the results for dA = dB in the calculus of
a single variable is in the case of vector fields with
something-like-the-differentials equal, and further both equal to zero?
I am reminded of an off-hand assertion once seen that "d/dt x.x =
2x.(dx/dt)" where x is a vector. It may be true, but if it doesn't make
one stop and say "Just a damn minute, how do we know we can do that?",
then one is not thinking.
(a) What do physicists care about mathematical rigour :)
(b) The "= 0" part is only useful and necessary since we _know_
div(curl()) = 0. If it was a matter of dA = dB, the 0 would be un-needed.
The 0 has nothing to do with the dA = dB part, just the B = curl(C) part,
so don't get hung up on that.
Also, note that in your A = B + constant above, A and B are _not_
constants, but functions.
(c) Yes, if we know it's true, we're much more likely to accept vague
handwaving.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Edward Green" |
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| Title: Re: Magnetic potential |
05 Apr 2006 12:13:59 AM |
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Timo Nieminen wrote:
On Wed, 4 Apr 2006, Edward Green wrote:
May one not be troubled that we have not merely
dA = dB
but in fact
dA = 0 = dB,
and fear we are talking nonsense, even in elementary calculus?
Ok. It apparently is still true in the calculus of a single variable
that we may infer from this condition that
A = B + constant
trivially, since A and B are both constants.
Also, note that in your A = B + constant above, A and B are _not_
constants, but functions.
Well, if we add the condition dA = dB = 0, and we are speaking of
functions of a single variable, then A and B are indeed constants.
When we go over from ordinary differentials to something-like
differentials in vector fields on several variables, then we may no
longer infer from the zero that A and B are constant, but "something
like" constant -- meaning, in this case, that their path integrals
around loops are zero, of course.
I've noticed before that our favorite old school theorems of vector
calculus are something-like the fundamental theorem of calculus:
*something* integrated over the inside is equal to *something else*
evaluated at the limits -- curl integrated over a surface is equal to
the path integral around the bounding curve, divergence integrated over
a volume is equal to an integral over the bounding surface.
If differential geometry provides us with a unifying way of looking at
such things and discovering new such relations, then I suppose it might
repay the pain of learning it!
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| User: "michaeld" |
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| Title: Re: Magnetic potential |
05 Apr 2006 05:05:29 PM |
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Edward Green wrote:
Timo Nieminen wrote:
On Wed, 4 Apr 2006, Edward Green wrote:
May one not be troubled that we have not merely
dA = dB
but in fact
dA = 0 = dB,
and fear we are talking nonsense, even in elementary calculus?
Ok. It apparently is still true in the calculus of a single variable
that we may infer from this condition that
A = B + constant
trivially, since A and B are both constants.
Also, note that in your A = B + constant above, A and B are _not_
constants, but functions.
Well, if we add the condition dA = dB = 0, and we are speaking of
functions of a single variable, then A and B are indeed constants.
When we go over from ordinary differentials to something-like
differentials in vector fields on several variables, then we may no
longer infer from the zero that A and B are constant, but "something
like" constant -- meaning, in this case, that their path integrals
around loops are zero, of course.
dA = 0 is the condition that A is closed. If the space has no
cohomology (basically that means "no holes") then A is exact - i.e. A
can be written in the form A = dH for some H and also the integral of A
over any surface is always zero. (And note that if A can be written
like that then dA = 0 automatically since d^2 is always zero. In other
words exact ==> closed always but the converse only holds when the
cohomology groups of the space vanish.)
I've noticed before that our favorite old school theorems of vector
calculus are something-like the fundamental theorem of calculus:
*something* integrated over the inside is equal to *something else*
evaluated at the limits -- curl integrated over a surface is equal to
the path integral around the bounding curve, divergence integrated over
a volume is equal to an integral over the bounding surface.
Yes, they're all versions of Stoke's theorem for differential forms. As
in:
http://planetmath.org/encyclopedia/GeneralStokesTheorem.html
If differential geometry provides us with a unifying way of looking at
such things and discovering new such relations, then I suppose it might
repay the pain of learning it!
Right. Certainly I think DG can be viewed as "multi-variable calculus
done right" and this is one good motivation for its development. The
concepts all make much more sense under the general umbrella. Plus
differential geometry opens up many avenues for exploration that don't
exist in mere multi-variable calculus.
Michael
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