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Edward Green science forum addict
Joined: 21 May 2005
Posts: 95

Posted: Thu Jun 22, 2006 12:18 am Post subject:
Pop my thought balloon



I stumbled upon this gedanken, and would appreciate it if somebody
would either pop it or confirm its surprising conclusion.
A spherical shell of positive charge is held at radius r_o prior to
time t_o, when it is allowed to expand by electrostatic repulsion.
Prior to release the field outside the shell is the coulomb field of a
point charge of equal total charge at the center. Subsequent to
release, Gauss's law, which applies to dynamic as well as static charge
distributions (?), combined with the spherical symmetry, ensures that
the electric field cannot change at any point beyond the instantaneous
radius of expansion. Since the electrical field is static there is no
magnetic field in these regions either, and hence, all that happens as
the shell expands is the sweeping out of regions of field and replacing
them by zero field.
Can this be true? Will the expanding shell of accelerated charge
really produce zero advance warning in the field? 

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Frreita science forum addict
Joined: 13 Jun 2005
Posts: 96

Posted: Thu Jun 22, 2006 12:23 am Post subject:
Re: Pop my thought balloon



"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:1150935496.810469.284700@c74g2000cwc.googlegroups.com...
Quote:  I stumbled upon this gedanken, and would appreciate it if somebody
would either pop it or confirm its surprising conclusion.
A spherical shell of positive charge is held at radius r_o prior to
time t_o, when it is allowed to expand by electrostatic repulsion.
Prior to release the field outside the shell is the coulomb field of a
point charge of equal total charge at the center. Subsequent to
release, Gauss's law, which applies to dynamic as well as static charge
distributions (?), combined with the spherical symmetry, ensures that
the electric field cannot change at any point beyond the instantaneous
radius of expansion. Since the electrical field is static there is no
magnetic field in these regions either, and hence, all that happens as
the shell expands is the sweeping out of regions of field and replacing
them by zero field.
Can this be true? Will the expanding shell of accelerated charge
really produce zero advance warning in the field?

the field is decreasing in intensity, therefore changing, wouldn't that
cause a magnetic field to be generated ? 

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Ben RudiakGould science forum Guru
Joined: 04 May 2005
Posts: 382

Posted: Thu Jun 22, 2006 12:20 pm Post subject:
Re: Pop my thought balloon



Edward Green wrote:
Quote:  Can this be true? Will the expanding shell of accelerated charge
really produce zero advance warning in the field?

I'm pretty sure that's correct. Here's one way to see it: it follows from
symmetry that the B field must be radial with no theta or phi dependence.
The curl of any such field is identically zero. Therefore J + @E/@t = 0,
i.e. the E field only changes inside the expanding shell of charge.
 Ben 

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Thomas Smid science forum addict
Joined: 30 Apr 2005
Posts: 91

Posted: Thu Jun 22, 2006 2:29 pm Post subject:
Re: Pop my thought balloon



Edward Green wrote:
Quote:  I stumbled upon this gedanken, and would appreciate it if somebody
would either pop it or confirm its surprising conclusion.
A spherical shell of positive charge is held at radius r_o prior to
time t_o, when it is allowed to expand by electrostatic repulsion.
Prior to release the field outside the shell is the coulomb field of a
point charge of equal total charge at the center. Subsequent to
release, Gauss's law, which applies to dynamic as well as static charge
distributions (?), combined with the spherical symmetry, ensures that
the electric field cannot change at any point beyond the instantaneous
radius of expansion. Since the electrical field is static there is no
magnetic field in these regions either, and hence, all that happens as
the shell expands is the sweeping out of regions of field and replacing
them by zero field.
Can this be true? Will the expanding shell of accelerated charge
really produce zero advance warning in the field?

It should be true if you assume that the electrostatic interaction is
instantaneous. However, according to classical electrodynamics, even
static fields propagate with the speed of light. This means that the
charge distribution will not appear as spherically symmetric anymore
because you see the more distant regions of the shell effectively at an
earlier instant than the ones closer to you (because the field takes a
longer time to travel). You would therefore see a changing electric
field as the shell expands. I personally do not subscribe to this
assumption of retarded static fields however as I think it is
inconsistent with the concept of a force (see my webpage
http://www.physicsmyths.org.uk/retard.htm ), so I personally would say
that your conclusion is correct.
Basically the same arguments can of course also be applied to the
gravitational field of a spherically expanding mass. Again, the
assumption of an instantaneous gravitational field would result in the
latter being unaffected by the expansion outside the mass, but if one
assumes a finite speed of gravity, the gravitational field would
change.
Thomas 

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Ben RudiakGould science forum Guru
Joined: 04 May 2005
Posts: 382

Posted: Thu Jun 22, 2006 6:37 pm Post subject:
Re: Pop my thought balloon



Thomas Smid wrote:
Quote:  It should be true if you assume that the electrostatic interaction is
instantaneous. However, according to classical electrodynamics, even
static fields propagate with the speed of light. This means that the
charge distribution will not appear as spherically symmetric anymore
because you see the more distant regions of the shell effectively at an
earlier instant than the ones closer to you (because the field takes a
longer time to travel). You would therefore see a changing electric
field as the shell expands.

No, as I said, Maxwell's equations predict no change in the field. This is
true whether you use retarded potentials, advanced potentials, or half and half.
Quote:  I personally do not subscribe to this
assumption of retarded static fields

Well, now you have one less reason to doubt it...
Quote:  Basically the same arguments can of course also be applied to the
gravitational field of a spherically expanding mass. Again, the
assumption of an instantaneous gravitational field would result in the
latter being unaffected by the expansion outside the mass, but if one
assumes a finite speed of gravity, the gravitational field would
change.

We have a theory with a finite speed of gravity (general relativity) and I'm
almost certain that it makes the same prediction as Maxwell's theory here.
 Ben 

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Timo Nieminen science forum Guru Wannabe
Joined: 12 May 2005
Posts: 244

Posted: Thu Jun 22, 2006 7:44 pm Post subject:
Re: Pop my thought balloon



On Thu, 22 Jun 2006, Ben RudiakGould wrote:
Quote:  Edward Green wrote:
Can this be true? Will the expanding shell of accelerated charge
really produce zero advance warning in the field?
I'm pretty sure that's correct. Here's one way to see it: it follows from
symmetry that the B field must be radial with no theta or phi dependence. The
curl of any such field is identically zero.

And since div.B = 0, B=0 (at least for isotropic, linear, homogeneous
media).
Quote:  Therefore J + @E/@t = 0, i.e. the
E field only changes inside the expanding shell of charge.

And then there are those haircombing theorems, that every radiation field
from a finite source must have at least 2 singularities. Given a
spherically symmetric source, there can't be anything that tells you what
directions those singularities are in, so there can't be any radiation.

Timo Nieminen  Home page: http://www.physics.uq.edu.au/people/nieminen/
Eprints: 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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Edward Green science forum addict
Joined: 21 May 2005
Posts: 95

Posted: Thu Jun 22, 2006 9:58 pm Post subject:
Re: Pop my thought balloon



Timo A. Nieminen wrote:
Quote:  On Thu, 22 Jun 2006, Ben RudiakGould wrote:
Edward Green wrote:
Can this be true? Will the expanding shell of accelerated charge
really produce zero advance warning in the field?
I'm pretty sure that's correct. Here's one way to see it: it follows from
symmetry that the B field must be radial with no theta or phi dependence. The
curl of any such field is identically zero.

Stronger than that, an isotropic radial B field would imply a magnetic
monopole, no?
Quote:  And since div.B = 0, B=0 (at least for isotropic, linear, homogeneous
media).

Maybe that's another way of saying the same thing.
Quote:  Therefore J + @E/@t = 0, i.e. the
E field only changes inside the expanding shell of charge.
And then there are those haircombing theorems, that every radiation field
from a finite source must have at least 2 singularities. Given a
spherically symmetric source, there can't be anything that tells you what
directions those singularities are in, so there can't be any radiation.

That's a pretty cool theorem!
Wait a sec... are you saying this result is more general? No radiation
from finite spherically symmetric sources?
You probably weren't following our tedious discussion, but this
gendanken stems from the question of whether two like charges allowed
to accelerate away from each other would radiate. Certainly some
advance disturbance must propagate in this less symmetric case, but I'm
not sure it will mathematically qualify as "radiation".
Any thoughts? 

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Timo Nieminen science forum Guru Wannabe
Joined: 12 May 2005
Posts: 244

Posted: Thu Jun 22, 2006 11:14 pm Post subject:
Re: Pop my thought balloon



On Fri, 22 Jun 2006, Edward Green wrote:
Quote:  Timo A. Nieminen wrote:
On Thu, 22 Jun 2006, Ben RudiakGould wrote:
Edward Green wrote:
Can this be true? Will the expanding shell of accelerated charge
really produce zero advance warning in the field?
I'm pretty sure that's correct. Here's one way to see it: it follows from
symmetry that the B field must be radial with no theta or phi dependence. The
curl of any such field is identically zero.
Stronger than that, an isotropic radial B field would imply a magnetic
monopole, no?
And since div.B = 0, B=0 (at least for isotropic, linear, homogeneous
media).
Maybe that's another way of saying the same thing.

Just so!
Quote:  Therefore J + @E/@t = 0, i.e. the
E field only changes inside the expanding shell of charge.
And then there are those haircombing theorems, that every radiation field
from a finite source must have at least 2 singularities. Given a
spherically symmetric source, there can't be anything that tells you what
directions those singularities are in, so there can't be any radiation.
That's a pretty cool theorem!
Wait a sec... are you saying this result is more general? No radiation
from finite spherically symmetric sources?

No EM radiation from such. Acoustic radiation, or elastic waves from such
sources, sure. Acoustic hairy balls are actually bald, so haircombing
theorems don't apply, and elastic hairy balls don't need to be combed 
their hair can stick straight out without any singularities. EM hairy
balls insist on being neat and tidy and combed, so must have parts or
whorls, requiring the symmetry to be broken.
Quote:  You probably weren't following our tedious discussion, but this
gendanken stems from the question of whether two like charges allowed
to accelerate away from each other would radiate. Certainly some
advance disturbance must propagate in this less symmetric case, but I'm
not sure it will mathematically qualify as "radiation".

Sure. Why not? Except for special charge distributions and uniform
acceleration, accelerated charges radiate. Run it in reverse  two charges
coming together  and you have a case where it's known to radiate
(Brehmsstrahlung).
Note that the quadrupole moment of the two charges will be changing (the
monopole moment will be constant, the dipole moment will be zero, and the
quadrupole moment will be the lowest order multipole moment that matters).

Timo Nieminen  Home page: http://www.physics.uq.edu.au/people/nieminen/
Eprints: 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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Edward Green science forum addict
Joined: 21 May 2005
Posts: 95

Posted: Fri Jun 23, 2006 11:25 pm Post subject:
Re: Pop my thought balloon



Timo Nieminen wrote:
Quote:  On Fri, 22 Jun 2006, Edward Green wrote:
Timo A. Nieminen wrote:
And then there are those haircombing theorems, that every radiation field
from a finite source must have at least 2 singularities. Given a
spherically symmetric source, there can't be anything that tells you what
directions those singularities are in, so there can't be any radiation.
That's a pretty cool theorem!
Wait a sec... are you saying this result is more general? No radiation
from finite spherically symmetric sources?
No EM radiation from such. Acoustic radiation, or elastic waves from such
sources, sure. Acoustic hairy balls are actually bald, so haircombing
theorems don't apply, and elastic hairy balls don't need to be combed 
their hair can stick straight out without any singularities. EM hairy
balls insist on being neat and tidy and combed, so must have parts or
whorls, requiring the symmetry to be broken.

Tantalizing.
No doubt this is related to the transversality of EM waves, and
problems with mapping the surface of a sphere neatly to a cartesian
coordinate system.
Quote:  You probably weren't following our tedious discussion, but this
gendanken stems from the question of whether two like charges allowed
to accelerate away from each other would radiate. Certainly some
advance disturbance must propagate in this less symmetric case, but I'm
not sure it will mathematically qualify as "radiation".
Sure. Why not?

OK. I was thinking that the signature of radiation is the far field
dependence on r, and hence there might be time varying disturbances in
the field which propagated to infinity but did not qualify as
radiation. If this model _didn't_ radiate (which seems to be the wrong
answer), this would answer the follow on question how it was
possible/what it meant to have field changing at infinity, but no
radiation.
Quote:  Except for special charge distributions and uniform
acceleration, accelerated charges radiate. Run it in reverse  two charges
coming together  and you have a case where it's known to radiate
(Brehmsstrahlung).

For whatever it is worth, I suggest that even if formally
microscopically reversible, radiation is a thermodynamically
irreversible process. The microscopic reversal of Brehmsstrahlung is
not two charges flying apart and radiating, it's two charges flying
apart and at the focus of incoming radiation miraculously prearranged
so to anihilate itself while increasing the acceleration of the
charges. It's not clear that this tells us about the empircal vs. the
magical timereversal.
Quote:  Note that the quadrupole moment of the two charges will be changing (the
monopole moment will be constant, the dipole moment will be zero, and the
quadrupole moment will be the lowest order multipole moment that matters).

The argument was that there is no change in the dipole moment. I did
notice there would be a change in the quadrapole moment. So at that
moment I should merely have said "Oh, but there is a change in the
quadrapole moment", and been done with it?
Annoying thought experiment #437: Suppose two like charges separating
so as to steadily increase their quadrapole moment radiate. Next, take
an expanding distribution of like charges arranged so that variation in
the octopole moment is the leading term. Claim that this radiates.
Proceed upward through the npole moments, arguing that each radiates,
until, in the limit, we have a spherically symmetric distribution of
separating charges, which is known not to radiate.
Well, all that really implies is that the radiation goes to zero as n
> oo. 

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Timo Nieminen science forum Guru Wannabe
Joined: 12 May 2005
Posts: 244

Posted: Sat Jun 24, 2006 12:07 am Post subject:
Re: Pop my thought balloon



On Sat, 23 Jun 2006, Edward Green wrote:
Quote:  Timo Nieminen wrote:
On Fri, 22 Jun 2006, Edward Green wrote:
Timo A. Nieminen wrote:
And then there are those haircombing theorems, that every radiation field
from a finite source must have at least 2 singularities. Given a
spherically symmetric source, there can't be anything that tells you what
directions those singularities are in, so there can't be any radiation.
That's a pretty cool theorem!
Wait a sec... are you saying this result is more general? No radiation
from finite spherically symmetric sources?
No EM radiation from such. Acoustic radiation, or elastic waves from such
sources, sure. Acoustic hairy balls are actually bald, so haircombing
theorems don't apply, and elastic hairy balls don't need to be combed 
their hair can stick straight out without any singularities. EM hairy
balls insist on being neat and tidy and combed, so must have parts or
whorls, requiring the symmetry to be broken.
Tantalizing.
No doubt this is related to the transversality of EM waves, and
problems with mapping the surface of a sphere neatly to a cartesian
coordinate system.

That's why the hairy ball needs to be combed; the hair in the far field
isn't allowed to stick out.
Quote:  If this model _didn't_ radiate (which seems to be the wrong
answer), this would answer the follow on question how it was
possible/what it meant to have field changing at infinity, but no
radiation.

There's still an ongoing controversy, despite being "definitively
answered" back in the 1960s, as to whether a uniformly accelerated charge
radiates. A big part of the reason why the controversy still continues is
just that: what is radiation?
Quote:  Except for special charge distributions and uniform
acceleration, accelerated charges radiate. Run it in reverse  two charges
coming together  and you have a case where it's known to radiate
(Brehmsstrahlung).
For whatever it is worth, I suggest that even if formally
microscopically reversible, radiation is a thermodynamically
irreversible process. The microscopic reversal of Brehmsstrahlung is
not two charges flying apart and radiating, it's two charges flying
apart and at the focus of incoming radiation miraculously prearranged
so to anihilate itself while increasing the acceleration of the
charges. It's not clear that this tells us about the empircal vs. the
magical timereversal.

Consider: two charges moving together, slowing down: Brehmsstrahlung. They
come to relative rest, and then they move apart. Your scenario of two
charges moving apart is merely the tail end of a conventional
Bremsstrahlung process. It isn't a timesymmetry argument, just that why
would you expect the radiation to stop at a particular time while the
charges are still accelerating?
Quote:  Note that the quadrupole moment of the two charges will be changing (the
monopole moment will be constant, the dipole moment will be zero, and the
quadrupole moment will be the lowest order multipole moment that matters).
The argument was that there is no change in the dipole moment. I did
notice there would be a change in the quadrapole moment. So at that
moment I should merely have said "Oh, but there is a change in the
quadrapole moment", and been done with it?
Annoying thought experiment #437: Suppose two like charges separating
so as to steadily increase their quadrapole moment radiate. Next, take
an expanding distribution of like charges arranged so that variation in
the octopole moment is the leading term. Claim that this radiates.
Proceed upward through the npole moments, arguing that each radiates,
until, in the limit, we have a spherically symmetric distribution of
separating charges, which is known not to radiate.
Well, all that really implies is that the radiation goes to zero as n
> oo.

Given a source of finite size, past a certain n depending on the size, the
radiation at a given wavelength becomes insignificant. (n approx kr is the
usual rule of thumb.)
The computational methods I use to calculate optical forces and torques on
things in optical traps depend intimately on this convergence property. A
more mathematically inclined colleague tells me that, at least for the
(unusual) algorithm I use, the method is known to converge in the far
field for sufficiently large (but it's unknown as to how large is
sufficient) n. Convergence in the near field is unkown. Rayleigh
hypothesis and all that. What I do know is that it agrees with experiments
(except for one of our experiments where there was a discrepancy of x100 
but that had approximations on both the theoretical and experimental
sides, so I'll reserve judgment until I can get some really plausible
numbers). Since I'm on the applied physics side, and not on the maths
side, that's good enough for me for the moment.

Timo Nieminen  Home page: http://www.physics.uq.edu.au/people/nieminen/
Eprints: 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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j.alexander.hawkins@gmail science forum beginner
Joined: 28 Jun 2006
Posts: 5

Posted: Wed Jun 28, 2006 6:45 am Post subject:
Re: Pop my thought balloon



Edward Green wrote:
Quote:  I stumbled upon this gedanken, and would appreciate it if somebody
would either pop it or confirm its surprising conclusion.
A spherical shell of positive charge is held at radius r_o prior to
time t_o, when it is allowed to expand by electrostatic repulsion.
Prior to release the field outside the shell is the coulomb field of a
point charge of equal total charge at the center. Subsequent to
release, Gauss's law, which applies to dynamic as well as static charge
distributions (?), combined with the spherical symmetry, ensures that
the electric field cannot change at any point beyond the instantaneous
radius of expansion. Since the electrical field is static there is no
magnetic field in these regions either, and hence, all that happens as
the shell expands is the sweeping out of regions of field and replacing
them by zero field.
no, its not the zero field. the field inside is not zero, the 
integral....
{
/ E. dS=0 (crappy ASCII closed surface integral of a vector
field)
}
Quote: 
Can this be true? Will the expanding shell of accelerated charge
really produce zero advance warning in the field?
no. keep in mind gauss law doesnt give you the E field. all it gives 
you is....
SUM (normal component of the E field over the surface you
chose)dot(infantesimalsurface area)
i think the discontinuity upsets you, but keep in mind, its an integral
of the radial (in this case) component of the electric field summed
over infantesimal surface elements.. etc. etc. that your getting. NOT
the field. the charge distribution your talking about will still
radiate. 

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