Effect of the self-field of a charged sphere on itself when hyperbolically accelerated?

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john mcandrew

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Dec 23, 2020, 11:06:54 AM12/23/20
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First, a Merry Christmas to the Christians and a celebratory solstice to everyone else! Now onto my question, and thanks in advance for any replies:

Let a continuous total charge +q be constrained to move freely on a spherical surface, referring to this from now on as a hollow charged sphere.
If this is statically placed in a constant electric field E, the surface charge will be pushed towards one end so that the total electric field inside returns to zero. Letting the sphere accelerate, I now have to take into account the self-field being retarded, affecting each charge element in the rest frame of the sphere by a retarded electric field E'. This looks pretty difficult to calculate because I need to know the distribution of the polarized charge around the sphere which in turn will be effected by the retarded self-field E' acting back on it etc. Which leads me to my question:

What is the charge distribution of a hyperbolically accelerated charged hollow sphere?

Intuitively, I'd expect this retarded self E' to lower the self energy of the polarized static sphere by depolarizing it, thus providing the necessary energy to redistribute the charge: perhaps the charge is evenly redistributed?

Thanks in advance again,

JMcA

Jos Bergervoet

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Dec 25, 2020, 6:08:29 AM12/25/20
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On 20/12/23 5:06 PM, john mcandrew wrote:
>
> First, a Merry Christmas to the Christians and a celebratory solstice to everyone else! Now onto my question, and thanks in advance for any replies:

Merry Christmas John, and to everyone else in the newsgroup!

> Let a continuous total charge +q be constrained to move freely on a spherical surface, referring to this from now on as a hollow charged sphere.

You have to give more information, especially about the mass (we could
have 3 mass contributions here: the mass of the supporting sphere, that
of the moving charge, and the electromagnetic mass of the surrounding
fields).

> If this is statically placed in a constant electric field E, the surface charge will be pushed towards one end so that the total electric field inside returns to zero.

For that case there is an exact solution (e.g. in Jackson). But that
is strictly speaking not unique. On top of it the charge can be
sloshing back and forward (of course these oscillations will die out
eventually, but still, the steady state will then only be reached
at t=infinity)

some time thes

> Letting the sphere accelerate, I now have to take into account the self-field being retarded,

But do you suddenly release the sphere? Or do you arrange for a more
"smooth" onset of the acelleration?

> .. affecting each charge element in the rest frame of the sphere by a retarded electric field E'. This looks pretty difficult to calculate because I need to know the distribution of the polarized charge around the sphere which in turn will be effected by the retarded self-field E' acting back on it etc. Which leads me to my question:
>
> What is the charge distribution of a hyperbolically accelerated charged hollow sphere?

Also here the solution will not be unique.. Especially if you release
the sphere quite suddenly. Then there certainly will be a "sloshing
mode" for the charge in the early stages.

> Intuitively, I'd expect this retarded self E' to lower the self energy of the polarized static sphere by depolarizing it, thus providing the necessary energy to redistribute the charge: perhaps the charge is evenly redistributed?

I would expect the same (after some steady state is reached) but I'm
not sure if this is easy to prove..


>
> Thanks in advance again,
>
> JMcA
>

--
Jos

john mcandrew

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Jan 3, 2021, 5:13:06 PMJan 3
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I had in mind the change in the electric field being 'negligible' over the sphere's radius to minimize the sloshing of the charge. I think the calculation shouldn't be too hard in first assuming the charge distribution in equilibrium is uniform, and then calculating the total electric force on any charge element dq. I'd expect this would be exactly what's required to maintain the constant charge density or 'rigidness'; so that the trailing edge is given an extra shove compared to the leading edge.

This sloshing of charge seems unavoidable when this charged sphere is moving in a constant magnetic field B; where the electric field in the rest frame of the sphere rotates the charge distribution at a constant angular velocity, if I'm not mistaken.

JMcA

Jos Bergervoet

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Jan 4, 2021, 7:30:03 AMJan 4
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> I had in mind the change in the electric field being 'negligible' over the sphere's radius to minimize the sloshing of the charge.

But you need no pre-existing field at all for the sloshing. It simply
will happen if you bring the charge out of balance.. And of course
it dampens out by radiating (or dissipating if your conductor has
resistance).

> I think the calculation shouldn't be too hard in first assuming the charge distribution in equilibrium is uniform, and then calculating the total electric force on any charge element dq. I'd expect this would be exactly what's required to maintain the constant charge density or 'rigidness'; so that the trailing edge is given an extra shove compared to the leading edge.

I also think the calculation is simple, once we agree that we can
avoid the sloshing. And that will be an oscilation with a Q-factor
only in the order of 1 (like any dipole antenna). So we can quickly
hand-wave it away!

> This sloshing of charge seems unavoidable when this charged sphere is moving in a constant magnetic field B; where the electric field in the rest frame of the sphere rotates the charge distribution at a constant angular velocity, if I'm not mistaken.

"Unavoidable" would only apply if there is incoming radiation of
a certain frequency, for which the sphere will be a receiving antenna.

And I would hesitate to call the sphere's accelerated frame a "rest
frame". But of course it has its own frame which is stationary in
Rindler coordinates.. Acc. to GR's equivalence principle we can view
the sphere as stationary in a frame with a constant gravity field.
And in your case you also have the constant E-field. In fact the two
fields should totally balance each other in the sphere's frame..

This means that there will be a stationary solution with the charge
pulled to one side of the sphere by the E-field and the sphere itself
pulled to the other side by the gravity field. (Assuming the sphere
has non-zero mass). Any additional sloshing will die out soon and just
leave us with this stationary solution.

NB: A pure E-field does not change under Lorentz boost parallel to
the field so it remains what it was, even in the accelerated frame.

--
Jos


dditional






john mcandrew

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Jan 6, 2021, 8:14:52 PMJan 6
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I know little about GR, but I would say that the gravitational field also pulls the charge to the other side, cancelling the polarizing effect of the E field on the charge, as well as accelerating the sphere depending upon its bare mass: The net effect being a uniformly distributed charge density on the surface of the sphere, with the field lines curved.

Here, I'm interpreting one version of GR's equivalence principle as: an accelerated frame in flat space-time is equivalent to a stationary frame with the equivalent necessary gravitational field.

JMcA

Jos Bergervoet

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Jan 12, 2021, 5:25:29 AMJan 12
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Amd it isn't needed. Since the equivalence principle in this case
returns everything to a stationary case with a constant gravity
field.

> but I would say that the gravitational field also pulls the charge to the other side, cancelling the polarizing effect of the E field on the charge,

No, gravity pulls everything to one side, the E-field pulls only
the charge to the other side, so you do get polarization.

> as well as accelerating the sphere depending upon its bare mass: The net effect being a uniformly distributed charge density on the surface of the sphere, with the field lines curved.

No, the net effect being a polarized sphere. But of course the
field lines are indeed curved. They would be curved in almost
any solution you construct from either polarized or uniform
charge on the sphere after adding the uniform external E-field.

>
> Here, I'm interpreting one version of GR's equivalence principle as: an accelerated frame in flat space-time is equivalent to a stationary frame with the equivalent necessary gravitational field.

Yes, that's the easy part. You then have to reason a bit further
to make sure that the initial uniform E-field remains the same
in this new frame (the same strength and still uniform).

And the solution is a charged sphere hovering. The lift from the
external E-field is balancing gravity. (If it isn't then we haven't
chosen the accelerated frame correctly, it should be co-moving
with the accelerated sphere.)

--
Jos

john mcandrew

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Jan 12, 2021, 7:12:51 PMJan 12
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I'm assuming you mean GR rather than the E field isn't needed, and Newtonian gravity is adequate?

> > but I would say that the gravitational field also pulls the charge to the other side, cancelling the polarizing effect of the E field on the charge,
> No, gravity pulls everything to one side, the E-field pulls only
> the charge to the other side, so you do get polarization.
> > as well as accelerating the sphere depending upon its bare mass: The net effect being a uniformly distributed charge density on the surface of the sphere, with the field lines curved.

> No, the net effect being a polarized sphere. But of course the
> field lines are indeed curved. They would be curved in almost
> any solution you construct from either polarized or uniform
> charge on the sphere after adding the uniform external E-field.

This is what confuses me: the polarizing of the surface charge by the E field means the corresponding energy density and therefore equivalent mass at each point also varies upon the spherical surface. Hence I'd expect the gravitational field to now have a varying effect upon the surface charge, causing it to at least polarize to a degree.

> >
> > Here, I'm interpreting one version of GR's equivalence principle as: an accelerated frame in flat space-time is equivalent to a stationary frame with the equivalent necessary gravitational field.
> Yes, that's the easy part. You then have to reason a bit further
> to make sure that the initial uniform E-field remains the same
> in this new frame (the same strength and still uniform).
>
> And the solution is a charged sphere hovering. The lift from the
> external E-field is balancing gravity. (If it isn't then we haven't
> chosen the accelerated frame correctly, it should be co-moving
> with the accelerated sphere.)

I agree that the charged sphere is hovering, but also that the charge remains uniformly distributed as in the equivalent accelerated frame without the gravitational field, which agrees with the equivalence principle here if I'm not mistaken.

Thanks,

JMcA

Jos Bergervoet

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Jan 13, 2021, 3:20:02 AMJan 13
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"turns everything into a stationary case with a constant gravity
field and a (co-linear) constant E-field," I should have written
perhaps.

>>> but I would say that the gravitational field also pulls the charge to the other side, cancelling the polarizing effect of the E field on the charge,
>> No, gravity pulls everything to one side, the E-field pulls only
>> the charge to the other side, so you do get polarization.
>>> as well as accelerating the sphere depending upon its bare mass: The net effect being a uniformly distributed charge density on the surface of the sphere, with the field lines curved.
>
>> No, the net effect being a polarized sphere. But of course the
>> field lines are indeed curved. They would be curved in almost
>> any solution you construct from either polarized or uniform
>> charge on the sphere after adding the uniform external E-field.
>
> This is what confuses me: the polarizing of the surface charge by the E field means the corresponding energy density and therefore equivalent mass at each point also varies upon the spherical surface. Hence I'd expect the gravitational field to now have a varying effect upon the surface charge, causing it to at least polarize to a degree.

It is already polarized anyway by the opposing directions of
the E-field and the gravity, but some extra polarizing by this
second-order effect is obviously possible.

>>>
>>> Here, I'm interpreting one version of GR's equivalence principle as: an accelerated frame in flat space-time is equivalent to a stationary frame with the equivalent necessary gravitational field.
>> Yes, that's the easy part. You then have to reason a bit further
>> to make sure that the initial uniform E-field remains the same
>> in this new frame (the same strength and still uniform).
>>
>> And the solution is a charged sphere hovering. The lift from the
>> external E-field is balancing gravity. (If it isn't then we haven't
>> chosen the accelerated frame correctly, it should be co-moving
>> with the accelerated sphere.)
>
> I agree that the charged sphere is hovering, but also that the charge remains uniformly distributed

That isn't "agreeing" since I clearly stated that the charge is
not uniformly distributed.

> as in the equivalent accelerated frame without the gravitational field, which agrees with the equivalence principle here if I'm not mistaken.

The equivalence principle will not change the charge. It may deform
the sphere by Lorentz contraction when going from one frame to the
other, but in thic case that won't change a non-uniform distribution
into a uniform one.

And the charge will be non-uniform. Except in the case when the sphere
has zero mass (and all mass is in the charged particles and in the
E-field density, and no mass in the rigid sphere. But that is
unphysical).

--
Jos

john mcandrew

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Jan 13, 2021, 6:26:32 PMJan 13
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On Friday, December 25, 2020 at 11:08:29 AM UTC, Jos Bergervoet wrote:
You're agreeing with me here, in that the charge will end up evenly distributed in the steady state when the sphere is hyperbolically accelerated: the applied E field polarizing the charge and the retarded E' field depolarizing it, cancelling one another exactly. But doesn't this imply the charge is also uniformly distributed in the equivalent case with a gravitational field replacing the accelerated frame?

JMcA

Jos Bergervoet

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Jan 14, 2021, 4:20:16 PMJan 14
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Agreeing with you that the lowest energy will eventually be
reached, also agreeing that depolarization will happen after
initial polarization, but it will not be depolarized back
to zero. After the E-field is (magically) switched on and the
sphere starts to accelerate, I would (intuitively) expect the
polarization vs. time to be something like this:

* *
* * * * * *
* * *
* * *
0*

Of course the 'sloshing' will consist of infinitely many
oscillations, decreasing exponentially in strength.

> in that the charge will end up evenly distributed in the steady state when the sphere is hyperbolically accelerated: the applied E field polarizing the charge and the retarded E' field depolarizing it, cancelling one another exactly.

It isn't what I would intuitively expect, I'd expect that as
long as the acceleration lasts, the charge is being pulled a
little bit ahead of the sphere's centre, so it remains
polarized all the time.

> But doesn't this imply the charge is also uniformly distributed in the equivalent case with a gravitational field replacing the accelerated frame?

My statement would imply that the sphere is *also polarized*
when hovering in the gravitational field. Also there, the
charge is pulled upwards by the E-field and will be off-center.

And in the latter case I am much more confident that this is
easy to prove! (Of course we then do have to prove in addition
that the equivalence principle actually applies here, but still
this reduces the original problem to proving two sub-problems.)

--
Jos

john mcandrew

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Jan 14, 2021, 7:25:49 PMJan 14
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I agree with the above for a sphere with a finite bare mass that dominates the electromagnetic mass. And I also agree with what you've written elsewhere in this thread:
https://groups.google.com/g/sci.physics.electromag/c/JTPBOZ3W4wU/m/bqoSBw5fCQAJ

"The equivalence principle will not change the charge. It may deform
the sphere by Lorentz contraction when going from one frame to the
other, but in thic case that won't change a non-uniform distribution
into a uniform one.

And the charge will be non-uniform. Except in the case when the sphere
has zero mass (and all mass is in the charged particles and in the
E-field density, and no mass in the rigid sphere. But that is
unphysical)."

Would I be correct in saying that you believe the polarizing and depolarizing effects on the surface charge approach cancelling one another as the electromagnetic mass >> bare mass?

JMcA

Jos Bergervoet

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Jan 17, 2021, 5:50:02 AMJan 17
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When we discussed what could "intuitively" be expected, I only
considered decent, simple, 19th-century electrostatics, with
physical objects that exist in everyday life!

With a mass-zero sphere (and mass-less charge carriers as well)
things might start to escape our range of intuition.. I would
not be sure whether the pull of gravity will actually change the
direction of field lines, for instance, if the system is hovering
in a constant gravity field. The problem is that, even if the
equivalence principle allows us to use that frame, I do not know
how Maxwell's equations might change in a non-inertial frame. (It
probably can be looked up on Wikipedia, but intuitively I would
just expect the unexpected.)

For safety, I would go back to the original frame, where we know
Maxwell, and try to solve it there. But the combination of E-
and B-fields and moving charges make this complicated of course.
The precise shape of the field is not obvious to begin with, and
since you ask whether some things might be precisely "cancelled",
you would really need precise information!

Alternatively, we can work in the sphere's own stationary frame
and *assume* that Maxwell is unaltered for electrostatics in a
gravity field. In that case your proposed solution with uniform
charge does not seem to be stable. There is an upward pull of the
E-field on the charge carriers (which have no mass so they are
not pulled down by gravity) and there must be a downward pull of
gravity on the E-field energy density around the sphere (which
contains all the mass now). So intuitively we might still expect
the charge to move off centre (upwards) and the E-field lines to
be drooping downwards.. but the latter is incompatible with the
starting assumption that electrostatics is unaltered by gravity.
(So probably it isn't!)

--
Jos

john mcandrew

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Jan 20, 2021, 6:35:33 PMJan 20
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Initially, I can imagine the sphere to have a large bare mass compared to the electromagnetic so that it's 'gently" accelerated. The surface will obviously be an equipotential with a non-uniform charge distribution giving rise to an internal 0 electric field. As the bare mass --> 0, the acceleration increases causing the retarded field to have an increasing effect on the charge distribution, noting that I only have to worry about the retarded electric field for steady state conditions in the sphere's frame where the surface remains an equipotential.

For hyperbolic motion, the LAD equation for a point charge has the Shott and radiation terms cancelling one another; the Shott term being the "acceleration energy" above. So likewise I'm confident of this cancelling for a charge sphere also, although to be sure I need to calculate it of course.

> Alternatively, we can work in the sphere's own stationary frame
> and *assume* that Maxwell is unaltered for electrostatics in a
> gravity field. In that case your proposed solution with uniform
> charge does not seem to be stable. There is an upward pull of the
> E-field on the charge carriers (which have no mass so they are
> not pulled down by gravity) and there must be a downward pull of
> gravity on the E-field energy density around the sphere (which
> contains all the mass now). So intuitively we might still expect
> the charge to move off centre (upwards) and the E-field lines to
> be drooping downwards.. but the latter is incompatible with the
> starting assumption that electrostatics is unaltered by gravity.
> (So probably it isn't!)

GR looks like a minefield to me even for the experts which is why I try to keep away from it. I'm adding the following two links to remind me later if I come back to this thread:

Paradox of radiation of charged particles in a gravitational field
https://en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field

Resolution by Rohrlich
https://en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field#Resolution_by_Rohrlich

JMcA

Jos Bergervoet

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Jan 23, 2021, 5:30:02 PMJan 23
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But that makes it intriguing to see which mines there are in any
given case! In the case here it is the EM-field's behavior in a non-
inertial frame, i.e. with constant gravity present. In that case
Maxwell's equations, and even simple Coulomb electrostatics, do not
apply unaltered!

As you can read in Wikipedia's Rohrlich paragraph below, "Maxwell
equations do not hold" and a bit further "the gravitational field
slightly distorts the Coulomb field".

Of course we also know that combined E- and B-fields, like a moving
wave packet, would not propagate according to plain Maxwell, because
that would be rectilinear motion of the packet, and we know that
light is bent downwards by gravity. But it seems even electrostatics
is affected..

> ...
All this does not answer your question whether the charge will be
uniformly distributed in the sphere's (non-inertial) frame, but the
fact that gravity "slightly distorts" the ordinary electrostatic
situation, as stated, would make me believe (intuitively) that the
charge is not uniform any more..

--
Jos


john mcandrew

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Mar 8, 2021, 9:06:47 PMMar 8
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The Figure at the end of section three in this paper shows the fields in the instantaneous rest frame of a hyperbolically accelerated charged spherical shell:

The fields and self-force of a constantly accelerating spherical shell
https://royalsocietypublishing.org/doi/10.1098/rspa.2013.0480

Do you agree that the direction of the electric field inside shown here is incorrect and needs to be reversed?

JMcA

john mcandrew

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Mar 9, 2021, 6:30:39 PMMar 9
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On second thoughts it looks right: its internal electric field cancels the applied electric field along the +x-axis giving a zero net electric field inside.

JMcA

Jos Bergervoet

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Mar 13, 2021, 1:30:03 PMMar 13
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Thanks for theinteresting reference! What is a bit strange, however, is
that the abstract says about their result "this is conjectured to be
exact", whereas the introduction describes the approach as "restricting
the motion to a simple case, and solving it exactly." So which is it?

What is very pleasing to see is that the author does actually define
what is meant by the rigid motion with constant acceleration, which
would of course otherwise lend itself for dozens of interpretations.
Here it is "moves such that there exists a frame in which the whole
shell is at rest at some moment", period.

As for the drawing of the field lines that you mention: I believe there
is no "applied" electric field! I did not see it mentioned in the text
before that point. I think the assumption is that the acceleration is
caused by other forces, left unspecified, and that the calculation
only looks at what field is then created by the shell itself.

And whether the drawing is correct I can't judge, I did not try to
follow the details of the calculation. But there's nothing in the
picture that looks impossible. The field looks divergence-free in
empty space, and you can always find a charge distribution on the
shell that gives such a field (starting with a double layer for the
case of a Faraday cage with external and internal surface charges
to do the two seperate jobs of creating those fields, and then just
uniting them to one layer.)

What worries me a little is that those field lines do not have zero
tangential component (along the shell surface). So why wouldn't the
charges move? It is a conducting sphere, I'd expect. So maybe you
are right and the drawing is at least a bit sloppy..

--
Jos

john mcandrew

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Mar 13, 2021, 6:30:29 PMMar 13
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It's exact for hyperbolic motion only.

> What is very pleasing to see is that the author does actually define
> what is meant by the rigid motion with constant acceleration, which
> would of course otherwise lend itself for dozens of interpretations.
> Here it is "moves such that there exists a frame in which the whole
> shell is at rest at some moment", period.
>
> As for the drawing of the field lines that you mention: I believe there
> is no "applied" electric field! I did not see it mentioned in the text
> before that point. I think the assumption is that the acceleration is
> caused by other forces, left unspecified, and that the calculation
> only looks at what field is then created by the shell itself.

This is my view also where the fields here only depend upon the acceleration of the charged sphere. Nevertheless, it struck me that this internal electric field appears in the calculation, and is equal but in the opposite direction to the needed E field to accelerate the charge.

> And whether the drawing is correct I can't judge, I did not try to
> follow the details of the calculation. But there's nothing in the
> picture that looks impossible. The field looks divergence-free in
> empty space, and you can always find a charge distribution on the
> shell that gives such a field (starting with a double layer for the
> case of a Faraday cage with external and internal surface charges
> to do the two seperate jobs of creating those fields, and then just
> uniting them to one layer.)

Also, note how the internal electric field isn't exactly uniform but slightly higher at the leading edge than the trailing edge. This maintains the rigidness of the sphere, but I'm not sure if the author has added this in, or whether this comes from the calculation itself.

> What worries me a little is that those field lines do not have zero
> tangential component (along the shell surface). So why wouldn't the
> charges move? It is a conducting sphere, I'd expect. So maybe you
> are right and the drawing is at least a bit sloppy..

The applied electric field E provides the additional necessary tangential component.

JMcA

john mcandrew

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Mar 16, 2021, 8:09:09 PMMar 16
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[snipped]

> > What worries me a little is that those field lines do not have zero
> > tangential component (along the shell surface). So why wouldn't the
> > charges move? It is a conducting sphere, I'd expect. So maybe you
> > are right and the drawing is at least a bit sloppy..

> The applied electric field E provides the additional necessary tangential component.

I think I'm wrong here after rethinking about it. In the calculation the author assumes each component charge is accelerated by the necessary constant force keeping it rigid with its neighbours, and so comes under a force having a component (Poincare non EM force) not explicitly shown, but assumed in the diagram: this keeps the charges from moving wrt one another. Jos's comment has also made me rethink my OP where I claimed the retarded E field depolarizes the the charge so that it becomes continuous, whereas now I think I'm also wrong here. I didn't think carefully enough about the physics of keeping a charged sphere with a large bare mass stationary in a constant electric field E, and then releasing it: does the charge on and hence the electric field from the sphere remain polarized when it accelerates?

Before I said yes, whereas now I find it embarrassingly obvious it's no, after realizing what's responsible for the redistribution of the charges and the electric field: the non EM forces of constraint keeping the charges confined to the surface of the sphere. Despite these non-EM forces of constraint doing no net work on the charges, they're still able to move them around and hence change the distribution of the EM field while keeping its total energy constant. Also, keeping the sphere stationary requires an external force which can either be EM and hence oppose the applied EM, or non-EM and applied to the bare mass. There's a complex interplay of how the non-EM forces and EM forces interact with one another for any constrained EM system, no matter how small the total bare mass is, and this can't be ignored.

JMcA

john mcandrew

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Mar 16, 2021, 9:34:48 PMMar 16
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This isn't correct. At the instant the external force is removed from the bare mass the infinitesimal charge there will move in the direction of the total local E field , opposite to the non-EM constraining force there; so there is an exchange of some energy/momentum in the non-EM forces and the EM field.

[snipped]

JMcA

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