108 views

Skip to first unread message

Dec 23, 2020, 11:06:54 AM12/23/20

to

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

Dec 25, 2020, 6:08:29 AM12/25/20

to

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!
>

> 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.

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.

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,

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

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?

not sure if this is easy to prove..

>

> Thanks in advance again,

>

> JMcA

>

Jos

Jan 3, 2021, 5:13:06 PMJan 3

to

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

Jan 4, 2021, 7:30:03 AMJan 4

to

> 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.

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.

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

Jan 6, 2021, 8:14:52 PMJan 6

to

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

Jan 12, 2021, 5:25:29 AMJan 12

to

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,

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.

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.

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

Jan 12, 2021, 7:12:51 PMJan 12

to

> > 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.)

Thanks,

JMcA

Jan 13, 2021, 3:20:02 AMJan 13

to

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.

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

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 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

Jan 13, 2021, 6:26:32 PMJan 13

to

On Friday, December 25, 2020 at 11:08:29 AM UTC, Jos Bergervoet wrote:

JMcA

Jan 14, 2021, 4:20:16 PMJan 14

to

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.

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?

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

Jan 14, 2021, 7:25:49 PMJan 14

to

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)."

JMcA

Jan 17, 2021, 5:50:02 AMJan 17

to

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

Jan 20, 2021, 6:35:33 PMJan 20

to

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!)

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

Jan 23, 2021, 5:30:02 PMJan 23

to

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..

> ...

> Resolution by Rohrlich

> https://en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field#Resolution_by_Rohrlich

All this does not answer your question whether the charge will be
> https://en.wikipedia.org/wiki/Paradox_of_radiation_of_charged_particles_in_a_gravitational_field#Resolution_by_Rohrlich

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

Mar 8, 2021, 9:06:47 PMMar 8

to

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

Mar 9, 2021, 6:30:39 PMMar 9

to

JMcA

Mar 13, 2021, 1:30:03 PMMar 13

to

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

Mar 13, 2021, 6:30:29 PMMar 13

to

> 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..

JMcA

Mar 16, 2021, 8:09:09 PMMar 16

to

> > 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.

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

Mar 16, 2021, 9:34:48 PMMar 16

to

[snipped]

JMcA

Reply all

Reply to author

Forward

0 new messages

Search

Clear search

Close search

Google apps

Main menu