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Paradox about Maxwell near an accumulation of charge

  1. Oct 18, 2012 #1
    Hi. I am trying to understand Maxwell's equations. I am used to Gaussian units. The following sums up what I do not understand. (Or at least one thing I do not understand ... )

    Let x0 be a position situated in a vacuum, with no photons in it, next to an infinite pane of glass. Ions arrive at the glass on the other side. What happens to the electric field at x0?
    Due to the Coulombic potential from the arriving ions, common sense tells us that the electrostatic potential gradient at x0 is changing, and that an electric field is therefore increasing in magnitude.
    However, Faraday's equation (microscopic, differential) reads
    dB/dt = -c ∇× E
    Since there is no current at x, the Ampere-Maxwell equation reads
    dE/dt = c ∇× B
    If both the electric and magnetic fields are regarded as locally determined, this means that the information about the E field has to propagate from the glass to x0 as a wave involving the B field.
    But we said the glass was infinite, say in the y and z direction, so dE/dt at the glass (caused by the current of ions arriving) could be constant in y and z. Then everywhere, ∇×E is zero (Ex is constant in y and z, Ey=Ez=0) so B should not be changing anywhere.
    It seems like under Maxwell, the location x0 will never experience the electric field from the nearby charge accumulation.
    What did I do wrong?

    To put it another way, are there situations where applying Maxwell's equations - or at least, the differential version of them - is not physically correct?

    The reason I ask this is that I am working on a problem involving a quasineutral plasma filament in a DC circuit. My real paradox is different:
    Ampere's equation J = c/4π ∇× B approximately holds, and it then follows from Faraday, and Gauss (Div E = 0) that
    dJ/dt = c2/4π ∇2 E.
    However, for Coulomb gauge magnetic potential A, it also follows from Ampere that
    J = -c/4π ∇2 A, so it follows (by equating expressions for dJ/dt) that
    2 E = -1/c ∇2 dA/dt
    from which it follows that the electrostatic gradient satisfies Laplace's equation. But therefore from dJ/dt = c2/4π ∇2 E, the p.d. has no influence on the current flowing!

    I'm not a physicist so it's not surprising that I got confused, but it's very important that I get to grips with this problem. Any help much appreciated.
     
    Last edited: Oct 18, 2012
  2. jcsd
  3. Oct 18, 2012 #2

    Simon Bridge

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    No. However, there are situations where it is conceptually easier to use the integral forms.

    Note:
    * ions arriving at the glass plate are an electric current
    * real ions may have their own non-zero magnetic dipole moments
    * the glass plate accumulates charge
    * the charges give rise to their own electric potentials
    * glass is a dielectric

    If you have position x0 near an infinite plane of uniform charge distribution (i.e. the charge has already accumulated) you'd be able to apply Maxwels equations OK right?
     
  4. Oct 18, 2012 #3
    Thanks v much for reply.

    Hmm no, I don't think I know how to apply Maxwell in that situation. We know Div E = 0 since there is a neighbourhood of x0 with no charge density, and we know -c Curl E = dB/dt, but I don't see where we'd be getting dB/dt to know E that way. Presumably dB/dt should be 0. Sorry if I am being very slow here. So that is an even simpler example where I do not understand.

    I may have unnecessarily complicated things by adding the glass. Say that the charges just stop moving when they reach a certain plane, for some unknown reason.
     
  5. Oct 18, 2012 #4
    OK I think after all maybe I chose a bad example. The infinite plane of charges creates a divergent integral of the Coulombic potential at x0, I think. And if you take a large sheet of charges, then that integral is probably tending to be constant in x.

    Presumably if you have a finite lump of charges and you are some distance away from it, propagation of E via curl E -> dB/dt -> curl B -> dE/dt is the correct inference. That strikes me as a bit strange though since why call it electrostatic interaction if it's propagated by electromagnetic waves. Is it correct?

    If you have any comment on the second paradox I'd be interested to hear what I am doing wrong there also.
     
  6. Oct 18, 2012 #5

    mfb

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    If your ions cover the whole space (at least perpendicular to their propagation), you run in unphysical situations as you have an infinite amount of charges.
    With a finite beam, the field is not uniform along the glass plate and you can get an electric field.
     
  7. Oct 18, 2012 #6

    Simon Bridge

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    The infinite plane of charge?
    Well if there were a spherically distributed charge dencity in the neighborhood of x0 the the field at that point would still be zero wouldn't it?

    So there is something wrong with your application of the equations.

    How about the even simpler case ... you have a ball of charge Q radius R centered at the origin, and you are on the x axis at x=x0 some distance x0>>R away from the ball.

    By your reasoning, there are no free charges close to x0, therefore ∇E=0 there.
    So far so good. But you go on to conclude that E=0 which is not correct. You have forgotten that the field has to be continuous and there are some places where there are free charges.

    What you actually have is that ##\vec{\nabla}\cdot\vec{E}=0:|\vec{r}|>R## and ##\nabla\vec{E}=3Q/4\pi R^3 \epsilon_0: |\vec{r}|\leq R## (see next post)

    Which pretty much means you need to know about the charge "over there" to know about the field "here"... you are going to say that this spoils the locality of the classical description of EM fields. (I suspect this is at the heart of your question.)

    You want the field to be entirely locally determined and you have discovered that you cannot do that. But that's not what we mean. It is the force that a charge experiences that is locally determined ... all the charge knows is that it experienced the local electric field pulling on it.

    The locally determined thingy here is more properly the electric potential.
    (Even more properly - it should be considered/used with the vector potential.)

    A charge rolls around on a "terrain" described by the electric potential much as a ball rolls around on a physical surface which is described by the gravitational potential at each point on the surface. The surface is flexible - at any point on the surface, the curvature, slope, and height, is determined by the way the surface is pulling on that point in each direction ... so it is entirely locally determined.

    However - in order to calculate the local curvature etc we need to know the source of the distortion. In the "rubber sheet" analogy - we need to know about what is pressing down on the sheet in order to calculate the height of the sheet at other locations. We also need to know about some properties of the sheet - like the distribution of elastic coefficients between mass elements, and, for that matter, the mass distribution. For the electric potential we need to know the permitivity and the charge distribution. These describe properties of the "fabric of space-time" that are important to electrodynamics just like mass-distribution and elasticity are important properties of the fabric of the rubber sheet for rolling marbles.

    Be wary though because the rubber sheet analogy is just that - an analogy. It has it's limitations - especially in GR, where you more usually see it. Hopefully you can see, now, how we can talk about force-fields being locally determined (by the potential).

    If you really want to get rid of every last shred of non-locality, though, you have to go to the full Quantum Field Theory... which is another can of wyrms. I'd concentrate on getting used to using Maxwell's equations first.
     
    Last edited: Oct 18, 2012
  8. Oct 18, 2012 #7

    Simon Bridge

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    (... from the redirect above)
    I wanted to elaborate on the ##\vec{\nabla}\cdot\vec{E}=0## thing ... if you look at it, for the example, then you end up with $$r^2\frac{d}{dr}E(r)+2rE(r)=0:r>R$$... which does not give you ##E(r=x_0)=0## ... so the reasoning is flawed from a purely mathematical standpoint.
    Just from noticing that there are no charges close to ##x_0##, and knowing the symmetry of the charges elsewhere, we see that at that point, ##E(x_0)=A/x_0^2##.

    Even without knowing - and taking the rectangular coordinates - and supposing the the position ##(x,y,z)=(x_0,0,0)## is centered in a region where ##\rho=0## then you still get:$$\vec{\nabla}\cdot\vec{E}=\frac{\partial}{ \partial x}E_x+\frac{\partial}{\partial y}E_y+\frac{\partial}{\partial z}E_z=0$$ ... which is true if each component of ##\vec{E}=(E_x,E_y,E_z)## is a constant ... but can also be true if, say $$\frac{\partial}{\partial x}E_x=-\frac{\partial}{\partial y}E_y \; ; \; \frac{\partial}{\partial z}E_z=0$$ ... and many other possible combinations.

    You should now realize how to apply Maxwell's equations from this example.
     
    Last edited: Oct 18, 2012
  9. Oct 18, 2012 #8

    Simon Bridge

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    To what extent does that matter? We use unphysical situations all the time ... for instance, it is a common exercise for students to compute the field due to an infinite plane of charge.

    The setup may have the charges coming with a uniform current density j from the -x direction and encountering an infinite plane barrier at x=0 where they accumulate as a uniform charge density σ(t) - and you need to compute the field at x=x0 > 0. presumably dσ/dt=j.

    This is very similar to the infinite plane of charge with a time-varying charge distribution.
     
  10. Oct 19, 2012 #9

    mfb

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    An infinite plane is fine - and leads to a constant field everywhere, even if that plane is moving. But what about an infinite volume of constant charge density? Your integrals diverge.
     
  11. Oct 19, 2012 #10

    Simon Bridge

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    Zarquon's knees! That's a good point - since at any time there is a volume of charge occupying everything x<0. I should be more alert.
     
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