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Free and bound charge

  1. Oct 24, 2013 #1
    Hi, I just need to ventilate and see your opinions.

    Charge can be partitioned into groups of particles of one or more elements. If they have more than one element it is a molecule, and its position is represented by for example the center of mass, seeing each particle as carrying both the property of mass and property of charge at the same time. Then, by Taylor expansion of the relative positions of the elements in the groups (the bound charges which seen as a group is a free charge), the H and D fields arise. Also, the center of mass position can be interpreted as the mass concentration in that point, so that the Nernst-Planck equation can be inserted into the expression. This gives an expression for Ohm's law. Do you have anything to add to this?
     
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  3. Oct 25, 2013 #2

    Simon Bridge

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    Some particles can carry charge. Such particles are said to be "charged".

    Some charged particles may have bound states that make up atoms. A single atom is an element.

    Not all bound states of charged particles are atoms.

    Atoms are usually neutral but may, in some situations, become "ionized" - where they gain or lose a number of electrons.

    A bound state of two or more atoms is called a molecule.

    The position of any macroscopic body may be defined as the position of it's the center of mass - with it's orientation being defined "about" the center of mass.

    You mean molecular ions or something?

    Free charges can be anything that carries a charge.
    Individual charges can have an intrinsic charge - otoh large scale objects can also carry a charge.

    You can have D and H fields without free charges.

    I think you need to be more careful with your terminology - and you need to be more explicit about the limits to the definitions you use.
     
  4. Oct 25, 2013 #3
    I mean elements of the group, not atoms! Like electrons or protons. I don't believe you about the D and H fields, without charged particles Maxwells equations do not apply. Like in Russakoff, there is always free charge (it can have charge 0) that is composed of bound charges. The "bound charges" are just the different groups, and each group is a free charge, so an atom/molecule is a free charge composed of bound protons and electrons, et cetera.
     
    Last edited: Oct 25, 2013
  5. Oct 25, 2013 #4

    DrDu

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  6. Oct 25, 2013 #5
    If he doesn't do spatial averaging, his equations are not smooth and thus curl is not defined? Curl only works on smooth differential manifolds? Something feels amiss. Also, why does he only include polarisation and not the higher order moments? Thanks anyway... Guess I'll have to search even more *sigh*. Actually no. I'll just write that it is a simple model and go with it. I wonder how this works together with plasma physics though, where spatial averaging is built into the model...
     
    Last edited: Oct 25, 2013
  7. Oct 25, 2013 #6

    DrDu

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    a) Polarisation is not a moment. So I don't understand what you mean with not including higher moments.
    b) Charge and current densities in matter are usually smooth functions, so polarisation and magnetisation are so, too.
    However, if you are only interested in the macroscopic fields, you are free to consider only the low wavenumber components of the field.

    This approach is especially useful in plasma physics.
    See e.g. the book by Nobel prize winner
    V. L. Ginzburg, The propagation of electromagnetic waves in plasma, Oxford, Pergamon, 1970
     
  8. Oct 25, 2013 #7
    I thought the plane wave approach to the field equations were only a partial solution. Okay I guess the wave equations for the particles in quantum physics is a smooth function. I was reading Classical multipole theory by Raab and Lange and they did not mention this in the non quantum chapter, how typical.
     
  9. Oct 25, 2013 #8

    DrDu

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    Even in classical mechanics I would regard non-smoothness to be rather a technical problem which can be overcome using distributions.
    I am not sure what you mean with plane wave approach yielding only a partial solution.
    Basically you can Fourier transform any charge- or currentdistribution or any field.
    This is independent on whether plane wave solutions are solutions of some free field equation or not.
     
  10. Oct 25, 2013 #9
    The approach I have seen is to write the free charges as
    [tex]\rho_f = \langle\sum_{g\in G}(\sum_{k\in g} q_k) \delta(\mathbf{x}-\mathbf{x}_g)\rangle[/tex]
    and free current as
    [tex]J_f = \langle\sum_{g\in G}(\sum_{k\in g} q_k)\frac{d \mathbf{x}_g}{d t} \delta(\mathbf{x}-\mathbf{x}_g)\rangle[/tex]
    where the partition of all particles is into groups g (so the union of all g in G is all the particles) that can consist of one or more particles and [tex]\mathbf{x}_g[/tex] is the center of mass of the group of particles. Is your point that this is wrong (since it ignores quantum effects), or is it the Taylor expansions that give rise to the D and H fields that is wrong (since it ignores large scale effects), or something else I didn't think of?

    The Fourier transform only admits certain classes of functions, and the plane wave is an Anszats is it not? That is why I am sceptical.
     
    Last edited: Oct 25, 2013
  11. Oct 25, 2013 #10

    DrDu

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    At least in QM, it is not possible to distinguish rigorously between free and bound charges. It turns out that this distinction is not necessary at all. A text-book example where you calculate the polarisation of free electrons is the calculation of the Lindhard dielectric function. http://en.wikipedia.org/wiki/Lindhard_theory
    That's also an example of a plasma in what you seem to be interested.

    As a basic theorem of applied physics, any physical useful function is Fourier transformable, whatever mathematicians object :-)
     
  12. Oct 25, 2013 #11

    Simon Bridge

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    I think you need to be more careful with your terminology - and you need to be more explicit about the limits to the definitions you use.

    The EM wave equation is a direct consequence of Maxwell's equations and describes an EM (D-H) field propagating in free space without any charges present - even though the equations themselves are commonly, as in your references, derived using charges, chargeless solutions to the equations exist and are physically meaningful.

    An example you will be familiar with is called "light".

    If the charge is bound then it is not free. You seem to be having trouble coming up with a consistent definition for "free charge".

    You mean you can have composite objects with a net zero charge? Then that's "net charge" not "charge".

    You can define a quantity and call it free charge and then set it's value to zero and so say you always have free charge even if it is zero but that's still the same as saying there are no free charges present.

    When people say there are no charges - they are saying the amount of charge is zero.

    It is possible, in the models that produce Maxwel's equations, to have a region of space with no physical matter inside it - a classical vacuum - therefore, no composite objects available to carry charge. Ergo - no charge... not even zero net charge. You must have heard of this??

    You appear to be using archaic terminology - this is not helpful.
    You are best advised to adopt more up-to-date models.
     
  13. Oct 26, 2013 #12

    Dale

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