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Constitution of a field?

  1. Dec 31, 2008 #1
    if an electric charge pops into space it's electromagnetic field is constituted everywhere in space instantaneously right? so what aspect of its influence propagates at the speed of light? i know if i perturb the charge then an electromagnetic wave will propagate at the speed of light but i'm considering before that, when the charge just appears?

    to be clearer: if a charge pops into space its influence propagates at the speed of light ( this i know from herr einstein ), but its field is constituted instantaneously. so what part/form/expression/aspect of its influence is it that propagates at the speed of light?
     
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  3. Dec 31, 2008 #2

    jtbell

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    I consider this to be an unanswerable question, at least in the context of classical electrodynamics, because a charge can't simply "pop into space." It would violate the continuity equation for electric charge and current, which can be derived from Maxwell's Equations.
     
  4. Dec 31, 2008 #3

    HallsofIvy

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    I was under the impression that an electron's "field" simply expressed its "influence" at each point. How can the field be "constituted" instantaneously (and what do you mean by "constituted").
     
  5. Jan 2, 2009 #4
    i buy that but then how about in qed? i mean for example i've heard virtual particles pop in and out of existence due to "quantum fluctuations" and if one of these virtual particle anti-particle pairs gets sucked into a black hole then the surviving shoots off into space. so then the question becomes how quickly does that particles field propagate?
    well i mean the particle and hence the field weren't there before and supposing the particle appears the concomitant field appears.
     
  6. Jan 2, 2009 #5

    jtbell

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    For the QED version you should ask in the Quantum Physics forum, not this one. :wink:
     
  7. Jan 2, 2009 #6

    Vanadium 50

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    JTBell is right - you don't have charges "pop" into place. Asking for the details of a process that can't occur is a little like asking "what's north of the North Pole?" - and there is usually a follow on question analogous to "Well, but if you could go north of the North Pole, what would you see?"

    While it is impossible to create a charge, it is possible to create a dipole: an object with one side that is relatively positive and another side that is relatively negative. This is done by moving charge from one spot to a nearby spot. Note that in QED this is all that happens - you never create a virtual electron without also creating a virtual positron next to it.

    One can ask how long it takes the fields to change in response to this dipole. At a distance r from the dipole, you see a change in the fields at t = r/c.
     
  8. Jan 2, 2009 #7

    jtbell

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    Now that I think of it, you can construct a classical analog to particle-antiparticle creation. Start with a sphere with a uniform positive charge distribution. It's initially stationary, then starts moving to the right. You can calculate (at least in principle) the electric field produced by this charge distribution, as a function of time. Outside the sphere, it starts as the familiar 1/r^2 field. When the sphere starts to move, the resulting changes in the field propagate outward at speed c.

    Now imagine a second sphere, identical to the first one, but with a negative charge distribution. It's initially stationary, then starts moving to the left. The field it produces is similar to the one from the positive sphere, except for the direction of the field while stationary, and the changes are different in direction when it starts to move, because the motion is in the opposite direction.

    Now superpose the two spheres onto each other initially. The charge densities and fields add, by the principle of superposition. Initially, the net charge density is zero everywhere, and so is the net field. As the spheres start to move, and separate from each other, regions of net positive and negative charge appear, and the net field becomes nonzero. It seems to me that the changes that make the net field nonzero should propagate outward at speed c, because the changes in the individual fields do.
     
  9. Jan 2, 2009 #8

    Vanadium 50

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    That's exactly right - you can do this with superposition. What you will find is that the two 1/r^2 terms almost exactly cancel. You will be left with a single, static 1/r^3 piece (field from a dipole) that propagates out at c, and a 1/r piece from the retarded terms that also propagates out at c. That latter piece is the radiation you get from a time-varying electric dipole.
     
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