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I Magnetostatics: What if "steady" currents were divergent?

  1. Oct 7, 2016 #1
    Why must steady currents be non-divergent in magnetostatics?

    Based on an article by Kirk T. McDonald (http://www.physics.princeton.edu/~mcdonald/examples/current.pdf), it appears that the answer is that by extrapolating the linear time dependence of the charge density from a constant divergence of current density, the charge density is unbounded in time.

    But surely it should be possible within a finite time interval that the divergence of currents is maintained, such that during that finite time interval, we could conceivably end up with a magnetostatic situation which could briefly sustain divergent current. Being magnetostatic, such a system would not radiate during that finite time interval since the magnetic field would be unchanging. However, because the electric potential is changing during this interval, there would be an extra electric field term that varies inversely with the distance from each source of changing charge density.

    Despite the temporarily constant currents, existing only for a finite time interval, only one of the fields, the electric field, is changing. The electric field term due to time-dependent charge density would be steady if the time dependence of the charge density is linear, but there is an electric scalar potential which produces an electric field (the negative gradient of the electric scalar potential) with linear time-dependence, which drops with the inverse square of the distance.

    I have been trying to look for a reliable source regarding this particular condition, which of course only lasts temporarily, although for how long could of course depend on the parameters of the system. A simple example would be a simple RC circuit with a very large value for resistance R and capacitance C with a very steady discharge rate for a significant time period. Another would be a LC circuit with a very high value of inductance L as well as capacitance C, which results in a very low resonant frequency. Given this leads to the possibility for non-closed currents existing for arbitrarily long periods of time without changing magnetic fields, and therefore no radiation (in a quasi-static approximation of course), I have been perplexed by the fact that in principle this could lead to Lorentz forces which do not necessarily cancel between two non-closed currents.

    Shouldn't there be an additional "missing force" which comes into play which addresses the obvious imbalance? What is that exactly?

    One thought that came to mind is that changing the electric scalar potential essentially results in opposite changes of mass for negative charges vs. positive charges, due to the effect on their potential energy due to the changing electric scalar potential. In the realistic case that the wire bearing the non-closed currents are electrically neutral, then the effect on the net electric potential energy of the wire is of course nil, and therefore so the effect on the total mass of the wire would also be nil. However, since the conduction electrons have a velocity relative to the atoms of the wire, and hence a drift velocity, the change of the momentum due to the equal and opposite changes of mass is non-zero.

    I suspect that this would be the resolution, but I need some help in confirming that this is indeed the explanation that resolves the issue of a "missing force".

    Sincerely,

    Kevin M.
     
  2. jcsd
  3. Oct 7, 2016 #2

    Dale

    Staff: Mentor

    Then it would not be magnetostatic. It would be magnetodynamic. What you envision is certainly physically possible, it is just called something else.
     
  4. Oct 7, 2016 #3
    So basically then "magnetodynamic" could be used to refer to certain scenarios where magnetic fields don't change. That is certainly awkward though. I would like to think there would have been a proper term to label this particular type of scenario, but I have never seen it.
     
  5. Oct 7, 2016 #4

    Dale

    Staff: Mentor

    Well, magnetodynamic isn't a standard term, I was just emphasizing the fact that the scenario you described would not be magnetostatic. Usually the term electrodynamics includes all non static situations.
     
  6. Oct 8, 2016 #5

    vanhees71

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    2016 Award

    In magnetostatics you have Ampere's Law (in Heaviside-Lorentz units),
    $$\vec{\nabla} \times \vec{H}=\frac{1}{c} \vec{j}.$$
    This implies necessarily
    $$\vec{\nabla} \cdot \vec{j} = c \vec{\nabla} \cdot (\vec{\nabla} \times \vec{H})=0,$$
    i.e., ##\vec{\nabla} \cdot \vec{j}## is a necessary constraint to make Ampere's law solvable. So it's a integrability constraint for the magnetostatic equations to be solvable at all!
     
  7. Oct 8, 2016 #6
    Hmm. Why didn't I see that? So essentially, this approach to magnetostatics implicitly assumes the absence of displacement current, and therefore it assumes that there is no changing electric field. Strictly speaking then, if I understand correctly, a system with no changing magnetic field is not necessarily magnetostatic if it amits a changing electric field whose curl is zero, as per my scenario.
     
  8. Oct 9, 2016 #7

    vanhees71

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    2016 Award

    In the vacuum the Maxwell equations read (in Heaviside-Lorentz units)
    $$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0,\\
    \vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
    The first two equations, Faraday's Law of induction and Gauss's Law for the magnetic field, are constraint equations and lead to the introduction of the potentials, but that's not necessary classical electrodynamics since it complicates things, because it introduces the gauge freedom (which however is at the heart of electromagnetism from a fundamental point of view).

    Now we take the divergence of the 3rd equation, the Ampere-Maxwell Law. This gives
    $$\partial_t \vec{\nabla} \cdot \vec{E}=-\vec{\nabla} \cdot \vec{j},$$
    and with the 4th equation, Gauss's Law for the electric field, this gives
    $$\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0,$$
    which is nothing else than the local form of electric-charge conservation, which is a necessary condition for the consistency of the Maxwell equations.

    Now let's investigate your scenario. Assuming that the magnetic field is time independent, also the current density should be time-indepenent, which implies that ##\partial_t \vec{E}=\text{const}##, i.e., you'd get ##\vec{E}(t,\vec{x})=t \vec{E}_0(\vec{x})##, and from Gauss's Law for the electric field that also ##\rho(t,\vec{x})=t \rho_0(\vec{x})##. Then you have
    $$\vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j} +\frac{1}{c} \vec{E}_0.$$
    The continuity equation gets
    $$\rho_0+\vec{\nabla} \cdot \vec{j}=0.$$
    This gives indeed a formal solution of the Maxwell equations, but the charge density and the electric field grow to infinity at late times. This is unphysical. That's why in magnetostatics one assumes that also the electric field and the charge distribution are time-independent.
     
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