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A Magnetostatics force equation for continuous current density

  1. Oct 16, 2016 #1
    In Jackson, the following equations for the vector potential, magnetostatic force and torque are derived


    ##\mathbf{m} = \frac{1}{{2}} \int \mathbf{x}' \times \mathbf{J}(\mathbf{x}') d^3 x'##

    ##\mathbf{A} = \frac{\mu_0}{4\pi} \frac{\mathbf{m} \times \mathbf{x}}{\left\lvert {\mathbf{x}} \right\rvert^3}##

    ##\mathbf{F} = \boldsymbol{\nabla}( \mathbf{m} \cdot \mathbf{B} ),##

    ##\mathbf{N} = \mathbf{m} \times \mathbf{B},##

    where $$\mathbf{B}$$ is an applied external magnetic field and $$\mathbf{m}$$ is the magnetic dipole for the current in question.

    These all follow from an analysis of localized current densities $$\mathbf{J}$$, evaluated far enough away from the current sources. I worked through the vector potential results, and made sense of his derivation (lots of sneaky tricks are required). I've also done the same for the force and torque derivations. While I now understand the mathematical steps he uses, there's a detail about the starting point of his derivation, where he writes

    ##\mathbf{F} = \int \mathbf{J}(\mathbf{x}) \times \mathbf{B}(\mathbf{x}) d^3 x##

    This is clearly the continuum generalization of the point particle Lorentz force equation, which for zero electric field is

    ##\mathbf{F} = q \mathbf{v} \times \mathbf{B}##

    For the point particle, this is the force on the particle when it is in the external magnetic field. i.e. this is the force at the position of the particle.

    However, for the continuum Force equation, it integrates over all space. How can we have a force that is applied to all space, as opposed to a force applied at a single point, or across a surface?
     
  2. jcsd
  3. Oct 16, 2016 #2

    Charles Link

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    I think the answer is you are computing the total forces on all currents and/or all magnetic moments and these currents and magnetic moments are distributed over all space.
     
  4. Oct 16, 2016 #3
    That seems a bit strange in general, but in the special case of localized currents seems like a reasonable thing to calculate.
     
  5. Oct 17, 2016 #4

    vanhees71

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    It's not strange. To the contrary, what's strange and in fact makes a lot of trouble, are point particles in electrodynamics. They are the strangers in a continuum theory. This becomes clear from the fact that a single-point charge's charge and current density are given by (in relativistic notation)
    $$j^{\mu}(x)=\int_{-\infty}^{\infty} \mathrm{d} \tau q \frac{\mathrm{d} y^{\mu}(\tau)}{\mathrm{d} \tau} \delta^{(4)}[x-\vec{y}(\tau)].$$
    The "force law" follows from Noether's theorem applied to the Lagrangian of interacting charge-current distributions.
     
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