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Coulomb gauge derivation (static field)

  1. Jan 20, 2016 #1
    Hi, I am a little confused of derivation of Coulomb Gauage.


    mfield_h2.png (2)

    First, prime notation is adopted to describe the magnetic field density source current.
    Non-prime notation is for position that we are specifically interested in (ex. the position magnetic force acts on).


    mfield15.png (23)
    if equation (13) is true, then divergence of vector potential is 0.


    Hence,
    mfield7.png (13)
    vecid1.png (14)


    we can use vector indentity (14) to express (13) in another form.
    mfield10.png (15)
    mfield11.png (16)
    mfield12.png (17)
    mfield13.png (18)

    at this step, I am lost. We can break down divergence of vector potential into
    two parts. I understand the first part is 0 because we assume that the field is static with respect to time,
    but I wonder why the second flux integral part is 0.



    (reference: http://ghebook.blogspot.ca/2011/06/energy-of-magnetic-field.html)
     
    Last edited: Jan 20, 2016
  2. jcsd
  3. Jan 26, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jan 27, 2016 #3

    vanhees71

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    2016 Award

    I can't read the website, but the 2nd integral should be 0 since the volume must enclose the full region, where the current density is non-zero, i.e., at the boundary of the volume ##\vec{J}=0##, and thus the 2nd integral vanishes. The first integral vanishes, because the integrability condition for the magnetostatic field is ##\vec{\nabla} \cdot \vec{J}=0##.

    The integrability condition for the full Maxwell equations is the continuity equation for the current density, i.e.,
    $$\partial_t \rho + \vec{\nabla} \cdot \vec{J}=0,$$
    which is the local form of electric-charge conservation and a consequence of the gauge invariance of electromagnetics.
     
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