Coulomb gauge derivation (static field)

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kidsasd987
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Hi, I am a little confused of derivation of Coulomb Gauage.
Equation for magnetic force between two current loops: F_m equals the closed line integral over loop c of I dl times the closed line integral over loop c prime of mu0/(4 pi) times I' dl' cross r-hat over R squared.
(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).
Vector identity written as curl of B-bar equals curl of curl of A-bar, then equals gradient of divergence of A-bar minus Laplacian of A-bar, which equals minus the Laplacian of A-bar.
(23)
if equation (13) is true, then divergence of vector potential is 0.Hence,
Equation: divergence of vector A equals 0, written with the nabla operator dotted with A.
(13)
Vector identity: divergence of (f times A-vector) equals f times the divergence of A-vector, plus A-vector dotted with the gradient of f.
(14)we can use vector indentity (14) to express (13) in another form.
Equation: the divergence of J' over R, with the gradient operator acting on the unprimed coordinates, equals J' dotted with the gradient of 1/R, which equals minus J' dotted with the primed gradient of 1/R.
(15)
Vector identity for the divergence of a product: the divergence of (J' divided by R) equals one over R times the divergence of J', plus J' dotted with the gradient of one over R.
(16)
Vector identity: divergence of (J' over R) equals (1/R) times the divergence of J' minus the gradient of (1/R) dotted with J'.
(17)
Equation: divergence of A equals mu0/(4 pi) times the integral over v' of (1/R) times the divergence of J' dv', minus the closed surface integral over s' of (J' dot da')/R.
(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:
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.