Coulomb gauge derivation (static field)

In summary, the conversation discusses the derivation of Coulomb Gauge and the use of vector identities to express it in different forms. The integrability condition for the magnetostatic field and the full Maxwell equations are also mentioned. The conversation ends with a question about why the second flux integral is 0.
  • #1
kidsasd987
143
4
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:
  • #3
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.
 

What is Coulomb gauge derivation?

Coulomb gauge derivation is a mathematical process used to obtain the electric potential and electric field equations for a static electric field.

Why is Coulomb gauge derivation important?

Coulomb gauge derivation is important because it provides a mathematical framework for understanding and analyzing the behavior of static electric fields, which are important in many areas of science and technology.

What are the key assumptions made in Coulomb gauge derivation?

The key assumptions made in Coulomb gauge derivation include assuming that the electric field is static (does not change over time), that there are no sources or sinks of electric charge, and that the electric field is divergence-free (has a net charge of zero).

How is Coulomb gauge derivation different from other gauge derivations?

Coulomb gauge derivation is unique in that it specifically focuses on the behavior of static electric fields, while other gauge derivations may consider different types of fields (such as time-varying fields) or different physical phenomena.

What are some real-world applications of Coulomb gauge derivation?

Coulomb gauge derivation has practical applications in many areas, including electrical engineering, physics, and materials science. It is used to analyze and design electric circuits, study the properties of materials, and predict the behavior of electric fields in various systems.

Similar threads

Replies
3
Views
2K
Replies
1
Views
997
Replies
3
Views
968
Replies
3
Views
22K
  • Electromagnetism
Replies
14
Views
1K
Replies
1
Views
765
  • Electromagnetism
2
Replies
51
Views
6K
Replies
3
Views
2K
Replies
8
Views
743
  • Electromagnetism
Replies
1
Views
2K
Back
Top