Gauge choice for a magnetic vector potential

Click For Summary
SUMMARY

This discussion focuses on the verification of gauge choices for the magnetic vector potential A, specifically examining conditions such as A x F(r,t) = 0 and the implications of integral constraints like ∫∫ A x F(r,t) dS = 0. The conversation highlights established gauge conditions including the Lorenz gauge and Coulomb gauge, as well as additional conditions found in quantum field theory (QFT) such as temporal gauge and axial gauge. The validity of these conditions is determined by ensuring that the resulting electric field (E) and magnetic field (B) solutions satisfy the physical problem at hand.

PREREQUISITES
  • Understanding of magnetic vector potentials and gauge theories
  • Familiarity with Lorenz and Coulomb gauge conditions
  • Basic knowledge of quantum field theory (QFT) and gauge fixing
  • Proficiency in vector calculus and integral operations
NEXT STEPS
  • Research the implications of gauge choices in quantum electrodynamics (QED)
  • Study the role of gauge fixing conditions in quantum field theory
  • Explore the mathematical framework of the Rξ gauges
  • Investigate the physical significance of the integral constraints on gauge conditions
USEFUL FOR

Physicists, particularly those specializing in electromagnetism and quantum field theory, as well as students and researchers interested in gauge theories and their applications in theoretical physics.

Lodeg
Messages
12
Reaction score
0
How do we verify whether a condition on the magnetic vector potential A constitutes a possible gauge choice ?
Specifically, could a relation in the form A x F(r,t) be a gauge , where F is an arbitrary vector field?
 
Physics news on Phys.org
How do we verify whether a condition on the magnetic vector potential A constitutes a possible gauge choice ?
Specifically, could a relation in the form: A x F(r,t) = 0, be a gauge , where F is an arbitrary vector field?
 
This looks a bit too much constrained, because it practically says that ##\vec{A}=\lambda \vec{F}##, but you can impose only one "scalar" condition like
$$\frac{1}{c} \partial_t \Phi + \vec{\nabla} \cdot \vec{A}=0 \qquad \text{(Lorenz gauge)},$$
$$\vec{\nabla} \cdot \vec{A}=0 \qquad \text{(Coulomb gauge)}.$$
 
In that case, would a condition like ∫∫ A x F(r,t) dS = 0 , be acceptable as a guage?
 
More generally, could a condition like ∫∫ L(A) x F(r,t) dS = 0 be a gauge, where L is a linear operator?
 
I've never seen gauge-constraints involving integrals. This looks very complicated. What do you think it may be good for?
 
This allows simplifying some expressions.

Are there any criteria to judge the validity of such a condition?

Are there any reference that mention different gauge conditions other than Coulomb and Lorenz conditions?
 
Well, if it helps you with a concrete example, the only validation is to check that the final solutions for the physical fields, ##\vec{E}## and ##\vec{B}##, really solve the problem. In the literature there are also more gauge fixing conditions, particularly in QFT. Some simple ones are temporal gauge, ##A^0=0##, or axial gauge ##A^3=0##, and also the ##R_{\xi}## gauges, based on the action principle rather than a specific constraint for the gauge fields. There are also some more less common ones. See

https://en.wikipedia.org/wiki/Gauge_fixing
 
Thank you very much
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
414
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
869
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
740
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
613