Why Does Vector Potential Align with Current in Infinite Distributions?

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SUMMARY

The discussion centers on the alignment of vector potential (A) with current density (J) in infinite current distributions, as presented in Griffiths' "Introduction to Electrodynamics." The expression A = (μ/4π) ∫(J / (r - r')) dV is valid for finite current distributions, yet examples 5.12 and 5.25 demonstrate that even in infinite distributions, the direction of A remains aligned with J. This indicates that the relationship holds true despite the gauge dependency of the vector potential, particularly in the context of Coulomb gauge.

PREREQUISITES
  • Understanding of vector potential in electromagnetism
  • Familiarity with current density and its mathematical representation
  • Knowledge of gauge theories, specifically Coulomb gauge
  • Proficiency in calculus and integral equations
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  • Study Griffiths' "Introduction to Electrodynamics" for detailed examples of vector potential
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Students and professionals in physics, particularly those specializing in electromagnetism, as well as educators seeking to clarify the relationship between vector potential and current density in various distributions.

Kolahal Bhattacharya
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We know in all ordinary cases the direction of vector potential will be mimic to the direction of current.As it is evident from the expression of A=(mu/4*pi) int{J /(r-r')}dV which is valid only for finite current distributions.
But in example 5.12 and in exercise 5.25 Griffiths has given situations where current distributions extend upto infinity.But,still the direction of A is the same as direction of J .Can you help me to reconcile these two apparently contrasting views?
 
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Kolahal Bhattacharya said:
We know in all ordinary cases the direction of vector potential will be mimic to the direction of current.
You do know it is gauge dependent, right (The formula you brought works for Coulomb)? I don't know about the example you talked about.
 

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