Vector Potential and Zero Divergence

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Discussion Overview

The discussion focuses on the conditions under which a vector field can be expressed as the curl of a vector potential, specifically addressing the possibility of having a vector potential with zero divergence. The scope includes theoretical aspects of vector potentials in electromagnetism and gauge transformations.

Discussion Character

  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant seeks to understand the proof that a vector potential can have zero divergence, referencing the equation B=∇×A.
  • Another participant explains the concept of gauge transformation, stating that if A = A' + ∇ξ, where ξ is an arbitrary scalar field, then both A and A' yield the same magnetic field B.
  • This participant further elaborates that by finding a ξ that satisfies ∇²ξ = -∇·A', one can perform a gauge transformation to achieve a vector potential A with zero divergence, known as the Coulomb gauge.

Areas of Agreement / Disagreement

Participants appear to agree on the mechanics of gauge transformations and the implications for vector potentials, but the initial participant's inquiry about the proof remains open and unresolved.

Contextual Notes

The discussion does not resolve the specific conditions or assumptions required for the existence of a vector potential with zero divergence, nor does it clarify the implications of the gauge transformation on the scalar potential.

hellsingfan
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I'm trying to understand when a vector field is equal to the curl of a vector potential. Why is it possible that there is always a vector potential with zero divergence?

relevant Equation:

B=∇χA

I'm trying to understand the proof that the above vector potential A can be one with zero divergence.
 
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We know that if we perform a gauge transformation A = A' + \triangledown \xi, where \xi is an arbitrary scalar field, then both A' and A result in the same observed magnetic field i.e. B = \triangledown \times A' = \triangledown \times A (and of course, as usual, we have to perform the associated gauge transformation of the scalar potential to keep the observed electric field the same).

Say we are given a vector potential A'. We can find a \xi that solves \triangledown ^{2}\xi = -\triangledown \cdot A'. Performing the gauge transformation A = A' + \triangledown \xi we see that \triangledown \cdot A = \triangledown \cdot A' + \triangledown ^{2}\xi = 0 hence we can fix this gauge (again, after performing the associated gauge transformation of the scalar potential) so that we have B = \triangledown \times A, \triangledown \cdot A = 0. This is called the Coulomb gauge.
 
Thank You!
 
No problem!
 

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