Scalar potential for magnetic field

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SUMMARY

The discussion centers on the scalar potential ∅ in the context of magnetic fields, specifically addressing the equation ∇²∅ = 0. It is established that ∅ must be a constant due to the implications of the Debye potentials in magnetic field calculations, as referenced in C.G. Gray's work from the American Journal of Physics. The conclusion drawn is that with the absence of magnetic charges and appropriate boundary conditions, the scalar potential φ equals zero everywhere, reinforcing the necessity of a constant scalar potential in this framework.

PREREQUISITES
  • Understanding of vector calculus, specifically divergence and Laplacian operators.
  • Familiarity with magnetic field theory and the concept of magnetic charges.
  • Knowledge of Debye potentials and their application in electromagnetism.
  • Ability to interpret mathematical expressions related to vector fields and potentials.
NEXT STEPS
  • Study the implications of the Debye decomposition of vector fields in electromagnetism.
  • Explore the mathematical properties of Laplace's equation and its boundary conditions.
  • Investigate the role of magnetic field lines and the absence of magnetic monopoles in classical physics.
  • Review C.G. Gray's 1978 paper in the American Journal of Physics for deeper insights into scalar potentials.
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Physicists, electrical engineers, and students studying electromagnetism, particularly those focusing on magnetic field theory and vector calculus applications.

Mr. Rho
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I have that ∇2∅ = 0 everywhere. ∅ is a scalar potential and must be finite everywhere.
Why is it that ∅ must be a constant?

I'm trying to understand magnetic field B in terms of the Debye potentials: B = Lψ+Lχ+∇∅. I get this from C.G.Gray, Am. J. Phys. 46 (1978) page 169. Here they found that Lχ=0 and ∇∅=0 and therefore gives no contribution to B.

any help?
 
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First of all the Debye decomposition of an arbitrary vector field is given as
$$\vec{V}=\vec{L} \psi + \vec{\nabla} \times (\vec{L} \chi)+\vec{\nabla} \phi.$$
By definition
$$\vec{L}=\vec{x} \times \vec{\nabla}.$$
First of all you have
$$\vec{\nabla} \cdot \vec{V}=0,$$
because
$$\vec{\nabla} \cdot (\vec{L} \psi)=\partial_j (\epsilon_{jkl} r_k \partial_l \psi)=\epsilon_{jkl} (\delta_{jk} \partial_l \psi +r_k \partial_j \partial_l \psi)=0$$
and
$$\vec{\nabla} \cdot (\vec{\nabla} \times \vec{L} \chi)=0.$$
For the magnetic field you additionally have the absence of magnetic charges,
$$\vec{\nabla} \cdot \vec{B}=0.$$
This implies
$$\Delta \phi=0$$
everywhere. With the appropriate boundary conditions this implies that ##\phi=0## everywhere.
 

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