Scalar potential for magnetic field

AI Thread Summary
The discussion centers on the scalar potential ∅ in relation to the magnetic field B, specifically through the Debye potentials. It is established that the equation ∇²∅ = 0 indicates that ∅ must be finite everywhere and leads to the conclusion that ∅ must be a constant under certain conditions. The contributions to B from the terms involving Lχ and ∇∅ are shown to be zero, reinforcing the significance of the scalar potential's constancy. The absence of magnetic charges further implies that Δ∅ = 0 everywhere, which, with appropriate boundary conditions, results in ∅ being zero everywhere. This highlights the relationship between scalar potentials and magnetic fields in the context of vector field decomposition.
Mr. Rho
Messages
14
Reaction score
1
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?
 
Physics news on Phys.org
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.
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top