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Scalar potential for magnetic field

  1. Mar 29, 2015 #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?
     
  2. jcsd
  3. Mar 30, 2015 #2

    vanhees71

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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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