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Vector potential of B

  1. Nov 14, 2004 #1
    vector potential of magnetic field

    Hi everyone.
    I have a question about the vector potential of magnetic field, A:
    we just set the divergance(A)=0 inorder to define the potential. I just wanted to know why we are allowed to do this.
    Thanks a lot.
    Somy :smile:
  2. jcsd
  3. Nov 14, 2004 #2


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    It is a mathematical fact that ANY vector [tex]\vec{v}[/tex] can be written as:
    Here, [tex]\Phi[/tex] is the associated scalar potential, whereas [tex]\vec{A}[/tex] is the associated vector potential.

    A hand-wavy argument for the allowance of the divergence-free condition on [tex]\vec{A}[/tex] is the following:
    An arbitrary [tex]\vec{v}[/tex] has 3 free components, whereas an arbitrary scalar potential [tex]\Phi[/tex] plus an arbitrary vector [tex]\vec{A}[/tex] has 4 free components.
    Hence, we should be able to express [tex]\vec{v}[/tex] in terms of [tex]\Phi,\vec{A}[/tex] by adding an additional constraint on [tex]\vec{A}[/tex]
    (In addition to the 3 component equations, of course)
    Last edited: Nov 14, 2004
  4. Nov 14, 2004 #3
    As in electrodynamics the fields are the important quantities and the potentials are only introduced to make life easier, you can do anything with A if it does not affect the fields. This is more or less the same situation as with the scalar potential V wich can be altered with a constant because that does not change the electric field; this is called a 'gauge freedom'.

    Why does adding a constant to V not alter the value of the electric field? Because the electric field is the gradient of the V and the gradient of a constant is zero. So if you call this constant v then:

    [tex]\vec{E}=-\nabla (V+v)=-\nabla V[/tex] because [tex]\nabla v=0[/tex] with v a constant.

    In the case of the magnetic field the relation between the vector potential and the magnetic field is given by:

    [tex]\vec{B}=\nabla \times \vec{A}[/tex]

    And you can look this up if you can't recall it but the curl of a gradient is zero. This is a mathematical fact just like the divergence of a constant is zero. So adding a gradient to A does not effect the magnetic field. Let's call this gradient [itex]\nabla a[/itex] and the original vector potential (without the extra gradient added) [itex]\vec{A}_0[/itex] then:

    [tex]\nabla \cdot \vec{A} = \nabla \cdot \vec{A}_0 + \nabla^2 a[/tex]

    And we can choose the value of the gradient such that the divergence of A vanishes without changing the magnetic field (i.e. by choosing [itex]\nabla^2 a= -\nabla \cdot \vec{A}_0[/itex])
  5. Nov 14, 2004 #4


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    The magnetic field has no sources, so its diverges is zero everywhere. If it had a divergence within any closed reason it would have a "magnetic charge" somewhere in there.
  6. Nov 14, 2004 #5
    The question is about the magnetic vector potential, not the magnetic field...
  7. Nov 15, 2004 #6
    Thank you da willem,
    The answer was clear and useful.
  8. Nov 15, 2004 #7


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    I think this is a more convincing proof and it is quite general,since it uses the Helmholtz theorem which is valid vor any vector field defined on [tex] R^3 [/tex].
    And i think selfAdjoint's remark is a correct one,since it's Gauss's law for magnetistatics that u exploit in setting [tex]\vec{B}=\nabla \times \vec{A}[/tex]
    And the condition you invoked is A GAUGE FIXING CONDITION which is actually a supplimentary first class constraint imposed at Hamiltonian level,in order to make the Legendre transformation invertible...Why does it have that form??Simply,because it is chosen to resemble the secondary first class constraint of the em field which is of course:[tex] -\nabla\vec\Pi=0 [/tex].
  9. Nov 15, 2004 #8
    I think it is obvious SelfAdjoint misunderstood the question, as he talks about the divergence of the magnetic field and not the divergence of the magnetic vector potential. His remark is correct, but not very appropriate in this thread...
  10. Nov 15, 2004 #9


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    I think the relevance and the purpose of such a forum is to give people as much as possible correct answers and to do that in a comprehendable way.It's maybe so that some of us,if not all,missinterpret the original questions and end up discussing other issues.
    And i strongly believe that's not a sin.I therefore leave it to the (presumably) interested reader to see whether my post "is appropriate" in that thread.
  11. Nov 15, 2004 #10
    Ok, maybe you're right. I as well don't think it is a sin to deviate from the original question, overall this is a good thing. But I just pointed out that the sender was refering to something else so the original post-er would not be confused... :smile:
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