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I When does the vector potential A affect the E field?

  1. Oct 28, 2017 #1
    This is probably fairly simple, but I have a hard time to grasp the concept of the vector potential A in electromgnetism. Especially, in the following equation for the electric field :

    [tex]\vec{E} = - \nabla V - \frac{\partial \vec{A}}{\partial t}[/tex]

    When does the second term is not 0 ?
     
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  3. Oct 28, 2017 #2

    Orodruin

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    When the vector potential is time-dependent.
     
  4. Oct 28, 2017 #3
    Aha, thanks, but that's obvious in the equation. I'm looking to what makes it time-dependent.

    EDIT : Well actually, I suppose this is when B is time-dependant ?
     
  5. Oct 28, 2017 #4

    Orodruin

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    You need charge distributions or currents that change over time. Otherwise everything can be treated as static.
     
  6. Oct 28, 2017 #5

    Dale

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    See Jefimenko's equations.
     
  7. Oct 29, 2017 #6

    vanhees71

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    As the most simple example take the free plane-wave modes of the electromagnetic field. In the radiation gauge, i.e., with the potentials fulfilling
    $$A^0=V=0, \quad \vec{\nabla} \cdot \vec{A}=0,$$
    You have
    $$\vec{A}(t,\vec{x})=\vec{A}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}),$$
    where ##\vec{A}_0 \cdot \vec{k}=0## and ##\omega=c |\vec{k}|##. Then
    $$\vec{E}=-\frac{1}{c} \partial_t \vec{A}=\mathrm{i} \omega/c \vec{A}_0 \exp(\cdots), \quad \vec{B}=\vec{\nabla} \times \vec{A} = \mathrm{i} \vec{k} \times \vec{A}_0 \exp(\cdots).$$
     
  8. Oct 29, 2017 #7
    If the underlying space is simply connected, the time derivative of the vector potential vanishes if the time derivative of the magnetic field vanishes:

    ##\mathrm{d}_\mathrm{s}B = 0 \Rightarrow B = \mathrm{d}_\mathrm{s} A##, where the subscript s stands for "spatial". Or in other notation ##\vec{\nabla} \times \vec{A} = \vec{B}##. You can pull the time derivative into ##\mathrm{d}_\mathrm{s}##.

    The time derivative of ##\vec{B}## vanishes if and only if ##\vec{j} + \frac{\partial \vec{E}}{\partial t} = 0## (in natural units with ##c=1##).

    If the underlying space is not simply connected, there does not a priori exist a vector potential, and hence the right side of your equation does not necessarily exist.
     
  9. Oct 30, 2017 #8
    Thanks for the answers and insights ! It does help.
     
  10. Nov 7, 2017 #9

    Jano L.

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    That's wrong. According to Maxwell's equations, the time derivative of B vanishes if and only if
    $$
    \nabla \times \mathbf{E} = \mathbf{0}.
    $$
     
  11. Nov 7, 2017 #10
    Sure, I would edit it, but somehow this isn't possible any longer. But thanks for correcting it.
     
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