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Is it theoretically possible to create any electric field

  1. Apr 29, 2012 #1
    So given any vector field V(x, y, z) in three dimensional space, is it possible to create an electric field that is described by V.
    Someone told me that according to Helmholtz theorem, a vector field (with certain constrains) can be expressed as a sum of a curless field and a divergenceless field. To me since charge density is basically the divergence of the electric field, and change in B field creates curl in electric field, it should be possible to create any electric field with these two elements. The procedure would be: given V, decompose it into a curless field and a divergenceless field; then put electric charges in space such that the charge density is described by the divergenceless field at all points; similary, do the same with ∂B/∂t at each point to create curl in E field that is described by the curless field; since E field obeys the superposition principle, everything should add and the resulting E field would be V.

    Now the thing is I'm not familiar with Helmholtz theorem and the conditions it requires. Can someone clarify?
  2. jcsd
  3. Apr 29, 2012 #2


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    I think, it doesn't help you much mixing a purely mathematical theorem with an electromagnetic interpretation. So let's first look at the purely mathematical theorem. The statement is that there exists a rotation free field [itex]\vec{V}_1[/itex] and a solenoidal (source-free) field [itex]\vec{V}_2[/itex] such that


    To prove this, you assume that the fields go to 0 sufficiently quickly for [itex]\vec{x} \rightarrow \infty[/itex] and you look at simply connected regions in space.

    Then there must exist a scalar field [itex]\Phi[/itex] and a vector field [itex]\vec{A}[/itex] such that

    [tex]\vec{V}_1=-\vec{\nabla} \Phi, \quad \vec{V}_2=\vec{\nabla} \times \vec{A}.[/tex]

    Obviously, with these ansatzes, you have

    [tex]\vec{\nabla} \times \vec{V}_1=0, \quad \vec{\nabla} \cdot \vec{V}_2=0.[/tex]

    It's also clear that [itex]\vec{A}[/itex] is only determined up to the gradient of a scalar field, which we shall use below to simplify the determination of the vector potential.

    Putting our ansatz into the statement of the theorem, we find

    [tex]\vec{\nabla} \cdot \vec{V}=\vec{\nabla} \cdot \vec{V}_1=-\Delta \Phi.[/tex]

    Now you should invoke your electromagnetic knowledge. Just use the Green's function for the Laplacian to find the solution

    [tex]\Phi(\vec{x})=\frac{1}{4 \pi} \int \mathrm{d}^3 \vec{x}' \frac{\vec{\nabla} \cdot \vec{V}(\vec{x}')}{|\vec{x}-\vec{x}'|}.[/tex]

    On the other hand you have

    [tex]\vec{\nabla} \times \vec{V}=\vec{\nabla} \times \vec{V}_2=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}.[/tex]

    It should be kept in mind that the latter step is valid only in Cartesian coordinates. Now we can choose a special gauge by setting the constraint

    [tex]\vec{\nabla} \cdot \vec{A}=0[/tex],

    which can always be achieved by adding an appropriate gradient field to [itex]\vec{A}[/itex], and [itex]\vec{V}_2=\vec{\nabla} \times \vec{A}[/itex] doesn't depend on any such gradient field. With this gauge contraint we have

    [tex]\Delta \vec{A}=-\vec{\nabla} \times \vec{V},[/tex]

    for which again you can immediately write the solution, using the Green's function,

    [tex]\vec{A}(\vec{x})=\frac{1}{4 \pi} \int \mathrm{d}^3 \vec{x}' \frac{\vec{\nabla} \times \vec{V}(\vec{x}')}{|\vec{x}-\vec{x}'|}.[/tex]

    You should read about this in some math book on vector calculus. There, you find more careful statements on the conditions (smoothness) on the fields, for which the Helmholtz-decomposition theorem holds.
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