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Derivation Of The Faraday Two-Form

  1. Jul 28, 2006 #1
    The Faraday two form is a two form on pairs of four dimensional vectors (3 space +1 time). It is given by(may I be forgiven for the notation):

    [tex]F = E_x dx\wedge dt + E_y dy\wedge dt + E_z dz\wedge dt + B_z dx\wedge dy + B_x dy\wedge dz + B_y dz\wedge dx[/tex]

    Most books then go on to say that for any closed manifold in four space [tex]\partial \sigma[/tex](may I be forgiven for the notation):

    [tex]\int_{\partial \sigma} F = 0[/tex]
    Or equivilently
    [tex]dF = 0[/tex]

    Thus far, I have been unable to find either a derivation of the quantity, or instead, a proof that the form is closed. I'm actually seriously doubting the latter, but I digress.

    Does anyone know of any good treatments of the topic?
    Last edited: Jul 28, 2006
  2. jcsd
  3. Jul 28, 2006 #2


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    You need to start with the fact that F is exact:
    = \underrightarrow{d} \underrightarrow{A}
    = \underrightarrow{dx^i} \underrightarrow{dx^j} \partial_i A_j
    Then it's closed,
    0 = \underrightarrow{d} \underrightarrow{F}
    = \underrightarrow{d} \underrightarrow{d} \underrightarrow{A}
    = \underrightarrow{dx^k} \underrightarrow{dx^i} \underrightarrow{dx^j} \partial_k \partial_i A_j
    because the basis 1-forms anticommute and the corresponding partial derivatives commute:
    \underrightarrow{d} \underrightarrow{d} = 0

    If you insist on starting with E and B as fundamental, instead of A, then you have to impose
    \underrightarrow{d} \underrightarrow{F} = 0
    as some of Maxwell's equations. Then you can get that F is exact over most (but not all!) manifolds, F=dA.
    Last edited: Jul 28, 2006
  4. Jul 28, 2006 #3
    What's [tex]A[/tex]? The magnetic potential?
  5. Jul 28, 2006 #4
    I believe so. Since you're integrating F over it, the submanifold [itex]\partial \sigma[/itex] needs to be a surface in (3,1)-space. The result you are quoting seems to be assuming that this submanifold is the boundary of a 3-diml. submanifold of the ambient space, i.e. [itex]\sigma[/itex]. The result then is an application of Stokes' Theorem: The integral of F on the boundary of \sigma is equal to the integral of dF (=0) on \sigma.

    Actually, I just looked this up in Ward and Wells' Twistor Geometry and Field Theory, and discovered that dF=0 is just a nifty way of writing the Maxwell Equations.
    Last edited: Jul 28, 2006
  6. Jul 28, 2006 #5


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    Electromagnetic potential -- it's a 1-form, with four components. The time component is the electric potential, and the spatial components are the magnetic potential.
  7. Jul 29, 2006 #6
    Right that makes sense. Electro dynamic potential fields:

    [tex]\mathbf{B} = \nabla \times \mathbf{A}[/tex]
    [tex]\mathbf{E} = -\nabla \phi - \partial_t \mathbf{A}[/tex]

    Define the one form:

    [tex]\acute{A} = A_x d\acute{x} + A_y d\acute{y} + A_z d\acute{z} - \phi d\acute{t}[/tex]

    [tex]d \equiv \acute{\nabla} = \partial_x d\acute{x} + \partial_y d\acute{y} + \partial_z d\acute{z} + \partial_t d\acute{t}[/tex]

    [tex] \acute{F} = d\acute{A} = \acute{\nabla}\wedge \acute{A}[/tex]
    Which comes out to be
    \acute{F} = (-\partial_x \phi -\partial_t A_x )d\acute{x} \wedge d\acute{t} + (-\partial_y \phi -\partial_t A_y )d\acute{y} \wedge d\acute{t} + (-\partial_z \phi -\partial_t A_z )d\acute{z} \wedge d\acute{t} \\
    + (\partial_y A_z -\partial_z A_y )d\acute{y} \wedge d\acute{z} + (\partial_z A_x -\partial_x A_z )d\acute{z} \wedge d\acute{x} + (\partial_x A_y -\partial_y A_x )d\acute{x} \wedge d\acute{y}

    Which is simply
    \acute{F} = E_x d\acute{x} \wedge d\acute{t} + E_y d\acute{y} \wedge d\acute{t} + E_z d\acute{z} \wedge d\acute{t} \\
    + B_x d\acute{y} \wedge d\acute{z} + B_y d\acute{z} \wedge d\acute{x} + B_z d\acute{x} \wedge d\acute{y}

    That makes things clearer, and at least the Faraday doesn't have to be very awkwardly inferred from handwaving about Faraday's Law. I was having serious doubts about its validity in the case of allowing the spatial loops to vary in size or position, but I suppose as long as the potentials remain valid, so will the Faraday. I suppose the one form [tex]\acute{A}[/tex] when integrated represents some kind of 4-d potential difference between two points in space and time. Dimensional analysis time.
    Last edited: Jul 29, 2006
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