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Exact (vector) differentials

  1. Jan 19, 2010 #1
    Can someone please tell me necessary and sufficient conditions on a differential [tex]d \mathbf F[/tex], where [tex]\mathbf F[/tex] is a vector field, for the differential to be exact?
  2. jcsd
  3. Jan 21, 2010 #2
    Well, maybe an easier (hopefully?) question is in order: Let [tex]\mathbf r[/tex] be the position vector in Euclidean 3-space. How do we know that [tex]\oint_C d\mathbf r = 0[/tex] for every closed curve [tex]C[/tex]?
  4. Jan 21, 2010 #3
    By dF do you mean the 1 form <F,>? <,> is the euclidean inner product.

    A necessary and sufficient condition for exactness of a 1 form is that its integral over any closed curve is zero. Choose any point in space and define f(x) = Integral from p to x of the form along any curve connecting p to x. The function is well defined because two different paths determine a closed loop and the integral over the closed loop is zero.

    the other way around is just the fundamental theorem of calculus.
  5. Jan 21, 2010 #4

    Ben Niehoff

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    I think the OP is asking about vector-valued forms, of the form

    [tex]\omega = \omega^a{}_\mu \vec e_a \; dx^\mu[/tex]

    One can think of this object either as a 1-form whose components are vectors, or as a vector whose components are 1-forms. I think in this case, the latter description is easier. Then, a vector of 1-forms is exact if and only if each of its component 1-forms is exact.

    In particular, for any vector field [itex]\vec F[/itex], the vector-valued 1-form [itex]d \vec F[/itex] is exact by definition.
  6. Jan 21, 2010 #5
    I'm not trying to hijack the thread, but how do you people visualize differential forms? I mean, a vector is an arrow, but what is a form? I guess a 1-form is a linear functional which takes the inner product between some vector and it's input vector? Much like a bra in the Dirac notation.
  7. Jan 21, 2010 #6
    I think you are right. The integral will be zero iff the components are exact. The arguments are the same.

    F is a vector of its component functions. dF is the vector of differentials of the component functions and so is exact.

    In dimension 3 this can be looked at another way.

    The integral of the vector field over a closed curve in 3 space is the integral of the normal component of its curl over any surface that the curve bounds. So if the curl is zero its integral over any closed curve is zero and it is exact. This a just a vector version of the Poincare lemma which says that the homology of Euclidean space is zero.

    In higher dimensions you don't have this nice duality between vector fields and forms but there is an analogous statement.
  8. Jan 21, 2010 #7
    Forms are difficult to visualize generally. Some cases are tractable.

    If you have a metric then a 1 form can be thought of as a vector field using the dual mapping. A different metric will determine a different vector field. For instance the differential of a function has different gradients depending on the metric.

    In 3 space a two form can be visualized through its kernel which is a1` dimensional vector space that has a natural orientation. This can be thought of as a vector field - I think.
    If you follow the flow of the field you can think of it as carrying the 2 form along the flow from point to point.

    But there is more to it. At each point the 2 form determines a measure on the tranversal 2 plane to the flow. So really the flow carries a measure along with it. If the form is closed this means - I think - that the measure is preserved by the flow. the reason is that if you draw a little square in a transversal plane a let if fill out a small solid as it flows along the vector field - the form integrates to zero over the entire solid - Stoke's Theorem - and its integral along the flow sides is zero because the tangent space to the flow side contains the flow vectors. This may be a little wrong but is substantially correct.

    A simlar type of reasoning can be used for 1 forms in 3 space - but then the kernel is not a flow so is probably harder to visualize.
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