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I Differential forms in GR

  1. Nov 2, 2016 #1
    The differential form of a function is
    [itex]\partial{f(x^1,...,x^n)}=\frac{\partial{f(x^1,...,x^n)}}{\partial{x^1}}dx^1+...+\frac{\partial{f(x^1,...,x^n)}}{\partial{x^n}}dx^n[/itex]


    Is there (especially in General Relativity) differential of higher orders, like [itex]\partial^2{f(x^1,...,x^n)}[/itex]? If so, how is it computed?
     
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  3. Nov 2, 2016 #2

    jedishrfu

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    This paper discusses Differential forms, Tensors and uses in General Relativity so I would say yes higher order forms are used and are useful in General Relativity.

    http://www.aei.mpg.de/~gielen/diffgeo.pdf
     
  4. Nov 3, 2016 #3

    Orodruin

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    But at the same time no, there are no forms that are the second derivative of a function (as the OP suggests) as the exterior derivative applied twice gives zero (the any-dimensional equivalent of curl(grad(f))=0).

    Of course there are other forms of higher order.
     
  5. Nov 3, 2016 #4

    stevendaryl

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    Yes, [itex]d^2[/itex] always produces zero. However, you can get something sort of conceptually similar to [itex]d^2[/itex]:
    1. Operate on [itex]F[/itex] with [itex]d[/itex] to produce [itex]dF[/itex].
    2. Take the Hodge dual, [itex]*dF[/itex].
    3. Operate on THAT with [itex]d[/itex], to produce [itex]d * dF[/itex]
    This isn't necessarily zero, and is sort of like a higher-order derivative. In 3-D, if [itex]F[/itex] is a scalar function, then [itex]* d * d F = \nabla^2 F[/itex].
     
  6. Nov 3, 2016 #5
    In General Relativity one frequently deal with differential forms, say W. What actually is it? I know it has to be a completely antisymmetric (0,p) tensor. But what are its components Wμ1...μp? Would it be some array of derivatives of a function?
     
  7. Nov 4, 2016 #6

    haushofer

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    The components don't have to be derivatives. They are just antisymmetric, which makes threm taylor-fit for integration over manifolds.
     
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