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Integration by parts in spacetime

  1. Nov 9, 2015 #1


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    In this paper
    we have p18 an integral on space time M. The author takes a 3 dimensional space like Cauchy surface ##\Sigma## which separates M in two regions, the future and the past of ##\Sigma##. He gets so the sum of two integrals on these regions. He writes then let us integrate each of them by parts. The fact that ##\Sigma## is a boundary for these regions is obvious. the vector n orthogonal to the boundary occurs in the result but i think that this uses the (- +++) metric. What is the formula used to integrate by parts in relativity?
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  3. Nov 9, 2015 #2


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    This is not really a matter of "integration by parts in spacetime" or "in relativity", it is just "integration by parts", a mathematical technique. The idea of integration by parts is that [itex]\int u dv= uv- \int v du[/itex]. Here, the integrals are of the form [itex]\int f \psi dvol[/itex]. What they are doing is taking [itex]u= \psi[/itex] and [itex]dv= f dvol[/itex] while using the physics fact that a conservative force is the derivative of the potential energy function so [itex]v= E_n[/itex] and [itex]du= \nabla_n\psi[/itex].
  4. Nov 9, 2015 #3


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    You can read that n is the future-pointing unit normal vector field on ##\Sigma##
    How does n appear in the result (it needs the metric)?
    Last edited: Nov 9, 2015
  5. Nov 10, 2015 #4


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    The result is
    ##\int_\Sigma (\Phi \nabla_n \Psi - \Psi \nabla_n \Phi) d\Sigma##
    Here ##\nabla_n## is the Lie derivative along n.
    Is there a rule for integation by parts with Lie derivatives?
  6. Nov 24, 2015 #5


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    Lie derivative obeys to Leibniz rule so I think will be a integration by parts, if boundary doesn't contribute I think it is ## \int_{\Sigma}\Phi\nabla_{n}\Psi\,d\,\sigma=-\int_{\Sigma}\Psi\nabla_{n}\Phi\,d\,\Sigma##
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