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Gravitational wave solution boundary conditions

  1. Dec 13, 2013 #1
    In linearized gravity we can one sets

    $$(1) \ \ g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$$

    where h is taken to be a small perturbation about the flat space metric. One common decomposition of h is to write the spatial part as

    $$ h_{i j} = 2 s_{ij} - 2\psi \delta_{ij} \ h_{0i} \equiv w_i \ h_00 = -2\phi$$

    There are certain gauge transformations that leave (1) invariant which can be used to simplify Einstein's equation; one choice, "the transverse gauge", makes ##\nabla \cdot w = 0## and ##\partial_i s^{ij} =0##. One can show that by expressing the (time-time) part Einstein field equations in terms of these fields in the transverse gauge yields for empty space yields

    $$\nabla^2 \psi = 0.$$

    Now at the beginning of section 7.4 "Gravitational Wave solutions" in Carroll's "Spacetime and Geometry", he states that the above equation with "well behaved boundary conditions" implies

    $$\psi = 0.$$

    I'm not sure what to make of this. What does he mean by well behaved boundary conditions, and why are these relevant for the gravitational wave solutions?
     
  2. jcsd
  3. Dec 14, 2013 #2
  4. Dec 14, 2013 #3

    Bill_K

    User Avatar
    Science Advisor

    Zero incoming waves. He wants to get rid of ψ, since it is only a gauge freedom, and for the homogeneous wave equation, zero incoming waves implies ψ must be zero everywhere.
     
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