# Gravitational wave solution boundary conditions

1. Dec 13, 2013

### center o bass

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. Dec 14, 2013

### Naty1

3. Dec 14, 2013

### Bill_K

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.