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VantagePoint72
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Homework Statement
In ”almost inertial” coordinates the energy momentum tensor of a straight cosmic string aligned along the z-axis is [itex]T_{\mu\nu} = \mu\delta(x)\delta(y)diag(1,0,0,-1)[/itex] where μ is a small positive constant. Look for a time-independent solution of the linearized Einstein equation, finding h11 = h22 = −λ as the only non-zero components of the perturbed metric tensor, where λ ≡ 8μ log(r/r0), r2 = x2 + y2, and r0 is an arbitrary length.
Homework Equations
The linearized EFEs: [itex]\partial^\rho\partial_\rho\bar{h}_{\mu\nu}=-16\pi T_{\mu\nu}[/itex] (in natural units) and their solution by a Green's function:
[itex]\bar{h}_{\mu\nu}(t,\vec{x})= 4 \int d^3\vec{x}' \frac{T_{\mu\nu}(t-|\vec{x}-\vec{x}'|,\vec{x}')}{|\vec{x}-\vec{x}'|}[/itex] (integrated over all space).
The Attempt at a Solution
The two non-zero components of [itex]\bar{h}[/itex] will clearly be the 00 and 33 components (which will be negatives of each other). Then we just have to convert [itex]\bar{h}[/itex] to [itex]h[/itex] which is straightforward. So, the problem can be reduced to determining the 00 component of the trace-reversed metric perturbation:
[itex]\bar{h}_{00}(t,\vec{x})= 4 \int d^3\vec{x}' \frac{\mu\delta(x')\delta(y')}{|\vec{x}-\vec{x}'|}[/itex]
This seems simple, but I'm running into an issue. Just one thing worth pointing out: I realize the solution is easily obtained just by comparing this to the electrostatic potential of an infinite line charge. However, in the interest of being able to handle other similar problems that don't lend themselves to such analogies, I want to know how to explicitly do the calculation. So, going ahead and integrating out the two delta functions:
[itex]\bar{h}_{00}(t,\vec{x})= 4\mu \int dz' \frac{1}{\sqrt(x^2 + y^2 + (z-z')^2)}[/itex]
which some judicious substitutions reduce to the problem of computing the integral:
[itex]\int_{-\infty}^\infty du \frac{1}{\sqrt(1 + u^2)}[/itex]
The natural way to proceed would be to make the obvious trig substitutions, except that, according to Wolfram Alpha this integral diverges. What's gone wrong? The solution to the linearized EFEs by a Green's function is completely general, so how can it yield a nonsensical answer to a well-posed physical question?