Linearized graviational field of a cosmic string

[VantagePoint72]

Homework Statement

In ”almost inertial” coordinates the energy momentum tensor of a straight cosmic string aligned along the z-axis is T_{\mu\nu} = \mu\delta(x)\delta(y)diag(1,0,0,-1) 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: \partial^\rho\partial_\rho\bar{h}_{\mu\nu}=-16\pi T_{\mu\nu} (in natural units) and their solution by a Green's function:

\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}'|} (integrated over all space).

The Attempt at a Solution

The two non-zero components of \bar{h} will clearly be the 00 and 33 components (which will be negatives of each other). Then we just have to convert \bar{h} to h which is straightforward. So, the problem can be reduced to determining the 00 component of the trace-reversed metric perturbation:

\bar{h}_{00}(t,\vec{x})= 4 \int d^3\vec{x}' \frac{\mu\delta(x')\delta(y')}{|\vec{x}-\vec{x}'|}

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:

\bar{h}_{00}(t,\vec{x})= 4\mu \int dz' \frac{1}{\sqrt(x^2 + y^2 + (z-z')^2)}

which some judicious substitutions reduce to the problem of computing the integral:

\int_{-\infty}^\infty du \frac{1}{\sqrt(1 + u^2)}

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?

 

[andrien]

why is that surprising.it is just because you are summing over all sources of z' which extends to infinity.

 

[VantagePoint72]

That doesn't mean the total contribution to the gravitational field is infinite. There are many comparable cases in electrostatics. The answer that should be obtained is provided with the question, and it is finite.

 

[andrien]

it is finite because there is an r₀ which corresponds to length of string. the z' integral will only extend over only the length of it which is r₀.If you take limit as ∞ then there is trouble because r₀ is then ∞ which will give divergent integral.By the way the integral does not contain any z,it is determined in xy plane.Got it.(I am busy,so i will be back after sometime perhaps 48 hours).

 

[VantagePoint72]

I'm pretty sure that's not true. When you do the analogous calculation of the potential of an infinite line charge, you are able to do the whole integral over the entire infinite length. In that case, r_0 corresponds to the arbitrary radial distance from the line that you choose to set the 0 potential at. I'm just not sure what it corresponds to here.

 

[andrien]

I'm pretty sure that's not true. When you do the analogous calculation of the potential of an infinite line charge, you are able to do the whole integral over the entire infinite length. In that case, r_0 corresponds to the arbitrary radial distance from the line that you choose to set the 0 potential at. I'm just not sure what it corresponds to here.

so why don't you just calculate the potential of an infinite line charge by using same method and see if the potential(not field) is finite for infinite line charge.By the way,using that r₀ is the length and large,I have been able to see the required result.