How is the deficit angle due to a relativistic cosmic string derived?

1. Sep 5, 2013

Dilatino

The presence of a cosmic string does not lead to gravitational attraction of a particle placed some distance away from it. But it affects the geometry of planes orthogonal to the cosmic string, such that the circumference of a circle traced out when moving around it at a distance r is given by

$$C = r(2\pi -\Delta)$$

which means that the flat plane is deformed to a cone.

How can the deficit angle due to a relativistic string

$$\Delta = \frac{8\pi G T_0}{c^4} = \frac{8\pi G \mu_0}{c^2}$$

be derived from general relativity?

Last edited: Sep 5, 2013
2. Sep 5, 2013

Bill_K

This page has a derivation. He starts with an infinite line source having mass density μ and tension T. Solves the linearized field equations (or claims to), and shows that the resulting metric can be transformed into a metric which is flat, but with an angular deficit.