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

  1. Sep 5, 2013 #1
    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. jcsd
  3. Sep 5, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    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.
     
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