Table of Contents
Fetching ...

Semiclassical cosmic string evaporation

Robert Penna

Abstract

We describe a new tunneling solution for the decay of a cosmic string into a burst of gravitational waves. We find the relevant instanton and compute the tunneling rate. Locally, our solution is just an analytic continuation of the Kerr metric (but there is no black hole in our solution). An interesting feature of our result is that there is a conical singularity in the initial state but there is no singularity in the final state. This demonstrates that singularities can disappear in quantum gravity in ways that are impossible in classical gravity.

Semiclassical cosmic string evaporation

Abstract

We describe a new tunneling solution for the decay of a cosmic string into a burst of gravitational waves. We find the relevant instanton and compute the tunneling rate. Locally, our solution is just an analytic continuation of the Kerr metric (but there is no black hole in our solution). An interesting feature of our result is that there is a conical singularity in the initial state but there is no singularity in the final state. This demonstrates that singularities can disappear in quantum gravity in ways that are impossible in classical gravity.
Paper Structure (1 section, 16 equations, 2 figures)

This paper contains 1 section, 16 equations, 2 figures.

Table of Contents

  1. Acknowledgements

Figures (2)

  • Figure 1: Plots of $\gamma(r)$ at Euclidean times $\tau = -20$ (solid), $-1.1$ (dashed), $-0.9$ (dot-dashed), and 0 (dotted). At large negative $\tau$, all of the $C$ energy is at $r=0$. The plotted solution has $\tilde{a}=1$ and $\alpha=2$.
  • Figure 2: The positive vertical axis is Lorentzian time ($t$) and the negative vertical axis is Euclidean time ($\tau$). The dashed line is the string. The solid black contour indicates the distribution of $C$ energy: $90\%$ of the $C$ energy is to the left of the contour. The decay of the string is mediated by a negative $C$ energy bubble near $\tau=-1$. This solution has $\tilde{a} = 1$ and $\alpha = 2$.