Cosmic string network

An isotropic network of long cosmic strings has average equation-of-state parameter $w=-1/3$ and energy density proportional to $a^{-2}$.

The equation of state averages the stress of a cosmic string over an isotropic network; it is not the definition of the individual defect.

An isotropic cosmic string network with $\rho =C/a^2$ has equation-of-state parameter $w=-1/3$. The Friedmann equation gives ${\dot{a}}^2=8\pi GC/3-K$. On the expanding branch with positive right side, $a=B(t-t_B)$, the comoving Hubble radius is constant, and the cosmological density parameter is constant. There is no growing flatness problem, but flatness is not an attractor. If this exact coasting solution extends all the way to $t_B$, then ${\int }_{t_B}^{t}d{t}^{\prime }/a(t^{\prime })$ diverges and there is no finite particle horizon. A finite starting time changes that global conclusion.

A closed universe containing only a cosmic string network can remain static. Its Friedmann acceleration equation vanishes identically because $P_s=-{\rho }_s/3$, and its Friedmann equation requires ${\rho }_s{a}^2=3K/(8\pi G)$. Once this conserved normalization is tuned, every positive radius gives a static solution. The usual Hubble-normalized cosmological density parameter is undefined at a static epoch.

At fixed string normalization $C={\rho }_s{a}^2$, scale-factor shifts move between static solutions and are neutral, with no restoring force. The acceleration equation gives $\ddot{a}=0$, but the Friedmann constraint also imposes ${\dot{a}}^2=8\pi GC/3-K$. At the static normalization it excludes a nonzero velocity. A slightly larger $C$ admits linearly expanding or contracting branches; a smaller $C$ admits no real closed solution. This distinguishes permitted perturbations from constraint violations.

Consider a spatially flat universe containing radiation and a cosmic string network. If their densities are both ${\rho }_{\mathrm{eq}}$ at scale factor $a_{\mathrm{eq}}$, thenIn conformal time $d\tau =dt/a$, the flat Friedmann equation becomesChoosing the Big Bang at $\tau =0$, its expanding solution is