Cosmic string network

An isotropic cosmic string network averages the directional stress and energy density of many long cosmic strings. For an ideal smooth string whose tension equals its rest energy per unit length, a network frozen into the expansion has negative averaged pressure,A comoving string segment then has physical length proportional to $a$, while its enclosing volume grows as $a^3$, giving $\rho \propto a^{-2}$. This is the nonmoving, noninteracting limit, not the general evolution law of a mobile network.

In units $c=1$, let $v^2$ be the energy-weighted mean squared transverse string velocity. For an isotropic smooth-string network the averaged equation of state isThe kinetic contribution raises the averaged pressure; for example $v^2=1/2$ gives $w=0$. Motion, reconnection and production of smaller loops therefore matter even when the initial network is statistically isotropic.

The corresponding long-string balance, neglecting friction against matter, can be writtenwhere $\mathcal{Q}\ge 0$ is energy removed from the long-string population into loops or other radiation channels. For $\mathcal{Q}=0$ this agrees with $\dot{\rho }+3H(\rho +P)=0$ and proves the equation-of-state relation. If $v$ is constant and losses vanish, the solution is $\rho \propto a^{-2(1+v^2)}$. The primary velocity-dependent model defines $\rho =\mu /L^2$ and evolves both its characteristic length $L$ and its velocity.^[ref]

A horizon-scaling cosmic string network is consequently distinct from a frozen fluid with fixed $w=-1/3$. The ideal frozen equation of state can be used in a specified cosmological toy model, but it cannot be inferred from isotropy alone.

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