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This question is inspired by the following comment:

the strings in string theory are relativistic and on a large enough piece of world sheet, the internal SO(1,1) Lorentz symmetry is preserved. That's why a string carries not only an energy density ρ but also a negative pressure p=−ρ in the direction along the string.

by Lubos at the end of his answer to the question "What is tension in String Theory?".

I don't see how having a $SO(1,1)$ symmetry for the worldsheet leads to a negative pressure. I have the following questions:

  1. Why does a string carry negative pressure?

  2. Can a gas of strings then be treated as a substance which has a negative equation of state:

    $$ w = \frac{p}{\rho} = -1 $$

  3. If this is possible then the next natural question is: what are the implications for the cosmological constant problem?

Background material to give some context for questions 2 and 3:

One part of the cosmological constant problem is an explanation of what form of matter could seed cosmological expansion. A cosmological constant term $\Lambda$ in the action for GR is equivalent to having a medium which satisfies the negative equation of state (relation between density and pressure). One simple example of matter with a negative equation of state is a homogenous, isotropic scalar field [for details see any book on cosmology with a chapter on inflation]. If strings carry negative pressure and if a string gas can be treated as matter with $w\sim-1$ then one would have a far more natural alternative to a scalar field.

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up vote 3 down vote accepted

Consider a patch of a world sheet and choose the coordinates so that it is extended in the $t$ and $z$ directions, setting $x=y=0$ - and similarly for the other transverse dimensions I omitted here. The spacetime dimensions is $D$.

The statement that the strings have tension equal to $T$ means that the energy density has to be $$T_{tt} = T \delta^{(D-2)} s(x,y,\dots).$$ However, the world sheet is required to preserve the $SO(1,1)$ symmetry rotating the $t,z$ axes, so it must be true that a part of the stress energy tensor, the components $$T_{tt}, T_{tz}, T_{zt}, T_{zz}$$ have to be proportional to the 1+1-dimensional metric tensor because multiples of the metric tensor are the only tensors that are invariant under the Lorentz transformations, $SO(1,1)$ in this case. It follows that the $T_{tz}$ and $T_{zt}$ components have to vanish while $$T_{zz} = -T \delta^{(D-2)} (x,y,\dots).$$ The doubly spatial components of the stress-energy tensor represent the pressure but don't forget that the doubly transverse components such as $T_{xx}$ continue to vanish.

If one has a gas of randomly oriented strings and he averages over all directions of the string, the average $T_{ii}$ will be $-T_{tt}/(D-1)$ where $(D-1)=3$ in our 4-dimensional spacetime so that $p=-\rho/3$. That's the standard pressure from cosmic strings - domain walls would have $p=-2\rho/3$ for very similar reasons.

(The appearance of $1/3$ is not hard to understand: all the rotated strings will have the same trace over the spatial part of $T_{ii}$. The $z$-oriented string has this trace equal to $-\rho/3$ so after the averaging over directions i.e. over $SO(d-1)=SO(3)$, when the spatial part becomes proportional to $\delta_{ij}$, the coefficient of this Kronecker delta inevitably has to become $-\rho/3$ to preserve the spatial trace.)

The negative pressure of strings is not too important if the strings remain small and compact but it is important e.g. in string gas cosmology,5&as_ylo=&as_vis=0

So your number $p=-\rho$ is incorrect; it should be $p=-\rho/(D-1)$ because the pressure only exists in the direction along the string, so the pressure in all directions get diluted by the averaging over directions. That's why the string gas has a different equation of state than the cosmological constant. Like any gas, string gas picks a preferred reference frame, unlike the cosmological constant.

The precision measurement are enough to exclude the possibility that dark energy boils down to a network of cosmic strings or domain walls because their $w$ is just too far from the "approximately observed" $w=-1$.

By the way, $w$ is defined as $p/\rho$ rather than $\rho/p$.

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thanks for the answer and the correct expression for $w$. – user346 Feb 3 '11 at 13:00
It was a pleasure, @space_cadet. – Luboš Motl Feb 3 '11 at 13:22

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