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.


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


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 $\rho/p$.

  • 2
    $\begingroup$ thanks for the answer and the correct expression for $w$. $\endgroup$
    – user346
    Feb 3 '11 at 13:00
  • $\begingroup$ It was a pleasure, @space_cadet. $\endgroup$ Feb 3 '11 at 13:22

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