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Is there an intuitive way to understand what negative pressure means in general relativity in the same way as positive pressure can be thought to be kinetic energy of gas particles? Dark energy has positive energy but negative pressure, is there other examples with negative pressure? If dark energy was caused by vacuum fluctuations of quantum fields, then why do the vacuum fluctuations cause negative pressure instead of positive?

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Is there other examples with negative pressure?

Any elastic object under tension has negative pressure. Soap bubbles have surface tension, which is a lower-dimensional equivalent of negative pressure.

If dark energy was caused by vacuum fluctuations of quantum fields, then why do the vacuum fluctuations cause negative pressure instead of positive?

Suppose the vacuum energy density is positive. This doesn't depend on the total amount of space, so the more space you have, the more total energy there is. That is, intuitively it takes work to expand space, so the vacuum carries negative pressure. It's just the opposite of how you get work when you expand a gas, so gases have positive pressure.

A more formal way to put it is that the vacuum energy-momentum is by definition Lorentz invariant, so the only option is that it's proportional to the Minkowski metric $\eta^{\mu\nu}$. So it has to have a relative sign between the timelike component (energy) and the spacelike component (pressure).

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  • $\begingroup$ Can we say that the negative pressure arises due to the following argument. Suppose that we have a volume filled with constant energy density. Therefore, the energy in the differential volume is $d E = \rho d V$. However, if we would change $dV$, we do work on the system so $d E = \rho d V = -pdV$ (since we are not adding heat and assuming canonical ensemble $\mu dN=0$ ). Since $\rho>0 \implies p<0$. So the vacuum acts with negative pressure and tries to counter the pull. Does this make sense? $\endgroup$ Mar 5, 2022 at 19:57
  • $\begingroup$ @AlexanderCska Yes, there's a perfectly legitimate way of understanding the negative pressure. $\endgroup$
    – knzhou
    Mar 5, 2022 at 20:24
  • $\begingroup$ thank you for your reply. How are $\mu dN=0$ and $ TdS=0$ justified in the case of cosmology. $\endgroup$ Mar 5, 2022 at 21:16
  • $\begingroup$ @AlexanderCska If you have followup questions, you should just ask a separate question! $\endgroup$
    – knzhou
    Mar 5, 2022 at 21:21
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Is there an intuitive way to understand what negative pressure means in general relativity

No, at least not with classical analogues which are very misleading . In a gas for example, pressure does not lead to an expansion: It's the pressure difference at the boundary that leads to an expansion or contraction.

The universe in cosmology has no boundaries, the universe is isotropic and homogeneous which means it looks everywhere the same in every direction.

The (accelerated) expansion of the universe is modeled by the Friedmann equations which do not have classical analogues. The easiest way to model dark energy is by means of $\Lambda$, the cosmological constant from Einstein's field equations, that occurs in both Friedmann equations.

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