The quote below has been taken from the Wikipedia article on cosmological constant, https://en.wikipedia.org/wiki/Cosmological_constant.

Einstein included the cosmological constant as a term in his field equations for general relativity because he was dissatisfied that otherwise his equations did not allow, apparently, for a static universe: gravity would cause a universe that was initially at dynamic equilibrium to contract. To counteract this possibility, Einstein added the cosmological constant. However, soon after Einstein developed his static theory, observations by Edwin Hubble indicated that the universe appears to be expanding; this was consistent with a cosmological solution to the original general relativity equations that had been found by the mathematician Friedmann, working on the Einstein equations of general relativity. Einstein reportedly referred to his failure to accept the validation of his equations—when they had predicted the expansion of the universe in theory, before it was demonstrated in observation of the cosmological redshift—as his "biggest blunder".

In fact, adding the cosmological constant to Einstein's equations does not lead to a static universe at equilibrium because the equilibrium is unstable: if the universe expands slightly, then the expansion releases vacuum energy, which causes yet more expansion. Likewise, a universe that contracts slightly will continue contracting.

Source: https://en.wikipedia.org/wiki/Cosmological_constant#History

My question is about where it says, "if the universe expands slightly, then the expansion releases vacuum energy".

I think the term "vacuum energy" above refers to the dark energy inherent to fabric of space. Why would the expansion release vacuum energy? Could you please guide me with it in simple terms?


1 Answer 1


It is not certain that dark energy is vacuum energy per se (at least not as a sole contributor), but vacuum energy is in fact inherent to the spacetime as you point out. Since the cosmological constant appears in the action as such:

\begin{equation} S = -\frac{1}{16\pi G} \int d^{4}x \, \sqrt{-g} \, (R - 2\Lambda) \end{equation}

it effectively acts as a constant vacuum potential energy density.

As for why expansion "releases" vacuum energy, that's a somewhat clunky way of saying that the expansion of the Universe and thus the increase of the physical volume of the spacetime will lead to an increase in vacuum potential energy. That is of course a trivial consequence of the way the cosmological constant was introduced above: if we have a constant vacuum potential density (i.e. $\frac{\Lambda}{8\pi G}$), an increase in physical volume will lead to a proportional increase in the potential energy that permeates it.

A more intuitive way to phrase this would be "the vacuum has a constant energy density given by the cosmological constant, so an increase in physical volume - and thus an increase in the "amount" of vacuum we have - leads to an increase in its total energy".

  • 2
    $\begingroup$ This is a good answer. It might also be useful for OP to note that other forms of energy density $\rho$ (e.g. matter, radiation, etc) drop off as volume increases, in contrast to $\rho_{\Lambda}$ $\endgroup$
    – Eletie
    Commented Jun 28, 2022 at 8:58
  • $\begingroup$ @rhomaios Thanks! You said, "the vacuum has a constant energy density given by the cosmological constant, so an increase in physical volume - and thus an increase in the "amount" of vacuum we have - leads to an increase in its total energy". Is there any other material other than vacuum whose energy remains constant as its size becomes larger? I find it quite counterintuitive that the vacuum could keep its energy density same while increasing it size. If it was some other material, the material would need to borrow extra energy from some external source to keep its energy density constant $\endgroup$
    – PG1995
    Commented Jun 29, 2022 at 2:55
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    $\begingroup$ From the Friedmann equations, the derived equation of state for the energy density $\rho$ in the Universe is $\dot{\rho} + 3H(\rho + P) = 0$ where $P$ the cosmological pressure of the contents at hand. We are assuming a perfect fluid which means the the pressure is proportional to energy density: $P = w\rho$. The value of the constant $w$ determines the type of matter you have. Non-relativistic "cold" matter is 0, highly relativistic matter and radiation is $1/3$ and the vacuum is -1. As you can see, all other forms of matter have $\dot{\rho} <0$ and only the vacuum has a constant $\rho$. $\endgroup$
    – rhomaios
    Commented Jun 29, 2022 at 6:24
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    $\begingroup$ As for an intuitive explanation for this, think of the cosmological pressure as a tendency for matter to "collapse" in on itself and thus exerting some pressure "inwards". The vacuum having a negative pressure means that it effectively experiences "anti-gravity" and is pressured "outwards" causing it to expand. $\endgroup$
    – rhomaios
    Commented Jun 29, 2022 at 6:31
  • $\begingroup$ Thank you! So, the gravity is property of matter and the anti-gravity is the property of space, and this anti-gravity is far, far weaker than the gravity. It took centuries to make some sense of gravity and I think it'd also take some amount of time to make sense of this anti-gravity. I'd speculate as time passes the rate of acceleration for the space expansion would decrease but would never go to zero if this is really some anti-gravity where space tries to run away from itself! $\endgroup$
    – PG1995
    Commented Jun 30, 2022 at 6:28

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