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?
Best Answer
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".