Would it be possible to extract energy from the universe's expansion by having two massive bodies orbit each other in such a way that their rate of infall due to emitting gravitational waves is exactly countered by the expansion of the universe, then collecting the energy from the waves to do work? If so could this method be used to generate energy well beyond the evaporation of the last black hole?


1 Answer 1


No, it can't be done.

The Hubble expansion isn't a force. You can extract kinetic energy from the relative motion of receding objects, but once that's gone it's gone.

The repulsion due to the cosmological constant is a force, but it's a conservative force like ordinary gravity, so there can be no net energy output from a periodic system like the one that you're envisioning.

If you allow your system to grow in size then it's more complicated.

In a Newtonian universe with a repulsive force similar to the cosmological constant, I'm pretty sure you can extract unlimited energy from it with an ever-growing system, because, although it's still a conservative force, the potential energy function has no minimum.

In general relativity, though, when the cosmological constant is positive, there's a cosmological horizon that limits the size to which any system can grow, and the maximum energy you can extract is finite. (See this answer.)

  • $\begingroup$ Can there be a setup where the volume inside the cosmological horizon grows unboundedly? $\endgroup$
    – A.V.S.
    Nov 21, 2019 at 8:25
  • $\begingroup$ @A.V.S. Not with a cosmological constant. If the dark energy is really a scalar field with some more complicated equation of state, then I don't know what's possible. $\endgroup$
    – benrg
    Nov 21, 2019 at 8:29
  • $\begingroup$ @benrg What do you mean in the last part with "using a device that fell closer and closer to the horizon while never reaching it". Is that possible? $\endgroup$
    – vengaq
    Oct 27, 2022 at 20:27
  • $\begingroup$ @vengaq I deleted the paragraph since I think it was wrong. I added a link to an answer (to one of your questions) where I derived a finite total extractable energy. $\endgroup$
    – benrg
    Oct 28, 2022 at 6:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.