4
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

At the horizon, the gravitational acceleration goes to infinity, and it's not obvious to me that you couldn't, in principle, classically, extract unlimited energy using a device that fell closer and closer to the horizon while never reaching it. But, at the very least, quantum gravitational effects will limit the energy you can get this way (and practical limitations much sooner than that).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can there be a setup where the volume inside the cosmological horizon grows unboundedly? $\endgroup$ – A.V.S. Nov 21 '19 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 '19 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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