Does dark energy get used up in the expansion of the universe?

Question

Now, I am a beginner in Cosmology, so I am not sure if this makes sense.

Since the universe is expanding at an accelerating rate, and thus distant objects are also accelerating away. In that sense, from our (non-comoving) frame of reference, they are gaining kinetic energy. Similarly, for the frame of reference of a faraway object looking back at Earth will see the Earth accelerating away and therefore gaining energy.

Does this therefore mean that dark energy is being converted to kinetic energy? Otherwise what is the source of this otherwise unaccounted increase in energy.

Answer 1

Dark energy does not get used up.

I'll first note that energy is tricky to define over cosmological distances, due to the global curvature of spacetime. In particular, you mention reference frames, but reference frames are only locally meaningful.

Nevertheless, there is a clear picture of how energy is conserved on scales much smaller than the Hubble scale. Indeed this picture is Newtonian, since sub-horizon cosmological dynamics are Newtonian (at least if dark energy is a cosmological constant). Consider the cosmological constant $\Lambda$, which is $8\pi G/c^2$ times the dark energy density. With respect to an observer at the origin, it sources an acceleration $$\ddot{\vec{r}} = \frac{\Lambda c^2}{3} \vec{r}.$$ But this can be written as the gradient of a potential $\Phi$, i.e. $\ddot{\vec{r}} = -\nabla\Phi$, where $$\Phi = -\frac{\Lambda c^2}{6}r^2.$$ Due to the negative sign, this is something like a hilltop that objects like galaxies are sliding off of. Gaining distance in this potential corresponds to losing gravitational potential energy, and that energy gets converted into kinetic energy.

Answer 2

Good morning,

First of all, thank you for your question. Secondly, I am not an expert in Cosmology but I do not think that your point of view is correct.

We now that from the Cosmological Principle the metric of the Universe at longer distances is $$ d s^2=c^2 d t^2-a^2(t)\left[\frac{d r^2}{1-k r^2}+r^2\left(d \theta^2+\sin ^2 \theta d \phi^2\right)\right] $$ The objetive is do some assumptions about the mass, energy and curvature of the Universe and obtaining the expression of $a(t).$ The dark energy appears introducing a modification in the Einstein's equations: $$ R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}+\Lambda g_{\mu \nu}=\frac{8 \pi G}{c^4} T_{\mu \nu} $$ the constant $\Lambda$ can be associated to a density of dark energy given by $$ \epsilon_{\Lambda}:=\frac{\Lambda c^2}{8 \pi G} $$ For example, a plane universe which is dominated by the dark energy has $$ a(t) \propto \exp \left(H_0 t\right) $$ BUT it is not necessary to introduce the dark energy to has a redshift or other effects derived from Universe expansion.

FOR EXAMPLE, in a plane Universe dominated by matter you will have $$ a(t)=a_0\left(\frac{3}{2} H_0 t\right)^{2 / 3} $$ so it is in expansion, but there is no dark energy.

This occurs because the Earth is not obtaining kinetic energy, its movements occurs because the space between it and the other galaxy is increasing. In fact, this displacement could be at more speed that $c$.