What type of energy is Dark Energy? Just as the title says. Is it kinetic/potential or some other type?


2 Answers 2


In Newtonian mechanics, we have potential energy and kinetic energy. All types of energy can be classified as one or the other -- never both or neither.

But when you deal with fields, this distinction doesn't really work. For example, a light wave (electromagnetic field) carries energy, and this energy doesn't fit neatly into either the PE or the KE category.

Dark energy appears in our models as a field, so the same thing applies to it.

Note also that energy is not globally conserved in cosmology.


To add to Ben Crowell's answer: the "energy" spoken of is most like that in a charged capacitor (the electic field therein has energy density $\frac{1}{2} \epsilon\,|\vec{E}|^2$) or within a ring of current (the magnetic field has energy density $\frac{1}{2} \mu\,|\vec{H}|^2$).

"Dark Energy" is a simply a term that Einstein added to his "raw" field equations as a kind of "fudge factor" (more about this below).

As Ben says, energy is not globally conserved in cosmology, BUT it is locally. Einstein's equations have the form:

$$\mathfrak{G} = \mathfrak{T} - \Lambda\, \mathfrak{g}$$

where $\mathfrak{G}$ is a 16 component "tensor" object describing the pure geometry of spacetime (in particular it quantifies how much the geometry deviates from Euclid's postulates) and $\mathfrak{T}$ is the stress energy tensor - which describes the distribution of "stuff" ("energy") in the universe. $\Lambda$ is the cosmological constant and $\Lambda\, \mathfrak{g}$ is the dark energy term. Now, if we take the tensor equivalent of the divergence of both sides, a general, mathematical (i.e. this is not physics) property of the geometry tensor $\mathfrak{G}$ is that its divergence vanishes. This has the effect of turning the divergence of the right hand side of the equation into a something like a continuity equation, that says that the rate flux of energy (over small time and length scales) into a volume equals the time rate of change of the total energy within that volume. So by writing $-\Lambda\, \mathfrak{g}$ on the right hand side, it looks as though you are including it in the "budget" to make local conservation of momentum / energy hold (simply by dint of the general mathematical property that the divergence of $\mathfrak{G}$ always vanishes) is the reason it is called an "energy". But the even this analogy is weak: for the divergence of $\Lambda\,\mathfrak{g}$ is nought anyway and does not affect local energy conservation if $\Lambda$ is truly constant. It is in some ways like an energy density that "drives" expansion / contraction of the universe, but it is quite unlike the other things that go into the stress energy tensor (whose divergences do not a prior vanish). From cosmological observations, $\Lambda$ is positive, and is equivalent to a negative pressure that drives the observed acceleration of the universe's expansion.

Einstein added the term to his original equations as an afterthought (it wasn't in his 1916 paper "Foundations of General Relativity Theory") to "hold back gravity's effect" because he understood his field equations implied an ever changing universe - the predominant idea in the 1910s was that the Universe was static: it had always been there roughly as it is now and would always be so. He was shocked when Hubble osberved the Universe's expansion, something that more resembled the behaviour of his "raw" equations. So he regretted his afterthought in in the light of Hubble's observation. Moreover, it was misguided on Einstein's part to try to achieve a stable universe in this way: a dark energy term could stabilise the equations, it is true, but it would be an unstable equilibrium, like a pencil standing with balanced forces on its point. The tiniest perturbation would beget a runaway expansion / contraction of the gravitational system.

Nowadays we find we need $\Lambda$ to match the field equations with the observed acceleration cosmological expansion. So it was "worse than Einstein thought": not only is the universe expanding, but that rate of expansion is accelerating, so the observed cosmological constant does the opposite of what Einstein meant it to do!

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    $\begingroup$ The fact that one has to include −Λg in that "budget" to make local conservation of momentum / energy hold (simply by dint of the general mathematical property that the divergence of G always vanishes) is the reason it is called an "energy". I don't think this is quite right. The Einstein field equations are not a statement of conservation of energy, and adding a cosmological constant term to them has no effect on local conservation of energy. The divergence of $\Lambda g$ is zero, since the covariant derivative of the metric is 0 -- that's the defining property of the covariant derivative. $\endgroup$
    – user4552
    Oct 28, 2014 at 23:50
  • $\begingroup$ @BenCrowell S**t! "defining property of the covariant derivative" - quite right. Somehow I was thinking of $\Lambda R$ the scalar curvature. Fixing now! $\endgroup$ Oct 28, 2014 at 23:53
  • $\begingroup$ It is true that the Einstein field equations indirectly imply local conservation of energy; they're not self-consistent if the stress-energy tensor has a nonvanishing divergence. But if anything, adding a cosmological constant term raises the danger of violating this requirement. If you don't take $\Lambda$ to be constant, then that's what happens. $\endgroup$
    – user4552
    Oct 28, 2014 at 23:57
  • $\begingroup$ @BenCrowell It's kind of fixed now. "they're not self-consistent if the stress-energy tensor has a nonvanishing divergence" - this is exactly why I said that $\mathfrak{G}$'s vanishing divergence is a mathematical, not really a physics property - now that's a dog's breakfast of a statement (an oxymoron, since we're describing physics) but I can't think of a sharper one right now. I guess I am trying to say that no matter what physics you encode, anything that sums to $\mathfrak{G}$ must have a vanishing divergence. $\endgroup$ Oct 29, 2014 at 0:05
  • $\begingroup$ @BenCrowell BTW I appreciate your correction. I guess I sometimes think of $\Lambda$ as something you add to half the scalar curvature on the left hand side. I was a while back intensely studying the mathematics and geometry of Einstein manifolds: as you would likely know is what one tends to think: it all gets lumped into $\mathfrak{R} = k\, \mathfrak{g}$. I was trying to get a grip on string theory: I learnt alot about Calabi-Yau manifolds in the process, but somehow I'm still just as clueless on ST! $\endgroup$ Oct 29, 2014 at 0:17

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