# Could Dark Energy be evidence that space equals energy, in a similar manner to $E=mc^2$?

I was reading another Phys.SE question and answer, and this quote jumped out at me:

As for dark energy, the current accepted model, Λ-CDM, treats dark energy like a constant energy density. That means as the universe expands, the amount of dark energy per unit volume remains constant.

It struck me that this effectively ties dark energy and space together in an equation form. Something about the "dark energy per unit volume remains constant" phrasing. (Alternatively this could tie space to matter, of course). And I was wondering if it's not "dark energy remains constant per unit volume", but instead, "space is equivalent to energy, and we measure that as something we call dark energy".

I was wondering if there were any current theories suggesting this already, or conversely if any experiments or observations would contradict this?

• This question physics.stackexchange.com/q/273468 is related to yours, imo. – user108787 Aug 10 '16 at 22:58
• What does "space is equivalent to energy" actually mean? – ACuriousMind Aug 11 '16 at 13:52
• @ACuriousMind: To use Snyder005's phrasing, that space has energy, rather than dark energy being something entirely separate that just happens to be show up with space. I don't know if it matters, except conceptually. – Dan Smolinske Aug 11 '16 at 15:22

This question at it heart seems to deal more with semantics than with physics, but is still interesting. First, I don't think it would be correct to say energy equals mass, but rather that mass has energy, with the amount available given by $E=mc^2$. In a similar manner, one could state that dark energy could be evidence that space itself has energy, the so called vacuum energy. This arises from your choice of how you write the Einstein field equations

There are two ways to write the Einstein field equations. The first is with the dark energy term on the same side as the stress energy tensor.

$G_{\mu\nu} = \frac{8\pi G}{c^4}\left( T_{\mu\nu} + T_{\mu\nu, \mathrm{vacuum}}\right)$

where $T_{\mu\nu, \mathrm{vacuum}} = - \frac{\Lambda c^4}{8 \pi G} g_{\mu\nu}$. In this case we treat it as some additional intrinsic mass-energy density, i.e. vacuum energy.

We could also move this term to the other side of the equation. (I will rewrite $G_{\mu\nu}$ for clarity).

$G_{\mu\nu} + \Lambda g_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$

Here the dark energy term appears as a constant times the metric. In this case we treat it as some sort of fundamental constant to "add" to the space curvature, the cosmological constant.

So to answer your questions, there is fundamentally no difference in the two interpretations as far as its effect on General Relativity. You can choose to treat the effects of dark energy as a fundamental property of space, or as some energy, but the they are one and the same. Any constant added to our spacetime is identical to spacetime having intrinsic energy.

As far as experiments making distinctions, most wouldn't care the underlying reason, since the effects manifest the same. I would say most observational experiments "treat" it as "physical energy", in the sense that they treat dark energy as in the first case to define it as a form of energy and then study its properties (see: the equation of state). Modified theories of gravity would be an extreme case, and definitely would treat it as the second case; dark energy simply arising from an incomplete description of gravity, i.e. $G_{\mu\nu}$ is incorrect and must be changed, but, in the correct limit, $G'_{\mu\nu} \approx G_{\mu\nu} + \Lambda g_{\mu\nu}$.