# Why is cosmological acceleration expressed in terms of an energy density?

In the articles that I have (tried to) read, acceleration ends up being expressed as a dimensionless constant (omega-lambda) or else occasionally in terms of a "dark" energy density. Presumably one can multiply that density by the volume of the visible universe (46 Gl.y. radius) to get some kind of estimate of total dark energy. But if the universe has mass, and there is acceleration, doesn't that imply some kind of energy expenditure on a time basis? Is there way to calculate the work done per unit time to drive the acceleration? This is presuming the rate of acceleration (a-double dot) is itself constant, and I gather that that is not certain but at least plausible.

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– Qmechanic Feb 1 '12 at 20:50

The density $\rho$ is a more logical quantity to describe the situation because according to the cosmological principle (which postulates the uniformity and isotropicity of the Universe at the long enough distance scales), the same physics (and expansion) applies to each cubic meter of space and we don't really need to know what the size of the Universe is (and even whether it is finite or not).

One could of course talk about the "density of work" needed to accelerate the matter etc. However, these discussions wouldn't be too useful because "work" is a useful concept if the energy is conserved. However, in cosmology, especially in an expanding Universe, the total energy isn't conserved. There isn't any nonzero definition of an energy that would be conserved while the Universe is expanding. See

http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html