Does expanding space cost energy?

Question

Does the cosmic inflation reduce the energy density (inversely) proportional to the volume, or does the inflation "cost" energy? Is space itself "something" created at the expense of energy?

Answer 1

One of the Friedmann equations is a "conservation" equation: $$\dot \rho_i= -3\frac{\dot a}{a} (\rho_i + p_i) \tag{1}$$ where $\rho_i, p_i$ describes energy density and pression for a particular "fluid" (dust, relativist particle, dark energy/cosmological constant). For each fluid, there is a relation between $p_i$ and $\rho_i$ (respectively $p_i=0, p_i =\frac{\rho_i}{3}$, $p_i = -\rho_i$). One found easily the expressions of the $\rho_i$ as function of $a$, $\rho_i = \rho_i(a)$.

Except for the dark energy/cosmological constant case ($\rho_i= Cte$), one sees that the density is not constant. If $\dot a>0$, then the densities of dust and relativistic particles, are decreasing.

Now, considering energies instead of energy densities (excluding gravitational energy), equation $(1)$ can be recasted in a thermodynamic-like relation: $$ d (\Delta U_i) = d(\rho_i \Delta V) = - p_i d(\Delta V)\tag{2}$$

where $\Delta V = (\frac{4 \pi a^3}{3}) \Delta x \Delta y \Delta z$ represents the physical volume at constants $\Delta x, \Delta y, \Delta z$.

$\Delta U_i$ is the internal energy, due to the i-th fluid, contained in the physical volume $\Delta V$

Considering $\dot a>0$, so that $d(\Delta V)>0$, one sees that the internal energy $\Delta U_i$ is constant for dust, decreases for relativist particles, and increases for dark energy/cosmological constant.

Now, to have a complete behaviour, you must make the sum $d(\Delta U) = \sum\limits_i d(\Delta U_i)$. However, as said before, I am not counting gravitational energy here.

Answer 2

Ok, this is going to be a math-free-ish answer. Let's get something straight right off the bat. Inflation is over. Inflation refers to the first ~$10^{-34}s$ of the universe after the Big Bang. What we have now is accelerated expansion. Second, asking if this accelerated expansion "costs energy" is not particularly meaningful. What you'll find is that most cosmologists you ask this question to will hem and haw and maybe stutter a bit. Not because they don't know the answer, it's because this question is so non-specific and interpretive that virtually any answer they give will be in some way correct.

However, that first question you ask is nicely posed. Does the expansion reduce the energy density proportional (although perhaps you meant inversely proportional) to the volume? The answer is sometimes. When it comes to matter (dust, dark matter, stars) the answer is a resounding yes. Three atoms in a box represents a specific amount of energy. The same three atoms in a larger box is virtually the same amount of energy, so the energy density decreases with the increase in volume. For radiation and relativistic particles, the answer is no. Because expansion of the universe also red-shifts radiation, which decreases the photon's energy, the radiation energy density decreases not only with an $R^3$-like volume, it decreases like $R^4$ because of the redshift (OK $R$ is not actually what we use, but it's close enough to make a point with). Dark energy energy density also does not decrease. In the standard cosmological model, $\Lambda CDM$, the energy density of dark matter does not change ever. In a way, this can be seen as the expansion costing negative energy. As the universe expands the total amount of dark energy increases.