I stumbled upon this question while preparing for the astronomy Olympiad.

It is question #17 in this pdf.

Given that dark energy is vacuum energy, and that the densities of dark energy, dark matter and normal matter in the universe are currently $\rho_\wedge = 6.7 \times 10^{-30} \;\text{g}\,\text{cm}^{-3}$, $\rho_{DM} =2.4 \times 10^{-30} \;\text{g}\,\text{cm}^{-3}$ and $\rho = 0.5 \times 10^{-30}\;\text{g}\,\text{cm}^{-3}$, what is the ratio of the density of dark energy at the time of the cosmic microwave background emission, to the current density of dark energy?

Any insight and some resources for similar problems would be helpful.

  • $\begingroup$ See the help center. Homework-type questions are off-topic. $\endgroup$
    – ProfRob
    Jun 29 '21 at 7:16

The concept behind this question is how the various components scale as the universe expands.

For matter (both normal and dark matter) this is obvious. Suppose we consider a cube of space of side $x$ then if the universe doubles in size this expands to a cube of size $2x$ and hence the volume increases by a factor of eight. Hence the density of the matter decreases by factor of eight.

We describe the expansion of the universe using a scale factor that is conventionally represented by the symbol $a$. We take $a=1$ at the current time, so in the future when everything in the universe is twice as far apart we would have $a=2$. Likewise if we go back in time to the moment when everything in the universe was half as far apart we would have $a=0.5$. Given this, it is hopefully obvious that the density of matter is inversely proportional to $a^3$:

$$ \rho_M(a) = \frac{\rho_{M_0}}{a^3} \tag{1}$$

where $\rho_{M_0}$ is the density of matter at the current time.

Equation (1) probably seems so obvious that you might wonder why I bothered with it. The reason is that other components do not scale in this way. For example radiation (and relativistic matter) exert a pressure and they do work as the universe expands. This means their energy density scales as $a^{-4}$ instead of $a^{-3}$.

$$ \rho_R(a) = \frac{\rho_{R_0}}{a^4} \tag{2}$$

And finally dark energy is (we think) a property of spacetime and does not change as the universe expands so:

$$ \rho_\Lambda(a) = \rho_{\Lambda_0} \tag{3} $$

The point of all this is that the proportions of the different types of matter/energy change as the universe expands. Just after the Big Bang when $a \ll 1$ the radiation density was dominant. Then as $a$ increase the matter density became dominant, and right now the dark energy density dominates. If you are interested in more details see the question How does the Hubble parameter change with the age of the universe?

We don't provide answers to homework type questions here, so I won't directly answer the question. However since the question only asks about $\rho_\Lambda$, and equation (3) tells us that is constant, the answer is rather obvious.


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