# How is dark energy calculated

This should be a very simple question. What would be the proper way to calculate Dark Energy in Joules at any point in history and that is consistent with the Standard Model? I'm thinking that knowing the mass-energy of matter (after estimating the Mass of the Universe):

$$E_m = mc^2\tag{1}$$

and knowing,

$$\rho = \frac{\rho_{m}} {\Omega_{m}}\tag{2}$$

$$\rho_{\lambda} = \rho \Omega_{\lambda}\tag{3}$$

of the universe $M$, I could get the total Energy of the Universe combining (1) and (2):

$$E_{tot} = \frac { mc^2}{ \Omega_m}$$

And then multiply by the $\Omega_{\lambda}$ factor in order to get the final dark energy.

$$DE = mc^2\cdot\frac {\Omega_{\lambda}}{ \Omega_m}$$

Of course, $\Omega_{\lambda}$ and $\Omega_m$ change overtime, so I should be able to find what they are at the given time and figure out what $DE$ is at that time. Does anyone see a problem with this calculations?

• The standard, concordance $\Lambda$CDM universe is infinite, and so too is its mass (not that one should be defining a global thing like mass in GR anyway, at least not without serious thought as to what it means). Unless I'm misunderstanding, your first, fourth, and fifth equations all read $\infty = \infty$. – user10851 Feb 3 '14 at 23:41

It's simpler, since the amount of dark energy remains the same while de Universe evolves. Then, you can calculate it today: $\rho_\Lambda(t)=\rho_\Lambda(today)$.
Today $\Omega_\Lambda=\rho_\Lambda/\rho_c$, where $\rho_c$ is the critical energy density today, and $\Omega_\Lambda$ is measured to have a value $\Omega_\Lambda\approx0.685$. The critical density today is given by $$\rho_c=\frac{3 H_0^2}{8\pi G}\approx4800\,\mathrm{eV\cdot cm^{-3}}$$ so that $$\rho_\Lambda=\Omega_\Lambda\rho_c\approx3300\,\mathrm{eV\cdot cm^{-3}}.$$
Remember that the $\rho$'s are not energies, but energy densities, i.e. energy over volume. You can's treat with total masses or total volumes. Is the Universe finite? You can only use energy densities, and of course, calculating dark energy in joules is impossible unless you define a volume $V$, such as the volume of your house or something like that.