According to the standard cosmology, our universe is expanding. During the process, the content of the Universe is also changing. For example, about 13.7 billion years ago, the Universe was dominated by

The answer is in the Friedmann's equations themselves. Recall the Robertson-Walker metric \[

\mathrm ds^2 = \mathrm dt^2 - a^2(t) \left( \frac{\mathrm dr^2}{1-k r^2} + r^2 \mathrm d\theta^2 + r^2 \sin^2\theta\mathrm d\phi^2 \right) \] where, $k$ represent the curvature of the space, and $a(t)$ is the scale factor that satisfies the Friedmann's equations state, \[

\begin{split}

\left(\frac{\dot a}{a} \right)^2 &= \frac{8\pi G}{3} \rho + \frac{\Lambda}{3} - \frac{k}{a^2} \\

\frac{\ddot a}{a} &= -\frac{4\pi G}{3} (\rho + 3p) + \frac{\Lambda}{3}

\end{split}

\] Take the derivative on the first equation and substitute the second one into it and then we get, \[

\dot \rho = -3\frac{\dot a}{a} (\rho + p ) \Rightarrow \mathrm d\rho = -3\frac{\mathrm da}{a}(\rho + p).

\] The volume of the Universe expands as $V = V_0 a^3$ or $\frac{\mathrm dV}{V} = 3 \frac{\mathrm da}{a}$. Here $\rho$ is the energy density of the matters (baryon matter, radiation and dark matter etc). Then the internal energy is $U = \rho V$ or $\mathrm dU = \mathrm d\rho V + \rho \mathrm dV$. Plug in the expression for $\mathrm dV$ and $\mathrm d\rho$. We arrive, \[

\mathrm dU + p \mathrm dV = 0! \] This is precisely the first law of thermal dynamics (for matters) for an adiabatic expansion. Therefore, the loss of radiation energy is due to the work of the CMB ratiation done to the Universe, which makes the Universe expands.

Now, what about the dark energy? Where does the dark energy come from? Simply consider the internal energy of the dark energy: $\mathrm d U_\Lambda = \rho_\Lambda \mathrm d V$. This is because the energy density of the dark energy is constant. Add the work term, $\mathrm dU_\Lambda + p_\Lambda \mathrm dV= (\rho_\Lambda + p_\Lambda) \mathrm dV$. Now we need a relation that relates the energy density and the pressure, which is called the equation of state (EoS). To obtain the EoS for dark energy, we can look at the second Friedmann's equation. If the cosmological constant is treated as some energy, the right-hand side of the equation should be written as \[

\frac{\ddot a}{a} = -\frac{4\pi G}{3} (\rho + 3p) -\frac{4\pi G}{3} (\rho_\Lambda + 3p_\Lambda)

\] Namely, $4\pi G (\rho_\Lambda + 3p_\Lambda) = -\Lambda$. But $\rho_\Lambda = \frac{\Lambda}{8\pi G}$. Therefore, $p_\Lambda = - \rho_\Lambda$. This is the EoS for the dark energy, which says the dark energy has negative pressure! Turn back to the thermal dynamical equation, clearly $\mathrm dU_\Lambda + p_\Lambda \mathrm dV = 0$. Again, the dark energy also obeys the first law of thermal dynamics. It turns out the dark energy was created from the negative pressure and the work done by the Universe.

*dark matter*; however, today it is the*dark energy*that rules the sky. According to the standard cosmological model, Lambda-CMD, the energy density of the dark energy $\rho_\Lambda = \frac{\Lambda}{8\pi G}$ ($\Lambda$ is Einstein's cosmological constant) does not change during the expansion of the Universe. That is why the dark energy is increasing as the Universe becomes larger. There is no known interaction between the dark energy and the ordinary matter (including dark matter) except for gravity - which according to the general relativity is merely the curvature of the Universe. There is no exchange of energy. So where does this large amount of energy come from? Does the first law of thermal dynamics (the law of energy conservation) still hold? Given the "dark" nature of the dark energy, one may turn to consider physical matters, such as photons - the cosmic microwave background radiation (CMB). As the Universe expands, the photons are redshifted, i.e., their frequencies $\nu$ become smaller $\nu/(1+z), z>0$. Since the Universe is extremely empty, the loss of photons due to reaction is negligible. Therefore, the total energy of the radiation $E = N_\gamma h \nu$ would decrease as the Universe expands. So, again, where does the energy go?Fig. 1, Left: an artist's impression of the chronology of the Universe. Right: the contents of the Universe according to WMAP project. Credit: NASA / WMAP Science Team |

\mathrm ds^2 = \mathrm dt^2 - a^2(t) \left( \frac{\mathrm dr^2}{1-k r^2} + r^2 \mathrm d\theta^2 + r^2 \sin^2\theta\mathrm d\phi^2 \right) \] where, $k$ represent the curvature of the space, and $a(t)$ is the scale factor that satisfies the Friedmann's equations state, \[

\begin{split}

\left(\frac{\dot a}{a} \right)^2 &= \frac{8\pi G}{3} \rho + \frac{\Lambda}{3} - \frac{k}{a^2} \\

\frac{\ddot a}{a} &= -\frac{4\pi G}{3} (\rho + 3p) + \frac{\Lambda}{3}

\end{split}

\] Take the derivative on the first equation and substitute the second one into it and then we get, \[

\dot \rho = -3\frac{\dot a}{a} (\rho + p ) \Rightarrow \mathrm d\rho = -3\frac{\mathrm da}{a}(\rho + p).

\] The volume of the Universe expands as $V = V_0 a^3$ or $\frac{\mathrm dV}{V} = 3 \frac{\mathrm da}{a}$. Here $\rho$ is the energy density of the matters (baryon matter, radiation and dark matter etc). Then the internal energy is $U = \rho V$ or $\mathrm dU = \mathrm d\rho V + \rho \mathrm dV$. Plug in the expression for $\mathrm dV$ and $\mathrm d\rho$. We arrive, \[

\mathrm dU + p \mathrm dV = 0! \] This is precisely the first law of thermal dynamics (for matters) for an adiabatic expansion. Therefore, the loss of radiation energy is due to the work of the CMB ratiation done to the Universe, which makes the Universe expands.

Now, what about the dark energy? Where does the dark energy come from? Simply consider the internal energy of the dark energy: $\mathrm d U_\Lambda = \rho_\Lambda \mathrm d V$. This is because the energy density of the dark energy is constant. Add the work term, $\mathrm dU_\Lambda + p_\Lambda \mathrm dV= (\rho_\Lambda + p_\Lambda) \mathrm dV$. Now we need a relation that relates the energy density and the pressure, which is called the equation of state (EoS). To obtain the EoS for dark energy, we can look at the second Friedmann's equation. If the cosmological constant is treated as some energy, the right-hand side of the equation should be written as \[

\frac{\ddot a}{a} = -\frac{4\pi G}{3} (\rho + 3p) -\frac{4\pi G}{3} (\rho_\Lambda + 3p_\Lambda)

\] Namely, $4\pi G (\rho_\Lambda + 3p_\Lambda) = -\Lambda$. But $\rho_\Lambda = \frac{\Lambda}{8\pi G}$. Therefore, $p_\Lambda = - \rho_\Lambda$. This is the EoS for the dark energy, which says the dark energy has negative pressure! Turn back to the thermal dynamical equation, clearly $\mathrm dU_\Lambda + p_\Lambda \mathrm dV = 0$. Again, the dark energy also obeys the first law of thermal dynamics. It turns out the dark energy was created from the negative pressure and the work done by the Universe.