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As we know, if an ideal gas expands in vacuum, as its energy is unchanged, the temperature remains the same. An ideal gas's energy does not depend on volume. In general, the energy is $kT$ times the total degrees of freedom, like in an ideal gas, the total degrees of freedom is $N$ particles plus three dimensions, $3N$.

Then if the total energy of the universe is $kT$ times the total degrees of freedoms of the universe, as the universe expands, its energy and entropy should not change, but if the temperature falls, the number of degrees of freedom should grow. It is quite puzzling to me that the universe is having more and more new degrees of freedom. It seems to be contradictory to the entropy argument.

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"its energy and entropy should not change": that's your mistake. Cross out energy and you have your answer - energy is not conserved cosmologically. –  Michael Brown Mar 5 '13 at 4:18
    
Yep! It comes surprising at you when you hear the truth for the first time but "Energy is not conserved" in our universe. –  Cheeku Mar 5 '13 at 7:18
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@TMS Not sure what you're talking about there. The measured value of the vacuum energy density (i.e. the cosmological constant) is positive. –  Michael Brown Mar 5 '13 at 11:20
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@Xiao-QiSun Energy is not conserved because the universe is expanding! Energy conservation is a consequence of time translation invariance, i.e. an experiment done yesterday would have the same outcome as the same experiment done today. This is no longer true on a cosmological scale. –  Michael Brown Mar 5 '13 at 11:21
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This is actually a little controversial, though the controversy really only boils down to semantics. The actual physics is not ambiguous. You can search this site for more discussion of energy non-conservation in cosmology, or you can read this for a completely correct discussion. :) –  Michael Brown Mar 5 '13 at 11:24
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2 Answers

Here's the answer Ludwig Boltzmann has given in 1884:

For reasons of extensivity the energy density of electromagnetic radiation can be written as $U(T,V)=u(T)V$. We furthermore know from classical electrodynamics that the pressure is one third of the energy density, $p(T)=u(T)/3$, which for instance follows by tracing over the Maxwell stress tensor. If we assume that the chemical potential vanishes, $\mu=0$ (that's a hard one to guess: Boltzmann didn't even know photons...), then (by Euler's equation) the energy is given $U=TS-PV$, and hence $$ S=\frac{U+PV}{T}=\frac{4}{3}V\frac{u(T)}{T} \ . $$ Next, from the differential ${\rm d}F=-S\,{\rm d}T-P\,{\rm d}V$ follows the Maxwell relation $$ \Big(\frac{\partial P}{\partial T}\Big)_V=\Big(\frac{\partial S}{\partial V}\Big)_T \ , $$ and inserting what we know about $P(T)$ and $S(T,V)$ leads to $$ \frac{1}{3}u'(T) = \Big(\frac{\partial P}{\partial T}\Big)_V=\Big(\frac{\partial S}{\partial V}\Big)_T = \frac{4}{3}\,\frac{u(T)}{T} \ . $$ This differential equation for $u(T)$ is easily solved and leads to $u(T)=aT^4$, with some constant $a$—and thus Boltzmann derived the law previously discovered experimentally by Stefan.

But we now also know the entropy: $S(T,V)=\frac{4}{3}aVT^3$.

Punchline: During the adiabatic expansion of the universe the entropy stays constant. Hence, the product $VT^3$ must stay constant, and thus the temperature of the cosmic background radiation decrases inversely with the universe's scale factor.

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That is very interesting. It seems that the photon's energy density is proportional to the fouth power of T is very important. And this comes from the fact that it is massless. This is amazing! –  Xiao-Qi Sun Mar 19 '13 at 4:13
    
This doesn't work. You can't just take a thermodynamic calculation like this and apply it to cosmology. For one thing, an expanding universe doesn't obey the first law of thermodynamics. –  Ben Crowell Aug 21 '13 at 18:00
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Generally speaking is not possible to assert if the energy is conserved in General Relativity (GR). There are several subtle points about the definition of the energy of the gravitational field and how this could introduce a concept of total energy (including gravitational energy), however, here I will discuss only the energy of the matter content.

In some cases one can prove that the total energy of the matter content is indeed conserved. The total energy momentum tensor (EMT) $T_{\mu\nu}$ must satisfy $$\nabla_\mu{}T^\mu{}_\nu = 0.$$ These conditions come from the Bianchi identities along with the Einstein's equations. Given some timelike vector field $v^\mu$ it is possible to define the energy momentum flow through the foliation defined by $v^\mu$ as $P^\mu = T^{\mu\nu}v_\nu$ (not all timelike vector fields define a global foliation, however, I will ignore this point here).

The vector field $P^\mu$ has its divergence given by $$\nabla_\mu{}P^\mu = \nabla_\mu{}T^{\mu\nu}v_\nu + T^{\mu\nu}\nabla_\mu{}v_\nu = T^{\mu\nu}\nabla_\mu{}v_\nu = T^{\mu\nu}\nabla_{(\mu}{}v_{\nu)},$$ where the parenthesis represent the symmetric part and the last equality comes from the symmetric property of the EMT. If for some reason $\nabla_\mu{}P^\mu = 0$, then the Stoke's theorem (plus some conditions on the manifold or on the EMT at infinity) guarantees that the total energy is conserved, i.e., $$\left.\int\mathrm{d}^3x\sqrt{\gamma}P^\nu{}v_\nu\right\vert_{t_1} = \left.\int\mathrm{d}^3x\sqrt{\gamma}P^\nu{}v_\nu\right\vert_{t_2},$$ where $\gamma$ is the determinant of the metric projected on the spatial hypersurface defined by $v^\nu$ and $t_1, t_2$ are two labels defining two different hypersurfaces.

If $v_\mu$ is a Killing vector field it satisfy $$\mathcal{L}_v g_{\mu\nu} = n^\alpha\nabla_\alpha{}g_{\mu\nu} + 2\nabla_{(\mu}v_{\nu)} = 2\nabla_{(\mu}v_{\nu)} = 0,$$ where we used the covariant derivative compatible with $g_{\mu\nu}$. This shows that, if there is a timelike Killing field then the total energy is conserved, however, the converse is not true, i.e., the following statement is not true: if the total energy momentum is conserved then there is a timelike Killing field.

In our current cosmological model, the universe (at zero order) is described by a Friedmann-Lamaître-Robertson-Walker (FLRW) metric which do not posses a timelike Killing vector. This is why the energy of a radiation-like fluid is not conserved, from the direct calculation of the EMT divergence in a FLRW model we have $$\dot{\rho} + 3H(\rho+p) = 0,$$ where $H=\dot{a}/a$ is the Hubble function, $a$ the scale factor, $\dot{f} = v^\mu\partial_\mu{}f$ the time derivative of a scalar function $f$, the EMT is given by $T_{\mu\nu} = \rho{}v_\mu{}v_\nu + p\gamma_{\mu\nu}$, $v_\mu$ is the field represent the fluid flow, $\gamma_{\mu\nu} = g_{\mu\nu} + v_{\mu}v_{\nu}$ is the spatial projector, $\rho$ the energy density in this frame and $p$ the isotropic pressure also in this frame. For a constant equation of state ($w = p/\rho$) we have $$\rho = \rho_0\left(\frac{a_0}{a}\right)^{3(1+w)},$$ where $a_0$ and $\rho_0$ are the scale factor and energy density calculated in a spatial section defined by $t_0$.

From the direct calculation of the total energy (in this frame) we have $$\int\mathrm{d}^3x\sqrt{\gamma}P^\nu{}v_\nu = \int\mathrm{d}^3x\sqrt{\gamma}\rho = \int\mathrm{d}^3x\sqrt{\gamma_0}\rho_0\left(\frac{a_0}{a}\right)^{3w},$$ where we used that $\dot{\sqrt{\gamma}} = 3H\sqrt{\gamma}$ and, therefore, $\sqrt{\gamma} = \sqrt{\gamma_0}(a_0/a)^3$ (this is true in FLRW, in general $3H$ is substituted by the expansion factor $\Theta \equiv \nabla_\mu{}v^\mu$). This shows that, for radiation ($w=1/3$) the energy decreases with $\propto a^{-1}$ when the universe expands. Note also that the presence of dark energy $w<-1/3$ make the total energy increases, e.g., the cosmological constant has $w=-1$ then the energy goes like $a^3$.

For the special case of dust $w=0$ the total energy is conserved. This is an example of what I said before, we can have total energy conservation without a Killing field, in this case this happens because $T_{\mu\nu} = \rho{}v_\mu{}v_\nu$ is orthogonal to $\nabla_\mu{}v_\nu = \mathcal{K}_{\mu\nu} = H\gamma_{\mu\nu}$, where $\mathcal{K}_{\mu\nu}$ is the extrinsic curvature that in FLRW is proportional to $\gamma_{\mu\nu}$.

Finally, we can only have thermodynamic equilibrium when we have a timelike Killing field with the exception that for radiation we just need a conformal Killing field to achieve equilibrium (see "Kinetic theory in the expanding universe" Bernstein 1988). In a FLRW universe we have a timelike conformal Killing field and thats why we have a well defined temperature for radiation, using the Bose-Einstein distribution (assuming kinetic equilibrium) we obtain that $T \propto a^{-1}$, thats why, in the thermodynamical viewpoint, the total energy is not conserved, the temperature drops when the universe expands.

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Your answer talks about "total energy," but what definition of energy are you using that would apply to a cosmological spacetime? Well known tensorial measures of energy such as the Komar mass and ADM mass don't work here, because this is not an asymptotically flat spacetime. If you're calculating something nontensorial, then it's coordinate-dependent, and it's not clear that it has any physical significance. –  Ben Crowell Aug 21 '13 at 18:05
    
@BenCrowell did you read the first paragraph of my answer? –  Sandro Vitenti Aug 29 '13 at 15:31
    
I did, but it doesn't clarify for me what quantity you're discussing. –  Ben Crowell Aug 29 '13 at 15:48
    
In your comment you give examples of possible energy definitions for the gravitational field, that is why I pointed to the first paragraph. If you are concerned with the definition of energy for the matter content, note that the energy density is defined as $\rho_{(v)} = T^{\mu\nu}v_\mu{}v_\nu$ for some global timelike vector field $v^\mu$. This means that we need to be able to define global spatial sections (Frobenius's Theorem) and that the energy definition is frame dependent as usual. Finally, in the answer I used the FLRW homogeneous and isotropic frame to define energy. –  Sandro Vitenti Aug 29 '13 at 18:52
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