According to this Wikipedia article, the universe is expanding adiabatically.

However, any system that is expanding adiabatically must be losing energy $\left(\mathrm{d}E = -p\,\mathrm{d}V\right) .$

Does not this violate conservation of energy, as the internal energy of the universe is decreasing?

For any finite system expanding adiabatically, the conservation of energy is not violated as the system transfers its internal energy into the rest of the universe through the form of work.

However, when the whole universe is taken as the system, assuming there is nothing outside it (I am not very sure about this statement), where is the work done going into?

Note: A constant energy free expansion model will imply energy density times the volume is constant, so $T^4 V = \text{constant}$, or $T \propto a^{-\frac{3}{4}}$.

But the adiabatic expansion model implies $TV^{\frac{1}{3}} = \text{constant}$, or $T \propto \frac{1}{a}$, where $a$ is the scale factor.

Please note that I am looking for an explanation that can be understood without much knowledge in General Relativity.

  • $\begingroup$ Note that this is not a duplicate of physics.stackexchange.com/questions/410392/… because that question is about possible loss of energy due to redshift, while this is about possible loss of energy due to work done by the expanding universe. $\endgroup$ Commented Apr 4, 2019 at 14:40

2 Answers 2


Well, there are a few ways to answer this question.

Suppose you draw an imaginary (comoving) box around some region of spacetime. The volume of this box expands over time, so you can think of the contents of this box as doing work on the rest of the universe, outside of the box.

Now, you might say this isn't satisfying if we want to consider the universe as a whole. In that case, you can think of the energy as all going into gravitational potential energy. This is a perfectly good picture in the Newtonian limit.

However, it turns out that in general relativity, it's very hard to make the notation of "gravitational potential energy" precise. For example, you can't talk about the gravitational potential energy density at a point, because you can always go into a freely falling frame there, where the observed gravitational field is zero. For this reason, relativity textbooks generally say that the gravitational potential energy is not defined at all; instead energy (defined as not including this extra ill-defined piece) simply isn't conserved in general relativity. The energy doesn't "go" anywhere, it just vanishes.

If you think this is unacceptable, remember that the only reason we elevated conservation of energy to an important principle in the 19th century was that it was observed to work in everyday situations. We never tested it in exotic situations like those with curved spacetime, so there's no reason to expect the principle to continue to hold up. At a deeper level, Noether's theorem tells us that energy conservation is related to time translation invariance, and we don't have that in an expanding universe.


Just a small addition. It is crucial here that the density of the vacuum energy is constant over time. This means that if a given volume of vacuum energy expands, its energy density remains constant, in contrast to what happens if gas expands adiabatically instead. This means that the work done to adiabatically expand the universe is just sufficient to keep the vacuum energy density constant.

  • $\begingroup$ If vacuum energy density remains constant, the $\textbf{Temperature must remain constant}$ (energy density $\propto T^4$), which does not happen. That would also imply that the energy of the universe is increasing. $\endgroup$ Commented Apr 6, 2019 at 9:07
  • $\begingroup$ energy density ∝T4 : this concerns a photon gas, not vacuum. Note also that the Wikipedia article you refer to "needs attention from an expert in physics. The specific problem is: The article discusses the total energy of the universe and applies the first law of thermodynamics to the entire universe, which is incorrect ...". Mentioned at the top. $\endgroup$
    – timm
    Commented Apr 6, 2019 at 9:39

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