From third law of thermodynamics, if temperature $T\to0$, the entropy $S\to0$. In another way we can also say that with the rise of temperature, the entropy will rise.

But we also know that our universe at the very beginning of the time was a very hot dense state of a fire ball and it was in the very low entropy state. So here we observe that temperature $T$ is high but entropy $S$ is low. Is it in the contradiction with the third law of thermodynamics? Or am I explaining/observing it wrong?

Or can we say that we have two system, both in ordered state (low entropy) but one with extremely high temperature (universe at early state) and one with extremely low temperature(Bose Einstein condensate)? How is it possible?

  • $\begingroup$ physics.stackexchange.com/q/101859 $\endgroup$
    – user126422
    Jan 25, 2017 at 19:30
  • $\begingroup$ No, you are not wrong, this is a rather large gap in our understanding and a paradox that remains unsolved. You could see a "good" side to it, in that it illustrates how different the Big Bang was to any physical phenomena that we observe today, so trying to imagine the Big Bang as a large scale version of some other event is probably wrong. $\endgroup$
    – user140606
    Jan 25, 2017 at 20:00

2 Answers 2


General relativity does some odd things to thermodynamics. This can be seen qualitatively with Hawking radiation. Suppose you have a black hole (BH) in a cosmic background of temperature $T$ such that the temperature of the black hole is $T~=~1/8\pi M$ in naturalized units of $\hbar~=~c$ $=~k~=~G~=~1$. The traditional way of thinking is that since the temperature of the BH and the background are the same you have equilibrium. However, if the black hole emits a photon by Hawking radiation it is clear that with $M~\rightarrow~M~-~\delta M$ the temperature increases. If the black hole absorbs a photon with $M~\rightarrow~M~+~\delta M$ the temperature decreases. Also in these two cases the BH now has a higher probability of emitting more photons or absorbing photons respectively. This means the BH has a higher probability in the two cases of quantum decay or growth into a monster supermassive BH. Equality of temperature is not equilibrium.

One way of looking at this is the effective specific heat of an event horizon has a negative value. Consequently increasing temperature lower entropy. This turns thermodynamics on its head. It is contrary to our usual understanding and further there is no clear meaning to the notion of equilibrium. Even in the distant future when all matter has accelerated away so that every region bounded by the cosmological event horizon is a vacuum, there is Hawking-Gibbon radiation where the cosmological horizon emits extremely long wavelength photons and recedes away --- essentially forever, or at a time equal to the quantum Poincare recurrence time of the universe at about $10^{10^{80}}$ years.

For the universe as a whole it is not as clear how thermodynamics plays a role. Clearly as the universe expands temperature decreases. At the same time matter and energy, or maybe most importantly quantum information, crosses the cosmological horizon where it is not accessible. This is a loss of information that equates to entropy increase. Again very low temperature is associated with higher entropy.


with the rise of temperature, the entropy will rise

But that is not the only way that entropy will rise. Remember the definition:

$$S = -k \log(\Omega)$$

Omega is a measure of the total number of possible states that a system can take, and that grows with physical boundaries.

Think about how many ways you can arrange a set of 4 lego 2-blocks on top of a 8-by-2 strip. Now thing about how many ways you can do the same on the green "lawn" piece. The later has higher potential entropy even at the same energy level, which in this case is zero.

This is also why the entropy of the universe is increasing in spite of it cooling off. The two effects are related to the same underlying measure, as it were, the age.


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