I'm really confused about the concept of entropy when applied to the whole universe. The often hear that the universe started with very low entropy and as the entropy increases the universe will slowly reach the heat death. The initial low entropy at the big bang is also used to explain the arrow of time.

I just do not understand these arguments. My main problem is with the concept of entropy. To me, the entropy make sense only and only if we define the macrostates we work with. If we want to talk about entropy of the whole universe we have to mention the macrostates we work with.

So when people talk about entropy of the universe, what macrostates do they have in mind?

Also, when we reach the supposed "heat death of the universe". I do not see any reason why we would not be able to use different macroscopic variable in which the universe would not look like as in its "heat death" state.

  • $\begingroup$ I'm not sure if this is what you're stuck on, but does the concept of continuous entropy make sense to you? What do you define as the "macroscopic variable" of a continuum? There are an uncountably infinite number of configurations. I think the fundamental issue is that you don't need microstates and macrostates to define entropy. From an information-theoretic standpoint at least, the entropy of $X$ is just $\mathbb{E}_{X}[\log \frac{1}{\mathbb{P}(X)}]$. I don't see a fundamental need to define macrostates or microstates to define entropy. $\endgroup$
    – user541686
    Aug 10, 2018 at 1:45
  • $\begingroup$ @Mehrdad The problem with entropy defined as $- \int p \log p$ is that it stays constant because of the Liouville's theorem. To increase the entropy over time you have to somehow deteriorate the knowledge about the system, the probability distributrion $p$. To do that you basically have to define something like macrostates. $\endgroup$
    – Tom
    Aug 10, 2018 at 6:32
  • $\begingroup$ I think what @Mehrdad is getting at is that if you want to quantify entropy you need to define micro/macro states. But at the end of the day entropy is just a measure of likelihood. The 2nd law basically just says the most likely thing will happen, or or system will move towards the most likely configuration (no matter what states you look at). So you don't really need to define macrostates to get at the essence of what entropy and the second law really are. $\endgroup$ Aug 10, 2018 at 9:21
  • $\begingroup$ Seems like anyone making a statement about universal entropy constantly increasing is speaking more tautologically than physically. This is, for any compliant definition of "entropy" in the context of some believed physical model, it's statistically expected that the entropy will tend to increase over time. $\endgroup$
    – Nat
    Aug 10, 2018 at 19:36

2 Answers 2


It sounds like you're under the impression that people are claiming to have a general formulation of thermodynamics that gives a complete and rigorous description of cosmology, generalizing all of the 19th-century laws of thermodynamics in an appropriate way. As far as I know, nobody in the field really makes that claim.

One reason I'm pretty sure we don't have anything like this is that the first law of thermodynamics is conservation of energy, which is usually stated in terms of a global quantity that stays the same. But general relativity does not have a global, scalar, conserved measure of energy that applies to cosmological spacetimes. (It does have things like local conservation of energy-momentum, and conservation of energy in asymptotically flat spacetimes, but that's not the same thing.)

The laws of thermodynamics also refer to temperature, but there is no really satisfactory relativistic definition of temperature.

There are the laws of black hole thermodynamics, but those don't really connect to the ordinary laws of thermodynamics in a comprehensive way, and the way that time-reversal asymmetry comes in seems to me to be qualitatively different from the way it works in standard thermodynamics. Basically you get time-reversal asymmetry in the second law of black hole thermodynamics simply because you define a horizon as a surface from which you can't reach future null infinity.

So when people (including me) say that the arrow of time comes from the fact that we had a low-entropy big bang -- well, I can only speak for myself, but I say that in a loose way, not believing that there is any fully systematic underlying theory.

So when people talk about entropy of the universe, what macrostates do they have in mind? [...] Also, when we reach the supposed "heat death of the universe". I do not see any reason why we would not be able to use different macroscopic variable in which the universe would not look like as in its "heat death" state.

I don't quite understand what objections you have in mind here, though. These seem like questions that are not qualitatively different from the ones we would ask about a steam engine, and the answers would be pretty much the same, wouldn't they? Maybe you could edit your answer to spell out in more detail what you have in mind. E.g., are you worried about how to handle the time variable, so that we can say that the macrostate is the state at a certain time? (If so, then I think the answer would be that you can take any Cauchy surface that you like.)

  • $\begingroup$ ...but...but...people sound so convincing that they know what they are talking about... $\endgroup$
    – Tom
    Aug 10, 2018 at 7:18
  • $\begingroup$ @Tom: The lack of a complete, rigorous, overarching theory doesn't necessarily mean that we can't say anything with assurance. For example, people can predict the results of various particle physics experiments without having fully characterized the Higgs (e.g., whether it's a single Higgs or multiple Higgses). The laws of black hole thermodynamics have rigorous proofs, but those proofs just have to make certain technical assumptions, which I believe make the results not applicable to cosmological spacetimes. One can understand the general outline without knowing every technical detail. $\endgroup$
    – user4552
    Aug 10, 2018 at 18:02

There are different definitions for "makro-state" floating arround. In my opinion the most natural one is that a makro-state is identified with a probability distribution $\rho$ which is defined on the mikro states of the system.

This definition works out for the universe as well: The mikrostates of the universe would be every possible little degree of freedom of every little particle that you can imagine. If the universe consists of n-particles (and we want to stay classical...) then the mikrostates would be points in the 6N-dimensional Phase space, containing 3 position and 3 momentum variables for each particle: $\vec{x}_1$, $\vec{x}_2$ .... and $\vec{p}_1$, $\vec{p}_2$ ... and so on. The Makrostate the universe is in would then be a probabillity distribution $\rho(\vec{x}_1, \vec{x}_2 ....., \vec{p}_1,\vec{p}_2.....)$ that tells you what combination of momentums and positions is more or less probable.

It makes sense to define entropy as a functional on this probability density as $S = \int \rho Ln (\rho)$: This way for the entire universe, because it is defined in the exact same way for all its subsystems (like for example a hot pot of tea).

To adress you second question: The assumption is now that entropy as a functional of $\rho$ will grow throughout the time evolution of $\rho$ (there are plausibility arguments why it as to grow, and that aside, this is has proven to be an excellent assumption when one tries to find the steady state of a system). If $\rho$ doesn't change anymore (it reached a steady state), the entropy should be at its maximum (the maximum in the space of all possible probability distributions). While this steady state appears rather quick in a teapot in which you pour in your milk, the universe is not (yet?) in it's steady state.

When will this steady state (called heat death) happen - There are different assumptions on when it will happen, or if it will happen at all. The problem is that systems that reach their steady state are either supposed to be isolated, or in contact with a reservoir, that together with the system then forms a closed system. For the universe, whose components undergo an accelleration whe yet don't know where it comes from, one can't claim that the observable univsere is an isolated system.

To adress your last question: Heat death is the steady state for the makrostate $\rho$ of the universe. If this makrostate $\rho$ doesn't change anymore, then any makroscopic variable $O = \int \rho O$ (integral over all mikrostates) doesn't change as well. Any makroscopic variable you can think of would be static as well.

  • 1
    $\begingroup$ But by Liouville's theorem such entropy stays constant. To increase entropy you have to define a mechanism how the information is lost and in my opinion that is what the definition of macrostates is about. $\endgroup$
    – Tom
    Aug 10, 2018 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.