# Entropy of the Universe

1. Dec 7, 2012

### vinven7

For a system that is completely isolated from its surroundings, basic thermodynamics requires that the quasi-static heat flux dQ and the entropy change dS be related by:
dQ = TdS
and since the system is isolated,
dQ=dS=0
Therefore, an isolated system should be isentropic, that is, it's entropy must remain a constant.

The UNiverse is itself an isolated system (please don't bring up branes and stuff like that, let's keep it classical) - so there is no heat input or output. Consequently, the entropy of the Universe should also be constant.
However, my understanding of the second law is that the entropy of the Universe is always increasing. Further, many cosmological models require that the entropy of the universe increase drastically immediately after big bang.

Can someone please explain if I am wrong somewhere and if so, how do I reconcile the two?

2. Dec 7, 2012

### Staff: Mentor

Your first sentence contains a very important word: quasi-static.

3. Dec 7, 2012

4. Dec 7, 2012

### Staff: Mentor

Are all processes in the universe quasi-static?

5. Dec 7, 2012

### vinven7

If we are talking about the Universe as a whole and treat it as our system, why would the processes within the system matter at all? Please correct me if I am wrong

6. Dec 7, 2012

### Darwin123

You are wrong because you assume that entropy can't be made. Entropy can and is made all the time by irreversible processes.

The processes inside the system can make entropy. If a system is isolated, then the entropy can not be moved out of the system. However, the system can still make entropy.

Entropy is an intensive property. Therefore, every bit of entropy is located somewhere in space. There is an entropy density describing how the density is distributed in space. The entropy density at any location can be changed either by moving entropy or by making it.

The second law of thermodynamics only says that entropy can't be destroyed. However, there is no law that says that entropy can't be made. There is no law that says that entropy can't be moved. The "trick" in many thermodynamic problems is distinguishing one from the other.

There are two things that a process can do with entropy: move it or make it. A lot of confusion comes about because people don't realize there are two options. So when the entropy of a system changes, it is one of the two options. Move it or make it.

There are two important types of process that are defined in terms of what they do to the entropy in a closed system. An adiabatic process can make entropy in a closed system, but it often doesn't move entropy in or out of the closed system. An isothermal process can move entropy in or out of the system, but it often doesn't make entropy.

There are two processes that are defined on what they do to total entropy. An irreversible process is by definition a process that makes entropy. A reversible process by definition a process that doesn't make entropy.

7. Dec 7, 2012

### vinven7

Thanks so much! That cleared my confusion

8. Dec 7, 2012

### Jano L.

vinven7,
in order to find out what happens to entropy, one must first define what it is. In thermodynamics, change of entropy is defined by integral

$$\Delta S = \int_1^2 \frac{dQ}{T}$$

where the integration is over trajectory in space of quasistatic states. The entropy is then function of the upper bound of the integral, or in other words, of the state variable.

But it is difficult to define heat and state variables for Universe. The Universe does not seem to be in equilibrium. So the thermodynamics definition of entropy does not seem to be applicable.

9. Dec 7, 2012

### Staff: Mentor

For an irreversible processes that takes place in a closed system, Clausius inequality states that

dS > dQ/T

where dQ and T represent the heat inflow and the corresponding temperature, respectively, at the boundary of the system. For the universe, which represents an isolated system, dQ = 0, so that dS > 0.

10. Dec 7, 2012

### vinven7

Thank you. This is awesome. In strict thermodynamic terms, how do we show that the universe is not an equilibrium system?

11. Dec 7, 2012

### Jano L.

Chestermiller,

the Clausius inequality applies to quasistatic process which creates entropy in the system. In order to call a process quasistatic, one has to have macroscopic quantities to define the state of the system and change them slowly.

What quantities should we use for macroscopic state of the Universe?

It does not make sense to apply thermodynamics in this naive way to whole Universe. Thermodynamics is about control of few variables and about equilibrium properties between them - and for Universe, this is out of question.

12. Dec 7, 2012

### Rap

A system in equilibrium is (macroscopically) unchanging in time, by definition. Pressure, temperature, etc. are all unchanging in time. Since we know things are changing in time, we know that the universe is not an equilibrium system.

13. Dec 7, 2012

### vinven7

By this definition the earth moving around the sun cannot be an equilibrium system - can it? If we assume that there was only the sun and the earth (leave everything else out)

14. Dec 8, 2012

### Staff: Mentor

When we talk about the "universe" in a thermodynamic sense, what we are really referring to is an isolated system which is comprised of two parts: a "system" portion plus a "surroundings" portion. The "system" plus the "surroundings" add up to what we call the "universe." But this only a sub-region of the actual UNIVERSE that the astrophysicists refer to. Still, the actual UNIVERSE is continually undergoing irreversible processes and, in the end, when it does reach thermodynamic equilibrium, its entropy will be greater than when it started.

I also want to add that the Clausius inequality does not refer to closed systems undergoing quasistatic processes, but rather to systems undergoing irreversible processes (which are not quasistatic). I also want to point out that the mathematical form in which the Clausius inequality typically appears is very imprecise. A more precise form is to say that, if B represents the boundary surface of a closed system that is undergoing an irreversible change from thermodynamic equilibrium state 1 to thermodynamic equilibrium state 2, and if we calculate the integral over time and over the surface B of the normal component of the inwardly directed heat flux divided by the local temperature at the surface, this integral will be less than the change in entropy for the system contained within B.

15. Dec 8, 2012

### Rap

That's an interesting question. For the case of the sun and the earth, the sun is not in equilibrium, it will burn up its fuel to a point where it becomes a red giant, etc. But suppose we had two iron balls orbiting each other. Unless they were at zero K, they would evaporate iron and so they are not in equilibrium, but I don't know what the final equilibrium state would be. If they were at zero K, I think there would still be "tidal forces" which would distort the balls, heating them up, causing the rate of rotation to decrease, and bringing them closer, so even that would not be equilibrium. I'm not sure if maybe the balls were rotating on their axes at just the right angular velocity, you could make the tidal forces not have a heating effect.

(Edit) - thinking about it, even if the tidal forces could be eliminated as a source of heating, there would be dissipation of energy by gravitational waves, and there would be no way around that. A single spinning iron ball at zero K would not emit gravitational waves, but then its angular momentum would just be part of its thermodynamic internal energy, so I guess the idea that thermal equilibrium is that all macro parameters are constant in time is still valid.

Is that true? I mean, if you have a container of stoichiometric hydrogen and oxygen at room temperature, if you don't explode it, the reaction to H2O will proceed slowly (quasistatically) and it's irreversible.

Last edited: Dec 8, 2012
16. Dec 8, 2012

### Studiot

Good morning vinven7,

I think your difficulty arises from a basic misconception, that Chestermiller has touched upon.

You observe that the universe is an isolated system, then try to apply dQ.

By definition of an isolated system dQ = 0. Period.

Remember that Q is the heat flux across the boundary, not the flux within parts of the system.
The energy of an isolated system is constant.
The entropy of an isolated system either remains constant or increases, it cannot decrease.

As regards the universe it is either infinite in which case the entropy is infinite. What does it mean to say that infinity is increasing or constant?

Or the universe is finite in which case the simple statements above apply.

Do you understand the proof that the entropy of an isolated system must increase or be constant?

17. Dec 8, 2012

### Staff: Mentor

Let me restate what I said a little more precisely: The Clausius inequality does not refer to closed systems undergoing exclusively quasistatic processes, but rather to systems undergoing irreversible processes (which are not restricted to being quasistatic). My point is that the key word here is "irreversible," not "quasistatic."

18. Dec 8, 2012

### Staff: Mentor

Studiot, I think you also meant to say that an isolated system is closed, so that no mass enters or leaves the system (dm = 0).

19. Dec 8, 2012

### Studiot

Perfectly true but I'm not sure I needed to say that.

An isolated system is one in which nothing, neither energy (whether in the form of heat or work or whatever) nor matter (mass) crosses the boundary.

A closed system is one in which energy (work ,heat etc) may be transferred across the boundary but matter (mass) may not.

So, of necessity an isolated system is also a closed one.

20. Dec 8, 2012

### Staff: Mentor

I'm not sure you needed to say that either. Still, when you were talking specifically about the definition of an isolated system, I though it might be worth mentioning. No big deal.