An isolated system $A$ has entropy $S_a>0$.

Next, the isolation of $A$ is temporarily violated, and it has entropy reduced $$S_b ~=~ S_a - S,\space\space\space S\leq S_a.$$

Is it true to say: the process of lowering entropy of a system requires work and energy?

I am not sure if energy of the system must be changed when entropy is reduced. However, energy certainly is required – changing entropy is work and uses energy?

  • 3
    $\begingroup$ Counter example: hot thing in cold bath. $\endgroup$
    – genneth
    Apr 21, 2012 at 2:07
  • $\begingroup$ ok, but you first have to put it into the bath? the system is isolated for some period of time. $\endgroup$
    – Mooncer
    Apr 21, 2012 at 2:10
  • $\begingroup$ @Steffen it takes no work to put it in the bath. $\endgroup$
    – Ron Maimon
    Apr 21, 2012 at 4:51

2 Answers 2


The thermodynamic equation is: $$dU = TdS -PdV +\mu dN,$$ where $T$ is the temperature, $S$ is the entropy, $P$ is the pressure, $V$ is the volume, $\mu$ is the chemical energy (which is important in systems which can exchange particles with some reservoir), and $N$ is the number of particles. $U$ is the energy and in this formulation is a function of $S,V,N$ so we can write $U(S,V,N)$. If you make $S$ smaller, then you have to decrease the energy $U$. I.e. if $dS<0$ then $dU$ is also $<0$. In fact, $dU = TdS$, that is, the temperature is just the ratio of small changes in energy and entropy.

So in order to decrease $S$, you will have to remove energy from the system. Thus the system does work on the reservoir, not vice versa. One way of accomplishing this is to put your system in contact with a temperature reservoir at a lower temperature. Then your system gives energy to the cold reservoir.

I think what you're getting at is the fact that when you do this, because the temperature of the reservoir is lower than the temperature of your system, the total amount of entropy in the universe has to go up. But the heat (work) is transferred to the cold reservoir, not the other way around.

Hey, it's easy to take energy out of very hot (high entropy) things. Just cool them down with whatever you happen to have.

  • $\begingroup$ There are other ways to reduce entropy? For example, if the system would be a gas, entropy could be reduced via separating fast and slow molecules [slow|fast] by inserting a partition in the middle of the box, and somehow moving the molecules. And all this is work and needs energy? $\endgroup$
    – Mooncer
    Apr 21, 2012 at 2:55
  • $\begingroup$ @Steffen: That's Maxwell's demon. And the "somehow" uses up energy. $\endgroup$ Apr 21, 2012 at 3:25
  • $\begingroup$ What you are describing with the word "work" is heat not work. The equation you give is best formulated as dS=... not dU=... (it makes a difference conceptually). $\endgroup$
    – Ron Maimon
    Apr 21, 2012 at 4:52
  • $\begingroup$ I agree about the difference between work and heat, though they can be measured in the same units. As far as formulating the equation the other way, solve for dS. $\endgroup$ Apr 21, 2012 at 5:22
  • $\begingroup$ As far as doing it with dS = (1/T)dU + ..., write it up and I'll vote for it. $\endgroup$ Apr 21, 2012 at 5:29

Let's take a look at the fundamental equation of thermodynamics: $$ dU = TdS - PdV + \sum_i \mu_i dN_i + \phi dQ + v dp + \dots, $$ where $U$ is internal energy, $T$ is temperature, $P$ is pressure, $V$ is volume, $\mu_i$ and $N_i$ are the chemical potential and number of molecules of various chemical species, $\phi$ and $Q$ are electric potential and charge, $v$ and $p$ are velocity and momentum, and the dots indicate that there are many other, more exotic pairs of variables that can be added to the end of this equation.

What we're interested in is a change in entropy, so let's rearrange the equation to reflect this: $$ dS = \frac{1}{T}dU + \frac{P}{T}dV - \sum_i \frac{\mu_i}{T} dN_i - \frac{\phi}{T} dQ - \frac{v}{T} dp - \dots $$ From this you can see that if you make a small change $dU$ in the energy, while keeping everything else constant, then the entropy will change by $\frac{1}{T}dU$. But also, if you make a small change in the volume while keeping everying else, including the energy, constant, the entropy will change by $\frac{P}{T}dV$. Similarly, you can change the entropy by changing the concentration of any chemical species, or the charge, or the momentum (i.e. by accelerating the system), or by making a change in any other extensive quantity.

The problem is that, in practice, it's not usually very easy to keep the energy constant while changing something else. It's easy enough to keep the temperature constant (you make the change isothermally, i.e. while keeping the system in contact with a heat bath), but that's not the same thing. It's also easy enough (in principle) to keep the entropy constant (you make the change adiabatically and do it very slowly). But generally, in most practical situations, if you try to change one of the other variables you'll also end up changing the energy a bit as well. For example, if you change the volume of a system you do work on it, and that changes the energy. But this is a mere practical issue - it's certainly possible in principle to change the entropy of a system without changing its energy.

Thermodynamics is often thought to be mostly about energy, but when you really get down to it, the role played by energy is no different from that played by any other conserved quantity. Out of all the extensive quantities, the only really special one is the entropy, since it isn't conserved. So for me, the above rearranged version of the fundamental equation is more fundamental than the "fundamental" one.

Another, somewhat unrelated point is that entropy only increases over time on average. For very small systems there are fluctuations, which mean that the entropy can temporarily decrease all by itself. It turns out that you can't use this phenomenon to do work, so the result that you can't build a perpetual motion machine isn't affected by this. To get a feel for fluctuations, consider Boltzmann's result that $$ S = k \log W, $$ where $k$ is Boltzmann's constant and $W$ is the number of possible microscopic states the system might be in, given the values for its volume, energy, chemical concentrations, etc. Einstein pointed out that you can invert this to $W = e^{S/k}$ and said (roughly) that the probability of a system fluctuating from a state with $W=W_1$ to $W=W_2$ should be $$ \frac{W_2}{W_1} = e^{\frac{S_2-S_1}{k}}. $$ If you plug some numbers into this you'll see that for systems of a macroscopic size will fluctuate by only tiny, unobservable amounts, whereas systems on the scale of molecules fluctuate quite a bit. If the system is isolated then these fluctuations will not change the amount of energy in the system, even though they sometimes temporarily reduce the entropy.

So there are two ways the entropy of a system can decrease without a change in energy: because of a change in another extensive quantity that happens to keep the energy constant; or, if it's a small isolated system, because of a thermal fluctuation.

  • $\begingroup$ What if a) fluid in box will be heated from one side, leading to observable convection b) a way of influencing the box will be found, to separate chemical species spatially into 2 groups. ? I now see that this is probably the case of non equilibrium thermodynamics? In both a) and b) work is done, i.e. energy is used on (or even added to) system to lower its entropy? $\endgroup$
    – Mooncer
    Apr 21, 2012 at 15:55
  • $\begingroup$ In case a), yes, the system's entropy is decreased because of an energy flow (but not that this is not an isolated system!). Case b) is interesting, because it's not necessarily about energy at all. If we're talking about a mixture of ideal gases, the energy of the mixed state is the same as the energy of the unmixed one; for other systems it's even possible for it to be lower. I go into some detail about it in my answer here: physics.stackexchange.com/questions/9411/energy-of-unmixing/… (continued in next comment) $\endgroup$
    – N. Virgo
    Apr 21, 2012 at 17:29
  • $\begingroup$ If you want to unmix some miscible fluids, the second law says you must have to increase the entropy of some other system by at least $\Delta S_\text{mixing}$. One way in which we often do this is by converting work into heat, and it can be easy to think this is the only way it can be done, but it isn't necessarily. (If I can think of a good counterexample I'll let you know.) $\endgroup$
    – N. Virgo
    Apr 21, 2012 at 17:32
  • $\begingroup$ Here's an example of unmixing without doing work. Take a mixture of two fluids with different freezing points, such as alcohol and water. Lower the temperature so that one of the liquids (water) freezes but the other doesn't. Remove the liquid and store it in a separate container. Return the temperature to the previous value. Of course, freezing and thawing do involve changes in energy, but only in the form of heat, not work. $\endgroup$
    – N. Virgo
    Apr 21, 2012 at 19:27
  • $\begingroup$ Yes my question is about non-isolated system $A$, i.e. about isolated system that gets open to heat flow for some time $[t_1,t_2]$. Its entropy is lower at $t_2$ than at $t_1$. I want to show that $A$ did receive energy. Now I see that this is the same problem as in non-equilibrium thermodynamics? For example, the reason why tornado is being created, is that it receives energy from the sun. Non-equilibrium systems in the sense of Prigodine work -- aren't they effect of differences in energy in the system... ?? $\endgroup$
    – Mooncer
    Apr 21, 2012 at 20:18

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