# Best example of energy-entropy competition? [closed]

What are the best examples in practical life of an energy-entropy competition which favors entropy over energy? My initial thought is a clogged drain -- too unlikely for the hair/spaghetti to align itself along the pipe -- but this is probably far from an optimal example. Curious to see what you got. Thanks.

• I'd say all living organisms seem to enter this cathegory. But it's hard to prove quantitatively. But obviously, when we eat, we don't do so to increase our energy (except when growing), we eat to diminish or keep our entropy low. – Raskolnikov Nov 27 '10 at 20:13
• Read as comment to Piotr's answer (not enough points to comment). I believe that "Brazil nut simulations" have been done with "Brazil nuts" that are denser (i.e. lead), as well as larger, than the other nuts, and they still rise to the top. I think they do that in this article: prl.aps.org/abstract/PRL/v58/i10/p1038_1 Packing is of course important, since it keeps the lead nuts from falling to the bottom! – Greg P Nov 27 '10 at 21:36
• Personally, I eat because I am either hungry or else I have sweet tooth. I don't recall ever eating in order to diminish my entropy :-) – Marek Nov 28 '10 at 6:42
• @Raskolnikov: I've heard several mentions of life and low-entropy, but I've yet to see one that goes any deeper than "a living being is an ordered system". I'd be curious to know what motivates this idea. – Malabarba Dec 25 '10 at 0:42
• @Bruce, eg measurement of entropy of some denaturation reactions. example: hardboiling an egg. – Georg Jan 18 '11 at 20:58

One example I know is so-called Brazil nut effect. When you place balls in a container of two different sizes and shake it, the larger ones will go up (even if they are denser than the smaller ones). So the final energy of system after introducing noise is clearly greater than the initial. I believe that the phenomena needs to be entropy-driven. However, I don't know a proof.

• This is not very correct. Put lead brazil nuts in, instead of normal ones. They will fall to the bottom. The effect is more related to packing than entropy (small shapes pack better than large ones). – Sklivvz Nov 27 '10 at 21:22
• @Sklivvz: Of course for a given ball size ratio there is a density threshold. No, it is not that related to packing - even average density (with air included) is larger at the top. – Piotr Migdal Nov 27 '10 at 21:36
• Read as comment to Piotr's answer (not enough points to comment). I believe that "Brazil nut simulations" have been done with "Brazil nuts" that are denser (i.e. lead), as well as larger, than the other nuts, and they still rise to the top. I think they do that in this article: prl.aps.org/abstract/PRL/v58/i10/p1038_1 Packing is of course important, since it keeps the lead nuts from falling to the bottom! – Greg P Nov 27 '10 at 21:36
• General question/comment (again, not enough points). What exactly (quantitatively) is meant when we say that "energy-entropy competition favors entropy"? In the appropriate ensemble, entropy and energy are always competing. In equilibrium a situation is reached that balances them exactly. Who is to say who "won"? If instead we are talking about a non-equilibrium situation, then what you are asking is for a process where entropy increases. Then there are plenty of examples - see for example Sklivvz's answer. Similarly, take any example of heat conduction from a hot object to a cold one, etc. – Greg P Nov 27 '10 at 21:45
• Thanks. The comment on denser nuts is crucial here, in order for the big nuts at top to be energetically less favorable. Maybe it should be called the "wing nut effect." (Ising model is surely not from everyday life.) By the way, the clogged drain rather similar -- the probable configuration of particles makes the lower-energy packings less favorable. – Eric Zaslow Nov 28 '10 at 14:04

The air in this room. Because of gravity, the lowest energy configuration is clearly one in which all the molecules lie on the floor. But from the entropic point of view they should be exploring all of their phase space and bouncing around the room.

I would like to say also that examples such as the clogged drain and so on should be viewed purely as analogies. Entropy is a real physical quantity that can be calculated. As an actual physics problem, the calculation of the entropy of the drain + hair + water system would be a monster and there is no telling what the result will be! Is it even in equilibrium?

The Brazil nut effect cited above is a non-equilibrium effect, so although it is interesting I don't think it necessarily means anything about energy-entropy competition. The balance of energy and entropy happens when a system (at constant temperature for example) reaches equilibrium, minimizing its free energy. The shaken Brazil nuts are a non-equilibrium problem, and thus are not minimizing their free energy! But very interesting nevertheless, all the more so since it is non-equilibrium.

• Every time I edit to say "the Brazil nut effect cited below" Piotr's post ends up above mine, and when I changed it back to "cited above" it appeared below :) – Greg P Nov 27 '10 at 20:39
• +1 for really simple & important example. When in comes to brazil nut effect, I don't get why it can't be an equilibrium? – Piotr Migdal Nov 27 '10 at 20:41
• @Greg: lol :) It depends on how people are viewing the answers: they can be ordered by votes (which is how most people view them), or by age, either direction. Right now your answer has a higher vote count than Piotr's so it's appearing above his. I find that it's better to say something like "...cited in Piotr's answer" or "in the other answers" because you can't count on "above" or "below" staying the same. – David Z Nov 27 '10 at 20:53
• As you shake the nuts, you add energy to the system that is dissipated in the nuts (thermalized). If we start in the equilibrium state (which by density arguments has the larger nuts on the bottom), shake the nuts to bring the large ones to the top, then let them cool to their original temperature, we have a non-equilibrium state. It would take years for the Brazil nuts to find their way spontaneously back to the bottom where they 'want' to be! Accordingly, the simulations of 'Brazil nut effect' are not equilibrium Monte Carlo sims but rather include steady input of energy (non-equilibrium). – Greg P Nov 27 '10 at 20:58
• @Greg: Take bouncing balls that don't dissipate energy - then large balls should be up while the system is in equilibrium. Or, more realistically, constantly shake your nuts. In case of stopping shaking you are right - it is some kind of metastable state (or rather - a local minimum). – Piotr Migdal Nov 27 '10 at 21:05

Blackbody radiation: anything hotter than its environment radiates energy thus increasing the entropy of the universe. Entropy wins :-)

The Sun :-) The Sun's energy does not increase the Earth's total energy! In fact, the Earth radiates almost exactly the same amount of energy as it receives from the Sun. What we really gain from the Sun is that we use the sun rays' low entropy to power life on Earth, and the Earth radiates high entropy microwaves in the night.

• @Sklivvz: Even though it is true I have no idea what it have to do with energy-entropy competition... – Piotr Migdal Nov 28 '10 at 10:51
• @Piotr Migdal, the first Google result for "energy entropy competition" is this article which talks about planetary atmospheres: mdpi.com/1099-4300/10/4/462 – Sklivvz Nov 28 '10 at 10:56
• @Piotr Migdal: The first example is an example where nature favors entropy over energy: energy decreases, entropy increases. The second is an example where nature favors energy over entropy: energy is constant, entropy decreases. – Sklivvz Nov 28 '10 at 10:59
• I don't think blackbody radiation increases entropy. Think of a gas inside a container of ideal mirrors. Let's say I drop a lot o photons inside this (isolated) system. From the point I put them it, up until the point when the photons get absorbed by the gas, the system was isolated. If blackbody radiation increased entropy, than absorption of photons would have to decrease entropy. If that were the case, than the system I mentioned above will have decreased its own entropy spontaneously. – Malabarba Dec 25 '10 at 0:39
• I don't understand your comment... A system radiating is increasing its entropy – Sklivvz Dec 25 '10 at 16:49

One of the nicest examples I know is the Kosterlitz-Thouless phase transition in the XY model. What is cool is that the transition is driven by the condensation of vortices which have an energy that diverges logarithmically with the size of the system. You would think they couldn't contribute at all because of this, but it turns out their entropy also diverges in the same way, so the free energy $F=E-TS \propto (c-k_BT) log(R)$ where $c$ is a parameter and $R$ is the size of the system. At sufficiently large $T$ the entropy terms wins and the system undergoes a transition through formation of vortices.

p.s. after rereading the question I realize my answer does not involve "practical life" but I'll leave it anyway.

Well, as a physicist working mainly in the field of statistical physics, I obviously have to mention Ising model. In this case the model is actually tractable (at least in two dimensions) and can tell us a whole lot about (not just) energy-entropy battle. It's obvious that the ground state (for ferromagnetic case) is all spins pointing one way (say up). Now, if you point some spins the other way then you are losing in terms of energy (something like number of neighbors times number of wrong spins) but actually you gain hugely in entropy (because of translational invariance of lattice models). This argument can be made very precise when working with polymer model (which is isomorphic to Ising model) and considering low-temperature cluster expansion.

I am sorry, but I am not really able to provide nice references. Wikipedia articles are pretty bad. Maybe I should spend some time bringing them up to the current knowledge (that is to say, knowledge since like 1970s) about cluster expansion. For now, if anyone is interested, just read the basic paper on the topic.

If you are willing to go down to microscopic scales, a nice example of "entropy winning" is the phenomenon of depletion forces. Large particles in a suspension of smaller ones feel an effective attractive force, even if the interaction between all particles is just hard-wall. The attractive force arises because the volume available to the smaller particles increases when the larger ones get sufficiently close, and hence their entropy increases. See e.g.

Sho Asakura and Fumio Oosawa,
"Interaction between particles suspended in solutions of macromolecules",
J. Pol. Sci. 33 (1958) 183-192.

Depletion forces can be measured directly and are quite important for biological systems.

In my view it's not really correct to talk about an "energy-entropy competition". I guess the reason people tend to think in those terms is because physicists like to speak of systems at a constant temperature minimising the Helmholtz free energy $A=U - TS$. This looks like it's an energy term ($U$) added to a term involving the entropy $S$, with the temperature acting as a kind of conversion factor. So it's easy to think that the system must be trying to minimise its energy and maximise its entropy at the same time, putting both factors in competition with each other.

But really that's not what's happening at all. What's happening is that the system we're interested in is exchanging energy with a heat bath in order to remain at a constant temperature, and the entropy of the combined system (system+heat bath) is increasing towards a maximum. If an amount $\Delta U$ of energy is transferred from the heat bath to the system of interest, causing a change $\Delta S$ in the entropy of the system of interest, then the total entropy has changed by $\Delta S - \Delta U/T$.

Now if we multiply this by $-T$ we get $\Delta U - T\Delta S$, which is equal to $\Delta A$, the change in Helmholtz free energy. Since the total entropy must be maximised, and since $T$ is assumed constant, $A$ must therefore be minimised. But now we can see that not only is the $TS$ term an entropy, but so is the $U$ one. The change in $U$ just represents the change in the entropy of the heat bath - it's just that multiplying it by $T$ has obscured that. The competition is not between the system's energy and its entropy, but between the system's entropy and the heat bath's entropy. The reason for multiplying by $-T$ is purely historical and has always struck me as rather unhelpful. Apart from anything else it's only permissible if $T$ is constant, whereas the $\Delta S - \Delta U/T$ formula works even if it isn't.

Having said all that I can now try to answer the question. What we need is an example of a system where the $TS$ term dominates over the $U$ one. One way to do this is just to prevent any heat from being transferred between the system and the heat bath. For any system that's completely isolated from its environment, the $\Delta U/T$ term will go to zero and only the $\Delta S$ term will be left. So any chemical reaction or other physical process performed in a thermally insulated container will be entirely dominated by the entropy term.

But perhaps a more satisfying example would be an endothermic chemical reaction such as cooking an egg. In this case the $\Delta U$ term is negative, so the system actually sucks in heat from its environment, reducing its entropy. This is offset by a greater increase in the entropy of the system - so we can say the entropy of the system has "won" over the entropy of the heat bath. Or in free energy terms, the $TS$ term has "won" over the $U$ one.