# Can entropy be regarded as energy dispersal?

In several answers here the claim has been made that thermodynamic entropy can be regarded as energy dispersion. See, in particular here, and here and here. This is apparently the pet theory of a chemistry professor, Frank Lambert. Apparently (at least according to his Wikipedia page) this definition has become popular in the chemistry community.

This seems like obvious nonsense to me, but I would like to try to keep an open mind in case I am missing something. Is this description actually consistent with the principles of thermodynamics and statistical mechanics?

• Why do feel this approach nonsense? Can you clarify that? – user36790 Nov 21 '15 at 18:22
• Hi @user36790- I was working on my response while you typed this- see below. – Rococo Nov 21 '15 at 18:28
• I agree that this is rubbish as a fundamental definition of entropy. Maybe it's useful to some people as a mental image of entropy, but it certainly has no general validity (except if you are ready to reinterpret in looser and looser ways the notions of energy and of dispersion, in order to suit each counterexample). Entropy is not a measure of disorder, nor is it a measure of energy dispersal. It has precise, general definitions both in Thermodynamics and in Statistical Physics. The latter may not be easy for students to grasp, but that does not mean that one should get rid of them... – Yvan Velenik Nov 22 '15 at 9:43

The minimal counterexample seems to me to be the following:

Take two materials, placed next to each other:

____________________ | | | | Material|Material | | 1 | 2 | ____________________

E1 _ _ _

E0 _ _

They have energy levels as indicated above- both have states at E0 and E1, but one has two excited states.

They start out isolated with the same average energy of $(E0+E1)/2$, distributed like:

E1 1/2 1/4 1/4

E0 1/2 1/2

Clearly, the energy distribution is perfectly uniform in this case. The entropy can be calculated by Gibb's formula, $S=-k_B\sum_i p_i \ln p_i$, and is $-k_B(3(1/2\ln1/2)+2(1/4\ln1/4))\approx 1.73 k_B$.

Now let these two materials exchange energy until they come to equilibium. The equilibrium state will be (by the fundamental posulate of statistical mechanics):

E1 1/3 1/3 1/3

E0 1/2 1/2

Calculating the entropy again, we see it has increased to $k_B(\ln 2 + \ln 3)\approx 1.79 k_B$. And the average energy is, plainly, no longer $(E0+E1)/2$ on each side. Material 1 has less energy and material 2 has more.

So what happened here is that we started with a perfectly uniform energy distribution, and found that the entropy increased as the energy distribution become more irregular, the opposite of what this formulation would claim. Moreover, this is a general feature of any two systems with different densities of states. By modifying the densities of states appropriately, one could make the energy distribution at equilibrium whatever is desired.

This example was in the microcanonical ensemble for simplicity, but this argument generalizes to an open system. In that case, for example, you could have two objects with identical initial energy go into a heat bath, and one would gain energy as it equilibrates while the other loses energy.

Entropy increase does correspond to energy dispersal when considering only one type of energy structure (i.e., if we started with two pieces of material 1 with different distributions, and allowed them to exchange energy). But this seems like a very limited example and one that is not suitable for making any general arguments about what entropy "is."

Edit: By request I will summarize the moral of the argument. For a system without quantum correlations, the usual definition of entropy is $S=-k_B\sum_i p_i \ln p_i$, where $i$ are the microstates of the system and $p_i$ is the probability that this microstate is occupied. This, for our purposes, is what entropy "is". This does measure a dispersal in a sense, but it is not dispersal of anything in space. Rather it is dispersal of the probability that the system can be found in a given configuration. Entropy is maximized when there is a low probability of many different microstates, rather than a high probability of being in just a few.

When your microstates correspond to energy levels in single particles, and each particle has the same structure of energy levels, then dispersal of probability in microstates does correspond to dispersal of energy. Maybe this is what Lambert has in mind. But in any other case, such as when you have two different types of objects with different microstates, this is no longer true. So, at least in my opinion, the idea of entropy as energy dispersion is much too limited and more likely to cause confusion than anything else when trying to understand physics in a fundamental way.

• Could you explain your example in a more easy way? It is beyond my intellect now:( – user36790 Nov 21 '15 at 18:37
• The main problem seems to be that Lambert is explaining what is meant by entropy rather than using "energy dispersal" as a foundational notion of entropy. (This comes from a cursory reading of his website.) I'm willing to bet that he would be fine with your example, because probably he would say that the dispersal happens over microstates rather than over subsystems (although who knows? he does use the language of "dispersed in space", so perhaps he really means "space"). (By the way, in your example, you're assuming high temperatures so that states 1 and 0 are equally populated?) – march Nov 21 '15 at 22:08
• Yeah my claim is not necessarily that he is wrong (I haven't bothered looking at his writings), but that people here seem to have gotten some wrong ideas from him regardless of what he knows or doesn't know. – Rococo Nov 22 '15 at 18:19
• Hi @user36790- I certainly agree with the content of that quote. But as Lambert says there, and as I say above, it is probability that is in a sense being dispersed, not energy. Describing energy as being spread among the microstates is nonsensical. So it seems to me that either he is repeating the same ideas everyone else has, and claiming they are novel, or he is saying something misleading or just wrong. Anyway, as long as we are all agreed on the physically correct statement that's all that I really care about. – Rococo Nov 23 '15 at 3:47
• It's probability that is dispersed; you've pointed it right & not dispersal of energy. BTW, I don't think this deserves downvote. +1 for a neutral discussion:) – user36790 Nov 23 '15 at 10:32