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Currently we're deriving the Gibbs expression for the entropy during an introductory thermal physics course. We're dealing with a system with $N$ different equally likely microstates that are "partially" distinguishable. Indistinguishable microstates are in the same macrostate (whatever that means).

Now, apparently, both the entropy of the micro- as well as the macrostates contribute to the total entropy of the system, \begin{equation} S=S_\mathrm{micro}+S_\mathrm{macro}. \end{equation} I can squeeze out the right answer all right, but I don't understand what I'm doing here.

So could anyone elaborate how I am supposed to interpret this? I always thought of entropy as a way of expressing in how many different states a system can arrange itself. I understand how microstates can contribute to this, but I don't understand how to rationalize that it must equal a certain theoretical expression.

My exercise then proceeds to mention that I should take the statistical average of the entropy over each macrostate $i$, $\langle S_i\rangle$ to get the contribution of all microstates to the total entropy. What? This really seems out of the blue. I'd very much like to gain an understanding of this.

And how should I differentiate between the contribution of the macrostates? I'm starting to believe I don't understand the difference anymore...

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  • $\begingroup$ I think you are not clear on what a macrostate is. A macrostate is a set (ensemble) of microstates, all of which are constrained to have the same value of some macroscopic property: volume, temperature ... In @Steeven excellent answer, the macroscopic property is: the total number of heads is $N$. The only time you consider "the contribution of the macrostates" is in deciding which one has the highest entropy, not in the calculation of entropy itself. In a particular ensemble, all states are in the same macrostate. $\endgroup$ – garyp Nov 8 '14 at 12:51
  • $\begingroup$ @garyp How do I fit the statistical average of the entropy over each macrostate into his explanation? $\endgroup$ – user55789 Nov 8 '14 at 12:59
  • $\begingroup$ I see. I missed that on my first reading. I'm not sure from the context of your problem, but it might be this: If, for example, your system is immersed in a bath of constant temperature, then the energy of the system is not exactly known. There is a distribution of energies among all of the macrostates. To find the entropy one may calculate the entropy for each macrostate of energy $E_i$, and then average over all $i$. In this case, the average would be weighted by the Boltzmann factor. $\endgroup$ – garyp Nov 8 '14 at 13:40
  • $\begingroup$ @garyp "There is a distribution of energies among all the macrostates." - The system is in a certain macrostate. Do you mean here that the system can have a certain amount of energy depending on the macrostate? $\endgroup$ – user55789 Nov 8 '14 at 13:44
  • $\begingroup$ @garyp Rereading what you said: "You don't consider the contribution of macrostates" in the calculation of entropy itself". My exercise gave that expression of $S_\mathrm{total}=S_\mathrm{micro}+S_\mathrm{macro}$ and I have to admit, most of my confusion stems from this expression. I did not come up with this relation myself. $\endgroup$ – user55789 Nov 8 '14 at 14:01
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Flip four coins. Coins have heads $H$ and tails $T$. What are the possible results?

1: HHHH
2: HHHT, HHTH, HTHH, THHH
3: HHTT, HTHT, THHT, HTTH, THTH, TTHH
4: HTTT, THTT, TTHT, TTTH
5: TTTT

There are five different outcomes. That is, five different macro-states.

Macro-state 3 happens for most different micro-states. It is the most probable outcome.

If the micro-states were not all equally probable as here, probabilities have to be included in finding the most probable macro-state.

The macro-state with the highest entropy is no. 3. In that case we know the least about how the micro-state outcome looks. There we have more chaos than in the other cases.

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  • $\begingroup$ I'm still confused about two things. It is given that all probabilities are equally probable (in my case as well). However, for calculation of the entropy of the microstates I suddenly do have to use probabilities. Big contradiction there. $\endgroup$ – user55789 Nov 8 '14 at 13:24
  • $\begingroup$ And I'm still not sure how I distinguish the entropy contribution of macrostates versus microstates. What's the fundamental difference? $\endgroup$ – user55789 Nov 8 '14 at 13:24
  • $\begingroup$ In your question you say "equally likely" micro-states. So, yes, I must assume that the probability for each micro-state is the same. I don't see where you then do have to use probabilities? $\endgroup$ – Steeven Nov 8 '14 at 16:26
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    $\begingroup$ I am not sure about the question of entropy of macro-states. Macro-states are defined as you want to define it. In my example Macro-states are the number of head and tails. Another situation could be all cases of less than half heads. That would be one macro-state while the other would be half or more heads. Macro-states are just a way to group the micro-states. And usually in physics Macro-states are defined as the actual property we can measure or see (temperature e.g.) $\endgroup$ – Steeven Nov 8 '14 at 16:32
  • $\begingroup$ @Steeven, is there a convention to formally define the set for the macrostate property the microstates follow? $\endgroup$ – Vass Jun 24 '18 at 14:11

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