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Entropy maximization and energy minimization are equivalent statements of the same thing, as I understand it. If the internal energy is fixed, entropy is maximized because of statiatical reasons. If the entropy is fixed, and the system can exchange energy with its environment, then the system will give energy to its environment to maximize the environments entropy (and hence total entropy).

But I haven't been able to understand why we would assume here that the entropy of a system is fixed. What physical mechanism causes this?

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  • $\begingroup$ What is the basis of your understanding in the first sentence? Can you provide a reference? $\endgroup$
    – Bob D
    Aug 26 '19 at 5:21
  • $\begingroup$ @BobD, I've added a link $\endgroup$
    – user56834
    Aug 26 '19 at 5:41
  • $\begingroup$ $dU = TdS-PdV$, so even if $U$ is constant and work is done on the system, $S$ is not constant. Why is it necessary that $S$ be constant if $U$ is constant? $\endgroup$ Aug 26 '19 at 6:21
  • $\begingroup$ I would not interpret this equation like so. It's better to re-write like this $dS=1/T(dU+pdV)$ then you will see that change in entropy can happen when system experiences change in internal energy (cooling,heating,etc) or when change in volume happens (squeezing under pressure, etc) $\endgroup$ Aug 26 '19 at 9:08
  • $\begingroup$ But once equilibrium is achieved, $dS$ is zero again. So what is the issue? Am I misunderstanding something? $\endgroup$ Aug 26 '19 at 9:56
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I think you are misunderstanding a few things. Both the entropy maximization principle and energy minimization principle are due to statistical reasons. Suppose we have some parameter of the system $X$ (e.g. volume, particle number, etc.) that is free to vary, and the energy is fixed at $U_0$. The system will evolve to a $X_0$ such that the entropy is maximized at the value $S_0$.

Now the energy minimization principle refers to this same point $X_0, S_0, U_0$. If we change $X$ away from $X_0$ while keeping $U_0$ fixed, $S$ will decrease. So if we want to bump $S$ back up to $S_0$ we need to add energy. So any point $X$ which has the same entropy $S_0$ must have at least as much energy as $U_0$. This is all the energy minimization principle is.

This is just an alternate way of characterizing the same point $X_0, S_0, U_0$. Again, if we move away from it while keeping $S_0$ the same, we need to add energy. It is a choice we make, not some physical mechanism.

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  • $\begingroup$ My understanding was: in an isolated system, energy is fixed, and the system will evolve keeping energy fixed but entropy increases. But we can have a different situation where energy can move in and out of the environment. My understanding was that in this case, literally (free) energy decreases over time, just as entropy increases over time. $\endgroup$
    – user56834
    Aug 26 '19 at 11:44
  • $\begingroup$ So I guess my question is, if there isnt a physical situation where entropy stays constant over time but energy decreases (thereby increasing total entropy by increasing entropy outside the system), then why would we ever encounter a situation where we need to know what the energy of a system would be after changing $X$, while keeping $S$ constant? $\endgroup$
    – user56834
    Aug 26 '19 at 11:51
  • $\begingroup$ @user56834, Look at the example of a marble rolling in a bowl with friction that is found in the Wikipedia article you linked to. In that case $X$ could be the amplitude of marble oscillations, and the equilibrium point $X_0$ where the marble is at the bottom is equally well described by max entropy (all mechanical energy converted to thermal energy) or min energy for a given value of entropy (which is entirely determined by the thermal energy). $\endgroup$
    – octonion
    Aug 26 '19 at 12:12
  • $\begingroup$ @user56834, if you want to you could put the marble-bowl system in a thermal bath so that the entropy stays constant as it goes to equilibrium, but as I was trying to explain, that isn't necessary. The min energy principle is true for the equilibrium state no matter how you get there $\endgroup$
    – octonion
    Aug 26 '19 at 12:17
  • $\begingroup$ Thanks this is clarifying. I think this is the part that I'm confused about: "if you want to you could put the marble-bowl system in a thermal bath so that the entropy stays constant as it goes to equilibrium". Why would the entropy stay constant if you put something in a thermal bath as it goes to equilibrium? $\endgroup$
    – user56834
    Aug 26 '19 at 12:25
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The energy maximisation or equivalently entropy maximisation is a property of an equilibrium state. So in the case of when there is energy exchange with the surrounding, clearly the system isn’t in equilibrium. Once all the exchange is complete, the energy/entropy is maximised.

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