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Consider the Wikipedia article "https://en.wikipedia.org/wiki/Principle_of_minimum_energy". It clearly says

  1. For an isolated system with fixed energy, the entropy is maximized.
  2. For a closed system with fixed entropy, the energy is minimized.

The problem comes when they provide a mathematical explanation under the header "Mathematical explanation" (in the same article). It first states that $$\bigg(\dfrac{\partial S}{\partial X}\bigg)_U=0,~~\bigg(\dfrac{\partial^2 S}{\partial X^2}\bigg)_U<0,$$ at equilibrium for an isolated system with fixed internal energy. Then the article uses some algebra to connect these terms to $$-\dfrac{1}{T}\bigg(\dfrac{\partial U}{\partial X}\bigg)_S~~\&~~-\dfrac{1}{T}\bigg(\dfrac{\partial^2 U}{\partial X^2}\bigg)_S~~\text{respectively}.$$ This is then used to claim the minimization of energy.

My confusion is that the entropy maximization requires internal energy constant (point 1 above). Thus $$\dfrac{\partial U}{\partial X}=\dfrac{\partial U^2}{\partial X^2}=0.$$

The first equality is okay, but for the minimization of the internal energy, we require that the second one be positive and not $0$.

I was not aware of a connection, mathematical in nature, between the minimization of internal energy and the maximization of entropy, till I came across this Wikipedia document. This then leads to all of this confusion. I don't see where I am going wrong. Also, this is the proof given in the thermodynamics books by R.H. Swendsen. Any help is highly appreciated.

  • Related/possible duplicates: https://physics.stackexchange.com/q/47253/50583, https://physics.stackexchange.com/q/338613/50583 – ACuriousMind Mar 02 '20 at 17:21
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  • @Sam My doubt is related to the mathematics used. I don't have any problem with the Philosophical/Physical understanding of entropy maximization and energy minimization – Shoham Sen Mar 03 '20 at 19:53
  • @AcuriousMind My question is not related to the post that you flagged. My question is related to the mathematics used in getting an equivalence between Entropy maximization and energy minimization. – Shoham Sen Mar 03 '20 at 19:55
  • I think the figure in the first answer here https://physics.stackexchange.com/questions/534173/why-does-the-minimum-energy-principle-work should clarify th mathematical side of your question. – GiorgioP-DoomsdayClockIsAt-90 Mar 30 '20 at 17:13
  • @GiorgioP No, it doesn't. It's because the post says that the system is at constant entropy then it claims that the entropy is increasing. Doesn't this violate the constant entropy claim? It's the same problem that I have with the math above (as identified by the author). – Shoham Sen Mar 31 '20 at 01:31
  • I think it should. It shows that there is a surface in the $S,U,X$ space which can be seen either as the graph of a function of $U,X$ or of a differnt function of $S,X$. Depending on which function is represented by the same surface, ine is facing a problem of maximum or a problem of minimum. It also makes clear the relation with the quantity which is kept constant in the partial derivative. – GiorgioP-DoomsdayClockIsAt-90 Apr 01 '20 at 05:58
  • @GiorgioP, If you claim that your figure represents the system of interest then for different X values, we have different values of U. Thus the internal energy of the system is not constant. This means that its not an isolated system with fixed internal energy. This is in contrast to point 1 above (highlighted). Conversely, for different X values, the figure has varying S and is thus not representative of a closed system with constant entropy. This is in contrast to point 2 above (highlighted). – Shoham Sen Apr 01 '20 at 17:50
  • @ShohamSen you should look at the figure in the right way. You have to consider the line which gives $S$ as a function of $X$ at fixed $U$ and the line which represents $U$ at fixed $S$, alway as a function of $X$. The first is the graph of a function with a maximum while the second is the graph having a minimum at the same $X$. – GiorgioP-DoomsdayClockIsAt-90 Apr 02 '20 at 20:36
  • @GiorgioP, yes I think you may be right. Thanks. – Shoham Sen Apr 03 '20 at 16:59

1 Answers1

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I will attempt an answer to this, although I'm not 100% sure I understand the question, I see that there is some confusion with respect to the derivation of the minimum energy principle, and I think I know where the confusion comes from.

What the article shows is that if the function $S$ has an extrema at a point $(U_0, X_0)$, for which it takes the value $S_0 = S(U_0, X_0)$ then the function $U$ has an extrema at the point $(S_0, X_0)$, for which it takes the value $U_0$. This extrema is a maximum for the function $S$ but a minimum for the function $U$ it is a minimum.

I think your confusion is related with the fact that writing down $$\frac{\partial S}{\partial X} \bigg|_U$$ seems to imply somehow that now the other function, $U$ is a constant, and when you take its derivative it should vanish, but this is not the case, using this notation for the derivative only means that $S$ is being considered as a function of $X$ and $U$, and not other variables. A different way to write this would be:

$$\frac{\partial S_{U,X}}{\partial X}$$ where the subindices just indicate as a function of which variables you are considering your function to depend on. Using this notation then, the derivation reads something like this, for the first derivative:

$$\frac{\partial S_{U,X}}{\partial X} = -\frac{\partial S_{U,X}}{\partial U} \frac{\partial U_{S, X}}{\partial X} = - T \frac{\partial U_{S, X}}{\partial X} $$ Where I've used the cyclic chain rule. From this equality you can see that the gradients of both functions are related, so that if one has a critical point then the other one does too, and at no point I have considered $U$ to be just a constant.

Edit after comment:

It seems that what is unclear is that the derivative $$\frac{\partial S_{U, X}}{\partial U}$$ is not zero for a fixed value of energy. Maybe this way of thinking about it would help:

If you have a function $F(x,y) = yx + x^2$ for example, and you take the derivative with respect to x then the result is $$\frac{\partial F_{x,y}}{\partial x}(x, y) = y + 2x$$. Lets say now that we want to study this function for a fixed value of x, namely $x_0$, then our function will be $$F(x_0, y) = yx_0 + x^2_0$$ and our derivative will be $$\frac{\partial F_{x,y}}{\partial x}(x_0, y) = y + 2x_0$$ which is not necessarily equal to zero.

A similar thing happens when you want to study the function $S(U_0, X)$ and its derivative with respect to $U$; $$\frac{\partial S_{U,X}}{\partial U}(U_0, X)$$ is not necessarily equal to zero.

Ignacio
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  • Well, Entropy maximization is when you are dealing with a system at constant energy. Thus when you write down $\dfrac{\partial S(U, X)}{\partial U}$, this term should be 0, as S is not changing with U.\ PS- I have modified the question a bit. Please let me know if its still unclear, also what I can do to change it. – Shoham Sen Mar 03 '20 at 20:16
  • @Shoham Sen I added some explanation to try to clarify why that derivative is not zero – Ignacio Mar 03 '20 at 22:10
  • appreciate the example, though I don't really think that it's an accurate description of whats going on. I guess my doubt can be boiled down to the following. If I look at the math, it would claim that whenever you have Entropy maximized, you also have Internal energy minimized since they are related as such, isn't that what the math says? However, the entropy Maximization principle claims that Entropy is maximized at fixed internal energy. Thus energy cannot be minimized. PS- thanks for the help, really appreciate it. – Shoham Sen Mar 03 '20 at 23:48