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From Clausius' inequality,
$\oint\frac{dQ}{T}\leq 0$
From this, we can show that $\frac{dQ}{T}\leq dS$

For an isolated system having adiabatic walls, $dQ=0$
So, $dS\geq 0\tag{1}$
So, a isolated system when move towards equilibrium state, its entropy increases (spontaneous process maximizes the entropy).

In Callen's Themodynamics and an Introduction to Thermostatics, the maximum entropy principle is given as

The equilibrium value of any unconstrained internal parameter is such as to maximize the entropy for the given value of the total internal energy.

Mathematically, for an isolated system
if $S(U,x)$, where $x$ is an extensive independent coordinate $\frac{\partial S}{\partial x}\Bigg\rvert_U=0$ and $\frac{\partial^2 S}{\partial x^2}\Bigg\rvert_U<0$

I have the following doubt-

We know that the Clausius' inequality and Entropy maximization principle both are the statements of Second Law of Thermodynamics. I am not able to prove Entropy maximization principle from the Clausius' inequality.
Like (1) is the consequence of Clausius' inequality, but it suggests that entropy in spontaneous process of isolated system increases (maximizes). But this shows that
$\frac{\partial S}{\partial x}\Bigg\rvert_U=0$ and $\frac{\partial^2 S}{\partial x^2}\Bigg\rvert_U<0$ or $\frac{\partial S}{\partial U}\Bigg\rvert_x=0$ and $\frac{\partial^2 S}{\partial U^2}\Bigg\rvert_x<0$ or both.
But entropy maximization principle tells that $\frac{\partial S}{\partial x}\Bigg\rvert_U=0$ and $\frac{\partial^2 S}{\partial x^2}\Bigg\rvert_U<0$ (there is a coordinate x for which system attains maximum entropy at a particular internal energy) holds for sure. Like why instead of maximum entropy at a praticular internal energy, it is not the case that system attains maximum entropy at a particular coordinate for sure?

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  • $\begingroup$ The argument is not correct. We imply the maximum entropy principle in microscopic configurational space with certain constrain, e.g $\langle E\rangle = U$, and $V$, $N$ are kept constants. The entropy will certainly grow with volume. There is no $\frac{\partial S}{\partial V} = 0$. ...... and also $\frac{\partial S}{\partial U} = \frac{1}{T}$ is not zero. $\endgroup$ – ytlu Apr 16 at 17:59
  • $\begingroup$ I recall that the argument is something like: $S(E, V, N)$ if we keep $V$ and $N$ constant, and under condition that $\langle E \rangle = U$. The maximum entropy (with Lagrangian multiplier) leads to the Maxwell relation $\frac{\partial S}{\partial E}\vert_{U} = \frac{1}{T}$. And the expansion of $S$ around the equilibrium energy $U$ renders the Boltzmann probability distribution. $\endgroup$ – ytlu Apr 16 at 18:13
  • $\begingroup$ But the concept of entropy maximization is introduced in Classical thermodynamic in Callen and other books also. I have not studied Statistical Mechanics till. I am not able to to prove how Clausius' inequality implies Entropy maximization principle. $\endgroup$ – Iti Apr 16 at 18:37
  • $\begingroup$ In thermal dynamics, the entropy is defined $dS = dQ / T$. And the second law, $\Delta S \ge 0$, which doesn't imply $\frac{\partial S}{\partial U} \vert _{V}= 0$. $\endgroup$ – ytlu Apr 16 at 18:43
  • $\begingroup$ For example of the ideal gas: $S(U, V, N) = NK \ln V + \frac{3}{2} NK \ln T$, that is $ NK \ln V + \frac{3}{2} NK \ln \left( \frac{2U}{3NK_b}\right)$. You may use to check you derivatives. $\endgroup$ – ytlu Apr 16 at 18:48
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The entropy is an extensive quantity $$ S = S(E, V, N) $$ and as well all its variables are extensive. That means they will grows linearly as the system grows big: $S \uparrow$ as $E \uparrow$ or $V \uparrow$ or $N \uparrow$. Therefore \begin{align} \frac{\partial S}{\partial E} &\ne 0;\\ \frac{\partial S}{\partial V} &\ne 0;\\ \frac{\partial S}{\partial N} &\ne 0;\\ \end{align}

But, $S$ entropy has to be maximized under the given constrains $E=U$. You then maximum the entropy using Lagrangian un-determine multiplier:

$$ \frac{\partial}{\partial E} \left\{ S - \lambda_E \left(E-U\right) \right\}=0 $$ which then gives $$ \frac{\partial S}{\partial E}\Big\vert_{E=U} = \lambda_E. $$ Compare with Maxwell relation, we conclude that $\lambda_E = \frac{1}{T}$.

Similarly, we maximize $S$ entropy under the given constrains $V=V_0$: $$ \frac{\partial}{\partial V} \left\{ S - \lambda_v \left(V-V_0\right) \right\}=0 $$ which then gives $$ \frac{\partial S}{\partial V}\Big\vert_{V=V_0} = \lambda_v \to \frac{P}{T}. $$


Let me address more about the 3 principles.
  1. What is the meaning of $\frac{\partial S}{\partial E}\Big\vert_{E=U} = \frac{1}{T}$. It is $\Delta F = 0$.

Lets rewrite this equation as : \begin{align} T \Delta S =& \Delta U;\\ \Delta U - T \Delta S =& 0;\\ \Delta F =& 0. \end{align}

For a constant temperature, the equilibrium of an isolated system is determined by the minimum of Helmholtz free energy $F = U - TS$. It quantify the well known two counter balance factors: minimum energy and maximum randomness.

  1. Maximum entropy (maximum configurations)

In thermodynamics, the equilibrim of a state is not determined by the maximum of entropy. Then when to apply the maximum entropy principle? The maximum entropy is used in statistical mechanics to determine the distribution function. For microcanonical ensemble. The maximum entropy (maximum configurational number) is the equal probability, every micorstates has equal accessing probability. And for canonical ensemble, the maximum entropy leads to the Boltzmann distribution $p(E) \propto e^{-\beta E}$, and thus the minimum of free energy $F = -KT \ln Z$.

  1. About the second law $\Delta S \ge 0$.

This relation is referring to the entropy change of the system or/and the surrounding during a thermal process. A thermal process always involve something exchanged with reservoirs. This law cannot apply to an isolated state. This is mentioned by Bod D. The idea that the thermal processes intent to grow larger the universal total entropy. The "maximization" of the universal entropy is nothing to do with the equilibrium rule of a thermal state, and not related to the statistical maximum entropy rule.

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  • $\begingroup$ what is the difference between E and U? Is E the total energy (internal energy(U)+energy due to motion of system)? $\endgroup$ – Iti Apr 17 at 4:38
  • $\begingroup$ No. $E$ is a adjustable virtual energy in the variational process, U is the thermal internal energy that you are currently interested in. All thermal relations is applied to U, under thermal equilibrium. $\endgroup$ – ytlu Apr 17 at 4:42
  • $\begingroup$ Just like when you are doing calculus $\frac{df}{dx}\big\vert_{x_0} = \frac{f(x)-f(x_0)}{x-x_0}$. The $E$ is similar to $x$ and $U$, $x_0$. $\endgroup$ – ytlu Apr 17 at 4:45
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    $\begingroup$ So, $U$ is the internal energy of the isolated system at the equilibrium. So, if the isolated system is not in equilibrium state then it undergoes spontaneous process to reach equilibrium state with internal energy E tends to U, so the constraint is $E=U$. Is it that what you mean by the constraint $E-U=0$? $\endgroup$ – Iti Apr 17 at 4:50
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    $\begingroup$ I have read your answer several times. The Lagrangian multiplier method you use is clear to me. In Callen, they show that if we pass a $U=constant$ plane in $S-U-V$ graph, then there is a volume for which entropy takes a maximum value (entropy maximization principle) for an isolated system. I have a doubt that i) how it follows from Clausius inequality? As both are the statements of second law so there should be equivalence between them? ii) In the answer l, for maximization of entropy $\frac{\partial S}{\partial E}=\lambda_E=1/T=0$, what does this imply? Please help I am very confused. $\endgroup$ – Iti Apr 17 at 5:10
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We know that the Clausius' inequality and Entropy maximization principle both are the statements of Second Law of Thermodynamics. I am not able to prove Entropy maximization principle from the Clausius' inequality.

The Clausius equality

$$\oint\frac{dQ}{T}\leq 0$$

applies to any real heat engine cycle, where $Q$ is the heat entering the system at any point during the cycle and $T$ is the temperature at the point of heat entry. Since heat enters the system in the Clausius's inequality, it does not apply to an isolated or adiabatic system. As such I'm not sure you can use the Clausius inequality to imply or prove the entropy maximization principle, which pertains to an isolated system.

On the other hand, it can be shown that the Clausius inequality leads to the increase in entropy principle of the second law, or

$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{sur}>0$$

The Clausius inequality means that for a real (irreversible) heat engine the entropy transferred to the surroundings by the system in the form of heat is larger than the entropy transferred to the engine from the hot reservoir in the form of heat, the difference being entropy generated in the system.

And since, for any cycle (reversible or not), we always have

$$\Delta S_{sys}=0$$

Then, for an irreversible cycle,

$$\Delta S_{sur}>0$$

Hope this helps.

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  • $\begingroup$ Yes, Clausius' inequality applies to any real heat engine. From this we can prove that $\frac{dq}{T}\leq dS$ and if the system is isolated or having adiabatic boundaries, then $dQ=0$, so then for any spontaneous process, $dS\geq 0$. This suggests that entropy maximizes to reach the equilibrium state if the system is isolated. But the doubt is that how the above idea translates to entropy maximization principle (The equilibrium value of any unconstrained internal parameter is such as to maximize the entropy for the given value of the total internal energy). $\endgroup$ – Iti Apr 17 at 4:15
  • $\begingroup$ As S maximizes and if it is a function of $U$ and $x$(extensive independent coordinate), then how $\frac{\partial S}{\partial x}\Bigg\rvert_U=0$ holds for sure, not $\frac{\partial S}{\partial U}\Bigg\rvert_x=0$? $\endgroup$ – Iti Apr 17 at 4:16
  • $\begingroup$ "But the doubt is that how the above idea translates to entropy maximization principle". My point is why should it? I have doubts that the entropy maximization principle (introduced in 1957) is a statement of the second law and on that basis the premise of equivalency may be faulty. But I confess to not being conversant in statistical thermodynamics, which the principle is based on. $\endgroup$ – Bob D Apr 17 at 14:18

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