Question 6.11.1: 1. Perform a second-order series expansion with respect to Δ...

  1. Perform a second-order series expansion with respect to ΔU and ΔV and show that the global condition for the concavity of entropy (6.9) as a function of internal energy and volume can be expressed as

S (U − ΔU, V − ΔV) + S (U+ΔU, V+ΔV) ≤ 2S (U, V)

\frac{\partial^2 S}{\partial U^2} ΔU^2+2\frac{\partial^2 S}{\partial U \partial V}ΔUΔV+\frac{\partial^2 S}{\partial V^2}ΔV^2≤0.

  1. Take into account the local concavity condition for the entropy (6.12) as a function of internal energy and deduce from the above result that,

\frac{\partial^2 S}{\partial U^2}≤0.      and      \frac{\partial^2 S}{\partial V^2}≤0.

\biggl(\frac{\partial^2 S}{\partial U^2}ΔU+\frac{\partial^2 S}{\partial U \partial V}ΔV\biggr)^2+\biggl(\frac{\partial^2 S}{\partial U^2}\frac{\partial^2 S}{\partial V^2}-\Bigl(\frac{\partial^2 S}{\partial U \partial V}\Bigr)^2\biggr)ΔV^2≥0.

  1. Deduce from this the local concavity condition for entropy (6.13) as a function of internal energy and volume.

\frac{\partial^2 S}{\partial U^2}\frac{\partial^2 S}{\partial V^2}-\Bigl(\frac{\partial^2 S}{\partial U \partial V}\Bigr)^2≥0.

  1. Henceforth, consider entropy S (U) as a function of internal energy U only. Then, the global condition of concavity for the entropy can be written as,

λ S (U_1) + (1 − λ) S (U_2) ≤ S (λU_1 + (1 − λ) U_2)     where      λ ∈ [ 0, 1 ]

Use the property that the inverse of a monotonously increasing function is also a monotonously increasing function to show graphically that the global concavity condition for entropy implies the global convexity condition for internal energy,

λU(S_1) + (1 − λ) U(S_2) ≥ U(λ S_1 + (1 − λ) S_2)      where       λ ∈ [ 0, 1 ]

  1. Show that the global conditions of entropy concavity and internal energy convexity can be written as,

S (U − ΔU) + S (U+ΔU) ≤ 2 S (U)

U(S − ΔS) + U(S+ΔS) ≥ 2U(S)

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  1. The global condition for the concavity of entropy (6.9) as a function of internal energy and volume is written,

S (U − ΔU, V − ΔV) + S (U+ΔU, V+ΔV) ≤ 2S (U, V)

The series expansion of entropy S (U+ΔU, V+ΔV) to second-order in ΔU and ΔV is given by,

S (U+ΔU, V+ΔV) \simeq S (U, V) +\frac{\partial S(U,V)}{\partial U}\Delta U+\frac{\partial S(U,V)}{\partial V}\Delta V+\frac{1}{2!}\bigl(\frac{\partial^2 S(U,V)}{\partial U^2}\Delta U^2+2\frac{\partial^2 S(U,V)}{\partial U \partial V}\Delta U\Delta V+\frac{\partial^2 S(U,V)}{\partial V^2}\Delta V^2 \bigr).

and the series expansion of entropy S (U − ΔU, V − ΔV) to second-order in ΔU and
ΔV by,

S (U-ΔU, V-ΔV) \simeq S (U, V) -\frac{\partial S(U,V)}{\partial U}\Delta U-\frac{\partial S(U,V)}{\partial V}\Delta V+\frac{1}{2!}\bigl(\frac{\partial^2 S(U,V)}{\partial U^2}\Delta U^2+2\frac{\partial^2 S(U,V)}{\partial U \partial V}\Delta U\Delta V+\frac{\partial^2 S(U,V)}{\partial V^2}\Delta V^2 \bigr).

Substituting these two series expansions into the global condition for the entropy concavity, the zeroth-order terms and the first-order terms cancel each other out and the inequality, to second-order, becomes,

\frac{\partial^2 S(U,V)}{\partial U^2}\Delta U^2+2\frac{\partial^2 S(U,V)}{\partial U \partial V}\Delta U\Delta V+\frac{\partial^2 S(U,V)}{\partial V^2}\Delta V^2≤0.

In order to simplify this expression, the dependence of the partial derivatives on the state variables U and V will no longer be indicated explicitly. Thus, the previous result is written,

\frac{\partial^2 S}{\partial U^2}\Delta U^2+2\frac{\partial^2 S}{\partial U \partial V}\Delta U\Delta V+\frac{\partial^2 S}{\partial V^2}\Delta V^2≤0.

  1. Multiplying this inequality by the local condition for entropy concavity (6.12) as a function of internal energy,

\frac{\partial^2 S}{\partial U^2}≤0.      and      \frac{\partial^2 S}{\partial V^2}≤0.

\frac{\partial^2 S}{\partial U^2}≤0.

yields the following inequality,

\bigl(\frac{\partial^2 S}{\partial U^2} \bigr)^2 ΔU^2+2\frac{\partial^2 S}{\partial U^2}\frac{\partial^2 S}{\partial U \partial V}ΔUΔV+\frac{\partial^2 S}{\partial U^2}\frac{\partial^2 S}{ \partial V^2}ΔV^2≥0.

It can be recast as,

\bigl(\frac{\partial^2 S}{\partial U^2}\Delta U+\frac{\partial^2 S}{\partial U\partial V}\Delta V \bigr)^2+\Bigl(\frac{\partial^2 S}{\partial U^2} \frac{\partial^2 S}{\partial V^2}-\bigl(\frac{\partial^2 S}{\partial U\partial V} \bigr)^2\Bigr)\Delta V^2\geq 0.

  1. The internal energy variation ΔU and volume variation ΔV can always be chosen such that the terms in the first brackets on the left-hand side vanish. Thus, in order for this inequality to be alway satisfied, the difference of terms in the brackets above has to be
    non-negative, i.e.

\frac{\partial^2 S}{\partial U^2}\frac{\partial^2 S}{\partial V^2}-\bigl(\frac{\partial^2 S}{\partial U\partial V} \bigr)^2≥ 0.

This corresponds to the local condition for entropy concavity (6.13) as a function of internal energy and volume.

\frac{\partial^2 S}{\partial U^2}\frac{\partial^2 S}{\partial V^2}-\Bigl(\frac{\partial^2 S}{\partial U \partial V}\Bigr)^2≥0.

  1. Since the inverse of the entropy function S (U) is monotonous and increasing, the inverses on both sides of the global condition for the entropy concavity (Fig. 6.13) have to satisfy the inequality,

S^{−1} (λ S_1 + (1 − λ) S_2) ≤ λU_1 + (1 − λ) U_2      and     λ ∈ [ 0, 1 ]

where S_1 = S (U_1), S_2 = S (U_2), U_1 = U(S_1).       and       U_2 = U(S_2). Moreover,
U(λ S_1 + (1 − λ) S_2) = S_1 (λ S_1 + (1 − λ) S_2).          where λ ∈ [ 0, 1 ]

These two equations imply the global condition for the internal energy convexity,

λU(S_1) + (1 − λ) U(S_2) ≥ U(λ S_1 + (1 − λ) S_2)     where       λ ∈ [ 0, 1 ]

  1. In the particular case where λ = 1/2, U_1 = U− ΔU, U_2 = U+ΔU, S_1 = S− ΔS and S_2 = S+ΔS, the global conditions for entropy concavity and internal energy convexity become,
\frac{1}{2}S (U − ΔU)+\frac{1}{2}S (U+ΔU) ≤ S (U)

\frac{1}{2}U(S − ΔS)+\frac{1}{2}U(S+ΔS) ≥ U(S).

and correspond to the inequalities stated in the problem, up to a factor of 2.

6.13

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