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In section 11 of Landau's statistical physics part 1, he wrote the following:

Let us suppose that a body is thermally isolated, and is subject to external conditions which vary sufficiently slowly. Such a process is said to be adiabatic. We shall show that, in an adiabatic process, the entropy of the body remains unchanged, i.e. the process is reversible.

We shall describe the external conditions by certain parameters which are given functions of time. For example, suppose that there is only one such parameter, which we denote by λ. The time derivative dS/dt go the entropy will depend in some manner on the rate of variation dλ/dt of the parameter λ. Since dλ/dt is small, we can expand dS/dt in powers of dλ/dt.

My question is that can dS/dt depends on high order time derivative of λ?

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Kevin Kwok
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  • Landau is presumably discussing equilibrium thermodynamics. In that case introduction of time is not necessary. Then $S(\lambda+d\lambda)=S(\lambda)+d\lambda ~S'(\lambda)+(d\lambda)^2S''(\lambda)/2+...$ and $dS=S(\lambda+d\lambda)-S(\lambda)$. – Deep May 10 '18 at 10:04
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    @Deep Actually Landau later argued that the zeroth and first order terms of the power series must vanish such that $\frac{d S}{d t}=A \Big( \frac{d\lambda}{d t} \Big)^2$ and thus $\frac{d S}{d \lambda}=A \frac{d\lambda}{d t}$. How can I go from your way to his conclusion? – Kevin Kwok May 12 '18 at 15:52
  • Thermodynamics is equipped with a formalism to calculate the rate of entropy changes due to irreversible processes. Hence, thermodynamics describes macroscopic systems where changes are slow enough that we can consider a limited number of collective degrees of freedom, not necessarily so slowly that everything is always at equilibrium https://physics.stackexchange.com/a/780147/226902. When the processes is so slow that we go trough a sequence of equilibrium states, the entropy does not change (and this is what L&L call "adiabatic") https://physics.stackexchange.com/a/524050/226902 – Quillo Nov 07 '23 at 08:23

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In Landau's adiabatic processes, $\mathrm{d}\lambda/\mathrm{d}t$ is a small quantity. As a good approximation, higher orders of this quantity are omitted.

Yufei
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