The condition of equilibrium in statistical mechanics is $\frac{\partial \rho}{\partial t}=0$ where $\rho$ is the phase space density. By virtue of Liouville's theorem, this is equivalent to the statement $\{\rho,H\}=0$. Mathematically, does this mean that $H$ must also be explicitly time-independent like $\rho$ is?
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asked Feb 14, 2020 at 17:05
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$\begingroup$ Also be careful with explicit vs. total time dependence... $\endgroup$– Qmechanic ♦Commented Feb 14, 2020 at 17:15
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$\begingroup$ @ValterMoretti Thanks. So it does not mean that H have to be time-independent. $\endgroup$– SolidificationCommented Feb 14, 2020 at 17:17
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$\begingroup$ Yes, see my explicit answer. $\endgroup$– Valter MorettiCommented Feb 14, 2020 at 17:24
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1 Answer
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If $H$ is time independent and satisfies $\{\rho, H\}=0$, also $H(t)=f(t)H+ g(t)$ does for every pairs of smooth maps $f,g$. Hence the answer is negative. $H$ can be time dependent.
answered Feb 14, 2020 at 17:22
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$\begingroup$ Does the derivation of Liouville's theorem hold for a general $H$ with explicit $t$- dependence? $\endgroup$ Commented Feb 14, 2020 at 17:28
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2$\begingroup$ Yes, in the proof of the Liouville theorem it is not required that $H$ does not explicitly depend on time. $\endgroup$ Commented Feb 14, 2020 at 17:31