Is the Integral of the Square of Phase Density Constant in Classical Mechanics?

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SUMMARY

The integral of the square of phase density, \(\int \rho(x,t)^{2} dx\), remains constant over time in non-dissipative systems, as established by Liouville's equation. This conclusion is supported by the paper by Y. Gu (PLA, 149, 95, 1990), which asserts that both the position and momentum space integrals of phase density are conserved. The relationship \(\int \rho(x,t)^2 dx = \int \rho(p,t)^2 dp\) demonstrates that the total phase-space density is invariant, reinforcing the constancy of the integral of the square of phase density.

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  • Understanding of Liouville's equation in classical mechanics
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  • Basic calculus, particularly integration techniques
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In the classical mechanics each system can be described by the phase density \rho(x,t)
, which is evolved by Liouville's equation.
Recently I read a paper: http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TVM-46S5C37-24M&_user=6104324&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000069295&_version=1&_urlVersion=0&_userid=6104324&md5=d41e737ddc2514dbbeca4a99047e66f7"(Y, Gu, PLA, 149, 95 (1990)). In it, it says that

\int\rho(x,t)^{2} dx is constant in time.
But, as far as I know, due to the conservation of particles, \int\rho(x,t) dx
is constant in time.
I don’t know how he got the conclusion?
 
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Could you help me with this question?Thanks! The paper you cited is talking about the non-dissipative systems. For such systems, the total phase-space density is conserved. That means not only the integral of the density over the position space is conserved, but also the integral of the density over the momentum space is conserved. Both the integrals are related by the expression \(\int \rho(x,t)^2 dx = \int \rho(p,t)^2 dp\), which implies that the integral of the phase-space density is constant in time.
 

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