Timeline for Particle density in phase space normalization under proliferation

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Dec 5, 2017 at 21:02	comment	added	Jack_Stiller10		Ok thanks. I want to use $u(x,t)$ as a probability density. Following your comment you mean I should normalize the following way? $\tilde{u} = \frac{1}{N \int u(x,t) dt} $ to achieve $ \int \tilde{u}(t,x)dx =1$ for all $t$ and thus having a probability density.
Dec 5, 2017 at 14:44	comment	added	Carlo Beenakker		if you want the "average per particle" you just divide by $\int u(x,t)dt$; if you want, say the total energy you don't divide; it all depends on what you want, the problem is perfectly well defined.
Dec 5, 2017 at 14:32	comment	added	Jack_Stiller10		Thanks for your comment. I totally agree with you. But how can I then use $u(x,t)$ for calculating the expectation value? Or to put it differently if we assume in addition a pair force how can I calculate the mean field force? If one considers for example the vlasov equation it is no problem as $u(t,x)$ satisfies conservation of mass.
Dec 5, 2017 at 13:24	comment	added	Carlo Beenakker		$u(x,t)$ is not a probability density, it is a particle density: the number of particles at time $t$ in the interval $(x,x+dx)$ equals $u(x,t)dx$; so the integral $\int u(x,t)dx$ can become larger than unity, it just means the total number of particles is greater than one.
Dec 5, 2017 at 8:10	review	First posts
Dec 5, 2017 at 8:46
Dec 5, 2017 at 8:09	history	asked	Jack_Stiller10	CC BY-SA 3.0