Register to reply 
Louiville's Theorem, a question 
Share this thread: 
#1
Jan2713, 09:21 PM

P: 5

So I noticed that when learning about Liouville's Theorem in class, that it was described in terms of ensemble: i.e; you have a 6N dimensional phase space for N particles moving in 3 dimensions and each system of the ensemble has a representative system point and that the density function in Liouville's theorem refers to the density of systems in the ensemble in a given microstate and that these points, since essentially they follow a global conservation law and move in phase space, are subject to the continuity equation for the points. This all makes sense, but I was wondering if we can equivalently reformulate Liouville's problem without considering an ensemble of systems.
What I mean is this: imagine you have a system of N particles, and for simplicity that they are capable of moving in only one dimension and thus they each have a spatial coordinate x and a momentum p. Let's consider now a 2 dimensional phase space whose axes are x and p (instead of a 2N dimensional phase space whose axes are the x_{i} and p_{i} of all of the N particles). And now instead of plotting ensembles in the phase space, we plot the momenta and position of each particle (even though this is practically impossible for a very large system such as that of a liter of gas, we could do this in theory and that's all that matters, I think). This gives us a space with N particles rather than N system points, and this still follows a global conservation in that the particles of the system can not be created nor destroyed. So it seems that a form of Liouville's theorem would hold for a new density function which rather than representing the density of system points in phase space, represents now the number of particles per unit area in phase space (in this 1D example) whose coordinates and moments are in the interval (x, p, x+dx, p+dp), much like that of the molecular distribution function. Is this indeed true? The reason that I ask is because if this holds, then if we look at the resulting equation where the density function d = d(x,p) and the "velocity" in the continuity equation is given by <[itex]\frac{dx}{dt}[/itex],[itex]\frac{dp}{dt}[/itex]>, or <v, F>, we get: 0 = [itex]\frac{∂d}{∂t}[/itex] + [itex]\frac{∂d}{∂x}[/itex]*v + [itex]\frac{∂d}{∂p}[/itex]*F, where F= F(x), which follows from ∂d/∂t + div (dv) = 0 where the del operator is given by ∇=[itex]\frac{∂}{∂x}[/itex]+[itex]\frac{∂}{∂p}[/itex] in phase space. and in the case where we have thermodynamic equilibrium and d does not depend explicitly on time, we have a separable first order partial differential equation whose solutions give the exponential dependence on F(x) and on v^{2}, i.e; the Boltzmann distribution. I thought this was more than a coincidence but I didn't find sources on this. Thoughts would be greatly appreciated. Thanks! 


Register to reply 
Related Discussions  
Mean Value Theorem Question  Calculus & Beyond Homework  9  
Question regarding Mean Value Theorem.  Calculus & Beyond Homework  1  
Louiville theorem  Differential Equations  1  
Mean Value Theorem Question  Calculus & Beyond Homework  7  
Mean value theorem question  Calculus  4 