# Classical statistical mechanics question

• facenian
In summary: This is what I ment.In summary, Liouville's theorem states that the material derivative of the phase space density function, rho, is zero, meaning that rho remains constant along a trajectory in phase space. However, this does not mean that if you fix your position in phase space and let time flow, rho will remain constant. This is because Liouville's theorem does not apply to partial derivatives with respect to position or momentum, which can change over time. The boundary conditions for a first-order partial differential equation determine the solution, and for this problem, rho=constant is the only solution that satisfies the boundary conditions and the differential equation.

#### facenian

Hello. I read this assertion in a book: if we take at an initial time t_0 a constant density distribution $\rho(p,q,t_0)$
in phase space, then this implies that \rho wil remain independent of time for all $t>t_0$ because by Liouville's theorem
$$\frac{\partial\rho}{\partial t}=-\sum(\frac{\partial\rho}{\partial q}\dot{q}+\frac{\partial\rho}{\partial p}\dot{p})=0$$
I don't understand this because we take $\frac{\partial\rho}{\partial q}=\frac{\partial\rho}{\partial p}=0$ only at t=t_0 which means that $\frac{\partial\rho}{\partial t}=0$ only at t=t_0 it doesn't mean that $\frac{\partial\rho}{\partial t}=0$ for t>t_0 therefore we can not assert that $\frac{\partial\rho}{\partial q}=0$ and $\frac{\partial\rho}{\partial q}=0$ for $t>t_0$

I just happened across this post and thought I'd respond (though its a bit late). Why do you take the partials of rho to be zero, as they are not necessarily. Liouville's theorem says that density is constant in time, not in space (over q's and p's)

Yes, a uniform distribution would correspond to the high-temperature Botlzmann distribution. Such a distribution is stationary.

You can use analyticity and induction to prove this. As you say only $$\dot{\rho}(t_0) = 0$$ immediately follows. Therefore at an infinitesimal time later you have $$\rho(t_0+dt) = \rho(t_0) + dt \dot{\rho}(t_0) = \rho(t_0)$$. You can probably see the rest.

homology said:
I just happened across this post and thought I'd respond (though its a bit late). Why do you take the partials of rho to be zero, as they are not necessarily. Liouville's theorem says that density is constant in time, not in space (over q's and p's)
Liouville's theorem does not say that rho is constant in time it says that the materilal(substancial) derivate of rho is zero. If you stay in one place in phase space then rho may vary in time.

C. H. Fleming said:
Yes, a uniform distribution would correspond to the high-temperature Botlzmann distribution. Such a distribution is stationary.

You can use analyticity and induction to prove this. As you say only $$\dot{\rho}(t_0) = 0$$ immediately follows. Therefore at an infinitesimal time later you have $$\rho(t_0+dt) = \rho(t_0) + dt \dot{\rho}(t_0) = \rho(t_0)$$. You can probably see the rest.
I think you are simply restating Liuville's theorem which says $\frac{d\rho}{dt}=0$ at all times, that is not what I asked.

By constant in time I meant that the total derivative $$d\rho/dt = 0$$. But the partials with respect to position need not be, that is: $$\partial \rho / \partial q \neq 0$$, ditto with respect to momenta. Are you asking for a proof of the theorem? Many books don't really prove the theorem but sort of wave their hands and throw up a continuity equation for the density.

You have a 1st order partial differential equation for the function rho (the velocities can be written in terms of the coordinates through derivatives of the Hamiltonian).

What is the boundary condition for a 1st order partial differential equation? The answer is values of the function on the hypersurface.

You are given the values on the hypersurface t=0.

So that should be enough to determine the solution completely.

Hence rho=constant is the solution for all time and all phase space.

Now there's always the question whether these Cauchy conditions are too much and no solution exists (for 1st order differential equation the Cauchy conditions will be just the value of the function on a hypersurface). However, for this problem you don't have that problem, since rho=constant is obviously a solution.

So rho=constant is the only solution that satisfies the boundary conditions and the differential equation.

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facenian said:
I think you are simply restating Liuville's theorem which says $\frac{d\rho}{dt}=0$ at all times, that is not what I asked.

I meant partial derivatives with respect to time. Replace all $$\dot{\rho}$$ with $$\frac{\partial \rho}{\partial t}$$ and consider fixed coordinates. This is not Liouville's theorem, it's just calculus. If the distribution cannot change at the initial time, and it cannot change a moment later, and it cannot change a moment later, ..., then it cannot change at all (barring non-analyticity).

homology said:
By constant in time I meant that the total derivative $$d\rho/dt = 0$$. But the partials with respect to position need not be, that is: $$\partial \rho / \partial q \neq 0$$, ditto with respect to momenta. Are you asking for a proof of the theorem? Many books don't really prove the theorem but sort of wave their hands and throw up a continuity equation for the density.

No, I'm not asking for a proof of the theorem, maybe I did not express myself clearly. I'm sorry but my english is not good

RedX said:
You have a 1st order partial differential equation for the function rho (the velocities can be written in terms of the coordinates through derivatives of the Hamiltonian).

What is the boundary condition for a 1st order partial differential equation? The answer is values of the function on the hypersurface.

You are given the values on the hypersurface t=0.

So that should be enough to determine the solution completely.

Hence rho=constant is the solution for all time and all phase space.

Now there's always the question whether these Cauchy conditions are too much and no solution exists (for 1st order differential equation the Cauchy conditions will be just the value of the function on a hypersurface). However, for this problem you don't have that problem, since rho=constant is obviously a solution.

So rho=constant is the only solution that satisfies the boundary conditions and the differential equation.

this sounds convincing, thank you

C. H. Fleming said:
This is not Liouville's theorem, it's just calculus. If the distribution cannot change at the initial time, and it cannot change a moment later, and it cannot change a moment later, ..., then it cannot change at all (barring non-analyticity).

Yes, I agree that what you are saying is just calculus and you don't need to perform a numerical integration to get that rho is constant in time in the sense your are taking it because we know by one of the fundamental theorems of calculus that a null derivative implies a constant function.

Consider this, rho is constant along a trayectory in phase space and that's what Liouville's theorem is all about. However this doen not mean that if you fix your position in phase space and let the time flow rho will remain constant, neither it means that if you freeze time at same instant t and wonder arround phase space rho wil remain constant as you move.

I think the argument given by RedX could be correct

## 1. What is classical statistical mechanics?

Classical statistical mechanics is a branch of physics that uses statistical methods to study the behavior of large systems of particles. It is based on classical mechanics, which describes the motion of particles at the macroscopic level, and statistical mechanics, which uses probability to describe the behavior of a large number of particles at the microscopic level.

## 2. What is the difference between classical and quantum statistical mechanics?

The main difference between classical and quantum statistical mechanics is the underlying principles and laws that govern the behavior of particles. Classical statistical mechanics is based on classical mechanics and does not take into account the quantum nature of particles. Quantum statistical mechanics, on the other hand, uses the principles of quantum mechanics to describe the behavior of particles at the microscopic level.

## 3. What are the main assumptions of classical statistical mechanics?

The main assumptions of classical statistical mechanics are that particles are distinguishable and that their interactions can be described by classical mechanics. This means that particles can be identified and tracked individually, and their interactions can be described using Newton's laws of motion.

## 4. How is classical statistical mechanics used in practical applications?

Classical statistical mechanics is used in many practical applications, such as thermodynamics, material science, and fluid dynamics. It is used to understand and predict the behavior of complex systems, such as gases, liquids, and solids, and to design and improve various technologies, such as engines, refrigerators, and materials for construction.

## 5. What are some limitations of classical statistical mechanics?

Classical statistical mechanics has some limitations, especially when it comes to studying systems at the atomic and subatomic levels. It does not take into account the wave-like nature of particles and cannot accurately describe phenomena such as quantum tunneling and entanglement. It also struggles with systems that have a high degree of complexity or non-equilibrium states.