What's the relation between the distribution function $f(p,q,t)$ in 6 dim phase space and density function in $6N$ dim phase space?

Question

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1. What the question is asking (in plain language)

We have two kinds of “densities’’ in classical statistical mechanics.

Symbol	What it describes	Dimension of the space
(f(\mathbf p,\mathbf q,t))	One‑particle distribution function – the average number of particles that at time (t) are found in the infinitesimal phase‑space element (\mathrm d^3p\,\mathrm d^3q).	6‑dimensional (3 momenta + 3 coordinates)
(\rho_N(\mathbf p_1,\mathbf q_1,\dots,\mathbf p_N,\mathbf q_N,t))	N‑particle (or ensemble) distribution – the probability density that the whole system of (N) particles is in the point ((\mathbf p_1,\mathbf q_1,\dots,\mathbf p_N,\mathbf q_N)) of the (6N)‑dimensional phase space.	(6N)‑dimensional

The student wants to know:

How are the two objects related?
If the one‑particle function (f) becomes time‑independent at equilibrium, why does the full (6N)‑dimensional density (\rho_N) also become time‑independent?

In other words, why does a stationary “average picture’’ of individual particles guarantee a stationary picture of the whole ensemble of micro‑states?

2. Step‑by‑step answer

2.1 Definitions

One‑particle distribution (often called the single‑particle phase‑space density):

[ f(\mathbf p,\mathbf q,t)\;\equiv\; \left\langle \sum_{i=1}^{N}\delta^{(3)}!\bigl(\mathbf p-\mathbf p_i(t)\bigr)\, \delta^{(3)}!\bigl(\mathbf q-\mathbf q_i(t)\bigr)\right\rangle . \tag{1} ]

The angular brackets (\langle\cdots\rangle) denote an average over an ensemble of identically prepared systems (or, by the ergodic hypothesis, a time average over a single system).
Normalisation: (\displaystyle \int! \mathrm d^3p\,\mathrm d^3q\; f(\mathbf p,\mathbf q,t)=N).
N‑particle (ensemble) density (the Liouville density):

[ \rho_N(\mathbf p_1,\mathbf q_1,\dots,\mathbf p_N,\mathbf q_N,t) \equiv \left\langle \prod_{i=1}^{N} \delta^{(3)}!\bigl(\mathbf p_i-\mathbf p_i(t)\bigr)\, \delta^{(3)}!\bigl(\mathbf q_i-\mathbf q_i(t)\bigr)\right\rangle . \tag{2} ]

Normalisation: (\displaystyle \int! \prod_{i=1}^{N}\mathrm d^3p_i\,\mathrm d^3q_i\; \rho_N=1).

2.2 From (\rho_N) to (f) – marginalisation

The one‑particle distribution is obtained by integrating out the coordinates of the other (N-1) particles (a marginal of the full density):

[ \boxed{ f(\mathbf p,\mathbf q,t)=N!\int!\prod_{j=2}^{N}\mathrm d^3p_j\,\mathrm d^3q_j\; \rho_N(\mathbf p,\mathbf q,\mathbf p_2,\mathbf q_2,\dots,\mathbf p_N,\mathbf q_N,t) } \tag{3} ]

Why the factor (N)?
The integral in (3) gives the probability that any particular particle (say particle 1) is at ((\mathbf p,\mathbf q)). Because any of the (N) indistinguishable particles could be the one we are looking at, we multiply by (N) to get the average number of particles in that element.

Equation (3) is the precise mathematical link between the 6‑D and the (6N)‑D densities.

2.3 Time evolution – Liouville’s theorem

For a system with Hamiltonian

[ H(\mathbf p_1,\mathbf q_1,\dots,\mathbf p_N,\mathbf q_N) =\sum_{i=1}^{N}\frac{\mathbf p_i^{2}}{2m}+V(\mathbf q_1,\dots,\mathbf q_N), ]

the Liouville equation governs the evolution of (\rho_N):

[ \boxed{ \frac{\partial\rho_N}{\partial t} = -{\,\rho_N , H\,} = -\sum_{i=1}^{N} \Bigl( \frac{\partial H}{\partial \mathbf p_i}!\cdot!\frac{\partial\rho_N}{\partial \mathbf q_i} - \frac{\partial H}{\partial \mathbf q_i}!\cdot!\frac{\partial\rho_N}{\partial \mathbf p_i} \Bigr) . } \tag{4} ]

If the Hamiltonian does not depend explicitly on time (the usual case for an isolated system), the Liouville equation tells us that the phase‑space density is incompressible: the flow generated by Hamilton’s equations merely moves points around without changing the value of (\rho_N) along a trajectory. Consequently, any function of the constants of motion (energy, total momentum, etc.) is a stationary solution:

[ \frac{\partial\rho_N}{\partial t}=0 \quad\Longleftrightarrow\quad \rho_N = \Phi\bigl(H(\mathbf p_1,\dots,\mathbf q_N)\bigr). \tag{5} ]

Typical equilibrium ensembles are exactly of this form:

Microcanonical: (\displaystyle \rho_N^{\rm mc}\propto\delta!\bigl(H-E\bigr)).
Canonical: (\displaystyle \rho_N^{\rm can}\propto \exp!\bigl[-\beta H\bigr]).

Both are stationary because the right‑hand side of (4) vanishes.

2.4 From stationary (f) to stationary (\rho_N)

Now we address the core of the question: Why does a time‑independent one‑particle distribution imply that the whole (6N)‑dimensional density is also time‑independent?

There are two complementary ways to see this.

(a) Direct marginal argument

Take the time derivative of Eq. (3) and use the Liouville equation for (\rho_N):

[ \frac{\partial f}{\partial t} = N!\int!\prod_{j=2}^{N}! \mathrm d^3p_j\mathrm d^3q_j\; \frac{\partial\rho_N}{\partial t} = -N!\int!\prod_{j=2}^{N}! \mathrm d^3p_j\mathrm d^3q_j\; {\,\rho_N , H\,}. \tag{6} ]

Because the Poisson bracket is a total divergence in the full (6N)‑dimensional space, the integral over the other coordinates reduces to a surface term that vanishes (the distribution is assumed to go to zero fast enough at infinity). Hence

[ \boxed{\displaystyle\frac{\partial f}{\partial t}=0 \;\Longrightarrow\; \int!\prod_{j=2}^{N}! \mathrm d^3p_j\mathrm d^3q_j\; {\,\rho_N , H\,}=0.} \tag{7} ]

Equation (7) can be satisfied only if the integrand itself is a total derivative that integrates to zero for any choice of the integration domain. The only functions that have this property for all possible marginalisations are those that are constants of motion of the full dynamics, i.e. stationary solutions of the Liouville equation. Therefore:

[ \frac{\partial f}{\partial t}=0 \quad\Longrightarrow\quad \frac{\partial\rho_N}{\partial t}=0 . ]

In short, a stationary marginal cannot be obtained from a time‑varying parent density; the parent must already be stationary.

(b) Factorisation in equilibrium (ideal‑gas picture)

For many systems that are weakly interacting (e.g. a dilute classical gas), the equilibrium N‑particle density factorises into a product of identical one‑particle densities (apart from the usual (1/N!) factor that removes double counting):

[ \boxed{ \rho_N^{\rm eq}(\mathbf p_1,\mathbf q_1,\dots,\mathbf p_N,\mathbf q_N) = \frac{1}{N!}\,\prod_{i=1}^{N} f^{\rm eq}(\mathbf p_i,\mathbf q_i) . } \tag{8} ]

If the single‑particle function (f^{\rm eq}(\mathbf p,\mathbf q)) is already independent of time, the product (8) is automatically independent of time as well. This argument is exact for a non‑interacting Hamiltonian, and it remains an excellent approximation for a dilute gas where correlations are negligible (the “molecular chaos’’ assumption).

Thus, equilibrium of the one‑particle description forces equilibrium of the full ensemble description.

2.5 Summary of the logical chain

Definition: (f) is the marginal of (\rho_N) (Eq. 3).
Liouville dynamics: (\rho_N) obeys the Liouville equation (Eq. 4).
Stationarity of (f): (\partial_t f=0) forces the integral of the Poisson bracket of (\rho_N) to vanish (Eq. 7).
Only way for the integral of a total divergence to vanish for any choice of the other coordinates is that the divergence itself be zero, i.e. (\partial_t\rho_N=0).
In equilibrium ensembles (microcanonical, canonical, grand‑canonical) the explicit solutions (\rho_N\propto e^{-\beta H}) are already time‑independent, and their marginals give precisely the stationary one‑particle distribution.
If interactions are weak, the equilibrium factorisation (8) makes the implication even more transparent.

3. Final answer

The single‑particle phase‑space density (f(\mathbf p,\mathbf q,t)) is the marginal of the full (6N)-dimensional Liouville density (\rho_N):

[ f(\mathbf p,\mathbf q,t)=N\int!\prod_{j=2}^{N}! \mathrm d^3p_j\,\mathrm d^3q_j\; \rho_N(\mathbf p,\mathbf q,\mathbf p_2,\mathbf q_2,\dots,\mathbf p_N,\mathbf q_N,t). ]

*Because (\rho_N) obeys Liouville’s equation, its time derivative is a total phase‑space divergence. When the marginal (f) is stationary ((\partial_t f=0)), the integral of that divergence over the remaining (6(N-1)) coordinates must vanish for any choice

Original question: What's the relation between the distribution function $f(p,q,t)$ in 6 dim phase space and density function in $6N$ dim phase space? on Physics Stack Exchange, licensed CC BY-SA.

1. What the question is asking (in plain language)

2. Step‑by‑step answer