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I am looking for help to taking a multi-particle system and defining a local equilibrium so that I can get a variational principle for the slightly simplified system.

For example, if I have the system,

\begin{align} S &= \int dt \sum_i \mathcal{L}(q_i, \dot{q}_i) \\ &= \int dt\ d^3x \sum_i \delta(q_i - x) \mathcal{L}(q_i, \dot{q}_i) \\ &= \sum_{\mathbb{B}_n} \int_{\mathbb{B}_n} dt\ d^3x \sum_i \delta(q_i - x) \mathcal{L}(q_i, \dot{q}_i) \end{align}

Then for each box $ \mathbb{B}_n $ I would approximate with the system being in local equilibrium and take the Stat Mech approach to writing the equilibrium system equation for each box.

My concern is that I wouldn't know how to write the local equilibrium Lagrangian to replace the $ \int_{\mathbb{B}_n} dt\ d^3x \sum_i \delta(q_i - x) \mathcal{L}(q_i, \dot{q}_i) $.

I read A Lagrangian formalism for nonequilibrium thermodynamics, but I am not entirely sure how to stick the Lagrangian into the action above because the state variables change.

Can you give me a reference or the next step with this problem?

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I will explain this in the order that you might want to read about it as opposed to the logical order, then again in logical order.

Variational Principle for Hamiltonian Fluids

From "Hamiltonian Fluid Dynamics" by P. J. Morrison, we learn how to write the variational form of fluid equations. We use Lagrangian coordinates and write the action as,

\begin{align} S[q] &= \int dt\ L[q, \dot{q}] \\ &= \int dt\ d^3a\ \left[ \frac{1}{2} \rho_0 \dot{q}^2 - \rho_0 U(\rho, s) \right] \end{align} where coordinates $ a = q(t=0) $, density $ \rho_0 = \rho(t=0) $ and $ U(\rho, s) $ is the thermodynamic internal energy transformed to use density instead of volume (see Specific Volume).

Local Equilibrium

From "Equilibrium Statistical Mechanics" by Plischke, we can compute the internal energy from the microcanonical ensemble.

\begin{align} \Omega(E, V, N) &= \frac{1}{h^{3N} N!} \int_{E < \mathcal{H}(x, p) \leq E + \delta E} d^{3N}x\ d^{3N}p \\ S(E, V, N) &= k_B \log \Omega(E, V, N) \\ S_{\text{ideal gas}} &= N k_B \log \frac{V}{N} + \frac{3N}{2} k_B \log \frac{4 \pi m E}{3 N h^2} + \frac{5}{2} N k_B \end{align}

For use in the Hamiltonian from Morrison, we actually want to transform the variables a bit, Morrison uses these transformations and yield the following properties, \begin{align} U &= \frac{E}{mN} ;\ \ \rho = \frac{mN}{V} ;\ \ s = \frac{S}{mN} \\ \frac{\partial U}{\partial \rho} &= \frac{1}{mN} \frac{\partial E}{\partial V} \frac{\partial V}{\partial \rho} = \frac{1}{mN} (-p) \frac{-mN}{\rho^2} = \frac{p}{\rho^2} \\ \frac{\partial U}{\partial s} &= \frac{1}{mN} \frac{\partial E}{\partial S} \frac{\partial S}{\partial s} = \frac{1}{mN} (T) mN = T \end{align}

which Morrison claims in his paper.

Many-body to Fluid

The many body action would look like, \begin{align} S[q_i] &= \int dt \sum_i L(q_i, \dot{q}_i) \\ &= \int dt\ d^3a\ \rho_0(a)\ L(q(a), \dot{q}(a)) \\ &= \int dt\ d^3a\ \rho_0\ \left( p \frac{\partial \mathcal{H}}{\partial p} - \mathcal{H} \right) \\ \end{align} where $ a $ has just become our continuum label and we assume that there are many particles in each region of label-space $ d^3a $ so we can make an equilibrium approximation.

Now, we break our dynamics up into average "large-scale" dynamics and local dynamics which shall be claimed to be in equilibrium. Frankly, I am not sure of how to do this super formally, so I invoke the physicist right of hand-waving!

\begin{align} \mathcal{H}(q, p) = \mathcal{H}_{avg}(\bar{q}, \bar{p}) + \mathcal{H}_{equil}(\Delta q, \Delta p) \end{align} where $ \mathcal{H}_{equil} $ could be computed with a Taylor series, but we can opt for an easier method. Simply compute the density of states for $ \mathcal{H}_{equil}(\Delta q, \Delta p) = \mathcal{H}(\bar{q} + \Delta q, \bar{p} + \Delta p) - \mathcal{H}_{avg}(\bar{q}, \bar{p}) $.

\begin{align} \Omega(E, V, N) &= \frac{1}{h^{3N} N!} \int_{E < \mathcal{H}_{equil}(\Delta q, \Delta p) \leq E + \delta E} d^{3N}\Delta q\ d^{3N} \Delta p \\ \end{align} which is just the normal density of states for our system with phase space coordinate translations and a gauge shift. Normal physical systems ought to be invariant to both.

Thus our final Hamiltonian can be taken as, \begin{align} \mathcal{H}(q, p) \rightarrow \mathcal{H}(q, p) + U(\rho, s) \end{align} where I dropped the over-bar denoting averages and $ U $ is computed as above.

Noting that $ \rho_0 d^3a = \rho d^3q $, we find our action to be, \begin{align} S &= \int dt\ d^3a\ \rho_0\ \left( p \frac{\partial \mathcal{H}}{\partial p} - \mathcal{H}(q, p) - U(\rho, s) \right) \\ &= \int dt\ d^3a\ \rho_0\ \left( \mathcal{L} - U(\rho, s) \right) \\ &= \int dt\ d^3q\ \rho \left( \mathcal{L} - U(\rho, s) \right) \end{align}

which reproduces Morrison's results. In practicing using this formalism, it seems that best practice is to treat $ q(a, t) $ as field activation and do the variation on field coordinates $ (a, t) $ (ie. using $ \int dt\ d^3a $ version). For example, see below.

Ideal Gas

As an example, we get the equations for an ideal gas. From previous definitions, \begin{align} s &= \frac{k_B}{m} \log m \rho + \frac{3}{2m} k_B \log \frac{4 \pi m^2 U}{2 h^2} + \frac{5 k_B}{2m} \\ U &= c \rho^{\gamma - 1} e^{\alpha s} = \left[\frac{h^{2} e^{- \frac{5}{3}}}{2 \pi m^{\frac{8}{3}}} \right] \rho^{\frac{1}{3}-1} e^{\frac{2 m}{3 k}s}\\ S &= \int dt\ d^3a\ \rho_0 \left( \frac{1}{2} \dot{q}^2 - U \right) \end{align}

With uniform density (ie. $ \nabla \rho_0 = 0 $) and the Jacobian $ \mathcal{J} = \frac{d^3q}{d^3a} $, \begin{align} \delta S &= \int dt\ d^3a\ \left[ \rho_0 \dot{q}_i \delta \dot{q}_i - \rho_0 \left( \frac{\partial U}{\partial \rho} \delta \rho + \frac{\partial U}{\partial s} \delta s \right) \right] \\ &= \int dt\ d^3a\ \left[ -\rho_0 \ddot{q}_i \delta q_i - \rho_0^2 \frac{\partial U}{\partial \rho} \frac{\partial \frac{1}{\mathcal{J}}}{\partial \frac{\partial q^k}{\partial a^j}} \delta \frac{\partial q^k}{\partial a^j} \right] \\ &= \int dt\ d^3a\ \left[ -\rho_0 \ddot{q}_k - \frac{\partial}{\partial a^j} \left( \rho^2 \frac{\partial U}{\partial \rho} \frac{\partial \mathcal{J}}{\partial \frac{\partial q^k}{\partial a^j}} \right) \right] \delta q_k \\ \end{align} which reproduces Morrison. More details shown here.

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