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In statistical mechanics, we usually think in terms of the Hamiltonian formalism. At a particular time $t$, the system is in a particular state, where "state" means the generalised coordinates and momenta for a potentially very large number of particles. (I'm interested primarily in classical systems for the sake of this question.) Since this state cannot be known precisely, we consider an ensemble of systems. By integrating each point in this ensemble forward in time (or, more often, by considering what would happen if we were able to perform such an integral), we deduce results about the ensemble's macroscopic behaviour. Using the Hamiltonian formalism is useful in particular because it gives us the concept of phase space volume, which is conserved under time evolution for an isolated system.

It seems to me that we could also consider ensembles within the Lagrangian formalism. In this case we would have a probability distribution over initial values of the coordinates (but not their velocities), and another distribution over the final values of the coordinates (but not their velocities). (Actually I guess these would need to be two jointly distributed random variables, since there could easily be correlations between the two.) This would then lead to a probability distribution over the paths the system takes to get from one to the other. I have never seen this Lagrangian approach mentioned in statistical mechanics. I'm curious about whether the idea has been pursued, and whether it leads to any useful results. In particular, I'm interested in whether the idea of phase space volume has any direct meaning in terms of such a Lagrangian ensemble.

N. Virgo
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    There is the Response-function / Martin-Siggia-Rose formalism which casts a Langevin description into a path-integral picture. See here for a simpler one-particle description. Not sure if this is what you are looking for. – Vijay Murthy Jan 18 '13 at 11:16
  • @VijayMurthy that looks interesting, and I'll look into it further. From those handwritten notes it looks like they're starting with some stochastic dynamics and then deriving something that looks like a path integral; whereas I'm hoping for something that starts with a classical Lagrangian and then derives a statistical ensemble based on it. But thanks, and I look forward to taking a closer look. – N. Virgo Jan 18 '13 at 11:27
  • @VijayMurthy For Wilson RG, all spin models get cast into continuous form which means you get Landau-Ginzburg-type Lagrangian. – Slaviks Mar 04 '13 at 19:23
  • @Slaviks, Thanks for the comment. One can write an action functional and need not do an RG. The OP asked for a Lagrangian description, not an RG. The MSR action functional can be written for particle systems too. So I dont get your comment. Perhaps I am missing something. – Vijay Murthy Mar 04 '13 at 20:56
  • As emphasized by Weinberg in his QFT textbook, the generator of time translations in quantum mechanics is the Hamiltonian, not the Lagrangian. The so-called Lagrangian formulation of quantum mechanics is not a proper Lagrangian formulation, because it is the Hamiltonian which is being really used: "It is the Hamiltonian formalism that is needed to calculate the S-matrix (whether by operator or path-integral methods) but it is not always easy to choose Hamiltonians that yield a Lorentz-invariant S-matrix." – juanrga Jul 13 '18 at 19:18
  • @juanrga the question is purely about classical physics. – N. Virgo Jul 14 '18 at 07:04
  • The generator of time translations in statistical mechanics (either classical or quantum) is the Liouvillian, which is a function of the Hamiltonian, never the Lagrangian. In classical statistical mechanics $\mathcal{L}={H, }$ and the state evolution is $\rho(t) = \exp[\mathcal{L}t] \rho_0$. The Liouvillian is a function of the Hamiltonian because the generator of time translations in quantum mechanics is the Hamiltonian. – juanrga Jul 17 '18 at 07:17
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    Note that the classical Liouvillian is just the first term in a power series expansion of the quantum Liouvillian in powers of $\hbar$. And the quantum Liouvillian follows from a mixed superposition of quantum pure states. – juanrga Jul 17 '18 at 07:22
  • @juanrga I understand that, more or less, but I am having trouble seeing the connection to my question. Could you be more explicit about that please? – N. Virgo Jul 17 '18 at 10:24

5 Answers5

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The transition between the Hamiltonian and Lagrangian formalisms in Mechanics can be accomplished by means of the Hamilton-Jacobi theory.

Consider for example a classical statistical ensemble on a phase space $(x,p)$ defined by:

A. The (initial) state of this ensemble is defined by a distribution function $f(x_0,p_0)$ satisfying the normalization condition:

$$\displaystyle{\int f(x_0,p_0) dx_0dp_0 = 1}$$

($(x_0,p_0)$ are the initial conditions)

B. The time evolution is governed by the Hamiltonian function $H(x,p, t)$.

According to the Hamilton-Jacobi theory, there exists Hamilton-Jacobi phase function $S(x_0, x_1, t_0, t_1)$ satisfying the Hamilton-Jacobi equation:

$$\displaystyle{\frac{\partial S}{\partial t}+H\left(x_1,\frac{\partial S}{\partial x_1}, t\right) = 0}$$

(where $(x_1,p_1)$ are the coordinates and momenta at time $t$)

The momenta can be derived from the Hamilton-Jacobi phase function:

$$\displaystyle{p_i = \frac{\partial S}{\partial x_i}}$$

The problem of expressing the state of the system in terms of the initial and final coordinates is rendered to a problem of transformation of probability distributions. We can define the state of the system in the initial and final coordinates as:

$$\displaystyle{F_t(x_0, x_1) = f\left(x_0,\frac{\partial S}{\partial x_1}(x_0, x_1, t) \right)}$$

The transformation Jacobian is given by:

$$ \displaystyle{dx_0 dp_0 = \frac{\partial^2 S}{\partial x_0\partial x_1}}dx_0 dx_1 $$

And the normalization condition:

$$\displaystyle{\int F_t(x_0, x_1) \frac{\partial^2 S}{\partial x_0\partial x_1}(x_0, x_1, t)dx_0dx_1 = 1}$$

In the general case, the joint distribution $F_t(x_0, x_1)$ will not be separable

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There is a field theory version of statistical physics. The temperature is like the imaginary time. In this way we can formulate theory by path integral with action determined by Lagrangian.

  • I know about this. I wish I understood it better, but I don't think it's what I'm looking for. At least, the connection between the two ideas is not obvious. – N. Virgo Mar 05 '13 at 04:59
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    Could you elaborate on this a bit more, it seems to be exactly related to a question I previously asked: http://physics.stackexchange.com/q/161559/ – Skyler Jan 30 '15 at 18:48
  • The field theory version of statistical mechanics comes when you go to Euclidean time and the Lagrangian simply becomes Hamiltonian there. That's how the connection is established in the first place. What I am saying is that there is no conceptual link established between the Lagrangian formalism and statistical mechanics, rather, the Lagrangian is converted to a Hamiltonian to make the transition to statistical mechanics. Correct me if I am wrong. –  May 18 '20 at 12:29
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I am not sure if this is what you are up to (it is related to what Xiao-Qi Sun said) to but I'll give it a try too ...

At the beginning of Chapter V.2 of his QFT Nutshell, Anthony Zee explains how classical statistical mechanics (characterized by the corresponding partition function involving the Hamilton function) in $d$- dimensional space is related to Eucledian field theory (characterized by the corresponding generating functional or path integral involving the Lagrangian).

To see this relationship, consider for example the Minkowskian path integral of a scalar field

$$ (1) \,\, \cal{Z} = \int\cal{D}\phi e^{(i/\hbar)\int d^dx[\frac{1}{2}(\partial\phi)^2-V(\phi)]} = \int\cal{D}\phi e^{(i/\hbar)\int d^dx\cal{L}(\phi)} = \int\cal{D}\phi e^{(i/\hbar)S(\phi)} $$

Upon Wick rotation, the Lagrange density $\cal{L}(\phi)$ turns into the energy density and the action $S(\phi)$ gets replaced by the energy functional $\cal E(\phi)$ of the field $\phi$

$$ (2) \,\, \cal{Z} = \int\cal{D}\phi e^{(-1/\hbar)\int d^d_Ex[\frac{1}{2}(\partial\phi)^2+V(\phi)]} = \int\cal{D}\phi e^{(-1/\hbar)\cal{E}(\phi)} $$

with

$$ \cal E(\phi) = \int d^d_Ex[\frac{1}{2}(\partial\phi)^2+V(\phi)] $$

This can now be compared to the classical statistical mechanics of an N-particle system with the Energy

$$ E(p,q) = \sum_i \frac{1}{2m}p_i^2+V(q_1,q_2,\cdots,q_N) $$

and the corresponding partition function

$$ Z = \prod_i\int dp_i dq_i e^{-\beta E(p,q)} $$

Integrating over the momenta $p_i$ one obtains the reduced partition function

$$ Z = \prod_i\int dq_i e^{-\beta V(q_1,q_2,\cdots,q_N)} $$

Following the usual procedure to obtain the field theory which corresponds to this reduced partiction function by letting $i\rightarrow x$, $q_i \rightarrow \phi(x)$ and identifying $\hbar = 1/\beta = k_B T$ it has exactly the same form as the Euclidian path integral (2).

So it can finally be seen that in this example, the (reduced) partition function of an N-particle system in d-dimensional space corresponds to the path integral of a scaler field in d-dimensional spacetime.

These arguments can be further generalized to obtain a path integral representation of the quantum partition funcction, finite temperature Feynman diagrams, etc too ...

If I understand this right, this line of thought relating statistical mechanics to field theory is for example applied in topics like the Nonequilibrium functional renormalization group or in AdS/CFT to relate the correlation functions on the CFT side to the string amplitudes on the AdS side.

Dilaton
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  • Thanks. I was asking about formulating classical statistical mechanics in terms of the classical Lagrangian, whereas this links classical statistical mechanics to the quantum Lagrangian approach (i.e. path integrals). But I (somewhat idly) wonder whether there's some kind of deep connection between these two ideas. – N. Virgo May 10 '13 at 10:20
  • (For some reason my inbox wasn't pinged when you answered. I wonder why.) – N. Virgo May 10 '13 at 10:20
  • @Nathaniel I wrote the first version of this post way after midnight, so I had to hide it away first until I have profread and completed it in the morning ... – Dilaton May 10 '13 at 12:17
  • I have recently learned that even for classical systems, like for example the Navier Stokes equations, one can derive some kind of "path" or functional integral. The integration is then among other things done over all solutions of the Navier-Stokes equations, instead of all pathes, the scalar fields correspond to the (components) of the velocity, etc And a corresponding (quite long and ugly) action can be derived in case too. – Dilaton May 10 '13 at 12:23
  • Since for example turbulence theory can be described by nonequilibrium statistical mechanics (in the MaxEnt formalism one would have the energy or enstrophy flux that are constant on an a scale invariant subrange as additional relevant variables to appear in the nonequilibrium distribution function), there must be a connection between the functional integrals of the Navier-Stokes equations and certain statistical mechanics partition functions too I think. But this I have not yet seen in to much detail up to now. – Dilaton May 10 '13 at 12:29
  • Mabe the relationship appears in this paper which should have a paragraph about turbulence too. I have not yet have time to look at it closer. – Dilaton May 10 '13 at 12:31
  • Are you talking about the work of Roddy Dewar by any chance? This talk of classical path integrals and MaxEnt puts me in mind of his work. – N. Virgo May 10 '13 at 12:43
  • @Nathaniel what exactly is the work of Roddy Dewar about? I dont know it ... I just know about the MaxEnt for both, classical and quantum mechanic systems from a nonequilibrium statistics course I have taken and this paper I found quite instructive and about how to finde a path integral formulation of the Navier Stokes equations I learned from this at the first time. The action of the Navier Stokes equation involves some crazy Grassmann and ghost fields too, ha ha ... – Dilaton May 10 '13 at 21:40
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The Hamilton formulation of classical dynamics gives rise to a very strong and important theorem in statistical mechanics that is the Liouville theorem. As you probably know already it states that the probability density $\rho(\mathbf{r}, \mathbf{p})$ to be around a given point $(\mathbf{r}, \mathbf{p})$ in phase space follows the equation of evolution:

$\frac{\partial \rho}{\partial t} = \{\rho, H \}$ where $\{ \cdot\}$ denotes the Poisson brackets.

This equation is equivalent to Hamilton equations of evolution for $(\mathbf{r}, \mathbf{p})$.

Now, when you look at macrovariables, it can be worked out (it has been done first by Zwandsig I think) that the Liouville equation (for the microvariables) gives rise to a Fokker-Planck equation for these macrovariables. It is in spirit very similar to the Liouville equation except that there is a stochastic component in it whose simplest characteristic is to add a second space derivative on the right hand side of the evolution equation.

Now, if you know your maths, you also know that any Fokker-Planck equation can be associated to a set of stochastic equations for the macrovariables under study (one very famous being the Langevin equation)...and we are back to something very close to the Hamilton equations but for macrovariables.

In case you were wondering if there is a minimum action principle for these stochastic equations, I am not aware of that. I think, they are very similar to Shrodinger equation in this respect. However what it means is that indeed the macrovariable propagators can be expressed as path integrals. The Wiener measure is one typical case.

Note that my answer is focused on Hamilton and Lagrangian dynamics in the classical sense where they were used to compute trajectories in time.

In classical statistical mechanics, you could find a Lagrangian approach akin to what is done in, say, QFT. This would be the Landau-Ginsburg approach of phase transitions and complex systems in general.

gatsu
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  • Thanks for the answer, but my question is about whether there is a formalism that uses the least action principle applied to the microvariables to directly construct an ensemble over phase space paths, without first deriving Hamiltonian equations of motion and then a Fokker-Planck equation. – N. Virgo Mar 24 '13 at 05:21
  • It sounds difficult for at least one reason. Least action principle is a functional of the generalized coordinates $\mathbf{q}(t)$ only so there is no such thing as phase space to begin with. Moreover, as far as I know, Lagrangian formalism is not suited to give the evolution of any function of $\mathbf{q}(t)$ so basically I don't quite see how it would work...but maybe you have something more precise in mind. Note however, that there is something called topological entropy that counts the number of possible paths in the system. – gatsu Mar 24 '13 at 10:48
  • The idea is that you would start with a joint distribution of $\mathbf{q}(t_0)$ and $\mathbf{q}(t_1)$, which should then uniquely specify the distribution over paths taken to get from initial to final points. That much is clear, but I'm interested in knowing what follows from such a line of reasoning. One possible application might be to give a concise answer to this question. – N. Virgo Mar 24 '13 at 13:06
  • Ok, I don't very much about this question but you might be interested in the work of Fabbrice Debbasch. Publication 20, although maybe too simple for you, may give you some hints on the directions to take. – gatsu Mar 24 '13 at 13:38
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I'm not sure this is quite what you're looking for, as classical statistical mechanics in equilibrium is, by definition, not a properly dynamical system, but min/maxing the function $$S[p_i] = -k_{B}\sum_{i}{\ln{p(x)}} + Z\sum_i{\big(p(x)-1\big)}+\beta\sum_i{p_iE_i-\langle E\rangle}+\mu\sum_i{N_ip_i-\langle N\rangle}+...$$ not only recovers the Boltzmann distribution, but naturally gives rise to the second law of thermodynamics and provides a unified way of dealing with any kind of ensemble. As such, this S is occasionally described as the Action for an equilibrium distribution

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    Can you explain what the terms in this equation mean, and/or give some kind of reference for further reading? It looks rather odd at a first glance - is $p(x)$ the same as $p_i$? (In which case $\sum_i(p(x)-1)$ is zero.) Is $\langle E \rangle$ defined differently from $\sum_i p_i E_i$? (If not, why do both appear in the equation?) – N. Virgo Nov 01 '15 at 00:54
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    Ah - now I get it. What you've written down is not $S[p_i]$ but $\partial S/\partial p_i$ with Lagrange multipliers. This is the MaxEnt approach to stat mech. Indeed I already know it, and the question is about its extension to systems in which there is time. – N. Virgo Nov 01 '15 at 00:56