Consider a constrained classical Lagrangian $ \mathcal{L}' = \mathcal{L}(q, \dot{q}) + \lambda f(q) $ where $ \lambda $ is the lagrange multiplier for the constraint. We can get a Hamiltonian for this system by the standard Legendre transform,

\begin{align} \mathcal{H}' = \dot{q} \frac{\partial}{\partial \dot{q}} \mathcal{L}' - \mathcal{L'} = \mathcal{H} - \lambda f(q) \end{align}

If I were trying to get the equations of motion, I would normally treat $ \lambda $ as a normal variable and get the equations of motion as usual.

Treating it like a normal variable, the microcanonical ensemble is,

\begin{align} \Omega(E) &= \int dq dp d\lambda~\delta \big( E - \mathcal{H} + \lambda f(q) \big) \end{align}

But if I inject the constain by hand into the microcanonical ensemble, I would expect an integral like, \begin{align} \Omega(E) &= \int dq dp~\delta \big( E - \mathcal{H} \big) ~\delta \big( f(q) \big) \end{align}

What is the connection? I assume the second integral is correct as it makes more sense to me, but how to I derive it? What about KKT type constrained dynamics like a ball falling onto a floor?

  • $\begingroup$ I'm no expert on the subject but I think that treating $\lambda$ as an independent degree of freedom here is wrong. It isn't really independent and in fact it isn't a coordinate in the sense that it doesn't add to the dimension of phase space and doesn't have a conjugate momenta. In short, I don't see a reason to sum over it. This doesn't resolve your issue though, and I'm curios too. $\endgroup$ – Yair M Sep 25 '17 at 4:52
  • $\begingroup$ @YairM I agree that the second one makes more sense. I was hoping to find a connection so I can use Stat Mech procedures on discrete time control systems with Lagrange multipliers that do act as dynamical variables and eventually try to find a connection between constraints and Thermodynamical legendre transforms by constraining at a statistical level. $\endgroup$ – aidan.plenert.macdonald Sep 25 '17 at 18:32

In the only field where I have seen this issue treated, polymers, the approach is your second take at it, i.e. introducing delta functions in the partition function [1]. Just to give you a feel for it, the simplest model of a polymer is a freely joint chain. If $r_i$ is the position of the $i$-th joint, the first constraint is to require that the distance between two consecutive joints is a constant, $\delta\big((r_{i+1}-r_i)^2 - a^2\big)$. The second one is that the motion of each joint shall be perpendicular to the link in the frame of the opposite joint, $\delta\big((p_{i+1}-p_i)\cdot(r_{i+1}-r_i)\big)$. Then the partition function is written

$$Z = \int \Pi_i dr_i dp_i \delta\big((r_{i+1}-r_i)^2 - a^2\big) \delta\big((p_{i+1}-p_i)\cdot(r_{i+1}-r_i)\big)\exp\left(-\beta\frac{p^2}{2m}\right)$$

Caveat: I do not make the claim that the problem is always treated that way, as I am not a specialist in this area, but working in the field of crystallography, I have some exposure to protein physics, and this is the only approach I have been exposed to.

[1] Martial Mazars. Statistical physics of the freely jointed chain. Phys. Rev. E, 53:6297–6319, Jun 1996.

| cite | improve this answer | |

In Section VI of Non-Hamiltonian molecular dynamics: Generalizing Hamiltonian phase space principles to non-Hamiltonian systems, a statistical treatment of Holonomic constraints is given. In section II.D, the microcanonical ensemble is described as,

\begin{align} f(x) = \prod_k \delta \big\{ \Lambda_k(q, p) - C_k \big\} \end{align} where $ \Lambda_k $ is a conserved quantity and $ C_k $ is the value we are controlling (like $ E $).

Then, Holonomic constraints are just another conserved quantity.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.