# Principle of least action and entropy

+ 1 like - 0 dislike
265 views

Is there any relations between the principle of least action and the the principle of increase of entropy?

asked Sep 4, 2016
retagged Sep 4, 2016

Hamiltonian or Lagrangian mechanics as described by the principle of least action are  kind of complementary to entropy as described by thermodynamics:

Hamiltonian/Lagrangian mechanics only describes reversible processes in accordance with Liouville's equation. The entropy remains constant.

Irreversibility and the second law of thermodynamics, characterized by an increase of entropy, emerge as a result of losing information by "coarse graining" the evolution of the system in phase space.

Thanks for your comment, @Dilaton ! Could you explain the concept of 'losing information' , please? Doesn't the increase of entropy mean the increase of information ? In my opinion , the entropy and information seem the same.@Dilaton

Increase of entropy means loss of information to unmodelled degrees of freedom.

Thanks, @Arnold

+ 1 like - 0 dislike

Hamiltonian or Lagrangian mechanics as described by the principle of least action is some  kind of complementary to the principle of increasing entropy (second law of thermodynamics) needed to describe described dissibative systems in  thermodynamics.

Hamiltonian/Lagrangian mechanics only describes the dynamics of reversible processes in accordance with Liouville's/von Neumann's equation

$\frac{\partial \rho}{\partial t} + iL\rho = 0$

where $\rho$ is the distribution function (density operator in QM) and $L$ is the Liouville operator involving the Poisson bracket (commutator in QM). The fine grained entropy $S = -k_B \ln \rho$ which is proportional to volume in phase space does not increase.

For large systems with many degrees of freedom, it is for practical reasons (initial conditions not exactly known, instability of the system, etc) not feasable to follow the trajectory in the corresponding very high-dimensional phase space. For large systems that are not in thermodynamic equilibrium, irreversibility and the second law of thermodynamics, characterized by an increase of the coarse grained entropy $S = -k_B \int d\Gamma f_N \ln f_s$ (where $f_s$ is a reduced distribution function) emerge as a result of losing information (to not modelled degrees of freedom) by "coarse graining" the evolution of the system in phase space. The evolution of the system does no longer follow the homogenous Liouville/von Neumann equation given above; as dissipative terms that appear on the right hand side, have to be taken into account in this case.

The amount of information needed to describe the dynamic evolution of a system depends on how far away it is from thermodynamic equilibrium. To describe a system way out of thermodynamic equilibrium, in principle  the full N-particle distribution function is needed to describe the state of the system by given by the mean values of appropriate relevant quantities. In the course of the evolution towards equilibrium, the number of (effective) degrees of freedom that have to be explicitely modelled reduces until the system has reached equilibrium. In thermodymanic equilibrium, the system is fully describe by its conserved quantities and all dissipative processes have stopped.

answered Sep 7, 2016 by (4,305 points)

How about phase transitions?

Nothing is special about phase transitions with respect to entropy, dissipation, or equilibrium.

+ 0 like - 0 dislike

No, there is no relations between the principle of least action and the the principle of increase of entropy.

answered Sep 5, 2016 by (32 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.