Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Meaning of Non-dissipative Dynamical System

+ 1 like - 0 dislike
83 views

What does it mean to say that a dynamical system is non-dissipative? I am particularly interested in an answer in the context of field theory or particle dynamics.

Also, how does this imply that we can (or that it is wise to) study such a system using a Lagrangian formalism?

This post imported from StackExchange Physics at 2016-11-27 19:05 (UTC), posted by SE-user Optimus Prime
asked Nov 27, 2016 in Theoretical Physics by Optimus Prime (80 points) [ no revision ]
retagged Nov 27, 2016
It basically means that is time-reversible. In a mechanical system this is equivalent to say that if you transform the conjugated momenta $p\to -p$ the system remains the same. Turns out that Lagrangian/Hamiltonian dynamics are theories specifically designed for non-dissipative systems. Although it is possible to deal with types of non-conservative interactions in these formalisms

This post imported from StackExchange Physics at 2016-11-27 19:05 (UTC), posted by SE-user caverac
@caverac is there a nuance between $p \rightarrow -p$ and $t \rightarrow -t$? Or is this also sufficient?

This post imported from StackExchange Physics at 2016-11-27 19:05 (UTC), posted by SE-user ComptonScattering
@ComptonScattering Changing $t\to -t$ leads to a change in the sign of the momenta. For example, for a single particle $p = mv = mdx/dt \to m dx/d(-t) = -p$

This post imported from StackExchange Physics at 2016-11-27 19:05 (UTC), posted by SE-user caverac
@caverac indeed, it also changes some other things though, such as inverting magnetic fields. I wondered if showing that the dynamics were well defined for $t \rightarrow -t$ is sufficient to say they are not dissipative.

This post imported from StackExchange Physics at 2016-11-27 19:05 (UTC), posted by SE-user ComptonScattering
@ComptonScattering Sorry, misunderstood your question. No, Imagine this problem $H(q,p) = p^2/2 + V(q,t^2)$, in this case the energy is not conserved

This post imported from StackExchange Physics at 2016-11-27 19:05 (UTC), posted by SE-user caverac

1 Answer

+ 0 like - 0 dislike

In a non-dissipative system there is no thermodynamically irreversible transformation of (mechanical) kinetic and potential energy into thermal energy or any other form of energy that decreases the ability of the system to perform work.

Dissipation occurs, for example, by mechanical friction, joule heating in resistors, viscous flow, turbulence or chemical reactions.

The Lagrange formalism can be extended to non-conservative forces (like friction) $Q_i$ by introducing them into the Lagrange equations $$\frac{d ∂L}{dt ∂\dot{q_i}}-\frac{∂L}{∂q_i}=Q_i$$

This post imported from StackExchange Physics at 2016-11-27 19:05 (UTC), posted by SE-user freecharly
answered Nov 27, 2016 by freecharly (0 points) [ no revision ]

Your answer

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):
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$\hbar$ysicsOver$\varnothing$low
Then 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.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...