# Meaning of Non-dissipative Dynamical System

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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
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

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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 (0 points)

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