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stafusa
stafusa
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Yes. The first two points are equivalent definitions of nonconservative forces (as shown, e.g., in Wikipedia) and therefore preclude, as described in the answer above to Question 1, the system from being conservative.
And yes again - being nonconservative implies a loss or injection of energy in the system that prevents it from "reverting" - returning to a previous configuration. For instance, a dissipative pendulum starting from rest at $3^\circ$ won't manage to climb back again to $3^\circ$ due to the energy lost to, say, the air drag in itsit's subjected to as it oscillates back and fort oscillation.
Also reversibility in the sense of time-reversal symmetry would break for nonconservative systems (see this question and this).

Yes. The first two points are equivalent definitions of nonconservative forces (as shown, e.g., in Wikipedia) and therefore preclude, as described in the answer above to Question 1, the system from being conservative.
And yes again - being nonconservative implies a loss or injection of energy in the system that prevents it from "reverting" - returning to a previous configuration. For instance, a dissipative pendulum starting from rest at $3^\circ$ won't manage to climb back again to $3^\circ$ due to the energy lost to, say, air drag in its back and fort oscillation.
Also reversibility in the sense of time-reversal symmetry would break for nonconservative systems (see this question and this).

Yes. The first two points are equivalent definitions of nonconservative forces (as shown, e.g., in Wikipedia) and therefore preclude, as described in the answer above to Question 1, the system from being conservative.
And yes again - being nonconservative implies a loss or injection of energy in the system that prevents it from "reverting" - returning to a previous configuration. For instance, a dissipative pendulum starting from rest at $3^\circ$ won't manage to climb back again to $3^\circ$ due to the energy lost to, say, the air drag it's subjected to as it oscillates back and fort.
Also reversibility in the sense of time-reversal symmetry would break for nonconservative systems (see this question and this).

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stafusa
stafusa
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Simply put: a conservative system conserves energy, a non-conservativenonconservative one doesn't.

Notice that a nonautonomous Hamiltonian $H(t,x,p)$ can be used for describing a dissipative (i.e., nonconservative) system, but one most often implicitly implies time independence and uses "conservative" and "Hamiltonian" interchangeably. Notice also that for many systems $H$ is just the system's mechanical energy - in this case, $H$ being independent of time is the same as the system's energy being constant.

Yes. First, a conservative force is a particular case of a conservative vector field (see, e.g., Wikipedia and this question). Second, the force has to be conservative in order for it to correspond to a meaningful and time-independent potential energy, which in turn you typically need for defining aan autonomous Hamiltonian, i.e., for having a conservative system (see also this question, this, this, this and this). Standard examples of dissipative forces are friction and drag.

AboutAs for

Yes. The first two points are equivalent definitions of nonconservative forces (as shown, e.g., in Wikipedia) and therefore preclude, as described in the answer above to Question 1, the system from being conservative. And
And yes again - being nonconservative implies a loss or injection of energy in the system that prevents it from "reverting" - returning to a previous configuration. AlsoFor instance, a dissipative pendulum starting from rest at $3^\circ$ won't manage to climb back again to $3^\circ$ due to the energy lost to, say, air drag in its back and fort oscillation.
Also reversibility in the sense of time-reversal symmetry would break for nonconservative systems (see this question and this).

All the above is quite pedestrian - for a more complete and sophisticated take, one can start by checking the sources linked.

In particular - it's important to remark - fundamental forces are conservative, so the dissipative forces we see are emergent phenomena (such as friction arising from electromagnetic interactions), or effective/phenomenological descriptions, or a consequence of considering open systems, etc.

And lastly, yes, as it I hope is now clear, there is a relative consensus of what a non-conservativenonconservative system is, even if it often goes unstated.

Simply put: a conservative system conserves energy, a non-conservative one doesn't.

Notice that a nonautonomous Hamiltonian $H(t,x,p)$ can be used for describing a nonconservative system, but one most often implicitly implies time independence and uses "conservative" and "Hamiltonian" interchangeably. Notice also that for many systems $H$ is just the system's mechanical energy - in this case, $H$ being independent of time is the same as the system's energy being constant.

Yes. First, a conservative force is a particular case of a conservative vector field (see, e.g., Wikipedia and this question). Second, the force has to be conservative in order for it to correspond to a meaningful and time-independent potential energy, which in turn you typically need for defining a autonomous Hamiltonian, i.e., for having a conservative system (see also this question, this, this, this and this).

About

Yes. The first two points are equivalent definitions of nonconservative forces (as shown, e.g., in Wikipedia) and therefore preclude, as described in the answer above to Question 1, the system from being conservative. And yes again - being nonconservative implies a loss or injection of energy in the system that prevents it from returning to a previous configuration. Also reversibility in the sense of time-reversal symmetry would break for nonconservative systems (see this question and this).

All the above is quite pedestrian - for a more complete and sophisticated take, one can start by checking the sources linked.

And lastly, yes, as I hope is now clear, there is a relative consensus of what a non-conservative system is, even if it often goes unstated.

Simply put: a conservative system conserves energy, a nonconservative one doesn't.

Notice that a nonautonomous Hamiltonian $H(t,x,p)$ can be used for describing a dissipative (i.e., nonconservative) system, but one most often implicitly implies time independence and uses "conservative" and "Hamiltonian" interchangeably. Notice also that for many systems $H$ is just the system's mechanical energy - in this case, $H$ being independent of time is the same as the system's energy being constant.

Yes. First, a conservative force is a particular case of a conservative vector field (see, e.g., Wikipedia and this question). Second, the force has to be conservative in order for it to correspond to a meaningful and time-independent potential energy, which in turn you typically need for defining an autonomous Hamiltonian, i.e., for having a conservative system (see also this question, this, this, this and this). Standard examples of dissipative forces are friction and drag.

As for

Yes. The first two points are equivalent definitions of nonconservative forces (as shown, e.g., in Wikipedia) and therefore preclude, as described in the answer above to Question 1, the system from being conservative.
And yes again - being nonconservative implies a loss or injection of energy in the system that prevents it from "reverting" - returning to a previous configuration. For instance, a dissipative pendulum starting from rest at $3^\circ$ won't manage to climb back again to $3^\circ$ due to the energy lost to, say, air drag in its back and fort oscillation.
Also reversibility in the sense of time-reversal symmetry would break for nonconservative systems (see this question and this).

All the above is quite pedestrian - for a more complete and sophisticated take, one can start by checking the sources linked.

In particular - it's important to remark - fundamental forces are conservative, so the dissipative forces we see are emergent phenomena (such as friction arising from electromagnetic interactions), or effective/phenomenological descriptions, or a consequence of considering open systems, etc.

And lastly, yes, as it I hope is now clear, there is a relative consensus of what a nonconservative system is, even if it often goes unstated.

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stafusa
stafusa
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Simply put: a conservative system conserves energy, a non-conservative one doesn't.

In a conservative system:

  • trajectories follow paths of constant energy - i.e., if you start the system with a given configuration and let it evolve according to its dynamics, the configuration (say, a particle's position and momentum) might change with time, but its energy remains constant = is conserved;
  • phase-space volumes are preserved - i.e., any arbitrary chunk of the phase space (a blob of initial configurations in the space of possible configurations) maintains a constant volume as it evolves according to the system dynamics; it might deform and even split as much as it wants, but its total volume won't change.

This second description is a statement of Liouville's theorem for Hamiltonian systems, which leads us (see this question) to yet another description of a conservative system, namely a system whose

  • Hamiltonian is autonomous - i.e., it's a function $H(x,p)$ that doesn't depend on time, but only on the phase-space variables $x$ and $p$.

Notice that a nonautonomous Hamiltonian $H(t,x,p)$ can be used for describing a nonconservative system, but one most often implicitly implies time independence and uses "conservative" and "Hamiltonian" interchangeably. Notice also that for many systems $H$ is just the system's mechanical energy - in this case, $H$ being independent of time is the same as the system's energy being constant.

For mechanical systems, we can also say that, in a conservative system:

Which brings us to your:

Question 1: is conservative vector field and/or conservative force related to a conservative system?

Yes. First, a conservative force is a particular case of a conservative vector field (see, e.g., Wikipedia and this question). Second, the force has to be conservative in order for it to correspond to a meaningful and time-independent potential energy, which in turn you typically need for defining a autonomous Hamiltonian, i.e., for having a conservative system (see also this question, this, this, this and this).

About

Question 2: does a non-conservative system not include any of the points in the list from the quote above? I.e., Is the work in a non-conservative system:

  • Dependent on path
  • Not Equal to the difference between the final and initial values of an energy function.
  • Completely irreversible.

Yes. The first two points are equivalent definitions of nonconservative forces (as shown, e.g., in Wikipedia) and therefore preclude, as described in the answer above to Question 1, the system from being conservative. And yes again - being nonconservative implies a loss or injection of energy in the system that prevents it from returning to a previous configuration. Also reversibility in the sense of time-reversal symmetry would break for nonconservative systems (see this question and this).

All the above is quite pedestrian - for a more complete and sophisticated take, one can start by checking the sources linked.

As for the expression "nonconservative system" being found more often in papers from 1980's and 90's, I'd guess it's mostly down to the research topic being more active back then.

And lastly, yes, as I hope is now clear, there is a relative consensus of what a non-conservative system is, even if it often goes unstated.