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I understand that one cannot assign a potential energy to all points in space in the presence of a non-conservative force field due to the work done by the force being dependent on the path taken. However, I have often come across this statement that energy is "irrecoverably" lost when work is being done in a non-conservative force field (look at the photo below for example). What does this mean exactly? What do we mean by "irrecoverable" here? Where does this extra work/energy actually go? Does this mean that some part of the energy fed into the must be wasted in the form of heat or sound (somewhat like how there's an upper limit on the theoretical efficiency of a Carnot engine)? I would really appreciate some more clarity on this notion, and I apologise if there are duplicates I couldn't find.

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  • $\begingroup$ I"understand that one cannot assign a potential energy to all points in space in the presence of a non-conservative force field". What is a non-conservative force field? $\endgroup$
    – Bob D
    Feb 1 at 13:34

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Work is lost at a macroscopic level, but nothing is lost if you look at the microscopic level.

"Lost work" at the macroscopic level correspond to an increase of internal energy (and often temperature), as a macroscopic manifestation of the agitation of molecules at the microscopic level.

This work is "lost" at the macroscopic level, since the full conversion back from the internal agitation energy to a macroscopic useful form of energy is not free. This could be a broad description of the $2^{nd}$ principle of thermodynamics; this is how nature works.

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  • $\begingroup$ So, in case of a non conservative force, the extra work done which does not manifest itself in the form of kinetic energy of (suppose) the rigid body under study, goes into say, heating it? $\endgroup$ Feb 1 at 11:54
  • $\begingroup$ That is correct. It can also be lost to the surroundings as heat. $\endgroup$ Feb 1 at 11:59
  • $\begingroup$ it manifest in the kinetic energy of molecules, roughly speaking agitation/random motion (temperature can be defined with the average kinetic energy of this random motion of molecules), that can't be seen at the macroscopic level. What we call heating at a macroscopic level is nothing more than the transmission of the agitation (and kinetic energy of the microscopic random motion) of molecules at the microscopic level $\endgroup$
    – basics
    Feb 1 at 12:15
  • $\begingroup$ No energy is lost. It's more about the most convenient description of a system. If we don't need/want to analyze (because not feasible, or impossible, or just we don't need it for what we're doing) a system describing each of its molecules, we use a macroscopic description, and within it we need thermodynamic concepts and variables, like temperature, internal energy, heat, dissipation. We wouldn't need these concepts if you perform a microscopic description in terms of position and velocity of the particles and their interactions through forces $\endgroup$
    – basics
    Feb 1 at 12:19
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The quote at best is a badly phrased sentence, at worst is a fundamental misunderstanding. Instead of being unrecoverable the entropy generated by friction and other irreversible processes are in fact can work if they are transported to a lower temperature. Count Rumford's experiment of cannon boring could boil water and boiling water could lift a lid, etc.

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I understand that one cannot assign a potential energy to all points in space in the presence of a non-conservative force field due to the work done by the force being dependent on the path taken.

This is not true. One can define a potential (and thus potential energy of a single test charge) in almost all points of space, derived from the field, even if it isn't conservative.

Every field can be decomposed uniquely into a conservative and a divergence-less part, and we can use the conservative part to define a potential and thus also potential energy of a test charge.

This is fine and it is actually how potential and potential energy is used in practice, for example, for real electric circuits. Total electric field is never conservative everywhere, because there is always some non-zero non-conservative field contribution due to moving charges somewhere, and one could always devise a complicated extended path that comes close to those charges, to make the line integral of total field dependent on that path.

Electric potential can be defined in many different but mathematically equivalent ways, but in the most important convention, the Coulomb gauge, it is equal to line integral of the conservative part of the electric field, and this leads to the familiar Coulombic formula for electric potential $V(\mathbf x) = \sum_a K\frac{q_a}{|\mathbf x- \mathbf r_a|}$ and potential energy of charge $b$:

$$ P.E. = q_b V(\mathbf r_b). $$

What is true is that in non-conservative field, particle's kinetic and potential energy do not obey any simple law like

$$ K.E. + P.E. = const. $$ which holds in conservative field, because in non-conservative field, total force on the particle $b$ is not described completely by gradient of this potential energy function.

The correct statement here would be that non-conservative field cannot be expressed as gradient of a potential, or no potential function can fully describe the non-conservative field.

However, I have often come across this statement that energy is "irrecoverably" lost when work is being done in a non-conservative force field (look at the photo below for example). What does this mean exactly?

The usual kind of energy - kinetic plus potential - is not guaranteed to be conserved in time, when total force on a particle is such a function of position that is a non-conservative field.

For example, if there is an electric field with lines of force circling around a cylinder in space, such field is non-conservative, and if a charged particle was constrained to move along those circles, its kinetic energy would change in time, and also sum of kinetic and potential energy would change in time.

However, this does not mean "irrecoverable loss of energy", because while those energies assigned to the particle are not conserved, total energy of the system, including the source of the non-conservative field, and EM field everywhere, can be conserved.

For example, electric field near and inside conductive path of an LC circuit with an inductor and a capacitor is not conservative, and current-forming particles' energies are not conserved, but this does not mean energy gets lost irrecoverably; energy just transforms from electric to magnetic form repeatedly, and non-conservative electric field takes part in that conversion.

That statement is correct when a particular interpretation of terms is adopted, such as when we talk about friction force, and mechanical energy only. Friction force is, in general, not even a function of position, it often depends on velocity or other details of the system which are not traced, and result of its action is often that mechanical energy gets "lost irrecoverably", as it turns into internal energy of the bodies experiencing friction. This is a different notion of a "non-conservative force", which does not really refer to properties of line integrals of a field, but refers to the fact that mechanical energy transforms into heat.

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  • $\begingroup$ That is brilliantly clear, thank you very much, I understand perfectly now! $\endgroup$ Feb 1 at 17:07
  • $\begingroup$ @Cognoscenti You're welcome! $\endgroup$ Feb 1 at 21:46
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In case of a Conservative force the work done by it is position dependent, it has no dependence on the path, so in a round trip if the system comes back to its initial configuration the work done by the conservative force is zero, its 'Potential energy'(usually defined for the position dependent/conservative forces only)doesn't change. The total energy is always conserved be the force is conservative or non-conservative. The workdone by the non-conservative force (like friction) in a round trip is not zero that doesn't mean any energy is 'lost', suppose we consider a isolated block-bench system even if a non-conservative force like friction is present between them,the total energy of the system will remain the same, no matter what. The friction increases the temperature of the block and the bench, the workdone by the friction force in a round trip is finite which heat up the system. What is temperature? it is simply the mean kinetic energy of the molecules of the block-bench system due to random motion. The total energy is still conserved . Now the 'irrevocable' term is used for this thermal energy in the context of 2nd law of thermodynamics, you can't 'cool down' or decrease the temperature of the bench-block system and turn that 'random' energy (mean kinetic energy of the molecules of the system arising from the random thermal motion of them) into the 'systematic' mechanical kinetic energy of the block or the bench as a whole without bringing any change to the surrounding.

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However, I have often come across this statement that energy is "irrecoverably" lost when work is being done in a non-conservative force field (look at the photo below for example). What does this mean exactly?

The kinetic energy (KE) and potential energy (PE) of an object can be put into two categories: (1) Macroscopic mechanical KE and PE, which is the energy of the motion and position of the object as a whole with respect to an external (to the object) frame of reference and (2) Microscopic KE and PE, i.e., "internal energy", which is the energy of the object at the molecular level, due to molecular motion and intermolecular forces.

For example, when a falling object and encounters air resistance, a non-conservative force, some of its mechanical gravitational PE is converted to internal molecular KE (instead of an increase in mechanical KE) due to air friction. The energy is not "lost". It is converted to microscopic kinetic energy. It is "irrecoverable" in that it cannot be completely converted back to mechanical energy.

Hope this helps.

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  • $\begingroup$ What about a magnetic field? $\endgroup$ Feb 1 at 17:05
  • $\begingroup$ @Cognoscenti Oops!. Forgot that one. Thanks $\endgroup$
    – Bob D
    Feb 1 at 17:07
  • $\begingroup$ Do you have any idea why do we sometimes define a potential energy for a magnetic field, say when a bar magnet is in a uniform field, sir? $\endgroup$ Feb 1 at 17:08
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    $\begingroup$ @Cognoscenti although magnetic forces are not conservative they are also non dissipative unlike, for example, friction. So there is potential energy for magnetic fields $\endgroup$
    – Bob D
    Feb 1 at 18:06
  • $\begingroup$ Got it, thanks! $\endgroup$ Feb 2 at 7:33

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