We know that gravitational, electrical and magnetic forces are conservative in nature. We also know that friction is fundamentally Electromagnetic in nature. How is it then a non-conservative force?
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1$\begingroup$ Please have a look to the answers to the following question: What exactly makes a force conservative?. They surely answer your question. $\endgroup$– DiracologyCommented Jan 31, 2018 at 17:06
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4$\begingroup$ Possible duplicate of What exactly makes a force conservative? $\endgroup$– sammy gerbilCommented Feb 2, 2018 at 17:10
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4 Answers
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All the forces that we know of are conservative.
Non-conservative forces are only an effective macroscopic description we use when we try and handle friction and its dissipation of energy.
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As you noted, you do have an electromagnetic interaction. Key point here is that not always is it conservative, and in fact gravity may also not be conservative because under certain circumstances, the interaction causes part of the potential or kinetic energy to be lost. Sliding over the ground is much like plastic collision with the ground, and that never conserves energy.
It can get rather complicated, so I'll sum up simply: try rubbing your hands together. You'll notice two things: one there is friction, the second your hands get warmer. That's where the energy goes. So in a way you can still attach a potential to electromagnetic interactions, and do all the calculations to find eventually the stable state would be to distribute that energy among the particles making up the objects, to loosely describe it.
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Physicists prefer to work with conservative forces because this results in path independence. What this means is that the net work done by a system over a closed path goes to zero:
\begin{equation} W = \oint_C \vec{F}\cdot d\vec{r} = 0 \end{equation}
This can only be true if the force $F$ does not depend explicitly on the velocity $\dot{q}$ of the system. We know that we can obtain the force on a particle from its potential energy
\begin{equation} F = - \frac{dV}{dq} \end{equation}
This treatment works for both Newtonian gravity and the Coulomb force. The above equations don't work if the potential $V$ has an explicit time dependence, however.
I want to be clear about something here: it is not correct to say that "conservative forces are more fundamental than non-conservative forces." Rather, the reality is that the mathematics of classical mechanics was historically developed to only deal with conservative forces.
Quantum mechanics was built from an extension of classical Hamiltonian mechanics, so the expectation that all subatomic interactions are conservative emerged from the limitations of the mathematical formalism. The fact that spontaneous-symmetry breaking occurs in beta decay is evidence of the limitations of extending classical expectations to the quantum world.
Non-conservative systems abound in nature. Any open thermodynamic system dissipates energy into its environment and failing to account for the missing degrees of freedom requires treating it as non-conservative. Finally, General Relativity is non-conservative because the Einstein field equations are non-linear.
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You seem to be mixed up about conservation of energy, and the conservation of forces in interactions.
The total energy is conserved, but the type changes, and follows the laws of thermodynamics so towards heat - in total for a closed system.
For a given interaction, various special conditions can tell us what is conserved. Mass is only conserved if nuclear or antimatter reactions don't occur, and can only be treated as constant below relativistic energies. Similar caveats for other fields and forces.
