0
$\begingroup$

I know all of the fundamental forces/fields(Gravitational, Electromagnetic, Strong and Weak) of the nature are conservative and we know every other derived force/field in the nature is just produced by one of the fundamental forces/fields.

But how non-conservative fields/forces can exist in a universe constructed of conservative fields/forces?(As we have non-conservative forces like: friction, magnetic force, etc.)

$\endgroup$
1
$\begingroup$

From a microscopic level there is no such thing as a "non-conservative force"- energy is conserved always (probably). "Non-conservative forces" are forces that do not conserve mechanical energy. Frictional force turns mechanical energy (mainly kinetic energy) into thermal energy (associated with random movement of particles). It's not disappearing- it's just changing from one form to another.

$\endgroup$
1
$\begingroup$

And to add something to @Chris's good answer, the non conservative friction force is just a macroscopic force which is statistically composed of all the microscopic interactions between surface causing friction and object, their particles (molecules, atoms and electrons). We can't keep track of all of those for each individual particle, too many.

Those forces cause increases in the kinetic energies of those particles. We describe that statistically as increasing thermodynamic temperature (and maybe volume, pressure etc), and causing heat. So the heat created, which is the reason it is not from a conservative force, is just kinetic energies of particles composing the body. The microscopic forces are all conservative, the semi random motion of particles is heat and we count that as the reason for non conservative. Thermodynamics still conserves total energy in the first law of thermodynamics: heat plus work done is total conserved energy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.