# Definition of non-conservative force [duplicate]

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In defining conservative force, we say that

"The potential energy difference is path independent."

However, as far as I understand, potential energy only exists when there is a force field.

People say one example of non-conservative force.

By definition, non-conservative force should be the one in which the difference in potential energy is path dependent. But where is potential energy for friction which is not a force field?

## marked as duplicate by Aaron Stevens, John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 15 at 16:40

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• Can you give a reference? Potential energy is something we can define for conservative forces, but you are right, if they are not conservative then there is no potential energy associated with them – Aaron Stevens Jul 15 at 15:45
• Hint: The correct definition refers to work, not potential energy. – Qmechanic Jul 15 at 15:50
• Possible duplicate of What causes a force field to be "non-conservative?" – Aaron Stevens Jul 15 at 15:57

## 1 Answer

I think you mean to say a conservative force $$\mathbf F$$ is one where we can define a potential energy $$U$$ such that $$\mathbf F=-\nabla U$$

Then the work done by that force is independent of the path and only depends on the endpoints of the path. In other words, the work is given by: $$W=\int\mathbf F\cdot\text d\mathbf l=\int(-\nabla U)\cdot\text d\mathbf l=U_{\text{start}}-U_{\text{end}}$$ by the fundamental theorem of calculus.

On the other hand, we cannot express a non-conservative force in terms of a potential energy. Therefore, we cannot apply the fundamental theorem of calculus to the work integral, and therefore there is a path dependence on the work.

• Thank you very much for edit my question (con - "with"). Thank you again. – Sebastiano Jul 19 at 8:08